http://drorbn.net/api.php?action=feedcontributions&user=Trefor&feedformat=atom Drorbn - User contributions [en] 2022-10-04T12:46:39Z User contributions MediaWiki 1.21.1 http://drorbn.net/index.php?title=0708-1300/Class_notes_for_Thursday,_March_27 0708-1300/Class notes for Thursday, March 27 2008-05-01T23:24:27Z <p>Trefor: </p> <hr /> <div>{{0708-1300/Navigation}}<br /> <br /> ==Handout==<br /> <br /> {{Home Link|classes/0102/AlgTop/Pathologies/index.html|Topological Pathologies in &lt;math&gt;{\mathbb R}^n&lt;/math&gt;}} (from {{Dror}}'s {{Home Link|classes/0102/AlgTop/|2001-2002 class}}).<br /> <br /> <br /> ===Class Notes===<br /> <br /> &lt;span style=&quot;color: red;&quot;&gt;The notes below are by the students and for the students. Hopefully they are useful, but they come with no guarantee of any kind.&lt;/span&gt;<br /> <br /> [[Image:0708-1300-March27_01.jpg|400px]]<br /> [[Image:0708-1300-March27_02.jpg|400px]]<br /> [[Image:0708-1300-March27_03.jpg|400px]]<br /> [[Image:0708-1300-March27_04.jpg|400px]]</div> Trefor http://drorbn.net/index.php?title=0708-1300/Class_notes_for_Tuesday,_February_26 0708-1300/Class notes for Tuesday, February 26 2008-04-26T18:34:40Z <p>Trefor: </p> <hr /> <div>{{0708-1300/Navigation}}<br /> <br /> ==A Homology Theory is a Monster==<br /> [[Image:0708-1300-AxiomsForHomology.png|thumb|center|540px|Page 183 of Bredon's book]]<br /> <br /> '''Bredon's Plan of Attack:''' State all, apply all, prove all.<br /> <br /> '''Our Route:''' Axiom by axiom - state, apply, prove. Thus everything we will do will be, or should be, labeled either &quot;'''S'''tate&quot; or &quot;'''P'''rove&quot; or &quot;'''A'''pply&quot;.<br /> <br /> ==Typed Notes==<br /> <br /> &lt;span style=&quot;color: red;&quot;&gt;The notes below are by the students and for the students. Hopefully they are useful, but they come with no guarantee of any kind.&lt;/span&gt;<br /> <br /> ===First Hour===<br /> <br /> <br /> <br /> Recall we had defined for a chain complex the associated homology groups: &lt;math&gt;H_p(C_*) :=\ ker\partial_p/im\partial_{p+1}&lt;/math&gt;<br /> <br /> From this we get the pth homology for a topological space &lt;math&gt;H_p(X)&lt;/math&gt;<br /> <br /> We have previously shown that <br /> <br /> ''1) &lt;math&gt;H_p(\cup X) = \oplus H_p(X_i)&lt;/math&gt; for disjoint unions of spaces &lt;math&gt;X_i&lt;/math&gt;<br /> <br /> 2) &lt;math&gt;H_p(pt) = \mathbb{Z}\delta_{p0}&lt;/math&gt;<br /> <br /> 3) &lt;math&gt;H_0(connected) = \mathbb{Z}&lt;/math&gt;<br /> <br /> 4) &lt;math&gt;H_1(connected) \cong \pi_1^{ab}(X)&lt;/math&gt; via the map<br /> <br /> &lt;math&gt;\phi:[\gamma]_{\pi_1^{ab}}\mapsto[\gamma]_{H_1}&lt;/math&gt;<br /> <br /> &lt;math&gt;\psi:\sigma\in C_1\mapsto[\gamma_{\sigma(0)}\sigma\bar{\gamma_{\sigma(1)}}]&lt;/math&gt; where &lt;math&gt;\sigma_x&lt;/math&gt; is a path connecting &lt;math&gt;x_0&lt;/math&gt; to x. <br /> <br /> <br /> We need to check the maps are in fact inverses of each other. <br /> <br /> Lets consider &lt;math&gt;\psi\circ\phi&lt;/math&gt;. We start with a closed path starting at &lt;math&gt;x_0&lt;/math&gt; (thought of as in the fundamental group). &lt;math&gt;\phi&lt;/math&gt; means we now think of it as a simplex in X with a point at &lt;math&gt;x_0&lt;/math&gt;. &lt;math&gt;\Psi&lt;/math&gt; now takes this to the path that parks at &lt;math&gt;x_0&lt;/math&gt; for a third of the time, goes around the loop and then parks for the remaining third of the time. Clearly this is homotopic this composition is homotopic to the identity. <br /> <br /> We now consider &lt;math&gt;\phi\circ\psi&lt;/math&gt;. Start with just a path &lt;math&gt;\sigma&lt;/math&gt;. Then &lt;math&gt;\psi&lt;/math&gt; makes a loop adding two paths from the &lt;math&gt;x_0&lt;/math&gt; to the start and finish of &lt;math&gt;\sigma&lt;/math&gt; forming a triangular like closed loop. We think of this loop as &lt;math&gt;\sigma'\in C_1&lt;/math&gt;<br /> <br /> Now, we start from c being &lt;math&gt;c = \sum a_i\sigma_i&lt;/math&gt; with &lt;math&gt;\partial c = 0&lt;/math&gt;. So get &lt;math&gt;\sum a_i(\partial \sigma_1) = \sum a_i(\sigma_i(1)-\sigma_i(0))&lt;/math&gt;<br /> <br /> So &lt;math&gt;\Psi(c) = [\gamma_{\sigma_i(0)}\sigma_i\bar{\gamma_{\sigma_i(1)}}]_{\pi_1}&lt;/math&gt; which maps to, under &lt;math&gt;\phi&lt;/math&gt;, &lt;math&gt;\sum a_i(\gamma_{\sigma_i(0)} + \sigma_i - \gamma_{\sigma_i(1)}) = \sum a_i\sigma_i = c&lt;/math&gt; ( in the homology gamma's cancel as &lt;math&gt;\partial c = 0&lt;/math&gt;)<br /> <br /> <br /> ''Axiomized Homology''<br /> <br /> We now will move to an approach where we prove that our defined homology satisfies a series of established homology axioms that will allow us to apply the machinery of general homology to our specific &quot;singular&quot; homology defined via simplexes. <br /> <br /> Axiom 0) Homology if a functor<br /> <br /> '''Definition''' The &quot;category of chain complexes&quot; is a category whose objects are chain complexes (of abelian groups) and morphisms which is a homomorphism between each abelian group in one chain and the corresponding group in the other chain such that the resulting diagram commutes. I.e, &lt;math&gt;Mor((C_p)_{p=0}^{\infty}, (D_p)^{\infty}_{p=0}) = \{(f_p:C_p\rightarrow D_p)_{p=0}^{\infty}\ |\ f_{p-1}\partial_p^C=\partial_p^D f_p\}&lt;/math&gt;<br /> <br /> <br /> Now, in our case, the chain complexes do in fact commute because &lt;math&gt;\partial&lt;/math&gt; is defined by pre-composition but f is defined by post-composition. Hence, associativity of composition yields commutativity. <br /> <br /> <br /> '''Claim'''<br /> <br /> Homology of chain complexes is a functor in the natural way. That is, if &lt;math&gt;f_p:C_p\rightarrow D_p&lt;/math&gt; for each p induces the functor &lt;math&gt;f_*:H_p(C_*)\rightarrow H_p(D_*)&lt;/math&gt;<br /> <br /> The proof is by &quot;diagram chasing&quot;. Well, let &lt;math&gt;c\in C_p&lt;/math&gt;, &lt;math&gt;\partial c =0&lt;/math&gt;. <br /> <br /> Let &lt;math&gt;f*[c] = [fc]&lt;/math&gt;. Now &lt;math&gt;\partial fc = f\partial c = 0&lt;/math&gt;. Furthermore, suppose &lt;math&gt;c = \partial b&lt;/math&gt;. Then, &lt;math&gt;f_*c = fc = \partial fb&lt;/math&gt; so therefore &lt;math&gt;fc = \partial\beta&lt;/math&gt; some &lt;math&gt;\beta&lt;/math&gt;. This shows &lt;math&gt;f_*&lt;/math&gt; is well defined. <br /> <br /> Thus, for &lt;math&gt;c\in H_p(C_*)&lt;/math&gt; get &lt;math&gt;fc\in H_p(D_*)&lt;/math&gt; via the well defined functor &lt;math&gt;f_*&lt;/math&gt;.<br /> <br /> <br /> ===Second Hour===<br /> <br /> '''1) Homotopy Axioms'''<br /> <br /> If &lt;math&gt;f,g:X\rightarrow Y&lt;/math&gt; are homotopic then &lt;math&gt;f_* = g_*: H_p(X)\rightarrow H_p(Y)&lt;/math&gt;<br /> <br /> <br /> Applications: If X and Y are homotopy equivalent then &lt;math&gt;H_*(X) \cong H_*(Y)&lt;/math&gt;<br /> <br /> ''Proof:''<br /> <br /> let &lt;math&gt;f:X\rightarrow Y&lt;/math&gt;, &lt;math&gt;g:Y\rightarrow X&lt;/math&gt; such that &lt;math&gt;f\circ g\sim id_y&lt;/math&gt; and &lt;math&gt;g\circ f \sim id_x&lt;/math&gt;. Well &lt;math&gt;f_*\circ g_* = id_{H(Y)}&lt;/math&gt; and &lt;math&gt;g_*\circ f_* = id_{H(X)}&lt;/math&gt;<br /> <br /> Hence, &lt;math&gt;f_*&lt;/math&gt; and &lt;math&gt;g_*&lt;/math&gt; are invertible maps of each other. ''Q.E.D''<br /> <br /> <br /> '''Definition'''<br /> <br /> Two morphisms &lt;math&gt;f,g:C_*\rightarrow D_*&lt;/math&gt; between chain complexes are ''homotopic'' if you can find maps &lt;math&gt;h_p:C_p\rightarrow D_{p+1}&lt;/math&gt; such that &lt;math&gt;f_p - g_p = \partial_{p+1} h_p + h_{p-1}\partial_p&lt;/math&gt;<br /> <br /> <br /> '''Claim 1'''<br /> <br /> Given H a homotopy connecting f&lt;math&gt;,g:X\rightarrow&lt;/math&gt; Y we can construct a chain homotopy between &lt;math&gt;f_*,g_*:C_*(X)\rightarrow C_*(&lt;/math&gt;Y)<br /> <br /> <br /> '''Claim 2'''<br /> <br /> If &lt;math&gt;f,g:C_*\rightarrow C_*&lt;/math&gt; are chain homotopic then they induce equal maps on homology. <br /> <br /> <br /> <br /> ''Proof of 2''<br /> <br /> Assume &lt;math&gt;[c]\in H_p(C_*)&lt;/math&gt;, that is, &lt;math&gt;\partial c =0&lt;/math&gt;<br /> <br /> &lt;math&gt;[f_* c] - [g_* c] = [(f_*-g_*)c] =[(\partial h + h\partial)c] = 0&lt;/math&gt; (as &lt;math&gt;\partial c = 0&lt;/math&gt; and homology ignores exact forms)<br /> <br /> Hence, at the level of homology they are the same. <br /> <br /> <br /> ''Proof of 1''<br /> <br /> Consider a simplex in X. Now consider its image, a simplex, in Y under g and f respectively. Because of the homotopy we can construct a triangular based cylinder in Y with the image under f at the top and the image under y at the bottom. <br /> <br /> Define &lt;math&gt;h\sigma&lt;/math&gt; = the above prism formed by &lt;math&gt;\sigma&lt;/math&gt; and the homotopy H. <br /> <br /> &lt;math&gt;(f_*-g_*)\sigma = h\partial\sigma + \partial h\sigma&lt;/math&gt;<br /> <br /> This, pictorially is correct but we need to be able to break up the prism, &lt;math&gt;\Delta_p\times I&lt;/math&gt; into a union of images of simplexes. <br /> <br /> Suppose p=0, i.e. a point. Hence &lt;math&gt;\Delta_0\times I&lt;/math&gt; is a line, which is a simplex. <br /> <br /> Suppose p=1 which yields a square. Adding a diagonal divides the square into two triangles, so is clearly a union of simplexes. <br /> <br /> Suppose p=2. We get a prism which has a triangle for a base and a top. Raise each vertex on the bottom to the top in turn. This makes the prism a union of three simplexes. <br /> <br /> In general for &lt;math&gt;\Delta_p\times I&lt;/math&gt; let &lt;math&gt;f_i = (l_i,0)&lt;/math&gt; and &lt;math&gt;g_i = (l_i,1)&lt;/math&gt; for vertexes &lt;math&gt;l_i&lt;/math&gt;<br /> <br /> Then, &lt;math&gt;h\sigma = \sum_{i=0}^p (-1)^i H\circ(\sigma\times I)\circ[f_0\cdots f_i g_i g_{i+1}\cdots g_p]&lt;/math&gt;<br /> <br /> which is in &lt;math&gt;C_{p+1}(Y)&lt;/math&gt;<br /> <br /> So have maps &lt;math&gt;Y\leftarrow_H X\times I\leftarrow \Delta_p\times I \leftarrow\Delta_{p+1}&lt;/math&gt;<br /> <br /> <br /> ''Claim: ''<br /> <br /> &lt;math&gt;\partial h +h\partial = f-g&lt;/math&gt;<br /> <br /> Loosely, &lt;math&gt;\partial h&lt;/math&gt; cuts each &lt;math&gt;[f_0\cdots f_i g_i g_{i+1}\cdots g_p]&lt;/math&gt; between the f_i and g_i and then deletes an entry. h\partial however does these in reverse order. Hence all that we are left with is &lt;math&gt;[f_0\cdots f_p] - [g_0\cdots g_p]&lt;/math&gt;</div> Trefor http://drorbn.net/index.php?title=0708-1300/Class_notes_for_Tuesday,_February_12 0708-1300/Class notes for Tuesday, February 12 2008-04-26T15:50:10Z <p>Trefor: </p> <hr /> <div>{{0708-1300/Navigation}}<br /> <br /> INCOMPLETE typed notes at bottom<br /> <br /> <br /> ===Class Notes===<br /> <br /> &lt;span style=&quot;color: red;&quot;&gt;The notes below are by the students and for the students. Hopefully they are useful, but they come with no guarantee of any kind.&lt;/span&gt;<br /> <br /> [[Image:0708-1300-Topofeb12p1.jpg|700px]]<br /> [[Image:0708-1300-Topofeb12p2.jpg|700px]]<br /> [[Image:0708-1300-Topofeb12p3.jpg|700px]]<br /> [[Image:0708-1300-Topofeb12p4.jpg|700px]]<br /> [[Image:0708-1300-Topofeb12p5.jpg|700px]]<br /> <br /> <br /> ==Typed Notes==<br /> <br /> Informal definition of homology:<br /> <br /> Given X (General topological spaces with no base point) then H_n(X) = {Boundaryless compact n-dim m submanifolds of X}/{Boundaries of (n+1)-dim submanifolds<br /> <br /> because &lt;math&gt;\partial\partial = \empty&lt;/math&gt;, the this makes sense. <br /> <br /> We will need to formalize this by using vector spaces generated by sets. <br /> <br /> We begin by computing the homology of some examples. These are correct in spirit but very wrong in the details, indeed, we havn't even defined what the right details are!<br /> <br /> Now, 0 dimensional submanifolds are merely points. 1 dimensional submanifolds are paths. So, if two points have a path between them, then the points contribute nothing to the zeroth homology. Hence H_0(S^1) = H_0(S^2) = 0 etc...<br /> <br /> Hence, H_0(connected) = &lt;pts&gt;/{all points are connected] = \mathbb{Z}<br /> <br /> H_1(S^1) = \mathbb{Z} as boundaryless paths are generated by loops around the circle, which are vacuously not the boundary of 2 manifolds. <br /> <br /> H_1(S^2) = &lt;closed paths&gt;/&lt;boundaries of disks &gt; = 0<br /> <br /> &lt;math&gt;H_2(S^2) = &lt;S^2&gt;&lt;/math&gt;/ nothing = &lt;math&gt;\mathbb{Z}&lt;/math&gt;<br /> <br /> <br /> In general we have <br /> <br /> &lt;math&gt;H_k(S^n) = \begin{matrix}<br /> \mathbb{Z}&amp;k=0\\<br /> 0&amp;0&lt;k&lt;n\\<br /> \mathbb{Z}&amp;k=n\\<br /> \end{matrix}&lt;/math&gt;<br /> <br /> Likewise, we know the zeroth and second homology of the torus is \mathbb{Z}<br /> <br /> We now need to compute the first homology of the torus. That is, we search for closed paths with no inner disk. <br /> <br /> We think of the torus generated by a square subset of the real plane with the usual identifications. Call one side a and the other b. <br /> Now consider a diagonal across the square, labeled c. Clearly we have a path with, c = a + b, which is the boundary of a triangle. <br /> <br /> Varying the diagonal into an arbitrary path which yields the same result we see that H_1(T^2)=\mathbb{Z}^2<br /> <br /> <br /> We get functorality as for f:X\rightarrow Y we get a map f^*:H_n(X)\rightarrow H_n(Y) as f maps boundaryless manifolds to boundaryless manifolds. <br /> <br /> <br /> ''Applications''<br /> <br /> Brouwer's theorem for D^{n+1}<br /> <br /> I.e., there is no retract D^{n+1}\rightarrow S^n<br /> <br /> Indeed, if there were such an retract we would get a commuting diagram and I = i :S^n\rightarrow S^n} where I is the identity on S^n and i is the natural inclusion into D^{n+1}. But after applying the functor this would get a commuting diagram where S^n goes to \mathbb{Z} and D^{n+1} goes to zero. But then have I* = 0 map, this is impossible. <br /> <br /> <br /> '''Definition:'''<br /> <br /> The p-simplex \Delta_p is is<br /> <br /> \mathbb{R}^{p+1}\ni\Delta_p:=\{(\lambda_0\cdots\lambda_p)\ |\ \sum_i^p \lambda_i = 1,\ \lambda_i\geq0\} = \{\sum\lambda_i e_i\ |\ \sum_i^p \lambda_i = 1,\ \lambda_i\geq0\}<br /> <br /> <br /> '''Example'''<br /> <br /> p=0 is just the point 1 on the real line. <br /> <br /> p=1 is the diagonal line between (1,0) and (0,1) in the real plane. here e_1 represents the vector from 0 to (0,1) and e_0 represents the vector from 0 to (1,0)<br /> <br /> p=2 forms a triangle between the tips of the three unit vectors in \mathbb{R}^3. <br /> <br /> In general, we have the convex full of p+1 points. <br /> <br /> <br /> Likewise, if (v_0,\cdots, v_p)\in\mathbb{R}^N then have map [v_0,\cdots, v_p]:\Delta_p\rightarrow\mathbb{R}^N defined by \sum\lambda_i e_i\mapsto \sum \lambda_i v_i<br /> <br /> <br /> Now, we want to define the boundary of a simplex. Loosely, we get a bunch of p-1 simplexes found by forgetting a vertex each time. <br /> <br /> '''Definition'''<br /> <br /> F_i^p:\Delta_{p-1}\rightarrow\Delta_p is the ith face of \Delta_p, namely, F_i^p = [e_0\cdots \hat{e_i}\cdots e_p]<br /> <br /> <br /> '''Definition'''<br /> <br /> Given X, \Delta_p(X)=C_p(X) = &lt;p\ simplexes\ in\ X&gt; = \{\sum a_i\sigma_i\ |\ a_i\in\mathbb{Z},\ \sigma_i:\Delta_p\rightarrow X\ is\ cont.\}<br /> <br /> Note: \sigma is not necessarily 1:1 as was implied in the earlier naive computations. Furthermore, we note that C_p(X) has an abelian group structure)<br /> <br /> Definition: \partial_p:C_p(X)\rightarrow Cc_{p-1}(X) a group homomorphism via \sum a_i\sigma_i\mapsto \sum a_i\partial_p(\sigma_i) where \partial_p\sigma = \sum_{i=0}^p (-1)^i\sigma\circ\F_i^p<br /> <br /> <br /> This results in a chain \cdots C_3(X)\rightarrow^{\partial_3}C_2(X)\rightarrow^{\partial_2}C_1(X)\rightarrow^{\partial_1}C_0(X) <br /> <br /> <br /> '''Definition'''<br /> <br /> Z_p(X) = ker\partial_p called &quot;p cycles in X&quot;<br /> <br /> B_p(X) = Im\partial_{p+1} called &quot;p boundaries in X&quot;<br /> <br /> <br /> We want to define H_p(X) = Z_p(X)/B_p(X)<br /> <br /> <br /> '''Lemma'''<br /> <br /> B_p(X)\subset Z_p(X) \Leftrightarrow im\partial{p+1}\subset \partial_p\Leftrightarrow \partial_p\circ\partial_{p+1} =0<br /> <br /> Notationally we simply write \partial^2 = 0<br /> <br /> <br /> Let \sigma\in C_{p+1}<br /> <br /> \sigma:\Delta_{p+1}\rightarrow X, \partial\sigma = \sum_{i=0}^p (-1)^i\sigma\circ F_i^{p}<br /> <br /> \partial\partial\sigma = \partial(\sum (-1)^i\sigma\circ F_i^p) = \sum_{i=0}^{p+1}\sum_{j=0}^p\sigma\circ F_i^p\circ F_j^{p-1}(-1)^{i+j}<br /> <br /> <br /> I.e., dropping e_m, e_n occurs twice, only with a different sign. Formally, <br /> <br /> F_i^p\circ F_j^{p-1} = [e_o\cdots \hat{e_j},\hat{e_i}\cdots e_{p+1}] for i&gt;j and <br /> <br /> F_i^p\circ F_j^{p-1} = [e_o\cdots \hat{e_i},\hat{e_j+1}\cdots e_{p+1}] for i\leq j<br /> <br /> So \partial\partial\sigma = \sum_{i=o}^{p+1}\sum_{j=0}^{p-1}(-1)^{i+j}\sigma\circ[\hat{e_j}\hat{e_i}] + \sum_{i=0}^{p+1}\sum_{j=i}^p (-1)^{i+j}\sigma\circ[\hat{e_i}\hat{e_{j+1}}]</div> Trefor http://drorbn.net/index.php?title=0708-1300/Class_notes_for_Tuesday,_March_25 0708-1300/Class notes for Tuesday, March 25 2008-04-11T19:15:23Z <p>Trefor: </p> <hr /> <div>{{0708-1300/Navigation}}<br /> <br /> ==Typed Notes==<br /> <br /> &lt;span style=&quot;color: red;&quot;&gt;The notes below are by the students and for the students. Hopefully they are useful, but they come with no guarantee of any kind.&lt;/span&gt;<br /> <br /> ===First Hour===<br /> <br /> '''Definition'''<br /> <br /> We define the CW chain complex via:<br /> <br /> &lt;math&gt;C_n^{CW}(K):=&lt;K_n&gt;&lt;/math&gt;<br /> <br /> and the boundary maps via &lt;math&gt;\partial:C_n^{CW}\rightarrow C^{CW}_{n-1}&lt;/math&gt; by &lt;math&gt;\partial\sigma=\sum_{\tau\in K_{n-1}}[\tau:\sigma]\tau&lt;/math&gt;<br /> <br /> where &lt;math&gt;[\tau:\sigma]&lt;/math&gt; is roughly the number of times that &lt;math&gt;\partial\sigma&lt;/math&gt; covers &lt;math&gt;\tau&lt;/math&gt;. <br /> <br /> <br /> Let's make this precise. <br /> <br /> For &lt;math&gt;f_{\sigma}:D_{\sigma}^n\rightarrow K&lt;/math&gt; (not quite an embedding) this restricts to a map &lt;math&gt;f_{\partial\sigma}:S_{\sigma}^n\rightarrow K&lt;/math&gt;. Given &lt;math&gt;\tau\in K_m&lt;/math&gt; let &lt;math&gt;p_{\tau}:K^n\rightarrow S^n = B^n\cup\{\infty\}&lt;/math&gt; such that &lt;math&gt;int(D^n_{\tau})\mapsto B^n&lt;/math&gt; and the rest maps to the point &lt;math&gt;\infty&lt;/math&gt;. <br /> <br /> <br /> Example:<br /> <br /> If you just have the segment consisting of two endpoints and the line connecting them, and call this &lt;math&gt;\tau&lt;/math&gt;, then &lt;math&gt;p_{\tau}&lt;/math&gt; takes the two end points to &lt;math&gt;\infty&lt;/math&gt; and the rest (the open interval) gets mapped to &lt;math&gt;B^1&lt;/math&gt;. Hence, we get a circle. <br /> <br /> <br /> We thus can now formally define &lt;math&gt;[\tau:\sigma]= deg(p_{\tau}\circ f_{\partial\sigma}:S^{n-1}\rightarrow S^{n-1})&lt;/math&gt;<br /> <br /> <br /> <br /> '''Theorem'''<br /> <br /> <br /> &lt;math&gt;(C^{CW}_*,\partial)&lt;/math&gt; is a chain complex; &lt;math&gt;\partial^2 = 0&lt;/math&gt; and &lt;math&gt;H_*^{CW}(K) = H_*(K)&lt;/math&gt;<br /> <br /> <br /> '''Examples:'''<br /> <br /> <br /> 1) &lt;math&gt;S^n = \{\infty\}\cup D^n&lt;/math&gt; for n&gt;1<br /> <br /> &lt;math&gt;f_{\partial\sigma}: S^{n-1}\rightarrow \infty&lt;/math&gt;<br /> <br /> Hence &lt;math&gt;C^{CW}_n = &lt;\sigma&gt;, C^{CW}_0 = &lt;\infty&gt;&lt;/math&gt; and all the rest are zero. Hence, &lt;math&gt;H_p(S^n) = \mathbb{Z}&lt;/math&gt; for p = 0 or n and is zero otherwise. <br /> <br /> <br /> 2) Consider the torus thought of as a square with the usual identifications and &lt;math&gt;\sigma&lt;/math&gt; is the interior. Hence, &lt;math&gt;C^{CW}_0 =\{p\}, C^{CW}_1&lt;/math&gt; is generated by the figure 8 with one loop labeled a and the other labeled b, and &lt;math&gt;C^{CW}_2&lt;/math&gt; is generated by the entire torus. <br /> <br /> Ie we get &lt;math&gt;\mathbb{Z}\rightarrow\mathbb{Z}^2\rightarrow\mathbb{Z}&lt;/math&gt;<br /> <br /> Now, &lt;math&gt;\partial a = [p:a]p&lt;/math&gt;<br /> <br /> but &lt;math&gt;\partial&lt;/math&gt; a takes the two endpoints of a (both p) and maps them to p. Neither point is mapped to &lt;math&gt;\infty&lt;/math&gt;. Hence, &lt;math&gt;deg\partial a:S^0\rightarrow S^0 = 0&lt;/math&gt;<br /> <br /> Note: This ought to be checked from the definition of degree but was just stated in class<br /> <br /> <br /> Now, &lt;math&gt;[\sigma:a]&lt;/math&gt; = the degree of the map that takes the square to the figure 8...and hence is &lt;math&gt;\pm 1\mp 1 = 0&lt;/math&gt;. <br /> <br /> Hence the boundary map is zero at all places, so &lt;math&gt;H_n(\mathbb{T}^2) = \mathbb{Z}&lt;/math&gt; if n = 0 or 2, &lt;math&gt;\mathbb{Z}^2&lt;/math&gt; if n = 1 and is zero otherwise. <br /> <br /> <br /> 3) Consider the Klein bottle thought of as a square with the usual identifications. Under &lt;math&gt;p_b\circ f_{\partial\sigma}&lt;/math&gt; takes this to a circle with side labled b. <br /> <br /> I.e., &lt;math&gt;&lt;\sigma&gt;\mapsto&lt;a,b&gt;\mapsto^0&lt;p&gt;&lt;/math&gt;<br /> <br /> &lt;math&gt;\partial\sigma = [a:\sigma]a+b:\sigma b = 0_a + 2b&lt;/math&gt;<br /> <br /> Where the sign may be negative. Or more eloquantly put: &quot;2b or -2b, that is the question&quot;<br /> <br /> The kernal of &lt;math&gt;&lt;a,b&gt;\rightarrow&lt;p&gt;&lt;/math&gt; is everything, so the homology is &lt;math&gt;H_1(K) = &lt;a,b&gt;/2b=0\cong\mathbb{Z}\oplus\mathbb{Z}/2&lt;/math&gt;, &lt;math&gt;H_0(K)=\mathbb{Z}&lt;/math&gt; and &lt;math&gt;H_2(K) = 0&lt;/math&gt;.<br /> <br /> <br /> 4) &lt;math&gt;\Sigma_g&lt;/math&gt; the &quot;g holed torus&quot; or &quot;surface of genus g&quot; is formed by the normal diagram with edges identified in sets of 4 such as &lt;math&gt;aba^{-1}b^{-1}&lt;/math&gt;<br /> <br /> So, get &lt;math&gt;&lt;\sigma&gt;\rightarrow^0&lt;a_1,\cdots, a_g, b_1,\cdots,b_g&gt;\rightarrow^0&lt;p&gt;&lt;/math&gt;<br /> <br /> Therefore, &lt;math&gt;H_p = \mathbb{Z}&lt;/math&gt; if p = 2 or 0, &lt;math&gt;\mathbb{Z}^{2g}&lt;/math&gt; if p = 1 and zero elsewhere. <br /> <br /> <br /> 5) &lt;math&gt;\mathbb{R}P^n= D^n\cup \mathbb{R}P^{n-1} = D^n\cup D^{n-1}\cup\cdots\cup D^0&lt;/math&gt;<br /> <br /> So, &lt;math&gt;C_*^{CW}(\mathbb{R}P^n) is &lt;D^n&gt;\rightarrow&lt;D^{n-1}&gt;\rightarrow\cdots\rightarrow &lt;D^0&gt;&lt;/math&gt;<br /> <br /> The boundary map alternates between 0 and 2 where it is 2 for &lt;math&gt;&lt;D^i&gt;\rightarrow&lt;D^{i-1}&gt;&lt;/math&gt; if i is even and 0 if it is odd. <br /> <br /> Hence the homology alternates between &lt;math&gt;\mathbb{Z}/2&lt;/math&gt; for odd p and 0 for even p. <br /> <br /> <br /> ===Second Hour===<br /> <br /> &lt;math&gt;\mathbb{C}P^n:\{[z_0,\cdots,z_n]:z_i\in\mathbb{C}\ not\ all\ zero\}=\mathbb{C}^{n+1}/z\sim\alpha z&lt;/math&gt; for &lt;math&gt;z\in\mathbb{C}^{n+1}&lt;/math&gt;, &lt;math&gt;\alpha\in\mathbb{C}-\{0\}&lt;/math&gt;<br /> <br /> &lt;math&gt;\mathbb{C}P^n=\{[z_0,\cdots,z_n]:z_n\neq 0\}\cup\{[\ldots,0]\} = \{[z_0,\cdots,z_{n-1},1]\}\cup\mathbb{C}P^{n-1}&lt;/math&gt;<br /> =&lt;math&gt; \mathbb{C}^n\cup\mathbb{C}P^{n-1} = \mathbb{R}^{2n}\cup\mathbb{C}P^{n-1} = B^{2n}\cup\mathbb{C}P^{n-1}&lt;/math&gt;<br /> <br /> <br /> I.e., &lt;math&gt;\mathbb{C}P^n = D^{2n}\cup D^{2n-1}\cup\cdots&lt;/math&gt;<br /> <br /> <br /> So, &lt;math&gt;C^{CW}_*(\mathbb{C}P^n)&lt;/math&gt; alternates between 0 (for odd p) and &lt;math&gt;\mathbb{Z}&lt;/math&gt; for even p and thus as this also as the homology. (and clearly trivial greater than p)</div> Trefor http://drorbn.net/index.php?title=0708-1300/Class_notes_for_Tuesday,_April_3 0708-1300/Class notes for Tuesday, April 3 2008-04-05T03:59:53Z <p>Trefor: </p> <hr /> <div>{{0708-1300/Navigation}}<br /> <br /> INCOMPLETE - Will be finished (most likely) Tomorrow<br /> <br /> ==Typed Notes==<br /> <br /> &lt;span style=&quot;color: red;&quot;&gt;The notes below are by the students and for the students. Hopefully they are useful, but they come with no guarantee of any kind.&lt;/span&gt;<br /> <br /> ===First Hour===<br /> <br /> '''Theorem 1: Borsuk-Ulam'''<br /> <br /> f:S^n\rightarrow\mathbb{R}^n has a point x\in S^n so f(x) = f(-x) \Leftrightarrow there does not exists a g:S^n\rightarrow S^{n-1} that is odd, i.e., g(-x)=-g(x)<br /> <br /> <br /> (\Leftarrow) If an f with no such fixed point existed then g(x) = (f(x) - f(-x))/||f(x) - f(-x)||<br /> <br /> <br /> '''Theorem 2:'''<br /> <br /> If g:S^n\rightarrow S^n is odd then g_*:H_n(S^n,\mathbb{Z}/2)\rightarrow H_n(S^n,\mathbb{Z}/2), ie g_*:\mathbb{Z}/2\rightarrow\mathbb{Z}/2, is the identity. <br /> <br /> <br /> '''Claim:'''<br /> <br /> Theorem 2 implies Borsuk-Ulam. <br /> <br /> Indeed, if g:S^n\rightarrow S^{n-1} is odd and exists then S^n\rightarrow^g S^{n-1}\hookrightarrow\S^n where the final map is the inclusion to the equation then the composition of these maps \tilde{g}:S^n\rightarrow S^n is odd. But, \tilde{g} lives on the equator only and so is homotopic to a constant. Hence, \tilde{g}_* = (const)_* = 0. This establishes the contradiction. <br /> <br /> ''Q.E.D''<br /> <br /> <br /> ''Proof of Theorem 2''<br /> <br /> We use the technique of a &quot;transfer sequence&quot; over \mathbb{Z}/2<br /> <br /> <br /> Let p:X\rightarrow B be a two sheeted covering<br /> <br /> Over \mathbb{Z}/2 we have the following short exact sequence:<br /> <br /> \rightarrow C_*(B)\rightarrow^{l_*}C_*(X)\rightarrow^{p_*}C_*(B)\rightarrow 0<br /> <br /> where l is the sum of two liftings from the base. <br /> <br /> The above induces a long exact sequence of homology groups. <br /> <br /> &lt;math&gt;\begin{matrix}<br /> 0&amp;\rightarrow&amp;C_*(\mathbb{R}P^n)&amp;\rightarrow &amp; C_*(S^n)&amp;\rightarrow^{p_*} &amp; C_*(\mathbb{R}P^n)&amp;\rightarrow&amp;0\\<br /> &amp;&amp;\downarrow^1&amp;&amp;\downarrow^{g_*}&amp;&amp;\downarrow^{g_*}&amp;&amp;\\<br /> 0&amp;\rightarrow&amp;C_*(\mathbb{R}P^n)&amp;\rightarrow &amp; C_*(S^n)&amp;\rightarrow^{p_*} &amp; C_*(\mathbb{R}P^n)&amp;\rightarrow&amp;0\\<br /> <br /> \end{matrix}&lt;/math&gt;<br /> <br /> <br /> The right square commutes irregardless of where g is even or odd and the left square commutes if the g is odd. <br /> <br /> <br /> We get the following (very) long exact sequence:<br /> <br /> &lt;math&gt;\begin{matrix}<br /> H_{n+1}(\mathbb{R}P^n) = 0 &amp; \rightarrow^{\partial}&amp; H_n(\mathbb{R}P^n) &amp; \rightarrow^{l_*} &amp; H_n(S^n) &amp; \rightarrow^{p_*} &amp; H_n(\mathbb{R}P^n)&amp; \rightarrow^{\partial}&amp; H_{n-1}(\mathbb{R}P^n) &amp; \rightarrow \cdots\\<br /> <br /> &amp;&amp;\downarrow^{g_*}&amp;&amp;\downarrow^{g_*}&amp;&amp;\downarrow^{g_*}&amp;&amp;\downarrow^{g_*}&amp;\\<br /> <br /> <br /> H_{n+1}(\mathbb{R}P^n) = 0 &amp; \rightarrow^{\partial}&amp;H_n(\mathbb{R}P^n)&amp;\rightarrow^{l_*} &amp; H_n(S^n) &amp; \rightarrow^{p_*} &amp; H_n(\mathbb{R}P^n)&amp;\rightarrow^{\partial}&amp;H_{n-1}(\mathbb{R}P^n)&amp;\rightarrow \cdots\\<br /> \end{matrix}&lt;/math&gt;<br /> <br /> <br /> <br /> &lt;math&gt;\begin{matrix}<br /> <br /> \cdots &amp; \rightarrow &amp; H_{2}(S^n) = 0 &amp; \rightarrow &amp; H_2(\mathbb{R}P^n) &amp; \rightarrow^{\partial} &amp; H_1(\mathbb{R}P^n) &amp; \rightarrow &amp; 0 &amp; \rightarrow \\<br /> <br /> &amp;&amp;\downarrow^{g_*}&amp;&amp;\downarrow^{g_*}&amp;&amp;\downarrow^{g_*}&amp;&amp;\downarrow^{g_*}&amp;\\<br /> <br /> \cdots &amp; \rightarrow &amp; H_{2}(S^n) = 0 &amp; \rightarrow &amp; H_2(\mathbb{R}P^n) &amp; \rightarrow^{\partial} &amp; H_1(\mathbb{R}P^n) &amp; \rightarrow &amp; 0 &amp; \rightarrow \\<br /> <br /> \end{matrix}&lt;/math&gt;<br /> <br /> <br /> &lt;math&gt;\begin{matrix}<br /> <br /> \rightarrow &amp; H_1(\mathbb{R}P^n)&amp; \rightarrow^{\partial} &amp; H_0(\mathbb{R}P^n)&amp; \rightarrow &amp; H_0(S^n) &amp; \rightarrow &amp; H_0(\mathbb{R}P^n) &amp; \rightarrow &amp; 0 \\<br /> &amp;\downarrow^{g_*}&amp;&amp;\downarrow^{g_*}&amp;&amp;\downarrow^{g_*}&amp;&amp;\downarrow^{g_*}&amp;&amp;\\<br /> \rightarrow &amp; H_1(\mathbb{R}P^n)&amp; \rightarrow^{\partial} &amp; H_0(\mathbb{R}P^n)&amp; \rightarrow &amp; H_0(S^n) &amp; \rightarrow &amp; H_0(\mathbb{R}P^n) &amp; \rightarrow &amp; 0 \\<br /> <br /> <br /> \end{matrix}&lt;/math&gt;<br /> <br /> <br /> The above three diagrams are meant to be connected end to end. <br /> <br /> It is difficult to write, but fairly easy to see, that by starting at the far right (the right of the bottom most diagram) and working left that:<br /> <br /> H_k (\mathbb{R}P^n,\mathbb{Z}/2) = \mathbb{Z}/2 \forall p&lt;n <br /> <br /> and that g_* is in fact the identity at all places. ''Q.E.D''</div> Trefor http://drorbn.net/index.php?title=Template:0708-1300/Navigation Template:0708-1300/Navigation 2008-04-01T17:18:05Z <p>Trefor: </p> <hr /> <div>{| cellpadding=&quot;0&quot; cellspacing=&quot;0&quot; style=&quot;clear: right; float: right&quot;<br /> |- align=right<br /> |&lt;div class=&quot;NavFrame&quot;&gt;&lt;div class=&quot;NavHead&quot;&gt;[[0708-1300]]/[[Template:0708-1300/Navigation|Navigation Panel]]&amp;nbsp;&amp;nbsp;&lt;/div&gt;<br /> &lt;div class=&quot;NavContent&quot;&gt;<br /> {| border=&quot;1px&quot; cellpadding=&quot;1&quot; cellspacing=&quot;0&quot; width=&quot;220&quot; style=&quot;margin: 0 0 1em 0.5em; font-size: small&quot;<br /> |-<br /> |colspan=3 align=center|[[Image:0708-1300-ClassPhoto.jpg|215px]]&lt;br&gt;[[0708-1300/Class Photo|Add your name / see who's in!]]<br /> |- align=center<br /> !#<br /> !Week of...<br /> !Links<br /> |-<br /> |colspan=3 align=center|'''Fall Semester'''<br /> |- align=left<br /> |align=center|1<br /> |Sep 10<br /> |[[0708-1300/About This Class|About]], [[0708-1300/Class notes for Tuesday, September 11|Tue]], [[0708-1300/Class notes for Thursday, September 13|Thu]]<br /> |- align=left<br /> |align=center|2<br /> |Sep 17<br /> |[[0708-1300/Class notes for Tuesday, September 18|Tue]], [[0708-1300/Homework Assignment 1|HW1]], [[0708-1300/Class notes for Thursday, September 20|Thu]]<br /> |- align=left<br /> |align=center|3<br /> |Sep 24<br /> |[[0708-1300/Class notes for Tuesday, September 25|Tue]], [[0708-1300/Class Photo|Photo]], [[0708-1300/Class notes for Thursday, September 27|Thu]]<br /> |- align=left<br /> |align=center|4<br /> |Oct 1<br /> |[[0708-1300/Questionnaire|Questionnaire]], [[0708-1300/Class notes for Tuesday, October 2|Tue]], [[0708-1300/Homework Assignment 2|HW2]], [[0708-1300/Class notes for Thursday, October 4|Thu]]<br /> |- align=left<br /> |align=center|5<br /> |Oct 8<br /> |Thanksgiving, [[0708-1300/Class notes for Tuesday, October_9|Tue]], [[0708-1300/Class notes for Thursday, October 11|Thu]]<br /> |- align=left<br /> |align=center|6<br /> |Oct 15<br /> |[[0708-1300/Class notes for Tuesday, October 16|Tue]], [[0708-1300/Homework Assignment 3|HW3]], [[0708-1300/Class notes for Thursday, October 18|Thu]]<br /> |- align=left<br /> |align=center|7<br /> |Oct 22<br /> |[[0708-1300/Class notes for Tuesday, October 23|Tue]], [[0708-1300/Class notes for Thursday, October 25|Thu]]<br /> |- align=left<br /> |align=center|8<br /> |Oct 29<br /> |[[0708-1300/Class notes for Tuesday, October 30|Tue]], [[0708-1300/Homework Assignment 4|HW4]], [[0708-1300/Class notes for Thursday, November 1|Thu]], [[0708-1300/the unit sphere in a Hilbert space is contractible|Hilbert sphere]]<br /> |- align=left<br /> |align=center|9<br /> |Nov 5<br /> |[[0708-1300/Class notes for Tuesday, November 6|Tue]],[[0708-1300/Class notes for Thursday, November 8|Thu]], [[0708-1300/Term Exam 1|TE1]]<br /> |- align=left<br /> |align=center|10<br /> |Nov 12<br /> |[[0708-1300/Class notes for Tuesday, November 13|Tue]], &lt;strike&gt;[[0708-1300/Class notes for Thursday, November 15|Thu]]&lt;/strike&gt;<br /> |- align=left<br /> |align=center|11<br /> |Nov 19<br /> |[[0708-1300/Class notes for Tuesday, November 20|Tue]], [[0708-1300/Homework Assignment 5|HW5]]<br /> |- align=left<br /> |align=center|12<br /> |Nov 26<br /> |[[0708-1300/Class notes for Tuesday, November 27|Tue]], [[0708-1300/Class notes for Thursday, November 29|Thu]]<br /> |- align=left<br /> |align=center|13<br /> |Dec 3<br /> |[[0708-1300/Class notes for Tuesday, December 4|Tue]], [[0708-1300/Class notes for Thursday, December 6|Thu]], [[0708-1300/Homework Assignment 6|HW6]]<br /> |-<br /> |colspan=3 align=center|'''Spring Semester'''<br /> |- align=left<br /> |align=center|14<br /> |Jan 7<br /> |[[0708-1300/Class notes for Tuesday, January 8|Tue]], [[0708-1300/Class notes for Thursday, January 10|Thu]], [[0708-1300/Homework Assignment 7|HW7]]<br /> |- align=left<br /> |align=center|15<br /> |Jan 14<br /> |[[0708-1300/Class notes for Tuesday, January 15|Tue]], [[0708-1300/Class notes for Thursday, January 17|Thu]]<br /> |- align=left<br /> |align=center|16<br /> |Jan 21<br /> |[[0708-1300/Class notes for Tuesday, January 22|Tue]], [[0708-1300/Class notes for Thursday, January 24|Thu]], [[0708-1300/Homework Assignment 8|HW8]]<br /> |- align=left<br /> |align=center|17<br /> |Jan 28<br /> |[[0708-1300/Class notes for Tuesday, January 29|Tue]]<br /> |- align=left<br /> |align=center|18<br /> |Feb 4<br /> |[[0708-1300/Class notes for Tuesday, February 5|Tue]]<br /> |- align=left<br /> |align=center|19<br /> |Feb 11<br /> |[[0708-1300/Term Exam 2|TE2]], [[0708-1300/Homework Assignment 9|HW9]], [[0708-1300/Class notes for Thursday, February 14|Thu]], Feb 17: last chance to drop class<br /> |- align=left<br /> |align=center|R<br /> |Feb 18<br /> |Reading week<br /> |- align=left<br /> |align=center|20<br /> |Feb 25<br /> |[[0708-1300/Class notes for Tuesday, February 26|Tue]], [[0708-1300/Homework Assignment 10|HW10]]<br /> |- align=left<br /> |align=center|21<br /> |Mar 3<br /> |<br /> |- align=left<br /> |align=center|22<br /> |Mar 10<br /> |[[0708-1300/Class notes for Tuesday, March 11|Tue]], [[0708-1300/Homework Assignment 11|HW11]]<br /> |- align=left<br /> |align=center|23<br /> |Mar 17<br /> |[[0708-1300/Class notes for Tuesday, March 18|Tue]]<br /> |- align=left<br /> |align=center|24<br /> |Mar 24<br /> |[[0708-1300/Class notes for Tuesday, March 25|Tue]], [[0708-1300/Homework Assignment 12|HW12]], [[0708-1300/Class notes for Thursday, March 27|Thu]]<br /> |- align=left<br /> |align=center|25<br /> |Mar 31<br /> |[[0708-1300/Democracy At Last|Referendum]],[[0708-1300/Class notes for Tuesday, April 1|Tue]]<br /> |- align=left<br /> |align=center|26<br /> |Apr 7<br /> |<br /> |- align=left<br /> |align=center|R<br /> |Apr 14<br /> |<br /> |- align=left<br /> |align=center|R<br /> |Apr 21<br /> |<br /> |- align=left<br /> |align=center|F<br /> |Apr 28<br /> |[[0708-1300/The Final Exam|Final]] (Fri, May 2)<br /> |- align=center<br /> |colspan=3|[[0708-1300/Register of Good Deeds|Register of Good Deeds]]<br /> |- align=center<br /> |colspan=3|[[0708-1300/Errata to Bredon's Book|Errata to Bredon's Book]]<br /> |}<br /> &lt;/div&gt;&lt;/div&gt;<br /> |}<br /> &lt;div align=center style=&quot;color: red; font-size: 150%; display: none;&quot;&gt;Announcements go here&lt;/div&gt;</div> Trefor http://drorbn.net/index.php?title=0708-1300/Class_notes_for_Tuesday,_April_1 0708-1300/Class notes for Tuesday, April 1 2008-04-01T17:17:05Z <p>Trefor: </p> <hr /> <div>{{0708-1300/Navigation}}<br /> <br /> ==Typed Notes==<br /> <br /> &lt;span style=&quot;color: red;&quot;&gt;The notes below are by the students and for the students. Hopefully they are useful, but they come with no guarantee of any kind.&lt;/span&gt;<br /> <br /> ===First Hour===<br /> <br /> '''Theorem'''<br /> <br /> If &lt;math&gt;\phi:I^k\rightarrow S^n&lt;/math&gt; is an embbeding then &lt;math&gt;\tilde{H}_*(S^n-\phi(I^k)) = 0&lt;/math&gt;<br /> <br /> <br /> ''Proof:''<br /> <br /> By induction on k. <br /> <br /> For k=0 this is easy, &lt;math&gt;I^k&lt;/math&gt; is a single point and &lt;math&gt;S^n&lt;/math&gt; - a point is just &lt;math&gt;\mathbb{R}^n&lt;/math&gt;<br /> <br /> Now suppose we know the theorem is true for k-1, let &lt;math&gt;\phi:I^k\rightarrow S^n&lt;/math&gt; be an embedding. Write the cube &lt;math&gt;I^k = C&lt;/math&gt; as &lt;math&gt;C=C_L\cup C_R&lt;/math&gt; where &lt;math&gt;C_L=\{x\in C:x_1\leq 1/2\}&lt;/math&gt; and &lt;math&gt;C_R=\{x\in C:x_1\geq 1/2\}&lt;/math&gt;<br /> <br /> &lt;math&gt;C_M:= C_L\cap C_R\cong I^{k-1}&lt;/math&gt;<br /> <br /> So, &lt;math&gt;(S^n-\phi(C_m)) = (S^n-\phi(C_l))\cup(S^n-\phi(C_R))&lt;/math&gt;<br /> <br /> where both sets in the union are open. Recall for sets &lt;math&gt;X = A\cup B&lt;/math&gt; where A and B are open we have the Mayer Vietoris Sequence. <br /> <br /> Note: It is usefull to &quot;open the black box&quot; and think about the Mayer Vietoris Sequence from the veiwpoint of singular homology, ie using maps of simplexes and the like. This was briefly sketched in class.<br /> <br /> We get a sequence &lt;math&gt;\rightarrow\tilde{H}_{p+1}(S^n-\phi(C_m))\rightarrow \tilde{H}_{p}(S^n-\phi(C))\rightarrow \tilde{H}_{p}(S^n-\phi(C_L))\oplus \tilde{H}_{p}(S^n-\phi(C_R))\rightarrow \tilde{H}_{p}(S^n-\phi(C_M))\rightarrow&lt;/math&gt;<br /> <br /> The first and last terms in this sequence (at least the small part of it we have written) vanish by induction hypothesis. <br /> <br /> Note: Technically we need to check at Mayer Vietoris sequence also works for reduced homology. <br /> <br /> We thus get the isomorphism: &lt;math&gt;\tilde{H}_{p}(S^n-\phi(C))\cong \tilde{H}_{p}(S^n-\phi(C_L))\oplus \tilde{H}_{p}(S^n-\phi(C_R))&lt;/math&gt;<br /> <br /> <br /> Assume &lt;math&gt;0\neq[\beta]\in \tilde{H}_{p}(S^n-\phi(C))<br /> &lt;/math&gt;<br /> As the above isormorphism is induced by the inclusion maps at the chain level, we get that &lt;math&gt;[\beta]\neq 0&lt;/math&gt; either in &lt;math&gt;\tilde{H}_{p}(S^n-\phi(C_L))&lt;/math&gt; or in &lt;math&gt;\tilde{H}_{p}(S^n-\phi(C_R))&lt;/math&gt;<br /> <br /> <br /> Now repeat, find a sequence of intervals &lt;math&gt;I_j&lt;/math&gt; such that <br /> <br /> 1) &lt;math&gt;\cap I_j = \{x\}&lt;/math&gt;<br /> <br /> 2) &lt;math&gt;[\beta]\neq 0&lt;/math&gt; in &lt;math&gt;\tilde{H}_{p}(S^n-\phi(I_j\times I^{k-1}))&lt;/math&gt;<br /> <br /> Pictorially, we cut C in half, find which half &lt;math&gt;[\beta]&lt;/math&gt; is non zero, then cut this half in half and find the one the &lt;math&gt;[\beta]&lt;/math&gt; is non zero in and again cut it in half, etc...<br /> <br /> But, &lt;math&gt;[\beta] = 0&lt;/math&gt; in &lt;math&gt;\tilde{H}_{p}(S^n-\phi(\{x\}\times I^{k-1}))&lt;/math&gt; by induction. Hence, &lt;math&gt;\exists\gamma\in C_{p+1}(S^n-\phi(\{x\}\times I^{k-1}))&lt;/math&gt; such that &lt;math&gt;\partial\gamma=\beta&lt;/math&gt;. <br /> <br /> But, &lt;math&gt;supp\gamma&lt;/math&gt; = { union of images of simplexes in U} is compact, so, &lt;math&gt;supp\gamma\subset S^n-\phi(I_j\times I^{k-1})&lt;/math&gt; for some j. But this contradicts assumption 2)<br /> <br /> ''Q.E.D''<br /> <br /> <br /> We now prove an analogous theorem for spheres opposed to cubes<br /> <br /> <br /> '''Theorem'''<br /> <br /> For &lt;math&gt;\phi:S^k\rightarrow S^n&lt;/math&gt; an embedding then &lt;math&gt;\tilde{H}_{p}(S^n-\phi(S^k)) = \mathbb{Z}&lt;/math&gt; if p = n-k-1 and zero otherwise. <br /> <br /> <br /> Intuitively,&lt;math&gt; S^n-S^k = S^{n-k-1}&lt;/math&gt; <br /> <br /> Recall we had seen this explicitly for &lt;math&gt;S^3-S^1&lt;/math&gt; when we wrote &lt;math&gt;S^3&lt;/math&gt; as the union of two tori (an inflated circle)<br /> <br /> <br /> ''Proof:''<br /> <br /> As in the previous proof, we use Mayer Vietoris. Let &lt;math&gt;S^k = D^k_-\cup D^k_+&lt;/math&gt;<br /> <br /> Then, &lt;math&gt;S^{k-1} = D^k_-\cap D^k_+&lt;/math&gt;<br /> &lt;math&gt;<br /> \tilde{H}_{p}(S^n-\phi(S^k))\rightarrow \tilde{H}_{p}(S^n-\phi(D^k_-))\oplus\tilde{H}_{p}(S^n-\phi(D^k_+))\rightarrow \tilde{H}_{p}(S^n-\phi(S^{k-1}))\rightarrow \tilde{H}_{p-1}(S^n-\phi(S^k))&lt;/math&gt;<br /> <br /> The first and last term written vanish by induction hypothesis and so have an isomorphism &lt;math&gt;\tilde{H}_{p}(S^n-\phi(D^k_-))\oplus\tilde{H}_{p}(S^n-\phi(D^k_+))\cong \tilde{H}_{p}(S^n-\phi(S^{k-1}))&lt;/math&gt;<br /> <br /> Hence, it is enough to prove the theorem for k=0. Well in this case we have &lt;math&gt;\tilde{H}_{p}(S^n-\{2\ pts\})&lt;/math&gt;<br /> <br /> Since &lt;math&gt;S^n-\{2\ pts\}\sim S^{n-1}&lt;/math&gt;, this is &lt;math&gt;\mathbb{Z}&lt;/math&gt; for p = n-1 and 0 otherwise<br /> <br /> ''Q.E.D''<br /> <br /> <br /> <br /> Take k=n-1<br /> &lt;math&gt;<br /> \tilde{H}_{0}(S^n-\phi(S^{n-1})) =\mathbb{Z}&lt;/math&gt; so &lt;math&gt;H_{0}(S^n-\phi(S^{n-1}))=\mathbb{Z}^2&lt;/math&gt;<br /> <br /> Hence the compliment of an n-1 sphere in &lt;math&gt;S^n&lt;/math&gt; has two connected components. This is the Jordan Curve Theorem.<br /> <br /> <br /> ===Second Hour===<br /> <br /> <br /> '''Excersize'''<br /> <br /> Consider the embedding of a line in &lt;math&gt;\mathbb{R}^3&lt;/math&gt; which wraps arround somewhat like a helix only each loop makes a link with each previous loop and in addition the helical object gets smaller and converges to two points at the end. <br /> <br /> The question is, should we put a little circle L about a part of the helical object can we make this L the boundary of a disk embedded in &lt;math&gt;R^3&lt;/math&gt;. <br /> <br /> Sketch of Answer: Consider the bubble which has L as the boundary and goes all the way arround the object at one end. This is not an embedding since there are two intersection points with the helical object. Cut two small holes from the bubble at each of these points. This is still not quite it since the boundary also has these two small holes. Hence connect the two holes via a tubular neighbourhood going along the helical object and we have the result. <br /> <br /> The real excersize here is to figure out how the Mayer Vietoris sequence actually determines this particular surface. <br /> <br /> <br /> '''Proposition'''<br /> <br /> An open path connected set in &lt;math&gt;\mathbb{R}^n&lt;/math&gt; is clopen connected. <br /> <br /> ''Proof'' This is easy, just using general topology<br /> <br /> <br /> Hence, for open sets in &lt;math&gt;\mathbb{R}^n&lt;/math&gt; path components are the same as clopen components<br /> <br /> <br /> <br /> '''Theorem'''<br /> <br /> &quot;Invariance of Domain&quot; or &quot;Openness in &lt;math&gt;\mathbb{R}^n&lt;/math&gt; is intrinsic&quot;<br /> <br /> <br /> If &lt;math&gt;V\subset\mathbb{R}^n&lt;/math&gt; is homeomorphic to an open set in &lt;math&gt;\mathbb{R}^n&lt;/math&gt; then it is an open set in &lt;math&gt;\mathbb{R}^n&lt;/math&gt;<br /> <br /> <br /> Here are two examples that illustrate why this might not be obvious<br /> <br /> 1) Replacing open with dense would not be true, for instance, the rationals in [0,1] are not dence in &lt;math&gt;\mathbb{R}&lt;/math&gt; but they ARE homeomorphic to the rationals which ARE dense in R. <br /> <br /> 2) Replacing open with closed is also not true, for instance, &lt;math&gt;\mathbb{R}^n&lt;/math&gt; is closed in &lt;math&gt;\mathbb{R}^n&lt;/math&gt;<br /> <br /> but it is also homeomorphic to the open ball which is not closed. <br /> <br /> <br /> ''Proof''<br /> <br /> Let &lt;math&gt;\phi:U\rightarrow B&lt;/math&gt;<br /> <br /> Consider a ball B in U, with boundary the circle S and we get a disk &lt;math&gt;D=B\cup S&lt;/math&gt;<br /> <br /> We have that &lt;math&gt;\phi(S)&lt;/math&gt; devides &lt;math&gt;\mathbb{R}^n_V&lt;/math&gt; into two connected components. Call them &lt;math&gt;A_1&lt;/math&gt; and &lt;math&gt;A_2&lt;/math&gt;. <br /> <br /> We have both &lt;math&gt;\phi(B)&lt;/math&gt; and &lt;math&gt;\mathbb{R}^m - \phi(D)&lt;/math&gt; but we need to show these ARE the two connected components. We first note that both are in fact connected. <br /> <br /> &lt;math&gt;(R^n -\phi(D))\cup\phi(B) = \mathbb{R}^n-\phi(S) = A_1\cup A_2&lt;/math&gt;<br /> <br /> Hence &lt;math&gt;\phi(B)&lt;/math&gt; is one of the components and so it is open. ''Q.E.D''<br /> <br /> <br /> <br /> '''Theorem''' Borsuk-Ulam<br /> <br /> If &lt;math&gt;g:S^n\rightarrow R^n&lt;/math&gt; is continuous then &lt;math&gt;\exists x\in S^n&lt;/math&gt; such that g(x) = g(-x)<br /> <br /> <br /> '''Corrollary''' The Ham Sandwich Theorem (or Salad Bowl Theorem, according to Dr. Bar Natan)<br /> <br /> Loosely this says that if we have n objects in a bounded domain of &lt;math&gt;\mathbb{R}^n&lt;/math&gt; such as items in a salad or sandwhich then we can find an n-1 dimensional hyperplane such that the hyperplane precisely cuts each item in half. <br /> <br /> Formally, If &lt;math&gt;\mu_1,\cdots,\mu_n&lt;/math&gt; are densities in &lt;math&gt;\mathbb{R}^n&lt;/math&gt; with bounded support then &lt;math&gt;\exists&lt;/math&gt; a hyperplane &lt;math&gt;H\subset\mathbb{R}^n&lt;/math&gt; deviding &lt;math&gt;\mathbb{R}^n&lt;/math&gt; into &lt;math&gt;H_+\cup H_-&lt;/math&gt; such that &lt;math&gt;\forall j&lt;/math&gt;, &lt;math&gt;\mu_j(H_+) = \mu_j(H_-)&lt;/math&gt;<br /> <br /> <br /> ''Proof:''<br /> <br /> Consider the set of sided hyperplanes in &lt;math&gt;\mathbb{R}^n&lt;/math&gt; which equals &lt;math&gt;S^{n-1}_v\times R_t&lt;/math&gt; where the first component determines the normal vector v and the second component determines how far away this plane is front the origin in the direction of v. <br /> <br /> Define &lt;math&gt;\bar{g}:S^{n-1}\times\mathbb{R}\rightarrow\mathbb{R}^n&lt;/math&gt; via<br /> <br /> &lt;math&gt;(v,t)\mapsto (\mu_1(H_+),\cdots, \mu_n(H_-))&lt;/math&gt;<br /> <br /> Now, if t&gt;&gt;0, clearly &lt;math&gt;\bar{g}(v,t) = 0&lt;/math&gt; and if t&lt;&lt;0 then &lt;math&gt;\bar{g}(v,t) = (\mu_1(\mathbb{R}^n),\cdots,\mu_n(\mathbb{R}^n))&lt;/math&gt;<br /> <br /> Hence we have a cylinder mapping into &lt;math&gt;\mathbb{R}^n&lt;/math&gt; such that it is constant along the top and bottom of the cylindar. Hence, we can pinch the top and bottom of the cylinder. We thus get a map g from &lt;math&gt;S^n&lt;/math&gt; to &lt;math&gt;\mathbb{R}^n&lt;/math&gt; and, applying Borsuk-Ulam, deduce there is a point such that g(x) = g(-x) and so &lt;math&gt;\bar{g}(v,t) = \bar{g}(-v,-t)&lt;/math&gt; for some point (v,t). Hence &lt;math&gt;\mu_i(H_+)=\mu_i(H_-)&lt;/math&gt; for all i<br /> <br /> ''Q.E.D''</div> Trefor http://drorbn.net/index.php?title=0708-1300/Class_notes_for_Tuesday,_April_1 0708-1300/Class notes for Tuesday, April 1 2008-04-01T17:16:49Z <p>Trefor: </p> <hr /> <div>{{0708-1300/Navigation}}<br /> <br /> ==Typed Notes==<br /> <br /> &lt;span style=&quot;color: red;&quot;&gt;The notes below are by the students and for the students. Hopefully they are useful, but they come with no guarantee of any kind.&lt;/span&gt;<br /> <br /> ===First Hour===<br /> <br /> '''Theorem'''<br /> <br /> If &lt;math&gt;\phi:I^k\rightarrow S^n&lt;/math&gt; is an embbeding then &lt;math&gt;\tilde{H}_*(S^n-\phi(I^k)) = 0&lt;/math&gt;<br /> <br /> <br /> ''Proof:''<br /> <br /> By induction on k. <br /> <br /> For k=0 this is easy, &lt;math&gt;I^k&lt;/math&gt; is a single point and &lt;math&gt;S^n&lt;/math&gt; - a point is just &lt;math&gt;\mathbb{R}^n&lt;/math&gt;<br /> <br /> Now suppose we know the theorem is true for k-1, let &lt;math&gt;\phi:I^k\rightarrow S^n&lt;/math&gt; be an embedding. Write the cube &lt;math&gt;I^k = C&lt;/math&gt; as &lt;math&gt;C=C_L\cup C_R&lt;/math&gt; where &lt;math&gt;C_L=\{x\in C:x_1\leq 1/2\}&lt;/math&gt; and &lt;math&gt;C_R=\{x\in C:x_1\geq 1/2\}&lt;/math&gt;<br /> <br /> &lt;math&gt;C_M:= C_L\cap C_R\cong I^{k-1}&lt;/math&gt;<br /> <br /> So, &lt;math&gt;(S^n-\phi(C_m)) = (S^n-\phi(C_l))\cup(S^n-\phi(C_R))&lt;/math&gt;<br /> <br /> where both sets in the union are open. Recall for sets &lt;math&gt;X = A\cup B&lt;/math&gt; where A and B are open we have the Mayer Vietoris Sequence. <br /> <br /> Note: It is usefull to &quot;open the black box&quot; and think about the Mayer Vietoris Sequence from the veiwpoint of singular homology, ie using maps of simplexes and the like. This was briefly sketched in class.<br /> <br /> We get a sequence &lt;math&gt;\rightarrow\tilde{H}_{p+1}(S^n-\phi(C_m))\rightarrow \tilde{H}_{p}(S^n-\phi(C))\rightarrow \tilde{H}_{p}(S^n-\phi(C_L))\oplus \tilde{H}_{p}(S^n-\phi(C_R))\rightarrow \tilde{H}_{p}(S^n-\phi(C_M))\rightarrow&lt;/math&gt;<br /> <br /> The first and last terms in this sequence (at least the small part of it we have written) vanish by induction hypothesis. <br /> <br /> Note: Technically we need to check at Mayer Vietoris sequence also works for reduced homology. <br /> <br /> We thus get the isomorphism: &lt;math&gt;\tilde{H}_{p}(S^n-\phi(C))\cong \tilde{H}_{p}(S^n-\phi(C_L))\oplus \tilde{H}_{p}(S^n-\phi(C_R))&lt;/math&gt;<br /> <br /> <br /> Assume &lt;math&gt;0\neq[\beta]\in \tilde{H}_{p}(S^n-\phi(C))<br /> &lt;/math&gt;<br /> As the above isormorphism is induced by the inclusion maps at the chain level, we get that &lt;math&gt;[\beta]\neq 0&lt;/math&gt; either in &lt;math&gt;\tilde{H}_{p}(S^n-\phi(C_L))&lt;/math&gt; or in &lt;math&gt;\tilde{H}_{p}(S^n-\phi(C_R))&lt;/math&gt;<br /> <br /> <br /> Now repeat, find a sequence of intervals &lt;math&gt;I_j&lt;/math&gt; such that <br /> <br /> 1) &lt;math&gt;\cap I_j = \{x\}&lt;/math&gt;<br /> <br /> 2) &lt;math&gt;[\beta]\neq 0&lt;/math&gt; in &lt;math&gt;\tilde{H}_{p}(S^n-\phi(I_j\times I^{k-1}))&lt;/math&gt;<br /> <br /> Pictorially, we cut C in half, find which half &lt;math&gt;[\beta]&lt;/math&gt; is non zero, then cut this half in half and find the one the &lt;math&gt;[\beta]&lt;/math&gt; is non zero in and again cut it in half, etc...<br /> <br /> But, &lt;math&gt;[\beta] = 0&lt;/math&gt; in &lt;math&gt;\tilde{H}_{p}(S^n-\phi(\{x\}\times I^{k-1}))&lt;/math&gt; by induction. Hence, &lt;math&gt;\exists\gamma\in C_{p+1}(S^n-\phi(\{x\}\times I^{k-1}))&lt;/math&gt; such that &lt;math&gt;\partial\gamma=\beta&lt;/math&gt;. <br /> <br /> But, &lt;math&gt;supp\gamma&lt;/math&gt; = { union of images of simplexes in U} is compact, so, &lt;math&gt;supp\gamma\subset S^n-\phi(I_j\times I^{k-1})&lt;/math&gt; for some j. But this contradicts assumption 2)<br /> <br /> ''Q.E.D''<br /> <br /> <br /> We now prove an analogous theorem for spheres opposed to cubes<br /> <br /> <br /> '''Theorem'''<br /> <br /> For &lt;math&gt;\phi:S^k\rightarrow S^n&lt;/math&gt; an embedding then &lt;math&gt;\tilde{H}_{p}(S^n-\phi(S^k)) = \mathbb{Z}&lt;/math&gt; if p = n-k-1 and zero otherwise. <br /> <br /> <br /> Intuitively,&lt;math&gt; S^n-S^k = S^{n-k-1}&lt;/math&gt; <br /> <br /> Recall we had seen this explicitly for &lt;math&gt;S^3-S^1&lt;/math&gt; when we wrote &lt;math&gt;S^3&lt;/math&gt; as the union of two tori (an inflated circle)<br /> <br /> <br /> ''Proof:''<br /> <br /> As in the previous proof, we use Mayer Vietoris. Let &lt;math&gt;S^k = D^k_-\cup D^k_+&lt;/math&gt;<br /> <br /> Then, &lt;math&gt;S^{k-1} = D^k_-\cap D^k_+&lt;/math&gt;<br /> &lt;math&gt;<br /> \tilde{H}_{p}(S^n-\phi(S^k))\rightarrow \tilde{H}_{p}(S^n-\phi(D^k_-))\oplus\tilde{H}_{p}(S^n-\phi(D^k_+))\rightarrow \tilde{H}_{p}(S^n-\phi(S^{k-1}))\rightarrow \tilde{H}_{p-1}(S^n-\phi(S^k))&lt;/math&gt;<br /> <br /> The first and last term written vanish by induction hypothesis and so have an isomorphism &lt;math&gt;\tilde{H}_{p}(S^n-\phi(D^k_-))\oplus\tilde{H}_{p}(S^n-\phi(D^k_+))\cong \tilde{H}_{p}(S^n-\phi(S^{k-1}))&lt;/math&gt;<br /> <br /> Hence, it is enough to prove the theorem for k=0. Well in this case we have &lt;math&gt;\tilde{H}_{p}(S^n-\{2\ pts\})&lt;/math&gt;<br /> <br /> Since &lt;math&gt;S^n-\{2\ pts\}\sim S^{n-1}&lt;/math&gt;, this is &lt;math&gt;\mathbb{Z}&lt;/math&gt; for p = n-1 and 0 otherwise<br /> <br /> ''Q.E.D''<br /> <br /> <br /> <br /> Take k=n-1<br /> &lt;math&gt;<br /> \tilde{H}_{0}(S^n-\phi(S^{n-1})) =\mathbb{Z}&lt;/math&gt; so &lt;math&gt;H_{0}(S^n-\phi(S^{n-1}))=\mathbb{Z}^2&lt;/math&gt;<br /> <br /> Hence the compliment of an n-1 sphere in &lt;math&gt;S^n&lt;/math&gt; has two connected components. This is the Jordan Curve Theorem.</div> Trefor http://drorbn.net/index.php?title=Template:0708-1300/Navigation Template:0708-1300/Navigation 2008-03-25T18:04:16Z <p>Trefor: </p> <hr /> <div>{| cellpadding=&quot;0&quot; cellspacing=&quot;0&quot; style=&quot;clear: right; float: right&quot;<br /> |- align=right<br /> |&lt;div class=&quot;NavFrame&quot;&gt;&lt;div class=&quot;NavHead&quot;&gt;[[0708-1300]]/[[Template:0708-1300/Navigation|Navigation Panel]]&amp;nbsp;&amp;nbsp;&lt;/div&gt;<br /> &lt;div class=&quot;NavContent&quot;&gt;<br /> {| border=&quot;1px&quot; cellpadding=&quot;1&quot; cellspacing=&quot;0&quot; width=&quot;220&quot; style=&quot;margin: 0 0 1em 0.5em; font-size: small&quot;<br /> |-<br /> |colspan=3 align=center|[[Image:0708-1300-ClassPhoto.jpg|215px]]&lt;br&gt;[[0708-1300/Class Photo|Add your name / see who's in!]]<br /> |- align=center<br /> !#<br /> !Week of...<br /> !Links<br /> |-<br /> |colspan=3 align=center|'''Fall Semester'''<br /> |- align=left<br /> |align=center|1<br /> |Sep 10<br /> |[[0708-1300/About This Class|About]], [[0708-1300/Class notes for Tuesday, September 11|Tue]], [[0708-1300/Class notes for Thursday, September 13|Thu]]<br /> |- align=left<br /> |align=center|2<br /> |Sep 17<br /> |[[0708-1300/Class notes for Tuesday, September 18|Tue]], [[0708-1300/Homework Assignment 1|HW1]], [[0708-1300/Class notes for Thursday, September 20|Thu]]<br /> |- align=left<br /> |align=center|3<br /> |Sep 24<br /> |[[0708-1300/Class notes for Tuesday, September 25|Tue]], [[0708-1300/Class Photo|Photo]], [[0708-1300/Class notes for Thursday, September 27|Thu]]<br /> |- align=left<br /> |align=center|4<br /> |Oct 1<br /> |[[0708-1300/Questionnaire|Questionnaire]], [[0708-1300/Class notes for Tuesday, October 2|Tue]], [[0708-1300/Homework Assignment 2|HW2]], [[0708-1300/Class notes for Thursday, October 4|Thu]]<br /> |- align=left<br /> |align=center|5<br /> |Oct 8<br /> |Thanksgiving, [[0708-1300/Class notes for Tuesday, October_9|Tue]], [[0708-1300/Class notes for Thursday, October 11|Thu]]<br /> |- align=left<br /> |align=center|6<br /> |Oct 15<br /> |[[0708-1300/Class notes for Tuesday, October 16|Tue]], [[0708-1300/Homework Assignment 3|HW3]], [[0708-1300/Class notes for Thursday, October 18|Thu]]<br /> |- align=left<br /> |align=center|7<br /> |Oct 22<br /> |[[0708-1300/Class notes for Tuesday, October 23|Tue]], [[0708-1300/Class notes for Thursday, October 25|Thu]]<br /> |- align=left<br /> |align=center|8<br /> |Oct 29<br /> |[[0708-1300/Class notes for Tuesday, October 30|Tue]], [[0708-1300/Homework Assignment 4|HW4]], [[0708-1300/Class notes for Thursday, November 1|Thu]], [[0708-1300/the unit sphere in a Hilbert space is contractible|Hilbert sphere]]<br /> |- align=left<br /> |align=center|9<br /> |Nov 5<br /> |[[0708-1300/Class notes for Tuesday, November 6|Tue]],[[0708-1300/Class notes for Thursday, November 8|Thu]], [[0708-1300/Term Exam 1|TE1]]<br /> |- align=left<br /> |align=center|10<br /> |Nov 12<br /> |[[0708-1300/Class notes for Tuesday, November 13|Tue]], &lt;strike&gt;[[0708-1300/Class notes for Thursday, November 15|Thu]]&lt;/strike&gt;<br /> |- align=left<br /> |align=center|11<br /> |Nov 19<br /> |[[0708-1300/Class notes for Tuesday, November 20|Tue]], [[0708-1300/Homework Assignment 5|HW5]]<br /> |- align=left<br /> |align=center|12<br /> |Nov 26<br /> |[[0708-1300/Class notes for Tuesday, November 27|Tue]], [[0708-1300/Class notes for Thursday, November 29|Thu]]<br /> |- align=left<br /> |align=center|13<br /> |Dec 3<br /> |[[0708-1300/Class notes for Tuesday, December 4|Tue]], [[0708-1300/Class notes for Thursday, December 6|Thu]], [[0708-1300/Homework Assignment 6|HW6]]<br /> |-<br /> |colspan=3 align=center|'''Spring Semester'''<br /> |- align=left<br /> |align=center|14<br /> |Jan 7<br /> |[[0708-1300/Class notes for Tuesday, January 8|Tue]], [[0708-1300/Class notes for Thursday, January 10|Thu]], [[0708-1300/Homework Assignment 7|HW7]]<br /> |- align=left<br /> |align=center|15<br /> |Jan 14<br /> |[[0708-1300/Class notes for Tuesday, January 15|Tue]], [[0708-1300/Class notes for Thursday, January 17|Thu]]<br /> |- align=left<br /> |align=center|16<br /> |Jan 21<br /> |[[0708-1300/Class notes for Tuesday, January 22|Tue]], [[0708-1300/Class notes for Thursday, January 24|Thu]], [[0708-1300/Homework Assignment 8|HW8]]<br /> |- align=left<br /> |align=center|17<br /> |Jan 28<br /> |[[0708-1300/Class notes for Tuesday, January 29|Tue]]<br /> |- align=left<br /> |align=center|18<br /> |Feb 4<br /> |[[0708-1300/Class notes for Tuesday, February 5|Tue]]<br /> |- align=left<br /> |align=center|19<br /> |Feb 11<br /> |[[0708-1300/Term Exam 2|TE2]], [[0708-1300/Homework Assignment 9|HW9]], [[0708-1300/Class notes for Thursday, February 14|Thu]], Feb 17: last chance to drop class<br /> |- align=left<br /> |align=center|R<br /> |Feb 18<br /> |Reading week<br /> |- align=left<br /> |align=center|20<br /> |Feb 25<br /> |[[0708-1300/Class notes for Tuesday, February 26|Tue]], [[0708-1300/Homework Assignment 10|HW10]]<br /> |- align=left<br /> |align=center|21<br /> |Mar 3<br /> |<br /> |- align=left<br /> |align=center|22<br /> |Mar 10<br /> |[[0708-1300/Class notes for Tuesday, March 11|Tue]], [[0708-1300/Homework Assignment 11|HW11]]<br /> |- align=left<br /> |align=center|23<br /> |Mar 17<br /> |[[0708-1300/Class notes for Tuesday, March 18|Tue]]<br /> |- align=left<br /> |align=center|24<br /> |Mar 24<br /> |HW12,[[0708-1300/Class notes for Tuesday, March 25|Tue]]<br /> |- align=left<br /> |align=center|25<br /> |Mar 31<br /> |<br /> |- align=left<br /> |align=center|26<br /> |Apr 7<br /> |<br /> |- align=left<br /> |align=center|R<br /> |Apr 14<br /> |<br /> |- align=left<br /> |align=center|R<br /> |Apr 21<br /> |<br /> |- align=left<br /> |align=center|F<br /> |Apr 28<br /> |[[0708-1300/The Final Exam|Final]] (Fri, May 2)<br /> |- align=center<br /> |colspan=3|[[0708-1300/Register of Good Deeds|Register of Good Deeds]]<br /> |- align=center<br /> |colspan=3|[[0708-1300/Errata to Bredon's Book|Errata to Bredon's Book]]<br /> |}<br /> &lt;/div&gt;&lt;/div&gt;<br /> |}<br /> &lt;div align=center style=&quot;color: red; font-size: 150%; display: none;&quot;&gt;Announcements go here&lt;/div&gt;</div> Trefor http://drorbn.net/index.php?title=0708-1300/Class_notes_for_Tuesday,_March_25 0708-1300/Class notes for Tuesday, March 25 2008-03-25T18:01:31Z <p>Trefor: </p> <hr /> <div>{{0708-1300/Navigation}}<br /> <br /> ==Typed Notes==<br /> <br /> &lt;span style=&quot;color: red;&quot;&gt;The notes below are by the students and for the students. Hopefully they are useful, but they come with no guarantee of any kind.&lt;/span&gt;<br /> <br /> ===First Hour===<br /> <br /> '''Definition'''<br /> <br /> We define the CW chain complex via:<br /> <br /> &lt;math&gt;C_n^{CW}(K):=&lt;K_n&gt;&lt;/math&gt;<br /> <br /> and the boundary maps via &lt;math&gt;\partial:C_n^{CW}\rightarrow C^{CW}_{n-1}&lt;/math&gt; by &lt;math&gt;\partial\sigma=\sum_{\tau\in K_{n-1}}[\tau:\sigma]\tau&lt;/math&gt;<br /> <br /> where &lt;math&gt;[\tau:\sigma]&lt;/math&gt; is roughly the number of times that &lt;math&gt;\partial\sigma&lt;/math&gt; covers &lt;math&gt;\tau&lt;/math&gt;. <br /> <br /> <br /> Let's make this precise. <br /> <br /> For &lt;math&gt;f_{\sigma}:D_{\sigma}^n\rightarrow K&lt;/math&gt; (not quite an embedding) this restricts to a map &lt;math&gt;f_{\partial\sigma}:S_{\sigma}^n\rightarrow K&lt;/math&gt;. Given &lt;math&gt;\tau\in K_m&lt;/math&gt; let &lt;math&gt;p_{\tau}:K^n\rightarrow S^n = B^n\cup\{\infty\}&lt;/math&gt; such that &lt;math&gt;int(D^n_{\tau})\mapsto B^n&lt;/math&gt; and the rest maps to the point &lt;math&gt;\infty&lt;/math&gt;. <br /> <br /> <br /> Example:<br /> <br /> If you just have the segment consisting of two endpoints and the line connecting them, and call this &lt;math&gt;\tau&lt;/math&gt;, then &lt;math&gt;p_{\tau}&lt;/math&gt; takes the two end points to &lt;math&gt;\infty&lt;/math&gt; and the rest (the open interval) gets mapped to B^1. Hence, we get a circle. <br /> <br /> <br /> We thus can now formally define &lt;math&gt;[\tau:\sigma]= deg(p_{\tau}\circ f_{\partial\sigma}:S^{n-1}\rightarrow S^{n-1})&lt;/math&gt;<br /> <br /> <br /> <br /> '''Theorem'''<br /> <br /> <br /> &lt;math&gt;(C^{CW}_*,\partial)&lt;/math&gt; is a chain complex; &lt;math&gt;\partial^2 = 0&lt;/math&gt; and &lt;math&gt;H_*^{CW}(K) = H_*(K)&lt;/math&gt;<br /> <br /> <br /> '''Examples:'''<br /> <br /> <br /> 1) &lt;math&gt;S^n = \{\infty\}\cup D^n&lt;/math&gt; for n&gt;1<br /> <br /> &lt;math&gt;f_{\partial\sigma}: S^{n-1}\rightarrow \infty&lt;/math&gt;<br /> <br /> Hence &lt;math&gt;C^{CW}_n = &lt;\sigma&gt;, C^{CW}_0 = &lt;\infty&gt;&lt;/math&gt; and all the rest are zero. Hence, &lt;math&gt;H_p(S^n) = \mathbb{Z}&lt;/math&gt; for p = 0 or n and is zero otherwise. <br /> <br /> <br /> 2) Consider the torus thought of as a square with the usual identifications and &lt;math&gt;\sigma&lt;/math&gt; is the interior. Hence, &lt;math&gt;C^{CW}_0 =\{p\}, C^{CW}_1&lt;/math&gt; is generated by the figure 8 with one loop labeled a and the other labeled b, and &lt;math&gt;C^{CW}_2&lt;/math&gt; is generated by the entire torus. <br /> <br /> Ie we get &lt;math&gt;\mathbb{Z}\rightarrow\mathbb{Z}^2\rightarrow\mathbb{Z}&lt;/math&gt;<br /> <br /> Now, &lt;math&gt;\partial a = [p:a]p&lt;/math&gt;<br /> <br /> but &lt;math&gt;\partial&lt;/math&gt; a takes the two endpoints of a (both p) and maps them to p. Neither point is mapped to &lt;math&gt;\infty&lt;/math&gt;. Hence, &lt;math&gt;deg\partial a:S^0\rightarrow S^0 = 0&lt;/math&gt;<br /> <br /> Note: This ought to be checked from the definition of degree but was just stated in class<br /> <br /> <br /> Now, &lt;math&gt;[\sigma:a]&lt;/math&gt; = the degree of the map that takes the square to the figure 8...and hence is &lt;math&gt;\pm 1\mp 1 = 0&lt;/math&gt;. <br /> <br /> Hence the boundary map is zero at all places, so &lt;math&gt;H_n(\mathbb{T}^2) = \mathbb{Z}&lt;/math&gt; if n = 0 or 2, &lt;math&gt;\mathbb{Z}^2&lt;/math&gt; if n = 1 and is zero otherwise. <br /> <br /> <br /> 3) Consider the Klein bottle thought of as a square with the usual identifications. Under &lt;math&gt;p_b\circ f_{\partial\sigma}&lt;/math&gt; takes this to a circle with side labled b. <br /> <br /> I.e., &lt;math&gt;&lt;\sigma&gt;\mapsto&lt;a,b&gt;\mapsto^0&lt;p&gt;&lt;/math&gt;<br /> <br /> &lt;math&gt;\partial\sigma = [a:\sigma]a+b:\sigma b = 0_a + 2b&lt;/math&gt;<br /> <br /> Where the sign may be negative. Or more eloquantly put: &quot;2b or -2b, that is the question&quot;<br /> <br /> The kernal of &lt;math&gt;&lt;a,b&gt;\rightarrow&lt;p&gt;&lt;/math&gt; is everything, so the homology is &lt;math&gt;H_1(K) = &lt;a,b&gt;/2b=0\cong\mathbb{Z}\oplus\mathbb{Z}/2&lt;/math&gt;, &lt;math&gt;H_0(K)=\mathbb{Z}&lt;/math&gt; and &lt;math&gt;H_2(K) = 0&lt;/math&gt;.<br /> <br /> <br /> 4) &lt;math&gt;\Sigma_g&lt;/math&gt; the &quot;g holed torus&quot; or &quot;surface of genus g&quot; is formed by the normal diagram with edges identified in sets of 4 such as &lt;math&gt;aba^{-1}b^{-1}&lt;/math&gt;<br /> <br /> So, get &lt;math&gt;&lt;\sigma&gt;\rightarrow^0&lt;a_1,\cdots, a_g, b_1,\cdots,b_g&gt;\rightarrow^0&lt;p&gt;&lt;/math&gt;<br /> <br /> Therefore, &lt;math&gt;H_p = \mathbb{Z}&lt;/math&gt; if p = 2 or 0, &lt;math&gt;\mathbb{Z}^{2g}&lt;/math&gt; if p = 1 and zero elsewhere. <br /> <br /> <br /> 5) &lt;math&gt;\mathbb{R}P^n= D^n\cup \mathbb{R}P^{n-1} = D^n\cup D^{n-1}\cup\cdots\cup D^0&lt;/math&gt;<br /> <br /> So, &lt;math&gt;C_*^{CW}(\mathbb{R}P^n) is &lt;D^n&gt;\rightarrow&lt;D^{n-1}&gt;\rightarrow\cdots\rightarrow &lt;D^0&gt;&lt;/math&gt;<br /> <br /> The boundary map alternates between 0 and 2 where it is 2 for &lt;math&gt;&lt;D^i&gt;\rightarrow&lt;D^{i-1}&gt;&lt;/math&gt; if i is even and 0 if it is odd. <br /> <br /> Hence the homology alternates between &lt;math&gt;\mathbb{Z}/2&lt;/math&gt; for odd p and 0 for even p. <br /> <br /> <br /> ===Second Hour===<br /> <br /> &lt;math&gt;\mathbb{C}P^n:\{[z_0,\cdots,z_n]:z_i\in\mathbb{C}\ not\ all\ zero\}=\mathbb{C}^{n+1}/z\sim\alpha z&lt;/math&gt; for &lt;math&gt;z\in\mathbb{C}^{n+1}&lt;/math&gt;, &lt;math&gt;\alpha\in\mathbb{C}-\{0\}&lt;/math&gt;<br /> <br /> &lt;math&gt;\mathbb{C}P^n=\{[z_0,\cdots,z_n]:z_n\neq 0\}\cup\{[\ldots,0]\} = \{[z_0,\cdots,z_{n-1},1]\}\cup\mathbb{C}P^{n-1}&lt;/math&gt;<br /> =&lt;math&gt; \mathbb{C}^n\cup\mathbb{C}P^{n-1} = \mathbb{R}^{2n}\cup\mathbb{C}P^{n-1} = B^{2n}\cup\mathbb{C}P^{n-1}&lt;/math&gt;<br /> <br /> <br /> I.e., &lt;math&gt;\mathbb{C}P^n = D^{2n}\cup D^{2n-1}\cup\cdots&lt;/math&gt;<br /> <br /> <br /> So, &lt;math&gt;C^{CW}_*(\mathbb{C}P^n)&lt;/math&gt; alternates between 0 (for odd p) and &lt;math&gt;\mathbb{Z}&lt;/math&gt; for even p and thus as this also as the homology. (and clearly trivial greater than p)</div> Trefor http://drorbn.net/index.php?title=WikiHelp WikiHelp 2008-03-22T18:48:29Z <p>Trefor: </p> <hr /> <div>==How to get started using Wiki==<br /> <br /> ===Where/How to Scan Notes===<br /> <br /> # Go to Robarts Library to the &quot;Design Space&quot; computers (first floor, to the left of the information desk, all the way at the back of the building). All of these computers have a scanner. It's best to go in the morning or late evening to avoid line-ups. <br /> # Click Start--&gt;Publishing Tools--&gt;ScanWizard 5. <br /> # If something doesn't work right, it might be that the scanner is not turned on.<br /> # I used the options Original--&gt;Photo, Scan Type--&gt;Grey, Purpose--&gt;Laser Print Fine, and saved the file directly in the My Documents (temporary files) directory, to erase later. This seems to make the size of the file small enough for uploading.<br /> <br /> ===How to add a link to the Navigation Panel===<br /> <br /> # This I've spend half an hour trying to figure this out, so maybe someone could add instructions? I want to link a &quot;Thu&quot; page to the navigation panel to add notes, maybe it's obvious, but I don't know how to do it.<br /> <br /> ANSWER: <br /> <br /> Step 1) Go to the Navigation page (click the link at the top of the panel) and then click edit. Here is a direct link: <br /> http://katlas.math.toronto.edu/drorbn/index.php?title=Template:0708-1300/Navigation&amp;action=edit<br /> <br /> Step 2) As you scroll down you will notice a repeating pattern:<br /> <br /> |align=center|22<br /> <br /> |Mar 10<br /> <br /> |[%[0708-1300/Class notes for Tuesday, March 11|Tue]], [%[0708-1300/Homework Assignment 11|HW11]]<br /> <br /> |- align=left<br /> <br /> |align=center|23<br /> <br /> <br /> If nothing has been added for a given week it might look like this: <br /> <br /> <br /> |align=center|22<br /> <br /> |Mar 10<br /> <br /> |<br /> <br /> |- align=left<br /> <br /> |align=center|23<br /> <br /> <br /> <br /> Scroll down until you are in the week with the correct date that you want to add a page. (they correspond to mondays i think). Where the | is, simply write something like [%[0708-1300/Class notes for Tuesday, March 11|Tue]] where of course you put in the correct date. If there already is a something after the | such as a homework assignment link, put a comma and then type [%[0708-1300/Class notes for Tuesday, March 11|Tue]] with the date corrected. Now go to http://katlas.org/drorbn/index.php?title=0708-1300/Class_notes_for_Tuesday%2C_March_11 and this is where the page you have just created can be found.<br /> <br /> As a final note I put the symbol &quot;%&quot; in several places above just to break the format so it wouldn't appear as links in this page. You of course have to delete the % at all places that I put in. <br /> <br /> Hope the helps!</div> Trefor http://drorbn.net/index.php?title=User_talk:Canghel User talk:Canghel 2008-03-22T17:51:39Z <p>Trefor: </p> <hr /> <div>Dear Canghel,<br /> <br /> I've deleted [[Image:0708-1300-Lect-25-09.pdf]] because it was not linked anywhere.<br /> <br /> --[[User:Drorbn|Drorbn]] 17:46, 28 September 2007 (EDT)<br /> <br /> <br /> I responded to your query in the wikihelp page<br /> <br /> Trefor</div> Trefor http://drorbn.net/index.php?title=WikiHelp WikiHelp 2008-03-22T17:48:58Z <p>Trefor: </p> <hr /> <div>==How to get started using Wiki==<br /> <br /> ===Where/How to Scan Notes===<br /> <br /> # Go to Robarts Library to the &quot;Design Space&quot; computers (first floor, to the left of the information desk, all the way at the back of the building). All of these computers have a scanner. It's best to go in the morning or late evening to avoid line-ups. <br /> # Click Start--&gt;Publishing Tools--&gt;ScanWizard 5. <br /> # If something doesn't work right, it might be that the scanner is not turned on.<br /> # I used the options Original--&gt;Photo, Scan Type--&gt;Grey, Purpose--&gt;Laser Print Fine, and saved the file directly in the My Documents (temporary files) directory, to erase later. This seems to make the size of the file small enough for uploading.<br /> <br /> ===How to add a link to the Navigation Panel===<br /> <br /> # This I've spend half an hour trying to figure this out, so maybe someone could add instructions? I want to link a &quot;Thu&quot; page to the navigation panel to add notes, maybe it's obvious, but I don't know how to do it.<br /> <br /> ANSWER: <br /> <br /> Step 1) Go to the Navigation page (click the link at the top of the panel) and then click edit. Here is a direct link: <br /> http://katlas.math.toronto.edu/drorbn/index.php?title=Template:0708-1300/Navigation&amp;action=edit<br /> <br /> Step 2)As you scroll down you will notice a repeating pattern:<br /> <br /> |align=center|22<br /> |Mar 10<br /> |[%[0708-1300/Class notes for Tuesday, March 11|Tue]], [%[0708-1300/Homework Assignment 11|HW11]]<br /> |- align=left<br /> |align=center|23<br /> <br /> If nothing has been added for a given week it might look like this: <br /> <br /> |align=center|22<br /> |Mar 10<br /> |<br /> |- align=left<br /> |align=center|23<br /> <br /> Scroll down until you are in the week with the correct date that you want to add a page. (they correspond to mondays i think). Where the | is, simply write something like [%[0708-1300/Class notes for Tuesday, March 11|Tue]] where of course you put in the correct date. If there already is a something after the | such as a homework assignment link, put a comma and then type [%[0708-1300/Class notes for Tuesday, March 11|Tue]] with the date corrected. Now go to http://katlas.org/drorbn/index.php?title=0708-1300/Class_notes_for_Tuesday%2C_March_11 and this is where the page you have just created can be found.<br /> <br /> As a final note I put the symbol &quot;%&quot; in several places above just to break the format so it wouldn't appear as links in this page. You of course have to delete the % at all places that I put in. <br /> <br /> Hope the helps!</div> Trefor http://drorbn.net/index.php?title=WikiHelp WikiHelp 2008-03-22T17:47:12Z <p>Trefor: </p> <hr /> <div>==How to get started using Wiki==<br /> <br /> ===Where/How to Scan Notes===<br /> <br /> # Go to Robarts Library to the &quot;Design Space&quot; computers (first floor, to the left of the information desk, all the way at the back of the building). All of these computers have a scanner. It's best to go in the morning or late evening to avoid line-ups. <br /> # Click Start--&gt;Publishing Tools--&gt;ScanWizard 5. <br /> # If something doesn't work right, it might be that the scanner is not turned on.<br /> # I used the options Original--&gt;Photo, Scan Type--&gt;Grey, Purpose--&gt;Laser Print Fine, and saved the file directly in the My Documents (temporary files) directory, to erase later. This seems to make the size of the file small enough for uploading.<br /> <br /> ===How to add a link to the Navigation Panel===<br /> <br /> # This I've spend half an hour trying to figure this out, so maybe someone could add instructions? I want to link a &quot;Thu&quot; page to the navigation panel to add notes, maybe it's obvious, but I don't know how to do it.<br /> <br /> ANSWER: <br /> <br /> Step 1) Go to the Navigation page (click the link at the top of the panel) and then click edit. Here is a direct link: <br /> http://katlas.math.toronto.edu/drorbn/index.php?title=Template:0708-1300/Navigation&amp;action=edit<br /> <br /> Step 2)As you scroll down you will notice a repeating pattern:<br /> <br /> |align=center|22<br /> |Mar 10<br /> |[[0708-1300/Class notes for Tuesday, March 11|Tue]], [[0708-1300/Homework Assignment 11|HW11]]<br /> |- align=left<br /> |align=center|23<br /> <br /> If nothing has been added for a given week it might look like this: <br /> <br /> |align=center|22<br /> |Mar 10<br /> |<br /> |- align=left<br /> |align=center|23<br /> <br /> Scroll down until you are in the week with the correct date that you want to add a page. (they correspond to mondays i think). Where the | is, simply write something like [[0708-1300/Class notes for Tuesday, March 11|Tue]] where of course you put in the correct date. If there already is a something after the | such as a homework assignment link, put a comma and then type [[0708-1300/Class notes for Tuesday, March 11|Tue]] with the date corrected. Now go to http://katlas.org/drorbn/index.php?title=0708-1300/Class_notes_for_Tuesday%2C_March_11 and this is where the page you have just created can be found.<br /> <br /> Hope the helps!</div> Trefor http://drorbn.net/index.php?title=User:Trefor User:Trefor 2008-03-18T17:44:45Z <p>Trefor: </p> <hr /> <div>{{0708-1300/Navigation}}<br /> <br /> ==Typed Notes==<br /> <br /> &lt;span style=&quot;color: red;&quot;&gt;The notes below are by the students and for the students. Hopefully they are useful, but they come with no guarantee of any kind.&lt;/span&gt;<br /> <br /> ===First Hour===<br /> <br /> Recall we had defined &lt;math&gt;\tilde{H}(X):= ker\epsilon_*&lt;/math&gt; where &lt;math&gt;\epsilon_*:X\rightarrow\{pt\}&lt;/math&gt;, &lt;math&gt;\tilde{H}(X,A) = H(X,A)&lt;/math&gt;<br /> <br /> For &lt;math&gt;X\neq\empty&lt;/math&gt;, &lt;math&gt;\tilde{H}_p(X) = \tilde{H}_p(X)&lt;/math&gt; for &lt;math&gt;p\neq 0&lt;/math&gt; and equals &lt;math&gt;G\oplus\tilde{H}_0(X)&lt;/math&gt; for &lt;math&gt;p=0&lt;/math&gt;<br /> <br /> <br /> This homology definition satisfies the axioms with the following changes: Exactness only for &lt;math&gt;X\neq\empty&lt;/math&gt; and the dimension axiom being &lt;math&gt;\tilde{H}_*(pt) = 0&lt;/math&gt;. Furthermore, instead of additivity we have, under mild conditions of &lt;math&gt;b_0\in X&lt;/math&gt; and &lt;math&gt;b_1\in Y&lt;/math&gt; (ie non empty) define &lt;math&gt;X\vee Y:= X\cup Y / b_0\sim b_1&lt;/math&gt; for a disjoint union. Then, &lt;math&gt;\tilde{H}(X\vee Y)\cong \tilde{H}(X)\oplus \tilde{H}(Y)<br /> &lt;/math&gt;<br /> <br /> We can actually get the above isomorphism in the following way. There are natural projection maps &lt;math&gt;p_x&lt;/math&gt; and &lt;math&gt;p_y&lt;/math&gt; from &lt;math&gt;X\vee Y&lt;/math&gt; to X and Y respectively that simply contract Y and X respectively to the glued base point. There are also natural inclusion maps &lt;math&gt;i_x&lt;/math&gt; and &lt;math&gt;i_y&lt;/math&gt; going the other way. Then, &lt;math&gt;(p_{x*},p_{y*})&lt;/math&gt; and &lt;math&gt;i_{x*}+i_{y*}&lt;/math&gt; are the two maps in the isomorphism. Proving they are in fact an isomorphism is a homework problem that uses excision to prove it. <br /> <br /> &lt;math&gt;\tilde{H}&lt;/math&gt; is &quot;kinda&quot; natural: <br /> <br /> We have a chain complex where &lt;math&gt;C_p = &lt;\sigma:\Delta_p\rightarrow X&gt;&lt;/math&gt; where &lt;math&gt;\Delta_p=\{x:\mathbb{R}^{n+1}_{\geq 0}\ :\ \sum x_i = 1\}&lt;/math&gt;<br /> <br /> We thus get that &lt;math&gt;\Delta_{-1} = \empty&lt;/math&gt; since &lt;math&gt;\sum x_1 = 0\neq 1&lt;/math&gt; vacuously. <br /> <br /> Therefore, &lt;math&gt;C_{-1}(X) = &lt;\sigma:\Delta_{-1}\rightarrow X&gt; = \mathbb{Z}&lt;/math&gt;<br /> <br /> So, &lt;math&gt;\tilde{C}_*(X,A) = \tilde{C}_*(X)/\tilde{C}_*(A) = C(X,A)&lt;/math&gt;<br /> <br /> <br /> <br /> Note: We have never actually specified that p is positive axiomatically. In fact, &lt;math&gt;\tilde{H}_p(S^n) = G&lt;/math&gt; for p=n and 0 for &lt;math&gt;p\neq n&lt;/math&gt; works fine for all p's. So, since the spaces we are going to be interested in are those that can be constructed from spheres we really will only encounter non trivial homologies for positive p. <br /> <br /> <br /> '''Degrees'''<br /> <br /> &lt;math&gt;(G=\mathbb{Z})&lt;/math&gt;<br /> <br /> &lt;math&gt;f:S^n\rightarrow S^n&lt;/math&gt; then get &lt;math&gt;f_*:\mathbb{Z}\cong \tilde{H}_n(S^n)\rightarrow \tilde{H}_n(S^n)\cong\mathbb{Z}<br /> &lt;/math&gt;<br /> <br /> We thus define:<br /> &lt;math&gt;deg(f):=d = f_*(1)&lt;/math&gt;<br /> <br /> &lt;math&gt;<br /> f_{1,2}:S^0\rightarrow S^0&lt;/math&gt; has<br /> <br /> &lt;math&gt;1) f_1 = I&lt;/math&gt;<br /> <br /> 2) &lt;math&gt;f_2&lt;/math&gt; = flip, ie &lt;math&gt;x_0\rightarrow -x_0&lt;/math&gt;<br /> <br /> &lt;math&gt;deg f_1 = deg I = 1&lt;/math&gt; in all dimensions<br /> <br /> deg &lt;math&gt;f_2&lt;/math&gt; = deg flip = -1<br /> <br /> <br /> '''Proposition'''<br /> <br /> Let &lt;math&gt;f:S^n\rightarrow S^n&lt;/math&gt; be &lt;math&gt;x_0\mapsto-x_0&lt;/math&gt; and &lt;math&gt;x_i\mapsto x_i&lt;/math&gt; for i&gt;0<br /> <br /> then def f= -1<br /> <br /> ''Proof:''<br /> <br /> We get two rows of the following sequence, with the induced maps from f going vertically between them:<br /> <br /> &lt;math&gt;\tilde{H}_{n-1}(S^{n-1})\leftarrow^{\partial}\tilde{H}_n(D^n,S^{n-1})\rightarrow^{i_*} H_n(S^n, D^n_+)\leftarrow^{j_*}H_n(S^n)&lt;/math&gt;<br /> <br /> <br /> The resulting diagram from the two rows of the above sequence and the maps induced by f between them in fact commute at all places, where the left square commutes as a result of the properties of the connecting homomorphism &lt;math&gt;\partial&lt;/math&gt;<br /> <br /> <br /> '''Propositions:'''<br /> <br /> 1) if &lt;math&gt;f\sim g: S^n\rightarrow S^n&lt;/math&gt; then deg f = deg g<br /> <br /> 2) &lt;math&gt;S^n\rightarrow^f s^n\rightarrow^g S^n&lt;/math&gt; then &lt;math&gt;deg (g\circ f) = deg (f) deg (g)<br /> &lt;/math&gt;<br /> <br /> 3) deg a where a is the antipodal map &lt;math&gt;x\mapsto -x&lt;/math&gt; has &lt;math&gt;deg a = (-1)^{n+1}&lt;/math&gt; on &lt;math&gt;S^n&lt;/math&gt;<br /> <br /> <br /> <br /> ===Second Hour===<br /> <br /> '''Corollary'''<br /> <br /> If n is even, a is not homotopic to I<br /> <br /> <br /> '''Corollary''' Every &lt;math&gt;f:S^2\rightarrow S^2&lt;/math&gt; has a fixed point, or an antipodal point. Ie. f(x) = x or f(x) = -x for some value or x. (Note this is believed true for 2n not just 2, but the follow proof appears needs some modification to work in dimensions 2n)<br /> <br /> <br /> ''Proof''<br /> <br /> Suppose f has no fixed points. Thus x and f(x) are distinct and define a great circle. Thus there is a shortest path from f(x) to -x. This uniquely defines a homotopy between f and a. Suppose f also had no antipodal points. Then the same great circle defines a unique homotopy between f and I. But I is not homotopic to a, a contradiction. ''Q.E.D''<br /> <br /> <br /> '''Corollary'''<br /> <br /> Every vector field on &lt;math&gt;S^{2n}&lt;/math&gt; has a zero, i.e., &quot;on earth there must be a windless points&quot; or &quot;you can't comb the hair on a coconut&quot;<br /> <br /> ''Proof'' A non zero vector field induces a homotopy of I to a which is impossible. <br /> <br /> <br /> '''Theorem'''<br /> <br /> If &lt;math&gt;f:S^n\rightarrow S^n \ni y_0&lt;/math&gt; is smooth (and every map may be approximated by one) and &lt;math&gt;y_0\in S^n&lt;/math&gt; is a regular value (which occurs almost everywhere by Sard's Theorem) and &lt;math&gt;f^{-1}(y_0) = \{x_1,\cdots, x_n\}&lt;/math&gt; then deg f = &lt;math&gt;\sum_{j=1}^k \pm 1 = \sum&lt;/math&gt; sign (det(&lt;math&gt;df_{x_i}&lt;/math&gt;)) <br /> <br /> .ie. we get +1 if it preserves orientation and -1 if it reverses it. The latter term is done using an identification of the coordinates near &lt;math&gt;x_i&lt;/math&gt; with coordinates near &lt;math&gt;y_0&lt;/math&gt; using an orientation preserving rotation of &lt;math&gt;S^n&lt;/math&gt;<br /> <br /> <br /> ''Examples''<br /> <br /> 1) &lt;math&gt;S^1\rightarrow S^1&lt;/math&gt; via &lt;math&gt;z\mapsto z^k&lt;/math&gt;<br /> <br /> This map wraps the circle around itself k times yielding k preimages for each point in the image, all with the same sign. <br /> <br /> So, &lt;math&gt;deg f = \sum_k +1 = k&lt;/math&gt;<br /> <br /> <br /> 2) Consider the map of a sphere where you place a plastic bag over a sphere, collect the bag at a pole, twist it once, rewrap the sphere, twist and rewrap again k times. Then the &lt;math&gt;deg f = +1 -1 +1-1\cdots = 0&lt;/math&gt;<br /> <br /> <br /> ''Proof of Theorem''<br /> <br /> <br /> 1) Let &lt;math&gt;T:\mathbb{R}^{n+1}\rightarrow\mathbb{R}^{n+1}&lt;/math&gt; be linear and norm preserving, &lt;math&gt;T\in M_{n+1\times n+1}&lt;/math&gt; and &lt;math&gt;T^{T}T = I&lt;/math&gt;. Then, deg T = det T<br /> <br /> Proof: Every rigid rotation is a product of reflections. <br /> <br /> <br /> 2) Let &lt;math&gt;A:\mathbb{R}^N\rightarrow\mathbb{R}^n&lt;/math&gt; be any &lt;math&gt;n\times n&lt;/math&gt; non singular matrix so that &lt;math&gt;A(\infty)=\infty&lt;/math&gt; so this induces a map &lt;math&gt;\tilde{A}:S^n\rightarrow S^n&lt;/math&gt;. Then, &lt;math&gt;deg\tilde{A} = sign(det A)&lt;/math&gt;<br /> <br /> Proof: Gaussian elimination results in making A a product of &quot;elementary matrices&quot; which come in three types: A matrix with 1's along the diagonal except one diagonal entry being &lt;math&gt;\lambda&lt;/math&gt;. A matrix which is the identity only with two rows interchanged. A matrix which is the identity with a &lt;math&gt;\lambda&lt;/math&gt; in some non diagonal location. <br /> <br /> The latter of these is clearly homotopic to the identity by simply turning the &lt;math&gt;\lambda&lt;/math&gt; off. <br /> <br /> The middle of these is just a reflection. The former of these, if &lt;math&gt;\lambda&gt;0&lt;/math&gt; it is clearly homotopic to the identity. But if &lt;math&gt;\lambda&lt;0&lt;/math&gt; then it is homotopic to a reflection. <br /> <br /> <br /> 3) &lt;math&gt;f:\mathbb{R}^n\rightarrow\mathbb{R}^n&lt;/math&gt; such that &lt;math&gt;f^{-1}(0) = 0&lt;/math&gt;, &lt;math&gt;df|_0 = A&lt;/math&gt; non singular, &lt;math&gt;f(\infty) = \infty&lt;/math&gt; so f defines &lt;math&gt;\tilde{f}:S^n\rightarrow S^n&lt;/math&gt; then deg f = sign(det &lt;math&gt;df|_0)&lt;/math&gt;<br /> <br /> Proof: Consider for &lt;math&gt;t\geq 1&lt;/math&gt;, &lt;math&gt;f_t(x) := tf(x/1)&lt;/math&gt;. Then, &lt;math&gt;f_1 = f&lt;/math&gt;, &lt;math&gt;f_{\infty}=A&lt;/math&gt;. This is a homotopy as it makes good sense for &lt;math&gt;t\in[0,\infty]&lt;/math&gt;. So, &lt;math&gt;deg f = deg\tilde{A}&lt;/math&gt;<br /> <br /> <br /> 4) All that remains to prove the theorem is the shift from (0,0) to &lt;math&gt;(x_0,y_0)&lt;/math&gt; which induces a rotation at &lt;math&gt;df|_{x_0}&lt;/math&gt;<br /> <br /> <br /> <br /> ==Homologies with non trivial negative p's?==<br /> <br /> In the axiomatic definition of homology there is no specification that the homology groups &lt;math&gt;H_p(X)&lt;/math&gt; must be necessarily trivial for negative p's. That said, in the singular homology that we are developing using CW complexes we only get non trivial homologies for positive p's. The question was raised in class of when you could possibly have non trivial homology at negative p's. <br /> <br /> There are infact homology theories that DO have non trivial homology for negative p. In particular, Dr. Putnam from the University of Victoria has defined a homology theory on Smale space and this theory does in fact have non trivial homology groups at negative p's. While Smale spaces have a complicated axiomized definition, very loosely they are a topological space equipped with a metric and a homeomorphism from the space to itself such that locally you can write the space as the direct sum of a section of the space that increases under the homeomorphism(in terms of the metric) and one that decreases. I.e., it is like you have a local coordinate system that tells you where the space is expanding and where it is contracting under the homeomorphism. The classic example of this are Shifts of Finite Type from Dynamical systems. I computed the homology of a particular Smale space consisting of a torus as the space and a particular homeomorphism of it, and found it to have a non trivial homology at &lt;math&gt;H_{-1}<br /> &lt;/math&gt;</div> Trefor http://drorbn.net/index.php?title=0708-1300/Class_notes_for_Tuesday,_March_18 0708-1300/Class notes for Tuesday, March 18 2008-03-18T17:39:47Z <p>Trefor: </p> <hr /> <div>{{0708-1300/Navigation}}<br /> <br /> ==Typed Notes==<br /> <br /> &lt;span style=&quot;color: red;&quot;&gt;The notes below are by the students and for the students. Hopefully they are useful, but they come with no guarantee of any kind.&lt;/span&gt;<br /> <br /> ===First Hour===<br /> <br /> Recall we had defined &lt;math&gt;\tilde{H}(X):= ker\epsilon_*&lt;/math&gt; where &lt;math&gt;\epsilon_*:X\rightarrow\{pt\}&lt;/math&gt;, &lt;math&gt;\tilde{H}(X,A) = H(X,A)&lt;/math&gt;<br /> <br /> For &lt;math&gt;X\neq\empty&lt;/math&gt;, &lt;math&gt;\tilde{H}_p(X) = \tilde{H}_p(X)&lt;/math&gt; for &lt;math&gt;p\neq 0&lt;/math&gt; and equals &lt;math&gt;G\oplus\tilde{H}_0(X)&lt;/math&gt; for &lt;math&gt;p=0&lt;/math&gt;<br /> <br /> <br /> This homology definition satisfies the axioms with the following changes: Exactness only for &lt;math&gt;X\neq\empty&lt;/math&gt; and the dimension axiom being &lt;math&gt;\tilde{H}_*(pt) = 0&lt;/math&gt;. Furthermore, instead of additivity we have, under mild conditions of &lt;math&gt;b_0\in X&lt;/math&gt; and &lt;math&gt;b_1\in Y&lt;/math&gt; (ie non empty) define &lt;math&gt;X\vee Y:= X\cup Y / b_0\sim b_1&lt;/math&gt; for a disjoint union. Then, &lt;math&gt;\tilde{H}(X\vee Y)\cong \tilde{H}(X)\oplus \tilde{H}(Y)<br /> &lt;/math&gt;<br /> <br /> We can actually get the above isomorphism in the following way. There are natural projection maps &lt;math&gt;p_x&lt;/math&gt; and &lt;math&gt;p_y&lt;/math&gt; from &lt;math&gt;X\vee Y&lt;/math&gt; to X and Y respectively that simply contract Y and X respectively to the glued base point. There are also natural inclusion maps &lt;math&gt;i_x&lt;/math&gt; and &lt;math&gt;i_y&lt;/math&gt; going the other way. Then, &lt;math&gt;(p_{x*},p_{y*})&lt;/math&gt; and &lt;math&gt;i_{x*}+i_{y*}&lt;/math&gt; are the two maps in the isomorphism. Proving they are in fact an isomorphism is a homework problem that uses excision to prove it. <br /> <br /> &lt;math&gt;\tilde{H}&lt;/math&gt; is &quot;kinda&quot; natural: <br /> <br /> We have a chain complex where &lt;math&gt;C_p = &lt;\sigma:\Delta_p\rightarrow X&gt;&lt;/math&gt; where &lt;math&gt;\Delta_p=\{x:\mathbb{R}^{n+1}_{\geq 0}\ :\ \sum x_i = 1\}&lt;/math&gt;<br /> <br /> We thus get that &lt;math&gt;\Delta_{-1} = \empty&lt;/math&gt; since &lt;math&gt;\sum x_1 = 0\neq 1&lt;/math&gt; vacuously. <br /> <br /> Therefore, &lt;math&gt;C_{-1}(X) = &lt;\sigma:\Delta_{-1}\rightarrow X&gt; = \mathbb{Z}&lt;/math&gt;<br /> <br /> So, &lt;math&gt;\tilde{C}_*(X,A) = \tilde{C}_*(X)/\tilde{C}_*(A) = C(X,A)&lt;/math&gt;<br /> <br /> <br /> <br /> Note: We have never actually specified that p is positive axiomatically. In fact, &lt;math&gt;\tilde{H}_p(S^n) = G&lt;/math&gt; for p=n and 0 for &lt;math&gt;p\neq n&lt;/math&gt; works fine for all p's. So, since the spaces we are going to be interested in are those that can be constructed from spheres we really will only encounter non trivial homologies for positive p. <br /> <br /> <br /> '''Degrees'''<br /> <br /> &lt;math&gt;(G=\mathbb{Z})&lt;/math&gt;<br /> <br /> &lt;math&gt;f:S^n\rightarrow S^n&lt;/math&gt; then get &lt;math&gt;f_*:\mathbb{Z}\cong \tilde{H}_n(S^n)\rightarrow \tilde{H}_n(S^n)\cong\mathbb{Z}<br /> &lt;/math&gt;<br /> <br /> We thus define:<br /> &lt;math&gt;deg(f):=d = f_*(1)&lt;/math&gt;<br /> <br /> &lt;math&gt;<br /> f_{1,2}:S^0\rightarrow S^0&lt;/math&gt; has<br /> <br /> &lt;math&gt;1) f_1 = I&lt;/math&gt;<br /> <br /> 2) &lt;math&gt;f_2&lt;/math&gt; = flip, ie &lt;math&gt;x_0\rightarrow -x_0&lt;/math&gt;<br /> <br /> &lt;math&gt;deg f_1 = deg I = 1&lt;/math&gt; in all dimensions<br /> <br /> deg &lt;math&gt;f_2&lt;/math&gt; = deg flip = -1<br /> <br /> <br /> '''Proposition'''<br /> <br /> Let &lt;math&gt;f:S^n\rightarrow S^n&lt;/math&gt; be &lt;math&gt;x_0\mapsto-x_0&lt;/math&gt; and &lt;math&gt;x_i\mapsto x_i&lt;/math&gt; for i&gt;0<br /> <br /> then def f= -1<br /> <br /> ''Proof:''<br /> <br /> We get two rows of the following sequence, with the induced maps from f going vertically between them:<br /> <br /> &lt;math&gt;\tilde{H}_{n-1}(S^{n-1})\leftarrow^{\partial}\tilde{H}_n(D^n,S^{n-1})\rightarrow^{i_*} H_n(S^n, D^n_+)\leftarrow^{j_*}H_n(S^n)&lt;/math&gt;<br /> <br /> <br /> The resulting diagram from the two rows of the above sequence and the maps induced by f between them in fact commute at all places, where the left square commutes as a result of the properties of the connecting homomorphism &lt;math&gt;\partial&lt;/math&gt;<br /> <br /> <br /> '''Propositions:'''<br /> <br /> 1) if &lt;math&gt;f\sim g: S^n\rightarrow S^n&lt;/math&gt; then deg f = deg g<br /> <br /> 2) &lt;math&gt;S^n\rightarrow^f s^n\rightarrow^g S^n&lt;/math&gt; then &lt;math&gt;deg (g\circ f) = deg (f) deg (g)<br /> &lt;/math&gt;<br /> <br /> 3) deg a where a is the antipodal map &lt;math&gt;x\mapsto -x&lt;/math&gt; has &lt;math&gt;deg a = (-1)^{n+1}&lt;/math&gt; on &lt;math&gt;S^n&lt;/math&gt;<br /> <br /> <br /> <br /> ===Second Hour===<br /> <br /> '''Corollary'''<br /> <br /> If n is even, a is not homotopic to I<br /> <br /> <br /> '''Corollary''' Every &lt;math&gt;f:S^2\rightarrow S^2&lt;/math&gt; has a fixed point, or an antipodal point. Ie. f(x) = x or f(x) = -x for some value or x. (Note this is believed true for 2n not just 2, but the follow proof appears needs some modification to work in dimensions 2n)<br /> <br /> <br /> ''Proof''<br /> <br /> Suppose f has no fixed points. Thus x and f(x) are distinct and define a great circle. Thus there is a shortest path from f(x) to -x. This uniquely defines a homotopy between f and a. Suppose f also had no antipodal points. Then the same great circle defines a unique homotopy between f and I. But I is not homotopic to a, a contradiction. ''Q.E.D''<br /> <br /> <br /> '''Corollary'''<br /> <br /> Every vector field on &lt;math&gt;S^{2n}&lt;/math&gt; has a zero, i.e., &quot;on earth there must be a windless points&quot; or &quot;you can't comb the hair on a coconut&quot;<br /> <br /> ''Proof'' A non zero vector field induces a homotopy of I to a which is impossible. <br /> <br /> <br /> '''Theorem'''<br /> <br /> If &lt;math&gt;f:S^n\rightarrow S^n \ni y_0&lt;/math&gt; is smooth (and every map may be approximated by one) and &lt;math&gt;y_0\in S^n&lt;/math&gt; is a regular value (which occurs almost everywhere by Sard's Theorem) and &lt;math&gt;f^{-1}(y_0) = \{x_1,\cdots, x_n\}&lt;/math&gt; then deg f = &lt;math&gt;\sum_{j=1}^k \pm 1 = \sum&lt;/math&gt; sign (det(&lt;math&gt;df_{x_i}&lt;/math&gt;)) <br /> <br /> .ie. we get +1 if it preserves orientation and -1 if it reverses it. The latter term is done using an identification of the coordinates near &lt;math&gt;x_i&lt;/math&gt; with coordinates near &lt;math&gt;y_0&lt;/math&gt; using an orientation preserving rotation of &lt;math&gt;S^n&lt;/math&gt;<br /> <br /> <br /> ''Examples''<br /> <br /> 1) &lt;math&gt;S^1\rightarrow S^1&lt;/math&gt; via &lt;math&gt;z\mapsto z^k&lt;/math&gt;<br /> <br /> This map wraps the circle around itself k times yielding k preimages for each point in the image, all with the same sign. <br /> <br /> So, &lt;math&gt;deg f = \sum_k +1 = k&lt;/math&gt;<br /> <br /> <br /> 2) Consider the map of a sphere where you place a plastic bag over a sphere, collect the bag at a pole, twist it once, rewrap the sphere, twist and rewrap again k times. Then the &lt;math&gt;deg f = +1 -1 +1-1\cdots = 0&lt;/math&gt;<br /> <br /> <br /> ''Proof of Theorem''<br /> <br /> <br /> 1) Let &lt;math&gt;T:\mathbb{R}^{n+1}\rightarrow\mathbb{R}^{n+1}&lt;/math&gt; be linear and norm preserving, &lt;math&gt;T\in M_{n+1\times n+1}&lt;/math&gt; and &lt;math&gt;T^{T}T = I&lt;/math&gt;. Then, deg T = det T<br /> <br /> Proof: Every rigid rotation is a product of reflections. <br /> <br /> <br /> 2) Let &lt;math&gt;A:\mathbb{R}^N\rightarrow\mathbb{R}^n&lt;/math&gt; be any &lt;math&gt;n\times n&lt;/math&gt; non singular matrix so that &lt;math&gt;A(\infty)=\infty&lt;/math&gt; so this induces a map &lt;math&gt;\tilde{A}:S^n\rightarrow S^n&lt;/math&gt;. Then, &lt;math&gt;deg\tilde{A} = sign(det A)&lt;/math&gt;<br /> <br /> Proof: Gaussian elimination results in making A a product of &quot;elementary matrices&quot; which come in three types: A matrix with 1's along the diagonal except one diagonal entry being &lt;math&gt;\lambda&lt;/math&gt;. A matrix which is the identity only with two rows interchanged. A matrix which is the identity with a &lt;math&gt;\lambda&lt;/math&gt; in some non diagonal location. <br /> <br /> The latter of these is clearly homotopic to the identity by simply turning the &lt;math&gt;\lambda&lt;/math&gt; off. <br /> <br /> The middle of these is just a reflection. The former of these, if &lt;math&gt;\lambda&gt;0&lt;/math&gt; it is clearly homotopic to the identity. But if &lt;math&gt;\lambda&lt;0&lt;/math&gt; then it is homotopic to a reflection. <br /> <br /> <br /> 3) &lt;math&gt;f:\mathbb{R}^n\rightarrow\mathbb{R}^n&lt;/math&gt; such that &lt;math&gt;f^{-1}(0) = 0&lt;/math&gt;, &lt;math&gt;df|_0 = A&lt;/math&gt; non singular, &lt;math&gt;f(\infty) = \infty&lt;/math&gt; so f defines &lt;math&gt;\tilde{f}:S^n\rightarrow S^n&lt;/math&gt; then deg f = sign(det &lt;math&gt;df|_0)&lt;/math&gt;<br /> <br /> Proof: Consider for &lt;math&gt;t\geq 1&lt;/math&gt;, &lt;math&gt;f_t(x) := tf(x/1)&lt;/math&gt;. Then, &lt;math&gt;f_1 = f&lt;/math&gt;, &lt;math&gt;f_{\infty}=A&lt;/math&gt;. This is a homotopy as it makes good sense for &lt;math&gt;t\in[0,\infty]&lt;/math&gt;. So, &lt;math&gt;deg f = deg\tilde{A}&lt;/math&gt;<br /> <br /> <br /> 4) All that remains to prove the theorem is the shift from (0,0) to &lt;math&gt;(x_0,y_0)&lt;/math&gt; which induces a rotation at &lt;math&gt;df|_{x_0}&lt;/math&gt;<br /> <br /> <br /> <br /> ==Homologies with non trivial negative p's?==<br /> <br /> In the axiomatic definition of homology there is no specification that the homology groups &lt;math&gt;H_p(X)&lt;/math&gt; must be necessarily trivial for negative p's. That said, in the singular homology that we are developing using CW complexes we only get non trivial homologies for positive p's. The question was raised in class of when you could possibly have non trivial homology at negative p's. <br /> <br /> There are infact homology theories that DO have non trivial homology for negative p. In particular, Dr. Putnam from the University of Victoria has defined a homology theory on Smale space and this theory does in fact have non trivial homology groups at negative p's. While Smale spaces have a complicated axiomized definition, very loosely they are a topological space equipped with a metric and a homeomorphism from the space to itself such that locally you can write the space as the direct sum of a section of the space that increases under the homeomorphism(in terms of the metric) and one that decreases. I.e., it is like you have a local coordinate system that tells you where the space is expanding and where it is contracting under the homeomorphism. The classic example of this are Shifts of Finite Type from Dynamical systems. I computed the homology of a particular Smale space consisting of a torus as the space and a particular homeomorphism of it, and found it to have a non trivial homology at &lt;math&gt;H_{-1}<br /> &lt;/math&gt;</div> Trefor http://drorbn.net/index.php?title=Template:0708-1300/Navigation Template:0708-1300/Navigation 2008-03-18T17:26:28Z <p>Trefor: </p> <hr /> <div>{| cellpadding=&quot;0&quot; cellspacing=&quot;0&quot; style=&quot;clear: right; float: right&quot;<br /> |- align=right<br /> |&lt;div class=&quot;NavFrame&quot;&gt;&lt;div class=&quot;NavHead&quot;&gt;[[0708-1300]]/[[Template:0708-1300/Navigation|Navigation Panel]]&amp;nbsp;&amp;nbsp;&lt;/div&gt;<br /> &lt;div class=&quot;NavContent&quot;&gt;<br /> {| border=&quot;1px&quot; cellpadding=&quot;1&quot; cellspacing=&quot;0&quot; width=&quot;220&quot; style=&quot;margin: 0 0 1em 0.5em; font-size: small&quot;<br /> |-<br /> |colspan=3 align=center|[[Image:0708-1300-ClassPhoto.jpg|215px]]&lt;br&gt;[[0708-1300/Class Photo|Add your name / see who's in!]]<br /> |- align=center<br /> !#<br /> !Week of...<br /> !Links<br /> |-<br /> |colspan=3 align=center|'''Fall Semester'''<br /> |- align=left<br /> |align=center|1<br /> |Sep 10<br /> |[[0708-1300/About This Class|About]], [[0708-1300/Class notes for Tuesday, September 11|Tue]], [[0708-1300/Class notes for Thursday, September 13|Thu]]<br /> |- align=left<br /> |align=center|2<br /> |Sep 17<br /> |[[0708-1300/Class notes for Tuesday, September 18|Tue]], [[0708-1300/Homework Assignment 1|HW1]], [[0708-1300/Class notes for Thursday, September 20|Thu]]<br /> |- align=left<br /> |align=center|3<br /> |Sep 24<br /> |[[0708-1300/Class notes for Tuesday, September 25|Tue]], [[0708-1300/Class Photo|Photo]], [[0708-1300/Class notes for Thursday, September 27|Thu]]<br /> |- align=left<br /> |align=center|4<br /> |Oct 1<br /> |[[0708-1300/Questionnaire|Questionnaire]], [[0708-1300/Class notes for Tuesday, October 2|Tue]], [[0708-1300/Homework Assignment 2|HW2]], [[0708-1300/Class notes for Thursday, October 4|Thu]]<br /> |- align=left<br /> |align=center|5<br /> |Oct 8<br /> |Thanksgiving, [[0708-1300/Class notes for Tuesday, October_9|Tue]], [[0708-1300/Class notes for Thursday, October 11|Thu]]<br /> |- align=left<br /> |align=center|6<br /> |Oct 15<br /> |[[0708-1300/Class notes for Tuesday, October 16|Tue]], [[0708-1300/Homework Assignment 3|HW3]], [[0708-1300/Class notes for Thursday, October 18|Thu]]<br /> |- align=left<br /> |align=center|7<br /> |Oct 22<br /> |[[0708-1300/Class notes for Tuesday, October 23|Tue]], [[0708-1300/Class notes for Thursday, October 25|Thu]]<br /> |- align=left<br /> |align=center|8<br /> |Oct 29<br /> |[[0708-1300/Class notes for Tuesday, October 30|Tue]], [[0708-1300/Homework Assignment 4|HW4]], [[0708-1300/Class notes for Thursday, November 1|Thu]], [[0708-1300/the unit sphere in a Hilbert space is contractible|Hilbert sphere]]<br /> |- align=left<br /> |align=center|9<br /> |Nov 5<br /> |[[0708-1300/Class notes for Tuesday, November 6|Tue]],[[0708-1300/Class notes for Thursday, November 8|Thu]], [[0708-1300/Term Exam 1|TE1]]<br /> |- align=left<br /> |align=center|10<br /> |Nov 12<br /> |[[0708-1300/Class notes for Tuesday, November 13|Tue]], &lt;strike&gt;[[0708-1300/Class notes for Thursday, November 15|Thu]]&lt;/strike&gt;<br /> |- align=left<br /> |align=center|11<br /> |Nov 19<br /> |[[0708-1300/Class notes for Tuesday, November 20|Tue]], [[0708-1300/Homework Assignment 5|HW5]]<br /> |- align=left<br /> |align=center|12<br /> |Nov 26<br /> |[[0708-1300/Class notes for Tuesday, November 27|Tue]], [[0708-1300/Class notes for Thursday, November 29|Thu]]<br /> |- align=left<br /> |align=center|13<br /> |Dec 3<br /> |[[0708-1300/Class notes for Tuesday, December 4|Tue]], [[0708-1300/Class notes for Thursday, December 6|Thu]], [[0708-1300/Homework Assignment 6|HW6]]<br /> |-<br /> |colspan=3 align=center|'''Spring Semester'''<br /> |- align=left<br /> |align=center|14<br /> |Jan 7<br /> |[[0708-1300/Class notes for Tuesday, January 8|Tue]], [[0708-1300/Class notes for Thursday, January 10|Thu]], [[0708-1300/Homework Assignment 7|HW7]]<br /> |- align=left<br /> |align=center|15<br /> |Jan 14<br /> |[[0708-1300/Class notes for Tuesday, January 15|Tue]], [[0708-1300/Class notes for Thursday, January 17|Thu]]<br /> |- align=left<br /> |align=center|16<br /> |Jan 21<br /> |[[0708-1300/Class notes for Tuesday, January 22|Tue]], [[0708-1300/Class notes for Thursday, January 24|Thu]], [[0708-1300/Homework Assignment 8|HW8]]<br /> |- align=left<br /> |align=center|17<br /> |Jan 28<br /> |[[0708-1300/Class notes for Tuesday, January 29|Tue]]<br /> |- align=left<br /> |align=center|18<br /> |Feb 4<br /> |[[0708-1300/Class notes for Tuesday, February 5|Tue]]<br /> |- align=left<br /> |align=center|19<br /> |Feb 11<br /> |[[0708-1300/Term Exam 2|TE2]], [[0708-1300/Homework Assignment 9|HW9]], [[0708-1300/Class notes for Thursday, February 14|Thu]], Feb 17: last chance to drop class<br /> |- align=left<br /> |align=center|R<br /> |Feb 18<br /> |Reading week<br /> |- align=left<br /> |align=center|20<br /> |Feb 25<br /> |[[0708-1300/Class notes for Tuesday, February 26|Tue]], [[0708-1300/Homework Assignment 10|HW10]]<br /> |- align=left<br /> |align=center|21<br /> |Mar 3<br /> |<br /> |- align=left<br /> |align=center|22<br /> |Mar 10<br /> |[[0708-1300/Class notes for Tuesday, March 11|Tue]], [[0708-1300/Homework Assignment 11|HW11]]<br /> |- align=left<br /> |align=center|23<br /> |Mar 17<br /> |[[0708-1300/Class notes for Tuesday, March 18|Tue]]<br /> |- align=left<br /> |align=center|24<br /> |Mar 24<br /> |HW12<br /> |- align=left<br /> |align=center|25<br /> |Mar 31<br /> |<br /> |- align=left<br /> |align=center|26<br /> |Apr 7<br /> |<br /> |- align=left<br /> |align=center|R<br /> |Apr 14<br /> |<br /> |- align=left<br /> |align=center|R<br /> |Apr 21<br /> |<br /> |- align=left<br /> |align=center|F<br /> |Apr 28<br /> |[[0708-1300/The Final Exam|Final]] (Fri, May 2)<br /> |- align=center<br /> |colspan=3|[[0708-1300/Register of Good Deeds|Register of Good Deeds]]<br /> |- align=center<br /> |colspan=3|[[0708-1300/Errata to Bredon's Book|Errata to Bredon's Book]]<br /> |}<br /> &lt;/div&gt;&lt;/div&gt;<br /> |}<br /> &lt;div align=center style=&quot;color: red; font-size: 150%; display: none;&quot;&gt;Announcements go here&lt;/div&gt;</div> Trefor http://drorbn.net/index.php?title=0708-1300/Class_notes_for_Tuesday,_March_18 0708-1300/Class notes for Tuesday, March 18 2008-03-18T17:25:52Z <p>Trefor: </p> <hr /> <div>{{0708-1300/Navigation}}<br /> <br /> ==Typed Notes==<br /> <br /> &lt;span style=&quot;color: red;&quot;&gt;The notes below are by the students and for the students. Hopefully they are useful, but they come with no guarantee of any kind.&lt;/span&gt;<br /> <br /> ===First Hour===<br /> <br /> Recall we had defined &lt;math&gt;\tilde{H}(X):= ker\epsilon_*&lt;/math&gt; where &lt;math&gt;\epsilon_*:X\rightarrow\{pt\}&lt;/math&gt;, &lt;math&gt;\tilde{H}(X,A) = H(X,A)&lt;/math&gt;<br /> <br /> For &lt;math&gt;X\neq\empty&lt;/math&gt;, &lt;math&gt;\tilde{H}_p(X) = \tilde{H}_p(X)&lt;/math&gt; for &lt;math&gt;p\neq 0&lt;/math&gt; and equals &lt;math&gt;G\oplus\tilde{H}_0(X)&lt;/math&gt; for &lt;math&gt;p=0&lt;/math&gt;<br /> <br /> <br /> This homology definition satisfies the axioms with the following changes: Exactness only for &lt;math&gt;X\neq\empty&lt;/math&gt; and the dimension axiom being &lt;math&gt;\tilde{H}_*(pt) = 0&lt;/math&gt;. Furthermore, instead of additivity we have, under mild conditions of &lt;math&gt;b_0\in X&lt;/math&gt; and &lt;math&gt;b_1\in Y&lt;/math&gt; (ie non empty) define &lt;math&gt;X\vee Y:= X\cup Y / b_0\sim b_1&lt;/math&gt; for a disjoint union. Then, &lt;math&gt;\tilde{H}(X\vee Y)\cong \tilde{H}(X)\oplus \tilde{H}(Y)<br /> &lt;/math&gt;<br /> <br /> We can actually get the above isomorphism in the following way. There are natural projection maps &lt;math&gt;p_x&lt;/math&gt; and &lt;math&gt;p_y&lt;/math&gt; from &lt;math&gt;X\vee Y&lt;/math&gt; to X and Y respectively that simply contract Y and X respectively to the glued base point. There are also natural inclusion maps &lt;math&gt;i_x&lt;/math&gt; and &lt;math&gt;i_y&lt;/math&gt; going the other way. Then, &lt;math&gt;(p_{x*},p_{y*})&lt;/math&gt; and &lt;math&gt;i_{x*}+i_{y*}&lt;/math&gt; are the two maps in the isomorphism. Proving they are in fact an isomorphism is a homework problem that uses excision to prove it. <br /> <br /> &lt;math&gt;\tilde{H}&lt;/math&gt; is &quot;kinda&quot; natural: <br /> <br /> We have a chain complex where &lt;math&gt;C_p = &lt;\sigma:\Delta_p\rightarrow X&gt;&lt;/math&gt; where &lt;math&gt;\Delta_p=\{x:\mathbb{R}^{n+1}_{\geq 0}\ :\ \sum x_i = 1\}&lt;/math&gt;<br /> <br /> We thus get that &lt;math&gt;\Delta_{-1} = \empty&lt;/math&gt; since &lt;math&gt;\sum x_1 = 0\neq 1&lt;/math&gt; vacuously. <br /> <br /> Therefore, &lt;math&gt;C_{-1}(X) = &lt;\sigma:\Delta_{-1}\rightarrow X&gt; = \mathbb{Z}&lt;/math&gt;<br /> <br /> So, &lt;math&gt;\tilde{C}_*(X,A) = \tilde{C}_*(X)/\tilde{C}_*(A) = C(X,A)&lt;/math&gt;<br /> <br /> <br /> <br /> Note: We have never actually specified that p is positive axiomatically. In fact, &lt;math&gt;\tilde{H}_p(S^n) = G&lt;/math&gt; for p=n and 0 for &lt;math&gt;p\neq n&lt;/math&gt; works fine for all p's. So, since the spaces we are going to be interested in are those that can be constructed from spheres we really will only encounter non trivial homologies for positive p. <br /> <br /> <br /> '''Degrees'''<br /> <br /> &lt;math&gt;(G=\mathbb{Z})&lt;/math&gt;<br /> <br /> &lt;math&gt;f:S^n\rightarrow S^n&lt;/math&gt; then get &lt;math&gt;f_*:\mathbb{Z}\cong \tilde{H}_n(S^n)\rightarrow \tilde{H}_n(S^n)\cong\mathbb{Z}<br /> &lt;/math&gt;<br /> <br /> We thus define:<br /> &lt;math&gt;deg(f):=d = f_*(1)&lt;/math&gt;<br /> <br /> &lt;math&gt;<br /> f_{1,2}:S^0\rightarrow S^0&lt;/math&gt; has<br /> <br /> &lt;math&gt;1) f_1 = I&lt;/math&gt;<br /> <br /> 2) &lt;math&gt;f_2&lt;/math&gt; = flip, ie &lt;math&gt;x_0\rightarrow -x_0&lt;/math&gt;<br /> <br /> &lt;math&gt;deg f_1 = deg I = 1&lt;/math&gt; in all dimensions<br /> <br /> deg &lt;math&gt;f_2&lt;/math&gt; = deg flip = -1<br /> <br /> <br /> '''Proposition'''<br /> <br /> Let &lt;math&gt;f:S^n\rightarrow S^n&lt;/math&gt; be &lt;math&gt;x_0\mapsto-x_0&lt;/math&gt; and &lt;math&gt;x_i\mapsto x_i&lt;/math&gt; for i&gt;0<br /> <br /> then def f= -1<br /> <br /> ''Proof:''<br /> <br /> We get two rows of the following sequence, with the induced maps from f going vertically between them:<br /> <br /> &lt;math&gt;\tilde{H}_{n-1}(S^{n-1})\leftarrow^{\partial}\tilde{H}_n(D^n,S^{n-1})\rightarrow^{i_*} H_n(S^n, D^n_+)\leftarrow^{j_*}H_n(S^n)&lt;/math&gt;<br /> <br /> <br /> The resulting diagram from the two rows of the above sequence and the maps induced by f between them in fact commute at all places, where the left square commutes as a result of the properties of the connecting homomorphism &lt;math&gt;\partial&lt;/math&gt;<br /> <br /> <br /> '''Propositions:'''<br /> <br /> 1) if &lt;math&gt;f\sim g: S^n\rightarrow S^n&lt;/math&gt; then deg f = deg g<br /> <br /> 2) &lt;math&gt;S^n\rightarrow^f s^n\rightarrow^g S^n&lt;/math&gt; then &lt;math&gt;deg (g\circ f) = deg (f) deg (g)<br /> &lt;/math&gt;<br /> <br /> 3) deg a where a is the antipodal map &lt;math&gt;x\mapsto -x&lt;/math&gt; has &lt;math&gt;deg a = (-1)^{n+1}&lt;/math&gt; on &lt;math&gt;S^n&lt;/math&gt;<br /> <br /> <br /> <br /> ===Second Hour===<br /> <br /> '''Corollary'''<br /> <br /> If n is even, a is not homotopic to I<br /> <br /> <br /> '''Corollary''' Every &lt;math&gt;f:S^2\rightarrow S^2&lt;/math&gt; has a fixed point, or an antipodal point. Ie. f(x) = x or f(x) = -x for some value or x. (Note this is believed true for 2n not just 2, but the follow proof appears needs some modification to work in dimensions 2n)<br /> <br /> <br /> ''Proof''<br /> <br /> Suppose f has no fixed points. Thus x and f(x) are distinct and define a great circle. Thus there is a shortest path from f(x) to -x. This uniquely defines a homotopy between f and a. Suppose f also had no antipodal points. Then the same great circle defines a unique homotopy between f and I. But I is not homotopic to a, a contradiction. ''Q.E.D''<br /> <br /> <br /> '''Corollary'''<br /> <br /> Every vector field on &lt;math&gt;S^{2n}&lt;/math&gt; has a zero, i.e., &quot;on earth there must be a windless points&quot; or &quot;you can't comb the hair on a coconut&quot;<br /> <br /> ''Proof'' A non zero vector field induces a homotopy of I to a which is impossible. <br /> <br /> <br /> '''Theorem'''<br /> <br /> If &lt;math&gt;f:S^n\rightarrow S^n \ni y_0&lt;/math&gt; is smooth (and every map may be approximated by one) and &lt;math&gt;y_0\in S^n&lt;/math&gt; is a regular value (which occurs almost everywhere by Sard's Theorem) and &lt;math&gt;f^{-1}(y_0) = \{x_1,\cdots, x_n\}&lt;/math&gt; then deg f = &lt;math&gt;\sum_{j=1}^k \pm 1 = \sum&lt;/math&gt; sign (det(&lt;math&gt;df_{x_i}&lt;/math&gt;)) <br /> <br /> .ie. we get +1 if it preserves orientation and -1 if it reverses it. The latter term is done using an identification of the coordinates near &lt;math&gt;x_i&lt;/math&gt; with coordinates near &lt;math&gt;y_0&lt;/math&gt; using an orientation preserving rotation of &lt;math&gt;S^n&lt;/math&gt;<br /> <br /> <br /> ''Examples''<br /> <br /> 1) &lt;math&gt;S^1\rightarrow S^1&lt;/math&gt; via &lt;math&gt;z\mapsto z^k&lt;/math&gt;<br /> <br /> This map wraps the circle around itself k times yielding k preimages for each point in the image, all with the same sign. <br /> <br /> So, &lt;math&gt;deg f = \sum_k +1 = k&lt;/math&gt;<br /> <br /> <br /> 2) Consider the map of a sphere where you place a plastic bag over a sphere, collect the bag at a pole, twist it once, rewrap the sphere, twist and rewrap again k times. Then the &lt;math&gt;deg f = +1 -1 +1-1\cdots = 0&lt;/math&gt;<br /> <br /> <br /> ''Proof of Theorem''<br /> <br /> <br /> 1) Let &lt;math&gt;T:\mathbb{R}^{n+1}\rightarrow\mathbb{R}^{n+1}&lt;/math&gt; be linear and norm preserving, &lt;math&gt;T\in M_{n+1\times n+1}&lt;/math&gt; and &lt;math&gt;T^{T}T = I&lt;/math&gt;. Then, deg T = det T<br /> <br /> Proof: Every rigid rotation is a product of reflections. <br /> <br /> <br /> 2) Let &lt;math&gt;A:\mathbb{R}^N\rightarrow\mathbb{R}^n&lt;/math&gt; be any &lt;math&gt;n\times n&lt;/math&gt; non singular matrix so that &lt;math&gt;A(\infty)=\infty&lt;/math&gt; so this induces a map &lt;math&gt;\tilde{A}:S^n\rightarrow S^n&lt;/math&gt;. Then, &lt;math&gt;deg\tilde{A} = sign(det A)&lt;/math&gt;<br /> <br /> Proof: Gaussian elimination results in making A a product of &quot;elementary matrices&quot; which come in three types: A matrix with 1's along the diagonal except one diagonal entry being &lt;math&gt;\lambda&lt;/math&gt;. A matrix which is the identity only with two rows interchanged. A matrix which is the identity with a &lt;math&gt;\lambda&lt;/math&gt; in some non diagonal location. <br /> <br /> The latter of these is clearly homotopic to the identity by simply turning the &lt;math&gt;\lambda&lt;/math&gt; off. <br /> <br /> The middle of these is just a reflection. The former of these, if &lt;math&gt;\lambda&gt;0&lt;/math&gt; it is clearly homotopic to the identity. But if &lt;math&gt;\lambda&lt;0&lt;/math&gt; then it is homotopic to a reflection. <br /> <br /> <br /> 3) &lt;math&gt;f:\mathbb{R}^n\rightarrow\mathbb{R}^n&lt;/math&gt; such that &lt;math&gt;f^{-1}(0) = 0&lt;/math&gt;, &lt;math&gt;df|_0 = A&lt;/math&gt; non singular, &lt;math&gt;f(\infty) = \infty&lt;/math&gt; so f defines &lt;math&gt;\tilde{f}:S^n\rightarrow S^n&lt;/math&gt; then deg f = sign(det &lt;math&gt;df|_0)&lt;/math&gt;<br /> <br /> Proof: Consider for &lt;math&gt;t\geq 1&lt;/math&gt;, &lt;math&gt;f_t(x) := tf(x/1)&lt;/math&gt;. Then, &lt;math&gt;f_1 = f&lt;/math&gt;, &lt;math&gt;f_{\infty}=A&lt;/math&gt;. This is a homotopy as it makes good sense for &lt;math&gt;t\in[0,\infty]&lt;/math&gt;. So, &lt;math&gt;deg f = deg\tilde{A}&lt;/math&gt;<br /> <br /> <br /> 4) All that remains to prove the theorem is the shift from (0,0) to &lt;math&gt;(x_0,y_0)&lt;/math&gt; which induces a rotation at &lt;math&gt;df|_{x_0}&lt;/math&gt;</div> Trefor http://drorbn.net/index.php?title=0708-1300/Class_notes_for_Tuesday,_March_18 0708-1300/Class notes for Tuesday, March 18 2008-03-18T17:16:43Z <p>Trefor: </p> <hr /> <div>==Typed Notes==<br /> <br /> ===First Hour===<br /> <br /> Recall we had defined &lt;math&gt;\tilde{H}(X):= ker\epsilon_*&lt;/math&gt; where &lt;math&gt;\epsilon_*:X\rightarrow\{pt\}&lt;/math&gt;, &lt;math&gt;\tilde{H}(X,A) = H(X,A)&lt;/math&gt;<br /> <br /> For &lt;math&gt;X\neq\empty&lt;/math&gt;, &lt;math&gt;\tilde{H}_p(X) = \tilde{H}_p(X)&lt;/math&gt; for &lt;math&gt;p\neq 0&lt;/math&gt; and equals &lt;math&gt;G\oplus\tilde{H}_0(X)&lt;/math&gt; for &lt;math&gt;p=0&lt;/math&gt;<br /> <br /> <br /> This homology definition satisfies the axioms with the following changes: Exactness only for &lt;math&gt;X\neq\empty&lt;/math&gt; and the dimension axiom being &lt;math&gt;\tilde{H}_*(pt) = 0&lt;/math&gt;. Furthermore, instead of additivity we have, under mild conditions of &lt;math&gt;b_0\in X&lt;/math&gt; and &lt;math&gt;b_1\in Y&lt;/math&gt; (ie non empty) define &lt;math&gt;X\vee Y:= X\cup Y / b_0\sim b_1&lt;/math&gt; for a disjoint union. Then, &lt;math&gt;\tilde{H}(X\vee Y)\cong \tilde{H}(X)\oplus \tilde{H}(Y)<br /> &lt;/math&gt;<br /> <br /> We can actually get the above isomorphism in the following way. There are natural projection maps &lt;math&gt;p_x&lt;/math&gt; and &lt;math&gt;p_y&lt;/math&gt; from &lt;math&gt;X\vee Y&lt;/math&gt; to X and Y respectively that simply contract Y and X respectively to the glued base point. There are also natural inclusion maps &lt;math&gt;i_x&lt;/math&gt; and &lt;math&gt;i_y&lt;/math&gt; going the other way. Then, &lt;math&gt;(p_{x*},p_{y*})&lt;/math&gt; and &lt;math&gt;i_{x*}+i_{y*}&lt;/math&gt; are the two maps in the isomorphism. Proving they are in fact an isomorphism is a homework problem that uses excision to prove it. <br /> <br /> &lt;math&gt;\tilde{H}&lt;/math&gt; is &quot;kinda&quot; natural: <br /> <br /> We have a chain complex where &lt;math&gt;C_p = &lt;\sigma:\Delta_p\rightarrow X&gt;&lt;/math&gt; where &lt;math&gt;\Delta_p=\{x:\mathbb{R}^{n+1}_{\geq 0}\ :\ \sum x_i = 1\}&lt;/math&gt;<br /> <br /> We thus get that &lt;math&gt;\Delta_{-1} = \empty&lt;/math&gt; since &lt;math&gt;\sum x_1 = 0\neq 1&lt;/math&gt; vacuously. <br /> <br /> Therefore, &lt;math&gt;C_{-1}(X) = &lt;\sigma:\Delta_{-1}\rightarrow X&gt; = \mathbb{Z}&lt;/math&gt;<br /> <br /> So, &lt;math&gt;\tilde{C}_*(X,A) = \tilde{C}_*(X)/\tilde{C}_*(A) = C(X,A)&lt;/math&gt;<br /> <br /> <br /> <br /> Note: We have never actually specified that p is positive axiomatically. In fact, &lt;math&gt;\tilde{H}_p(S^n) = G&lt;/math&gt; for p=n and 0 for &lt;math&gt;p\neq n&lt;/math&gt; works fine for all p's. So, since the spaces we are going to be interested in are those that can be constructed from spheres we really will only encounter non trivial homologies for positive p. <br /> <br /> <br /> '''Degrees'''<br /> <br /> &lt;math&gt;(G=\mathbb{Z})&lt;/math&gt;<br /> <br /> &lt;math&gt;f:S^n\rightarrow S^n&lt;/math&gt; then get &lt;math&gt;f_*:\mathbb{Z}\cong \tilde{H}_n(S^n)\rightarrow \tilde{H}_n(S^n)\cong\mathbb{Z}<br /> &lt;/math&gt;<br /> <br /> We thus define:<br /> &lt;math&gt;deg(f):=d = f_*(1)&lt;/math&gt;<br /> <br /> &lt;math&gt;<br /> f_{1,2}:S^0\rightarrow S^0&lt;/math&gt; has<br /> <br /> &lt;math&gt;1) f_1 = I&lt;/math&gt;<br /> <br /> 2) &lt;math&gt;f_2&lt;/math&gt; = flip, ie &lt;math&gt;x_0\rightarrow -x_0&lt;/math&gt;<br /> <br /> &lt;math&gt;deg f_1 = deg I = 1&lt;/math&gt; in all dimensions<br /> <br /> deg &lt;math&gt;f_2&lt;/math&gt; = deg flip = -1<br /> <br /> <br /> '''Proposition'''<br /> <br /> Let &lt;math&gt;f:S^n\rightarrow S^n&lt;/math&gt; be &lt;math&gt;x_0\mapsto-x_0&lt;/math&gt; and &lt;math&gt;x_i\mapsto x_i&lt;/math&gt; for i&gt;0<br /> <br /> then def f= -1<br /> <br /> ''Proof:''<br /> <br /> We get two rows of the following sequence, with the induced maps from f going vertically between them:<br /> <br /> &lt;math&gt;\tilde{H}_{n-1}(S^{n-1})\leftarrow^{\partial}\tilde{H}_n(D^n,S^{n-1})\rightarrow^{i_*} H_n(S^n, D^n_+)\leftarrow^{j_*}H_n(S^n)&lt;/math&gt;<br /> <br /> <br /> The resulting diagram from the two rows of the above sequence and the maps induced by f between them in fact commute at all places, where the left square commutes as a result of the properties of the connecting homomorphism &lt;math&gt;\partial&lt;/math&gt;<br /> <br /> <br /> '''Propositions:'''<br /> <br /> 1) if &lt;math&gt;f\sim g: S^n\rightarrow S^n&lt;/math&gt; then deg f = deg g<br /> <br /> 2) &lt;math&gt;S^n\rightarrow^f s^n\rightarrow^g S^n&lt;/math&gt; then &lt;math&gt;deg (g\circ f) = deg (f) deg (g)<br /> &lt;/math&gt;<br /> <br /> 3) deg a where a is the antipodal map &lt;math&gt;x\mapsto -x&lt;/math&gt; has &lt;math&gt;deg a = (-1)^{n+1}&lt;/math&gt; on &lt;math&gt;S^n&lt;/math&gt;<br /> <br /> <br /> <br /> ===Second Hour===<br /> <br /> '''Corollary'''<br /> <br /> If n is even, a is not homotopic to I<br /> <br /> <br /> '''Corollary''' Every f:S^2\rightarrow S^2 has a fixed point, or an antipodal point. Ie. f(x) = x or f(x) = -x for some value or x. (Note this is believe true for 2n not just 2, but the follow proof appears needs some modification to work in dimensions 2n)<br /> <br /> <br /> ''Proof''<br /> <br /> Suppose f has no fixed points. Thus x and f(x) are distinct and define a great circle. Thus there is a shortest path from f(x) to -x. This uniquely defines a homotopy between f and a. Suppose f also had no antipodal points. Then the same great circle defines a unique homotopy between f and I. But I is not homotopic to a, a contradiction. ''Q.E.D''<br /> <br /> <br /> '''Corollary'''<br /> <br /> Every vector field on S^{2n} has a zero, i.e., &quot;on earth there must be a windless points&quot; or &quot;you can't comb the hair on a coconut&quot;<br /> <br /> ''Proof'' A non zero vector field induces a homotopy of I to a which is impossible. <br /> <br /> <br /> '''Theorem'''<br /> <br /> If f:S^n\rightarrow S^n \ni y_0 is smooth (and every map may be approximated by one) and y_0\in S^n is a regular value (which occurs almost everywhere by Sard's Theorem) and f^{-1}(y_0) = \{x_1,\cdots, x_n} then deg f = \sum_{j=1}^k \pm 1 = \sum sign (det)(df_{x_i})) <br /> <br /> .ie. we get +1 if it preserves orientation and -1 if it reverses it. The latter term is done using an identification of the coordinates near x_i w coordinates near y+0 using an orientation preserving rotation of S^n<br /> <br /> <br /> ''Examples''<br /> <br /> 1) S^1\rightarrow S^1 via z\mapsto z^k<br /> <br /> This map wraps the circle around itself k times yielding k preimages for each point in the image, all with the same sign. <br /> <br /> So, deg f = \sum_k +1 = k<br /> <br /> <br /> 2) Consider the map of a sphere where you place a plastic bag over a sphere, collect the bag at a pole, twist it once, rewrap the sphere, twist and rewrap again k times. Then the deg f = +1 -1 +1-1\cdots = 0<br /> <br /> <br /> ''Proof of Theorem''<br /> <br /> <br /> 1) Let T:\mathbb{R}^{n+1}\rightarrow\mathbb{R}^{n+1} be linear and norm preserving, T\in M_{n+1\times n+1} and T^{T}T = I. Then, deg T = det T<br /> <br /> Proof: Every rigid rotation is a product of reflections. <br /> <br /> <br /> 2) Let A:\mathbb{R}^N\rightarrow\mathbb{R}^n be any n\times n non singular matrix so that A(\infty)=\infty so this induces a map \tilde{A}:S^n\rightarrow S^n. Then, deg\tilde{A} = sign(det A)<br /> <br /> Proof: Gaussian elimination results in making A a product of &quot;elementary matrices&quot; which come in three types: A matrix with 1's along the diagonal except one diagonal entry being \lambda. A matrix which is the identity only with two rows interchanged. A matrix with is the identity with a \lambda in some non diagonal location. <br /> <br /> The latter of these is clearly homotopic to the identity by simply turning the \lambda off. <br /> <br /> The middle of these is just a reflection. The former of these, if \lambda&gt;0 it is clearly homotopic to the identity. But if \lambda&lt;0 then it is homotopic to a reflection. <br /> <br /> <br /> 3) f:\mathbb{R}^n\rightarrow\mathbb{R}^n such that f^{-1}(0) = 0, df|_0 = A non singular, f(\infty) = \infty so f defines \tilde{f}:S^n\rightarrow S^n then deg f = sign(det df|_0)<br /> <br /> Proof: Consider for t\geq 1, f_t(x) := tf(x/1). Then, f_1 = f, f_{\infty}=A. This is a homotopy as it makes good sense for t\in[0,\infty]. So, deg f = deg\tilde{A}<br /> <br /> <br /> 4) All that remains to prove the theorem is the shift from (0,0) to (x_0,y_0) which induces a rotation at &lt;math&gt;df|_{x_0}&lt;/math&gt;</div> Trefor http://drorbn.net/index.php?title=0708-1300/Class_notes_for_Tuesday,_February_12 0708-1300/Class notes for Tuesday, February 12 2008-03-09T22:35:35Z <p>Trefor: </p> <hr /> <div>INCOMPLETE<br /> <br /> <br /> ==Typed Notes==<br /> <br /> Informal definition of homology:<br /> <br /> Given X (General topological spaces with no base point) then H_n(X) = {Boundaryless compact n-dim m submanifolds of X}/{Boundaries of (n+1)-dim submanifolds<br /> <br /> because &lt;math&gt;\partial\partial = \empty&lt;/math&gt;, the this makes sense. <br /> <br /> We will need to formalize this by using vector spaces generated by sets. <br /> <br /> We begin by computing the homology of some examples. These are correct in spirit but very wrong in the details, indeed, we havn't even defined what the right details are!<br /> <br /> Now, 0 dimensional submanifolds are merely points. 1 dimensional submanifolds are paths. So, if two points have a path between them, then the points contribute nothing to the zeroth homology. Hence H_0(S^1) = H_0(S^2) = 0 etc...<br /> <br /> Hence, H_0(connected) = &lt;pts&gt;/{all points are connected] = \mathbb{Z}<br /> <br /> H_1(S^1) = \mathbb{Z} as boundaryless paths are generated by loops around the circle, which are vacuously not the boundary of 2 manifolds. <br /> <br /> H_1(S^2) = &lt;closed paths&gt;/&lt;boundaries of disks &gt; = 0<br /> <br /> &lt;math&gt;H_2(S^2) = &lt;S^2&gt;&lt;/math&gt;/ nothing = &lt;math&gt;\mathbb{Z}&lt;/math&gt;<br /> <br /> <br /> In general we have <br /> <br /> &lt;math&gt;H_k(S^n) = \begin{matrix}<br /> \mathbb{Z}&amp;k=0\\<br /> 0&amp;0&lt;k&lt;n\\<br /> \mathbb{Z}&amp;k=n\\<br /> \end{matrix}&lt;/math&gt;<br /> <br /> Likewise, we know the zeroth and second homology of the torus is \mathbb{Z}<br /> <br /> We now need to compute the first homology of the torus. That is, we search for closed paths with no inner disk. <br /> <br /> We think of the torus generated by a square subset of the real plane with the usual identifications. Call one side a and the other b. <br /> Now consider a diagonal across the square, labeled c. Clearly we have a path with, c = a + b, which is the boundary of a triangle. <br /> <br /> Varying the diagonal into an arbitrary path which yields the same result we see that H_1(T^2)=\mathbb{Z}^2<br /> <br /> <br /> We get functorality as for f:X\rightarrow Y we get a map f^*:H_n(X)\rightarrow H_n(Y) as f maps boundaryless manifolds to boundaryless manifolds. <br /> <br /> <br /> ''Applications''<br /> <br /> Brouwer's theorem for D^{n+1}<br /> <br /> I.e., there is no retract D^{n+1}\rightarrow S^n<br /> <br /> Indeed, if there were such an retract we would get a commuting diagram and I = i :S^n\rightarrow S^n} where I is the identity on S^n and i is the natural inclusion into D^{n+1}. But after applying the functor this would get a commuting diagram where S^n goes to \mathbb{Z} and D^{n+1} goes to zero. But then have I* = 0 map, this is impossible. <br /> <br /> <br /> '''Definition:'''<br /> <br /> The p-simplex \Delta_p is is<br /> <br /> \mathbb{R}^{p+1}\ni\Delta_p:=\{(\lambda_0\cdots\lambda_p)\ |\ \sum_i^p \lambda_i = 1,\ \lambda_i\geq0\} = \{\sum\lambda_i e_i\ |\ \sum_i^p \lambda_i = 1,\ \lambda_i\geq0\}<br /> <br /> <br /> '''Example'''<br /> <br /> p=0 is just the point 1 on the real line. <br /> <br /> p=1 is the diagonal line between (1,0) and (0,1) in the real plane. here e_1 represents the vector from 0 to (0,1) and e_0 represents the vector from 0 to (1,0)<br /> <br /> p=2 forms a triangle between the tips of the three unit vectors in \mathbb{R}^3. <br /> <br /> In general, we have the convex full of p+1 points. <br /> <br /> <br /> Likewise, if (v_0,\cdots, v_p)\in\mathbb{R}^N then have map [v_0,\cdots, v_p]:\Delta_p\rightarrow\mathbb{R}^N defined by \sum\lambda_i e_i\mapsto \sum \lambda_i v_i<br /> <br /> <br /> Now, we want to define the boundary of a simplex. Loosely, we get a bunch of p-1 simplexes found by forgetting a vertex each time. <br /> <br /> '''Definition'''<br /> <br /> F_i^p:\Delta_{p-1}\rightarrow\Delta_p is the ith face of \Delta_p, namely, F_i^p = [e_0\cdots \hat{e_i}\cdots e_p]<br /> <br /> <br /> '''Definition'''<br /> <br /> Given X, \Delta_p(X)=C_p(X) = &lt;p\ simplexes\ in\ X&gt; = \{\sum a_i\sigma_i\ |\ a_i\in\mathbb{Z},\ \sigma_i:\Delta_p\rightarrow X\ is\ cont.\}<br /> <br /> Note: \sigma is not necessarily 1:1 as was implied in the earlier naive computations. Furthermore, we note that C_p(X) has an abelian group structure)<br /> <br /> Definition: \partial_p:C_p(X)\rightarrow Cc_{p-1}(X) a group homomorphism via \sum a_i\sigma_i\mapsto \sum a_i\partial_p(\sigma_i) where \partial_p\sigma = \sum_{i=0}^p (-1)^i\sigma\circ\F_i^p<br /> <br /> <br /> This results in a chain \cdots C_3(X)\rightarrow^{\partial_3}C_2(X)\rightarrow^{\partial_2}C_1(X)\rightarrow^{\partial_1}C_0(X) <br /> <br /> <br /> '''Definition'''<br /> <br /> Z_p(X) = ker\partial_p called &quot;p cycles in X&quot;<br /> <br /> B_p(X) = Im\partial_{p+1} called &quot;p boundaries in X&quot;<br /> <br /> <br /> We want to define H_p(X) = Z_p(X)/B_p(X)<br /> <br /> <br /> '''Lemma'''<br /> <br /> B_p(X)\subset Z_p(X) \Leftrightarrow im\partial{p+1}\subset \partial_p\Leftrightarrow \partial_p\circ\partial_{p+1} =0<br /> <br /> Notationally we simply write \partial^2 = 0<br /> <br /> <br /> Let \sigma\in C_{p+1}<br /> <br /> \sigma:\Delta_{p+1}\rightarrow X, \partial\sigma = \sum_{i=0}^p (-1)^i\sigma\circ F_i^{p}<br /> <br /> \partial\partial\sigma = \partial(\sum (-1)^i\sigma\circ F_i^p) = \sum_{i=0}^{p+1}\sum_{j=0}^p\sigma\circ F_i^p\circ F_j^{p-1}(-1)^{i+j}<br /> <br /> <br /> I.e., dropping e_m, e_n occurs twice, only with a different sign. Formally, <br /> <br /> F_i^p\circ F_j^{p-1} = [e_o\cdots \hat{e_j},\hat{e_i}\cdots e_{p+1}] for i&gt;j and <br /> <br /> F_i^p\circ F_j^{p-1} = [e_o\cdots \hat{e_i},\hat{e_j+1}\cdots e_{p+1}] for i\leq j<br /> <br /> So \partial\partial\sigma = \sum_{i=o}^{p+1}\sum_{j=0}^{p-1}(-1)^{i+j}\sigma\circ[\hat{e_j}\hat{e_i}] + \sum_{i=0}^{p+1}\sum_{j=i}^p (-1)^{i+j}\sigma\circ[\hat{e_i}\hat{e_{j+1}}]</div> Trefor http://drorbn.net/index.php?title=0708-1300/Class_notes_for_Tuesday,_February_12 0708-1300/Class notes for Tuesday, February 12 2008-03-09T01:05:00Z <p>Trefor: </p> <hr /> <div>==Typed Notes==<br /> <br /> Informal definition of homology:<br /> <br /> Given X (General topological spaces with no base point) then H_n(X) = {Boundaryless compact n-dim m submanifolds of X}/{Boundaries of (n+1)-dim submanifolds<br /> <br /> because &lt;math&gt;\partial\partial = \empty&lt;/math&gt;, the this makes sense. <br /> <br /> We will need to formalize this by using vector spaces generated by sets. <br /> <br /> We begin by computing the homology of some examples. These are correct in spirit but very wrong in the details, indeed, we havn't even defined what the right details are!<br /> <br /> Now, 0 dimensional submanifolds are merely points. 1 dimensional submanifolds are paths. So, if two points have a path between them, then the points contribute nothing to the zeroth homology. Hence H_0(S^1) = H_0(S^2) = 0 etc...<br /> <br /> Hence, H_0(connected) = &lt;pts&gt;/{all points are connected] = \mathbb{Z}<br /> <br /> H_1(S^1) = \mathbb{Z} as boundaryless paths are generated by loops around the circle, which are vacuously not the boundary of 2 manifolds. <br /> <br /> H_1(S^2) = &lt;closed paths&gt;/&lt;boundaries of disks &gt; = 0<br /> <br /> &lt;math&gt;H_2(S^2) = &lt;S^2&gt;&lt;/math&gt;/ nothing = &lt;math&gt;\mathbb{Z}&lt;/math&gt;<br /> <br /> <br /> In general we have <br /> <br /> &lt;math&gt;H_k(S^n) = \begin{matrix}<br /> \mathbb{Z}&amp;k=0\\<br /> 0&amp;0&lt;k&lt;n\\<br /> \mathbb{Z}&amp;k=n\\<br /> \end{matrix}&lt;/math&gt;<br /> <br /> Likewise, we know the zeroth and second homology of the torus is \mathbb{Z}<br /> <br /> We now need to compute the first homology of the torus. That is, we search for closed paths with no inner disk. <br /> <br /> We think of the torus generated by a square subset of the real plane with the usual identifications. Call one side a and the other b. <br /> Now consider a diagonal across the square, labeled c. Clearly we have a path with, c = a + b, which is the boundary of a triangle. <br /> <br /> Varying the diagonal into an arbitrary path which yields the same result we see that H_1(T^2)=\mathbb{Z}^2<br /> <br /> <br /> We get functorality as for f:X\rightarrow Y we get a map f^*:H_n(X)\rightarrow H_n(Y) as f maps boundaryless manifolds to boundaryless manifolds. <br /> <br /> <br /> ''Applications''<br /> <br /> Brouwer's theorem for D^{n+1}<br /> <br /> I.e., there is no retract D^{n+1}\rightarrow S^n<br /> <br /> Indeed, if there were such an retract we would get a commuting diagram and I = i :S^n\rightarrow S^n} where I is the identity on S^n and i is the natural inclusion into D^{n+1}. But after applying the functor this would get a commuting diagram where S^n goes to \mathbb{Z} and D^{n+1} goes to zero. But then have I* = 0 map, this is impossible. <br /> <br /> <br /> '''Definition:'''<br /> <br /> The p-simplex \Delta_p is is<br /> <br /> \mathbb{R}^{p+1}\ni\Delta_p:=\{(\lambda_0\cdots\lambda_p)\ |\ \sum_i^p \lambda_i = 1,\ \lambda_i\geq0\} = \{\sum\lambda_i e_i\ |\ \sum_i^p \lambda_i = 1,\ \lambda_i\geq0\}<br /> <br /> <br /> '''Example'''<br /> <br /> p=0 is just the point 1 on the real line. <br /> <br /> p=1 is the diagonal line between (1,0) and (0,1) in the real plane. here e_1 represents the vector from 0 to (0,1) and e_0 represents the vector from 0 to (1,0)<br /> <br /> p=2 forms a triangle between the tips of the three unit vectors in \mathbb{R}^3. <br /> <br /> In general, we have the convex full of p+1 points. <br /> <br /> <br /> Likewise, if (v_0,\cdots, v_p)\in\mathbb{R}^N then have map [v_0,\cdots, v_p]:\Delta_p\rightarrow\mathbb{R}^N defined by \sum\lambda_i e_i\mapsto \sum \lambda_i v_i<br /> <br /> <br /> Now, we want to define the boundary of a simplex. Loosely, we get a bunch of p-1 simplexes found by forgetting a vertex each time. <br /> <br /> '''Definition'''<br /> <br /> F_i^p:\Delta_{p-1}\rightarrow\Delta_p is the ith face of \Delta_p, namely, F_i^p = [e_0\cdots \hat{e_i}\cdots e_p]<br /> <br /> <br /> '''Definition'''<br /> <br /> Given X, \Delta_p(X)=C_p(X) = &lt;p\ simplexes\ in\ X&gt; = \{\sum a_i\sigma_i\ |\ a_i\in\mathbb{Z},\ \sigma_i:\Delta_p\rightarrow X\ is\ cont.\}<br /> <br /> Note: \sigma is not necessarily 1:1 as was implied in the earlier naive computations. Furthermore, we note that C_p(X) has an abelian group structure)<br /> <br /> Definition: \partial_p:C_p(X)\rightarrow Cc_{p-1}(X) a group homomorphism via \sum a_i\sigma_i\mapsto \sum a_i\partial_p(\sigma_i) where \partial_p\sigma = \sum_{i=0}^p (-1)^i\sigma\circ\F_i^p<br /> <br /> <br /> This results in a chain \cdots C_3(X)\rightarrow^{\partial_3}C_2(X)\rightarrow^{\partial_2}C_1(X)\rightarrow^{\partial_1}C_0(X) <br /> <br /> <br /> '''Definition'''<br /> <br /> Z_p(X) = ker\partial_p called &quot;p cycles in X&quot;<br /> <br /> B_p(X) = Im\partial_{p+1} called &quot;p boundaries in X&quot;<br /> <br /> <br /> We want to define H_p(X) = Z_p(X)/B_p(X)<br /> <br /> <br /> '''Lemma'''<br /> <br /> B_p(X)\subset Z_p(X) \Leftrightarrow im\partial{p+1}\subset \partial_p\Leftrightarrow \partial_p\circ\partial_{p+1} =0<br /> <br /> Notationally we simply write \partial^2 = 0<br /> <br /> <br /> Let \sigma\in C_{p+1}<br /> <br /> \sigma:\Delta_{p+1}\rightarrow X, \partial\sigma = \sum_{i=0}^p (-1)^i\sigma\circ F_i^{p}</div> Trefor http://drorbn.net/index.php?title=0708-1300/Class_notes_for_Tuesday,_February_26 0708-1300/Class notes for Tuesday, February 26 2008-03-08T03:28:24Z <p>Trefor: </p> <hr /> <div>{{0708-1300/Navigation}}<br /> <br /> ==A Homology Theory is a Monster==<br /> [[Image:0708-1300-AxiomsForHomology.png|thumb|center|540px|Page 183 of Bredon's book]]<br /> <br /> '''Bredon's Plan of Attack:''' State all, apply all, prove all.<br /> <br /> '''Our Route:''' Axiom by axiom - state, apply, prove. Thus everything we will do will be, or should be, labeled either &quot;'''S'''tate&quot; or &quot;'''P'''rove&quot; or &quot;'''A'''pply&quot;.<br /> <br /> ==Typed Notes==<br /> <br /> &lt;span style=&quot;color: red;&quot;&gt;The notes below are by the students and for the students. Hopefully they are useful, but they come with no guarantee of any kind.&lt;/span&gt;<br /> <br /> ===First Hour===<br /> <br /> <br /> <br /> Recall we had defined for a chain complex the associated homology groups: &lt;math&gt;H_p(C_*) :=\ ker\partial_p/im\partial_{p+1}&lt;/math&gt;<br /> <br /> From this we get the pth homology for a topological space &lt;math&gt;H_p(X)&lt;/math&gt;<br /> <br /> We have previously shown that <br /> <br /> ''1) &lt;math&gt;H_p(\cup X) = \oplus H_p(X_i)&lt;/math&gt; for disjoint unions of spaces &lt;math&gt;X_i&lt;/math&gt;<br /> <br /> 2) &lt;math&gt;H_p(pt) = \mathbb{Z}\delta_{p0}&lt;/math&gt;<br /> <br /> 3) &lt;math&gt;H_0(connected) = \mathbb{Z}&lt;/math&gt;<br /> <br /> 4) &lt;math&gt;H_1(connected) \cong \pi_1^{ab}(X)&lt;/math&gt; via the map<br /> <br /> &lt;math&gt;\phi:[\gamma]_{\pi_1^{ab}}\mapsto[\gamma]_{H_1}&lt;/math&gt;<br /> <br /> &lt;math&gt;\psi:\sigma\in C_1\mapsto[\gamma_{\sigma(0)}\sigma\bar{\gamma_{\sigma(1)}}]&lt;/math&gt; where &lt;math&gt;\sigma_x&lt;/math&gt; is a path connecting &lt;math&gt;x_0&lt;/math&gt; to x. <br /> <br /> <br /> We need to check the maps are in fact inverses of each other. <br /> <br /> Lets consider &lt;math&gt;\psi\circ\phi&lt;/math&gt;. We start with a closed path starting at &lt;math&gt;x_0&lt;/math&gt; (thought of as in the fundamental group). &lt;math&gt;\phi&lt;/math&gt; means we now think of it as a simplex in X with a point at &lt;math&gt;x_0&lt;/math&gt;. &lt;math&gt;\Psi&lt;/math&gt; now takes this to the path that parks at &lt;math&gt;x_0&lt;/math&gt; for a third of the time, goes around the loop and then parks for the remaining third of the time. Clearly this is homotopic this composition is homotopic to the identity. <br /> <br /> We now consider &lt;math&gt;\phi\circ\psi&lt;/math&gt;. Start with just a path &lt;math&gt;\sigma&lt;/math&gt;. Then &lt;math&gt;\psi&lt;/math&gt; makes a loop adding two paths from the &lt;math&gt;x_0&lt;/math&gt; to the start and finish of &lt;math&gt;\sigma&lt;/math&gt; forming a triangular like closed loop. We think of this loop as &lt;math&gt;\sigma'\in C_1&lt;/math&gt;<br /> <br /> Now, we start from c being &lt;math&gt;c = \sum a_i\sigma_i&lt;/math&gt; with &lt;math&gt;\partial c = 0&lt;/math&gt;. So get &lt;math&gt;\sum a_i(\partial \sigma_1) = \sum a_i(\sigma_i(1)-\sigma_i(0))&lt;/math&gt;<br /> <br /> So &lt;math&gt;\Psi(c) = [\gamma_{\sigma_i(0)}\sigma_i\bar{\gamma_{\sigma_i(1)}}]_{\pi_1}&lt;/math&gt; which maps to, under &lt;math&gt;\phi&lt;/math&gt;, &lt;math&gt;\sum a_i(\gamma_{\sigma_i(0)} + \sigma_i - \gamma_{\sigma_i(1)}) = \sum a_i\sigma_i = c&lt;/math&gt; ( in the homology gamma's cancel as &lt;math&gt;\partial c = 0&lt;/math&gt;)<br /> <br /> <br /> ''Axiomized Homology''<br /> <br /> We now will move to an approach where we prove that our defined homology satisfies a series of established homology axioms that will allow us to apply the machinery of general homology to our specific &quot;singular&quot; homology defined via simplexes. <br /> <br /> Axiom 0) Homology if a function<br /> <br /> '''Definition''' The &quot;category of chain complexes&quot; is a category whose objects are chain complexes (of abelian groups) and morphisms which is a homomorphism between each abelian group in one chain and the corresponding group in the other chain such that the resulting diagram commutes. I.e, &lt;math&gt;Mor((C_p)_{p=0}^{\infty}, (D_p)^{\infty}_{p=0}) = \{(f_p:C_p\rightarrow D_p)_{p=0}^{\infty}\ |\ f_{p-1}\partial_p^C=\partial_p^D f_p\}&lt;/math&gt;<br /> <br /> <br /> Now, in our case, the chain complexes do in fact commute because &lt;math&gt;\partial&lt;/math&gt; is defined by pre-composition but f is defined by post-composition. Hence, associativity of composition yields commutativity. <br /> <br /> <br /> '''Claim'''<br /> <br /> Homology of chain complexes is a functor in the natural way. That is, if &lt;math&gt;f_p:C_p\rightarrow D_p&lt;/math&gt; for each p induces the functor &lt;math&gt;f_*:H_p(C_*)\rightarrow H_p(D_*)&lt;/math&gt;<br /> <br /> The proof is by &quot;diagram chasing&quot;. Well, let &lt;math&gt;c\in C_p&lt;/math&gt;, &lt;math&gt;\partial c =0&lt;/math&gt;. <br /> <br /> Let &lt;math&gt;f*[c] = [fc]&lt;/math&gt;. Now &lt;math&gt;\partial fc = f\partial c = 0&lt;/math&gt;. Furthermore, suppose &lt;math&gt;c = \partial b&lt;/math&gt;. Then, &lt;math&gt;f_*c = fc = \partial fb&lt;/math&gt; so therefore &lt;math&gt;fc = \partial\beta&lt;/math&gt; some &lt;math&gt;\beta&lt;/math&gt;. This shows &lt;math&gt;f_*&lt;/math&gt; is well defined. <br /> <br /> Thus, for &lt;math&gt;c\in H_p(C_*)&lt;/math&gt; get &lt;math&gt;fc\in H_p(D_*)&lt;/math&gt; via the well defined functor &lt;math&gt;f_*&lt;/math&gt;.<br /> <br /> <br /> ===Second Hour===<br /> <br /> '''1) Homotopy Axioms'''<br /> <br /> If &lt;math&gt;f,g:X\rightarrow Y&lt;/math&gt; are homotopic then &lt;math&gt;f_* = g_*: H_p(X)\rightarrow H_p(Y)&lt;/math&gt;<br /> <br /> <br /> Applications: If X and Y are homotopy equivalent then &lt;math&gt;H_*(X) \cong H_*(Y)&lt;/math&gt;<br /> <br /> ''Proof:''<br /> <br /> let &lt;math&gt;f:X\rightarrow Y&lt;/math&gt;, &lt;math&gt;g:Y\rightarrow X&lt;/math&gt; such that &lt;math&gt;f\circ g\sim id_y&lt;/math&gt; and &lt;math&gt;g\circ f \sim id_x&lt;/math&gt;. Well &lt;math&gt;f_*\circ g_* = id_{H(Y)}&lt;/math&gt; and &lt;math&gt;g_*\circ f_* = id_{H(X)}&lt;/math&gt;<br /> <br /> Hence, &lt;math&gt;f_*&lt;/math&gt; and &lt;math&gt;g_*&lt;/math&gt; are invertible maps of each other. ''Q.E.D''<br /> <br /> <br /> '''Definition'''<br /> <br /> Two morphisms &lt;math&gt;f,g:C_*\rightarrow D_*&lt;/math&gt; between chain complexes are ''homotopic'' if you can find maps &lt;math&gt;h_p:C_p\rightarrow D_{p+1}&lt;/math&gt; such that &lt;math&gt;f_p - g_p = \partial_{p+1} h_p = h_{p-1}\partial_p&lt;/math&gt;<br /> <br /> <br /> '''Claim 1'''<br /> <br /> Given H a homotopy connecting f&lt;math&gt;,g:X\rightarrow&lt;/math&gt; Y we can construct a chain homotopy between &lt;math&gt;f_*,g_*:C_*(X)\rightarrow C_*(&lt;/math&gt;Y)<br /> <br /> <br /> '''Claim 2'''<br /> <br /> If &lt;math&gt;f,g:C_*\rightarrow C_*&lt;/math&gt; are chain homotopic then they induce equal maps on homology. <br /> <br /> <br /> <br /> ''Proof of 2''<br /> <br /> Assume &lt;math&gt;[c]\in H_p(C_*)&lt;/math&gt;, that is, &lt;math&gt;\partial c =0&lt;/math&gt;<br /> <br /> &lt;math&gt;[f_* c] - [g_* c] = [(f_*-g_*)] =[(\partial h + h\partial)c] = 0&lt;/math&gt; (as &lt;math&gt;\partial c = 0&lt;/math&gt; and homology ignores exact forms)<br /> <br /> Hence, at the level of homology they are the same. <br /> <br /> <br /> ''Proof of 1''<br /> <br /> Consider a simplex in X. Now consider its image, a simplex, in Y under g and f respectively. Because of the homotopy we can construct a triangular based cylinder in Y with the image under f at the top and the image under y at the bottom. <br /> <br /> Define &lt;math&gt;h\sigma&lt;/math&gt; = the above prism formed by &lt;math&gt;\sigma&lt;/math&gt; and the homotopy H. <br /> <br /> &lt;math&gt;(f_*-g_*)\sigma = h\partial\sigma = \partial h\sigma&lt;/math&gt;<br /> <br /> This, pictorially is correct but we need to be able to break up the prism, &lt;math&gt;\Delta_p\times I&lt;/math&gt; into a union of images of simplexes. <br /> <br /> Suppose p=0, i.e. a point. Hence &lt;math&gt;\Delta_0\times I&lt;/math&gt; is a line, which is a simplex. <br /> <br /> Suppose p=1 which yields a square. Adding a diagonal divides the square into two triangles, so is clearly a union of simplexes. <br /> <br /> Suppose p=2. We get a prism which has a triangle for a base and a top. Raise each vertex on the bottom to the top in turn. This makes the prism a union of three simplexes. <br /> <br /> In general for &lt;math&gt;\Delta_p\times I&lt;/math&gt; let &lt;math&gt;f_i = (l_i,0)&lt;/math&gt; and &lt;math&gt;g_i = (l_i,1)&lt;/math&gt; for vertexes &lt;math&gt;l_i&lt;/math&gt;<br /> <br /> Then, &lt;math&gt;h\sigma = \sum_{i=0}^p (-1)^i H\circ(\sigma\times I)\circ[f_0\cdots f_i g_i g_{i+1}\cdots g_p]&lt;/math&gt;<br /> <br /> which is in &lt;math&gt;C_{p+1}(Y)&lt;/math&gt;<br /> <br /> So have maps &lt;math&gt;Y\leftarrow_H X\times I\leftarrow \Delta_p\times I \leftarrow\Delta_{p+1}&lt;/math&gt;<br /> <br /> <br /> ''Claim: ''<br /> <br /> &lt;math&gt;\partial h +h\partial = f-g&lt;/math&gt;<br /> <br /> Loosely, &lt;math&gt;\partial h&lt;/math&gt; cuts each &lt;math&gt;[f_0\cdots f_i g_i g_{i+1}\cdots g_p]&lt;/math&gt; between the f_i and g_i and then deletes an entry. h\partial however does these in reverse order. Hence all that we are left with is &lt;math&gt;[f_0\cdots f_p] - [g_0\cdots g_p]&lt;/math&gt;</div> Trefor http://drorbn.net/index.php?title=User:Trefor User:Trefor 2008-03-08T03:27:31Z <p>Trefor: </p> <hr /> <div>{{0708-1300/Navigation}}<br /> NOTE: This page is used as a placeholder for incomplete typed class notes. <br /> <br /> <br /> {{0708-1300/Navigation}}<br /> <br /> ==A Homology Theory is a Monster==<br /> [[Image:0708-1300-AxiomsForHomology.png|thumb|center|540px|Page 183 of Bredon's book]]<br /> <br /> '''Bredon's Plan of Attack:''' State all, apply all, prove all.<br /> <br /> '''Our Route:''' Axiom by axiom - state, apply, prove. Thus everything we will do will be, or should be, labeled either &quot;'''S'''tate&quot; or &quot;'''P'''rove&quot; or &quot;'''A'''pply&quot;.<br /> <br /> ==Typed Notes==<br /> <br /> &lt;span style=&quot;color: red;&quot;&gt;The notes below are by the students and for the students. Hopefully they are useful, but they come with no guarantee of any kind.&lt;/span&gt;<br /> <br /> ===First Hour===<br /> <br /> <br /> <br /> Recall we had defined for a chain complex the associated homology groups: &lt;math&gt;H_p(C_*) :=\ ker\partial_p/im\partial_{p+1}&lt;/math&gt;<br /> <br /> From this we get the pth homology for a topological space &lt;math&gt;H_p(X)&lt;/math&gt;<br /> <br /> We have previously shown that <br /> <br /> ''1) &lt;math&gt;H_p(\cup X) = \oplus H_p(X_i)&lt;/math&gt; for disjoint unions of spaces &lt;math&gt;X_i&lt;/math&gt;<br /> <br /> 2) &lt;math&gt;H_p(pt) = \mathbb{Z}\delta_{p0}&lt;/math&gt;<br /> <br /> 3) &lt;math&gt;H_0(connected) = \mathbb{Z}&lt;/math&gt;<br /> <br /> 4) &lt;math&gt;H_1(connected) \cong \pi_1^{ab}(X)&lt;/math&gt; via the map<br /> <br /> &lt;math&gt;\phi:[\gamma]_{\pi_1^{ab}}\mapsto[\gamma]_{H_1}&lt;/math&gt;<br /> <br /> &lt;math&gt;\psi:\sigma\in C_1\mapsto[\gamma_{\sigma(0)}\sigma\bar{\gamma_{\sigma(1)}}]&lt;/math&gt; where &lt;math&gt;\sigma_x&lt;/math&gt; is a path connecting &lt;math&gt;x_0&lt;/math&gt; to x. <br /> <br /> <br /> We need to check the maps are in fact inverses of each other. <br /> <br /> Lets consider &lt;math&gt;\psi\circ\phi&lt;/math&gt;. We start with a closed path starting at &lt;math&gt;x_0&lt;/math&gt; (thought of as in the fundamental group). &lt;math&gt;\phi&lt;/math&gt; means we now think of it as a simplex in X with a point at &lt;math&gt;x_0&lt;/math&gt;. &lt;math&gt;\Psi&lt;/math&gt; now takes this to the path that parks at &lt;math&gt;x_0&lt;/math&gt; for a third of the time, goes around the loop and then parks for the remaining third of the time. Clearly this is homotopic this composition is homotopic to the identity. <br /> <br /> We now consider &lt;math&gt;\phi\circ\psi&lt;/math&gt;. Start with just a path &lt;math&gt;\sigma&lt;/math&gt;. Then &lt;math&gt;\psi&lt;/math&gt; makes a loop adding two paths from the &lt;math&gt;x_0&lt;/math&gt; to the start and finish of &lt;math&gt;\sigma&lt;/math&gt; forming a triangular like closed loop. We think of this loop as &lt;math&gt;\sigma'\in C_1&lt;/math&gt;<br /> <br /> Now, we start from c being &lt;math&gt;c = \sum a_i\sigma_i&lt;/math&gt; with &lt;math&gt;\partial c = 0&lt;/math&gt;. So get &lt;math&gt;\sum a_i(\partial \sigma_1) = \sum a_i(\sigma_i(1)-\sigma_i(0))&lt;/math&gt;<br /> <br /> So &lt;math&gt;\Psi(c) = [\gamma_{\sigma_i(0)}\sigma_i\bar{\gamma_{\sigma_i(1)}}]_{\pi_1}&lt;/math&gt; which maps to, under &lt;math&gt;\phi&lt;/math&gt;, &lt;math&gt;\sum a_i(\gamma_{\sigma_i(0)} + \sigma_i - \gamma_{\sigma_i(1)}) = \sum a_i\sigma_i = c&lt;/math&gt; ( in the homology gamma's cancel as &lt;math&gt;\partial c = 0&lt;/math&gt;)<br /> <br /> <br /> ''Axiomized Homology''<br /> <br /> We now will move to an approach where we prove that our defined homology satisfies a series of established homology axioms that will allow us to apply the machinery of general homology to our specific &quot;singular&quot; homology defined via simplexes. <br /> <br /> Axiom 0) Homology if a function<br /> <br /> '''Definition''' The &quot;category of chain complexes&quot; is a category whose objects are chain complexes (of abelian groups) and morphisms which is a homomorphism between each abelian group in one chain and the corresponding group in the other chain such that the resulting diagram commutes. I.e, &lt;math&gt;Mor((C_p)_{p=0}^{\infty}, (D_p)^{\infty}_{p=0}) = \{(f_p:C_p\rightarrow D_p)_{p=0}^{\infty}\ |\ f_{p-1}\partial_p^C=\partial_p^D f_p\}&lt;/math&gt;<br /> <br /> <br /> Now, in our case, the chain complexes do in fact commute because &lt;math&gt;\partial&lt;/math&gt; is defined by pre-composition but f is defined by post-composition. Hence, associativity of composition yields commutativity. <br /> <br /> <br /> '''Claim'''<br /> <br /> Homology of chain complexes is a functor in the natural way. That is, if &lt;math&gt;f_p:C_p\rightarrow D_p&lt;/math&gt; for each p induces the functor &lt;math&gt;f_*:H_p(C_*)\rightarrow H_p(D_*)&lt;/math&gt;<br /> <br /> The proof is by &quot;diagram chasing&quot;. Well, let &lt;math&gt;c\in C_p&lt;/math&gt;, &lt;math&gt;\partial c =0&lt;/math&gt;. <br /> <br /> Let &lt;math&gt;f*[c] = [fc]&lt;/math&gt;. Now &lt;math&gt;\partial fc = f\partial c = 0&lt;/math&gt;. Furthermore, suppose &lt;math&gt;c = \partial b&lt;/math&gt;. Then, &lt;math&gt;f_*c = fc = \partial fb&lt;/math&gt; so therefore &lt;math&gt;fc = \partial\beta&lt;/math&gt; some &lt;math&gt;\beta&lt;/math&gt;. This shows &lt;math&gt;f_*&lt;/math&gt; is well defined. <br /> <br /> Thus, for &lt;math&gt;c\in H_p(C_*)&lt;/math&gt; get &lt;math&gt;fc\in H_p(D_*)&lt;/math&gt; via the well defined functor &lt;math&gt;f_*&lt;/math&gt;.<br /> <br /> <br /> ===Second Hour===<br /> <br /> '''1) Homotopy Axioms'''<br /> <br /> If &lt;math&gt;f,g:X\rightarrow Y&lt;/math&gt; are homotopic then &lt;math&gt;f_* = g_*: H_p(X)\rightarrow H_p(Y)&lt;/math&gt;<br /> <br /> <br /> Applications: If X and Y are homotopy equivalent then &lt;math&gt;H_*(X) \cong H_*(Y)&lt;/math&gt;<br /> <br /> ''Proof:''<br /> <br /> let &lt;math&gt;f:X\rightarrow Y&lt;/math&gt;, &lt;math&gt;g:Y\rightarrow X&lt;/math&gt; such that &lt;math&gt;f\circ g\sim id_y&lt;/math&gt; and &lt;math&gt;g\circ f \sim id_x&lt;/math&gt;. Well &lt;math&gt;f_*\circ g_* = id_{H(Y)}&lt;/math&gt; and &lt;math&gt;g_*\circ f_* = id_{H(X)}&lt;/math&gt;<br /> <br /> Hence, &lt;math&gt;f_*&lt;/math&gt; and &lt;math&gt;g_*&lt;/math&gt; are invertible maps of each other. ''Q.E.D''<br /> <br /> <br /> '''Definition'''<br /> <br /> Two morphisms &lt;math&gt;f,g:C_*\rightarrow D_*&lt;/math&gt; between chain complexes are ''homotopic'' if you can find maps &lt;math&gt;h_p:C_p\rightarrow D_{p+1}&lt;/math&gt; such that &lt;math&gt;f_p - g_p = \partial_{p+1} h_p = h_{p-1}\partial_p&lt;/math&gt;<br /> <br /> <br /> '''Claim 1'''<br /> <br /> Given H a homotopy connecting f&lt;math&gt;,g:X\rightarrow&lt;/math&gt; Y we can construct a chain homotopy between &lt;math&gt;f_*,g_*:C_*(X)\rightarrow C_*(&lt;/math&gt;Y)<br /> <br /> <br /> '''Claim 2'''<br /> <br /> If &lt;math&gt;f,g:C_*\rightarrow C_*&lt;/math&gt; are chain homotopic then they induce equal maps on homology. <br /> <br /> <br /> <br /> ''Proof of 2''<br /> <br /> Assume &lt;math&gt;[c]\in H_p(C_*)&lt;/math&gt;, that is, &lt;math&gt;\partial c =0&lt;/math&gt;<br /> <br /> &lt;math&gt;[f_* c] - [g_* c] = [(f_*-g_*)] =[(\partial h + h\partial)c] = 0&lt;/math&gt; (as &lt;math&gt;\partial c = 0&lt;/math&gt; and homology ignores exact forms)<br /> <br /> Hence, at the level of homology they are the same. <br /> <br /> <br /> ''Proof of 1''<br /> <br /> Consider a simplex in X. Now consider its image, a simplex, in Y under g and f respectively. Because of the homotopy we can construct a triangular based cylinder in Y with the image under f at the top and the image under y at the bottom. <br /> <br /> Define &lt;math&gt;h\sigma&lt;/math&gt; = the above prism formed by &lt;math&gt;\sigma&lt;/math&gt; and the homotopy H. <br /> <br /> &lt;math&gt;(f_*-g_*)\sigma = h\partial\sigma = \partial h\sigma&lt;/math&gt;<br /> <br /> This, pictorially is correct but we need to be able to break up the prism, &lt;math&gt;\Delta_p\times I&lt;/math&gt; into a union of images of simplexes. <br /> <br /> Suppose p=0, i.e. a point. Hence &lt;math&gt;\Delta_0\times I&lt;/math&gt; is a line, which is a simplex. <br /> <br /> Suppose p=1 which yields a square. Adding a diagonal divides the square into two triangles, so is clearly a union of simplexes. <br /> <br /> Suppose p=2. We get a prism which has a triangle for a base and a top. Raise each vertex on the bottom to the top in turn. This makes the prism a union of three simplexes. <br /> <br /> In general for &lt;math&gt;\Delta_p\times I&lt;/math&gt; let &lt;math&gt;f_i = (l_i,0)&lt;/math&gt; and &lt;math&gt;g_i = (l_i,1)&lt;/math&gt; for vertexes &lt;math&gt;l_i&lt;/math&gt;<br /> <br /> Then, &lt;math&gt;h\sigma = \sum_{i=0}^p (-1)^i H\circ(\sigma\times I)\circ[f_0\cdots f_i g_i g_{i+1}\cdots g_p]&lt;/math&gt;<br /> <br /> which is in &lt;math&gt;C_{p+1}(Y)&lt;/math&gt;<br /> <br /> So have maps &lt;math&gt;Y\leftarrow_H X\times I\leftarrow \Delta_p\times I \leftarrow\Delta_{p+1}&lt;/math&gt;<br /> <br /> <br /> ''Claim: ''<br /> <br /> &lt;math&gt;\partial h +h\partial = f-g&lt;/math&gt;<br /> <br /> Loosely, &lt;math&gt;\partial h&lt;/math&gt; cuts each &lt;math&gt;[f_0\cdots f_i g_i g_{i+1}\cdots g_p]&lt;/math&gt; between the f_i and g_i and then deletes an entry. h\partial however does these in reverse order. Hence all that we are left with is &lt;math&gt;[f_0\cdots f_p] - [g_0\cdots g_p]&lt;/math&gt;</div> Trefor http://drorbn.net/index.php?title=User:Trefor User:Trefor 2008-03-07T23:46:54Z <p>Trefor: </p> <hr /> <div>{{0708-1300/Navigation}}<br /> NOTE: This page is used as a placeholder for incomplete typed class notes. <br /> <br /> <br /> {{0708-1300/Navigation}}<br /> <br /> ==A Homology Theory is a Monster==<br /> [[Image:0708-1300-AxiomsForHomology.png|thumb|center|540px|Page 183 of Bredon's book]]<br /> <br /> '''Bredon's Plan of Attack:''' State all, apply all, prove all.<br /> <br /> '''Our Route:''' Axiom by axiom - state, apply, prove. Thus everything we will do will be, or should be, labeled either &quot;'''S'''tate&quot; or &quot;'''P'''rove&quot; or &quot;'''A'''pply&quot;.<br /> <br /> ==Typed Notes==<br /> <br /> &lt;span style=&quot;color: red;&quot;&gt;The notes below are by the students and for the students. Hopefully they are useful, but they come with no guarantee of any kind.&lt;/span&gt;<br /> <br /> ===First Hour===<br /> <br /> <br /> <br /> Recall we had defined for a chain complex the associated homology groups: &lt;math&gt;H_p(C_*) :=\ ker\partial_p/im\partial_{p+1}&lt;/math&gt;<br /> <br /> From this we get the pth homology for a topological space &lt;math&gt;H_p(X)&lt;/math&gt;<br /> <br /> We have previously shown that <br /> <br /> ''1) &lt;math&gt;H_p(\cup X) = \oplus H_p(X_i)&lt;/math&gt; for disjoint unions of spaces &lt;math&gt;X_i&lt;/math&gt;<br /> <br /> 2) &lt;math&gt;H_p(pt) = \mathbb{Z}\delta_{p0}&lt;/math&gt;<br /> <br /> 3) &lt;math&gt;H_0(connected) = \mathbb{Z}&lt;/math&gt;<br /> <br /> 4) &lt;math&gt;H_1(connected) \cong \pi_1^{ab}(X)&lt;/math&gt; via the map<br /> <br /> &lt;math&gt;\phi:[\gamma]_{\pi_1^{ab}}\mapsto[\gamma]_{H_1}&lt;/math&gt;<br /> <br /> &lt;math&gt;\psi:\sigma\in C_1\mapsto[\gamma_{\sigma(0)}\sigma\bar{\gamma_{\sigma(1)}}]&lt;/math&gt; where &lt;math&gt;\sigma_x&lt;/math&gt; is a path connecting &lt;math&gt;x_0&lt;/math&gt; to x. <br /> <br /> <br /> We need to check the maps are in fact inverses of each other. <br /> <br /> Lets consider &lt;math&gt;\psi\circ\phi&lt;/math&gt;. We start with a closed path starting at &lt;math&gt;x_0&lt;/math&gt; (thought of as in the fundamental group). &lt;math&gt;\phi&lt;/math&gt; means we now think of it as a simplex in X with a point at &lt;math&gt;x_0&lt;/math&gt;. &lt;math&gt;\Psi&lt;/math&gt; now takes this to the path that parks at &lt;math&gt;x_0&lt;/math&gt; for a third of the time, goes around the loop and then parks for the remaining third of the time. Clearly this is homotopic this composition is homotopic to the identity. <br /> <br /> We now consider &lt;math&gt;\phi\circ\psi&lt;/math&gt;. Start with just a path &lt;math&gt;\sigma&lt;/math&gt;. Then &lt;math&gt;\psi&lt;/math&gt; makes a loop adding two paths from the &lt;math&gt;x_0&lt;/math&gt; to the start and finish of &lt;math&gt;\sigma&lt;/math&gt; forming a triangular like closed loop. We think of this loop as &lt;math&gt;\sigma'\in C_1&lt;/math&gt;<br /> <br /> Now, we start from c being &lt;math&gt;c = \sum a_i\sigma_i&lt;/math&gt; with &lt;math&gt;\partial c = 0&lt;/math&gt;. So get &lt;math&gt;\sum a_i(\partial \sigma_1) = \sum a_i(\sigma_i(1)-\sigma_i(0))&lt;/math&gt;<br /> <br /> So &lt;math&gt;\Psi(c) = [\gamma_{\sigma_i(0)}\sigma_i\bar{\gamma_{\sigma_i(1)}}]_{\pi_1}&lt;/math&gt; which maps to, under &lt;math&gt;\phi&lt;/math&gt;, &lt;math&gt;\sum a_i(\gamma_{\sigma_i(0)} + \sigma_i - \gamma_{\sigma_i(1)}) = \sum a_i\sigma_i = c&lt;/math&gt; ( in the homology gamma's cancel as &lt;math&gt;\partial c = 0&lt;/math&gt;)<br /> <br /> <br /> ''Axiomized Homology''<br /> <br /> We now will move to an approach where we prove that our defined homology satisfies a series of established homology axioms that will allow us to apply the machinery of general homology to our specific &quot;singular&quot; homology defined via simplices. <br /> <br /> Axiom 0) Homology if a function<br /> <br /> '''Definition''' The &quot;category of chain complexes&quot; is a category whose objects are chain complexes (of abelian groups) and morphisms which is a homomorphism between each abelian group in one chain and the corresponding group in the other chain such that the resulting diagram commutes. I.e, &lt;math&gt;Mor((C_p)_{p=0}^{\infty}, (D_p)^{\infty}_{p=0}) = \{(f_p:C_p\rightarrow D_p)_{p=0}^{\infty}\ |\ f_{p-1}\partial_p^C=\partial_p^D f_p\}&lt;/math&gt;<br /> <br /> <br /> Now, in our case, the chain complexes do in fact commute because &lt;math&gt;\partial&lt;/math&gt; is defined by pre-composition but f is defined by post-composition. Hence, associativity of composition yields commutativity. <br /> <br /> <br /> '''Claim'''<br /> <br /> Homology of chain complexes is a functor in the natural way. That is, if &lt;math&gt;f_p:C_p\rightarrow D_p&lt;/math&gt; for each p induces the functor &lt;math&gt;f_*:H_p(C_*)\rightarrow H_p(D_*)&lt;/math&gt;<br /> <br /> The proof is by &quot;diagram chasing&quot;. Well, let &lt;math&gt;c\in C_p&lt;/math&gt;, &lt;math&gt;\partial c =0&lt;/math&gt;. <br /> <br /> Let &lt;math&gt;f*[c] = [fc]&lt;/math&gt;. Now &lt;math&gt;\partial fc = f\partial c = 0&lt;/math&gt;. Furthermore, suppose &lt;math&gt;c = \partial b&lt;/math&gt;. Then, &lt;math&gt;f_*c = fc = \partial fb&lt;/math&gt; so therefore &lt;math&gt;fc = \partial\beta&lt;/math&gt; some &lt;math&gt;\beta&lt;/math&gt;. This shows &lt;math&gt;f_*&lt;/math&gt; is well defined. <br /> <br /> Thus, for &lt;math&gt;c\in H_p(C_*)&lt;/math&gt; get &lt;math&gt;fc\in H_p(D_*)&lt;/math&gt; via the well defined functor &lt;math&gt;f_*&lt;/math&gt;.<br /> <br /> <br /> ===Second Hour===<br /> <br /> '''1) Homotopy Axioms'''<br /> <br /> If f,g:X\rightarrow Y are homotopic then g_* = g_*: H_p(X)\rightarrow H_p(Y)<br /> <br /> <br /> Applications: If X and Y are homotopy equivalent then H_*(X) \cong H_*(Y)<br /> <br /> ''Proof:''<br /> <br /> let f:X\rightarrow Y, g:Y\rightarrow X such that f\circ g\sim id_y and g\circ f \sim id_x. Well f_*\circ g_* = id_{H(Y)} and g_*\circ f_* = id_{H(X)}<br /> <br /> Hence, f_* and g_* are invertible maps of each other. ''Q.E.D''<br /> <br /> <br /> '''Definition'''<br /> <br /> Two morphisms f,g:C_*\rightarrow D_* between chain complexes are ''homotopic'' if you can find maps h_p:C_p\rightarrow D_{p+1} such that f_p - g_p = \partial_{p+1} h_p = h_{p-1}\partial_p<br /> <br /> <br /> '''Claim 1'''<br /> <br /> Given H a homotopy connecting f,g:X\rightarrow Y we can construct a chain homotopy between f_*,g_*:C_*(X)\rightarrow C_*(Y)<br /> <br /> <br /> '''Claim 2'''<br /> <br /> If f,g:C_*\rightarrow C_* are chain homotopic then they induce equal maps on homology. <br /> <br /> <br /> <br /> ''Proof of 2''<br /> <br /> Assume [c]\in H_p(C_*), that is, \partial c =0<br /> <br /> [f_* c] - [g_* c] = [(f_*-g_*)] =[(\partial h + h\partial)c] = 0 (as \partial c = 0 and homology ignores exact forms)<br /> <br /> Hence, at the level of homology they are the same. <br /> <br /> <br /> ''Proof of 1''<br /> <br /> Consider a simplex in X. Now consider its image, a simplex, in Y under g and f respectively. Because of the homotopy we can construct a triangular based cylinder in Y with the image under f at the top and the image under y at the bottom. <br /> <br /> Define h\sigma = the above prism formed by \sigma and the homotopy H. <br /> <br /> (f_*-g_*)\sigma = h\partial\sigma = \partial h\sigma<br /> <br /> This, pictorially is correct but we need to be able to break up the prism into a union of images of simplexes. <br /> <br /> Suppose</div> Trefor http://drorbn.net/index.php?title=0708-1300/Class_notes_for_Tuesday,_February_26 0708-1300/Class notes for Tuesday, February 26 2008-03-07T02:59:44Z <p>Trefor: /* A Homology Theory is a Monster */</p> <hr /> <div>{{0708-1300/Navigation}}<br /> <br /> ==A Homology Theory is a Monster==<br /> [[Image:0708-1300-AxiomsForHomology.png|thumb|center|540px|Page 183 of Bredon's book]]<br /> <br /> '''Bredon's Plan of Attack:''' State all, apply all, prove all.<br /> <br /> '''Our Route:''' Axiom by axiom - state, apply, prove. Thus everything we will do will be, or should be, labeled either &quot;'''S'''tate&quot; or &quot;'''P'''rove&quot; or &quot;'''A'''pply&quot;.<br /> <br /> ==Typed Notes==<br /> <br /> &lt;span style=&quot;color: red;&quot;&gt;The notes below are by the students and for the students. Hopefully they are useful, but they come with no guarantee of any kind.&lt;/span&gt;<br /> <br /> ===First Hour===<br /> <br /> <br /> <br /> Recall we had defined for a chain complex the associated homology groups: &lt;math&gt;H_p(C_*) :=\ ker\partial_p/im\partial_{p+1}&lt;/math&gt;<br /> <br /> From this we get the pth homology for a topological space &lt;math&gt;H_p(X)&lt;/math&gt;<br /> <br /> We have previously shown that <br /> <br /> 1) &lt;math&gt;H_p(\cup X) = \oplus H_p(X_i)&lt;/math&gt; for disjoint unions of spaces &lt;math&gt;X_i&lt;/math&gt;<br /> <br /> 2) &lt;math&gt;H_p(pt) = \mathbb{Z}\delta_{p0}&lt;/math&gt;<br /> <br /> 3) &lt;math&gt;H_0(connected) = \mathbb{Z}&lt;/math&gt;<br /> <br /> 4) &lt;math&gt;H_1(connected) \cong \pi_1^{ab}(X)&lt;/math&gt; via the map<br /> <br /> &lt;math&gt;\phi:[\gamma]_{\pi_1^{ab}}\mapsto[\gamma]_{H_1}&lt;/math&gt;<br /> <br /> &lt;math&gt;\psi:\sigma\in C_1\mapsto[\gamma_{\sigma(0)}\sigma\bar{\gamma_{\sigma(1)}}]&lt;/math&gt; where &lt;math&gt;\sigma_x&lt;/math&gt; is a path connecting &lt;math&gt;x_0&lt;/math&gt; to x. <br /> <br /> <br /> We need to check the maps are in fact inverses of each other. <br /> <br /> Lets consider &lt;math&gt;\psi\circ\phi&lt;/math&gt;. We start with a closed path starting at &lt;math&gt;x_0&lt;/math&gt; (thought of as in the fundamental group). &lt;math&gt;\phi&lt;/math&gt; means we now think of it as a simplex in X with a point at &lt;math&gt;x_0&lt;/math&gt;. &lt;math&gt;\Psi&lt;/math&gt; now takes this to the path that parks at &lt;math&gt;x_0&lt;/math&gt; for a third of the time, goes around the loop and then parks for the remaining third of the time. Clearly this is homotopic this composition is homotopic to the identity. <br /> <br /> We now consider &lt;math&gt;\phi\circ\psi&lt;/math&gt;. Start with just a path &lt;math&gt;\sigma&lt;/math&gt;. Then &lt;math&gt;\psi&lt;/math&gt; makes a loop adding two paths from the &lt;math&gt;x_0&lt;/math&gt; to the start and finish of &lt;math&gt;\sigma&lt;/math&gt; forming a triangular like closed loop. We think of this loop as &lt;math&gt;\sigma'\in C_1&lt;/math&gt;<br /> <br /> Now, we start from c being &lt;math&gt;c = \sum a_i\sigma_i&lt;/math&gt; with &lt;math&gt;\partial c = 0&lt;/math&gt;. So get &lt;math&gt;\sum a_i(\partial \sigma_1) = \sum a_i(\sigma_i(1)-\sigma_i(0))&lt;/math&gt;<br /> <br /> So &lt;math&gt;\Psi(c) = [\gamma_{\sigma_i(0)}\sigma_i\bar{\gamma_{\sigma_i(1)}}]_{\pi_1}&lt;/math&gt; which maps to, under &lt;math&gt;\phi&lt;/math&gt;, &lt;math&gt;\sum a_i(\gamma_{\sigma_i(0)} + \sigma_i - \gamma_{\sigma_i(1)}) = \sum a_i\sigma_i = c&lt;/math&gt; ( in the homology gamma's cancel as &lt;math&gt;\partial c = 0&lt;/math&gt;)<br /> <br /> <br /> ''Axiomized Homology''<br /> <br /> We now will move to an approach where we prove that our defined homology satisfies a series of established homology axioms that will allow us to apply the machinery of general homology to our specific &quot;singular&quot; homology defined via simplices. <br /> <br /> Axiom 0) Homology if a function<br /> <br /> '''Definition''' The &quot;category of chain complexes&quot; is a category whose objects are chain complexes (of abelian groups) and morphisms which is a homomorphism between each abelian group in one chain and the corresponding group in the other chain such that the resulting diagram commutes. I.e, &lt;math&gt;Mor((C_p)_{p=0}^{\infty}, (D_p)^{\infty}_{p=0}) = \{(f_p:C_p\rightarrow D_p)_{p=0}^{\infty}\ |\ f_{p-1}\partial_p^C=\partial_p^D f_p\}&lt;/math&gt;<br /> <br /> <br /> Now, in our case, the chain complexes do in fact commute because &lt;math&gt;\partial&lt;/math&gt; is defined by pre-composition but f is defined by post-composition. Hence, associativity of composition yields commutativity. <br /> <br /> <br /> '''Claim'''<br /> <br /> Homology of chain complexes is a functor in the natural way. That is, if &lt;math&gt;f_p:C_p\rightarrow D_p&lt;/math&gt; for each p induces the functor &lt;math&gt;f_*:H_p(C_*)\rightarrow H_p(D_*)&lt;/math&gt;<br /> <br /> The proof is by &quot;diagram chasing&quot;. Well, let &lt;math&gt;c\in C_p&lt;/math&gt;, &lt;math&gt;\partial c =0&lt;/math&gt;. <br /> <br /> Let &lt;math&gt;f*[c] = [fc]&lt;/math&gt;. Now &lt;math&gt;\partial fc = f\partial c = 0&lt;/math&gt;. Furthermore, suppose &lt;math&gt;c = \partial b&lt;/math&gt;. Then, &lt;math&gt;f_*c = fc = \partial fb&lt;/math&gt; so therefore &lt;math&gt;fc = \partial\beta&lt;/math&gt; some &lt;math&gt;\beta&lt;/math&gt;. This shows &lt;math&gt;f_*&lt;/math&gt; is well defined. <br /> <br /> Thus, for &lt;math&gt;c\in H_p(C_*)&lt;/math&gt; get &lt;math&gt;fc\in H_p(D_*)&lt;/math&gt; via the well defined functor &lt;math&gt;f_*&lt;/math&gt;.</div> Trefor http://drorbn.net/index.php?title=0708-1300/Class_notes_for_Thursday,_January_24 0708-1300/Class notes for Thursday, January 24 2008-02-10T02:15:41Z <p>Trefor: /* Typed Notes */</p> <hr /> <div><br /> {{0708-1300/Navigation}}<br /> <br /> ==Typed Notes==<br /> <br /> &lt;span style=&quot;color: red;&quot;&gt;The notes below are by the students and for the students. Hopefully they are useful, but they come with no guarantee of any kind.&lt;/span&gt;<br /> <br /> <br /> '''Proof of Van Kampen'''<br /> <br /> Let &lt;math&gt;G_i = \pi_1(U_i)&lt;/math&gt;<br /> <br /> &lt;math&gt;H = \pi_1(U_1\cap U_2)&lt;/math&gt;<br /> <br /> &lt;math&gt;G=\pi_1(U_1\cup U_2)&lt;/math&gt;<br /> <br /> We aim to show that &lt;math&gt;G=G_1*_H G_2&lt;/math&gt;<br /> <br /> <br /> Hence, we want to define two maps:<br /> <br /> &lt;math&gt;\Phi:G_1*_H G_2\rightarrow G &lt;/math&gt; and<br /> <br /> <br /> &lt;math&gt;\Psi:G\rightarrow G_1*_H G_2&lt;/math&gt;<br /> <br /> such that they are inverses of each other. <br /> <br /> <br /> Now, recall the commuting diagram: <br /> <br /> &lt;math&gt;\begin{matrix}<br /> &amp;\ \ \ \ U_1&amp;&amp;\\<br /> &amp;\nearrow^{i_1}&amp;\searrow^{j_1}&amp;\\<br /> U_1\cap U_2&amp;&amp;&amp;U_1\cup U_2 = X\\<br /> &amp;\searrow_{i_2}&amp;\nearrow^{j_2}&amp;\\<br /> &amp;\ \ \ \ U_2&amp;&amp;\\<br /> \end{matrix}&lt;/math&gt;<br /> <br /> <br /> Further, let &lt;math&gt;b_i&lt;/math&gt; alternate between 1 and 2 for successive i's. <br /> <br /> Hence we define &lt;math&gt;\Phi&lt;/math&gt; via for &lt;math&gt;\alpha_i\in G_{b_i}&lt;/math&gt;, <br /> <br /> [&lt;math&gt;\alpha_1][\alpha_2]\ldots[\alpha_n]\rightarrow [j_{b_1 *}\alpha_1\cdot\ldots\cdot j_{b_n *}\alpha_n]&lt;/math&gt;<br /> <br /> Clearly this is well defined. We need to check the relations in <br /> &lt;math&gt;G_1*_H G_2&lt;/math&gt; indeed hold. Well, the identity element corresponds to the identity path so the relation that removes identities holds. Furthermore, the concatenation of paths is a sum after &lt;math&gt;\Phi&lt;/math&gt; and the definition necessitates the third relation holds. <br /> <br /> <br /> Now for &lt;math&gt;\Psi&lt;/math&gt;:<br /> <br /> Elements in G correspond with paths &lt;math&gt;\gamma&lt;/math&gt; in &lt;math&gt;U_1\cup U_2&lt;/math&gt;<br /> <br /> <br /> On consider such a &lt;math&gt;\gamma&lt;/math&gt;. The Lesbegue Lemma let us break &lt;math&gt;\gamma&lt;/math&gt; up so &lt;math&gt;\gamma = \gamma_1\ldots\gamma_N&lt;/math&gt; such that each &lt;math&gt;\gamma_i&lt;/math&gt; is in just &lt;math&gt;G_{b_i}&lt;/math&gt;<br /> <br /> Now, &lt;math&gt;\gamma_i&lt;/math&gt; does not go from base point to base point, so we can't consider it as a loop itself. Let &lt;math&gt;x_i&lt;/math&gt; denote the endpoint of &lt;math&gt;\gamma_i&lt;/math&gt; (and hence &lt;math&gt;x_{i-1}&lt;/math&gt; is the beginning point) and further, &lt;math&gt;x_0 = b&lt;/math&gt; the base point. <br /> <br /> Choose paths &lt;math&gt;\eta_i&lt;/math&gt; connecting &lt;math&gt;x_i&lt;/math&gt; to b such that if &lt;math&gt;x_i\in U_{b_i}&lt;/math&gt; then &lt;math&gt;\eta_i\sub U_{b_i}&lt;/math&gt;<br /> <br /> Hence, &lt;math&gt;\gamma\sim\bar{\eta_0}\gamma_1\eta_1\bar{\eta_1}\gamma_2\eta_2\ldots \mapsto [\bar{\eta_0}\gamma_1\eta_1][\bar{\eta_1}\gamma_2\eta_2]\ldots&lt;/math&gt;<br /> <br /> where each section is entirely in &lt;math&gt;G_{b_i}&lt;/math&gt;. This above mapping is &lt;math&gt;\Psi&lt;/math&gt;<br /> <br /> We need to show that &lt;math&gt;\Psi&lt;/math&gt; is well defined. I.e., <br /> <br /> 1) &lt;math&gt;\Psi&lt;/math&gt; is independent of the subdivision<br /> <br /> 2) &lt;math&gt;\Psi&lt;/math&gt; is independent of choice of the &lt;math&gt;\eta_i&lt;/math&gt;'s<br /> <br /> 3) If &lt;math&gt;\gamma_1\sim\gamma_2&lt;/math&gt; then &lt;math&gt;\Psi(\gamma_1) = \Psi(\gamma_2)&lt;/math&gt;<br /> <br /> <br /> For 1) it is enough to show that one can add or remove a single subdivision point. Suppose you add a new subdivision point c in between two others a and b. Traveling between a and b is the same as going from a to c then to the basepoint and back and then continuing on to b. But this is precisely what happens when you add a subdivision point. Likewise for removing points. <br /> <br /> <br /> For 2), Suppose &lt;math&gt;\eta_i&lt;/math&gt; goes from the basepoint to &lt;math&gt;x_i&lt;/math&gt;. Add a subdivision point y right beside &lt;math&gt;x_i&lt;/math&gt; and remove the one at &lt;math&gt;x_i&lt;/math&gt;. Then add again the original point &lt;math&gt;x_i&lt;/math&gt;. The trick is that the new &lt;math&gt;\eta_i'&lt;/math&gt; is the one used when adding y. The map from the new &lt;math&gt;x_i&lt;/math&gt; to the basepoint is this &lt;math&gt;\eta_i'&lt;/math&gt; with the infinitesimal connection between y and &lt;math&gt;x_i&lt;/math&gt; added. The subdivision point y is then removed leaving our original configuration of basepoints, only with the path from &lt;math&gt;x_i&lt;/math&gt; to b now being &lt;math&gt;\eta_i'&lt;/math&gt; instead of the original &lt;math&gt;\eta_i&lt;/math&gt;<br /> <br /> <br /> For 3) we consider the homotopy between &lt;math&gt;\gamma_1&lt;/math&gt; and &lt;math&gt;\gamma_2&lt;/math&gt; thought of as a square with &lt;math&gt;\gamma_1&lt;/math&gt; on the bottom and &lt;math&gt;\gamma_2&lt;/math&gt; on the top. We use the Lesbegue Lemma to subdivide the square into many subsquares such that each one is entirely in &lt;math&gt;U_{b_i}&lt;/math&gt;<br /> <br /> We further modify the homotopy H to a new homotopy &lt;math&gt;\tilde{H}&lt;/math&gt; such that each grid point gets mapped to b. <br /> <br /> This can be thought of as &quot;pinching&quot; each gird point and pulling it to b or, alternatively, as tossing a &quot;handkerchief on a bed of nails&quot; <br /> <br /> If we let &lt;math&gt;\gamma_1&lt;/math&gt; be broken in to section &lt;math&gt;\alpha_1,\ldots, \alpha_n&lt;/math&gt; along the bottom then &lt;math&gt;\tilde{H}&lt;/math&gt; first lifts &lt;math&gt;\alpha_1&lt;/math&gt; to the top and right sides of the subsquare such that the gridpoint goes to b. &lt;math&gt;\tilde{H}&lt;/math&gt; then moves the next square up in an analogous manner until we are at the top with &lt;math&gt;\gamma_2&lt;/math&gt;.</div> Trefor http://drorbn.net/index.php?title=0708-1300/Class_notes_for_Tuesday,_January_22 0708-1300/Class notes for Tuesday, January 22 2008-02-10T02:10:03Z <p>Trefor: /* Second Hour */</p> <hr /> <div>{{0708-1300/Navigation}}<br /> <br /> ==Pictures for a Van-Kampen Computation==<br /> {{In|<br /> n = 1 |<br /> in = &lt;nowiki&gt;&lt;&lt; KnotTheory&lt;/nowiki&gt;}}<br /> <br /> &lt;tt&gt;Loading KnotTheory version of January 13, 2008, 20:30:12.1353.&lt;br&gt;<br /> Read more at http://katlas.org/wiki/KnotTheory.&lt;/tt&gt;<br /> <br /> {{Graphics|<br /> n = 2 |<br /> in = &lt;nowiki&gt;TubePlot[TorusKnot[8, 3]]&lt;/nowiki&gt; |<br /> img= 0708-1300-T83.png}}<br /> <br /> {{In|<br /> n = 3 |<br /> in = &lt;nowiki&gt;TC[r1_, t1_,r2_,t2_ ] := {<br /> (r1 +r2 Cos[2Pi t2])Cos[2Pi t1],<br /> (r1 +r2 Cos[2Pi t2])Sin[2Pi t1],<br /> r2 Sin[2Pi t2]<br /> };&lt;/nowiki&gt;}}<br /> <br /> {{In|<br /> n = 4 |<br /> in = &lt;nowiki&gt;InflatedTorus[p_, q_, b_] := ParametricPlot3D[<br /> TC[<br /> 2, p t - q s,<br /> 1 + b(p^2 + q^2)s(1 - (p^2 + q^2)s), q t + p s<br /> ],<br /> {t, 0, 1}, {s, 0, 1/(p^2 + q^2)},<br /> PlotPoints -&gt; {6(p^2 + q^2) + 1, 7},<br /> DisplayFunction -&gt; Identity<br /> ];&lt;/nowiki&gt;}}<br /> <br /> {{Graphics|<br /> n = 5 |<br /> in = &lt;nowiki&gt;GraphicsArray[{{InflatedTorus[3,8,1], InflatedTorus[3,8,-1]}}]&lt;/nowiki&gt; |<br /> img= 0708-1300-InflatedTori.png |<br /> width = 640px}}<br /> <br /> ==Typed Notes==<br /> <br /> ===First Hour===<br /> <br /> '''Today's Agenda:''' <br /> <br /> 1) More Examples of Van-Kampen Theorem<br /> <br /> 2) More Diagrams<br /> <br /> 3) Proof of Van-Kampen (was not done)<br /> <br /> <br /> We began by recalling the examples from last class. I will not repeat that here, merely making a few additional comments that came out:<br /> <br /> <br /> '''Notation:'''<br /> <br /> Technically, &lt;math&gt;A*_H B&lt;/math&gt; is poor notion as it implies that knowledge of A, B and H is sufficient to construct &lt;math&gt;A*_H B&lt;/math&gt;. In fact, we ALSO need to know the maps from H into A and B respectively in order for &lt;math&gt;A*_H B&lt;/math&gt; to be defined. <br /> <br /> <br /> '''Aside'''<br /> <br /> Last class we simply wrote down the schematic for the two holed torus as an octagon with the identifications on the edges given last class. We now consider how one arrives at this schematic. <br /> <br /> To create the two holed torus one begins with two tori. One then cuts out a small open disk from each torus and then glues the two boundaries together. Let us consider what this looks like when considering a torus as the normal schematic with a square in the plane with the normal identification of the sides. Removing an open disk is equivalent to removing the inside of a loop starting at one of the corners and finishing at that same corner. This is equivalent to making a pentagon with sides &lt;math&gt;aba^{-1}b^{-1}c&lt;/math&gt; where c is the added edge. <br /> <br /> Consider two such pentagons, gluing along the edge c forms precisely the octagon we had for the two holed torus last class. <br /> <br /> [[Image:0708-1300_notes_22-01-08a.jpg|200px]]<br /> <br /> '''Proposition'''<br /> <br /> Letting &lt;math&gt;\Sigma_g&lt;/math&gt; denote the g holed torus, then &lt;math&gt;\Sigma_g\neq\Sigma_{g'}&lt;/math&gt;<br /> <br /> (Note, I used the symbol &lt;math&gt;\neq&lt;/math&gt; to as the normal \ncong command doesn't seem to work. Take its meaning in context.)<br /> <br /> <br /> Aside: Consider a functor from the category of groups to the category of Abelian groups via<br /> <br /> &lt;math&gt;G\mapsto G^{ab} = G/(ab=ba)&lt;/math&gt;<br /> <br /> If we have a (homo)morphism from &lt;math&gt;G\rightarrow H&lt;/math&gt; then the functor takes &lt;math&gt;H\rightarrow H^{ab}&lt;/math&gt; and yields a map &lt;math&gt;G^{ab}\rightarrow H^{ab}&lt;/math&gt; such that everything commutes. <br /> <br /> Hence we know that &lt;math&gt;\pi_1^{ab}(\Sigma_g) \cong \mathbb{Z}^{2g}\neq\mathbb{Z}^{2g'} \cong\pi_1^{ab}(\Sigma_{g'})&lt;/math&gt;<br /> <br /> Of course, we need to know that in fact &lt;math&gt;\mathbb{Z}^m\neq\mathbb{Z}^n&lt;/math&gt; if &lt;math&gt;m\neq n&lt;/math&gt;<br /> <br /> As such, since the abelianizations are not isomorphic,neither are the original groups and the spaces themselves are not homeomorphic. <br /> <br /> <br /> '''Example'''<br /> <br /> &lt;math&gt;\pi_1(\mathbb{R}P^2)&lt;/math&gt; is &lt;math&gt;\pi_1&lt;/math&gt; of the space which can be written as a disk with two antipodal points on the boundary circle on it with the identification that the top path a (going clockwise along the boundary) is glued to the bottom path (also going clock wise). But &lt;math&gt;\pi_1&lt;/math&gt; of this is just &lt;math&gt;&lt;a&gt;/(a^2 = e) \cong \mathbb{Z}/2&lt;/math&gt;<br /> <br /> <br /> '''Claim:'''<br /> <br /> Puncturing an n-manifold, &lt;math&gt;n\geq 3&lt;/math&gt;, does not change &lt;math&gt;\pi_1(M)&lt;/math&gt;. I.e., if &lt;math&gt;p\in M^n&lt;/math&gt; then &lt;math&gt;\pi_1(M)\cong\pi_1(M-\{p\})&lt;/math&gt;<br /> <br /> <br /> ''Proof:''<br /> <br /> Let &lt;math&gt;U_1 = M-\{p\}&lt;/math&gt;<br /> <br /> &lt;math&gt;U_2&lt;/math&gt; = a coordinate patch about p. <br /> <br /> Then &lt;math&gt;U_1\cap U_2 = B^n-\{p\}\cong S^{n-1}&lt;/math&gt;<br /> <br /> If n=3, &lt;math&gt;\pi_1(S^2) = \{e\}&lt;/math&gt; as we have computed before. <br /> <br /> Hence, &lt;math&gt;\pi_1(M) = \pi_1(U_1)*_{\{\}}\{\} = \pi_1(U_1)&lt;/math&gt;<br /> <br /> <br /> Now, &lt;math&gt;\pi_1(S^3)\cong\pi_1(S^3-\{p\}) = \pi_1(B^3) = \{e\}&lt;/math&gt;<br /> <br /> Continuing inductively the theorem holds for all n. <br /> <br /> <br /> '''Aside:'''<br /> <br /> If X is connected and &lt;math&gt;b_1,\ b_2\in X&lt;/math&gt; then &lt;math&gt;\pi_1(X,b_1) = \pi_1(X,b_2)&lt;/math&gt;. I.e., it does not matter which base point we choose in a connected space, the fundamental group is invariant of this. <br /> <br /> ''Proof''<br /> <br /> Consider a path &lt;math&gt;\eta&lt;/math&gt; from &lt;math&gt;b_1&lt;/math&gt; to &lt;math&gt;b_2&lt;/math&gt;. The returning path is denoted &lt;math&gt;\bar{\eta}&lt;/math&gt;<br /> <br /> Consider a loop from &lt;math&gt;b_2&lt;/math&gt; called &lt;math&gt;\gamma&lt;/math&gt;. <br /> <br /> Then get a loop from &lt;math&gt;b_1&lt;/math&gt; via &lt;math&gt;\gamma\mapsto \bar{\eta}\gamma\eta&lt;/math&gt;<br /> <br /> Similarly about &lt;math&gt;b_2, \gamma\mapsto\eta\gamma\bar{\eta}&lt;/math&gt;<br /> <br /> <br /> Considering the composition we get &lt;math&gt;\eta\bar{\eta}\gamma\eta\bar{\eta}\sim\gamma&lt;/math&gt;.<br /> <br /> ===Second Hour===<br /> <br /> '''Claim'''<br /> <br /> &lt;math&gt;S^3&lt;/math&gt; is the union (with common boundary) of two solid tori &lt;math&gt;S^1\times D^1&lt;/math&gt;<br /> <br /> The natural way to add two tori with common boundary would be two glue the boundaries of two disks (making &lt;math&gt;S^2&lt;/math&gt;) together for each angle going around the torus thus yielding &lt;math&gt;S^1\times S^2&lt;/math&gt;. Clearly this is not the same as &lt;math&gt;S^3&lt;/math&gt; as the fundamental groups differ. <br /> <br /> Instead consider the following description. Look at a torus in the zx plane, this looks like two disks with the z axis in between them such that rotating these two disks about the z axis will yield the torus. <br /> <br /> Lets now add in the second torus into this picture. We first draw a horizontal line between the two disks. We then &quot;blow&quot; up from beneath so the horizontal line is slightly curved. We imagine continuing to blow yielding larger and larger loops between the two disks until it &quot;pops&quot; forming the pure horizontal line consisting of the loop at infinity. Do the same for the bottom. Hence, the boundaries of the two tori drawn this way clearly are the same, and between the two cover the entire zx plane (and &quot;point at infinity). Rotating this picture about the z axis yields all of S^1 as the union of these two sets. <br /> <br /> [[Image:0708-1300_notes_22-01-08b.jpg|200px]]<br /> <br /> '''Claim:'''<br /> <br /> &lt;math&gt;\pi_1(S^3) = \{e\}&lt;/math&gt;<br /> <br /> &lt;math&gt;U_1&lt;/math&gt; = the normal solid torus thickened a bit and under &lt;math&gt;\pi_1&lt;/math&gt; yields &lt;math&gt;&lt;\alpha&gt;&lt;/math&gt; <br /> <br /> &lt;math&gt;U_2&lt;/math&gt; = the other solid torus, also thickened a bit, under &lt;math&gt;\pi_1&lt;/math&gt; yields &lt;math&gt;&lt;\beta&gt;&lt;/math&gt;<br /> <br /> <br /> &lt;math&gt;U_1\cap U_2&lt;/math&gt; is a normal torus only with slightly thick walls opposed to infinitely thin ones (homotopically the same)<br /> <br /> So, &lt;math&gt;\pi_1(U_1\cap U_2)\cong\mathbb{Z}^2 \cong &lt;a,b&gt;/ab=ba&lt;/math&gt;<br /> <br /> So, &lt;math&gt;\pi_1(S^3) \cong\mathbb{Z}*_{\mathbb{Z}\times\mathbb{Z}}\mathbb{Z}&lt;/math&gt;<br /> <br /> However, we still need to describe &lt;math&gt;i_{1*}&lt;/math&gt; and &lt;math&gt;i_{2*}&lt;/math&gt;<br /> <br /> Do do this let me describe a,b,&lt;math&gt;\alpha&lt;/math&gt; and &lt;math&gt;\beta&lt;/math&gt; explicitly.<br /> <br /> Considering the description of the two tori given above, we let a go around the outside of one of the two disks in the plane and b go from a point on the boundary of the same disk, following the rotation about the z axis, to a point on the boundary of the other dis. &lt;math&gt;\alpha&lt;/math&gt; is similar to b, but thought of as being on the boundary of the OTHER torus. &lt;math&gt;\beta&lt;/math&gt; consists of the path along the z axis. <br /> <br /> Hence, <br /> <br /> &lt;math&gt;i_{1*}:&lt;/math&gt; <br /> <br /> &lt;math&gt;a\rightarrow e&lt;/math&gt;<br /> <br /> &lt;math&gt;b\rightarrow \alpha&lt;/math&gt;<br /> <br /> &lt;math&gt;i_{2*}:&lt;/math&gt;<br /> <br /> &lt;math&gt; a\rightarrow\beta&lt;/math&gt;<br /> <br /> &lt;math&gt;b\rightarrow e&lt;/math&gt; (as it is contractible)<br /> <br /> <br /> Hence, &lt;math&gt;\pi_1(S^3) = F(\alpha, \beta)/(e=\beta, \alpha = e)&lt;/math&gt;<br /> <br /> Hence, &lt;math&gt;\pi_1(S^3) = \{e\}&lt;/math&gt;<br /> <br /> <br /> '''Example'''<br /> <br /> Define the &quot;Torus knot &lt;math&gt;T_{p,q}&lt;/math&gt;&quot; where p and q are relatively prime integers. The knot &lt;math&gt;T_{8,3}&lt;/math&gt; is given above. We can think of this in the following ways:<br /> <br /> 1) &lt;math&gt;T_{p,q}&lt;/math&gt; is the knot that wraps around the torus p times one way and q times the other way. <br /> <br /> 2) Formally, let &lt;math&gt;\sigma:S^1\times S^1\rightarrow\mathbb{R}^3&lt;/math&gt; be the standard embedding of a torus. Let &lt;math&gt;\gamma:[0,1]\rightarrow S^1\times S^1&lt;/math&gt; be &lt;math&gt;t\rightarrow (e^{2\pi i pt}, e^{i2\pi qt})&lt;/math&gt;<br /> <br /> Then &lt;math&gt;T_{p,q}&lt;/math&gt; is &lt;math&gt;\sigma\circ\gamma&lt;/math&gt;<br /> <br /> <br /> 3) Recall that the torus can be thought of as the image of the mapping &lt;math&gt;\mathbb{R}^2\rightarrow\mathbb{R}^2/\mathbb{Z}^2&lt;/math&gt;<br /> <br /> Consider the rectangle in the real plane: ([0,p],[0,q]) and consider the path which is the diagonal line from the corner (0,0) to the corner (p,q)<br /> <br /> No two points on this line are the same under the mapping down to the torus. If they were, then &lt;math&gt;\Delta y&lt;/math&gt; and &lt;math&gt;\Delta x&lt;/math&gt; would be integers and hence &lt;math&gt;\Delta y/\Delta x&lt;/math&gt; would be the slope of the line. But the slope of the line is q/p which is already in lowest common terms by assumption. <br /> <br /> <br /> <br /> Lets compute the fundamental group of the compliment of the torus knot. <br /> <br /> &lt;math&gt;\pi_1(\mathbb{R}^3-T_{p,q})\cong\pi_1(S^3-T_{p,q})&lt;/math&gt;<br /> <br /> &lt;math&gt;U_1&lt;/math&gt;: inflated bagel, constrained by &lt;math&gt;T_{p,q}&lt;/math&gt;<br /> <br /> &lt;math&gt;U_2&lt;/math&gt;: inflated bubble constrained by &lt;math&gt;T_{p,q}&lt;/math&gt;<br /> <br /> (See top of page for pictures)<br /> <br /> The intersection &lt;math&gt;U_1\cap U_2&lt;/math&gt; looks somewhat like a belt. It has some thickness to it and is wrapped around the torus, eventually forming a loop. Hence it looks like a squashed disk cross a circle. Hence, under &lt;math&gt;\pi_1&lt;/math&gt; this is just &lt;math&gt;\mathbb{Z}\cong&lt;\gamma&gt;&lt;/math&gt; where &lt;math&gt;\gamma&lt;/math&gt; is the path parallel to &lt;math&gt;T_{p,q}&lt;/math&gt;<br /> <br /> <br /> We thus get the maps, <br /> <br /> &lt;math&gt;i_{1*}: \gamma\mapsto\alpha^p&lt;/math&gt;<br /> <br /> &lt;math&gt;i_{2*}:\gamma\mapsto\beta^q&lt;/math&gt;<br /> <br /> Hence, &lt;math&gt;\pi_1(T_{p,q}^c) = &lt;\alpha,\beta&gt;/\alpha^p = \beta^q&lt;/math&gt;<br /> <br /> Thus, &lt;math&gt;T_{p,q}\neq T_{p',q'}&lt;/math&gt;<br /> <br /> <br /> '''Diagrams:'''<br /> <br /> <br /> Recall our diagram from last class: <br /> <br /> <br /> &lt;math&gt;\begin{matrix}<br /> &amp;\ \ \ \ U_1&amp;&amp;\\<br /> &amp;\nearrow^{i_1}&amp;\searrow^{j_1}&amp;\\<br /> U_1\cap U_2&amp;&amp;&amp;U_1\cup U_2\\<br /> &amp;\searrow_{i_2}&amp;\nearrow^{j_2}&amp;\\<br /> &amp;\ \ \ \ U_2&amp;&amp;\\<br /> \end{matrix}&lt;/math&gt;<br /> <br /> <br /> &lt;math&gt;U_1\cup U_2&lt;/math&gt; can be ''defined'' as the object such that the above diagram commutes and should the following commute: <br /> <br /> &lt;math&gt;\begin{matrix}<br /> &amp;\ \ \ \ U_1&amp;&amp;\\<br /> &amp;\nearrow^{i_1}&amp;\searrow^{j_1}&amp;\\<br /> U_1\cap U_2&amp;&amp;&amp;Y\\<br /> &amp;\searrow_{i_2}&amp;\nearrow^{j_2}&amp;\\<br /> &amp;\ \ \ \ U_2&amp;&amp;\\<br /> \end{matrix}&lt;/math&gt;<br /> <br /> Then there is a unique map between &lt;math&gt;U_1\cup U_2&lt;/math&gt; and Y such that the composed diagram commutes. <br /> <br /> <br /> Indeed, the same is true for general categories. <br /> <br /> For <br /> <br /> &lt;math&gt;\begin{matrix}<br /> &amp;\ \ \ \ G_1&amp;&amp;\\<br /> &amp;\nearrow^{i_1}&amp;\searrow^{j_1}&amp;\\<br /> H&amp;&amp;&amp;P\\<br /> &amp;\searrow_{i_2}&amp;\nearrow^{j_2}&amp;\\<br /> &amp;\ \ \ \ G_2&amp;&amp;\\<br /> \end{matrix}&lt;/math&gt;<br /> <br /> commuting, P is defined as an object such that if also <br /> <br /> &lt;math&gt;\begin{matrix}<br /> &amp;\ \ \ \ G_1&amp;&amp;\\<br /> &amp;\nearrow^{i_1}&amp;\searrow^{j_1}&amp;\\<br /> H&amp;&amp;&amp;Q\\<br /> &amp;\searrow_{i_2}&amp;\nearrow^{j_2}&amp;\\<br /> &amp;\ \ \ \ G_2&amp;&amp;\\<br /> \end{matrix}&lt;/math&gt;<br /> <br /> were to commute then there is a unique morphism from P to Q such that the composed diagram computes. <br /> <br /> <br /> In the category of groups, this &quot;pushforward&quot; P is unique and is isomorphic to &lt;math&gt;G_1*_H G_2&lt;/math&gt;</div> Trefor http://drorbn.net/index.php?title=0708-1300/Class_notes_for_Tuesday,_November_27 0708-1300/Class notes for Tuesday, November 27 2008-02-09T22:00:56Z <p>Trefor: /* Second Hour */</p> <hr /> <div>{{0708-1300/Navigation}}<br /> ==Today's Agenda==<br /> * The planimeter with a picture from http://whistleralley.com/planimeter/planimeter.htm but our very own plane geometry and Stokes' theorem.<br /> * Completion of the proof of Stokes' theorem.<br /> * Completion of the discussion of the two- and three-dimensional cases of Stokes' theorem.<br /> * With luck, a discussion of de-Rham cohomology, homotopy invariance and Poincaré's lemma.<br /> <br /> <br /> ==Class Notes==<br /> <br /> &lt;span style=&quot;color: red;&quot;&gt;The notes below are by the students and for the students. Hopefully they are useful, but they come with no guarantee of any kind.&lt;/span&gt;<br /> <br /> ===First Hour===<br /> <br /> '''Planimeter'''<br /> <br /> A planimeter consists of two rods connected with a join where the end of one rod is fixed (but free to rotate) and the opposing end of the second rod traces out the boundary of some surface on the plane. I.e., the planimeter is kind of like a 1 legged roach. At the join of the two rods is a wheel which rotates (and measures the rotation) when the rod tracing the boundary moves in the normal direction and simply slides back an forth when moved in a tangential direction. <br /> <br /> Now we recall from plane geometry that we can locate points in the polar form &lt;math&gt;(r,\theta)&lt;/math&gt; and have the equations &lt;math&gt;x = rcos\theta&lt;/math&gt; and &lt;math&gt;y = rsin\theta&lt;/math&gt;<br /> <br /> Hence, <br /> <br /> &lt;math&gt;dx = cos\theta dr - rsin\theta d\theta&lt;/math&gt;<br /> <br /> &lt;math&gt;dy = sin\theta dr + rcos\theta d\theta&lt;/math&gt;<br /> <br /> Hence &lt;math&gt;dx\wedge dy = r(cos^2\theta + sin^2\theta)dr\wedge d\theta = rdr\wedge d\theta&lt;/math&gt;<br /> <br /> <br /> <br /> <br /> Now, the planimeter is essentially a 1 form corresponding to the speed of the wheel. We consider a diagram where the angle from the horizontal at the fixed end of the planimeter to the measuring end is &lt;math&gt;\theta&lt;/math&gt; and the angle from the horizontal to the first rod (the one connected to the fixed point) is &lt;math&gt;\theta + \phi&lt;/math&gt;. Hence &lt;math&gt;r = 2cos\phi&lt;/math&gt; and &lt;math&gt;dr = -2sin\phi d\phi&lt;/math&gt;<br /> <br /> With a little plane geometry we can see that &lt;math&gt;\omega = cos2\phi d(\theta + \phi)&lt;/math&gt;<br /> <br /> Computing, <br /> <br /> &lt;math&gt;d\omega = -2 sin2\phi d\phi\wedge(d\theta + d\phi) = -4 sin\phi cos\phi d\phi\wedge d\theta = 2cos\phi dr\wedge d\theta = rdr\wedge d\theta = dx\wedge dy&lt;/math&gt;<br /> <br /> <br /> Now applying stokes theorem, the the planimeter integrates &lt;math&gt;\omega&lt;/math&gt; over the boundary of our surface and hence this is just the integral of &lt;math&gt;d\omega&lt;/math&gt; over the surface. But this is just the integral of the area form. <br /> <br /> Hence the planimeter measure the area of a surface. <br /> <br /> <br /> '''Back to Stokes Theorem'''<br /> <br /> <br /> Firstly recall that &lt;math&gt;\partial M&lt;/math&gt; is oriented so that if you prepend the outward normal to its orientation you get the orientation of M<br /> <br /> Alternatively we recall that neighborhoods of points on the boundary look like the half space. Hence we can choose to restrict our attention to atlas's where all charts look like &lt;math&gt;H= \{x\in\mathbb{R}^n:x_1\leq 0\}&lt;/math&gt;<br /> <br /> We can see that these orientations are the same, i.e., just prepend the outward normal to the half space. <br /> <br /> <br /> ''Proof of Stokes''<br /> <br /> <br /> We have now defined all the terms. WLOG &lt;math&gt;\omega&lt;/math&gt; is supported in one chart (by linearity)<br /> <br /> For a compactly supported n-1 form on H need to show that &lt;math&gt;\int_{\partial H}\omega = \int_H d\omega&lt;/math&gt;<br /> <br /> <br /> We let &lt;math&gt;\omega = \sum f_i dx_1\wedge\cdots\wedge \hat{dx_i}\wedge\cdots\wedge dx_n&lt;/math&gt; (where the hat means it is omitted)<br /> <br /> &lt;math&gt;d\omega = \sum (-1)^{i-1} \frac{\partial f_i}{\partial x_i} dx_1\wedge\cdots\wedge dx_n&lt;/math&gt;<br /> <br /> <br /> So, &lt;math&gt;\int_H d\omega = \sum \int_{[x_1\leq 0]}(-1)^{i-1}\frac{\partial f_i}{\partial x_i}dx_1\wedge\cdots\wedge dx_n = \sum (-1)^{i-1}\int_{[x_1\leq 0]}\frac{\partial f_i}{\partial x_i}&lt;/math&gt; <br /> <br /> via fundamental theorem of calculus and that the &lt;math&gt;f_i&lt;/math&gt;'s are compactly supported we get <br /> <br /> &lt;math&gt;= \int_{[x_1\leq 0]} \frac{\partial f_1}{\partial x_1} = \int_{[x_1=0]} f_1&lt;/math&gt;<br /> <br /> <br /> Hence with the standard inclusion of &lt;math&gt;\partial H = \mathbb{R}^{n-1}_{x_2\cdots x_n}&lt;/math&gt; we get <br /> <br /> <br /> &lt;math&gt;\int_{\partial H}\omega = \int_{\mathbb{R}^{n-1}_{x_2\cdots x_n}}\iota^*(\sum f_i dx_1\wedge\cdots\wedge \hat{dx_i}\wedge\cdots\wedge dx_n) = \int_{[x_1=0]}f_1&lt;/math&gt;<br /> <br /> <br /> Thus these are the same and the theorem is proved ''Q.E.D.''<br /> <br /> <br /> '''Real Plane'''<br /> <br /> Consider &lt;math&gt;\Omega^0(\mathbb{R}^2)\rightarrow^d\Omega^1(\mathbb{R}^2)\rightarrow^d\Omega^2(\mathbb{R}^2)&lt;/math&gt;<br /> <br /> <br /> Forms in &lt;math&gt;\Omega^1(\mathbb{R}^2)&lt;/math&gt; look like &lt;math&gt;Fdx +Gdy&lt;/math&gt; and map under d to &lt;math&gt;(G_x - F_y)dx\wedge dy&lt;/math&gt;<br /> <br /> Hence applying Stokes' Theorem:<br /> <br /> &lt;math&gt;\int_{\partial D}Fdx + Gdy = \int_D (G_x-F_y)dxdy&lt;/math&gt;<br /> <br /> This is known as Greens Theorem<br /> <br /> <br /> In complex analysis we also have a similar result Cauchy's Theorem where the integral of an analytic function around a closed path is zero. This is because analytic functions obey the Cauchy-Riemann equations and hence &lt;math&gt;G_x-F_y&lt;/math&gt; is identically zero.<br /> <br /> ===Second Hour===<br /> <br /> <br /> '''Example 2'''<br /> <br /> &lt;math&gt;\Omega^k(\mathbb{R}^3)&lt;/math&gt;<br /> <br /> <br /> Recall previous we had consider the spaces &lt;math&gt;\Omega^k(\mathbb{R}^3)&lt;/math&gt; and showed that &lt;math&gt;\Omega^0&lt;/math&gt; and &lt;math&gt;\Omega^ 3&lt;/math&gt; corresponded with functions and that &lt;math&gt;\Omega^2&lt;/math&gt; and &lt;math&gt;\Omega^1&lt;/math&gt; corresponded with triples of functions (i.e. vector fields). We also showed that the d functions between these spaces are the gradient, curl and divergence functions from vector calculus. <br /> <br /> We are now interested in integrating, using Stokes Theorem, forms in these spaces. <br /> <br /> <br /> First, note that to a 0 manifold, assigning an orientation to the manifold is just assigning a plus or minus sign to the manifold as a result of it having a trivial basis. <br /> <br /> This is consistent with 0 manifolds being the boundary of 1 manifolds. Indeed, <br /> <br /> &lt;math&gt;\int_{\pm p_0}\omega_0 = \sum\pm f(p_i)&lt;/math&gt;<br /> <br /> <br /> Now consider a path &lt;math&gt;\gamma:[0,1]\rightarrow\mathbb{R}^3&lt;/math&gt;<br /> <br /> &lt;math&gt;\int_{\gamma}\omega_1 = \int_{[0,1]}\gamma^*\omega_1 = \int_{[0,1]}\sum f_i d\gamma^*(x_i) = \int_{[0,1]}\sum f_i d\gamma_i&lt;/math&gt;<br /> <br /> &lt;math&gt;= \int_{[0,1]}\sum f_i\dot{\gamma}_i dt = \int_{\gamma}\vec F\cdot \vec T_{\gamma}&lt;/math&gt;<br /> <br /> <br /> Now lets compute &lt;math&gt;\omega_2(v,w)&lt;/math&gt;<br /> <br /> First, &lt;math&gt;dx_2\wedge dx_3 (v,w) = v_2 w_3 - v_3 w_2&lt;/math&gt;<br /> <br /> Likewise for each component of &lt;math&gt;\omega_2&lt;/math&gt; we thus get <br /> <br /> &lt;math&gt;\omega_2(v,w) = \vec G(p)\cdot (v\times w)&lt;/math&gt; where &lt;math&gt;\vec G(p)&lt;/math&gt; is the vector of coefficients of &lt;math&gt;\omega_2&lt;/math&gt;<br /> <br /> Now we know that &lt;math&gt;v\times w&lt;/math&gt; is a vector perpendicular to v and w with magnitude equal to the area of the defined parallelogram. So, <br /> <br /> &lt;math&gt;\int_{\Sigma}\omega_2 = \int_{\Sigma} \vec G\cdot \vec n d\sigma&lt;/math&gt;<br /> <br /> where &lt;math&gt;\vec n&lt;/math&gt; denotes the normal vector and &lt;math&gt;d\sigma&lt;/math&gt; is the area form and &lt;math&gt;\Sigma&lt;/math&gt; is a surface<br /> <br /> <br /> Now for &lt;math&gt;\omega_3&lt;/math&gt;, <br /> <br /> <br /> &lt;math&gt;\int_D \omega_3 = \int_D g&lt;/math&gt;<br /> <br /> <br /> now, &lt;math&gt;f(\gamma(1)) - f(\gamma(0)) = \int_{\gamma} (grad\ f)\cdot\vec T&lt;/math&gt;<br /> <br /> and &lt;math&gt;\int_D div\ G = \int_{\partial D} G\cdot\vec n d\sigma&lt;/math&gt; <br /> <br /> <br /> This is Gauss' Divergence Theorem. <br /> <br /> We can think about this as saying that the flow from each point in a domain, when summed up, will be just the flow out of the boundary of the domain. <br /> <br /> <br /> We also get Stokes' Theorem: <br /> <br /> <br /> &lt;math&gt;\int_{\partial\Sigma} F\cdot\vec T = \int_{\Sigma} curl\ F\cdot\vec n d\sigma&lt;/math&gt;<br /> <br /> <br /> ''End of Example''<br /> <br /> <br /> We recall that since &lt;math&gt;d^2 = 0&lt;/math&gt;, if &lt;math&gt;\omega = d\lambda&lt;/math&gt; then &lt;math&gt;d\omega = 0&lt;/math&gt;. But is the converse true? The following Lemma says 'yes', if the domain is &lt;math&gt;\mathbb{R}^n&lt;/math&gt;<br /> <br /> <br /> '''Poincare's Lemma'''<br /> <br /> On &lt;math&gt;\mathbb{R}^n, d\omega = 0&lt;/math&gt; iff &lt;math&gt;\exists\lambda&lt;/math&gt; such that &lt;math&gt;\omega = d\lambda&lt;/math&gt;<br /> <br /> <br /> This is NOT true for general M, as our homework assignment showed since we had a form &lt;math&gt;d\theta&lt;/math&gt; that had &lt;math&gt;d(d\theta) = 0&lt;/math&gt; but was not d of a form. <br /> <br /> <br /> Likewise, on &lt;math&gt;\mathbb{R}^n-\{0\}&lt;/math&gt; we have<br /> &lt;math&gt;<br /> \omega = \frac{1}{||x||^{\alpha}}\sum_{i=1}^{n}x_i dx_1\wedge\cdots\wedge\hat{dx_i}\wedge\cdots\wedge dx_n \in\Omega^{n-1}(\mathbb{R}^n-\{0\})&lt;/math&gt;<br /> <br /> <br /> Claim: <br /> <br /> For appropriate &lt;math&gt;\alpha,\ d\omega = 0&lt;/math&gt; but &lt;math&gt;\exists&lt;/math&gt; no &lt;math&gt;\lambda&lt;/math&gt; such that &lt;math&gt;d\lambda = \omega&lt;/math&gt;<br /> <br /> This is in our next homework assignment. <br /> <br /> <br /> Now, if there was such a &lt;math&gt;\lambda&lt;/math&gt;, &lt;math&gt;\int_{\Sigma}\omega = \int_{\sigma}d\lambda = \int_{\partial\Sigma}\lambda = 0&lt;/math&gt;<br /> <br /> If &lt;math&gt;\partial\Sigma = \empty&lt;/math&gt; (such as any sphere)<br /> <br /> But, &lt;math&gt;\int_{S^2}\omega = 4\pi&lt;/math&gt;<br /> <br /> <br /> <br /> '''Definition'''<br /> <br /> &lt;math&gt;Z^k(M) := ker d|_{\Omega^k(M)}&lt;/math&gt;<br /> <br /> &lt;math&gt;<br /> B^k(M) := im d|_{\Omega^{k-1}(M)}&lt;/math&gt;<br /> <br /> <br /> Clearly &lt;math&gt;B^k\subset Z^k&lt;/math&gt; so the following definition makes sense:<br /> <br /> <br /> '''Definition''' (de-Rham Cohomology)<br /> <br /> &lt;math&gt;H^k(M):= Z^k(M)/B^k(M)&lt;/math&gt;<br /> <br /> <br /> '''Claim'''<br /> <br /> &lt;math&gt;H^k(\mathbb{R}^n) = 0&lt;/math&gt; yet &lt;math&gt;H^1(S^1)\neq 0&lt;/math&gt;. <br /> <br /> Also, &lt;math&gt;H^{n-1}(\mathbb{R}^n - \{x\})\neq 0&lt;/math&gt;<br /> <br /> ===More fun with the planimeter===<br /> <br /> You can build your own low-tech planimeter out of lego!<br /> <br /> [[image:0708-1300planimeter1.jpg|center|540px]]<br /> <br /> Here my left hand is holding the fixed point steady while the right hand traces the shape. The net number of times the wheel turns while the shape is traced out indicates the area. <br /> <br /> '''The empirical test:'''<br /> the two shapes below have the same area, and while tracing them the wheel turned almost exactly the same number of times (about 2.3 rotations). Moreover, tracing smaller shapes caused it to turn fewer times, larger shapes more.<br /> <br /> [[image:0708-1300planimeter2.jpg|center|540px]]<br /> <br /> '''The problem:''' I couldn't do better than this. Calibrating it (calculating how many rotations corresponds to exactly what area) was a nightmare. Mathematically it was possible, but my planimeter is not accurate enough to agree with my math. If you wish to build your own more accurate one, you might want to try using a thinner wheel or one that grips the table better so you don't accidently lose any turning motion.</div> Trefor http://drorbn.net/index.php?title=0708-1300/Class_notes_for_Tuesday,_November_27 0708-1300/Class notes for Tuesday, November 27 2008-02-09T21:34:47Z <p>Trefor: /* First Hour */</p> <hr /> <div>{{0708-1300/Navigation}}<br /> ==Today's Agenda==<br /> * The planimeter with a picture from http://whistleralley.com/planimeter/planimeter.htm but our very own plane geometry and Stokes' theorem.<br /> * Completion of the proof of Stokes' theorem.<br /> * Completion of the discussion of the two- and three-dimensional cases of Stokes' theorem.<br /> * With luck, a discussion of de-Rham cohomology, homotopy invariance and Poincaré's lemma.<br /> <br /> <br /> ==Class Notes==<br /> <br /> &lt;span style=&quot;color: red;&quot;&gt;The notes below are by the students and for the students. Hopefully they are useful, but they come with no guarantee of any kind.&lt;/span&gt;<br /> <br /> ===First Hour===<br /> <br /> '''Planimeter'''<br /> <br /> A planimeter consists of two rods connected with a join where the end of one rod is fixed (but free to rotate) and the opposing end of the second rod traces out the boundary of some surface on the plane. I.e., the planimeter is kind of like a 1 legged roach. At the join of the two rods is a wheel which rotates (and measures the rotation) when the rod tracing the boundary moves in the normal direction and simply slides back an forth when moved in a tangential direction. <br /> <br /> Now we recall from plane geometry that we can locate points in the polar form &lt;math&gt;(r,\theta)&lt;/math&gt; and have the equations &lt;math&gt;x = rcos\theta&lt;/math&gt; and &lt;math&gt;y = rsin\theta&lt;/math&gt;<br /> <br /> Hence, <br /> <br /> &lt;math&gt;dx = cos\theta dr - rsin\theta d\theta&lt;/math&gt;<br /> <br /> &lt;math&gt;dy = sin\theta dr + rcos\theta d\theta&lt;/math&gt;<br /> <br /> Hence &lt;math&gt;dx\wedge dy = r(cos^2\theta + sin^2\theta)dr\wedge d\theta = rdr\wedge d\theta&lt;/math&gt;<br /> <br /> <br /> <br /> <br /> Now, the planimeter is essentially a 1 form corresponding to the speed of the wheel. We consider a diagram where the angle from the horizontal at the fixed end of the planimeter to the measuring end is &lt;math&gt;\theta&lt;/math&gt; and the angle from the horizontal to the first rod (the one connected to the fixed point) is &lt;math&gt;\theta + \phi&lt;/math&gt;. Hence &lt;math&gt;r = 2cos\phi&lt;/math&gt; and &lt;math&gt;dr = -2sin\phi d\phi&lt;/math&gt;<br /> <br /> With a little plane geometry we can see that &lt;math&gt;\omega = cos2\phi d(\theta + \phi)&lt;/math&gt;<br /> <br /> Computing, <br /> <br /> &lt;math&gt;d\omega = -2 sin2\phi d\phi\wedge(d\theta + d\phi) = -4 sin\phi cos\phi d\phi\wedge d\theta = 2cos\phi dr\wedge d\theta = rdr\wedge d\theta = dx\wedge dy&lt;/math&gt;<br /> <br /> <br /> Now applying stokes theorem, the the planimeter integrates &lt;math&gt;\omega&lt;/math&gt; over the boundary of our surface and hence this is just the integral of &lt;math&gt;d\omega&lt;/math&gt; over the surface. But this is just the integral of the area form. <br /> <br /> Hence the planimeter measure the area of a surface. <br /> <br /> <br /> '''Back to Stokes Theorem'''<br /> <br /> <br /> Firstly recall that &lt;math&gt;\partial M&lt;/math&gt; is oriented so that if you prepend the outward normal to its orientation you get the orientation of M<br /> <br /> Alternatively we recall that neighborhoods of points on the boundary look like the half space. Hence we can choose to restrict our attention to atlas's where all charts look like &lt;math&gt;H= \{x\in\mathbb{R}^n:x_1\leq 0\}&lt;/math&gt;<br /> <br /> We can see that these orientations are the same, i.e., just prepend the outward normal to the half space. <br /> <br /> <br /> ''Proof of Stokes''<br /> <br /> <br /> We have now defined all the terms. WLOG &lt;math&gt;\omega&lt;/math&gt; is supported in one chart (by linearity)<br /> <br /> For a compactly supported n-1 form on H need to show that &lt;math&gt;\int_{\partial H}\omega = \int_H d\omega&lt;/math&gt;<br /> <br /> <br /> We let &lt;math&gt;\omega = \sum f_i dx_1\wedge\cdots\wedge \hat{dx_i}\wedge\cdots\wedge dx_n&lt;/math&gt; (where the hat means it is omitted)<br /> <br /> &lt;math&gt;d\omega = \sum (-1)^{i-1} \frac{\partial f_i}{\partial x_i} dx_1\wedge\cdots\wedge dx_n&lt;/math&gt;<br /> <br /> <br /> So, &lt;math&gt;\int_H d\omega = \sum \int_{[x_1\leq 0]}(-1)^{i-1}\frac{\partial f_i}{\partial x_i}dx_1\wedge\cdots\wedge dx_n = \sum (-1)^{i-1}\int_{[x_1\leq 0]}\frac{\partial f_i}{\partial x_i}&lt;/math&gt; <br /> <br /> via fundamental theorem of calculus and that the &lt;math&gt;f_i&lt;/math&gt;'s are compactly supported we get <br /> <br /> &lt;math&gt;= \int_{[x_1\leq 0]} \frac{\partial f_1}{\partial x_1} = \int_{[x_1=0]} f_1&lt;/math&gt;<br /> <br /> <br /> Hence with the standard inclusion of &lt;math&gt;\partial H = \mathbb{R}^{n-1}_{x_2\cdots x_n}&lt;/math&gt; we get <br /> <br /> <br /> &lt;math&gt;\int_{\partial H}\omega = \int_{\mathbb{R}^{n-1}_{x_2\cdots x_n}}\iota^*(\sum f_i dx_1\wedge\cdots\wedge \hat{dx_i}\wedge\cdots\wedge dx_n) = \int_{[x_1=0]}f_1&lt;/math&gt;<br /> <br /> <br /> Thus these are the same and the theorem is proved ''Q.E.D.''<br /> <br /> <br /> '''Real Plane'''<br /> <br /> Consider &lt;math&gt;\Omega^0(\mathbb{R}^2)\rightarrow^d\Omega^1(\mathbb{R}^2)\rightarrow^d\Omega^2(\mathbb{R}^2)&lt;/math&gt;<br /> <br /> <br /> Forms in &lt;math&gt;\Omega^1(\mathbb{R}^2)&lt;/math&gt; look like &lt;math&gt;Fdx +Gdy&lt;/math&gt; and map under d to &lt;math&gt;(G_x - F_y)dx\wedge dy&lt;/math&gt;<br /> <br /> Hence applying Stokes' Theorem:<br /> <br /> &lt;math&gt;\int_{\partial D}Fdx + Gdy = \int_D (G_x-F_y)dxdy&lt;/math&gt;<br /> <br /> This is known as Greens Theorem<br /> <br /> <br /> In complex analysis we also have a similar result Cauchy's Theorem where the integral of an analytic function around a closed path is zero. This is because analytic functions obey the Cauchy-Riemann equations and hence &lt;math&gt;G_x-F_y&lt;/math&gt; is identically zero.<br /> <br /> ===Second Hour===<br /> <br /> <br /> '''Example 2'''<br /> <br /> &lt;math&gt;\Omega^k(\mathbb{R}^3)&lt;/math&gt;<br /> <br /> <br /> Recall previous we had consider the spaces &lt;math&gt;\Omega^k(\mathbb{R}^3)&lt;/math&gt; and showed that &lt;math&gt;\Omega^0&lt;/math&gt; and &lt;math&gt;\Omega^ 3&lt;/math&gt; corresponded with functions and that &lt;math&gt;\Omega^2&lt;/math&gt; and &lt;math&gt;\Omega^1&lt;/math&gt; corresponded with triples of functions (i.e. vector fields). We also showed that the d function^s between these spaces are the gradient, curl and divergence functions from vector calculus. <br /> <br /> We are now interested in integrating, using Stokes Theorem, forms in these spaces. <br /> <br /> <br /> First, note that to a 0 manifold, assigning an orientation to the manifold is just assigning a plus or minus sign to the manifold as a result of it having a trivial basis. <br /> <br /> This is consistent with 0 manifolds being the boundary of 1 manifolds. Indeed, <br /> <br /> &lt;math&gt;\int_{\pm p_0}\omega_0 = \sum\pm f(p_i)&lt;/math&gt;<br /> <br /> <br /> Now consider a path &lt;math&gt;\gamma:[0,1]\rightarrow\mathbb{R}^3&lt;/math&gt;<br /> <br /> &lt;math&gt;\int_{\gamma}\omega_1 = \int_{[0,1]}\gamma^*\omega_1 = \int_{[0,1]}\sum f_i d\gamma^*(x_i) = \int_{[0,1]}\sum f_i d\gamma_i&lt;/math&gt;<br /> <br /> &lt;math&gt;= \int_{[0,1]}\sum f_i\dot{\gamma}_i dt = \int_{\gamma}\vec F\cdot \vec T_{\gamma}&lt;/math&gt;<br /> <br /> <br /> Now lets compute &lt;math&gt;\omega_2(v,w)&lt;/math&gt;<br /> <br /> First, &lt;math&gt;dx_2\wedge dx_3 (v,w) = v_2 w_3 - v_3 w_2&lt;/math&gt;<br /> <br /> Likewise for each component of &lt;math&gt;\omega_2&lt;/math&gt; we thus get <br /> <br /> &lt;math&gt;\omega_2(v,w) = \vec G(p)\cdot (v\times w)&lt;/math&gt; where &lt;math&gt;\vec G(p)&lt;/math&gt; is the vector of coefficients of &lt;math&gt;\omega_2&lt;/math&gt;<br /> <br /> Now we know that &lt;math&gt;v\times w&lt;/math&gt; is a vector perpendicular to v and w with magnitude equal to the area of the defined parallelogram. So, <br /> <br /> &lt;math&gt;\int_{\Sigma}\omega_2 = \int_{\Sigma} \vec G\cdot \vec n d\sigma&lt;/math&gt;<br /> <br /> where &lt;math&gt;\vec n&lt;/math&gt; denotes the normal vector and &lt;math&gt;d\sigma&lt;/math&gt; is the area form and &lt;math&gt;\Sigma&lt;/math&gt; is a surface<br /> <br /> <br /> Now for &lt;math&gt;\omega_3&lt;/math&gt;, <br /> <br /> <br /> &lt;math&gt;\int_D \omega_3 = \int_D g&lt;/math&gt;<br /> <br /> <br /> now, &lt;math&gt;f(\gamma(1)) - f(\gamma(0)) = \int_{\gamma} (grad\ f)\cdot\vec T&lt;/math&gt;<br /> <br /> and &lt;math&gt;\int_D div\ G = \int_{\partial D} G\cdot\vec n d\sigma&lt;/math&gt; <br /> <br /> <br /> This is Gauss' Divergence Theorem. <br /> <br /> We can think about this as saying that the flow from each point in a domain, when summed up, will be just the flow out of the boundary of the domain. <br /> <br /> <br /> We also get Stokes' Theorem: <br /> <br /> <br /> &lt;math&gt;\int_{\partial\Sigma} F\cdot\vec T = \int_{\Sigma} curl\ F\cdot\vec n d\sigma&lt;/math&gt;<br /> <br /> <br /> ''End of Example''<br /> <br /> <br /> We recall that since &lt;math&gt;d^2 = 0&lt;/math&gt;, if &lt;math&gt;\omega = d\lambda&lt;/math&gt; then &lt;math&gt;d\omega = 0&lt;/math&gt;. But is the converse true? The following Lemma says 'yes', if the domain is &lt;math&gt;\mathbb{R}^n&lt;/math&gt;<br /> <br /> <br /> '''Poincare's Lemma'''<br /> <br /> On &lt;math&gt;\mathbb{R}^n, d\omega = 0&lt;/math&gt; iff &lt;math&gt;\exists\lambda&lt;/math&gt; such that &lt;math&gt;\omega = d\lambda&lt;/math&gt;<br /> <br /> <br /> This is NOT true for general M, as our homework assignment showed since we had a form &lt;math&gt;d\theta&lt;/math&gt; that had &lt;math&gt;d(d\theta) = 0&lt;/math&gt; but was not d of a form. <br /> <br /> <br /> Likewise, on &lt;math&gt;\mathbb{R}^n-\{0\}&lt;/math&gt; we have<br /> &lt;math&gt;<br /> \omega = \frac{1}{||x||^{\alpha}}\sum_{i=1}^{n}x_i dx_1\wedge\cdots\wedge\hat{dx_i}\wedge\cdots\wedge dx_n \in\Omega^{n-1}(\mathbb{R}^n-\{0\})&lt;/math&gt;<br /> <br /> <br /> Claim: <br /> <br /> For appropriate &lt;math&gt;\alpha,\ d\omega = 0&lt;/math&gt; but &lt;math&gt;\exists&lt;/math&gt; no &lt;math&gt;\lambda&lt;/math&gt; such that &lt;math&gt;d\lambda = \omega&lt;/math&gt;<br /> <br /> This is in our next homework assignment. <br /> <br /> <br /> Now, if there was such a &lt;math&gt;\lambda&lt;/math&gt;, &lt;math&gt;\int_{\Sigma}\omega = \int_{\sigma}d\lambda = \int_{\partial\Sigma}\lambda = 0&lt;/math&gt;<br /> <br /> If &lt;math&gt;\partial\Sigma = \empty&lt;/math&gt; (such as any sphere)<br /> <br /> But, &lt;math&gt;\int_{S^2}\omega = 4\pi&lt;/math&gt;<br /> <br /> <br /> <br /> '''Definition'''<br /> <br /> &lt;math&gt;Z^k(M) := ker d|_{\Omega^k(M)}&lt;/math&gt;<br /> <br /> &lt;math&gt;<br /> B^k(M) := im d|_{\Omega^{k-1}(M)}&lt;/math&gt;<br /> <br /> <br /> Clearly &lt;math&gt;B^k\subset Z^k&lt;/math&gt; so the following definition makes sense:<br /> <br /> <br /> '''Definition''' (de-Rham Cohomology)<br /> <br /> &lt;math&gt;H^k(M):= Z^k(M)/B^k(M)&lt;/math&gt;<br /> <br /> <br /> '''Claim'''<br /> <br /> &lt;math&gt;H^k(\mathbb{R}^n) = 0&lt;/math&gt; yet &lt;math&gt;H^1(S^1)\neq 0&lt;/math&gt;. <br /> <br /> Also, &lt;math&gt;H^{n-1}(\mathbb{R}^n - \{x\})\neq 0&lt;/math&gt;<br /> <br /> ===More fun with the planimeter===<br /> <br /> You can build your own low-tech planimeter out of lego!<br /> <br /> [[image:0708-1300planimeter1.jpg|center|540px]]<br /> <br /> Here my left hand is holding the fixed point steady while the right hand traces the shape. The net number of times the wheel turns while the shape is traced out indicates the area. <br /> <br /> '''The empirical test:'''<br /> the two shapes below have the same area, and while tracing them the wheel turned almost exactly the same number of times (about 2.3 rotations). Moreover, tracing smaller shapes caused it to turn fewer times, larger shapes more.<br /> <br /> [[image:0708-1300planimeter2.jpg|center|540px]]<br /> <br /> '''The problem:''' I couldn't do better than this. Calibrating it (calculating how many rotations corresponds to exactly what area) was a nightmare. Mathematically it was possible, but my planimeter is not accurate enough to agree with my math. If you wish to build your own more accurate one, you might want to try using a thinner wheel or one that grips the table better so you don't accidently lose any turning motion.</div> Trefor http://drorbn.net/index.php?title=0708-1300/Class_notes_for_Tuesday,_November_20 0708-1300/Class notes for Tuesday, November 20 2008-02-09T20:11:37Z <p>Trefor: /* Second Hour */</p> <hr /> <div>{{0708-1300/Navigation}}<br /> <br /> ==Typed Notes==<br /> <br /> &lt;span style=&quot;color: red;&quot;&gt;The notes below are by the students and for the students. Hopefully they are useful, but they come with no guarantee of any kind.&lt;/span&gt;<br /> <br /> <br /> ===First Hour===<br /> <br /> Recall we are ultimately attempting to understand and prove Stokes theorem. Currently we are investigating the meaning of &lt;math&gt;d\omega&lt;/math&gt;. <br /> <br /> Recall we had that &lt;math&gt;d(\sum f_I dx^I) = \sum (df_I)\wedge dx^I = \sum_{I,j}\frac{\partial f_I}{\partial x^j} dx^j\wedge dx^I = \sum dx^j\frac{\partial\omega}{\partial x_j}&lt;/math&gt;<br /> <br /> <br /> Now we want to compute &lt;math&gt;d\omega&lt;/math&gt; on the parallelepiped formed from k+1 tangent vectors. For instance let us suppose k= 2 then are interested in the parallelepiped formed from the three basis vectors &lt;math&gt;v_1, v_2, v_3&lt;/math&gt;. <br /> <br /> Feeding in the parallelepiped we get &lt;math&gt;\sum_I df_I(v_1)dx^I(v_2,v_3) - \sum_I df_I(v_2)dx^I(v_1,v_3) + \sum_I df_I(v_3)dx^I(v_1,v_2) &lt;/math&gt;&lt;math&gt;= \sum_I (v_1 f_I)dx^I(v_2,v_3) - \sum_I (v_2 f_I)dx^I(v_1,v_3) + \sum_I (v_3 f_I)dx^I(v_1,v_2)&lt;/math&gt; <br /> <br /> &lt;math&gt;= (v_1 w)(v_2,v_3) - (v_2 w)(v_1,v_3) + (v_3 w)(v_1,v_2)&lt;/math&gt;<br /> <br /> Now, &lt;math&gt;v_1 f = lim\frac{f(p+\epsilon v_1) - f(p)}{\epsilon}&lt;/math&gt;, or, loosely &lt;math&gt;f(p+v_1) - f(p)&lt;/math&gt;<br /> <br /> So this corresponds to the difference between &lt;math&gt;\omega&lt;/math&gt; calculated on each of the two faces parallel to the &lt;math&gt;v_3,v_2&lt;/math&gt; plane. <br /> <br /> Hence, our &lt;math&gt;d\omega&lt;/math&gt; on the parallelepiped is just the sum of parallelograms making the boundary of the parallelepiped counted with some signs. <br /> <br /> <br /> We can see the loose idea of how the proof of stokes theorem is going to work: dividing the manifold up into little parallelepiped like this, &lt;math&gt;d\omega&lt;/math&gt; will just be the faces of the parallelepipeds and when summing over the whole manifold all of faces will cancel except those on the boundary thus just leaving the integral of &lt;math&gt;\omega&lt;/math&gt; along the boundary. <br /> <br /> <br /> We note that this is similar to the proof of the fundamental theorem of calculus, where we take an integral and compute the value of f' at many little subintervals. But the value of f' is just the difference of f' at the boundary of each sub interval so when we add everything up everything cancels except the values of the function at the endpoint of the big interval. <br /> <br /> <br /> '''Claim'''<br /> <br /> d exists if &lt;math&gt;M = \mathbb{R}^n&lt;/math&gt; and is unique.<br /> <br /> Define &lt;math&gt;d(\omega) = d(\sum f_I dx^I) :=\sum_{j,I}\frac{\partial f_I}{\partial x_j}dx^j\wedge dx^I&lt;/math&gt;<br /> <br /> ''Proof''<br /> <br /> We need to check that this satisfies the properties 1) - 3) from last lecture: <br /> <br /> 1) &lt;math&gt;df(\partial_i) = \sum_j \frac{\partial f}{\partial x_j} dx_j(\partial_i) = \sum_j \frac{\partial f}{\partial x_j} \delta_{ji} = \partial_i f&lt;/math&gt; and so satisfies property 1)<br /> <br /> <br /> ''Note''<br /> <br /> We now adopt ''Einstein Summation Convention'' which means that if in a term there is an index that is repeated, once as a subscript and once as a superscript, it is meant as implicit that we are summing over this index. This just cleans up the notation so we don't have to have sums everywhere.<br /> <br /> <br /> 2) &lt;math&gt;d(df_I dx^I) = d(\frac{\partial f_I}{\partial x^j} dx^j\wedge dx^I) = \frac{\partial^2 f_I}{\partial x^j \partial x^{j'}}dx^{j'}\wedge dx^j\wedge dx^I = 0&lt;/math&gt; because the mixed partial is symmetric under exchange of indices but the wedge product is antisymmetric under exchange of indices. That is, each term cancels with the one where j and j' are exchanged. <br /> <br /> <br /> 3) let &lt;math&gt;\omega=f_I dx^I&lt;/math&gt; and &lt;math&gt;\lambda = g_J dx^J&lt;/math&gt; then &lt;math&gt;\omega\wedge\lambda = f_I g_J dx^I\wedge dx^J&lt;/math&gt;<br /> <br /> so &lt;math&gt;d(f_I g_J dx^I\wedge dx^J) =\sum_j \frac{\partial f_I g_I}{\partial x^j} dx^j\wedge dx^I\wedge dx^J = \sum_j (\frac{\partial f_I}{\partial x^j}g_J + f_I\frac{\partial g_I}{\partial x^j})dx^j\wedge dx^I\wedge dx^J = d\omega\wedge\lambda \pm\omega\wedge d\lambda&lt;/math&gt;<br /> <br /> <br /> Via assignment 3 this is unique. <br /> <br /> ''Q.E.D.''<br /> <br /> <br /> Now, we can extend this definition on manifolds by using coordinate charts. <br /> <br /> <br /> '''Claim'''<br /> <br /> Properties 1-3 imply that on any M, d is ''local''. That is, if &lt;math&gt;\omega|_U = \lambda|_U&lt;/math&gt; then &lt;math&gt;d\omega|_U = d\lambda|_U&lt;/math&gt;<br /> <br /> ''Proof:'' Exercise. <br /> <br /> <br /> '''Definition'''<br /> <br /> For &lt;math&gt;\omega\in\Omega^k(M)&lt;/math&gt; the &lt;math&gt;supp\ \omega = \overline{\{p\in M\ :\ w|_p\neq0\}}&lt;/math&gt;<br /> <br /> Then ''&lt;math&gt;\omega&lt;/math&gt; has compact support'' if the supp &lt;math&gt;\omega&lt;/math&gt; is compact. <br /> <br /> Define &lt;math&gt;\Omega^*_c(M)&lt;/math&gt; := the compactly supported &lt;math&gt;w\in\Omega^*(M)&lt;/math&gt;<br /> <br /> <br /> '''Definition'''<br /> <br /> For &lt;math&gt;\omega\in\Omega^n_c(\mathbb{R}^n)&lt;/math&gt;, &lt;math&gt;\omega = fdx^1\wedge\cdots\wedge dx^n&lt;/math&gt;<br /> we define &lt;math&gt;\int_{\mathbb{R}^n}: \Omega^n_c(\mathbb{R}^n)\rightarrow\mathbb{R}&lt;/math&gt; by <br /> <br /> &lt;math&gt;\int_{\mathbb{R}^n}\omega := \int_{\mathbb{R}^n}f&lt;/math&gt;<br /> <br /> <br /> I.e., &lt;math&gt;\int_{\mathbb{R}^n}fdx^1\wedge\cdots\wedge dx^n = \int_{\mathbb{R}^n}fdx^1\ldots dx^n&lt;/math&gt;<br /> <br /> === Second Hour===<br /> <br /> In general if we have a diffeomorphism &lt;math&gt;\phi:\mathbb{R}^n \rightarrow\mathbb{R}^n&lt;/math&gt; then the normal integral <br /> <br /> &lt;math&gt;\int \phi^*f = \int f\circ\phi&lt;/math&gt; is not equal to &lt;math&gt;\int f&lt;/math&gt;<br /> <br /> However we claim that this IS true for differential forms. I.e., <br /> <br /> <br /> '''Claim'''<br /> <br /> &lt;math&gt;\int \phi^*\omega = \pm\int \omega&lt;/math&gt; as forms<br /> <br /> This is very important because it essentially means we can integrate in whatever charts we like and get the same thing. <br /> <br /> <br /> ''Proof''<br /> <br /> &lt;math&gt;\phi^*\omega = \phi^*(fdx^1\wedge\ldots\wedge dx^n) = \phi^* f\phi^* dx^1\wedge\ldots\wedge \phi^* dx^n&lt;/math&gt;<br /> <br /> Now &lt;math&gt;\phi^* (dg) = d\phi^* g&lt;/math&gt; by chain rule and so we extend to &lt;math&gt;\phi^*(d\omega) = d(\phi^*\omega)&lt;/math&gt;<br /> <br /> Hence, <br /> <br /> &lt;math&gt;\phi^*\omega = (f\circ\phi) d(x^1\circ\phi)\wedge\ldots\wedge d(x^n\circ\phi) = (f\circ\phi) d\phi^1\wedge\ldots\wedge d\phi^n&lt;/math&gt;<br /> <br /> &lt;math&gt;= (f\circ\phi)\left(\sum_{i_1} \frac{\partial\phi^1}{\partial y^{i_1}}dy^{i_1}\right)\wedge\ldots\wedge\left(\sum_{i_n}\frac{\partial\phi^n}{\partial y^{i_n}}dy^{i_n}\right) = (f\circ\phi)\sum_{i_1,\ldots,i_n =1}^n \frac{\partial\phi^1}{\partial y^{i_1}}\ldots\frac{\partial\phi^n}{\partial y^{i_n}}dy^{i_1}\wedge\ldots\wedge dy^{i_n}&lt;/math&gt;<br /> <br /> but the wedge product is zero unless &lt;math&gt;(i_1,\ldots,i_n)=\sigma\in S^n&lt;/math&gt; and in that case yields &lt;math&gt;(-1)^{\sigma} dy^1\wedge\ldots\wedge dy^n&lt;/math&gt;<br /> <br /> <br /> Hence we get, <br /> <br /> &lt;math&gt;=(f\circ\phi)\sum_{\sigma\in S^n}(-1)^{\sigma}\Pi_{\alpha} \frac{\partial\phi^{\alpha}}{\partial y^{\sigma(\alpha)}}dy^1\wedge\ldots\wedge dy^n = (f\circ\phi) det (d\phi)dy^1\wedge\ldots\wedge dy^n = (f\circ\phi)J_{\phi}dy^1\wedge\ldots\wedge dy^n&lt;/math&gt;<br /> <br /> where &lt;math&gt;J_{\phi}&lt;/math&gt; is the determinant of the Jacobian matrix. <br /> <br /> <br /> Hence, &lt;math&gt;\int_{\mathbb{R}^n} \phi^*\omega = \int_{\mathbb{R}^n}(f\circ\phi)J_{\phi}dy^1\wedge\ldots\wedge dy^n = \int_{Old\ sense} (f\circ\phi)J_{\phi} =\pm \int (f\circ\phi)|J_{\phi}| = \pm\int f = \pm\int \omega&lt;/math&gt;<br /> <br /> ''Q.E.D''<br /> <br /> <br /> If we restrict our attention to just the ''orientation preserving'' &lt;math&gt;\phi&lt;/math&gt;'s so that &lt;math&gt;J_{\phi}&gt;0&lt;/math&gt; then we will always get the + in the end. <br /> <br /> <br /> '''Definition'''<br /> <br /> An ''orientation'' of M is an assignment of the charts to &lt;math&gt;\pm 1&lt;/math&gt;, &lt;math&gt;\phi\mapsto S_{\phi}&lt;/math&gt;<br /> <br /> Then if the domains on &lt;math&gt;\phi&lt;/math&gt; and &lt;math&gt;\psi&lt;/math&gt; overlap then &lt;math&gt;S_{\psi} = (sign J_{\psi^{-1}\phi})S_{\phi}&lt;/math&gt;<br /> <br /> By definition, M is orientable if we can find an orientation. <br /> <br /> <br /> '''Examples'''<br /> <br /> 1) &lt;math&gt;\mathbb{R}^n&lt;/math&gt;. We declare the identity positive and then all other designations follow from this. <br /> <br /> <br /> 2) The finite cylinder &lt;math&gt;S^1\times I&lt;/math&gt;<br /> <br /> We can put two charts on the cylinder by considering two rectangles which overlap by a little bit that cover the whole cylinder. If we denote one of these as positive, the overlap makes the other positive. We compare any other chart to these two. <br /> <br /> <br /> 3) Consider the mobius strip and the same attempted charts as for the cylinder. If we label one section positive, the other section must be positive due to the overlap on one side, but must be negative due to the overlap on the other side. Thus the mobius strip is not orientable. <br /> <br /> <br /> <br /> '''Definition'''<br /> <br /> An ''orientation of a vector space V'' is an equivalence class of ordered bases &lt;math&gt;v_{\alpha}&lt;/math&gt; of V, where &lt;math&gt;v_{\alpha}\sim w_{\beta}&lt;/math&gt; if the determinant of the transition matrix between these two bases is positive. <br /> <br /> <br /> '''Definition'''<br /> <br /> An ''orientation of M'' is a continuous choice of orientations for &lt;math&gt;T_p M&lt;/math&gt; for any p. We haven't technically defined what it means to be continuous in this sense but the meaning is clear. <br /> <br /> <br /> '''Definition'''<br /> <br /> let &lt;math&gt;M^n&lt;/math&gt; be an oriented manifold and let &lt;math&gt;\omega\in\Omega^n_c(M)&lt;/math&gt;. Let &lt;math&gt;\phi_{\alpha}:U_{\alpha}\mapsto\mathbb{R}^n&lt;/math&gt; be a collection of positive charts that cover M. Let &lt;math&gt;\lambda_{\alpha}&lt;/math&gt; be a partition of unity subordinate to this cover then <br /> <br /> &lt;math&gt;\int_M\omega = \int_M 1\omega = \int_M \sum_{\alpha}\lambda_{\alpha}\omega = \sum_{\alpha}\int_{U_{\alpha}}\lambda_{\alpha}\omega = \sum_{\alpha}\int_{\mathbb{R}^n} (\phi^{-1}_{\alpha})^*(\lambda_{\alpha}\omega)&lt;/math&gt;<br /> <br /> Note all the intermediate steps were merely properties we would LIKE the integral to have, the actual definition is the equality of the left most and right most expressions. <br /> <br /> <br /> '''Theorem'''<br /> <br /> &lt;math&gt;\int_M\omega&lt;/math&gt; is independent of the choices.</div> Trefor http://drorbn.net/index.php?title=0708-1300/Class_notes_for_Tuesday,_November_20 0708-1300/Class notes for Tuesday, November 20 2008-02-09T19:47:30Z <p>Trefor: /* First Hour */</p> <hr /> <div>{{0708-1300/Navigation}}<br /> <br /> ==Typed Notes==<br /> <br /> &lt;span style=&quot;color: red;&quot;&gt;The notes below are by the students and for the students. Hopefully they are useful, but they come with no guarantee of any kind.&lt;/span&gt;<br /> <br /> <br /> ===First Hour===<br /> <br /> Recall we are ultimately attempting to understand and prove Stokes theorem. Currently we are investigating the meaning of &lt;math&gt;d\omega&lt;/math&gt;. <br /> <br /> Recall we had that &lt;math&gt;d(\sum f_I dx^I) = \sum (df_I)\wedge dx^I = \sum_{I,j}\frac{\partial f_I}{\partial x^j} dx^j\wedge dx^I = \sum dx^j\frac{\partial\omega}{\partial x_j}&lt;/math&gt;<br /> <br /> <br /> Now we want to compute &lt;math&gt;d\omega&lt;/math&gt; on the parallelepiped formed from k+1 tangent vectors. For instance let us suppose k= 2 then are interested in the parallelepiped formed from the three basis vectors &lt;math&gt;v_1, v_2, v_3&lt;/math&gt;. <br /> <br /> Feeding in the parallelepiped we get &lt;math&gt;\sum_I df_I(v_1)dx^I(v_2,v_3) - \sum_I df_I(v_2)dx^I(v_1,v_3) + \sum_I df_I(v_3)dx^I(v_1,v_2) &lt;/math&gt;&lt;math&gt;= \sum_I (v_1 f_I)dx^I(v_2,v_3) - \sum_I (v_2 f_I)dx^I(v_1,v_3) + \sum_I (v_3 f_I)dx^I(v_1,v_2)&lt;/math&gt; <br /> <br /> &lt;math&gt;= (v_1 w)(v_2,v_3) - (v_2 w)(v_1,v_3) + (v_3 w)(v_1,v_2)&lt;/math&gt;<br /> <br /> Now, &lt;math&gt;v_1 f = lim\frac{f(p+\epsilon v_1) - f(p)}{\epsilon}&lt;/math&gt;, or, loosely &lt;math&gt;f(p+v_1) - f(p)&lt;/math&gt;<br /> <br /> So this corresponds to the difference between &lt;math&gt;\omega&lt;/math&gt; calculated on each of the two faces parallel to the &lt;math&gt;v_3,v_2&lt;/math&gt; plane. <br /> <br /> Hence, our &lt;math&gt;d\omega&lt;/math&gt; on the parallelepiped is just the sum of parallelograms making the boundary of the parallelepiped counted with some signs. <br /> <br /> <br /> We can see the loose idea of how the proof of stokes theorem is going to work: dividing the manifold up into little parallelepiped like this, &lt;math&gt;d\omega&lt;/math&gt; will just be the faces of the parallelepipeds and when summing over the whole manifold all of faces will cancel except those on the boundary thus just leaving the integral of &lt;math&gt;\omega&lt;/math&gt; along the boundary. <br /> <br /> <br /> We note that this is similar to the proof of the fundamental theorem of calculus, where we take an integral and compute the value of f' at many little subintervals. But the value of f' is just the difference of f' at the boundary of each sub interval so when we add everything up everything cancels except the values of the function at the endpoint of the big interval. <br /> <br /> <br /> '''Claim'''<br /> <br /> d exists if &lt;math&gt;M = \mathbb{R}^n&lt;/math&gt; and is unique.<br /> <br /> Define &lt;math&gt;d(\omega) = d(\sum f_I dx^I) :=\sum_{j,I}\frac{\partial f_I}{\partial x_j}dx^j\wedge dx^I&lt;/math&gt;<br /> <br /> ''Proof''<br /> <br /> We need to check that this satisfies the properties 1) - 3) from last lecture: <br /> <br /> 1) &lt;math&gt;df(\partial_i) = \sum_j \frac{\partial f}{\partial x_j} dx_j(\partial_i) = \sum_j \frac{\partial f}{\partial x_j} \delta_{ji} = \partial_i f&lt;/math&gt; and so satisfies property 1)<br /> <br /> <br /> ''Note''<br /> <br /> We now adopt ''Einstein Summation Convention'' which means that if in a term there is an index that is repeated, once as a subscript and once as a superscript, it is meant as implicit that we are summing over this index. This just cleans up the notation so we don't have to have sums everywhere.<br /> <br /> <br /> 2) &lt;math&gt;d(df_I dx^I) = d(\frac{\partial f_I}{\partial x^j} dx^j\wedge dx^I) = \frac{\partial^2 f_I}{\partial x^j \partial x^{j'}}dx^{j'}\wedge dx^j\wedge dx^I = 0&lt;/math&gt; because the mixed partial is symmetric under exchange of indices but the wedge product is antisymmetric under exchange of indices. That is, each term cancels with the one where j and j' are exchanged. <br /> <br /> <br /> 3) let &lt;math&gt;\omega=f_I dx^I&lt;/math&gt; and &lt;math&gt;\lambda = g_J dx^J&lt;/math&gt; then &lt;math&gt;\omega\wedge\lambda = f_I g_J dx^I\wedge dx^J&lt;/math&gt;<br /> <br /> so &lt;math&gt;d(f_I g_J dx^I\wedge dx^J) =\sum_j \frac{\partial f_I g_I}{\partial x^j} dx^j\wedge dx^I\wedge dx^J = \sum_j (\frac{\partial f_I}{\partial x^j}g_J + f_I\frac{\partial g_I}{\partial x^j})dx^j\wedge dx^I\wedge dx^J = d\omega\wedge\lambda \pm\omega\wedge d\lambda&lt;/math&gt;<br /> <br /> <br /> Via assignment 3 this is unique. <br /> <br /> ''Q.E.D.''<br /> <br /> <br /> Now, we can extend this definition on manifolds by using coordinate charts. <br /> <br /> <br /> '''Claim'''<br /> <br /> Properties 1-3 imply that on any M, d is ''local''. That is, if &lt;math&gt;\omega|_U = \lambda|_U&lt;/math&gt; then &lt;math&gt;d\omega|_U = d\lambda|_U&lt;/math&gt;<br /> <br /> ''Proof:'' Exercise. <br /> <br /> <br /> '''Definition'''<br /> <br /> For &lt;math&gt;\omega\in\Omega^k(M)&lt;/math&gt; the &lt;math&gt;supp\ \omega = \overline{\{p\in M\ :\ w|_p\neq0\}}&lt;/math&gt;<br /> <br /> Then ''&lt;math&gt;\omega&lt;/math&gt; has compact support'' if the supp &lt;math&gt;\omega&lt;/math&gt; is compact. <br /> <br /> Define &lt;math&gt;\Omega^*_c(M)&lt;/math&gt; := the compactly supported &lt;math&gt;w\in\Omega^*(M)&lt;/math&gt;<br /> <br /> <br /> '''Definition'''<br /> <br /> For &lt;math&gt;\omega\in\Omega^n_c(\mathbb{R}^n)&lt;/math&gt;, &lt;math&gt;\omega = fdx^1\wedge\cdots\wedge dx^n&lt;/math&gt;<br /> we define &lt;math&gt;\int_{\mathbb{R}^n}: \Omega^n_c(\mathbb{R}^n)\rightarrow\mathbb{R}&lt;/math&gt; by <br /> <br /> &lt;math&gt;\int_{\mathbb{R}^n}\omega := \int_{\mathbb{R}^n}f&lt;/math&gt;<br /> <br /> <br /> I.e., &lt;math&gt;\int_{\mathbb{R}^n}fdx^1\wedge\cdots\wedge dx^n = \int_{\mathbb{R}^n}fdx^1\ldots dx^n&lt;/math&gt;<br /> <br /> === Second Hour===<br /> <br /> In general if we have a diffeomorphism &lt;math&gt;\phi:\mathbb{R}^n \rightarrow\mathbb{R}^n&lt;/math&gt; then the normal integral <br /> <br /> &lt;math&gt;\int \phi^*f = \int f\circ\phi&lt;/math&gt; is not equal to &lt;math&gt;\int f&lt;/math&gt;<br /> <br /> However we claim that this IS true for differential forms. I.e., <br /> <br /> <br /> '''Claim'''<br /> <br /> &lt;math&gt;\int \phi^*\omega = \pm\int \omega&lt;/math&gt; as forms<br /> <br /> This is very important because it essentially means we can integrate in whatever charts we like and get the same thing. <br /> <br /> <br /> ''Proof''<br /> <br /> &lt;math&gt;\phi^*\omega = \phi^*(fdx^1\wedge\ldots\wedge dx^n) = \phi^* f\phi^* dx^1\wedge\ldots\wedge \phi^* dx^n&lt;/math&gt;<br /> <br /> Now &lt;math&gt;\phi^* (dg) = d\phi^* g&lt;/math&gt; by chain rule and so we extend to &lt;math&gt;\phi^*(d\omega) = d(\phi^*\omega)&lt;/math&gt;<br /> <br /> Hence, <br /> <br /> &lt;math&gt;\phi^*\omega = (f\circ\phi) d(x^1\circ\phi)\wedge\ldots\wedge d(x^n\circ\phi) = (f\circ\phi) d\phi^1\wedge\ldots\wedge d\phi^n&lt;/math&gt;<br /> <br /> &lt;math&gt;= (f\circ\phi)\left(\sum_{i_1} \frac{\partial\phi^1}{\partial y^{i_1}}\right)\wedge\ldots\wedge\left(\sum_{i_n}\frac{\partial\phi^n}{\partial y^{i_n}}\right) = (f\circ\phi)\sum_{i_1,\ldots,i_n =1}^n \frac{\partial\phi^1}{\partial y^{i_1}}\ldots\frac{\partial\phi^n}{\partial y^{i_n}}dy^{i_1}\wedge\ldots\wedge dy^{i_n}&lt;/math&gt;<br /> <br /> but the wedge product is zero unless &lt;math&gt;(i_1,\ldots,i_n)=\sigma\in S^n&lt;/math&gt; and in that case yields &lt;math&gt;(-1)^{\sigma} dy^1\wedge\ldots\wedge dy^n&lt;/math&gt;<br /> <br /> <br /> Hence we get, <br /> <br /> &lt;math&gt;=(f\circ\phi)\sum_{\sigma\in S^n}(-1)^{\sigma}\Pi_{\alpha} \frac{\partial\phi^{\alpha}}{\partial y^{\sigma(\alpha)}}dy^1\wedge\ldots\wedge dy^n = (f\circ\phi) det (d\phi)dy^1\wedge\ldots\wedge dy^n = (f\circ\phi)J_{\phi}dy^1\wedge\ldots\wedge dy^n&lt;/math&gt;<br /> <br /> where &lt;math&gt;J_{\phi}&lt;/math&gt; is the determinant of the Jacobian matrix. <br /> <br /> <br /> Hence, &lt;math&gt;\int_{\mathbb{R}^n} \phi^*\omega = \int_{\mathbb{R}^n}(f\circ\phi)J_{\phi}dy^1\wedge\ldots\wedge dy^n = \int_{Old\ sense} (f\circ\phi)J_{\phi} =\pm \int (f\circ\phi)|J_{\phi}| = \pm\int f = \pm\int \omega&lt;/math&gt;<br /> <br /> ''Q.E.D''<br /> <br /> <br /> If we restrict our attention to just the ''orientation preserving'' &lt;math&gt;\phi&lt;/math&gt;'s so that &lt;math&gt;J_{\phi}&gt;0&lt;/math&gt; then we will always get the + in the end. <br /> <br /> <br /> '''Definition'''<br /> <br /> An ''orientation'' of M is an assignment of the charts to &lt;math&gt;\pm 1&lt;/math&gt;, &lt;math&gt;\phi\mapsto S_{\phi}&lt;/math&gt;<br /> <br /> Then if the domains on &lt;math&gt;\phi&lt;/math&gt; and &lt;math&gt;\psi&lt;/math&gt; overlap then &lt;math&gt;S_{\psi} = (sign J_{\psi^{-1}\phi})S_{\phi}&lt;/math&gt;<br /> <br /> By definition, M is orientable if we can find an orientation. <br /> <br /> <br /> '''Examples'''<br /> <br /> 1) &lt;math&gt;\mathbb{R}^n&lt;/math&gt;. We declare the identity positive and then all other designations follow from this. <br /> <br /> <br /> 2) The finite cylinder &lt;math&gt;S^1\times I&lt;/math&gt;<br /> <br /> We can put two charts on the cylinder by considering two rectangles which overlap by a little bit that cover the whole cylinder. If we denote one of these as positive, the overlap makes the other positive. We compare any other chart to these two. <br /> <br /> <br /> 3) Consider the mobius strip and the same attempted charts as for the cylinder. If we label one section positive, the other section must be positive due to the overlap on one side, but must be negative due to the overlap on the other side. Thus the mobius strip is not orientable. <br /> <br /> <br /> <br /> '''Definition'''<br /> <br /> An ''orientation of a vector space V'' is an equivalence class of ordered bases &lt;math&gt;v_{\alpha}&lt;/math&gt; of V, where &lt;math&gt;v_{\alpha}\sim w_{\beta}&lt;/math&gt; if the determinant of the transition matrix between these two bases is positive. <br /> <br /> <br /> '''Definition'''<br /> <br /> An ''orientation of M'' is a continuous choice of orientations for &lt;math&gt;T_p M&lt;/math&gt; for any p. We haven't technically defined what it means to be continuous in this sense but the meaning is clear. <br /> <br /> <br /> '''Definition'''<br /> <br /> let &lt;math&gt;M^n&lt;/math&gt; be an oriented manifold and let &lt;math&gt;\omega\in\Omega^n_c(M)&lt;/math&gt;. Let &lt;math&gt;\phi_{\alpha}:U_{\alpha}\mapsto\mathbb{R}^n&lt;/math&gt; be a collection of positive charts that cover M. Let &lt;math&gt;\lambda_{\alpha}&lt;/math&gt; be a partition of unity subordinate to this cover then <br /> <br /> &lt;math&gt;\int_M\omega = \int_M 1\omega = \int_M \sum_{\alpha}\lambda_{\alpha}\omega = \sum_{\alpha}\int_{U_{\alpha}}\lambda_{\alpha}\omega = \sum_{\alpha}\int_{\mathbb{R}^n} (\phi^{-1}_{\alpha})^*(\lambda_{\alpha}\omega)&lt;/math&gt;<br /> <br /> Note all the intermediate steps were merely properties we would LIKE the integral to have, the actual definition is the equality of the left most and right most expressions. <br /> <br /> <br /> '''Theorem'''<br /> <br /> &lt;math&gt;\int_M\omega&lt;/math&gt; is independent of the choices.</div> Trefor http://drorbn.net/index.php?title=0708-1300/Class_notes_for_Tuesday,_November_13 0708-1300/Class notes for Tuesday, November 13 2008-02-09T03:37:38Z <p>Trefor: /* Second Hour */</p> <hr /> <div>{{0708-1300/Navigation}}<br /> <br /> ==Typed Notes==<br /> <br /> &lt;span style=&quot;color: red;&quot;&gt;The notes below are by the students and for the students. Hopefully they are useful, but they come with no guarantee of any kind.&lt;/span&gt;<br /> <br /> <br /> ===First Hour===<br /> <br /> <br /> We begin with a review of last class. Since no one has typed up the notes for last class yet, I will do the review here. <br /> <br /> Recall had an association &lt;math&gt;M\rightarrow\Omega^k(M)&lt;/math&gt; which was the &quot;k forms on M&quot; which equaled &lt;math&gt;\{w:\ p\in M\rightarrow A^k(T_p M)\}&lt;/math&gt; <br /> <br /> where <br /> <br /> &lt;math&gt;A^k(V):=\{w:\underbrace{V\times\ldots\times V}_{k\ times} \rightarrow\mathbb{R}\}&lt;/math&gt; <br /> which is <br /> <br /> 1) Multilinear<br /> <br /> 2) Alternating<br /> <br /> We had proved that :<br /> <br /> 1) &lt;math&gt;A^k(V)&lt;/math&gt; is a vector space<br /> <br /> 2) there was a wedge product &lt;math&gt;\wedge:A^k(V)\times A^l(V)\rightarrow A^{k+l}(V)&lt;/math&gt; via &lt;math&gt;\omega\wedge\lambda(v_1,\ldots,v_{k+l}) = <br /> <br /> &lt;/math&gt;&lt;math&gt;\frac{1}{k!l!}\sum_{\sigma\in S_{k+l}}(-1)^{\sigma} \omega(v_{\sigma(1)},\ldots,v_{\sigma(k)})\lambda(v_{\sigma(k+1)},\ldots,v_{\sigma(k+l)})&lt;/math&gt;<br /> that is <br /> <br /> a) bilinear<br /> <br /> b) associative<br /> <br /> c) supercommutative, i.e., &lt;math&gt;\omega\wedge\lambda = (-1)^{deg(w)deg(\lambda)}\lambda\wedge\omega&lt;/math&gt;<br /> <br /> <br /> From these definitions we can define for &lt;math&gt;\omega\in\Omega^k(M)&lt;/math&gt; and &lt;math&gt;\lambda\in\Omega^l(M)&lt;/math&gt; that &lt;math&gt;\omega\wedge\lambda\in\Omega^{k+l}(M)&lt;/math&gt; with the same properties as above. <br /> <br /> <br /> ''Claim''<br /> <br /> <br /> If &lt;math&gt;\omega_1,\ldots,\omega_n&lt;/math&gt; is a basis of &lt;math&gt;A^1(V) = V^* &lt;/math&gt; then &lt;math&gt; \{\omega_I = w_{i_1}\wedge\ldots\wedge\omega_{i_k}\ :\ I=(i_1,\ldots,i_k)\ with\ 1\leq i_1&lt;\ldots&lt;i_k\leq n\}&lt;/math&gt; is a basis of &lt;math&gt;A^k(V)&lt;/math&gt; and &lt;math&gt;dim A^k(V) = nCk&lt;/math&gt;<br /> <br /> <br /> If &lt;math&gt;\omega_1,\ldots,\omega_n\in\Omega^1(M)&lt;/math&gt; and &lt;math&gt;w_1 |_p,\ldots,\omega_n |_p&lt;/math&gt; a basis of &lt;math&gt;(T_p M)^*\ \forall p\in M&lt;/math&gt; then any &lt;math&gt;\lambda&lt;/math&gt; can be written as &lt;math&gt;\lambda = \sum_I a_I(p)\omega_I&lt;/math&gt; where &lt;math&gt;a_I:M\rightarrow\mathbb{R}&lt;/math&gt; are smooth. <br /> <br /> <br /> The equivalence of these is left as an exercise. <br /> <br /> <br /> ''Example''<br /> <br /> Let us investigate &lt;math&gt;\Omega^*(\mathbb{R}^3)&lt;/math&gt; (the * just means &quot;anything&quot;).<br /> <br /> <br /> Now, &lt;math&gt;(T_p(\mathbb{R}^3))^* = &lt;dx_1,dx_2,dx_3&gt;&lt;/math&gt; where &lt;math&gt;x_i:\mathbb{R}^3\rightarrow\mathbb{R}&lt;/math&gt; and so &lt;math&gt;dx_i|_p:T_p\mathbb{R}^3\rightarrow T_{x_i(p)}\mathbb{R} =\mathbb{R}&lt;/math&gt;<br /> <br /> <br /> Hence, &lt;math&gt;dx_i|_p\in (T_p\mathbb{R}^3)^*&lt;/math&gt;<br /> <br /> <br /> Now, &lt;math&gt;dx_j(\frac{\partial}{\partial x}) = \frac{\partial}{\partial x_i}x_j =\delta_{ij}&lt;/math&gt; and hence we get a basis. <br /> <br /> <br /> So, &lt;math&gt;\Omega^1(\mathbb{R}^3) = \{g_1 dx_1 + g_2 dx_2 + g_3 dx_3\}\approx\{g_1,g_2,g_3\}\approx&lt;/math&gt; {vector fields on &lt;math&gt;\mathbb{R}^3&lt;/math&gt;}<br /> <br /> <br /> where the &lt;math&gt;g_i:\mathbb{R}^3\rightarrow\mathbb{R}&lt;/math&gt; are smooth.<br /> <br /> <br /> &lt;math&gt;\Omega^0(\mathbb{R}^3) = \{f:\mathbb{R}^3\rightarrow\mathbb{R}\}&lt;/math&gt;<br /> <br /> <br /> This is because to each point p we associate something that takes zero copies of the tangent space into the real numbers. Thus to each p we associate a number.<br /> <br /> <br /> &lt;math&gt;\Omega^3(\mathbb{R}^3) = \{kdx_1\wedge dx_2\wedge dx_3\} \approx&lt;/math&gt; {functions} where again the k is just a smooth function from &lt;math&gt;\mathbb{R}^3&lt;/math&gt; to &lt;math&gt;\mathbb{R}&lt;/math&gt;. <br /> <br /> <br /> &lt;math&gt;\Omega^2(\mathbb{R}^3) = \{h_1 dx_2\wedge dx_3 + h_2 dx_3\wedge dx_1 + h_3 dx_1\wedge dx_2\}\approx \{h_1,h_2,h_3\}\approx&lt;/math&gt; {vector fields}<br /> <br /> <br /> <br /> ''Aside''<br /> <br /> <br /> Recall our earlier discussion of how points and things like points (curves, equivalence classes of curves) pushfoward while things dual to points (functions) pullback and that things dual to functions (such as derivations) push forward. See earlier for the precise definitions. <br /> <br /> <br /> Now differential forms pull back, i.e., for &lt;math&gt;\phi:M\rightarrow N&lt;/math&gt; then &lt;math&gt;\phi^*(\lambda)\in\Omega^k(M)\leftarrow\lambda\in\Omega^k(N)&lt;/math&gt; <br /> via <br /> &lt;math&gt;\phi^*(\lambda)(v_1,\ldots,v_k)=\lambda(\phi_* v_1,\ldots \phi_* v_k)&lt;/math&gt;<br /> <br /> <br /> The pullback preserves all the properties discussed above and is well defined. In particular, it is compatible with the wedge product via &lt;math&gt;\phi^*(\omega\wedge\lambda)=\phi^*(\omega)\wedge\phi^*(\lambda)&lt;/math&gt;<br /> <br /> <br /> '''Theorem-Definition'''<br /> <br /> <br /> Given M, &lt;math&gt;\exists&lt;/math&gt; ! linear map &lt;math&gt;d:\Omega^k(M)\rightarrow\Omega^{k+1}(M)&lt;/math&gt; satisfies <br /> <br /> 1) If &lt;math&gt;f\in\Omega^0(M)&lt;/math&gt; then &lt;math&gt;df(X) = X(f)&lt;/math&gt; for &lt;math&gt;X\in TM&lt;/math&gt;<br /> <br /> 2) &lt;math&gt;d^2 = 0&lt;/math&gt;. I.e. if &lt;math&gt;d_k:\Omega^k\rightarrow\Omega^{k+1}&lt;/math&gt; and &lt;math&gt;d_{k+1}:\Omega^{k+1}\rightarrow\Omega^{k+2}&lt;/math&gt; then &lt;math&gt;d_{k+1}\circ d_k = 0&lt;/math&gt;. <br /> <br /> 3) &lt;math&gt;d(\omega\wedge\lambda) = d\omega\wedge\lambda + (-1)^{deg\omega}\omega\wedge d\lambda&lt;/math&gt;<br /> <br /> ===Second Hour===<br /> <br /> Some notes about the above definition:<br /> <br /> 1) When we restrict our ''d'' to functions we just get the old meaning for d. <br /> <br /> 2) Philosophically, there is a duality between differential forms and manifolds and that duality is given by integration. In this duality, d is the adjoint of the boundary operator on manifolds. For manifolds, the boundary of the boundary is empty and hence it is reasonable that &lt;math&gt;d^2=0&lt;/math&gt; on differential forms. <br /> <br /> 3) To remember the formula in 3 given above and others like it, it helps to keep in mind what objects are &quot;odd&quot; and what are &quot;even&quot; and thus when commuting such operators we will get the signs as you would expect from multiplying objects that are either odd or even. <br /> <br /> <br /> ''Example''<br /> <br /> Let us aim for a formula for d on &lt;math&gt;\Omega^*(\mathbb{R}^n)&lt;/math&gt;. <br /> <br /> Lets compute &lt;math&gt;d(\sum_I f_I dx_I)&lt;/math&gt; where &lt;math&gt;dx_I = dx_{i_1}\wedge\ldots\wedge dx_{i_k}&lt;/math&gt; and &lt;math&gt;I = (i_1,\ldots,i_k)&lt;/math&gt;<br /> <br /> Then, &lt;math&gt;d(\sum_I f_I dx_I) = \sum_I d(f_I \wedge dx_{i_1}\wedge\ldots\wedge dx_{i_k}) = \sum_I df_I\wedge(dx_I) + f_I\wedge d(dx_I)&lt;/math&gt;<br /> <br /> The last term vanishes because of (2) in the theorem (proving uniqueness!) <br /> <br /> <br /> Now, as an aside, we claim that for &lt;math&gt;f\in\Omega^0(M),\ df = \sum_{j=1}^{n}\frac{\partial f}{\partial x_j} dx_j&lt;/math&gt;<br /> <br /> Indeed, we know &lt;math&gt;(df)(\frac{\partial}{\partial x_i}) = \frac{\partial}{\partial x_i}f&lt;/math&gt; <br /> <br /> However, &lt;math&gt;(\sum_{j=1}^{n}\frac{\partial f}{\partial x_j} dx_j)(\frac{\partial}{\partial x_i}) = \sum_j \frac{\partial f}{\partial x_j}\delta_{ij} = \frac{\partial f}{\partial x_i}&lt;/math&gt; which is the same. <br /> <br /> <br /> Returning, we thus get &lt;math&gt;d(\sum_I f_I dx_I) = \sum_{I,j} \frac{\partial f_I}{\partial x_j} dx_j\wedge dx_I&lt;/math&gt; <br /> <br /> Thus our d takes functions to vector fields by &lt;math&gt;f\mapsto (\frac{\partial f}{\partial x_1},\frac{\partial f}{\partial x_2},\frac{\partial f}{\partial x_3})&lt;/math&gt;<br /> <br /> This is just the grad operator from calculus and we can see that the d operator appropriately takes things from &lt;math&gt;\Omega^0(M)&lt;/math&gt; to &lt;math&gt;\Omega^1(M)&lt;/math&gt;. <br /> <br /> <br /> Now let us compute &lt;math&gt;d(h_1 dx_2\wedge dx_3+h_2 dx_3\wedge dx_1+h_3 dx_1\wedge dx_2) =\frac{\partial h_1}{\partial x_1} dx_1\wedge dx_2\wedge dx_3 + \frac{\partial h_1}{\partial x_2} dx_2\wedge dx_2\wedge dx_3 + \frac{\partial h_1}{\partial x_3} dx_3\wedge dx_2\wedge dx_3&lt;/math&gt; + 6 more terms representing the 3 partials of each of the last 2 terms. <br /> <br /> As each &lt;math&gt;dx_i\wedge dx_i&lt;/math&gt; term vanishes we are left with just, <br /> <br /> &lt;math&gt;= (\frac{\partial h_1}{\partial x_1} + \frac{\partial h_2}{\partial x_2} + \frac{\partial h_3}{\partial x_3})dx_1\wedge dx_2\wedge dx_3&lt;/math&gt;<br /> <br /> I.e., d takes &lt;math&gt;(h_1,h_2,h_3)\mapsto \sum_i\frac{\partial h_i}{\partial x_i}&lt;/math&gt; <br /> <br /> this is just the div operator from calculus and appropriately takes vector fields to functions and represents the d from &lt;math&gt;\Omega^2(M)&lt;/math&gt; to &lt;math&gt;\Omega^3(M)&lt;/math&gt;. <br /> <br /> <br /> We are left with computing d from &lt;math&gt;\Omega^1(M)&lt;/math&gt; to &lt;math&gt;\Omega^2(M)&lt;/math&gt;<br /> <br /> Computing, &lt;math&gt;d(g_1 dx_1 + g_2 dx_2 + g_3 dx_3) = (\frac{\partial g_3}{\partial x_2} - \frac{\partial g_2}{\partial x_3})dx_2\wedge dx_3 + (\frac{\partial g_1}{\partial x_3} - \frac{\partial g_3}{\partial x_1})dx_3\wedge dx_1 + (\frac{\partial g_2}{\partial x_1} - \frac{\partial g_1}{\partial x_2})dx_1\wedge dx_2&lt;/math&gt;<br /> <br /> I.e., we just have the curl operator. <br /> <br /> Note that the well known calculus laws that curl grad = 0 and div curl = 0 are just the expression that &lt;math&gt;d^2 =0&lt;/math&gt;. <br /> <br /> To provide some physical insight to the meanings of these operators:<br /> <br /> 1) The gradient represents the direction of maximum descent. I.e. if you had a function on the plane the graph would look like the surface of a mountain range and the direction that water would run would be the gradient. <br /> <br /> 2) In a say compressible fluid, the divergence corresponds to the difference between in the inflow and outflow of fluid in some small epsilon box around a point. <br /> <br /> 3) The curl corresponds to the rotation vector for a ball. Ie consider a ball (of equal density to the liquid about it) going down a river. In the &lt;math&gt;x_2&lt;/math&gt;, &lt;math&gt;x_1&lt;/math&gt; plane the tenancy for it to rotate clockwise would be given by &lt;math&gt;\frac{\partial g_2}{\partial x_1} - \frac{\partial g_1}{\partial x_2}&lt;/math&gt;</div> Trefor http://drorbn.net/index.php?title=0708-1300/Class_notes_for_Tuesday,_November_13 0708-1300/Class notes for Tuesday, November 13 2008-02-09T03:33:55Z <p>Trefor: /* Second Hour */</p> <hr /> <div>{{0708-1300/Navigation}}<br /> <br /> ==Typed Notes==<br /> <br /> &lt;span style=&quot;color: red;&quot;&gt;The notes below are by the students and for the students. Hopefully they are useful, but they come with no guarantee of any kind.&lt;/span&gt;<br /> <br /> <br /> ===First Hour===<br /> <br /> <br /> We begin with a review of last class. Since no one has typed up the notes for last class yet, I will do the review here. <br /> <br /> Recall had an association &lt;math&gt;M\rightarrow\Omega^k(M)&lt;/math&gt; which was the &quot;k forms on M&quot; which equaled &lt;math&gt;\{w:\ p\in M\rightarrow A^k(T_p M)\}&lt;/math&gt; <br /> <br /> where <br /> <br /> &lt;math&gt;A^k(V):=\{w:\underbrace{V\times\ldots\times V}_{k\ times} \rightarrow\mathbb{R}\}&lt;/math&gt; <br /> which is <br /> <br /> 1) Multilinear<br /> <br /> 2) Alternating<br /> <br /> We had proved that :<br /> <br /> 1) &lt;math&gt;A^k(V)&lt;/math&gt; is a vector space<br /> <br /> 2) there was a wedge product &lt;math&gt;\wedge:A^k(V)\times A^l(V)\rightarrow A^{k+l}(V)&lt;/math&gt; via &lt;math&gt;\omega\wedge\lambda(v_1,\ldots,v_{k+l}) = <br /> <br /> &lt;/math&gt;&lt;math&gt;\frac{1}{k!l!}\sum_{\sigma\in S_{k+l}}(-1)^{\sigma} \omega(v_{\sigma(1)},\ldots,v_{\sigma(k)})\lambda(v_{\sigma(k+1)},\ldots,v_{\sigma(k+l)})&lt;/math&gt;<br /> that is <br /> <br /> a) bilinear<br /> <br /> b) associative<br /> <br /> c) supercommutative, i.e., &lt;math&gt;\omega\wedge\lambda = (-1)^{deg(w)deg(\lambda)}\lambda\wedge\omega&lt;/math&gt;<br /> <br /> <br /> From these definitions we can define for &lt;math&gt;\omega\in\Omega^k(M)&lt;/math&gt; and &lt;math&gt;\lambda\in\Omega^l(M)&lt;/math&gt; that &lt;math&gt;\omega\wedge\lambda\in\Omega^{k+l}(M)&lt;/math&gt; with the same properties as above. <br /> <br /> <br /> ''Claim''<br /> <br /> <br /> If &lt;math&gt;\omega_1,\ldots,\omega_n&lt;/math&gt; is a basis of &lt;math&gt;A^1(V) = V^* &lt;/math&gt; then &lt;math&gt; \{\omega_I = w_{i_1}\wedge\ldots\wedge\omega_{i_k}\ :\ I=(i_1,\ldots,i_k)\ with\ 1\leq i_1&lt;\ldots&lt;i_k\leq n\}&lt;/math&gt; is a basis of &lt;math&gt;A^k(V)&lt;/math&gt; and &lt;math&gt;dim A^k(V) = nCk&lt;/math&gt;<br /> <br /> <br /> If &lt;math&gt;\omega_1,\ldots,\omega_n\in\Omega^1(M)&lt;/math&gt; and &lt;math&gt;w_1 |_p,\ldots,\omega_n |_p&lt;/math&gt; a basis of &lt;math&gt;(T_p M)^*\ \forall p\in M&lt;/math&gt; then any &lt;math&gt;\lambda&lt;/math&gt; can be written as &lt;math&gt;\lambda = \sum_I a_I(p)\omega_I&lt;/math&gt; where &lt;math&gt;a_I:M\rightarrow\mathbb{R}&lt;/math&gt; are smooth. <br /> <br /> <br /> The equivalence of these is left as an exercise. <br /> <br /> <br /> ''Example''<br /> <br /> Let us investigate &lt;math&gt;\Omega^*(\mathbb{R}^3)&lt;/math&gt; (the * just means &quot;anything&quot;).<br /> <br /> <br /> Now, &lt;math&gt;(T_p(\mathbb{R}^3))^* = &lt;dx_1,dx_2,dx_3&gt;&lt;/math&gt; where &lt;math&gt;x_i:\mathbb{R}^3\rightarrow\mathbb{R}&lt;/math&gt; and so &lt;math&gt;dx_i|_p:T_p\mathbb{R}^3\rightarrow T_{x_i(p)}\mathbb{R} =\mathbb{R}&lt;/math&gt;<br /> <br /> <br /> Hence, &lt;math&gt;dx_i|_p\in (T_p\mathbb{R}^3)^*&lt;/math&gt;<br /> <br /> <br /> Now, &lt;math&gt;dx_j(\frac{\partial}{\partial x}) = \frac{\partial}{\partial x_i}x_j =\delta_{ij}&lt;/math&gt; and hence we get a basis. <br /> <br /> <br /> So, &lt;math&gt;\Omega^1(\mathbb{R}^3) = \{g_1 dx_1 + g_2 dx_2 + g_3 dx_3\}\approx\{g_1,g_2,g_3\}\approx&lt;/math&gt; {vector fields on &lt;math&gt;\mathbb{R}^3&lt;/math&gt;}<br /> <br /> <br /> where the &lt;math&gt;g_i:\mathbb{R}^3\rightarrow\mathbb{R}&lt;/math&gt; are smooth.<br /> <br /> <br /> &lt;math&gt;\Omega^0(\mathbb{R}^3) = \{f:\mathbb{R}^3\rightarrow\mathbb{R}\}&lt;/math&gt;<br /> <br /> <br /> This is because to each point p we associate something that takes zero copies of the tangent space into the real numbers. Thus to each p we associate a number.<br /> <br /> <br /> &lt;math&gt;\Omega^3(\mathbb{R}^3) = \{kdx_1\wedge dx_2\wedge dx_3\} \approx&lt;/math&gt; {functions} where again the k is just a smooth function from &lt;math&gt;\mathbb{R}^3&lt;/math&gt; to &lt;math&gt;\mathbb{R}&lt;/math&gt;. <br /> <br /> <br /> &lt;math&gt;\Omega^2(\mathbb{R}^3) = \{h_1 dx_2\wedge dx_3 + h_2 dx_3\wedge dx_1 + h_3 dx_1\wedge dx_2\}\approx \{h_1,h_2,h_3\}\approx&lt;/math&gt; {vector fields}<br /> <br /> <br /> <br /> ''Aside''<br /> <br /> <br /> Recall our earlier discussion of how points and things like points (curves, equivalence classes of curves) pushfoward while things dual to points (functions) pullback and that things dual to functions (such as derivations) push forward. See earlier for the precise definitions. <br /> <br /> <br /> Now differential forms pull back, i.e., for &lt;math&gt;\phi:M\rightarrow N&lt;/math&gt; then &lt;math&gt;\phi^*(\lambda)\in\Omega^k(M)\leftarrow\lambda\in\Omega^k(N)&lt;/math&gt; <br /> via <br /> &lt;math&gt;\phi^*(\lambda)(v_1,\ldots,v_k)=\lambda(\phi_* v_1,\ldots \phi_* v_k)&lt;/math&gt;<br /> <br /> <br /> The pullback preserves all the properties discussed above and is well defined. In particular, it is compatible with the wedge product via &lt;math&gt;\phi^*(\omega\wedge\lambda)=\phi^*(\omega)\wedge\phi^*(\lambda)&lt;/math&gt;<br /> <br /> <br /> '''Theorem-Definition'''<br /> <br /> <br /> Given M, &lt;math&gt;\exists&lt;/math&gt; ! linear map &lt;math&gt;d:\Omega^k(M)\rightarrow\Omega^{k+1}(M)&lt;/math&gt; satisfies <br /> <br /> 1) If &lt;math&gt;f\in\Omega^0(M)&lt;/math&gt; then &lt;math&gt;df(X) = X(f)&lt;/math&gt; for &lt;math&gt;X\in TM&lt;/math&gt;<br /> <br /> 2) &lt;math&gt;d^2 = 0&lt;/math&gt;. I.e. if &lt;math&gt;d_k:\Omega^k\rightarrow\Omega^{k+1}&lt;/math&gt; and &lt;math&gt;d_{k+1}:\Omega^{k+1}\rightarrow\Omega^{k+2}&lt;/math&gt; then &lt;math&gt;d_{k+1}\circ d_k = 0&lt;/math&gt;. <br /> <br /> 3) &lt;math&gt;d(\omega\wedge\lambda) = d\omega\wedge\lambda + (-1)^{deg\omega}\omega\wedge d\lambda&lt;/math&gt;<br /> <br /> ===Second Hour===<br /> <br /> Some notes about the above definition:<br /> <br /> 1) When we restrict our ''d'' to functions we just get the old meaning for d. <br /> <br /> 2) Philosophically, there is a duality between differential forms and manifolds and that duality is given by integration. In this duality, d is the adjoint of the boundary operator on manifolds. For manifolds, the boundary of the boundary is empty and hence it is reasonable that &lt;math&gt;d^2=0&lt;/math&gt; on differential forms. <br /> <br /> 3) To remember the formula in 3 given above and others like it, it helps to keep in mind what objects are &quot;odd&quot; and what are &quot;even&quot; and thus when commuting such operators we will get the signs as you would expect from multiplying objects that are either odd or even. <br /> <br /> <br /> ''Example''<br /> <br /> Let us aim for a formula for d on &lt;math&gt;\Omega^*(\mathbb{R}^n)&lt;/math&gt;. <br /> <br /> Lets compute &lt;math&gt;d(\sum_I f_I dx_I)&lt;/math&gt; where &lt;math&gt;dx_I = dx_{i_1}\wedge\ldots\wedge dx_{i_k}&lt;/math&gt; and &lt;math&gt;I = (i_1,\ldots,i_k)&lt;/math&gt;<br /> <br /> Then, &lt;math&gt;d(\sum_I f_I dx_I) = \sum_I d(f_I \wedge dx_{i_1}\wedge\ldots\wedge dx_{i_k}) = \sum_I df_I\wedge(dx_I) + f_I\wedge d(dx_I)&lt;/math&gt;<br /> <br /> The last term vanishes because of (2) in the theorem (proving uniqueness!) <br /> <br /> <br /> Now, as an aside, we claim that for &lt;math&gt;f\in\Omega^0(M),\ df = \sum_{j=1}^{n}\frac{\partial f}{\partial x_j} dx_j&lt;/math&gt;<br /> <br /> Indeed, we know &lt;math&gt;(df)(\frac{\partial}{\partial x_i}) = \frac{\partial}{\partial x_i}f&lt;/math&gt; <br /> <br /> However, &lt;math&gt;(\sum_{j=1}^{n}\frac{\partial f}{\partial x_j} dx_j)(\frac{\partial}{\partial x_i}) = \sum_j \frac{\partial f}{\partial x_j}\delta_{ij} = \frac{\partial f}{\partial x_i}&lt;/math&gt; which is the same. <br /> <br /> <br /> Returning, we thus get &lt;math&gt;d(\sum_I f_I dx_I) = \sum_{I,j} \frac{\partial f_I}{\partial x_j} dx_j\wedge dx_I&lt;/math&gt; <br /> <br /> Thus our d takes functions to vector fields by &lt;math&gt;f\mapsto (\frac{\partial f}{\partial x_1},\frac{\partial f}{\partial x_2},\frac{\partial f}{\partial x_3})&lt;/math&gt;<br /> <br /> This is just the grad operator from calculus and we can see that the d operator appropriately takes things from &lt;math&gt;\Omega^0(M)&lt;/math&gt; to &lt;math&gt;\Omega^1(M)&lt;/math&gt;. <br /> <br /> <br /> Now let us compute &lt;math&gt;d(h_1 dx_2\wedge dx_3+h_2 dx_3\wedge dx_1+h_3 dx_1\wedge dx_2) =\frac{\partial h_1}{\partial x_1} dx_1\wedge dx_2\wedge dx_3 + \frac{\partial h_1}{\partial x_2} dx_2\wedge dx_2\wedge dx_3 + \frac{\partial h_1}{\partial x_3} dx_3\wedge dx_2\wedge dx_3&lt;/math&gt; + 6 more terms representing the 3 partials of each of the last 2 terms. <br /> <br /> As each &lt;math&gt;dx_i\wedge dx_i&lt;/math&gt; term vanishes we are left with just, <br /> <br /> &lt;math&gt;= (\frac{\partial h_1}{\partial x_1} + \frac{\partial h_2}{\partial x_2} + \frac{\partial h_3}{\partial x_3})dx_1\wedge dx_2\wedge dx_3&lt;/math&gt;<br /> <br /> I.e., d takes &lt;math&gt;(h_1,h_2,h_3)\mapsto \sum_i\frac{\partial h_i}{\partial x_i}&lt;/math&gt; <br /> <br /> this is just the div operator from calculus and appropriately takes vector fields to functions and represents the d from &lt;math&gt;\Omega^2(M)&lt;/math&gt; to &lt;math&gt;\Omega^3(M)&lt;/math&gt;. <br /> <br /> <br /> We are left with computing d from &lt;math&gt;\Omega^1(M)&lt;/math&gt; to &lt;math&gt;\Omega^2(M)&lt;/math&gt;<br /> <br /> Computing, &lt;math&gt;d(g_1 dx_1 + g_2 dx_2 + g_3 dx_3) = (\frac{\partial g_3}{\partial x_2} - \frac{\partial g_2}{\partial x_3})dx_2\wedge dx_3 + (\frac{\partial g_1}{\partial x_3} - \frac{\partial g_3}{\partial x_1})dx_3\wedge dx_1 + (\frac{\partial g_2}{\partial x_1} - \frac{\partial g_1}{\partial x_2})dx_1\wedge dx_2&lt;/math&gt;<br /> <br /> I.e., we just have the curl operator. <br /> <br /> Note that the well known calculus laws that curl grad = 0 and div curl = 0 are just the expression that &lt;math&gt;d^2 =0&lt;/math&gt;. <br /> <br /> To provide some physical insight to the meanings of these operators:<br /> <br /> 1) The gradient represents the direction of maximum descent. I.e. if you had a function on the plane the graph would look like the surface of a mountain range and the direction that water would run would be the gradient. <br /> <br /> 2) In a say compressible fluid, the divergence corresponds to the difference between in the inflow and outflow of fluid in some small epsilon box around a point. <br /> <br /> 3) The curl corresponds to the rotation vector for a ball. Ie consider a ball (of equal density to the liquid about it) going down a river. In the x_2, x_1 plane the tenancy for it to rotate clockwise would be given by &lt;math&gt;\frac{\partial g_2}{\partial x_1} - \frac{\partial g_1}{\partial x_2}&lt;/math&gt;</div> Trefor http://drorbn.net/index.php?title=0708-1300/Class_notes_for_Tuesday,_November_13 0708-1300/Class notes for Tuesday, November 13 2008-02-09T03:07:52Z <p>Trefor: /* First Hour */</p> <hr /> <div>{{0708-1300/Navigation}}<br /> <br /> ==Typed Notes==<br /> <br /> &lt;span style=&quot;color: red;&quot;&gt;The notes below are by the students and for the students. Hopefully they are useful, but they come with no guarantee of any kind.&lt;/span&gt;<br /> <br /> <br /> ===First Hour===<br /> <br /> <br /> We begin with a review of last class. Since no one has typed up the notes for last class yet, I will do the review here. <br /> <br /> Recall had an association &lt;math&gt;M\rightarrow\Omega^k(M)&lt;/math&gt; which was the &quot;k forms on M&quot; which equaled &lt;math&gt;\{w:\ p\in M\rightarrow A^k(T_p M)\}&lt;/math&gt; <br /> <br /> where <br /> <br /> &lt;math&gt;A^k(V):=\{w:\underbrace{V\times\ldots\times V}_{k\ times} \rightarrow\mathbb{R}\}&lt;/math&gt; <br /> which is <br /> <br /> 1) Multilinear<br /> <br /> 2) Alternating<br /> <br /> We had proved that :<br /> <br /> 1) &lt;math&gt;A^k(V)&lt;/math&gt; is a vector space<br /> <br /> 2) there was a wedge product &lt;math&gt;\wedge:A^k(V)\times A^l(V)\rightarrow A^{k+l}(V)&lt;/math&gt; via &lt;math&gt;\omega\wedge\lambda(v_1,\ldots,v_{k+l}) = <br /> <br /> &lt;/math&gt;&lt;math&gt;\frac{1}{k!l!}\sum_{\sigma\in S_{k+l}}(-1)^{\sigma} \omega(v_{\sigma(1)},\ldots,v_{\sigma(k)})\lambda(v_{\sigma(k+1)},\ldots,v_{\sigma(k+l)})&lt;/math&gt;<br /> that is <br /> <br /> a) bilinear<br /> <br /> b) associative<br /> <br /> c) supercommutative, i.e., &lt;math&gt;\omega\wedge\lambda = (-1)^{deg(w)deg(\lambda)}\lambda\wedge\omega&lt;/math&gt;<br /> <br /> <br /> From these definitions we can define for &lt;math&gt;\omega\in\Omega^k(M)&lt;/math&gt; and &lt;math&gt;\lambda\in\Omega^l(M)&lt;/math&gt; that &lt;math&gt;\omega\wedge\lambda\in\Omega^{k+l}(M)&lt;/math&gt; with the same properties as above. <br /> <br /> <br /> ''Claim''<br /> <br /> <br /> If &lt;math&gt;\omega_1,\ldots,\omega_n&lt;/math&gt; is a basis of &lt;math&gt;A^1(V) = V^* &lt;/math&gt; then &lt;math&gt; \{\omega_I = w_{i_1}\wedge\ldots\wedge\omega_{i_k}\ :\ I=(i_1,\ldots,i_k)\ with\ 1\leq i_1&lt;\ldots&lt;i_k\leq n\}&lt;/math&gt; is a basis of &lt;math&gt;A^k(V)&lt;/math&gt; and &lt;math&gt;dim A^k(V) = nCk&lt;/math&gt;<br /> <br /> <br /> If &lt;math&gt;\omega_1,\ldots,\omega_n\in\Omega^1(M)&lt;/math&gt; and &lt;math&gt;w_1 |_p,\ldots,\omega_n |_p&lt;/math&gt; a basis of &lt;math&gt;(T_p M)^*\ \forall p\in M&lt;/math&gt; then any &lt;math&gt;\lambda&lt;/math&gt; can be written as &lt;math&gt;\lambda = \sum_I a_I(p)\omega_I&lt;/math&gt; where &lt;math&gt;a_I:M\rightarrow\mathbb{R}&lt;/math&gt; are smooth. <br /> <br /> <br /> The equivalence of these is left as an exercise. <br /> <br /> <br /> ''Example''<br /> <br /> Let us investigate &lt;math&gt;\Omega^*(\mathbb{R}^3)&lt;/math&gt; (the * just means &quot;anything&quot;).<br /> <br /> <br /> Now, &lt;math&gt;(T_p(\mathbb{R}^3))^* = &lt;dx_1,dx_2,dx_3&gt;&lt;/math&gt; where &lt;math&gt;x_i:\mathbb{R}^3\rightarrow\mathbb{R}&lt;/math&gt; and so &lt;math&gt;dx_i|_p:T_p\mathbb{R}^3\rightarrow T_{x_i(p)}\mathbb{R} =\mathbb{R}&lt;/math&gt;<br /> <br /> <br /> Hence, &lt;math&gt;dx_i|_p\in (T_p\mathbb{R}^3)^*&lt;/math&gt;<br /> <br /> <br /> Now, &lt;math&gt;dx_j(\frac{\partial}{\partial x}) = \frac{\partial}{\partial x_i}x_j =\delta_{ij}&lt;/math&gt; and hence we get a basis. <br /> <br /> <br /> So, &lt;math&gt;\Omega^1(\mathbb{R}^3) = \{g_1 dx_1 + g_2 dx_2 + g_3 dx_3\}\approx\{g_1,g_2,g_3\}\approx&lt;/math&gt; {vector fields on &lt;math&gt;\mathbb{R}^3&lt;/math&gt;}<br /> <br /> <br /> where the &lt;math&gt;g_i:\mathbb{R}^3\rightarrow\mathbb{R}&lt;/math&gt; are smooth.<br /> <br /> <br /> &lt;math&gt;\Omega^0(\mathbb{R}^3) = \{f:\mathbb{R}^3\rightarrow\mathbb{R}\}&lt;/math&gt;<br /> <br /> <br /> This is because to each point p we associate something that takes zero copies of the tangent space into the real numbers. Thus to each p we associate a number.<br /> <br /> <br /> &lt;math&gt;\Omega^3(\mathbb{R}^3) = \{kdx_1\wedge dx_2\wedge dx_3\} \approx&lt;/math&gt; {functions} where again the k is just a smooth function from &lt;math&gt;\mathbb{R}^3&lt;/math&gt; to &lt;math&gt;\mathbb{R}&lt;/math&gt;. <br /> <br /> <br /> &lt;math&gt;\Omega^2(\mathbb{R}^3) = \{h_1 dx_2\wedge dx_3 + h_2 dx_3\wedge dx_1 + h_3 dx_1\wedge dx_2\}\approx \{h_1,h_2,h_3\}\approx&lt;/math&gt; {vector fields}<br /> <br /> <br /> <br /> ''Aside''<br /> <br /> <br /> Recall our earlier discussion of how points and things like points (curves, equivalence classes of curves) pushfoward while things dual to points (functions) pullback and that things dual to functions (such as derivations) push forward. See earlier for the precise definitions. <br /> <br /> <br /> Now differential forms pull back, i.e., for &lt;math&gt;\phi:M\rightarrow N&lt;/math&gt; then &lt;math&gt;\phi^*(\lambda)\in\Omega^k(M)\leftarrow\lambda\in\Omega^k(N)&lt;/math&gt; <br /> via <br /> &lt;math&gt;\phi^*(\lambda)(v_1,\ldots,v_k)=\lambda(\phi_* v_1,\ldots \phi_* v_k)&lt;/math&gt;<br /> <br /> <br /> The pullback preserves all the properties discussed above and is well defined. In particular, it is compatible with the wedge product via &lt;math&gt;\phi^*(\omega\wedge\lambda)=\phi^*(\omega)\wedge\phi^*(\lambda)&lt;/math&gt;<br /> <br /> <br /> '''Theorem-Definition'''<br /> <br /> <br /> Given M, &lt;math&gt;\exists&lt;/math&gt; ! linear map &lt;math&gt;d:\Omega^k(M)\rightarrow\Omega^{k+1}(M)&lt;/math&gt; satisfies <br /> <br /> 1) If &lt;math&gt;f\in\Omega^0(M)&lt;/math&gt; then &lt;math&gt;df(X) = X(f)&lt;/math&gt; for &lt;math&gt;X\in TM&lt;/math&gt;<br /> <br /> 2) &lt;math&gt;d^2 = 0&lt;/math&gt;. I.e. if &lt;math&gt;d_k:\Omega^k\rightarrow\Omega^{k+1}&lt;/math&gt; and &lt;math&gt;d_{k+1}:\Omega^{k+1}\rightarrow\Omega^{k+2}&lt;/math&gt; then &lt;math&gt;d_{k+1}\circ d_k = 0&lt;/math&gt;. <br /> <br /> 3) &lt;math&gt;d(\omega\wedge\lambda) = d\omega\wedge\lambda + (-1)^{deg\omega}\omega\wedge d\lambda&lt;/math&gt;<br /> <br /> ===Second Hour===<br /> <br /> Some notes about the above definition:<br /> <br /> 1) When we restrict our ''d'' to functions we just get the old meaning for d. <br /> <br /> 2) Philosophically, there is a duality between differential forms and manifolds and that duality is given by integration. In this duality, d is the adjoint of the boundary operator on manifolds. For manifolds, the boundary of the boundary is empty and hence it is reasonable that &lt;math&gt;d^2=0&lt;/math&gt; on differential forms. <br /> <br /> 3) To remember the formula in 3 given above and others like it, it helps to keep in mind what objects are &quot;odd&quot; and what are &quot;even&quot; and thus when commuting such operators we will get the signs as you would expect from multiplying objects that are either odd or even. <br /> <br /> <br /> ''Example''<br /> <br /> Let us aim for a formula for d on &lt;math&gt;\Omega^*(\mathbb{R}^n)&lt;/math&gt;. <br /> <br /> Lets compute &lt;math&gt;d(\sum_I f_I dx_I)&lt;/math&gt; where &lt;math&gt;dx_I = dx_{i_1}\wedge\ldots\wedge dx_{i_k}&lt;/math&gt; and &lt;math&gt;I = (i_1,\ldots,i_k)&lt;/math&gt;<br /> <br /> Then, &lt;math&gt;d(\sum_I f_I dx_I) = \sum_I d(f_I \wedge dx_{i_1}\wedge\ldots\wedge dx_{i_k}) = \sum_I df_I\wedge(dx_I) + f_I\wedge d(dx_I)&lt;/math&gt;<br /> <br /> The last term vanishes because of (2) in the theorem (proving uniqueness!) <br /> <br /> <br /> Now, as an aside, we claim that for &lt;math&gt;f\in\Omega^0(M),\ df = \sum_{j=1}^{n}\frac{\partial f}{\partial x_j} dx_j&lt;/math&gt;<br /> <br /> Indeed, we know &lt;math&gt;(df)(\frac{\partial}{\partial x_i}) = \frac{\partial}{\partial x_i}f&lt;/math&gt; <br /> <br /> However, &lt;math&gt;(\sum_{j=1}^{n}\frac{\partial f}{\partial x_j} dx_j)(\frac{\partial}{\partial x_i}) = \sum_j \frac{\partial f}{\partial x_j}\delta_{ij} = \frac{\partial f}{\partial x_i}&lt;/math&gt; which is the same. <br /> <br /> <br /> Returning, we thus get &lt;math&gt;d(\sum_I f_I dx_I) = \sum_{I,j} \frac{\partial f_I}{\partial x_j} dx_j\wedge dx_I&lt;/math&gt; <br /> <br /> Thus our d takes functions to vector fields by &lt;math&gt;f\mapsto (\frac{\partial f}{\partial x_1},\frac{\partial f}{\partial x_2},\frac{\partial f}{\partial x_3})&lt;/math&gt;<br /> <br /> This is just the grad operator from calculus and we can see that the d operator appropriately takes things from &lt;math&gt;\Omega^0(M)&lt;/math&gt; to &lt;math&gt;\Omega^1(M)&lt;/math&gt;. <br /> <br /> <br /> Now let us compute &lt;math&gt;d(h_1 dx_2\wedge dx_3+h_2 dx_3\wedge dx_1+h_3 dx_1\wedge dx_2) =\frac{\partial h_1}{\partial x_1} dx_1\wedge dx_2\wedge dx_3 + \frac{\partial h_1}{\partial x_2} dx_3\wedge dx_2\wedge dx_3 + \frac{\partial h_1}{\partial x_3} dx_3\wedge dx_2\wedge dx_3&lt;/math&gt; + 6 more terms representing the 3 partials of each of the last 2 terms. <br /> <br /> As each &lt;math&gt;dx_i\wedge dx_i&lt;/math&gt; term vanishes we are left with just, <br /> <br /> &lt;math&gt;= (\frac{\partial h_1}{\partial x_1} + \frac{\partial h_2}{\partial x_2} + \frac{\partial h_3}{\partial x_3})dx_1\wedge dx_2\wedge dx_3&lt;/math&gt;<br /> <br /> I.e., d takes &lt;math&gt;(h_1,h_2,h_3)\mapsto \sum_i\frac{\partial h_i}{\partial x_i}&lt;/math&gt; <br /> <br /> this is just the div operator from calculus and appropriately takes vector fields to functions and represents the d from &lt;math&gt;\Omega^2(M)&lt;/math&gt; to &lt;math&gt;\Omega^3(M)&lt;/math&gt;. <br /> <br /> <br /> We are left with computing d from &lt;math&gt;\Omega^1(M)&lt;/math&gt; to &lt;math&gt;\Omega^2(M)&lt;/math&gt;<br /> <br /> Computing, &lt;math&gt;d(g_1 dx_1 + g_2 dx_2 + g_3 dx_3) = (\frac{\partial g_3}{\partial x_2} - \frac{\partial g_2}{\partial x_3})dx_2\wedge dx_3 + (\frac{\partial g_1}{\partial x_3} - \frac{\partial g_3}{\partial x_1})dx_3\wedge dx_1 + (\frac{\partial g_2}{\partial x_1} - \frac{\partial g_1}{\partial x_2})dx_1\wedge dx_2&lt;/math&gt;<br /> <br /> I.e., we just have the curl operator. <br /> <br /> Note that the well known calculus laws that curl grad = 0 and div curl = 0 are just the expression that &lt;math&gt;d^2 =0&lt;/math&gt;. <br /> <br /> To provide some physical insight to the meanings of these operators:<br /> <br /> 1) The gradient represents the direction of maximum descent. I.e. if you had a function on the plane the graph would look like the surface of a mountain range and the direction that water would run would be the gradient. <br /> <br /> 2) In a say compressible fluid, the divergence corresponds to the difference between in the inflow and outflow of fluid in some small epsilon box around a point. <br /> <br /> 3) The curl corresponds to the rotation vector for a ball. Ie consider a ball (of equal density to the liquid about it) going down a river. In the x_2, x_1 plane the tenancy for it to rotate clockwise would be given by &lt;math&gt;\frac{\partial g_2}{\partial x_1} - \frac{\partial g_1}{\partial x_2}&lt;/math&gt;</div> Trefor http://drorbn.net/index.php?title=0708-1300/Class_notes_for_Tuesday,_November_13 0708-1300/Class notes for Tuesday, November 13 2008-02-09T02:53:16Z <p>Trefor: /* First Hour */</p> <hr /> <div>{{0708-1300/Navigation}}<br /> <br /> ==Typed Notes==<br /> <br /> &lt;span style=&quot;color: red;&quot;&gt;The notes below are by the students and for the students. Hopefully they are useful, but they come with no guarantee of any kind.&lt;/span&gt;<br /> <br /> <br /> ===First Hour===<br /> <br /> <br /> We begin with a review of last class. Since no one has typed up the notes for last class yet, I will do the review here. <br /> <br /> Recall had an association &lt;math&gt;M\rightarrow\Omega^k(M)&lt;/math&gt; which was the &quot;k forms on M&quot; which equaled &lt;math&gt;\{w:\ p\in M\rightarrow A^k(T_p M)\}&lt;/math&gt; <br /> <br /> where <br /> <br /> &lt;math&gt;A^k(V):=\{w:\underbrace{V\times\ldots\times V}_{k\ times} \rightarrow\mathbb{R}\}&lt;/math&gt; <br /> which is <br /> <br /> 1) Multilinear<br /> <br /> 2) Alternating<br /> <br /> We had proved that :<br /> <br /> 1) &lt;math&gt;A^k(V)&lt;/math&gt; is a vector space<br /> <br /> 2) there was a wedge product &lt;math&gt;\wedge:A^k(V)\times A^l(V)\rightarrow A^{k+l}(V)&lt;/math&gt; via &lt;math&gt;\omega\wedge\lambda(v_1,\ldots,v_{k+l}) = <br /> <br /> &lt;/math&gt;&lt;math&gt;\frac{1}{k!l!}\sum_{\sigma\in S_{k+l}}(-1)^{\sigma} \omega(v_{\sigma(1)},\ldots,v_{\sigma(k)})\lambda(v_{\sigma(k+1)},\ldots,v_{\sigma(k+l)})&lt;/math&gt;<br /> that is <br /> <br /> a) bilinear<br /> <br /> b) associative<br /> <br /> c) supercommutative, i.e., &lt;math&gt;\omega\wedge\lambda = (-1)^{deg(w)deg(\lambda)}\lambda\wedge\omega&lt;/math&gt;<br /> <br /> <br /> From these definitions we can define for &lt;math&gt;\omega\in\Omega^k(M)&lt;/math&gt; and &lt;math&gt;\lambda\in\Omega^l(M)&lt;/math&gt; that &lt;math&gt;\omega\wedge\lambda\in\Omega^{k+l}(M)&lt;/math&gt; with the same properties as above. <br /> <br /> <br /> ''Claim''<br /> <br /> <br /> If &lt;math&gt;\omega_1,\ldots,\omega_n&lt;/math&gt; is a basis of &lt;math&gt;A^1(V) = V^* &lt;/math&gt; then &lt;math&gt; \{\omega_I = w_{i_1}\wedge\ldots\wedge\omega_{i_k}\ :\ I=(i_1,\ldots,i_k)\ with\ 1\leq i_1&lt;\ldots&lt;i_k\leq n\}&lt;/math&gt; is a basis of &lt;math&gt;A^k(V)&lt;/math&gt; and &lt;math&gt;dim A^k(V) = nCk&lt;/math&gt;<br /> <br /> <br /> If &lt;math&gt;\omega_1,\ldots,\omega_n\in\Omega^1(M)&lt;/math&gt; and &lt;math&gt;w_1 |_p,\ldots,\omega_n |_p&lt;/math&gt; a basis of &lt;math&gt;(T_p M)^*\ \forall p\in M&lt;/math&gt; then any &lt;math&gt;\lambda&lt;/math&gt; can be written as &lt;math&gt;\lambda = \sum_I a_I(p)\omega_I&lt;/math&gt; where &lt;math&gt;a_I:M\rightarrow\mathbb{R}&lt;/math&gt; are smooth. <br /> <br /> <br /> The equivalence of these is left as an exercise. <br /> <br /> <br /> ''Example''<br /> <br /> Let us investigate &lt;math&gt;\Omega^*(\mathbb{R}^3)&lt;/math&gt; (the * just means &quot;anything&quot;).<br /> <br /> <br /> Now, &lt;math&gt;(T_p(\mathbb{R}^3))^* = &lt;dx_1,dx_2,dx_3&gt;&lt;/math&gt; where &lt;math&gt;x_i:\mathbb{R}^3\rightarrow\mathbb{R}&lt;/math&gt; and so &lt;math&gt;dx_i|_p:T_p\mathbb{R}^3\rightarrow T_{x_i(p)}\mathbb{R} =\mathbb{R}&lt;/math&gt;<br /> <br /> <br /> Hence, &lt;math&gt;dx_i|_p\in (T_p\mathbb{R}^3)^*&lt;/math&gt;<br /> <br /> <br /> Now, &lt;math&gt;dx_j(\frac{\partial}{\partial x}) = \frac{\partial}{\partial x_i}x_j =\delta_{ij}&lt;/math&gt; and hence we get a basis. <br /> <br /> <br /> So, &lt;math&gt;\Omega^1(\mathbb{R}^3) = \{g_1 dx_1 + g_2 dx_2 + g_3 dx_3\}\approx\{g_1,g_2,g_3\}\approx&lt;/math&gt; {vector fields on &lt;math&gt;\mathbb{R}^3&lt;/math&gt;}<br /> <br /> <br /> where the &lt;math&gt;g_i:\mathbb{R}^3\rightarrow\mathbb{R}&lt;/math&gt; are smooth.<br /> <br /> <br /> &lt;math&gt;\Omega^0(\mathbb{R}^3) = \{f:\mathbb{R}^3\rightarrow\mathbb{R}\}&lt;/math&gt;<br /> <br /> <br /> This is because to each point p we associate something that takes zero copies of the tangent space into the real numbers. Thus to each p we associate a number.<br /> <br /> <br /> &lt;math&gt;\Omega^3(\mathbb{R}^3) = \{kdx_1\wedge dx_2\wedge dx_3\} \approx&lt;/math&gt; {functions} where again the k is just a smooth function from &lt;math&gt;\mathbb{R}^3&lt;/math&gt; to &lt;math&gt;\mathbb{R}&lt;/math&gt;. <br /> <br /> <br /> &lt;math&gt;\Omega^2(\mathbb{R}^3) = \{h_1 dx_2\wedge dx_3 + h_2 dx_3\wedge dx_1 + h_3 dx_1\wedge dx_2\}\approx \{h_1,h_2,h_3\}\approx&lt;/math&gt; {vector fields}<br /> <br /> <br /> <br /> ''Aside''<br /> <br /> <br /> Recall our earlier discussion of how points and things like points (curves, equivalence classes of curves) pushfoward while things dual to points (functions) pullback and that things dual to functions (such as derivations) push forward. See earlier for the precise definitions. <br /> <br /> <br /> Now differential forms pull back, i.e., for &lt;math&gt;\phi:M\rightarrow N&lt;/math&gt; then &lt;math&gt;\phi^*(\lambda)\in\Omega^k(M)\leftarrow\lambda\in\Omega^k(N)&lt;/math&gt; <br /> via <br /> &lt;math&gt;\phi^*(\lambda)(v_1,\ldots,v_k)=\lambda(\phi_* v_1,\ldots \phi_* v_k)&lt;/math&gt;<br /> <br /> <br /> The pullback preserves all the properties discussed above and is well defined. In particular, it is compatible with the wedge product via &lt;math&gt;\phi^*(\omega\wedge\lambda)=\phi^*(\omega)\wedge\phi^*(\lambda)&lt;/math&gt;<br /> <br /> <br /> '''Theorem-Definition'''<br /> <br /> <br /> Given M, &lt;math&gt;\exists&lt;/math&gt; ! linear map &lt;math&gt;d:\Omega^*(M)\rightarrow\Omega^{k*+1}(M)&lt;/math&gt; satisfies <br /> <br /> 1) If &lt;math&gt;f\in\Omega^0(M)&lt;/math&gt; then &lt;math&gt;df(X) = X(f)&lt;/math&gt; for &lt;math&gt;X\in TM&lt;/math&gt;<br /> <br /> 2) &lt;math&gt;d^2 = 0&lt;/math&gt;. I.e. if &lt;math&gt;d_k:\Omega^k\rightarrow\Omega^{k+1}&lt;/math&gt; and &lt;math&gt;d_{k+1}:\Omega^{k+1}\rightarrow\Omega^{k+2}&lt;/math&gt; then &lt;math&gt;d_{k+1}\circ d_k = 0&lt;/math&gt;. <br /> <br /> 3) &lt;math&gt;d(\omega\wedge\lambda) = d\omega\wedge\lambda + (-1)^{deg\omega}\omega\wedge d\lambda&lt;/math&gt;<br /> <br /> ===Second Hour===<br /> <br /> Some notes about the above definition:<br /> <br /> 1) When we restrict our ''d'' to functions we just get the old meaning for d. <br /> <br /> 2) Philosophically, there is a duality between differential forms and manifolds and that duality is given by integration. In this duality, d is the adjoint of the boundary operator on manifolds. For manifolds, the boundary of the boundary is empty and hence it is reasonable that &lt;math&gt;d^2=0&lt;/math&gt; on differential forms. <br /> <br /> 3) To remember the formula in 3 given above and others like it, it helps to keep in mind what objects are &quot;odd&quot; and what are &quot;even&quot; and thus when commuting such operators we will get the signs as you would expect from multiplying objects that are either odd or even. <br /> <br /> <br /> ''Example''<br /> <br /> Let us aim for a formula for d on &lt;math&gt;\Omega^*(\mathbb{R}^n)&lt;/math&gt;. <br /> <br /> Lets compute &lt;math&gt;d(\sum_I f_I dx_I)&lt;/math&gt; where &lt;math&gt;dx_I = dx_{i_1}\wedge\ldots\wedge dx_{i_k}&lt;/math&gt; and &lt;math&gt;I = (i_1,\ldots,i_k)&lt;/math&gt;<br /> <br /> Then, &lt;math&gt;d(\sum_I f_I dx_I) = \sum_I d(f_I \wedge dx_{i_1}\wedge\ldots\wedge dx_{i_k}) = \sum_I df_I\wedge(dx_I) + f_I\wedge d(dx_I)&lt;/math&gt;<br /> <br /> The last term vanishes because of (2) in the theorem (proving uniqueness!) <br /> <br /> <br /> Now, as an aside, we claim that for &lt;math&gt;f\in\Omega^0(M),\ df = \sum_{j=1}^{n}\frac{\partial f}{\partial x_j} dx_j&lt;/math&gt;<br /> <br /> Indeed, we know &lt;math&gt;(df)(\frac{\partial}{\partial x_i}) = \frac{\partial}{\partial x_i}f&lt;/math&gt; <br /> <br /> However, &lt;math&gt;(\sum_{j=1}^{n}\frac{\partial f}{\partial x_j} dx_j)(\frac{\partial}{\partial x_i}) = \sum_j \frac{\partial f}{\partial x_j}\delta_{ij} = \frac{\partial f}{\partial x_i}&lt;/math&gt; which is the same. <br /> <br /> <br /> Returning, we thus get &lt;math&gt;d(\sum_I f_I dx_I) = \sum_{I,j} \frac{\partial f_I}{\partial x_j} dx_j\wedge dx_I&lt;/math&gt; <br /> <br /> Thus our d takes functions to vector fields by &lt;math&gt;f\mapsto (\frac{\partial f}{\partial x_1},\frac{\partial f}{\partial x_2},\frac{\partial f}{\partial x_3})&lt;/math&gt;<br /> <br /> This is just the grad operator from calculus and we can see that the d operator appropriately takes things from &lt;math&gt;\Omega^0(M)&lt;/math&gt; to &lt;math&gt;\Omega^1(M)&lt;/math&gt;. <br /> <br /> <br /> Now let us compute &lt;math&gt;d(h_1 dx_2\wedge dx_3+h_2 dx_3\wedge dx_1+h_3 dx_1\wedge dx_2) =\frac{\partial h_1}{\partial x_1} dx_1\wedge dx_2\wedge dx_3 + \frac{\partial h_1}{\partial x_2} dx_3\wedge dx_2\wedge dx_3 + \frac{\partial h_1}{\partial x_3} dx_3\wedge dx_2\wedge dx_3&lt;/math&gt; + 6 more terms representing the 3 partials of each of the last 2 terms. <br /> <br /> As each &lt;math&gt;dx_i\wedge dx_i&lt;/math&gt; term vanishes we are left with just, <br /> <br /> &lt;math&gt;= (\frac{\partial h_1}{\partial x_1} + \frac{\partial h_2}{\partial x_2} + \frac{\partial h_3}{\partial x_3})dx_1\wedge dx_2\wedge dx_3&lt;/math&gt;<br /> <br /> I.e., d takes &lt;math&gt;(h_1,h_2,h_3)\mapsto \sum_i\frac{\partial h_i}{\partial x_i}&lt;/math&gt; <br /> <br /> this is just the div operator from calculus and appropriately takes vector fields to functions and represents the d from &lt;math&gt;\Omega^2(M)&lt;/math&gt; to &lt;math&gt;\Omega^3(M)&lt;/math&gt;. <br /> <br /> <br /> We are left with computing d from &lt;math&gt;\Omega^1(M)&lt;/math&gt; to &lt;math&gt;\Omega^2(M)&lt;/math&gt;<br /> <br /> Computing, &lt;math&gt;d(g_1 dx_1 + g_2 dx_2 + g_3 dx_3) = (\frac{\partial g_3}{\partial x_2} - \frac{\partial g_2}{\partial x_3})dx_2\wedge dx_3 + (\frac{\partial g_1}{\partial x_3} - \frac{\partial g_3}{\partial x_1})dx_3\wedge dx_1 + (\frac{\partial g_2}{\partial x_1} - \frac{\partial g_1}{\partial x_2})dx_1\wedge dx_2&lt;/math&gt;<br /> <br /> I.e., we just have the curl operator. <br /> <br /> Note that the well known calculus laws that curl grad = 0 and div curl = 0 are just the expression that &lt;math&gt;d^2 =0&lt;/math&gt;. <br /> <br /> To provide some physical insight to the meanings of these operators:<br /> <br /> 1) The gradient represents the direction of maximum descent. I.e. if you had a function on the plane the graph would look like the surface of a mountain range and the direction that water would run would be the gradient. <br /> <br /> 2) In a say compressible fluid, the divergence corresponds to the difference between in the inflow and outflow of fluid in some small epsilon box around a point. <br /> <br /> 3) The curl corresponds to the rotation vector for a ball. Ie consider a ball (of equal density to the liquid about it) going down a river. In the x_2, x_1 plane the tenancy for it to rotate clockwise would be given by &lt;math&gt;\frac{\partial g_2}{\partial x_1} - \frac{\partial g_1}{\partial x_2}&lt;/math&gt;</div> Trefor http://drorbn.net/index.php?title=0708-1300/Class_notes_for_Tuesday,_January_15 0708-1300/Class notes for Tuesday, January 15 2008-02-08T23:29:02Z <p>Trefor: /* Second Hour */</p> <hr /> <div>{{0708-1300/Navigation}}<br /> <br /> ==Class Notes==<br /> &lt;span style=&quot;color: red;&quot;&gt;The notes below are by the students and for the students. Hopefully they are useful, but they come with no guarantee of any kind.&lt;/span&gt;<br /> <br /> ===First Hour===<br /> <br /> We begin by reformulating our previous lemma into a more general form:<br /> <br /> <br /> '''Lemma 1'''<br /> <br /> Let &lt;math&gt;p:(X,x_0)\rightarrow(B,b_0)&lt;/math&gt; be a covering map. Then every family of paths &lt;math&gt;\gamma:Y\times I\rightarrow B &lt;/math&gt; such that &lt;math&gt;\forall y\in Y\ \gamma(y,0)=b_0&lt;/math&gt; has a unique lift &lt;math&gt;\tilde{\gamma}:Y\times I\rightarrow X&lt;/math&gt; such that <br /> <br /> 0) &lt;math&gt;\forall y\ \tilde{\gamma}(y,0) = x_0&lt;/math&gt;<br /> <br /> 1) &lt;math&gt;\gamma = p\circ\tilde{\gamma}&lt;/math&gt;<br /> <br /> <br /> Claim 1: &lt;math&gt;Ind[\gamma]&lt;/math&gt; is well defined, hence &lt;math&gt;\pi_1(S^1)\cong\mathbb{Z}&lt;/math&gt;<br /> &lt;math&gt;(\gamma:I\rightarrow S^1, \gamma(0)=1, ind [\gamma] = \tilde{\gamma}(1))&lt;/math&gt;<br /> <br /> <br /> ''Proof of Claim 1''<br /> <br /> We construct the homotopy H between two such gamma's which, schematically, is a square with &lt;math&gt;\gamma_1&lt;/math&gt; along the bottom, &lt;math&gt;\gamma_0&lt;/math&gt; along the top and along the side the parameter Y where the homotopy is just horizontal lines between &lt;math&gt;\gamma_1&lt;/math&gt; and &lt;math&gt;\gamma_0.&lt;/math&gt; <br /> <br /> Then, the lemma implies the existence of a homotopy &lt;math&gt;\tilde{H}&lt;/math&gt; which is schematically a square with &lt;math&gt;\tilde{\gamma_1}&lt;/math&gt; along the bottom, &lt;math&gt;\tilde{\gamma_0}&lt;/math&gt; along the top, x=0 on the left and side and the right hand side taking values in &lt;math&gt;p^{-1}(1)&lt;/math&gt;<br /> <br /> <br /> ''Proof of Lemma 1''<br /> <br /> Recall the proof for the version of the lemma when we were dealing with a single path, not a family. We now just need to extend this proof to the case of having a family of paths. We know that for each &lt;math&gt;y\in Y&lt;/math&gt; we can get a &lt;math&gt;\tilde{\gamma}&lt;/math&gt;<br /> <br /> We need to show that in fact the result is continuous. As continuity is a local property, we simply need to prove this in a neighborhood about &lt;math&gt;y_0&lt;/math&gt;. <br /> <br /> Now by a &quot;good&quot; open set in B, we mean that the preimage under p CAN be written as a product. <br /> <br /> Hence, we choose a neighborhood about the first interval in &lt;math&gt;\mathbb{R}&lt;/math&gt; extending from &lt;math&gt;y_0&lt;/math&gt; (see proof of one path case for explanation) and this gets mapped to a &quot;good&quot; open set in B. As it is the image of an interval, it is compact in an open set, so can put a small neighborhood about the interval such that the image of the neighborhood is in the &quot;good&quot; open set in B. <br /> <br /> Then, duplicating the proof of the earlier version of the lemma, this establishes continuity in the neighborhood.<br /> <br /> <br /> '''Applications of &lt;math&gt;\pi_1(S^1)\cong\mathbb{Z}&lt;/math&gt;'''<br /> <br /> <br /> 1) We get again the proof of non existence of a retract &lt;math&gt;r:D^2\rightarrow S^1&lt;/math&gt;<br /> <br /> Indeed, assume we DID have such a retract. We would have the following commuting diagram. <br /> <br /> &lt;math&gt;\begin{matrix}<br /> D^2&amp;\rightarrow^{r}&amp;S^1\\<br /> \uparrow &amp;\nearrow_{I}&amp; \\<br /> S^1&amp;&amp;\\<br /> \end{matrix}&lt;/math&gt;<br /> <br /> Applying the functor &lt;math&gt;\pi_1&lt;/math&gt; would yield the diagram<br /> <br /> &lt;math&gt;\begin{matrix}<br /> \{0\}&amp;\rightarrow^{r}&amp;\mathbb{Z}\\<br /> \uparrow &amp;\nearrow_{I}&amp; \\<br /> \mathbb{Z}&amp;&amp;\\<br /> \end{matrix}&lt;/math&gt;<br /> <br /> But clearly this does not exist, as the only map from &lt;math&gt;\{0\}&lt;/math&gt; is the trivial one. <br /> <br /> <br /> Recall the non existence of such retracts implies ''Brouwer's Theorem''<br /> <br /> <br /> 2) '''Fundamental Theorem of Algebra'''<br /> <br /> <br /> if &lt;math&gt;p\in\mathbb{C}[z]&lt;/math&gt; is a polynomial with degree greater than zero then &lt;math&gt;\exists z_0\in\mathbb{C}&lt;/math&gt; such that &lt;math&gt;p(z_0) =0&lt;/math&gt;<br /> <br /> <br /> ''Proof:''<br /> <br /> Suppose not, i.e., there exists a polynomial p that has no roots. dividing by the coefficient of the highest order term, &lt;math&gt;p = z^n +&lt;/math&gt; lower order terms. <br /> <br /> <br /> Define a homotopy of paths &lt;math&gt;S^1\rightarrow S^1&lt;/math&gt; by first setting &lt;math&gt;q(z) = p(z)/||p(z)||]\in S^1&lt;/math&gt;<br /> <br /> Now the homotopy starts with a base of q on the point 0. During the first half of the homotopy we grow the size of the loop on which q is acting. At the halfway point, the loop is so large such that the first term in the polynomial dominates the lower order terms. Then in the second half of the homotopy, the lower order terms are shut off. More precisely we consider the polynomial &lt;math&gt;p_{\alpha}(z) = z^n + \alpha&lt;/math&gt;(lower order terms) and let &lt;math&gt;\alpha&lt;/math&gt; tend from 1 to 0 in the second half of the homotopy. <br /> <br /> Hence, at the end of the homotopy we are left with &lt;math&gt;z^n/||z^n||&lt;/math&gt; on a large loop. This is the generator we had previously, and is not the identity. Hence, we have a constant homotopic to something non trivial and this establishes the contradiction. <br /> <br /> <br /> 3) '''Boruk-Ulam'''<br /> <br /> If &lt;math&gt;f:S^2\rightarrow \mathbb{R}^2&lt;/math&gt; is continuous then &lt;math&gt;\exists p\in S^2&lt;/math&gt; such that &lt;math&gt;f(p) = f(-p)&lt;/math&gt;<br /> <br /> <br /> Corollary: <br /> <br /> 1) &lt;math&gt;S^2&lt;/math&gt; is not a subset of &lt;math&gt;\mathbb{R}^2&lt;/math&gt; (cannot be embedded) as f is not 1:1<br /> <br /> 2) If &lt;math&gt;S^2 =A_1\cup A_2\cup A_3&lt;/math&gt;, with &lt;math&gt;A_i&lt;/math&gt; closed then at least one &lt;math&gt;A_i&lt;/math&gt; contains a pair of antipodes.<br /> <br /> <br /> '''Example:'''<br /> <br /> This is NOT true with 4 sections. For instance consider a triangular based pyramid inscribed inside a sphere with each face a different colour. Imagine a light bulb at the center of the sphere so that on the sphere we get 4 sections with 4 different colours from each side of the pyramid. Clearly none of them contains a pair of antipodes, yet are closed with a union of the whole sphere.<br /> <br /> ===Second Hour===<br /> <br /> ''Proof of Corollary:''<br /> <br /> Consider &lt;math&gt;f:S^2\rightarrow\mathbb{R}^2&lt;/math&gt; by &lt;math&gt;p\mapsto&lt;/math&gt; (dist to &lt;math&gt;A_1&lt;/math&gt;, dist to &lt;math&gt;A_2&lt;/math&gt;)<br /> <br /> If f(p) = f(-p) then, for both possible cases of f(p) being zero or positive, we get that p and -p are in the same &lt;math&gt;A_i&lt;/math&gt;. <br /> <br /> <br /> '''Definition:'''<br /> <br /> 1) &lt;math&gt;\gamma:S^1\rightarrow S^1&lt;/math&gt; is even if &lt;math&gt;\gamma(-x) = \gamma(x)\ \forall x&lt;/math&gt;<br /> <br /> 2) &lt;math&gt;\gamma&lt;/math&gt; is odd if &lt;math&gt;\gamma(-x) = -\gamma(x)\ \forall x&lt;/math&gt;<br /> <br /> <br /> '''Lemma:'''<br /> <br /> 1) If &lt;math&gt;\gamma&lt;/math&gt; is even then ind &lt;math&gt;\gamma&lt;/math&gt; is even<br /> <br /> 2) If &lt;math&gt;\gamma&lt;/math&gt; is odd then ind &lt;math&gt;\gamma&lt;/math&gt; is odd<br /> <br /> <br /> ''Proof of Borsuk-Ulam:''<br /> <br /> <br /> Assume &lt;math&gt;f:S^2\rightarrow\mathbb{R}^2&lt;/math&gt; has no p with f(p)=f(-p)<br /> <br /> Define &lt;math&gt;g:S^2\rightarrow S^1&lt;/math&gt; by &lt;math&gt;g(p) = (f(p)-f(-p))/||f(p)-f(-p)||&lt;/math&gt;<br /> <br /> &lt;math&gt;\gamma = g|_{equator}:S^1\rightarrow S^1&lt;/math&gt; is odd. Therefore, &lt;math&gt;[\gamma]\in\pi_1(S^1)&lt;/math&gt; is non zero. <br /> <br /> But &lt;math&gt;\gamma&lt;/math&gt; is homotopic to zero, a contradiction (the homotopy is a series of circles of decreasing radius where a point on the equator is fixed and its antipodal point is moved in an arc towards it)<br /> <br /> <br /> ''Proof of part 1 of lemma:''<br /> <br /> <br /> Suppose &lt;math&gt;\gamma:S^1\rightarrow S^1&lt;/math&gt; is even<br /> <br /> <br /> We can make a commuting diagram where &lt;math&gt;\gamma:S^1\rightarrow S^1&lt;/math&gt; is the same as going from &lt;math&gt;S^1\rightarrow S^1&lt;/math&gt; first via &lt;math&gt;z\mapsto z^2&lt;/math&gt; and then from &lt;math&gt;S^1\rightarrow S^1&lt;/math&gt; via &lt;math&gt;\lambda(z) = \gamma(\sqrt{z})&lt;/math&gt;<br /> <br /> <br /> We now consider this diagram under the functor &lt;math&gt;\pi_1&lt;/math&gt;<br /> <br /> We get &lt;math&gt;\mathbb{Z}\rightarrow^{\times 2}\mathbb{Z}\rightarrow^{\lambda_*}\mathbb{Z}&lt;/math&gt;<br /> <br /> which commutes with &lt;math&gt;\mathbb{Z}\rightarrow^{\gamma_*}\mathbb{Z}&lt;/math&gt; along the bottom<br /> <br /> &lt;math&gt;[\gamma] = \gamma_*[\gamma_1] = \lambda_*(2[\gamma_1])= 2\lambda_*([\gamma_1])&lt;/math&gt; which is even. <br /> <br /> (Note, by &lt;math&gt;\gamma_1:S^1\rightarrow S^1&lt;/math&gt; we mean the identity)<br /> <br /> <br /> Unfortunately, there is no simple adaption of this proof to deal with the odd case and so we introduce some more machinery first. <br /> <br /> <br /> '''Definition'''<br /> <br /> A ''topological group'' is a topological space G with a group structure such that all group operations are continuous<br /> <br /> <br /> '''Examples:'''<br /> <br /> 1) &lt;math&gt;S^1&lt;/math&gt; (with rotation giving the group structure)<br /> <br /> 2) SO(3) (See a previous lecture for more info)<br /> <br /> <br /> If G is a topological group there are two &quot;products&quot; we can define on &lt;math&gt;\pi_1(G)&lt;/math&gt;:<br /> <br /> <br /> 1) &lt;math&gt;[\gamma][\lambda] = [\gamma\circ\lambda]&lt;/math&gt; with &lt;math&gt;\circ&lt;/math&gt; being simply concatenation<br /> <br /> 2) &lt;math&gt;[\gamma]*[\lambda] = [\gamma *\lambda]&lt;/math&gt; where &lt;math&gt;(\gamma*\lambda)(t) = \gamma(t)\cdot\lambda(t)&lt;/math&gt;<br /> <br /> <br /> Claim:<br /> <br /> These two are the same and, in fact, are commutative. I will not attempt to describe the homotopy schematic here. <br /> <br /> <br /> We return to the proof of the odd case of the lemma:<br /> <br /> <br /> Assume &lt;math&gt;\gamma&lt;/math&gt; is odd. Then &lt;math&gt;\gamma*\gamma_1&lt;/math&gt; is even so &lt;math&gt;2\mathbb{Z}\in[\gamma*\gamma_1]=[\gamma\circ\gamma_1] = [\gamma] + [\gamma_1] = [\gamma] + 1&lt;/math&gt;. Therefore, &lt;math&gt;[\gamma]&lt;/math&gt; is odd. <br /> <br /> Note: The addition above is done in &lt;math&gt;\pi_1(S^1)&lt;/math&gt; which is just &lt;math&gt;\mathbb{Z}&lt;/math&gt;<br /> <br /> ''Q.E.D.''<br /> <br /> <br /> '''Claim:'''<br /> <br /> If &lt;math&gt;f,g:X\rightarrow Y&lt;/math&gt; are homotopic then &lt;math&gt;f_* = g_*:\pi_1(X)\rightarrow\pi_1(Y)&lt;/math&gt;<br /> <br /> <br /> Consider a loop &lt;math&gt;\gamma&lt;/math&gt; in X. Consider the images of this in Y under f and g. They look like two loops with the same base point and the homotopy H collapses them down to the same loop. <br /> <br /> Then, &lt;math&gt;f\circ\gamma&lt;/math&gt; is homotopic to &lt;math&gt;g\circ\gamma&lt;/math&gt; using &lt;math&gt;H\circ\gamma&lt;/math&gt; as the homotopy. <br /> <br /> <br /> '''Theorem'''<br /> <br /> 1) &lt;math&gt;\sim&lt;/math&gt; is an equivalence relation. <br /> <br /> 2) It is an ''ideal'' in the category of topological spaces. That is, &lt;math&gt;f\sim g \Rightarrow f\circ h\sim g\circ h&lt;/math&gt; and &lt;math&gt;h\circ f \sim h\circ g&lt;/math&gt; for h's such that these make sense. <br /> <br /> <br /> '''Definition'''<br /> <br /> X and Y are &quot;homotopy equivalent&quot; if they are isomorphic in {Topological Spaces} / {homotopy of maps}<br /> <br /> I.e., &lt;math&gt;\exists f,g&lt;/math&gt; so that &lt;math&gt;f\circ g\sim I_Y&lt;/math&gt; and &lt;math&gt;g\circ f\sim I_X&lt;/math&gt; for &lt;math&gt;f:X\rightarrow Y&lt;/math&gt; and &lt;math&gt;g:Y\rightarrow X&lt;/math&gt;<br /> <br /> <br /> '''Example'''<br /> <br /> 1) &lt;math&gt;\mathbb{R}^n&lt;/math&gt; is homotopy equivalent to a point via f which takes &lt;math&gt;\mathbb{R}^n&lt;/math&gt; to a point and g is the zero map<br /> <br /> 2) An annulus is homotopy equivalent to &lt;math&gt;S^1&lt;/math&gt;. Indeed, consider the map f which collapses the annulus to the circle in its middle, and g is just the identity. <br /> <br /> 3) Likewise for the figure &quot;thick A&quot; which is a subset of the plane in the shape of an A with everything being thick. There is a circle around the hole in the A. The map f just takes A to this circle and g is the identity. <br /> <br /> 4) The mobius band is also equivalent to a circle where f just collapses to the boundary circle. <br /> <br /> <br /> This last example gives us some idea of the limitation, or &quot;lack of subtlety&quot; to our invariants given that we would all agree there is something fundamentally different between the mobius band, the annulus and the circle, homotopy equivalence is not sensitive to this difference. <br /> <br /> <br /> '''Claim'''<br /> <br /> If X and Y are homotopically equivalent then &lt;math&gt;\pi_1(X)\cong\pi_1(Y)&lt;/math&gt;<br /> <br /> ''Proof:''<br /> <br /> Consider &lt;math&gt;X\rightarrow^f Y&lt;/math&gt; and &lt;math&gt;Y\rightarrow^g X&lt;/math&gt; forming a commuting diagram and lets consider its image under &lt;math&gt;\pi_1&lt;/math&gt;<br /> <br /> Then we get &lt;math&gt;\pi_1(X)\rightarrow^{f_*} \pi_1(Y)&lt;/math&gt; and &lt;math&gt;\pi_1 (Y)\rightarrow^{g_*} \pi_1(X&lt;/math&gt;) as a commuting diagram. <br /> <br /> We know that &lt;math&gt;f\circ g&lt;/math&gt; is homotopic to the identity, and thus we also get &lt;math&gt;f_*\circ g_*&lt;/math&gt; homotopic to the identity<br /> <br /> ''Q.E.D.''<br /> <br /> <br /> '''Theorem''' <br /> <br /> &lt;math&gt;\pi_1(X\times Y)\cong\pi_1(X)\times\pi_1(Y)&lt;/math&gt;<br /> <br /> <br /> Example: &lt;math&gt;\pi_1(\mathbb{T}^2) = \pi_1(S^1\times S^1) = \mathbb{Z}\times\mathbb{Z} = \mathbb{Z}^2&lt;/math&gt;</div> Trefor http://drorbn.net/index.php?title=0708-1300/Class_notes_for_Tuesday,_January_15 0708-1300/Class notes for Tuesday, January 15 2008-02-08T23:27:01Z <p>Trefor: /* Second Hour */</p> <hr /> <div>{{0708-1300/Navigation}}<br /> <br /> ==Class Notes==<br /> &lt;span style=&quot;color: red;&quot;&gt;The notes below are by the students and for the students. Hopefully they are useful, but they come with no guarantee of any kind.&lt;/span&gt;<br /> <br /> ===First Hour===<br /> <br /> We begin by reformulating our previous lemma into a more general form:<br /> <br /> <br /> '''Lemma 1'''<br /> <br /> Let &lt;math&gt;p:(X,x_0)\rightarrow(B,b_0)&lt;/math&gt; be a covering map. Then every family of paths &lt;math&gt;\gamma:Y\times I\rightarrow B &lt;/math&gt; such that &lt;math&gt;\forall y\in Y\ \gamma(y,0)=b_0&lt;/math&gt; has a unique lift &lt;math&gt;\tilde{\gamma}:Y\times I\rightarrow X&lt;/math&gt; such that <br /> <br /> 0) &lt;math&gt;\forall y\ \tilde{\gamma}(y,0) = x_0&lt;/math&gt;<br /> <br /> 1) &lt;math&gt;\gamma = p\circ\tilde{\gamma}&lt;/math&gt;<br /> <br /> <br /> Claim 1: &lt;math&gt;Ind[\gamma]&lt;/math&gt; is well defined, hence &lt;math&gt;\pi_1(S^1)\cong\mathbb{Z}&lt;/math&gt;<br /> &lt;math&gt;(\gamma:I\rightarrow S^1, \gamma(0)=1, ind [\gamma] = \tilde{\gamma}(1))&lt;/math&gt;<br /> <br /> <br /> ''Proof of Claim 1''<br /> <br /> We construct the homotopy H between two such gamma's which, schematically, is a square with &lt;math&gt;\gamma_1&lt;/math&gt; along the bottom, &lt;math&gt;\gamma_0&lt;/math&gt; along the top and along the side the parameter Y where the homotopy is just horizontal lines between &lt;math&gt;\gamma_1&lt;/math&gt; and &lt;math&gt;\gamma_0.&lt;/math&gt; <br /> <br /> Then, the lemma implies the existence of a homotopy &lt;math&gt;\tilde{H}&lt;/math&gt; which is schematically a square with &lt;math&gt;\tilde{\gamma_1}&lt;/math&gt; along the bottom, &lt;math&gt;\tilde{\gamma_0}&lt;/math&gt; along the top, x=0 on the left and side and the right hand side taking values in &lt;math&gt;p^{-1}(1)&lt;/math&gt;<br /> <br /> <br /> ''Proof of Lemma 1''<br /> <br /> Recall the proof for the version of the lemma when we were dealing with a single path, not a family. We now just need to extend this proof to the case of having a family of paths. We know that for each &lt;math&gt;y\in Y&lt;/math&gt; we can get a &lt;math&gt;\tilde{\gamma}&lt;/math&gt;<br /> <br /> We need to show that in fact the result is continuous. As continuity is a local property, we simply need to prove this in a neighborhood about &lt;math&gt;y_0&lt;/math&gt;. <br /> <br /> Now by a &quot;good&quot; open set in B, we mean that the preimage under p CAN be written as a product. <br /> <br /> Hence, we choose a neighborhood about the first interval in &lt;math&gt;\mathbb{R}&lt;/math&gt; extending from &lt;math&gt;y_0&lt;/math&gt; (see proof of one path case for explanation) and this gets mapped to a &quot;good&quot; open set in B. As it is the image of an interval, it is compact in an open set, so can put a small neighborhood about the interval such that the image of the neighborhood is in the &quot;good&quot; open set in B. <br /> <br /> Then, duplicating the proof of the earlier version of the lemma, this establishes continuity in the neighborhood.<br /> <br /> <br /> '''Applications of &lt;math&gt;\pi_1(S^1)\cong\mathbb{Z}&lt;/math&gt;'''<br /> <br /> <br /> 1) We get again the proof of non existence of a retract &lt;math&gt;r:D^2\rightarrow S^1&lt;/math&gt;<br /> <br /> Indeed, assume we DID have such a retract. We would have the following commuting diagram. <br /> <br /> &lt;math&gt;\begin{matrix}<br /> D^2&amp;\rightarrow^{r}&amp;S^1\\<br /> \uparrow &amp;\nearrow_{I}&amp; \\<br /> S^1&amp;&amp;\\<br /> \end{matrix}&lt;/math&gt;<br /> <br /> Applying the functor &lt;math&gt;\pi_1&lt;/math&gt; would yield the diagram<br /> <br /> &lt;math&gt;\begin{matrix}<br /> \{0\}&amp;\rightarrow^{r}&amp;\mathbb{Z}\\<br /> \uparrow &amp;\nearrow_{I}&amp; \\<br /> \mathbb{Z}&amp;&amp;\\<br /> \end{matrix}&lt;/math&gt;<br /> <br /> But clearly this does not exist, as the only map from &lt;math&gt;\{0\}&lt;/math&gt; is the trivial one. <br /> <br /> <br /> Recall the non existence of such retracts implies ''Brouwer's Theorem''<br /> <br /> <br /> 2) '''Fundamental Theorem of Algebra'''<br /> <br /> <br /> if &lt;math&gt;p\in\mathbb{C}[z]&lt;/math&gt; is a polynomial with degree greater than zero then &lt;math&gt;\exists z_0\in\mathbb{C}&lt;/math&gt; such that &lt;math&gt;p(z_0) =0&lt;/math&gt;<br /> <br /> <br /> ''Proof:''<br /> <br /> Suppose not, i.e., there exists a polynomial p that has no roots. dividing by the coefficient of the highest order term, &lt;math&gt;p = z^n +&lt;/math&gt; lower order terms. <br /> <br /> <br /> Define a homotopy of paths &lt;math&gt;S^1\rightarrow S^1&lt;/math&gt; by first setting &lt;math&gt;q(z) = p(z)/||p(z)||]\in S^1&lt;/math&gt;<br /> <br /> Now the homotopy starts with a base of q on the point 0. During the first half of the homotopy we grow the size of the loop on which q is acting. At the halfway point, the loop is so large such that the first term in the polynomial dominates the lower order terms. Then in the second half of the homotopy, the lower order terms are shut off. More precisely we consider the polynomial &lt;math&gt;p_{\alpha}(z) = z^n + \alpha&lt;/math&gt;(lower order terms) and let &lt;math&gt;\alpha&lt;/math&gt; tend from 1 to 0 in the second half of the homotopy. <br /> <br /> Hence, at the end of the homotopy we are left with &lt;math&gt;z^n/||z^n||&lt;/math&gt; on a large loop. This is the generator we had previously, and is not the identity. Hence, we have a constant homotopic to something non trivial and this establishes the contradiction. <br /> <br /> <br /> 3) '''Boruk-Ulam'''<br /> <br /> If &lt;math&gt;f:S^2\rightarrow \mathbb{R}^2&lt;/math&gt; is continuous then &lt;math&gt;\exists p\in S^2&lt;/math&gt; such that &lt;math&gt;f(p) = f(-p)&lt;/math&gt;<br /> <br /> <br /> Corollary: <br /> <br /> 1) &lt;math&gt;S^2&lt;/math&gt; is not a subset of &lt;math&gt;\mathbb{R}^2&lt;/math&gt; (cannot be embedded) as f is not 1:1<br /> <br /> 2) If &lt;math&gt;S^2 =A_1\cup A_2\cup A_3&lt;/math&gt;, with &lt;math&gt;A_i&lt;/math&gt; closed then at least one &lt;math&gt;A_i&lt;/math&gt; contains a pair of antipodes.<br /> <br /> <br /> '''Example:'''<br /> <br /> This is NOT true with 4 sections. For instance consider a triangular based pyramid inscribed inside a sphere with each face a different colour. Imagine a light bulb at the center of the sphere so that on the sphere we get 4 sections with 4 different colours from each side of the pyramid. Clearly none of them contains a pair of antipodes, yet are closed with a union of the whole sphere.<br /> <br /> ===Second Hour===<br /> <br /> ''Proof of Corollary:''<br /> <br /> Consider &lt;math&gt;f:S^2\rightarrow\mathbb{R}^2&lt;/math&gt; by &lt;math&gt;p\mapsto&lt;/math&gt; (dist to &lt;math&gt;A_1&lt;/math&gt;, dist to &lt;math&gt;A_2&lt;/math&gt;)<br /> <br /> If f(p) = f(-p) then, for both possible cases of f(p) being zero or positive, we get that p and -p are in the same &lt;math&gt;A_i&lt;/math&gt;. <br /> <br /> <br /> '''Definition:'''<br /> <br /> 1) &lt;math&gt;\gamma:S^1\rightarrow S^1&lt;/math&gt; is even if &lt;math&gt;\gamma(-x) = \gamma(x)\ \forall x&lt;/math&gt;<br /> <br /> 2) &lt;math&gt;\gamma&lt;/math&gt; is odd if &lt;math&gt;\gamma(-x) = -\gamma(x)\ \forall x&lt;/math&gt;<br /> <br /> <br /> '''Lemma:'''<br /> <br /> 1) If &lt;math&gt;\gamma&lt;/math&gt; is even then ind &lt;math&gt;\gamma&lt;/math&gt; is even<br /> <br /> 2) If &lt;math&gt;\gamma&lt;/math&gt; is odd then ind &lt;math&gt;\gamma&lt;/math&gt; is odd<br /> <br /> <br /> ''Proof of Borsuk-Ulam:''<br /> <br /> <br /> Assume &lt;math&gt;f:S^2\rightarrow\mathbb{R}^2&lt;/math&gt; has no p with f(p)=f(-p)<br /> <br /> Define &lt;math&gt;g:S^2\rightarrow S^1&lt;/math&gt; by &lt;math&gt;g(p) = (f(p)-f(-p))/||f(p)-f(-p)||&lt;/math&gt;<br /> <br /> &lt;math&gt;\gamma = g|_{equator}:S^1\rightarrow S^1&lt;/math&gt; is odd. Therefore, &lt;math&gt;[\gamma]\in\pi_1(S^1)&lt;/math&gt; is non zero. <br /> <br /> But &lt;math&gt;\gamma&lt;/math&gt; is homotopic to zero, a contradiction (the homotopy is a series of circles of decreasing radius where a point on the equator is fixed and its antipodal point is moved in an arc towards it)<br /> <br /> <br /> ''Proof of part 1 of lemma:''<br /> <br /> <br /> Suppose &lt;math&gt;\gamma:S^1\rightarrow S^1&lt;/math&gt; is even<br /> <br /> <br /> We can make a commuting diagram where &lt;math&gt;\gamma:S^1\rightarrow S^1&lt;/math&gt; is the same as going from &lt;math&gt;S^1\rightarrow S^1&lt;/math&gt; first via &lt;math&gt;z\mapsto z^2&lt;/math&gt; and then from &lt;math&gt;S^1\rightarrow S^1&lt;/math&gt; via &lt;math&gt;\lambda(z) = \gamma(\sqrt{z})&lt;/math&gt;<br /> <br /> <br /> We now consider this diagram under the functor &lt;math&gt;\pi_1&lt;/math&gt;<br /> <br /> We get &lt;math&gt;\mathbb{Z}\rightarrow^{\times 2}\mathbb{Z}\rightarrow^{\lambda_*}\mathbb{Z}&lt;/math&gt;<br /> <br /> which commutes with &lt;math&gt;\mathbb{Z}\rightarrow^{\gamma_*}\mathbb{Z}&lt;/math&gt; along the bottom<br /> <br /> &lt;math&gt;[\gamma] = \gamma_*[\gamma_1] = \lambda_*(2[\gamma_1])= 2\lambda_*([\gamma_1])&lt;/math&gt; which is even. <br /> <br /> (Note, by &lt;math&gt;\gamma_1:S^1\rightarrow S^1&lt;/math&gt; we mean the identity)<br /> <br /> <br /> Unfortunately, there is no simple adaption of this proof to deal with the odd case and so we introduce some more machinery first. <br /> <br /> <br /> '''Definition'''<br /> <br /> A ''topological group'' is a topological space G with a group structure such that all group operations are continuous<br /> <br /> <br /> '''Examples:'''<br /> <br /> 1) &lt;math&gt;S^1&lt;/math&gt; (with rotation giving the group structure)<br /> <br /> 2) SO(3) (See a previous lecture for more info)<br /> <br /> <br /> If G is a topological group there are two &quot;products&quot; we can define on &lt;math&gt;\pi_1(G)&lt;/math&gt;:<br /> <br /> <br /> 1) &lt;math&gt;[\gamma][\lambda] = [\gamma\circ\lambda]&lt;/math&gt; with &lt;math&gt;\circ&lt;/math&gt; being simply concatenation<br /> <br /> 2) &lt;math&gt;[\gamma]*[\lambda] = [\gamma *\lambda]&lt;/math&gt; where &lt;math&gt;(\gamma*\lambda)(t) = \gamma(t)\cdot\lambda(t)&lt;/math&gt;<br /> <br /> <br /> Claim:<br /> <br /> These two are the same and, in fact, are commutative. I will not attempt to describe the homotopy schematic here. <br /> <br /> <br /> We return to the proof of the odd case of the lemma:<br /> <br /> <br /> Assume &lt;math&gt;\gamma&lt;/math&gt; is odd. Then &lt;math&gt;\gamma*\gamma_1&lt;/math&gt; is even so &lt;math&gt;2\mathbb{Z}\in[\gamma*\gamma_1]=[\gamma\circ\gamma_1] = [\gamma] + [\gamma_1] = [\gamma] + 1&lt;/math&gt;. Therefore, &lt;math&gt;[\gamma]&lt;/math&gt; is odd. <br /> <br /> Note: The addition above is done in &lt;math&gt;\pi_1(S^1)&lt;/math&gt; which is just &lt;math&gt;\mathbb{Z}&lt;/math&gt;<br /> <br /> ''Q.E.D.''<br /> <br /> <br /> '''Claim:'''<br /> <br /> If &lt;math&gt;f,g:X\rightarrow Y&lt;/math&gt; are homotopic then &lt;math&gt;f_* = g_*:\pi_1(X)\rightarrow\pi_1(Y)&lt;/math&gt;<br /> <br /> <br /> Consider a loop &lt;math&gt;\gamma&lt;/math&gt; in X. Consider the images of this in Y under f and g. They look like two loops with the same base point and the homotopy H collapses them down to the same loop. <br /> <br /> Then, &lt;math&gt;f\circ\gamma&lt;/math&gt; is homotopic to &lt;math&gt;g\circ\gamma&lt;/math&gt; using &lt;math&gt;H\circ\gamma&lt;/math&gt; as the homotopy. <br /> <br /> <br /> '''Theorem'''<br /> <br /> 1) &lt;math&gt;\sim&lt;/math&gt; is an equivalence relation. <br /> <br /> 2) It is an ''ideal'' in the category of topological spaces. That is, &lt;math&gt;f\sim g \Rightarrow f\circ h\sim g\circ h&lt;/math&gt; and &lt;math&gt;h\circ f \sim h\circ g&lt;/math&gt; for h's such that these make sense. <br /> <br /> <br /> '''Definition'''<br /> <br /> X and Y are &quot;homotopy equivalent&quot; if they are isomorphic in {Topological Spaces} / {homotopy of maps}<br /> <br /> I.e., &lt;math&gt;\exists f,g&lt;/math&gt; so that &lt;math&gt;f\circ g\sim I_Y&lt;/math&gt; and &lt;math&gt;g\circ f\sim I_X&lt;/math&gt; for &lt;math&gt;f:X\rightarrow Y&lt;/math&gt; and &lt;math&gt;g:Y\rightarrow X&lt;/math&gt;<br /> <br /> <br /> '''Example'''<br /> <br /> 1) &lt;math&gt;\mathbb{R}^n&lt;/math&gt; is homotopy equivalent to a point via f which takes &lt;math&gt;\mathbb{R}^n&lt;/math&gt; to a point and g is the zero map<br /> <br /> 2) An annulus is homotopy equivalent to &lt;math&gt;S^1&lt;/math&gt;. Indeed, consider the map f which collapses the annulus to the circle in its middle, and g is just the identity. <br /> <br /> 3) Likewise for the figure &quot;thick A&quot; which is a subset of the plane in the shape of an A with everything being thick. There is a circle around the hole in the A. The map f just takes A to this circle and g is the identity. <br /> <br /> 4) The mobius band is also equivalent to a circle where f just collapses to the boundary circle. <br /> <br /> <br /> This last example gives us some idea of the limitation, or &quot;lack of subtlety&quot; to our invariants given that we would all agree there is something fundamentally different between the mobius band, the annulus and the circle, homotopy equivalence is not sensitive to this difference. <br /> <br /> <br /> '''Claim'''<br /> <br /> If X and Y are homotopically equivalent then &lt;math&gt;\pi_1(X)\cong\pi_1(Y)&lt;/math&gt;<br /> <br /> ''Proof:''<br /> <br /> Consider &lt;math&gt;X\rightarrow^f Y&lt;/math&gt; and &lt;math&gt;Y\rightarrow^g X&lt;/math&gt; forming a commuting diagram and lets consider its image under &lt;math&gt;\pi_1&lt;/math&gt;<br /> <br /> Then we get &lt;math&gt;\pi_1(X)\rightarrow^{f_*} \pi_1(Y)&lt;/math&gt; and &lt;math&gt;\pi_1 (Y)\rightarrow^{g_*} \pi_1(X&lt;/math&gt;) as a commuting diagram. <br /> <br /> We know that &lt;math&gt;f\circ g&lt;/math&gt; is homotopic to the identity, and thus we also get &lt;math&gt;f_*\circ g_*&lt;/math&gt; homotopic to the identity<br /> <br /> ''Q.E.D.''<br /> <br /> <br /> '''Theorem''' <br /> <br /> &lt;math&gt;\pi_1(X+Y)\cong\pi_1(X)+\pi_1(Y)&lt;/math&gt;<br /> <br /> <br /> Example: &lt;math&gt;\pi_1(\mathbb{T}^2) = \pi_1(S^1\times S^1) = \mathbb{Z}\times\mathbb{Z} = \mathbb{Z}^2&lt;/math&gt;</div> Trefor http://drorbn.net/index.php?title=0708-1300/Class_notes_for_Tuesday,_January_15 0708-1300/Class notes for Tuesday, January 15 2008-02-08T22:06:11Z <p>Trefor: /* Second Hour */</p> <hr /> <div>{{0708-1300/Navigation}}<br /> <br /> ==Class Notes==<br /> &lt;span style=&quot;color: red;&quot;&gt;The notes below are by the students and for the students. Hopefully they are useful, but they come with no guarantee of any kind.&lt;/span&gt;<br /> <br /> ===First Hour===<br /> <br /> We begin by reformulating our previous lemma into a more general form:<br /> <br /> <br /> '''Lemma 1'''<br /> <br /> Let &lt;math&gt;p:(X,x_0)\rightarrow(B,b_0)&lt;/math&gt; be a covering map. Then every family of paths &lt;math&gt;\gamma:Y\times I\rightarrow B &lt;/math&gt; such that &lt;math&gt;\forall y\in Y\ \gamma(y,0)=b_0&lt;/math&gt; has a unique lift &lt;math&gt;\tilde{\gamma}:Y\times I\rightarrow X&lt;/math&gt; such that <br /> <br /> 0) &lt;math&gt;\forall y\ \tilde{\gamma}(y,0) = x_0&lt;/math&gt;<br /> <br /> 1) &lt;math&gt;\gamma = p\circ\tilde{\gamma}&lt;/math&gt;<br /> <br /> <br /> Claim 1: &lt;math&gt;Ind[\gamma]&lt;/math&gt; is well defined, hence &lt;math&gt;\pi_1(S^1)\cong\mathbb{Z}&lt;/math&gt;<br /> &lt;math&gt;(\gamma:I\rightarrow S^1, \gamma(0)=1, ind [\gamma] = \tilde{\gamma}(1))&lt;/math&gt;<br /> <br /> <br /> ''Proof of Claim 1''<br /> <br /> We construct the homotopy H between two such gamma's which, schematically, is a square with &lt;math&gt;\gamma_1&lt;/math&gt; along the bottom, &lt;math&gt;\gamma_0&lt;/math&gt; along the top and along the side the parameter Y where the homotopy is just horizontal lines between &lt;math&gt;\gamma_1&lt;/math&gt; and &lt;math&gt;\gamma_0.&lt;/math&gt; <br /> <br /> Then, the lemma implies the existence of a homotopy &lt;math&gt;\tilde{H}&lt;/math&gt; which is schematically a square with &lt;math&gt;\tilde{\gamma_1}&lt;/math&gt; along the bottom, &lt;math&gt;\tilde{\gamma_0}&lt;/math&gt; along the top, x=0 on the left and side and the right hand side taking values in &lt;math&gt;p^{-1}(1)&lt;/math&gt;<br /> <br /> <br /> ''Proof of Lemma 1''<br /> <br /> Recall the proof for the version of the lemma when we were dealing with a single path, not a family. We now just need to extend this proof to the case of having a family of paths. We know that for each &lt;math&gt;y\in Y&lt;/math&gt; we can get a &lt;math&gt;\tilde{\gamma}&lt;/math&gt;<br /> <br /> We need to show that in fact the result is continuous. As continuity is a local property, we simply need to prove this in a neighborhood about &lt;math&gt;y_0&lt;/math&gt;. <br /> <br /> Now by a &quot;good&quot; open set in B, we mean that the preimage under p CAN be written as a product. <br /> <br /> Hence, we choose a neighborhood about the first interval in &lt;math&gt;\mathbb{R}&lt;/math&gt; extending from &lt;math&gt;y_0&lt;/math&gt; (see proof of one path case for explanation) and this gets mapped to a &quot;good&quot; open set in B. As it is the image of an interval, it is compact in an open set, so can put a small neighborhood about the interval such that the image of the neighborhood is in the &quot;good&quot; open set in B. <br /> <br /> Then, duplicating the proof of the earlier version of the lemma, this establishes continuity in the neighborhood.<br /> <br /> <br /> '''Applications of &lt;math&gt;\pi_1(S^1)\cong\mathbb{Z}&lt;/math&gt;'''<br /> <br /> <br /> 1) We get again the proof of non existence of a retract &lt;math&gt;r:D^2\rightarrow S^1&lt;/math&gt;<br /> <br /> Indeed, assume we DID have such a retract. We would have the following commuting diagram. <br /> <br /> &lt;math&gt;\begin{matrix}<br /> D^2&amp;\rightarrow^{r}&amp;S^1\\<br /> \uparrow &amp;\nearrow_{I}&amp; \\<br /> S^1&amp;&amp;\\<br /> \end{matrix}&lt;/math&gt;<br /> <br /> Applying the functor &lt;math&gt;\pi_1&lt;/math&gt; would yield the diagram<br /> <br /> &lt;math&gt;\begin{matrix}<br /> \{0\}&amp;\rightarrow^{r}&amp;\mathbb{Z}\\<br /> \uparrow &amp;\nearrow_{I}&amp; \\<br /> \mathbb{Z}&amp;&amp;\\<br /> \end{matrix}&lt;/math&gt;<br /> <br /> But clearly this does not exist, as the only map from &lt;math&gt;\{0\}&lt;/math&gt; is the trivial one. <br /> <br /> <br /> Recall the non existence of such retracts implies ''Brouwer's Theorem''<br /> <br /> <br /> 2) '''Fundamental Theorem of Algebra'''<br /> <br /> <br /> if &lt;math&gt;p\in\mathbb{C}[z]&lt;/math&gt; is a polynomial with degree greater than zero then &lt;math&gt;\exists z_0\in\mathbb{C}&lt;/math&gt; such that &lt;math&gt;p(z_0) =0&lt;/math&gt;<br /> <br /> <br /> ''Proof:''<br /> <br /> Suppose not, i.e., there exists a polynomial p that has no roots. dividing by the coefficient of the highest order term, &lt;math&gt;p = z^n +&lt;/math&gt; lower order terms. <br /> <br /> <br /> Define a homotopy of paths &lt;math&gt;S^1\rightarrow S^1&lt;/math&gt; by first setting &lt;math&gt;q(z) = p(z)/||p(z)||]\in S^1&lt;/math&gt;<br /> <br /> Now the homotopy starts with a base of q on the point 0. During the first half of the homotopy we grow the size of the loop on which q is acting. At the halfway point, the loop is so large such that the first term in the polynomial dominates the lower order terms. Then in the second half of the homotopy, the lower order terms are shut off. More precisely we consider the polynomial &lt;math&gt;p_{\alpha}(z) = z^n + \alpha&lt;/math&gt;(lower order terms) and let &lt;math&gt;\alpha&lt;/math&gt; tend from 1 to 0 in the second half of the homotopy. <br /> <br /> Hence, at the end of the homotopy we are left with &lt;math&gt;z^n/||z^n||&lt;/math&gt; on a large loop. This is the generator we had previously, and is not the identity. Hence, we have a constant homotopic to something non trivial and this establishes the contradiction. <br /> <br /> <br /> 3) '''Boruk-Ulam'''<br /> <br /> If &lt;math&gt;f:S^2\rightarrow \mathbb{R}^2&lt;/math&gt; is continuous then &lt;math&gt;\exists p\in S^2&lt;/math&gt; such that &lt;math&gt;f(p) = f(-p)&lt;/math&gt;<br /> <br /> <br /> Corollary: <br /> <br /> 1) &lt;math&gt;S^2&lt;/math&gt; is not a subset of &lt;math&gt;\mathbb{R}^2&lt;/math&gt; (cannot be embedded) as f is not 1:1<br /> <br /> 2) If &lt;math&gt;S^2 =A_1\cup A_2\cup A_3&lt;/math&gt;, with &lt;math&gt;A_i&lt;/math&gt; closed then at least one &lt;math&gt;A_i&lt;/math&gt; contains a pair of antipodes.<br /> <br /> <br /> '''Example:'''<br /> <br /> This is NOT true with 4 sections. For instance consider a triangular based pyramid inscribed inside a sphere with each face a different colour. Imagine a light bulb at the center of the sphere so that on the sphere we get 4 sections with 4 different colours from each side of the pyramid. Clearly none of them contains a pair of antipodes, yet are closed with a union of the whole sphere.<br /> <br /> ===Second Hour===<br /> <br /> ''Proof of Corollary:''<br /> <br /> Consider &lt;math&gt;f:S^2\rightarrow\mathbb{R}^2&lt;/math&gt; by &lt;math&gt;p\mapsto&lt;/math&gt; (dist to &lt;math&gt;A_1&lt;/math&gt;, dist to &lt;math&gt;A_2&lt;/math&gt;)<br /> <br /> If f(p) = f(-p) then, for both possible cases of f(p) being zero or positive, we get that p and -p are in the same &lt;math&gt;A_i&lt;/math&gt;. <br /> <br /> <br /> '''Definition:'''<br /> <br /> 1) &lt;math&gt;\gamma:S^1\rightarrow S^1&lt;/math&gt; is even if &lt;math&gt;\gamma(-x) = \gamma(x)\ \forall x&lt;/math&gt;<br /> <br /> 2) &lt;math&gt;\gamma&lt;/math&gt; is odd if &lt;math&gt;\gamma(-x) = -\gamma(x)\ \forall x&lt;/math&gt;<br /> <br /> <br /> '''Lemma:'''<br /> <br /> 1) If &lt;math&gt;\gamma&lt;/math&gt; is even then ind &lt;math&gt;\gamma&lt;/math&gt; is even<br /> <br /> 2) If &lt;math&gt;\gamma&lt;/math&gt; is odd then ind &lt;math&gt;\gamma&lt;/math&gt; is odd<br /> <br /> <br /> ''Proof of Borsuk-Ulam:''<br /> <br /> <br /> Assume &lt;math&gt;f:S^2\rightarrow\mathbb{R}^2&lt;/math&gt; has no p with f(p)=f(-p)<br /> <br /> Define &lt;math&gt;g:S^2\rightarrow S^1&lt;/math&gt; by &lt;math&gt;g(p) = (f(p)-f(-p))/||f(p)-f(-p)||&lt;/math&gt;<br /> <br /> &lt;math&gt;\gamma = g|_{equator}:S^1\rightarrow S^1&lt;/math&gt; is odd. Therefore, &lt;math&gt;[\gamma]\in\pi_1(S^1)&lt;/math&gt; is non zero. <br /> <br /> But &lt;math&gt;\gamma&lt;/math&gt; is homotopic to zero, a contradiction (the homotopy is a series of circles of decreasing radius where a point on the equator is fixed and its antipodal point is moved in an arc towards it)<br /> <br /> <br /> ''Proof of part 1 of lemma:''<br /> <br /> <br /> Suppose &lt;math&gt;\gamma:S^1\rightarrow S^1&lt;/math&gt; is even<br /> <br /> <br /> We can make a commuting diagram where &lt;math&gt;\gamma:S^1\rightarrow S^1&lt;/math&gt; is the same as going from &lt;math&gt;S^1\rightarrow S^1&lt;/math&gt; first via &lt;math&gt;z\mapsto z^2&lt;/math&gt; and then from &lt;math&gt;S^1\rightarrow S^1&lt;/math&gt; via &lt;math&gt;\lambda(z) = \gamma(\sqrt{z})&lt;/math&gt;<br /> <br /> <br /> We now consider this diagram under the functor &lt;math&gt;\pi_1&lt;/math&gt;<br /> <br /> We get &lt;math&gt;\mathbb{Z}\rightarrow^{\times 2}\mathbb{Z}\rightarrow^{\lambda_*}\mathbb{Z}&lt;/math&gt;<br /> <br /> which commutes with &lt;math&gt;\mathbb{Z}\rightarrow^{\gamma_*}\mathbb{Z}&lt;/math&gt; along the bottom<br /> <br /> &lt;math&gt;[\gamma] = \gamma_*[\gamma_1] = \lambda_*(2[\gamma_1])= 2\lambda_*([\gamma_1])&lt;/math&gt; which is even. <br /> <br /> (Note, by &lt;math&gt;\gamma_1:S^1\rightarrow S^1&lt;/math&gt; we mean the identity)<br /> <br /> <br /> Unfortunately, there is no simple adaption of this proof to deal with the odd case and so we introduce some more machinery first. <br /> <br /> <br /> '''Definition'''<br /> <br /> A ''topological group'' is a topological space G with a group structure such that all group operations are continuous<br /> <br /> <br /> '''Examples:'''<br /> <br /> 1) &lt;math&gt;S^1&lt;/math&gt; (with rotation giving the group structure)<br /> <br /> 2) SO(3) (See a previous lecture for more info)<br /> <br /> <br /> If G is a topological group there are two &quot;products&quot; we can define on &lt;math&gt;\pi_1(G)&lt;/math&gt;:<br /> <br /> <br /> 1) &lt;math&gt;[\gamma][\lambda] = [\gamma\circ\lambda]&lt;/math&gt; with &lt;math&gt;\circ&lt;/math&gt; being simply concatenation<br /> <br /> 2) &lt;math&gt;[\gamma]*[\lambda] = [\gamma *\lambda]&lt;/math&gt; where &lt;math&gt;(\gamma*\lambda)(t) = \gamma(t)\cdot\lambda(t)&lt;/math&gt;<br /> <br /> <br /> Claim:<br /> <br /> These two are the same and, in fact, are commutative. I will not attempt to describe the homotopy schematic here. <br /> <br /> <br /> We return to the proof of the odd case of the lemma:<br /> <br /> <br /> Assume &lt;math&gt;\gamma&lt;/math&gt; is odd. Then &lt;math&gt;\gamma*\gamma_1&lt;/math&gt; is even so &lt;math&gt;2\mathbb{Z}\in[\gamma*\gamma_1]=[\gamma\circ\gamma_1] = [\gamma] + [\gamma_1] = [\gamma] + 1&lt;/math&gt;. Therefore, &lt;math&gt;[\gamma]&lt;/math&gt; is odd. <br /> <br /> Note: The addition above is done in &lt;math&gt;\pi_1(S^1)&lt;/math&gt; which is just &lt;math&gt;\mathbb{Z}&lt;/math&gt;<br /> <br /> ''Q.E.D.''<br /> <br /> <br /> '''Claim:'''<br /> <br /> If &lt;math&gt;f,g:X\rightarrow Y&lt;/math&gt; are homotopic then &lt;math&gt;f_* = g_*:\pi_1(X)\rightarrow\pi_1(Y)&lt;/math&gt;<br /> <br /> <br /> Consider a loop &lt;math&gt;\gamma&lt;/math&gt; in X. Consider the images of this in Y under f and g. They look like two loops with the same base point and the homotopy H collapses them down to the same loop. <br /> <br /> Then, &lt;math&gt;f\circ\gamma&lt;/math&gt; is homotopic to &lt;math&gt;g\circ\gamma&lt;/math&gt; using &lt;math&gt;H\circ\gamma&lt;/math&gt; as the homotopy. <br /> <br /> <br /> '''Theorem'''<br /> <br /> 1) &lt;math&gt;\sim&lt;/math&gt; is an equivalence relation. <br /> <br /> 2) It is an ''ideal'' in the category of topological spaces. That is, &lt;math&gt;f\sim g \Rightarrow f\circ h\sim g\circ h&lt;/math&gt; and &lt;math&gt;h\circ f \sim h\circ g&lt;/math&gt; for h's such that these make sense. <br /> <br /> <br /> '''Definition'''<br /> <br /> X and Y are &quot;homotopy equivalent&quot; if they are isomorphic in {Topological Spaces} / {homotopy of maps}<br /> <br /> I.e., &lt;math&gt;\exists f,g&lt;/math&gt; so that &lt;math&gt;f\circ g\sim I_Y&lt;/math&gt; and &lt;math&gt;g\circ f\sim I_X&lt;/math&gt; for &lt;math&gt;f:X\rightarrow Y&lt;/math&gt; and &lt;math&gt;g:Y\rightarrow X&lt;/math&gt;<br /> <br /> <br /> '''Example'''<br /> <br /> 1) &lt;math&gt;\mathbb{R}^n&lt;/math&gt; is homotopy equivalent to a point via f which takes &lt;math&gt;\mathbb{R}^n&lt;/math&gt; to a point and g is the zero map<br /> <br /> 2) An annulus is homotopy equivalent to &lt;math&gt;S^1&lt;/math&gt;. Indeed, consider the map f which collapses the annulus to the circle in its middle, and g is just the identity. <br /> <br /> 3) Likewise for the figure &quot;thick A&quot; which is a subset of the plane in the shape of an A with everything being thick. There is a circle around the hole in the A. The map f just takes A to this circle and g is the identity. <br /> <br /> 4) The mobius band is also equivalent to a circle where f just collapses to the boundary circle. <br /> <br /> <br /> This last example gives us some idea of the limitation, or &quot;lack of subtlety&quot; to our invariants given that we would all agree there is something fundamentally different between the mobius band, the annulus and the circle, homotopy equivalence is not sensitive to this difference. <br /> <br /> <br /> '''Claim'''<br /> <br /> If X and Y are homotopically equivalent then &lt;math&gt;\pi_1(X)\cong\pi_1(Y)&lt;/math&gt;<br /> <br /> ''Proof:''<br /> <br /> Consider &lt;math&gt;X\rightarrow^f Y&lt;/math&gt; and &lt;math&gt;Y\rightarrow^g X&lt;/math&gt; forming a commuting diagram and lets consider its image under &lt;math&gt;\pi_1&lt;/math&gt;<br /> <br /> Then we get &lt;math&gt;\pi_1(X)\rightarrow^{f_*} \pi_1(Y)&lt;/math&gt; and &lt;math&gt;\pi_1 (Y)\rightarrow^{g_8} \pi_1(X&lt;/math&gt;) as a commuting diagram. <br /> <br /> We know that &lt;math&gt;f\circ g&lt;/math&gt; is homotopic to the identity, and thus we also get &lt;math&gt;f_*\circ g_*&lt;/math&gt; homotopic to the identity<br /> <br /> ''Q.E.D.''<br /> <br /> <br /> '''Theorem''' <br /> <br /> &lt;math&gt;\pi_1(X+Y)\cong\pi_1(X)+\pi_1(Y)&lt;/math&gt;<br /> <br /> <br /> Example: &lt;math&gt;\pi_1(\mathbb{T}^2) = \pi_1(S^1\times S^1) = \mathbb{Z}\times\mathbb{Z} = \mathbb{Z}^2&lt;/math&gt;</div> Trefor http://drorbn.net/index.php?title=0708-1300/Class_notes_for_Tuesday,_January_15 0708-1300/Class notes for Tuesday, January 15 2008-02-08T21:18:17Z <p>Trefor: </p> <hr /> <div>{{0708-1300/Navigation}}<br /> <br /> ==Class Notes==<br /> &lt;span style=&quot;color: red;&quot;&gt;The notes below are by the students and for the students. Hopefully they are useful, but they come with no guarantee of any kind.&lt;/span&gt;<br /> <br /> ===First Hour===<br /> <br /> We begin by reformulating our previous lemma into a more general form:<br /> <br /> <br /> '''Lemma 1'''<br /> <br /> Let &lt;math&gt;p:(X,x_0)\rightarrow(B,b_0)&lt;/math&gt; be a covering map. Then every family of paths &lt;math&gt;\gamma:Y\times I\rightarrow B &lt;/math&gt; such that &lt;math&gt;\forall y\in Y\ \gamma(y,0)=b_0&lt;/math&gt; has a unique lift &lt;math&gt;\tilde{\gamma}:Y\times I\rightarrow X&lt;/math&gt; such that <br /> <br /> 0) &lt;math&gt;\forall y\ \tilde{\gamma}(y,0) = x_0&lt;/math&gt;<br /> <br /> 1) &lt;math&gt;\gamma = p\circ\tilde{\gamma}&lt;/math&gt;<br /> <br /> <br /> Claim 1: &lt;math&gt;Ind[\gamma]&lt;/math&gt; is well defined, hence &lt;math&gt;\pi_1(S^1)\cong\mathbb{Z}&lt;/math&gt;<br /> &lt;math&gt;(\gamma:I\rightarrow S^1, \gamma(0)=1, ind [\gamma] = \tilde{\gamma}(1))&lt;/math&gt;<br /> <br /> <br /> ''Proof of Claim 1''<br /> <br /> We construct the homotopy H between two such gamma's which, schematically, is a square with &lt;math&gt;\gamma_1&lt;/math&gt; along the bottom, &lt;math&gt;\gamma_0&lt;/math&gt; along the top and along the side the parameter Y where the homotopy is just horizontal lines between &lt;math&gt;\gamma_1&lt;/math&gt; and &lt;math&gt;\gamma_0.&lt;/math&gt; <br /> <br /> Then, the lemma implies the existence of a homotopy &lt;math&gt;\tilde{H}&lt;/math&gt; which is schematically a square with &lt;math&gt;\tilde{\gamma_1}&lt;/math&gt; along the bottom, &lt;math&gt;\tilde{\gamma_0}&lt;/math&gt; along the top, x=0 on the left and side and the right hand side taking values in &lt;math&gt;p^{-1}(1)&lt;/math&gt;<br /> <br /> <br /> ''Proof of Lemma 1''<br /> <br /> Recall the proof for the version of the lemma when we were dealing with a single path, not a family. We now just need to extend this proof to the case of having a family of paths. We know that for each &lt;math&gt;y\in Y&lt;/math&gt; we can get a &lt;math&gt;\tilde{\gamma}&lt;/math&gt;<br /> <br /> We need to show that in fact the result is continuous. As continuity is a local property, we simply need to prove this in a neighborhood about &lt;math&gt;y_0&lt;/math&gt;. <br /> <br /> Now by a &quot;good&quot; open set in B, we mean that the preimage under p CAN be written as a product. <br /> <br /> Hence, we choose a neighborhood about the first interval in &lt;math&gt;\mathbb{R}&lt;/math&gt; extending from &lt;math&gt;y_0&lt;/math&gt; (see proof of one path case for explanation) and this gets mapped to a &quot;good&quot; open set in B. As it is the image of an interval, it is compact in an open set, so can put a small neighborhood about the interval such that the image of the neighborhood is in the &quot;good&quot; open set in B. <br /> <br /> Then, duplicating the proof of the earlier version of the lemma, this establishes continuity in the neighborhood.<br /> <br /> <br /> '''Applications of &lt;math&gt;\pi_1(S^1)\cong\mathbb{Z}&lt;/math&gt;'''<br /> <br /> <br /> 1) We get again the proof of non existence of a retract &lt;math&gt;r:D^2\rightarrow S^1&lt;/math&gt;<br /> <br /> Indeed, assume we DID have such a retract. We would have the following commuting diagram. <br /> <br /> &lt;math&gt;\begin{matrix}<br /> D^2&amp;\rightarrow^{r}&amp;S^1\\<br /> \uparrow &amp;\nearrow_{I}&amp; \\<br /> S^1&amp;&amp;\\<br /> \end{matrix}&lt;/math&gt;<br /> <br /> Applying the functor &lt;math&gt;\pi_1&lt;/math&gt; would yield the diagram<br /> <br /> &lt;math&gt;\begin{matrix}<br /> \{0\}&amp;\rightarrow^{r}&amp;\mathbb{Z}\\<br /> \uparrow &amp;\nearrow_{I}&amp; \\<br /> \mathbb{Z}&amp;&amp;\\<br /> \end{matrix}&lt;/math&gt;<br /> <br /> But clearly this does not exist, as the only map from &lt;math&gt;\{0\}&lt;/math&gt; is the trivial one. <br /> <br /> <br /> Recall the non existence of such retracts implies ''Brouwer's Theorem''<br /> <br /> <br /> 2) '''Fundamental Theorem of Algebra'''<br /> <br /> <br /> if &lt;math&gt;p\in\mathbb{C}[z]&lt;/math&gt; is a polynomial with degree greater than zero then &lt;math&gt;\exists z_0\in\mathbb{C}&lt;/math&gt; such that &lt;math&gt;p(z_0) =0&lt;/math&gt;<br /> <br /> <br /> ''Proof:''<br /> <br /> Suppose not, i.e., there exists a polynomial p that has no roots. dividing by the coefficient of the highest order term, &lt;math&gt;p = z^n +&lt;/math&gt; lower order terms. <br /> <br /> <br /> Define a homotopy of paths &lt;math&gt;S^1\rightarrow S^1&lt;/math&gt; by first setting &lt;math&gt;q(z) = p(z)/||p(z)||]\in S^1&lt;/math&gt;<br /> <br /> Now the homotopy starts with a base of q on the point 0. During the first half of the homotopy we grow the size of the loop on which q is acting. At the halfway point, the loop is so large such that the first term in the polynomial dominates the lower order terms. Then in the second half of the homotopy, the lower order terms are shut off. More precisely we consider the polynomial &lt;math&gt;p_{\alpha}(z) = z^n + \alpha&lt;/math&gt;(lower order terms) and let &lt;math&gt;\alpha&lt;/math&gt; tend from 1 to 0 in the second half of the homotopy. <br /> <br /> Hence, at the end of the homotopy we are left with &lt;math&gt;z^n/||z^n||&lt;/math&gt; on a large loop. This is the generator we had previously, and is not the identity. Hence, we have a constant homotopic to something non trivial and this establishes the contradiction. <br /> <br /> <br /> 3) '''Boruk-Ulam'''<br /> <br /> If &lt;math&gt;f:S^2\rightarrow \mathbb{R}^2&lt;/math&gt; is continuous then &lt;math&gt;\exists p\in S^2&lt;/math&gt; such that &lt;math&gt;f(p) = f(-p)&lt;/math&gt;<br /> <br /> <br /> Corollary: <br /> <br /> 1) &lt;math&gt;S^2&lt;/math&gt; is not a subset of &lt;math&gt;\mathbb{R}^2&lt;/math&gt; (cannot be embedded) as f is not 1:1<br /> <br /> 2) If &lt;math&gt;S^2 =A_1\cup A_2\cup A_3&lt;/math&gt;, with &lt;math&gt;A_i&lt;/math&gt; closed then at least one &lt;math&gt;A_i&lt;/math&gt; contains a pair of antipodes.<br /> <br /> <br /> '''Example:'''<br /> <br /> This is NOT true with 4 sections. For instance consider a triangular based pyramid inscribed inside a sphere with each face a different colour. Imagine a light bulb at the center of the sphere so that on the sphere we get 4 sections with 4 different colours from each side of the pyramid. Clearly none of them contains a pair of antipodes, yet are closed with a union of the whole sphere.<br /> <br /> ===Second Hour===<br /> <br /> ''Proof of Corollary:''<br /> <br /> Consider &lt;math&gt;f:S^2\rightarrow\mathbb{R}&lt;/math&gt; by &lt;math&gt;p\mapsto&lt;/math&gt; dist to &lt;math&gt;A_1&lt;/math&gt; + dist to &lt;math&gt;A_2&lt;/math&gt;<br /> <br /> If f(p) = f(-p) then, for both possible cases of f(p) being zero or positive, we get that p and -p are in the same &lt;math&gt;A_i&lt;/math&gt;. <br /> <br /> <br /> '''Definition:'''<br /> <br /> 1) &lt;math&gt;\gamma:S^1\rightarrow S^1&lt;/math&gt; is even if &lt;math&gt;\gamma(-x) = \gamma(x)\ \forall x&lt;/math&gt;<br /> <br /> 2) &lt;math&gt;\gamma&lt;/math&gt; is odd if &lt;math&gt;\gamma(-x) = -\gamma(x)\ \forall x&lt;/math&gt;<br /> <br /> <br /> '''Lemma:'''<br /> <br /> 1) If &lt;math&gt;\gamma&lt;/math&gt; is even then ind &lt;math&gt;\gamma&lt;/math&gt; is even<br /> <br /> 2) If &lt;math&gt;\gamma&lt;/math&gt; is odd then ind &lt;math&gt;\gamma&lt;/math&gt; is odd<br /> <br /> <br /> ''Proof of Borsuk-Ulam:''<br /> <br /> <br /> Assume &lt;math&gt;f:S^2\rightarrow\mathbb{R}&lt;/math&gt; has no p with f(p)=f(-p)<br /> <br /> Define &lt;math&gt;g:S^2\rightarrow S^1&lt;/math&gt; by &lt;math&gt;g(p) = (f(p)-f(-p))/||f(p)-f(-p)||&lt;/math&gt;<br /> <br /> &lt;math&gt;\gamma = g|_{equator}:S^1\rightarrow S^1&lt;/math&gt; is odd. Therefore, &lt;math&gt;[\gamma]\in\pi_1(S^1)&lt;/math&gt; is non zero. <br /> <br /> But &lt;math&gt;\gamma&lt;/math&gt; is homotopic to zero, a contradiction (the homotopy is a series of circles of decreasing radius where a point on the equator is fixed and its antipodal point is moved in an arc towards it)<br /> <br /> <br /> ''Proof of part 1 of lemma:''<br /> <br /> <br /> Suppose &lt;math&gt;\gamma:S^1\rightarrow S^1&lt;/math&gt; is even<br /> <br /> <br /> We can make a commuting diagram where &lt;math&gt;\gamma:S^1\rightarrow S^1&lt;/math&gt; is the same as going from &lt;math&gt;S^1\rightarrow S^1&lt;/math&gt; first via &lt;math&gt;z\mapsto z^2&lt;/math&gt; and then from &lt;math&gt;S^1\rightarrow S^1&lt;/math&gt; via &lt;math&gt;\lambda(z) = \gamma(\sqrt{z})&lt;/math&gt;<br /> <br /> <br /> We now consider this diagram under the functor &lt;math&gt;\pi_1&lt;/math&gt;<br /> <br /> We get &lt;math&gt;\mathbb{Z}\rightarrow^{\times 2}\mathbb{Z}\rightarrow^{\lambda_*}\mathbb{Z}&lt;/math&gt;<br /> <br /> which commutes with &lt;math&gt;\mathbb{Z}\rightarrow^{\gamma_*}\mathbb{Z}&lt;/math&gt; along the bottom<br /> <br /> &lt;math&gt;[\gamma] = \gamma_*[\gamma_1] = \lambda_*(2[\gamma_1])= 2\lambda_*([\gamma_1])&lt;/math&gt; which is even. <br /> <br /> (Note, by &lt;math&gt;\gamma_1:S^1\rightarrow S^1&lt;/math&gt; we mean the identity)<br /> <br /> <br /> Unfortunately, there is no simple adaption of this proof to deal with the odd case and so we introduce some more machinery first. <br /> <br /> <br /> '''Definition'''<br /> <br /> A ''topological group'' is a topological space G with a group structure such that all group operations are continuous<br /> <br /> <br /> '''Examples:'''<br /> <br /> 1) &lt;math&gt;S^1&lt;/math&gt; (with rotation giving the group structure)<br /> <br /> 2) SO(3) (See a previous lecture for more info)<br /> <br /> <br /> If G is a topological group there are two &quot;products&quot; we can define on &lt;math&gt;\pi_1(G)&lt;/math&gt;:<br /> <br /> <br /> 1) &lt;math&gt;[\gamma][\lambda] = [\gamma\circ\lambda]&lt;/math&gt; with &lt;math&gt;\circ&lt;/math&gt; being simply concatenation<br /> <br /> 2) &lt;math&gt;[\gamma]*[\lambda] = [\gamma *\lambda]&lt;/math&gt; where &lt;math&gt;(\gamma*\lambda)(t) = \gamma(t)\cdot\lambda(t)&lt;/math&gt;<br /> <br /> <br /> Claim:<br /> <br /> These two are the same and, in fact, are commutative. I will not attempt to describe the homotopy schematic here. <br /> <br /> <br /> We return to the proof of the odd case of the lemma:<br /> <br /> <br /> Assume &lt;math&gt;\gamma&lt;/math&gt; is odd. Then &lt;math&gt;\gamma*\gamma_1&lt;/math&gt; is even so &lt;math&gt;2\mathbb{Z}\in[\gamma*\gamma_1]=[\gamma\circ\gamma_1] = [\gamma] + [\gamma_1] = [\gamma] + 1&lt;/math&gt;. Therefore, &lt;math&gt;[\gamma]&lt;/math&gt; is odd. <br /> <br /> Note: The addition above is done in &lt;math&gt;\pi_1(S^1)&lt;/math&gt; which is just &lt;math&gt;\mathbb{Z}&lt;/math&gt;<br /> <br /> ''Q.E.D.''<br /> <br /> <br /> '''Claim:'''<br /> <br /> If &lt;math&gt;f,g:X\rightarrow Y&lt;/math&gt; are homotopic then &lt;math&gt;f_* = g_*:\pi_1(X)\rightarrow\pi_1(Y)&lt;/math&gt;<br /> <br /> <br /> Consider a loop &lt;math&gt;\gamma&lt;/math&gt; in X. Consider the images of this in Y under f and g. They look like two loops with the same base point and the homotopy H collapses them down to the same loop. <br /> <br /> Then, &lt;math&gt;f\circ\gamma&lt;/math&gt; is homotopic to &lt;math&gt;g\circ\gamma&lt;/math&gt; using &lt;math&gt;H\circ\gamma&lt;/math&gt; as the homotopy. <br /> <br /> <br /> '''Theorem'''<br /> <br /> 1) &lt;math&gt;\sim&lt;/math&gt; is an equivalence relation. <br /> <br /> 2) It is an ''ideal'' in the category of topological spaces. That is, &lt;math&gt;f\sim g \Rightarrow f\circ h\sim g\circ h&lt;/math&gt; and &lt;math&gt;h\circ f \sim h\circ g&lt;/math&gt; for h's such that these make sense. <br /> <br /> <br /> '''Definition'''<br /> <br /> X and Y are &quot;homotopy equivalent&quot; if they are isomorphic in {Topological Spaces} / {homotopy of maps}<br /> <br /> I.e., &lt;math&gt;\exists f,g&lt;/math&gt; so that &lt;math&gt;f\circ g\sim I_Y&lt;/math&gt; and &lt;math&gt;g\circ f\sim I_X&lt;/math&gt; for &lt;math&gt;f:X\rightarrow Y&lt;/math&gt; and &lt;math&gt;g:Y\rightarrow X&lt;/math&gt;<br /> <br /> <br /> '''Example'''<br /> <br /> 1) &lt;math&gt;\mathbb{R}^n&lt;/math&gt; is homotopy equivalent to a point via f which takes &lt;math&gt;\mathbb{R}^n&lt;/math&gt; to a point and g is the zero map<br /> <br /> 2) An annulus is homotopy equivalent to &lt;math&gt;S^1&lt;/math&gt;. Indeed, consider the map f which collapses the annulus to the circle in its middle, and g is just the identity. <br /> <br /> 3) Likewise for the figure &quot;thick A&quot; which is a subset of the plane in the shape of an A with everything being thick. There is a circle around the hole in the A. The map f just takes A to this circle and g is the identity. <br /> <br /> 4) The mobius band is also equivalent to a circle where f just collapses to the boundary circle. <br /> <br /> <br /> This last example gives us some idea of the limitation, or &quot;lack of subtlety&quot; to our invariants given that we would all agree there is something fundamentally different between the mobius band, the annulus and the circle, homotopy equivalence is not sensitive to this difference. <br /> <br /> <br /> '''Claim'''<br /> <br /> If X and Y are homotopically equivalent then &lt;math&gt;\pi_1(X)\cong\pi_1(Y)&lt;/math&gt;<br /> <br /> ''Proof:''<br /> <br /> Consider &lt;math&gt;X\rightarrow^f Y&lt;/math&gt; and &lt;math&gt;Y\rightarrow^g X&lt;/math&gt; forming a commuting diagram and lets consider its image under &lt;math&gt;\pi_1&lt;/math&gt;<br /> <br /> Then we get &lt;math&gt;\pi_1(X)\rightarrow^{f_*} \pi_1(Y)&lt;/math&gt; and &lt;math&gt;\pi_1 (Y)\rightarrow^{g_8} \pi_1(X&lt;/math&gt;) as a commuting diagram. <br /> <br /> We know that &lt;math&gt;f\circ g&lt;/math&gt; is homotopic to the identity, and thus we also get &lt;math&gt;f_*\circ g_*&lt;/math&gt; homotopic to the identity<br /> <br /> ''Q.E.D.''<br /> <br /> <br /> '''Theorem''' <br /> <br /> &lt;math&gt;\pi_1(X+Y)\cong\pi_1(X)+\pi_1(Y)&lt;/math&gt;<br /> <br /> <br /> Example: &lt;math&gt;\pi_1(\mathbb{T}^2) = \pi_1(S^1\times S^1) = \mathbb{Z}\times\mathbb{Z} = \mathbb{Z}^2&lt;/math&gt;</div> Trefor http://drorbn.net/index.php?title=0708-1300/Class_notes_for_Tuesday,_January_15 0708-1300/Class notes for Tuesday, January 15 2008-02-08T21:17:40Z <p>Trefor: </p> <hr /> <div>{{0708-1300/Navigation}}<br /> <br /> ==Class Notes==<br /> &lt;span style=&quot;color: red;&quot;&gt;The notes below are by the students and for the students. Hopefully they are useful, but they come with no guarantee of any kind.&lt;/span&gt;<br /> <br /> ===First Hour===<br /> <br /> We begin by reformulating our previous lemma into a more general form:<br /> <br /> <br /> '''Lemma 1'''<br /> <br /> Let &lt;math&gt;p:(X,x_0)\rightarrow(B,b_0)&lt;/math&gt; be a covering map. Then every family of paths &lt;math&gt;\gamma:Y\times I\rightarrow B &lt;/math&gt; such that &lt;math&gt;\forall y\in Y\ \gamma(y,0)=b_0&lt;/math&gt; has a unique lift &lt;math&gt;\tilde{\gamma}:Y\times I\rightarrow X&lt;/math&gt; such that <br /> <br /> 0) &lt;math&gt;\forall y\ \tilde{\gamma}(y,0) = x_0&lt;/math&gt;<br /> <br /> 1) &lt;math&gt;\gamma = p\circ\tilde{\gamma}&lt;/math&gt;<br /> <br /> <br /> Claim 1: &lt;math&gt;Ind[\gamma]&lt;/math&gt; is well defined, hence &lt;math&gt;\pi_1(S^1)\cong\mathbb{Z}&lt;/math&gt;<br /> &lt;math&gt;(\gamma:I\rightarrow S^1, \gamma(0)=1, ind \gamma = \tilde{\gamma}(1))&lt;/math&gt;<br /> <br /> <br /> ''Proof of Claim 1''<br /> <br /> We construct the homotopy H between two such gamma's which, schematically, is a square with &lt;math&gt;\gamma_1&lt;/math&gt; along the bottom, &lt;math&gt;\gamma_0&lt;/math&gt; along the top and along the side the parameter Y where the homotopy is just horizontal lines between &lt;math&gt;\gamma_1&lt;/math&gt; and &lt;math&gt;\gamma_0.&lt;/math&gt; <br /> <br /> Then, the lemma implies the existence of a homotopy &lt;math&gt;\tilde{H}&lt;/math&gt; which is schematically a square with &lt;math&gt;\tilde{\gamma_1}&lt;/math&gt; along the bottom, &lt;math&gt;\tilde{\gamma_0}&lt;/math&gt; along the top, x=0 on the left and side and the right hand side taking values in &lt;math&gt;p^{-1}(1)&lt;/math&gt;<br /> <br /> <br /> ''Proof of Lemma 1''<br /> <br /> Recall the proof for the version of the lemma when we were dealing with a single path, not a family. We now just need to extend this proof to the case of having a family of paths. We know that for each &lt;math&gt;y\in Y&lt;/math&gt; we can get a &lt;math&gt;\tilde{\gamma}&lt;/math&gt;<br /> <br /> We need to show that in fact the result is continuous. As continuity is a local property, we simply need to prove this in a neighborhood about &lt;math&gt;y_0&lt;/math&gt;. <br /> <br /> Now by a &quot;good&quot; open set in B, we mean that the preimage under p CAN be written as a product. <br /> <br /> Hence, we choose a neighborhood about the first interval in &lt;math&gt;\mathbb{R}&lt;/math&gt; extending from &lt;math&gt;y_0&lt;/math&gt; (see proof of one path case for explanation) and this gets mapped to a &quot;good&quot; open set in B. As it is the image of an interval, it is compact in an open set, so can put a small neighborhood about the interval such that the image of the neighborhood is in the &quot;good&quot; open set in B. <br /> <br /> Then, duplicating the proof of the earlier version of the lemma, this establishes continuity in the neighborhood.<br /> <br /> <br /> '''Applications of &lt;math&gt;\pi_1(S^1)\cong\mathbb{Z}&lt;/math&gt;'''<br /> <br /> <br /> 1) We get again the proof of non existence of a retract &lt;math&gt;r:D^2\rightarrow S^1&lt;/math&gt;<br /> <br /> Indeed, assume we DID have such a retract. We would have the following commuting diagram. <br /> <br /> &lt;math&gt;\begin{matrix}<br /> D^2&amp;\rightarrow^{r}&amp;S^1\\<br /> \uparrow &amp;\nearrow_{I}&amp; \\<br /> S^1&amp;&amp;\\<br /> \end{matrix}&lt;/math&gt;<br /> <br /> Applying the functor &lt;math&gt;\pi_1&lt;/math&gt; would yield the diagram<br /> <br /> &lt;math&gt;\begin{matrix}<br /> \{0\}&amp;\rightarrow^{r}&amp;\mathbb{Z}\\<br /> \uparrow &amp;\nearrow_{I}&amp; \\<br /> \mathbb{Z}&amp;&amp;\\<br /> \end{matrix}&lt;/math&gt;<br /> <br /> But clearly this does not exist, as the only map from &lt;math&gt;\{0\}&lt;/math&gt; is the trivial one. <br /> <br /> <br /> Recall the non existence of such retracts implies ''Brouwer's Theorem''<br /> <br /> <br /> 2) '''Fundamental Theorem of Algebra'''<br /> <br /> <br /> if &lt;math&gt;p\in\mathbb{C}[z]&lt;/math&gt; is a polynomial with degree greater than zero then &lt;math&gt;\exists z_0\in\mathbb{C}&lt;/math&gt; such that &lt;math&gt;p(z_0) =0&lt;/math&gt;<br /> <br /> <br /> ''Proof:''<br /> <br /> Suppose not, i.e., there exists a polynomial p that has no roots. dividing by the coefficient of the highest order term, &lt;math&gt;p = z^n +&lt;/math&gt; lower order terms. <br /> <br /> <br /> Define a homotopy of paths &lt;math&gt;S^1\rightarrow S^1&lt;/math&gt; by first setting &lt;math&gt;q(z) = p(z)/||p(z)||]\in S^1&lt;/math&gt;<br /> <br /> Now the homotopy starts with a base of q on the point 0. During the first half of the homotopy we grow the size of the loop on which q is acting. At the halfway point, the loop is so large such that the first term in the polynomial dominates the lower order terms. Then in the second half of the homotopy, the lower order terms are shut off. More precisely we consider the polynomial &lt;math&gt;p_{\alpha}(z) = z^n + \alpha&lt;/math&gt;(lower order terms) and let &lt;math&gt;\alpha&lt;/math&gt; tend from 1 to 0 in the second half of the homotopy. <br /> <br /> Hence, at the end of the homotopy we are left with &lt;math&gt;z^n/||z^n||&lt;/math&gt; on a large loop. This is the generator we had previously, and is not the identity. Hence, we have a constant homotopic to something non trivial and this establishes the contradiction. <br /> <br /> <br /> 3) '''Boruk-Ulam'''<br /> <br /> If &lt;math&gt;f:S^2\rightarrow \mathbb{R}^2&lt;/math&gt; is continuous then &lt;math&gt;\exists p\in S^2&lt;/math&gt; such that &lt;math&gt;f(p) = f(-p)&lt;/math&gt;<br /> <br /> <br /> Corollary: <br /> <br /> 1) &lt;math&gt;S^2&lt;/math&gt; is not a subset of &lt;math&gt;\mathbb{R}^2&lt;/math&gt; (cannot be embedded) as f is not 1:1<br /> <br /> 2) If &lt;math&gt;S^2 =A_1\cup A_2\cup A_3&lt;/math&gt;, with &lt;math&gt;A_i&lt;/math&gt; closed then at least one &lt;math&gt;A_i&lt;/math&gt; contains a pair of antipodes.<br /> <br /> <br /> '''Example:'''<br /> <br /> This is NOT true with 4 sections. For instance consider a triangular based pyramid inscribed inside a sphere with each face a different colour. Imagine a light bulb at the center of the sphere so that on the sphere we get 4 sections with 4 different colours from each side of the pyramid. Clearly none of them contains a pair of antipodes, yet are closed with a union of the whole sphere.<br /> <br /> ===Second Hour===<br /> <br /> ''Proof of Corollary:''<br /> <br /> Consider &lt;math&gt;f:S^2\rightarrow\mathbb{R}&lt;/math&gt; by &lt;math&gt;p\mapsto&lt;/math&gt; dist to &lt;math&gt;A_1&lt;/math&gt; + dist to &lt;math&gt;A_2&lt;/math&gt;<br /> <br /> If f(p) = f(-p) then, for both possible cases of f(p) being zero or positive, we get that p and -p are in the same &lt;math&gt;A_i&lt;/math&gt;. <br /> <br /> <br /> '''Definition:'''<br /> <br /> 1) &lt;math&gt;\gamma:S^1\rightarrow S^1&lt;/math&gt; is even if &lt;math&gt;\gamma(-x) = \gamma(x)\ \forall x&lt;/math&gt;<br /> <br /> 2) &lt;math&gt;\gamma&lt;/math&gt; is odd if &lt;math&gt;\gamma(-x) = -\gamma(x)\ \forall x&lt;/math&gt;<br /> <br /> <br /> '''Lemma:'''<br /> <br /> 1) If &lt;math&gt;\gamma&lt;/math&gt; is even then ind &lt;math&gt;\gamma&lt;/math&gt; is even<br /> <br /> 2) If &lt;math&gt;\gamma&lt;/math&gt; is odd then ind &lt;math&gt;\gamma&lt;/math&gt; is odd<br /> <br /> <br /> ''Proof of Borsuk-Ulam:''<br /> <br /> <br /> Assume &lt;math&gt;f:S^2\rightarrow\mathbb{R}&lt;/math&gt; has no p with f(p)=f(-p)<br /> <br /> Define &lt;math&gt;g:S^2\rightarrow S^1&lt;/math&gt; by &lt;math&gt;g(p) = (f(p)-f(-p))/||f(p)-f(-p)||&lt;/math&gt;<br /> <br /> &lt;math&gt;\gamma = g|_{equator}:S^1\rightarrow S^1&lt;/math&gt; is odd. Therefore, &lt;math&gt;[\gamma]\in\pi_1(S^1)&lt;/math&gt; is non zero. <br /> <br /> But &lt;math&gt;\gamma&lt;/math&gt; is homotopic to zero, a contradiction (the homotopy is a series of circles of decreasing radius where a point on the equator is fixed and its antipodal point is moved in an arc towards it)<br /> <br /> <br /> ''Proof of part 1 of lemma:''<br /> <br /> <br /> Suppose &lt;math&gt;\gamma:S^1\rightarrow S^1&lt;/math&gt; is even<br /> <br /> <br /> We can make a commuting diagram where &lt;math&gt;\gamma:S^1\rightarrow S^1&lt;/math&gt; is the same as going from &lt;math&gt;S^1\rightarrow S^1&lt;/math&gt; first via &lt;math&gt;z\mapsto z^2&lt;/math&gt; and then from &lt;math&gt;S^1\rightarrow S^1&lt;/math&gt; via &lt;math&gt;\lambda(z) = \gamma(\sqrt{z})&lt;/math&gt;<br /> <br /> <br /> We now consider this diagram under the functor &lt;math&gt;\pi_1&lt;/math&gt;<br /> <br /> We get &lt;math&gt;\mathbb{Z}\rightarrow^{\times 2}\mathbb{Z}\rightarrow^{\lambda_*}\mathbb{Z}&lt;/math&gt;<br /> <br /> which commutes with &lt;math&gt;\mathbb{Z}\rightarrow^{\gamma_*}\mathbb{Z}&lt;/math&gt; along the bottom<br /> <br /> &lt;math&gt;[\gamma] = \gamma_*[\gamma_1] = \lambda_*(2[\gamma_1])= 2\lambda_*([\gamma_1])&lt;/math&gt; which is even. <br /> <br /> (Note, by &lt;math&gt;\gamma_1:S^1\rightarrow S^1&lt;/math&gt; we mean the identity)<br /> <br /> <br /> Unfortunately, there is no simple adaption of this proof to deal with the odd case and so we introduce some more machinery first. <br /> <br /> <br /> '''Definition'''<br /> <br /> A ''topological group'' is a topological space G with a group structure such that all group operations are continuous<br /> <br /> <br /> '''Examples:'''<br /> <br /> 1) &lt;math&gt;S^1&lt;/math&gt; (with rotation giving the group structure)<br /> <br /> 2) SO(3) (See a previous lecture for more info)<br /> <br /> <br /> If G is a topological group there are two &quot;products&quot; we can define on &lt;math&gt;\pi_1(G)&lt;/math&gt;:<br /> <br /> <br /> 1) &lt;math&gt;[\gamma][\lambda] = [\gamma\circ\lambda]&lt;/math&gt; with &lt;math&gt;\circ&lt;/math&gt; being simply concatenation<br /> <br /> 2) &lt;math&gt;[\gamma]*[\lambda] = [\gamma *\lambda]&lt;/math&gt; where &lt;math&gt;(\gamma*\lambda)(t) = \gamma(t)\cdot\lambda(t)&lt;/math&gt;<br /> <br /> <br /> Claim:<br /> <br /> These two are the same and, in fact, are commutative. I will not attempt to describe the homotopy schematic here. <br /> <br /> <br /> We return to the proof of the odd case of the lemma:<br /> <br /> <br /> Assume &lt;math&gt;\gamma&lt;/math&gt; is odd. Then &lt;math&gt;\gamma*\gamma_1&lt;/math&gt; is even so &lt;math&gt;2\mathbb{Z}\in[\gamma*\gamma_1]=[\gamma\circ\gamma_1] = [\gamma] + [\gamma_1] = [\gamma] + 1&lt;/math&gt;. Therefore, &lt;math&gt;[\gamma]&lt;/math&gt; is odd. <br /> <br /> Note: The addition above is done in &lt;math&gt;\pi_1(S^1)&lt;/math&gt; which is just &lt;math&gt;\mathbb{Z}&lt;/math&gt;<br /> <br /> ''Q.E.D.''<br /> <br /> <br /> '''Claim:'''<br /> <br /> If &lt;math&gt;f,g:X\rightarrow Y&lt;/math&gt; are homotopic then &lt;math&gt;f_* = g_*:\pi_1(X)\rightarrow\pi_1(Y)&lt;/math&gt;<br /> <br /> <br /> Consider a loop &lt;math&gt;\gamma&lt;/math&gt; in X. Consider the images of this in Y under f and g. They look like two loops with the same base point and the homotopy H collapses them down to the same loop. <br /> <br /> Then, &lt;math&gt;f\circ\gamma&lt;/math&gt; is homotopic to &lt;math&gt;g\circ\gamma&lt;/math&gt; using &lt;math&gt;H\circ\gamma&lt;/math&gt; as the homotopy. <br /> <br /> <br /> '''Theorem'''<br /> <br /> 1) &lt;math&gt;\sim&lt;/math&gt; is an equivalence relation. <br /> <br /> 2) It is an ''ideal'' in the category of topological spaces. That is, &lt;math&gt;f\sim g \Rightarrow f\circ h\sim g\circ h&lt;/math&gt; and &lt;math&gt;h\circ f \sim h\circ g&lt;/math&gt; for h's such that these make sense. <br /> <br /> <br /> '''Definition'''<br /> <br /> X and Y are &quot;homotopy equivalent&quot; if they are isomorphic in {Topological Spaces} / {homotopy of maps}<br /> <br /> I.e., &lt;math&gt;\exists f,g&lt;/math&gt; so that &lt;math&gt;f\circ g\sim I_Y&lt;/math&gt; and &lt;math&gt;g\circ f\sim I_X&lt;/math&gt; for &lt;math&gt;f:X\rightarrow Y&lt;/math&gt; and &lt;math&gt;g:Y\rightarrow X&lt;/math&gt;<br /> <br /> <br /> '''Example'''<br /> <br /> 1) &lt;math&gt;\mathbb{R}^n&lt;/math&gt; is homotopy equivalent to a point via f which takes &lt;math&gt;\mathbb{R}^n&lt;/math&gt; to a point and g is the zero map<br /> <br /> 2) An annulus is homotopy equivalent to &lt;math&gt;S^1&lt;/math&gt;. Indeed, consider the map f which collapses the annulus to the circle in its middle, and g is just the identity. <br /> <br /> 3) Likewise for the figure &quot;thick A&quot; which is a subset of the plane in the shape of an A with everything being thick. There is a circle around the hole in the A. The map f just takes A to this circle and g is the identity. <br /> <br /> 4) The mobius band is also equivalent to a circle where f just collapses to the boundary circle. <br /> <br /> <br /> This last example gives us some idea of the limitation, or &quot;lack of subtlety&quot; to our invariants given that we would all agree there is something fundamentally different between the mobius band, the annulus and the circle, homotopy equivalence is not sensitive to this difference. <br /> <br /> <br /> '''Claim'''<br /> <br /> If X and Y are homotopically equivalent then &lt;math&gt;\pi_1(X)\cong\pi_1(Y)&lt;/math&gt;<br /> <br /> ''Proof:''<br /> <br /> Consider &lt;math&gt;X\rightarrow^f Y&lt;/math&gt; and &lt;math&gt;Y\rightarrow^g X&lt;/math&gt; forming a commuting diagram and lets consider its image under &lt;math&gt;\pi_1&lt;/math&gt;<br /> <br /> Then we get &lt;math&gt;\pi_1(X)\rightarrow^{f_*} \pi_1(Y)&lt;/math&gt; and &lt;math&gt;\pi_1 (Y)\rightarrow^{g_8} \pi_1(X&lt;/math&gt;) as a commuting diagram. <br /> <br /> We know that &lt;math&gt;f\circ g&lt;/math&gt; is homotopic to the identity, and thus we also get &lt;math&gt;f_*\circ g_*&lt;/math&gt; homotopic to the identity<br /> <br /> ''Q.E.D.''<br /> <br /> <br /> '''Theorem''' <br /> <br /> &lt;math&gt;\pi_1(X+Y)\cong\pi_1(X)+\pi_1(Y)&lt;/math&gt;<br /> <br /> <br /> Example: &lt;math&gt;\pi_1(\mathbb{T}^2) = \pi_1(S^1\times S^1) = \mathbb{Z}\times\mathbb{Z} = \mathbb{Z}^2&lt;/math&gt;</div> Trefor http://drorbn.net/index.php?title=0708-1300/Class_notes_for_Tuesday,_January_8 0708-1300/Class notes for Tuesday, January 8 2008-02-08T19:00:04Z <p>Trefor: </p> <hr /> <div>{{0708-1300/Navigation}}<br /> <br /> ==Algebraic Topology==<br /> <br /> Temporarily, at least, we will no longer assume that all structures are manifolds and that all functions are smooth. However, functions will still be assumed to be continuous.<br /> <br /> We will also, temporarily, assume that all spaces are pointed spaces. That is, a space X will be assumed to have a distinguished point x, whether mentioned explicitly or not.<br /> <br /> General idea of algebraic topology: to find &quot;functors&quot; from topology to algebra.<br /> <br /> '''Informal definition:'''<br /> A category C consists of<br /> <br /> 1) a collection Obj(C) of objects<br /> <br /> 2) a class hom(C) of morphisms between these objects, so that for each X&lt;sub&gt;1&lt;/sub&gt; and X&lt;sub&gt;2&lt;/sub&gt; in Obj(C) we get a set mor{X&lt;sub&gt;1&lt;/sub&gt;, X&lt;sub&gt;2&lt;/sub&gt;}<br /> <br /> 3) Composition laws:<br /> <br /> a) a binary function on hom(C): mor(X&lt;sub&gt;1&lt;/sub&gt;, X&lt;sub&gt;2&lt;/sub&gt;) x mor (X&lt;sub&gt;2&lt;/sub&gt;, X&lt;sub&gt;3&lt;/sub&gt; &amp;rarr; mor (X&lt;sub&gt;1&lt;/sub&gt;, X&lt;sub&gt;3&lt;/sub&gt;)<br /> <br /> b) Identities (for every X, there is an I&lt;sub&gt;X&lt;/sub&gt; in mor(X,X))<br /> <br /> 4) Compatibility laws:<br /> <br /> a) associativity of composition<br /> <br /> b) identities are in fact identities.<br /> <br /> <br /> <br /> &quot;Functors&quot; preserve categories.<br /> <br /> '''Example:'''<br /> Recall that the Brouwer Fixed Point Theorem is implied by the statement that there is no retract from the n-dimensional disc D&lt;sup&gt;n&lt;/sup&gt; to the (n-1)-dimensional sphere S&lt;sup&gt;n-1&lt;/sup&gt;. Later, we will prove this last statement in its categorical reformulation: we will find a functor H such that H(D&lt;sup&gt;&lt;/sup&gt;) = {0} and H(S&lt;sup&gt;n-1&lt;/sup&gt;) = '''Z'''. So the existence of a retract would imply that there are some functions f: '''Z''' &amp;rarr; {0} and g: {0} &amp;rarr; '''Z''' such that gf is the identity. Clearly this is false, so no retract exists.<br /> <br /> <br /> '''Definition:'''<br /> Given (X, x&lt;sub&gt;0&lt;/sub&gt;) (which, by the above comment, we will sometimes write simply as X) we define the &quot;fundamental group&quot; or &quot;Poincaré group&quot; of X:<br /> <br /> &amp;pi;&lt;sub&gt;1&lt;/sub&gt; (X, x&lt;sub&gt;0&lt;/sub&gt;) := { [&amp;gamma;]: &amp;gamma; : [0,1] &amp;rarr; X, &amp;gamma;(0) = &amp;gamma;(1) = x&lt;sub&gt;0&lt;/sub&gt;}<br /> <br /> <br /> Now we have to:<br /> <br /> 1) Define homotopy and prove that it's an equivalence relation<br /> <br /> 2) Define a binary operation, show that it's well-defined, and then show that the fundamental group is in fact a group.<br /> <br /> 3) Demonstrate functoriality<br /> <br /> <br /> '''Definition:'''<br /> If f, g : (X, x&lt;sub&gt;0&lt;/sub&gt;) &amp;rarr; (Y, y&lt;sub&gt;0&lt;/sub&gt;), we say that &quot;f is homotopic to g&quot; (and we write f ~ g) if there exists a function H : [0,1] &amp;times; X &amp;rarr; Y such that H restricted to {0}&amp;times;X is f, H restricted to {1}&amp;times;X is g, and H at any time t is in our category (ie H(t, x&lt;sub&gt;0&lt;/sub&gt;) = y&lt;sub&gt;0&lt;/sub&gt;.)<br /> <br /> <br /> We know now go through four boring claims, labeled BC1 - BC4. Many have simple visual proofs which I will not reproduce here.<br /> <br /> <br /> '''BC1:'''<br /> <br /> ~ is an equivalence relation.<br /> <br /> '''Proof of BC1:'''<br /> <br /> a) f ~ f for all functions f. Indeed, set H(t, x) = f(x) for all x.<br /> <br /> b) If f ~ g (by a function H) then g ~ f. Indeed, set H'(t,x) = H(1-t, x).<br /> <br /> c) If f ~ g (by a function H&lt;sub&gt;1&lt;/sub&gt;) and g ~ h (by a function H&lt;sub&gt;2&lt;/sub&gt;, then f ~ h. Indeed, set H(t, x) = H&lt;sub&gt;1&lt;/sub&gt;(2t,x) if t &amp;lt; ½ and H(t,x) = H&lt;sub&gt;2&lt;/sub&gt;(2t-1, x) if t &amp;gt; ½.<br /> <br /> <br /> '''Definition:'''<br /> We define a binary operation on the equivalence classes of paths:<br /> <br /> [&amp;gamma;] &amp;bull; [&amp;gamma;'] = [&amp;gamma; &amp;bull; &amp;gamma;']<br /> <br /> where the &amp;bull; represents the concatenation of paths. Intuitively, we follow the first path at twice its normal speed, then follow the second, again at twice its normal speed.<br /> <br /> '''BC2:'''<br /> This is well-defined,<br /> <br /> '''BC3:'''<br /> &amp;pi;&lt;sub&gt;1&lt;/sub&gt; (X, x&lt;sub&gt;0&lt;/sub&gt;) is a group.<br /> <br /> '''Proofette of BC3:'''<br /> <br /> Associativity: Let &amp;gamma;&lt;sub&gt;0&lt;/sub&gt;, &amp;gamma;&lt;sub&gt;1&lt;/sub&gt; and &amp;gamma;&lt;sub&gt;2&lt;/sub&gt; be paths. <br /> Note that while (&amp;gamma;&lt;sub&gt;0&lt;/sub&gt; &amp;bull; &amp;gamma;&lt;sub&gt;1&lt;/sub&gt;) &amp;bull; &amp;gamma;&lt;sub&gt;2&lt;/sub&gt; is not the same path as &amp;gamma;&lt;sub&gt;0&lt;/sub&gt; &amp;bull; (&amp;gamma;&lt;sub&gt;1&lt;/sub&gt; &amp;bull; &amp;gamma;&lt;sub&gt;2&lt;/sub&gt;), they are homotopic as paths.<br /> <br /> Identity: [e] = [x&lt;sub&gt;0&lt;/sub&gt;], where x&lt;sub&gt;0&lt;/sub&gt; is the constant path.<br /> <br /> Inverse: [&amp;gamma;]&lt;sup&gt;-1&lt;/sup&gt; = [&lt;math&gt;\tilde{\gamma}&lt;/math&gt;], where &lt;math&gt;\tilde{\gamma}&lt;/math&gt;(t) = &amp;gamma;(1-t). <br /> <br /> '''BC4:''' &amp;pi;&lt;sub&gt;1&lt;/sub&gt; is a functor.<br /> <br /> &lt;math&gt; f_* &lt;/math&gt; [&amp;gamma;] = [&lt;math&gt; f_* &lt;/math&gt; &amp;gamma;] = [f &amp;bull; &amp;gamma;] defines &lt;math&gt; f_* &lt;/math&gt; : &amp;pi;&lt;sub&gt;1&lt;/sub&gt;(f) : &amp;pi;&lt;sub&gt;1&lt;/sub&gt;(X) &amp;rarr; &amp;pi;&lt;sub&gt;1&lt;/sub&gt;(Y).<br /> <br /> To check (all are clear):<br /> <br /> 1) This is well-defined.<br /> <br /> 2) It respects compositions.<br /> <br /> 3) It respects the identity.<br /> <br /> 4) &lt;math&gt; (f\circ g)_* &lt;/math&gt; = &lt;math&gt; f_*\circ g_* &lt;/math&gt;.<br /> <br /> '''Examples:'''<br /> <br /> 1) &amp;pi;&lt;sub&gt;1&lt;/sub&gt;(x&lt;sub&gt;0&lt;/sub&gt;, x&lt;sub&gt;0&lt;/sub&gt;) = {e}.<br /> <br /> 2) &amp;pi;&lt;sub&gt;1&lt;/sub&gt; ('''R'''&lt;sup&gt;n&lt;/sup&gt;, 0) = {e}<br /> <br /> 3) &amp;pi;&lt;sub&gt;1&lt;/sub&gt; (S&lt;sup&gt;1&lt;/sup&gt;, 1) = '''Z'''.</div> Trefor http://drorbn.net/index.php?title=0708-1300/Class_notes_for_Tuesday,_January_8 0708-1300/Class notes for Tuesday, January 8 2008-02-08T18:30:15Z <p>Trefor: </p> <hr /> <div>{{0708-1300/Navigation}}<br /> <br /> ==Algebraic Topology==<br /> <br /> Temporarily, at least, we will no longer assume that all structures are manifolds and that all functions are smooth. However, functions will still be assumed to be continuous.<br /> <br /> We will also, temporarily, assume that all spaces are pointed spaces. That is, a space X will be assumed to have a distinguished point x, whether mentioned explicitly or not.<br /> <br /> General idea of algebraic topology: to find &quot;functors&quot; from topology to algebra.<br /> <br /> '''Informal definition:'''<br /> A category C consists of<br /> <br /> 1) a collection Obj(C) of objects<br /> <br /> 2) a class hom(C) of morphisms between these objects, so that for each X&lt;sub&gt;1&lt;/sub&gt; and X&lt;sub&gt;2&lt;/sub&gt; in Obj(C) we get a set mor{X&lt;sub&gt;1&lt;/sub&gt;, X&lt;sub&gt;2&lt;/sub&gt;}<br /> <br /> 3) Composition laws:<br /> <br /> a) a binary function on hom(C): mor(X&lt;sub&gt;1&lt;/sub&gt;, X&lt;sub&gt;2&lt;/sub&gt;) x mor (X&lt;sub&gt;2&lt;/sub&gt;, X&lt;sub&gt;3&lt;/sub&gt; &amp;rarr; mor (X&lt;sub&gt;1&lt;/sub&gt;, X&lt;sub&gt;3&lt;/sub&gt;)<br /> <br /> b) Identities (for every X, there is an I&lt;sub&gt;X&lt;/sub&gt; in mor(X,X))<br /> <br /> 4) Compatibility laws:<br /> <br /> a) associativity of composition<br /> <br /> b) identities are in fact identities.<br /> <br /> <br /> <br /> &quot;Functors&quot; preserve categories.<br /> <br /> '''Example:'''<br /> Recall that the Brouwer Fixed Point Theorem is implied by the statement that there is no retract from the n-dimensional disc D&lt;sup&gt;n&lt;/sup&gt; to the (n-1)-dimensional sphere S&lt;sup&gt;n-1&lt;/sup&gt;. Later, we will prove this last statement in its categorical reformulation: we will find a functor H such that H(D&lt;sup&gt;&lt;/sup&gt;) = {0} and H(S&lt;sup&gt;n-1&lt;/sup&gt;) = '''Z'''. So the existence of a retract would imply that there are some functions f: '''Z''' &amp;rarr; {0} and g: {0} &amp;rarr; '''Z''' such that gf is the identity. Clearly this is false, so no retract exists.<br /> <br /> <br /> '''Definition:'''<br /> Given (X, x&lt;sub&gt;0&lt;/sub&gt;) (which, by the above comment, we will sometimes write simply as X) we define the &quot;fundamental group&quot; or &quot;Poincaré group&quot; of X:<br /> <br /> &amp;pi;&lt;sub&gt;1&lt;/sub&gt; (X, x&lt;sub&gt;0&lt;/sub&gt;) := { [&amp;gamma;]: &amp;gamma; : [0,1] &amp;rarr; X, &amp;gamma;(0) = &amp;gamma;(1) = x&lt;sub&gt;0&lt;/sub&gt;}<br /> <br /> <br /> Now we have to:<br /> <br /> 1) Define homotopy and prove that it's an equivalence relation<br /> <br /> 2) Define a binary operation, show that it's well-defined, and then show that the fundamental group is in fact a group.<br /> <br /> 3) Demonstrate functoriality<br /> <br /> <br /> '''Definition:'''<br /> If f, g : (X, x&lt;sub&gt;0&lt;/sub&gt;) &amp;rarr; (Y, y&lt;sub&gt;0&lt;/sub&gt;), we say that &quot;f is homotopic to g&quot; (and we write f ~ g) if there exists a function H : [0,1] &amp;times; X &amp;rarr; Y such that H restricted to {0}&amp;times;X is f, H restricted to {1}&amp;times;X is g, and H at any time t is in our category (ie H(t, x&lt;sub&gt;0&lt;/sub&gt;) = y&lt;sub&gt;0&lt;/sub&gt;.)<br /> <br /> <br /> We know now go through four boring claims, labeled BC1 - BC4. Many have simple visual proofs which I will not reproduce here.<br /> <br /> <br /> '''BC1:'''<br /> <br /> ~ is an equivalence relation.<br /> <br /> '''Proof of BC1:'''<br /> <br /> a) f ~ f for all functions f. Indeed, set H(t, x) = f(x) for all x.<br /> <br /> b) If f ~ g (by a function H) then g ~ f. Indeed, set H'(t,x) = H(1-t, x).<br /> <br /> c) If f ~ g (by a function H&lt;sub&gt;1&lt;/sub&gt;) and g ~ h (by a function H&lt;sub&gt;2&lt;/sub&gt;, then f ~ h. Indeed, set H(t, x) = H&lt;sub&gt;1&lt;/sub&gt;(2t,x) if t &amp;lt; ½ and H(t,x) = H&lt;sub&gt;2&lt;/sub&gt;(2t-1, x) if t &amp;gt; ½.<br /> <br /> <br /> '''Definition:'''<br /> We define a binary operation on the equivalence classes of paths:<br /> <br /> [&amp;gamma;] &amp;bull; [&amp;gamma;'] = [&amp;gamma; &amp;bull; &amp;gamma;']<br /> <br /> where the &amp;bull; represents the concatenation of paths. Intuitively, we follow the first path at twice its normal speed, then follow the second, again at twice its normal speed.<br /> <br /> '''BC2:'''<br /> This is well-defined,<br /> <br /> '''BC3:'''<br /> &amp;pi;&lt;sub&gt;1&lt;/sub&gt; (X, x&lt;sub&gt;0&lt;/sub&gt;) is a group.<br /> <br /> '''Proofette of BC3:'''<br /> <br /> Associativity: Let &amp;gamma;&lt;sub&gt;0&lt;/sub&gt;, &amp;gamma;&lt;sub&gt;1&lt;/sub&gt; and &amp;gamma;&lt;sub&gt;2&lt;/sub&gt; be paths. <br /> Note that while (&amp;gamma;&lt;sub&gt;0&lt;/sub&gt; &amp;bull; &amp;gamma;&lt;sub&gt;1&lt;/sub&gt;) &amp;bull; &amp;gamma;&lt;sub&gt;2&lt;/sub&gt; is not the same path as &amp;gamma;&lt;sub&gt;0&lt;/sub&gt; &amp;bull; (&amp;gamma;&lt;sub&gt;1&lt;/sub&gt; &amp;bull; &amp;gamma;&lt;sub&gt;2&lt;/sub&gt;), they are homotopic as paths.<br /> <br /> Identity: [e] = [x&lt;sub&gt;0&lt;/sub&gt;], where x&lt;sub&gt;0&lt;/sub&gt; is the constant path.<br /> <br /> Inverse: [&amp;gamma;]&lt;sup&gt;-1&lt;/sup&gt; = [&lt;math&gt;\tilde{\gamma}&lt;/math&gt;], where &lt;math&gt;\tilde{\gamma}&lt;/math&gt;(t) = &amp;gamma;(t). <br /> <br /> '''BC4:''' &amp;pi;&lt;sub&gt;1&lt;/sub&gt; is a functor.<br /> <br /> &lt;math&gt; f_* &lt;/math&gt; [&amp;gamma;] = [&lt;math&gt; f_* &lt;/math&gt; &amp;gamma;] = [f &amp;bull; &amp;gamma;] defines &lt;math&gt; f_* &lt;/math&gt; : &amp;pi;&lt;sub&gt;1&lt;/sub&gt;(f) : &amp;pi;&lt;sub&gt;1&lt;/sub&gt;(X) &amp;rarr; &amp;pi;&lt;sub&gt;1&lt;/sub&gt;(Y).<br /> <br /> To check (all are clear):<br /> <br /> 1) This is well-defined.<br /> <br /> 2) It respects compositions.<br /> <br /> 3) It respects the identity.<br /> <br /> 4) &lt;math&gt; (gf)_* &lt;/math&gt; = &lt;math&gt; f_* &lt;/math&gt; &amp;bull; &lt;math&gt; g_* &lt;/math&gt;.<br /> <br /> '''Examples:'''<br /> <br /> 1) &amp;pi;&lt;sub&gt;1&lt;/sub&gt;(x&lt;sub&gt;0&lt;/sub&gt;, x&lt;sub&gt;0&lt;/sub&gt;) = {e}.<br /> <br /> 2) &amp;pi;&lt;sub&gt;1&lt;/sub&gt; ('''R'''&lt;sup&gt;n&lt;/sup&gt;, 0) = {e}<br /> <br /> 3) &amp;pi;&lt;sub&gt;1&lt;/sub&gt; (S&lt;sup&gt;1&lt;/sup&gt;, 1) = '''Z'''.</div> Trefor http://drorbn.net/index.php?title=0708-1300/Class_notes_for_Thursday,_December_6 0708-1300/Class notes for Thursday, December 6 2008-02-08T16:18:53Z <p>Trefor: </p> <hr /> <div>{{0708-1300/Navigation}}<br /> ===Class Notes===<br /> <br /> &lt;span style=&quot;color: red;&quot;&gt;The notes below are by the students and for the students. Hopefully they are useful, but they come with no guarantee of any kind.&lt;/span&gt;<br /> <br /> <br /> [[Image:0708-1300-Dec06_01.jpg|400px]]<br /> [[Image:0708-1300-Dec06_02.jpg|400px]]<br /> [[Image:0708-1300-Dec06_03.jpg|400px]]</div> Trefor http://drorbn.net/index.php?title=0708-1300/Class_notes_for_Tuesday,_January_22 0708-1300/Class notes for Tuesday, January 22 2008-02-06T23:44:54Z <p>Trefor: /* Second Hour */</p> <hr /> <div>{{0708-1300/Navigation}}<br /> <br /> ==Pictures for a Van-Kampen Computation==<br /> {{In|<br /> n = 1 |<br /> in = &lt;nowiki&gt;&lt;&lt; KnotTheory&lt;/nowiki&gt;}}<br /> <br /> &lt;tt&gt;Loading KnotTheory version of January 13, 2008, 20:30:12.1353.&lt;br&gt;<br /> Read more at http://katlas.org/wiki/KnotTheory.&lt;/tt&gt;<br /> <br /> {{Graphics|<br /> n = 2 |<br /> in = &lt;nowiki&gt;TubePlot[TorusKnot[8, 3]]&lt;/nowiki&gt; |<br /> img= 0708-1300-T83.png}}<br /> <br /> {{In|<br /> n = 3 |<br /> in = &lt;nowiki&gt;TC[r1_, t1_,r2_,t2_ ] := {<br /> (r1 +r2 Cos[2Pi t2])Cos[2Pi t1],<br /> (r1 +r2 Cos[2Pi t2])Sin[2Pi t1],<br /> r2 Sin[2Pi t2]<br /> };&lt;/nowiki&gt;}}<br /> <br /> {{In|<br /> n = 4 |<br /> in = &lt;nowiki&gt;InflatedTorus[p_, q_, b_] := ParametricPlot3D[<br /> TC[<br /> 2, p t - q s,<br /> 1 + b(p^2 + q^2)s(1 - (p^2 + q^2)s), q t + p s<br /> ],<br /> {t, 0, 1}, {s, 0, 1/(p^2 + q^2)},<br /> PlotPoints -&gt; {6(p^2 + q^2) + 1, 7},<br /> DisplayFunction -&gt; Identity<br /> ];&lt;/nowiki&gt;}}<br /> <br /> {{Graphics|<br /> n = 5 |<br /> in = &lt;nowiki&gt;GraphicsArray[{{InflatedTorus[3,8,1], InflatedTorus[3,8,-1]}}]&lt;/nowiki&gt; |<br /> img= 0708-1300-InflatedTori.png |<br /> width = 640px}}<br /> <br /> ==Typed Notes==<br /> <br /> ===First Hour===<br /> <br /> '''Today's Agenda:''' <br /> <br /> 1) More Examples of Van-Kampen Theorem<br /> <br /> 2) More Diagrams<br /> <br /> 3) Proof of Van-Kampen (was not done)<br /> <br /> <br /> We began by recalling the examples from last class. I will not repeat that here, merely making a few additional comments that came out:<br /> <br /> <br /> '''Notation:'''<br /> <br /> Technically, &lt;math&gt;A*_H B&lt;/math&gt; is poor notion as it implies that knowledge of A, B and H is sufficient to construct &lt;math&gt;A*_H B&lt;/math&gt;. In fact, we ALSO need to know the maps from H into A and B respectively in order for &lt;math&gt;A*_H B&lt;/math&gt; to be defined. <br /> <br /> <br /> '''Aside'''<br /> <br /> Last class we simply wrote down the schematic for the two holed torus as an octagon with the identifications on the edges given last class. We now consider how one arrives at this schematic. <br /> <br /> To create the two holed torus one begins with two tori. One then cuts out a small open disk from each torus and then glues the two boundaries together. Let us consider what this looks like when considering a torus as the normal schematic with a square in the plane with the normal identification of the sides. Removing an open disk is equivalent to removing the inside of a loop starting at one of the corners and finishing at that same corner. This is equivalent to making a pentagon with sides &lt;math&gt;aba^{-1}b^{-1}c&lt;/math&gt; where c is the added edge. <br /> <br /> Consider two such pentagons, gluing along the edge c forms precisely the octagon we had for the two holed torus last class. <br /> <br /> [[Image:0708-1300_notes_22-01-08a.jpg|200px]]<br /> <br /> '''Proposition'''<br /> <br /> Letting &lt;math&gt;\Sigma_g&lt;/math&gt; denote the g holed torus, then &lt;math&gt;\Sigma_g\neq\Sigma_{g'}&lt;/math&gt;<br /> <br /> (Note, I used the symbol &lt;math&gt;\neq&lt;/math&gt; to as the normal \ncong command doesn't seem to work. Take its meaning in context.)<br /> <br /> <br /> Aside: Consider a functor from the category of groups to the category of Abelian groups via<br /> <br /> &lt;math&gt;G\mapsto G^{ab} = G/(ab=ba)&lt;/math&gt;<br /> <br /> If we have a (homo)morphism from &lt;math&gt;G\rightarrow H&lt;/math&gt; then the functor takes &lt;math&gt;H\rightarrow H^{ab}&lt;/math&gt; and yields a map &lt;math&gt;G^{ab}\rightarrow H^{ab}&lt;/math&gt; such that everything commutes. <br /> <br /> Hence we know that &lt;math&gt;\pi_1^{ab}(\Sigma_g) \cong \mathbb{Z}^{2g}\neq\mathbb{Z}^{2g'} \cong\pi_1^{ab}(\Sigma_{g'})&lt;/math&gt;<br /> <br /> Of course, we need to know that in fact &lt;math&gt;\mathbb{Z}^m\neq\mathbb{Z}^n&lt;/math&gt; if &lt;math&gt;m\neq n&lt;/math&gt;<br /> <br /> As such, since the abelianizations are not isomorphic,neither are the original groups and the spaces themselves are not homeomorphic. <br /> <br /> <br /> '''Example'''<br /> <br /> &lt;math&gt;\pi_1(\mathbb{R}P^2)&lt;/math&gt; is &lt;math&gt;\pi_1&lt;/math&gt; of the space which can be written as a disk with two antipodal points on the boundary circle on it with the identification that the top path a (going clockwise along the boundary) is glued to the bottom path (also going clock wise). But &lt;math&gt;\pi_1&lt;/math&gt; of this is just &lt;math&gt;&lt;a&gt;/(a^2 = e) \cong \mathbb{Z}/2&lt;/math&gt;<br /> <br /> <br /> '''Claim:'''<br /> <br /> Puncturing an n-manifold, &lt;math&gt;n\geq 3&lt;/math&gt;, does not change &lt;math&gt;\pi_1(M)&lt;/math&gt;. I.e., if &lt;math&gt;p\in M^n&lt;/math&gt; then &lt;math&gt;\pi_1(M)\cong\pi_1(M-\{p\})&lt;/math&gt;<br /> <br /> <br /> ''Proof:''<br /> <br /> Let &lt;math&gt;U_1 = M-\{p\}&lt;/math&gt;<br /> <br /> &lt;math&gt;U_2&lt;/math&gt; = a coordinate patch about p. <br /> <br /> Then &lt;math&gt;U_1\cap U_2 = B^n-\{p\}\cong S^{n-1}&lt;/math&gt;<br /> <br /> If n=3, &lt;math&gt;\pi_1(S^2) = \{e\}&lt;/math&gt; as we have computed before. <br /> <br /> Hence, &lt;math&gt;\pi_1(M) = \pi_1(U_1)*_{\{\}}\{\} = \pi_1(U_1)&lt;/math&gt;<br /> <br /> <br /> Now, &lt;math&gt;\pi_1(S^3)\cong\pi_1(S^3-\{p\}) = \pi_1(B^3) = \{e\}&lt;/math&gt;<br /> <br /> Continuing inductively the theorem holds for all n. <br /> <br /> <br /> '''Aside:'''<br /> <br /> If X is connected and &lt;math&gt;b_1,\ b_2\in X&lt;/math&gt; then &lt;math&gt;\pi_1(X,b_1) = \pi_1(X,b_2)&lt;/math&gt;. I.e., it does not matter which base point we choose in a connected space, the fundamental group is invariant of this. <br /> <br /> ''Proof''<br /> <br /> Consider a path &lt;math&gt;\eta&lt;/math&gt; from &lt;math&gt;b_1&lt;/math&gt; to &lt;math&gt;b_2&lt;/math&gt;. The returning path is denoted &lt;math&gt;\bar{\eta}&lt;/math&gt;<br /> <br /> Consider a loop from &lt;math&gt;b_2&lt;/math&gt; called &lt;math&gt;\gamma&lt;/math&gt;. <br /> <br /> Then get a loop from &lt;math&gt;b_1&lt;/math&gt; via &lt;math&gt;\gamma\mapsto \bar{\eta}\gamma\eta&lt;/math&gt;<br /> <br /> Similarly about &lt;math&gt;b_2, \gamma\mapsto\eta\gamma\bar{\eta}&lt;/math&gt;<br /> <br /> <br /> Considering the composition we get &lt;math&gt;\eta\bar{\eta}\gamma\eta\bar{\eta}\sim\gamma&lt;/math&gt;.<br /> <br /> ===Second Hour===<br /> <br /> '''Claim'''<br /> <br /> &lt;math&gt;S^3&lt;/math&gt; is the union (with common boundary) of two solid tori &lt;math&gt;S^1\times D^1&lt;/math&gt;<br /> <br /> The natural way to add two tori with common boundary would be two glue the boundaries of two disks (making &lt;math&gt;S^2&lt;/math&gt;) together for each angle going around the torus thus yielding &lt;math&gt;S^1\times S^2&lt;/math&gt;. Clearly this is not the same as &lt;math&gt;S^3&lt;/math&gt; as the fundamental groups differ. <br /> <br /> Instead consider the following description. Look at a torus in the zx plane, this looks like two disks with the z axis in between them such that rotating these two disks about the z axis will yield the torus. <br /> <br /> Lets now add in the second torus into this picture. We first draw a horizontal line between the two disks. We then &quot;blow&quot; up from beneath so the horizontal line is slightly curved. We imagine continuing to blow yielding larger and larger loops between the two disks until it &quot;pops&quot; forming the pure horizontal line consisting of the loop at infinity. Do the same for the bottom. Hence, the boundaries of the two tori drawn this way clearly are the same, and between the two cover the entire zx plane (and &quot;point at infinity). Rotating this picture about the z axis yields all of S^1 as the union of these two sets. <br /> <br /> [[Image:0708-1300_notes_22-01-08b.jpg|200px]]<br /> <br /> '''Claim:'''<br /> <br /> &lt;math&gt;\pi_1(S^3) = \{e\}&lt;/math&gt;<br /> <br /> &lt;math&gt;U_1&lt;/math&gt; = the normal solid torus thickened a bit and under &lt;math&gt;\pi_1&lt;/math&gt; yields &lt;math&gt;&lt;\alpha&gt;&lt;/math&gt; <br /> <br /> &lt;math&gt;U_2&lt;/math&gt; = the other solid torus, also thickened a bit, under &lt;math&gt;\pi_1&lt;/math&gt; yields &lt;math&gt;&lt;\beta&gt;&lt;/math&gt;<br /> <br /> <br /> &lt;math&gt;U_1\cap U_2&lt;/math&gt; is a normal torus only with slightly thick walls opposed to infinitely thin ones (homotopically the same)<br /> <br /> So, &lt;math&gt;\pi_1(U_1\cap U_2)\cong\mathbb{Z}^2 \cong &lt;a,b&gt;/ab=ba&lt;/math&gt;<br /> <br /> So, &lt;math&gt;\pi_1(S^3) \cong\mathbb{Z}*_{\mathbb{Z}\times\mathbb{Z}}\mathbb{Z}&lt;/math&gt;<br /> <br /> However, we still need to describe &lt;math&gt;i_{1*}&lt;/math&gt; and &lt;math&gt;i_{2*}&lt;/math&gt;<br /> <br /> Do do this let me describe a,b,&lt;math&gt;\alpha&lt;/math&gt; and &lt;math&gt;\beta&lt;/math&gt; explicitly.<br /> <br /> Considering the description of the two tori given above, we let a go around the outside of one of the two disks in the plane and b go from a point on the boundary of the same disk, following the rotation about the z axis, to a point on the boundary of the other dis. &lt;math&gt;\alpha&lt;/math&gt; is similar to b, but thought of as being on the boundary of the OTHER torus. &lt;math&gt;\beta&lt;/math&gt; consists of the path along the z axis. <br /> <br /> Hence, <br /> <br /> &lt;math&gt;i_{1*}:&lt;/math&gt; <br /> <br /> &lt;math&gt;a\rightarrow e&lt;/math&gt;<br /> <br /> &lt;math&gt;b\rightarrow \alpha&lt;/math&gt;<br /> <br /> &lt;math&gt;i_{2*}:&lt;/math&gt;<br /> <br /> &lt;math&gt; a\rightarrow\beta&lt;/math&gt;<br /> <br /> &lt;math&gt;b\rightarrow e&lt;/math&gt; (as it is contractible)<br /> <br /> <br /> Hence, &lt;math&gt;\pi_1(S^3) = F(\alpha, \beta)/(e=\beta, \alpha = e)&lt;/math&gt;<br /> <br /> Hence, &lt;math&gt;\pi_1(S^3) = \{e\}&lt;/math&gt;<br /> <br /> <br /> '''Example'''<br /> <br /> Define the &quot;Torus knot &lt;math&gt;T_{p,q}&lt;/math&gt;&quot; where p and q are relatively prime integers. The knot &lt;math&gt;T_{8,3}&lt;/math&gt; is given above. We can think of this in the following ways:<br /> <br /> 1) &lt;math&gt;T_{p,q}&lt;/math&gt; is the knot that wraps around the torus p times one way and q times the other way. <br /> <br /> 2) Formally, let &lt;math&gt;\sigma:S^1\times S^1\mathbb{R}^3&lt;/math&gt; be standard embedding of a torus. Let &lt;math&gt;\gamma:[0,1]\rightarrow S^1\times S^1&lt;/math&gt; be &lt;math&gt;t\rightarrow (e^{2\pi i pt}, e^{i2\pi qt})&lt;/math&gt;<br /> <br /> Then &lt;math&gt;T_{p,q}&lt;/math&gt; is &lt;math&gt;\sigma\circ\gamma&lt;/math&gt;<br /> <br /> <br /> 3) Recall that the torus can be thought of as the image of the mapping &lt;math&gt;\mathbb{R}^2\rightarrow\mathbb{R}^2/\mathbb{Z}^2&lt;/math&gt;<br /> <br /> Consider the rectangle in the real plane: ([0,p],[0,q]) and consider the path which is the diagonal line from the corner (0,0) to the corner (p,q)<br /> <br /> No two points on this line are the same under the mapping down to the torus. If they were, then &lt;math&gt;\Delta y&lt;/math&gt; and &lt;math&gt;\Delta x&lt;/math&gt; would be integers and hence &lt;math&gt;\Delta y/\Delta x&lt;/math&gt; would be the slope of the line. But the slope of the line is q/p which is already in lowest common terms by assumption. <br /> <br /> <br /> <br /> Lets compute the fundamental group of the compliment of the torus knot. <br /> <br /> &lt;math&gt;\pi_1(\mathbb{R}^3-T_{p,q})\cong\pi_1(S^3-T_{p,q})&lt;/math&gt;<br /> <br /> &lt;math&gt;U_1&lt;/math&gt;: inflated bagel, constrained by &lt;math&gt;T_{p,q}&lt;/math&gt;<br /> <br /> &lt;math&gt;U_2&lt;/math&gt;: inflated bubble constrained by &lt;math&gt;T_{p,q}&lt;/math&gt;<br /> <br /> (See top of page for pictures)<br /> <br /> The intersection &lt;math&gt;U_1\cap U_2&lt;/math&gt; looks somewhat like a belt. It has some thickness to it and is wrapped around the torus, eventually forming a loop. Hence it looks like a squashed disk cross a circle. Hence, under &lt;math&gt;\pi_1&lt;/math&gt; this is just &lt;math&gt;\mathbb{Z}\cong&lt;\gamma&gt;&lt;/math&gt; where &lt;math&gt;\gamma&lt;/math&gt; is the path parallel to &lt;math&gt;T_{p,q}&lt;/math&gt;<br /> <br /> <br /> We thus get the maps, <br /> <br /> &lt;math&gt;i_{1*}: \gamma\mapsto\alpha^p&lt;/math&gt;<br /> <br /> &lt;math&gt;i_{2*}:\gamma\mapsto\beta^q&lt;/math&gt;<br /> <br /> Hence, &lt;math&gt;\pi_1(T_{p,q}^c) = &lt;\alpha,\beta&gt;/\alpha^p = \beta^q&lt;/math&gt;<br /> <br /> Thus, &lt;math&gt;T_{p,q}\neq T_{p',q'}&lt;/math&gt;<br /> <br /> <br /> '''Diagrams:'''<br /> <br /> <br /> Recall our diagram from last class: <br /> <br /> <br /> &lt;math&gt;\begin{matrix}<br /> &amp;\ \ \ \ U_1&amp;&amp;\\<br /> &amp;\nearrow^{i_1}&amp;\searrow^{j_1}&amp;\\<br /> U_1\cap U_2&amp;&amp;&amp;U_1\cup U_2\\<br /> &amp;\searrow_{i_2}&amp;\nearrow^{j_2}&amp;\\<br /> &amp;\ \ \ \ U_2&amp;&amp;\\<br /> \end{matrix}&lt;/math&gt;<br /> <br /> <br /> &lt;math&gt;U_1\cup U_2&lt;/math&gt; can be ''defined'' as the object such that the above diagram commutes and should the following commute: <br /> <br /> &lt;math&gt;\begin{matrix}<br /> &amp;\ \ \ \ U_1&amp;&amp;\\<br /> &amp;\nearrow^{i_1}&amp;\searrow^{j_1}&amp;\\<br /> U_1\cap U_2&amp;&amp;&amp;Y\\<br /> &amp;\searrow_{i_2}&amp;\nearrow^{j_2}&amp;\\<br /> &amp;\ \ \ \ U_2&amp;&amp;\\<br /> \end{matrix}&lt;/math&gt;<br /> <br /> Then there is a unique map between &lt;math&gt;U_1\cup U_2&lt;/math&gt; and Y such that the composed diagram commutes. <br /> <br /> <br /> Indeed, the same is true for general categories. <br /> <br /> For <br /> <br /> &lt;math&gt;\begin{matrix}<br /> &amp;\ \ \ \ G_1&amp;&amp;\\<br /> &amp;\nearrow^{i_1}&amp;\searrow^{j_1}&amp;\\<br /> H&amp;&amp;&amp;P\\<br /> &amp;\searrow_{i_2}&amp;\nearrow^{j_2}&amp;\\<br /> &amp;\ \ \ \ G_2&amp;&amp;\\<br /> \end{matrix}&lt;/math&gt;<br /> <br /> commuting, P is defined as an object such that if also <br /> <br /> &lt;math&gt;\begin{matrix}<br /> &amp;\ \ \ \ G_1&amp;&amp;\\<br /> &amp;\nearrow^{i_1}&amp;\searrow^{j_1}&amp;\\<br /> H&amp;&amp;&amp;Q\\<br /> &amp;\searrow_{i_2}&amp;\nearrow^{j_2}&amp;\\<br /> &amp;\ \ \ \ G_2&amp;&amp;\\<br /> \end{matrix}&lt;/math&gt;<br /> <br /> were to commute then there is a unique morphism from P to Q such that the composed diagram computes. <br /> <br /> <br /> In the category of groups, this &quot;pushforward&quot; P is unique and is isomorphic to &lt;math&gt;G_1*_H G_2&lt;/math&gt;</div> Trefor http://drorbn.net/index.php?title=0708-1300/Class_notes_for_Tuesday,_January_22 0708-1300/Class notes for Tuesday, January 22 2008-02-06T23:43:36Z <p>Trefor: /* Second Hour */</p> <hr /> <div>{{0708-1300/Navigation}}<br /> <br /> ==Pictures for a Van-Kampen Computation==<br /> {{In|<br /> n = 1 |<br /> in = &lt;nowiki&gt;&lt;&lt; KnotTheory&lt;/nowiki&gt;}}<br /> <br /> &lt;tt&gt;Loading KnotTheory version of January 13, 2008, 20:30:12.1353.&lt;br&gt;<br /> Read more at http://katlas.org/wiki/KnotTheory.&lt;/tt&gt;<br /> <br /> {{Graphics|<br /> n = 2 |<br /> in = &lt;nowiki&gt;TubePlot[TorusKnot[8, 3]]&lt;/nowiki&gt; |<br /> img= 0708-1300-T83.png}}<br /> <br /> {{In|<br /> n = 3 |<br /> in = &lt;nowiki&gt;TC[r1_, t1_,r2_,t2_ ] := {<br /> (r1 +r2 Cos[2Pi t2])Cos[2Pi t1],<br /> (r1 +r2 Cos[2Pi t2])Sin[2Pi t1],<br /> r2 Sin[2Pi t2]<br /> };&lt;/nowiki&gt;}}<br /> <br /> {{In|<br /> n = 4 |<br /> in = &lt;nowiki&gt;InflatedTorus[p_, q_, b_] := ParametricPlot3D[<br /> TC[<br /> 2, p t - q s,<br /> 1 + b(p^2 + q^2)s(1 - (p^2 + q^2)s), q t + p s<br /> ],<br /> {t, 0, 1}, {s, 0, 1/(p^2 + q^2)},<br /> PlotPoints -&gt; {6(p^2 + q^2) + 1, 7},<br /> DisplayFunction -&gt; Identity<br /> ];&lt;/nowiki&gt;}}<br /> <br /> {{Graphics|<br /> n = 5 |<br /> in = &lt;nowiki&gt;GraphicsArray[{{InflatedTorus[3,8,1], InflatedTorus[3,8,-1]}}]&lt;/nowiki&gt; |<br /> img= 0708-1300-InflatedTori.png |<br /> width = 640px}}<br /> <br /> ==Typed Notes==<br /> <br /> ===First Hour===<br /> <br /> '''Today's Agenda:''' <br /> <br /> 1) More Examples of Van-Kampen Theorem<br /> <br /> 2) More Diagrams<br /> <br /> 3) Proof of Van-Kampen (was not done)<br /> <br /> <br /> We began by recalling the examples from last class. I will not repeat that here, merely making a few additional comments that came out:<br /> <br /> <br /> '''Notation:'''<br /> <br /> Technically, &lt;math&gt;A*_H B&lt;/math&gt; is poor notion as it implies that knowledge of A, B and H is sufficient to construct &lt;math&gt;A*_H B&lt;/math&gt;. In fact, we ALSO need to know the maps from H into A and B respectively in order for &lt;math&gt;A*_H B&lt;/math&gt; to be defined. <br /> <br /> <br /> '''Aside'''<br /> <br /> Last class we simply wrote down the schematic for the two holed torus as an octagon with the identifications on the edges given last class. We now consider how one arrives at this schematic. <br /> <br /> To create the two holed torus one begins with two tori. One then cuts out a small open disk from each torus and then glues the two boundaries together. Let us consider what this looks like when considering a torus as the normal schematic with a square in the plane with the normal identification of the sides. Removing an open disk is equivalent to removing the inside of a loop starting at one of the corners and finishing at that same corner. This is equivalent to making a pentagon with sides &lt;math&gt;aba^{-1}b^{-1}c&lt;/math&gt; where c is the added edge. <br /> <br /> Consider two such pentagons, gluing along the edge c forms precisely the octagon we had for the two holed torus last class. <br /> <br /> [[Image:0708-1300_notes_22-01-08a.jpg|200px]]<br /> <br /> '''Proposition'''<br /> <br /> Letting &lt;math&gt;\Sigma_g&lt;/math&gt; denote the g holed torus, then &lt;math&gt;\Sigma_g\neq\Sigma_{g'}&lt;/math&gt;<br /> <br /> (Note, I used the symbol &lt;math&gt;\neq&lt;/math&gt; to as the normal \ncong command doesn't seem to work. Take its meaning in context.)<br /> <br /> <br /> Aside: Consider a functor from the category of groups to the category of Abelian groups via<br /> <br /> &lt;math&gt;G\mapsto G^{ab} = G/(ab=ba)&lt;/math&gt;<br /> <br /> If we have a (homo)morphism from &lt;math&gt;G\rightarrow H&lt;/math&gt; then the functor takes &lt;math&gt;H\rightarrow H^{ab}&lt;/math&gt; and yields a map &lt;math&gt;G^{ab}\rightarrow H^{ab}&lt;/math&gt; such that everything commutes. <br /> <br /> Hence we know that &lt;math&gt;\pi_1^{ab}(\Sigma_g) \cong \mathbb{Z}^{2g}\neq\mathbb{Z}^{2g'} \cong\pi_1^{ab}(\Sigma_{g'})&lt;/math&gt;<br /> <br /> Of course, we need to know that in fact &lt;math&gt;\mathbb{Z}^m\neq\mathbb{Z}^n&lt;/math&gt; if &lt;math&gt;m\neq n&lt;/math&gt;<br /> <br /> As such, since the abelianizations are not isomorphic,neither are the original groups and the spaces themselves are not homeomorphic. <br /> <br /> <br /> '''Example'''<br /> <br /> &lt;math&gt;\pi_1(\mathbb{R}P^2)&lt;/math&gt; is &lt;math&gt;\pi_1&lt;/math&gt; of the space which can be written as a disk with two antipodal points on the boundary circle on it with the identification that the top path a (going clockwise along the boundary) is glued to the bottom path (also going clock wise). But &lt;math&gt;\pi_1&lt;/math&gt; of this is just &lt;math&gt;&lt;a&gt;/(a^2 = e) \cong \mathbb{Z}/2&lt;/math&gt;<br /> <br /> <br /> '''Claim:'''<br /> <br /> Puncturing an n-manifold, &lt;math&gt;n\geq 3&lt;/math&gt;, does not change &lt;math&gt;\pi_1(M)&lt;/math&gt;. I.e., if &lt;math&gt;p\in M^n&lt;/math&gt; then &lt;math&gt;\pi_1(M)\cong\pi_1(M-\{p\})&lt;/math&gt;<br /> <br /> <br /> ''Proof:''<br /> <br /> Let &lt;math&gt;U_1 = M-\{p\}&lt;/math&gt;<br /> <br /> &lt;math&gt;U_2&lt;/math&gt; = a coordinate patch about p. <br /> <br /> Then &lt;math&gt;U_1\cap U_2 = B^n-\{p\}\cong S^{n-1}&lt;/math&gt;<br /> <br /> If n=3, &lt;math&gt;\pi_1(S^2) = \{e\}&lt;/math&gt; as we have computed before. <br /> <br /> Hence, &lt;math&gt;\pi_1(M) = \pi_1(U_1)*_{\{\}}\{\} = \pi_1(U_1)&lt;/math&gt;<br /> <br /> <br /> Now, &lt;math&gt;\pi_1(S^3)\cong\pi_1(S^3-\{p\}) = \pi_1(B^3) = \{e\}&lt;/math&gt;<br /> <br /> Continuing inductively the theorem holds for all n. <br /> <br /> <br /> '''Aside:'''<br /> <br /> If X is connected and &lt;math&gt;b_1,\ b_2\in X&lt;/math&gt; then &lt;math&gt;\pi_1(X,b_1) = \pi_1(X,b_2)&lt;/math&gt;. I.e., it does not matter which base point we choose in a connected space, the fundamental group is invariant of this. <br /> <br /> ''Proof''<br /> <br /> Consider a path &lt;math&gt;\eta&lt;/math&gt; from &lt;math&gt;b_1&lt;/math&gt; to &lt;math&gt;b_2&lt;/math&gt;. The returning path is denoted &lt;math&gt;\bar{\eta}&lt;/math&gt;<br /> <br /> Consider a loop from &lt;math&gt;b_2&lt;/math&gt; called &lt;math&gt;\gamma&lt;/math&gt;. <br /> <br /> Then get a loop from &lt;math&gt;b_1&lt;/math&gt; via &lt;math&gt;\gamma\mapsto \bar{\eta}\gamma\eta&lt;/math&gt;<br /> <br /> Similarly about &lt;math&gt;b_2, \gamma\mapsto\eta\gamma\bar{\eta}&lt;/math&gt;<br /> <br /> <br /> Considering the composition we get &lt;math&gt;\eta\bar{\eta}\gamma\eta\bar{\eta}\sim\gamma&lt;/math&gt;.<br /> <br /> ===Second Hour===<br /> <br /> '''Claim'''<br /> <br /> &lt;math&gt;S^3&lt;/math&gt; is the union (with common boundary) of two solid tori &lt;math&gt;S^1\times D^1&lt;/math&gt;<br /> <br /> The natural way to add two tori with common boundary would be two glue the boundaries of two disks (making &lt;math&gt;S^2&lt;/math&gt;) together for each angle going around the torus thus yielding &lt;math&gt;S^1\times S^2&lt;/math&gt;. Clearly this is not the same as &lt;math&gt;S^3&lt;/math&gt; as the fundamental groups differ. <br /> <br /> Instead consider the following description. Look at a torus in the zx plane, this looks like two disks with the z axis in between them such that rotating these two disks about the z axis will yield the torus. <br /> <br /> Lets now add in the second torus into this picture. We first draw a horizontal line between the two disks. We then &quot;blow&quot; up from beneath so the horizontal line is slightly curved. We imagine continuing to blow yielding larger and larger loops between the two disks until it &quot;pops&quot; forming the pure horizontal line consisting of the loop at infinity. Do the same for the bottom. Hence, the boundaries of the two tori drawn this way clearly are the same, and between the two cover the entire zx plane (and &quot;point at infinity). Rotating this picture about the z axis yields all of S^1 as the union of these two sets. <br /> <br /> [[Image:0708-1300_notes_22-01-08b.jpg|200px]]<br /> <br /> '''Claim:'''<br /> <br /> &lt;math&gt;\pi_1(S^3) = \{e\}&lt;/math&gt;<br /> <br /> &lt;math&gt;U_1&lt;/math&gt; = the normal solid torus thickened a bit and under &lt;math&gt;\pi_1&lt;/math&gt; yields &lt;math&gt;&lt;\alpha&gt;&lt;/math&gt; <br /> <br /> &lt;math&gt;U_2&lt;/math&gt; = the other solid torus, also thickened a bit, under &lt;math&gt;\pi_1&lt;/math&gt; yields &lt;math&gt;&lt;\beta&gt;&lt;/math&gt;<br /> &lt;math&gt;<br /> <br /> <br /> U_1\cap U_2&lt;/math&gt; is a normal torus only with slightly thick walls opposed to infinitely thin ones (homotopically the same)<br /> <br /> So, &lt;math&gt;\pi_1(U_1\cap U_2)\cong\mathbb{Z}^2 \cong &lt;a,b&gt;/ab=ba&lt;/math&gt;<br /> <br /> So, &lt;math&gt;\pi_1(S^3) \cong\mathbb{Z}*_{\mathbb{Z}\times\mathbb{Z}}\mathbb{Z}&lt;/math&gt;<br /> <br /> However, we still need to describe &lt;math&gt;i_{1*}&lt;/math&gt; and &lt;math&gt;i_{2*}&lt;/math&gt;<br /> <br /> Do do this let me describe a,b,&lt;math&gt;\alpha&lt;/math&gt; and &lt;math&gt;\beta&lt;/math&gt; explicitly.<br /> <br /> Considering the description of the two tori given above, we let a go around the outside of one of the two disks in the plane and b go from a point on the boundary of the same disk, following the rotation about the z axis, to a point on the boundary of the other dis. &lt;math&gt;\alpha&lt;/math&gt; is similar to b, but thought of as being on the boundary of the OTHER torus. &lt;math&gt;\beta&lt;/math&gt; consists of the path along the z axis. <br /> <br /> Hence, <br /> <br /> &lt;math&gt;i_{1*}:&lt;/math&gt; <br /> <br /> &lt;math&gt;a\rightarrow e&lt;/math&gt;<br /> <br /> &lt;math&gt;b\rightarrow \alpha&lt;/math&gt;<br /> <br /> &lt;math&gt;i_{2*}:&lt;/math&gt;<br /> <br /> &lt;math&gt; a\rightarrow\beta&lt;/math&gt;<br /> <br /> &lt;math&gt;b\rightarrow e&lt;/math&gt; (as it is contractible)<br /> <br /> <br /> Hence, &lt;math&gt;\pi_1(S^3) = F(\alpha, \beta)/(e=\beta, \alpha = e)&lt;/math&gt;<br /> <br /> Hence, &lt;math&gt;\pi_1(S^3) = \{e\}&lt;/math&gt;<br /> <br /> <br /> '''Example'''<br /> <br /> Define the &quot;Torus knot &lt;math&gt;T_{p,q}&lt;/math&gt;&quot; where p and q are relatively prime integers. The knot &lt;math&gt;T_{8,3}&lt;/math&gt; is given above. We can think of this in the following ways:<br /> <br /> 1) &lt;math&gt;T_{p,q}&lt;/math&gt; is the knot that wraps around the torus p times one way and q times the other way. <br /> <br /> 2) Formally, let &lt;math&gt;\sigma:S^1\times S^1\mathbb{R}^3&lt;/math&gt; be standard embedding of a torus. Let &lt;math&gt;\gamma:[0,1]\rightarrow S^1\times S^1&lt;/math&gt; be &lt;math&gt;t\rightarrow (e^{2\pi i pt}, e^{i2\pi qt})&lt;/math&gt;<br /> <br /> Then &lt;math&gt;T_{p,q}&lt;/math&gt; is &lt;math&gt;\sigma\circ\gamma&lt;/math&gt;<br /> <br /> <br /> 3) Recall that the torus can be thought of as the image of the mapping &lt;math&gt;\mathbb{R}^2\rightarrow\mathbb{R}^2/\mathbb{Z}^2&lt;/math&gt;<br /> <br /> Consider the rectangle in the real plane: ([0,p],[0,q]) and consider the path which is the diagonal line from the corner (0,0) to the corner (p,q)<br /> <br /> No two points on this line are the same under the mapping down to the torus. If they were, then &lt;math&gt;\Delta y&lt;/math&gt; and &lt;math&gt;\Delta x&lt;/math&gt; would be integers and hence &lt;math&gt;\Delta y/\Delta x&lt;/math&gt; would be the slope of the line. But the slope of the line is q/p which is already in lowest common terms by assumption. <br /> <br /> <br /> <br /> Lets compute the fundamental group of the compliment of the torus knot. <br /> <br /> &lt;math&gt;\pi_1(\mathbb{R}^3-T_{p,q})\cong\pi_1(S^3-T_{p,q})&lt;/math&gt;<br /> <br /> &lt;math&gt;U_1&lt;/math&gt;: inflated bagel, constrained by &lt;math&gt;T_{p,q}&lt;/math&gt;<br /> <br /> &lt;math&gt;U_2&lt;/math&gt;: inflated bubble constrained by &lt;math&gt;T_{p,q}&lt;/math&gt;<br /> <br /> (See top of page for pictures)<br /> <br /> The intersection &lt;math&gt;U_1\cap U_2&lt;/math&gt; looks somewhat like a belt. It has some thickness to it and is wrapped around the torus, eventually forming a loop. Hence it looks like a squashed disk cross a circle. Hence, under &lt;math&gt;\pi_1&lt;/math&gt; this is just &lt;math&gt;\mathbb{Z}\cong&lt;\gamma&gt;&lt;/math&gt; where &lt;math&gt;\gamma&lt;/math&gt; is the path parallel to &lt;math&gt;T_{p,q}&lt;/math&gt;<br /> <br /> <br /> We thus get the maps, <br /> <br /> &lt;math&gt;i_{1*}: \gamma\mapsto\alpha^p&lt;/math&gt;<br /> <br /> &lt;math&gt;i_{2*}:\gamma\mapsto\beta^q&lt;/math&gt;<br /> <br /> Hence, &lt;math&gt;\pi_1(T_{p,q}^c) = &lt;\alpha,\beta&gt;/\alpha^p = \beta^q&lt;/math&gt;<br /> <br /> Thus, &lt;math&gt;T_{p,q}\neq T_{p',q'}&lt;/math&gt;<br /> <br /> <br /> '''Diagrams:'''<br /> <br /> <br /> Recall our diagram from last class: <br /> <br /> <br /> &lt;math&gt;\begin{matrix}<br /> &amp;\ \ \ \ U_1&amp;&amp;\\<br /> &amp;\nearrow^{i_1}&amp;\searrow^{j_1}&amp;\\<br /> U_1\cap U_2&amp;&amp;&amp;U_1\cup U_2\\<br /> &amp;\searrow_{i_2}&amp;\nearrow^{j_2}&amp;\\<br /> &amp;\ \ \ \ U_2&amp;&amp;\\<br /> \end{matrix}&lt;/math&gt;<br /> <br /> <br /> &lt;math&gt;U_1\cup U_2&lt;/math&gt; can be ''defined'' as the object such that the above diagram commutes and should the following commute: <br /> <br /> &lt;math&gt;\begin{matrix}<br /> &amp;\ \ \ \ U_1&amp;&amp;\\<br /> &amp;\nearrow^{i_1}&amp;\searrow^{j_1}&amp;\\<br /> U_1\cap U_2&amp;&amp;&amp;Y\\<br /> &amp;\searrow_{i_2}&amp;\nearrow^{j_2}&amp;\\<br /> &amp;\ \ \ \ U_2&amp;&amp;\\<br /> \end{matrix}&lt;/math&gt;<br /> <br /> Then there is a unique map between &lt;math&gt;U_1\cup U_2&lt;/math&gt; and Y such that the composed diagram commutes. <br /> <br /> <br /> Indeed, the same is true for general categories. <br /> <br /> For <br /> <br /> &lt;math&gt;\begin{matrix}<br /> &amp;\ \ \ \ G_1&amp;&amp;\\<br /> &amp;\nearrow^{i_1}&amp;\searrow^{j_1}&amp;\\<br /> H&amp;&amp;&amp;P\\<br /> &amp;\searrow_{i_2}&amp;\nearrow^{j_2}&amp;\\<br /> &amp;\ \ \ \ G_2&amp;&amp;\\<br /> \end{matrix}&lt;/math&gt;<br /> <br /> commuting, P is defined as an object such that if also <br /> <br /> &lt;math&gt;\begin{matrix}<br /> &amp;\ \ \ \ G_1&amp;&amp;\\<br /> &amp;\nearrow^{i_1}&amp;\searrow^{j_1}&amp;\\<br /> H&amp;&amp;&amp;Q\\<br /> &amp;\searrow_{i_2}&amp;\nearrow^{j_2}&amp;\\<br /> &amp;\ \ \ \ G_2&amp;&amp;\\<br /> \end{matrix}&lt;/math&gt;<br /> <br /> were to commute then there is a unique morphism from P to Q such that the composed diagram computes. <br /> <br /> <br /> In the category of groups, this &quot;pushforward&quot; P is unique and is isomorphic to &lt;math&gt;G_1*_H G_2&lt;/math&gt;</div> Trefor http://drorbn.net/index.php?title=0708-1300/Class_notes_for_Thursday,_January_17 0708-1300/Class notes for Thursday, January 17 2008-02-06T23:16:08Z <p>Trefor: </p> <hr /> <div>{{0708-1300/Navigation}}<br /> <br /> '''Van-Kampen's Theorem'''<br /> <br /> Let X be a point pointed topological space such that &lt;math&gt;X = U_1\cup U_2&lt;/math&gt; where &lt;math&gt;U_1&lt;/math&gt; and &lt;math&gt;U_2&lt;/math&gt; are open and the base point b is in the (connected) intersection. <br /> <br /> Then, &lt;math&gt;\pi_1(X) = \pi_1(U_1)*_{\pi_1(U_1\cap U_2)}\pi_1(U_2)&lt;/math&gt;<br /> <br /> <br /> <br /> &lt;math&gt;\begin{matrix}<br /> &amp;\ \ \ \ U_1&amp;&amp;\\<br /> &amp;\nearrow^{i_1}&amp;\searrow^{j_1}&amp;\\<br /> U_1\cap U_2&amp;&amp;&amp;U_1\cup U_2 = X\\<br /> &amp;\searrow_{i_2}&amp;\nearrow^{j_2}&amp;\\<br /> &amp;\ \ \ \ U_2&amp;&amp;\\<br /> \end{matrix}&lt;/math&gt;<br /> <br /> where all the i's and j's are inclusions. <br /> <br /> <br /> Lets consider the image of this under the functor &lt;math&gt;\pi_1&lt;/math&gt;<br /> <br /> <br /> &lt;math&gt;\begin{matrix}<br /> &amp;\ \ \ \ \pi_1(U_1)&amp;&amp;\\<br /> &amp;\nearrow^{i_{1*}}&amp;\searrow^{j_{1*}}&amp;\\<br /> \pi_1(U_1\cap U_2)&amp;&amp;&amp; \pi_1(X)\\<br /> &amp;\searrow_{i_{2*}}&amp;\nearrow^{j_{2*}}&amp;\\<br /> &amp;\ \ \ \ \pi(U_2)&amp;&amp;\\<br /> \end{matrix}&lt;/math&gt;<br /> <br /> <br /> Now consider the situation as groups:<br /> <br /> <br /> &lt;math&gt;\begin{matrix}<br /> &amp;\ \ \ \ G_1&amp;&amp;\\<br /> &amp;\nearrow_{\varphi_1}&amp;\searrow&amp;\\<br /> H&amp;&amp;&amp;G_1*_H G_2\\<br /> &amp;\searrow_{\varphi_2}&amp;\nearrow&amp;\\<br /> &amp;\ \ \ \ G_2&amp;&amp;\\<br /> \end{matrix}&lt;/math&gt;<br /> <br /> <br /> Where &lt;math&gt;G_1 *_H G_2 = &lt;/math&gt;{ words with letters alternating between being in &lt;math&gt;G_1&lt;/math&gt; and &lt;math&gt;G_2&lt;/math&gt;, ignoring e } / See Later<br /> <br /> Considering just the set without the identification, we note this is a group with the operation being concatenation of words followed by reduction. <br /> <br /> Ex: &lt;math&gt;a_1b_1a_2 + a_3b_2a_4 = a_1b_1ab_2a_4&lt;/math&gt; where &lt;math&gt;a = a_2a_3&lt;/math&gt;<br /> <br /> <br /> '''Claim:'''<br /> <br /> This is really a group. <br /> <br /> <br /> So far, we have only defined the &quot;free group of &lt;math&gt;G_1&lt;/math&gt; and &lt;math&gt;G_2&lt;/math&gt;&quot;. We now consider the identification (denoted above by 'See Later') which is <br /> <br /> &lt;math&gt;\forall h\in H, \phi_1(h) = \phi_2(h&lt;/math&gt;)<br /> <br /> With this identification we have properly defined &lt;math&gt;G_1 *_H G_2&lt;/math&gt;<br /> <br /> <br /> Note: &lt;math&gt;G_1 *_H G_2&lt;/math&gt; is equivalent to { words in &lt;math&gt;G_1\cap G_2\}/ (e_1 = \{\}, e_2 = \{\}, g,h\in G_i, g\cdot h = gh)&lt;/math&gt;<br /> <br /> <br /> '''Example 0'''<br /> <br /> &lt;math&gt;\pi_1(S^n) &lt;/math&gt; for &lt;math&gt; n\geq 2&lt;/math&gt;<br /> <br /> We can think of &lt;math&gt;S^n&lt;/math&gt; as the union of two slightly overlapping open hemispheres which leaves the intersection as a band about the equator. As long as &lt;math&gt;n\geq 2&lt;/math&gt; this is connected (but fails for &lt;math&gt;S^1&lt;/math&gt;)<br /> <br /> So, &lt;math&gt;\pi_1(S^n) = \pi_1(U_1)*_{\pi(U_1\cap U_2)}\pi_1(U_2)&lt;/math&gt;<br /> <br /> But, since the hemispheres themselves are contractible, &lt;math&gt;\pi_1(U_1) = \pi_1(U_2) = \{e\}&lt;/math&gt;<br /> <br /> Hence, &lt;math&gt;\pi_1(S^n) = \{e\}&lt;/math&gt;<br /> <br /> <br /> <br /> '''Example 1'''<br /> <br /> Let us consider &lt;math&gt;\pi_1&lt;/math&gt; of a a figure eight. Let &lt;math&gt;U_1&lt;/math&gt; denote everything above a line slightly beneath the intersection and &lt;math&gt;U_2&lt;/math&gt; everything below a line slightly above the intersection point. <br /> <br /> Now both &lt;math&gt;U_1&lt;/math&gt; and &lt;math&gt;U_2&lt;/math&gt; are homotopically equivalent to a loop and so &lt;math&gt;\pi_1(U_1) = \pi_2(U_2) = \mathbb{Z}&lt;/math&gt;. We can think of these being the groups generated by a loop going around once, I.e., isomorphic to &lt;math&gt;&lt;\alpha&gt;&lt;/math&gt; and &lt;math&gt;&lt;\beta&gt;&lt;/math&gt; respectively. <br /> <br /> The intersection is an X, contractible to a point and so &lt;math&gt;\pi_1(U_1\cap U_2) = \{e\}&lt;/math&gt;<br /> <br /> So &lt;math&gt;\pi_1&lt;/math&gt;(figure 8)&lt;math&gt; = &lt;\alpha&gt;*_{\{\}}&lt;\beta&gt; = F(\alpha,\beta)&lt;/math&gt; the free group generated by &lt;math&gt;\alpha&lt;/math&gt; and &lt;math&gt;\beta&lt;/math&gt;<br /> <br /> <br /> This is non abelian <br /> <br /> <br /> '''Example 2'''<br /> <br /> &lt;math&gt;\pi_1(\mathbb{T}^2)&lt;/math&gt;<br /> <br /> We consider &lt;math&gt;\mathbb{T}^2&lt;/math&gt; in the normal way as a square with the normal identifications on the sides. We then consider two concentric squares inside this and define &lt;math&gt;U_1&lt;/math&gt; as everything inside the larger square and &lt;math&gt;U_2&lt;/math&gt; as everything outside the smaller square. <br /> <br /> Clearly &lt;math&gt;U_1&lt;/math&gt; is contractible, and hence &lt;math&gt;\pi_1(U_1) = \{e\}&lt;/math&gt;<br /> <br /> <br /> Now, the intersection of &lt;math&gt;U_1&lt;/math&gt; and &lt;math&gt;U_2&lt;/math&gt; is equivalent to an annulus and so &lt;math&gt;\pi_1(U_1\cap U_2) = \mathbb{Z} = &lt;\gamma&gt;&lt;/math&gt; where &lt;math&gt;\gamma&lt;/math&gt; is just a loop in the annulus. <br /> <br /> Now considering &lt;math&gt;U_2&lt;/math&gt;, we note that each of the four outer corner points in the big square are identified, and when we identify edges we are left with something equivalent to a figure 8. <br /> <br /> Hence &lt;math&gt;\pi_1(U_2) = F(\alpha, \beta)&lt;/math&gt; as in example 1<br /> <br /> <br /> Hence, &lt;math&gt;\pi_1(\mathbb{T}^2) = \{e\}*F(\alpha,\beta)/(i_{1*}(\gamma) = i_{2*}(\gamma))&lt;/math&gt;<br /> <br /> Now, &lt;math&gt;i_{1*}(\gamma) = e&lt;/math&gt; <br /> <br /> and<br /> <br /> &lt;math&gt;i_{2*}(\gamma) = \alpha\beta\alpha^{-1}\beta^{-1}&lt;/math&gt;<br /> <br /> I.e., &lt;math&gt;\pi_1(\mathbb{T}^2) = F(\alpha,\beta)/ e = \alpha\beta\alpha^{-1}\beta^{-1}&lt;/math&gt;<br /> <br /> &lt;math&gt; = F(\alpha,\beta)/(\alpha\beta = \beta\alpha)&lt;/math&gt;<br /> <br /> This is just the Free Abelian group on two symbols and, <br /> <br /> &lt;math&gt;= \{\alpha^n\beta^m\} = \mathbb{Z}^2&lt;/math&gt;<br /> <br /> Hence, &lt;math&gt;\pi_1(\mathbb{T}^2) = \mathbb{Z}^2&lt;/math&gt;<br /> <br /> <br /> '''Example 3'''<br /> <br /> <br /> The two holed torus: &lt;math&gt;\Sigma_2&lt;/math&gt;<br /> <br /> Consider the schematic for this surface, consising of an octagon with edges labeled &lt;math&gt;a_1,b_1,a_1^{-1},b_1^{-1},a_2,b_2,a_2^{-1},b_2^{-1}&lt;/math&gt;<br /> <br /> As in the previous example, consider two concentric circles inside the octagon. Let everything inside the larger circle be &lt;math&gt;U_1&lt;/math&gt; and everything outside the smaller circle be &lt;math&gt;U_2&lt;/math&gt;. <br /> <br /> Clearly &lt;math&gt;\pi_1(U_1) = \{e\}&lt;/math&gt; as before. <br /> <br /> &lt;math&gt;\pi_1(U_1\cap U_2) = &lt;\gamma&gt;&lt;/math&gt; as before. <br /> <br /> Now, &lt;math&gt;U_2&lt;/math&gt; this times when doing the identifications looks like a clover (4 loops intersecting at one point)<br /> <br /> Completely analogously to before, we see that &lt;math&gt;\pi_1(U_2) = F(\alpha_1, \beta_1, \alpha_2, \beta_2)&lt;/math&gt;<br /> <br /> Again, &lt;math&gt;i_{1*}(\gamma) = e&lt;/math&gt;<br /> <br /> &lt;math&gt;i_{2*}(\gamma) = \alpha_1\beta_1\alpha_1^{-1}\alpha_2\beta_2\alpha_1^{-1}\beta_2^{-1}&lt;/math&gt;<br /> <br /> <br /> Therefore, <br /> <br /> &lt;math&gt;\pi_{\Sigma_2} = F(\alpha_1, \beta_1, \alpha_2, \beta_2)/(e =\alpha_1\beta_1\alpha_1^{-1}\alpha_2\beta_2\alpha_1^{-1}\beta_2^{-1})&lt;/math&gt;<br /> <br /> The ''abelianization'' of this group is <br /> <br /> &lt;math&gt;\pi_1^{ab}(\Sigma_2) = \pi_1(\Sigma_2)/ gh=hg = F.A.G (\alpha_1,\alpha_2,\beta_1,\beta^2) = \mathbb{Z}^4 \neq \mathbb{Z}^2&lt;/math&gt;<br /> <br /> <br /> In case someone might want diagrams for the examples above:<br /> <br /> [[Image:0708-1300_notes_17-01-08c.jpg|200px]]</div> Trefor http://drorbn.net/index.php?title=User:Trefor User:Trefor 2008-02-04T20:05:37Z <p>Trefor: </p> <hr /> <div>{{0708-1300/Navigation}}<br /> NOTE: This page is used as a placeholder for incomplete typed class notes. <br /> <br /> <br /> ==Typed Notes==<br /> <br /> &lt;span style=&quot;color: red;&quot;&gt;The notes below are by the students and for the students. Hopefully they are useful, but they come with no guarantee of any kind.&lt;/span&gt;<br /> <br /> <br /> We start with a one further comment on Van-Kampen:<br /> <br /> Let &lt;math&gt;X = \bigcup_{\alpha\in A}U_{\alpha}&lt;/math&gt;, with &lt;math&gt;U_{\alpha}&lt;/math&gt; open and with &lt;math&gt;\forall \alpha,\beta&lt;/math&gt; then &lt;math&gt;U_{\alpha}\cap U_{\beta}&lt;/math&gt; connected.<br /> <br /> Then, &lt;math&gt;\pi_1(X) = *\pi_1(U_{\alpha})/(\gamma\in\pi_1(U_{\alpha}\cap U_{\beta}),\ and\ j_{\alpha\beta*}(\gamma) = j_{\beta\alpha*}(\gamma))&lt;/math&gt;<br /> <br /> We actually also need &lt;math&gt;\forall\alpha,\beta,\gamma,\delta&lt;/math&gt; then &lt;math&gt;U_{\alpha}\cap U_{\beta}\cap U_{\gamma}\cap U_{\delta}&lt;/math&gt; to be connected. Recall we previously had a square grid in our homotopy, so needed the four sections around a grid point to intersect in a connected way. <br /> <br /> However, we can employ a clever trick to reduce this to only needing triples of &lt;math&gt;U_{\alpha_i}&lt;/math&gt; to be connected. That is, instead of covering it was squares on top of each other, cover it with rectangles arranged in the way bricks are laid in a wall.<br /> <br /> <br /> '''Definition'''<br /> &lt;math&gt;<br /> p:X\rightarrow B&lt;/math&gt; is a &quot;covering&quot; if, for some fixed discrete set F, every &lt;math&gt;b\in B&lt;/math&gt; has a neighborhood U such that &lt;math&gt;p^{-1}(U)\cong F\times U&lt;/math&gt;. We have the natural map &lt;math&gt;\pi:F\times U\rightarrow U&lt;/math&gt;<br /> <br /> <br /> Now a naive dream would be to classify all such covering spaces. <br /> <br /> <br /> '''Example 1'''<br /> <br /> The double annulus depicted above is a covering space for the single annulus<br /> <br /> <br /> '''Example 2'''<br /> <br /> Consider a series of rectangular planes stacked on top of each other. Make a small cut in each plane, and then glue one side of the cut to the opposite side in the plane above. Finally glue the remaining unglued sides in the very top and bottom planes to each other. <br /> <br /> Analogously this can be done in higher dimensions. <br /> <br /> <br /> '''Theorem'''<br /> <br /> For a &quot;decent&quot; B (a topological condition to be discussed later) there is a bijection between connected coverings of B and subgroups of &lt;math&gt;\pi_1(B)&lt;/math&gt;<br /> <br /> given by &lt;math&gt;x\mapsto p_*\pi_1(X)&lt;/math&gt; included in &lt;math&gt;\pi_1(B)&lt;/math&gt;<br /> <br /> <br /> '''Example 3'''<br /> <br /> Consider &lt;math&gt;\mathbb{R}P^3&lt;/math&gt;<br /> <br /> Now, &lt;math&gt;\pi_1(\mathbb{R}P^3) \cong\pi_1(\mathbb{R}P^3 - \{pt\})\cong \pi_1&lt;/math&gt;(equator with identification of antipodes)&lt;math&gt;\cong\mathbb{Z}_2&lt;/math&gt; as we have previously computed. <br /> <br /> There are only two subgroups of &lt;math&gt;\mathbb{Z}_2&lt;/math&gt;, namely {e} and &lt;math&gt;\mathbb{Z}_2&lt;/math&gt;<br /> <br /> <br /> Lets consider the case where the subgroup is &lt;math&gt;\mathbb{Z}_2&lt;/math&gt;. Then &lt;math&gt;I:\mathbb{R}P^3\rightarrow\mathbb{Z}P^3&lt;/math&gt; the identity covering with F merely a point. So &lt;math&gt;p_*\pi_1(X) = \pi(X)&lt;/math&gt;<br /> <br /> <br /> For {e}, &lt;math&gt;p:S^3\rightarrow\mathbb{R}P^3&lt;/math&gt; is the other covering. <br /> <br /> <br /> ''Aside:''<br /> &lt;math&gt;<br /> \mathbb{R}P^3 = SO(3)&lt;/math&gt;<br /> <br /> Consider a belt. A point on it can be associated with three vectors. One vector is tangent to the belt in the direction of one end. The other vector is normal to the belt. And the third is normal to both of these. <br /> <br /> Now fix the orientation of the end points. Hence, we can think of the belt as a path in SO(3) as a homotopy. Pulling the belt tight is the identity homotopy. <br /> <br /> We note that if one twists the orientation of the end 360 degree, then this is non trivial and there is no way to return this to the identity while holding the ends fixed in orientation. However, if you twists the orientation of an end point 720 degrees then in fact you CAN &quot;untwist&quot; the belt without changing the orientation of the endpoints!</div> Trefor http://drorbn.net/index.php?title=Template:0708-1300/Navigation Template:0708-1300/Navigation 2008-02-01T20:22:33Z <p>Trefor: </p> <hr /> <div>{| cellpadding=&quot;0&quot; cellspacing=&quot;0&quot; style=&quot;clear: right; float: right&quot;<br /> |- align=right<br /> |&lt;div class=&quot;NavFrame&quot;&gt;&lt;div class=&quot;NavHead&quot;&gt;[[0708-1300]]/[[Template:0708-1300/Navigation|Navigation Panel]]&amp;nbsp;&amp;nbsp;&lt;/div&gt;<br /> &lt;div class=&quot;NavContent&quot;&gt;<br /> {| border=&quot;1px&quot; cellpadding=&quot;1&quot; cellspacing=&quot;0&quot; width=&quot;220&quot; style=&quot;margin: 0 0 1em 0.5em; font-size: small&quot;<br /> |-<br /> |colspan=3 align=center|[[Image:0708-1300-ClassPhoto.jpg|215px]]&lt;br&gt;[[0708-1300/Class Photo|Add your name / see who's in!]]<br /> |- align=center<br /> !#<br /> !Week of...<br /> !Links<br /> |-<br /> |colspan=3 align=center|'''Fall Semester'''<br /> |- align=left<br /> |align=center|1<br /> |Sep 10<br /> |[[0708-1300/About This Class|About]], [[0708-1300/Class notes for Tuesday, September 11|Tue]], [[0708-1300/Class notes for Thursday, September 13|Thu]]<br /> |- align=left<br /> |align=center|2<br /> |Sep 17<br /> |[[0708-1300/Class notes for Tuesday, September 18|Tue]], [[0708-1300/Homework Assignment 1|HW1]], [[0708-1300/Class notes for Thursday, September 20|Thu]]<br /> |- align=left<br /> |align=center|3<br /> |Sep 24<br /> |[[0708-1300/Class notes for Tuesday, September 25|Tue]], [[0708-1300/Class Photo|Photo]], [[0708-1300/Class notes for Thursday, September 27|Thu]]<br /> |- align=left<br /> |align=center|4<br /> |Oct 1<br /> |[[0708-1300/Questionnaire|Questionnaire]], [[0708-1300/Class notes for Tuesday, October 2|Tue]], [[0708-1300/Homework Assignment 2|HW2]], [[0708-1300/Class notes for Thursday, October 4|Thu]]<br /> |- align=left<br /> |align=center|5<br /> |Oct 8<br /> |Thanksgiving, [[0708-1300/Class notes for Tuesday, October_9|Tue]], [[0708-1300/Class notes for Thursday, October 11|Thu]]<br /> |- align=left<br /> |align=center|6<br /> |Oct 15<br /> |[[0708-1300/Class notes for Tuesday, October 16|Tue]], [[0708-1300/Homework Assignment 3|HW3]], [[0708-1300/Class notes for Thursday, October 18|Thu]]<br /> |- align=left<br /> |align=center|7<br /> |Oct 22<br /> |[[0708-1300/Class notes for Tuesday, October 23|Tue]], [[0708-1300/Class notes for Thursday, October 25|Thu]]<br /> |- align=left<br /> |align=center|8<br /> |Oct 29<br /> |[[0708-1300/Class notes for Tuesday, October 30|Tue]], [[0708-1300/Homework Assignment 4|HW4]], [[0708-1300/Class notes for Thursday, November 1|Thu]], [[0708-1300/the unit sphere in a Hilbert space is contractible|Hilbert sphere]]<br /> |- align=left<br /> |align=center|9<br /> |Nov 5<br /> |[[0708-1300/Class notes for Tuesday, November 6|Tue]],[[0708-1300/Class notes for Thursday, November 8|Thu]], [[0708-1300/Term Exam 1|TE1]]<br /> |- align=left<br /> |align=center|10<br /> |Nov 12<br /> |[[0708-1300/Class notes for Tuesday, November 13|Tue]], &lt;strike&gt;[[0708-1300/Class notes for Thursday, November 15|Thu]]&lt;/strike&gt;<br /> |- align=left<br /> |align=center|11<br /> |Nov 19<br /> |[[0708-1300/Class notes for Tuesday, November 20|Tue]], [[0708-1300/Homework Assignment 5|HW5]]<br /> |- align=left<br /> |align=center|12<br /> |Nov 26<br /> |[[0708-1300/Class notes for Tuesday, November 27|Tue]], [[0708-1300/Class notes for Thursday, November 29|Thu]]<br /> |- align=left<br /> |align=center|13<br /> |Dec 3<br /> |[[0708-1300/Class notes for Tuesday, December 4|Tue]], [[0708-1300/Homework Assignment 6|HW6]]<br /> |-<br /> |colspan=3 align=center|'''Spring Semester'''<br /> |- align=left<br /> |align=center|14<br /> |Jan 7<br /> |[[0708-1300/Class notes for Tuesday, January 8|Tue]], [[0708-1300/Class notes for Thursday, January 10|Thu]], [[0708-1300/Homework Assignment 7|HW7]]<br /> |- align=left<br /> |align=center|15<br /> |Jan 14<br /> |[[0708-1300/Class notes for Tuesday, January 15|Tue]], [[0708-1300/Class notes for Thursday, January 17|Thu]]<br /> |- align=left<br /> |align=center|16<br /> |Jan 21<br /> |[[0708-1300/Class notes for Tuesday, January 22|Tue]], [[0708-1300/Class notes for Thursday, January 24|Thu]],[[0708-1300/Homework Assignment 8|HW8]]<br /> |- align=left<br /> |align=center|17<br /> |Jan 28<br /> |[[0708-1300/Class notes for Tuesday, January 29|Tue]]<br /> |- align=left<br /> |align=center|18<br /> |Feb 4<br /> |HW9<br /> |- align=left<br /> |align=center|19<br /> |Feb 11<br /> |TE2; Feb 17: last chance to drop class<br /> |- align=left<br /> |align=center|R<br /> |Feb 18<br /> |<br /> |- align=left<br /> |align=center|20<br /> |Feb 25<br /> |HW10<br /> |- align=left<br /> |align=center|21<br /> |Mar 3<br /> |<br /> |- align=left<br /> |align=center|22<br /> |Mar 10<br /> |HW11<br /> |- align=left<br /> |align=center|23<br /> |Mar 17<br /> |<br /> |- align=left<br /> |align=center|24<br /> |Mar 24<br /> |HW12<br /> |- align=left<br /> |align=center|25<br /> |Mar 31<br /> |<br /> |- align=left<br /> |align=center|26<br /> |Apr 7<br /> |<br /> |- align=center<br /> |colspan=3|[[0708-1300/Register of Good Deeds|Register of Good Deeds]]<br /> |- align=center<br /> |colspan=3|[[0708-1300/Errata to Bredon's Book|Errata to Bredon's Book]]<br /> |}<br /> &lt;/div&gt;&lt;/div&gt;<br /> |}<br /> &lt;div align=center style=&quot;color: red; font-size: 150%; display: none;&quot;&gt;Announcements go here&lt;/div&gt;</div> Trefor http://drorbn.net/index.php?title=0708-1300/Class_notes_for_Thursday,_January_24 0708-1300/Class notes for Thursday, January 24 2008-02-01T20:21:41Z <p>Trefor: </p> <hr /> <div><br /> {{0708-1300/Navigation}}<br /> <br /> ==Typed Notes==<br /> <br /> &lt;span style=&quot;color: red;&quot;&gt;The notes below are by the students and for the students. Hopefully they are useful, but they come with no guarantee of any kind.&lt;/span&gt;<br /> <br /> <br /> '''Proof of Van Kampen'''<br /> <br /> Let &lt;math&gt;G_i = \pi_1(U_i)&lt;/math&gt;<br /> <br /> &lt;math&gt;H = \pi_1(Y_1\cap U_2)&lt;/math&gt;<br /> <br /> &lt;math&gt;G=\pi_1(U_1\cup U_2)&lt;/math&gt;<br /> <br /> We aim to show that &lt;math&gt;G=G_1*_H G_2&lt;/math&gt;<br /> <br /> <br /> Hence, we want to define two maps:<br /> <br /> &lt;math&gt;\Phi:G_1*_H G_2\rightarrow G &lt;/math&gt; and<br /> <br /> <br /> &lt;math&gt;\Psi:G\rightarrow G_1*_H G_2&lt;/math&gt;<br /> <br /> such that they are inverses of each other. <br /> <br /> <br /> Now, recall the commuting diagram: <br /> <br /> &lt;math&gt;\begin{matrix}<br /> &amp;\ \ \ \ U_1&amp;&amp;\\<br /> &amp;\nearrow^{i_1}&amp;\searrow^{j_1}&amp;\\<br /> U_1\cap U_2&amp;&amp;&amp;U_1\cup U_2 = X\\<br /> &amp;\searrow_{i_2}&amp;\nearrow^{j_2}&amp;\\<br /> &amp;\ \ \ \ U_2&amp;&amp;\\<br /> \end{matrix}&lt;/math&gt;<br /> <br /> <br /> Further, let &lt;math&gt;b_i&lt;/math&gt; alternate between 1 and 2 for successive i's. <br /> <br /> Hence we define &lt;math&gt;\Phi&lt;/math&gt; via for &lt;math&gt;\alpha_i\in G_{b_i}&lt;/math&gt;, <br /> <br /> [&lt;math&gt;\alpha_1][\alpha_2]\ldots[\alpha_n]\rightarrow [j_{b_1 *}\alpha_1\cdot\ldots\cdot j_{b_n *}\alpha_n]&lt;/math&gt;<br /> <br /> Clearly this is well defined. We need to check the relations in <br /> &lt;math&gt;G_1*_H G_2&lt;/math&gt; indeed hold. Well, the identity element corresponds to the identity path so the relation that removes identities holds. Furthermore, the concatenation of paths is a sum after &lt;math&gt;\Phi&lt;/math&gt; and the definition necessitates the third relation holds. <br /> <br /> <br /> Now for &lt;math&gt;\Psi&lt;/math&gt;:<br /> <br /> Elements in G correspond with paths &lt;math&gt;\gamma&lt;/math&gt; in &lt;math&gt;U_1\cup U_2&lt;/math&gt;<br /> <br /> <br /> On consider such a &lt;math&gt;\gamma&lt;/math&gt;. The Lesbegue Lemma let us break &lt;math&gt;\gamma&lt;/math&gt; up so &lt;math&gt;\gamma = \gamma_1\ldots\gamma_N&lt;/math&gt; such that each &lt;math&gt;\gamma_i&lt;/math&gt; is in just &lt;math&gt;G_{b_i}&lt;/math&gt;<br /> <br /> Now, &lt;math&gt;\gamma_i&lt;/math&gt; does not go from base point to base point, so we can't consider it as a loop itself. Let &lt;math&gt;x_i&lt;/math&gt; denote the endpoint of &lt;math&gt;\gamma_i&lt;/math&gt; (and hence &lt;math&gt;x_{i-1}&lt;/math&gt; is the beginning point) and further, &lt;math&gt;x_0 = b&lt;/math&gt; the base point. <br /> <br /> Choose paths &lt;math&gt;\eta_i&lt;/math&gt; connecting &lt;math&gt;x_i&lt;/math&gt; to b such that if &lt;math&gt;x_i\in U_{b_i}&lt;/math&gt; then &lt;math&gt;\eta_i\sub U_{b_i}&lt;/math&gt;<br /> <br /> Hence, &lt;math&gt;\gamma\sim\bar{\eta_0}\gamma_1\eta_1\bar{\eta_1}\gamma_2\eta_2\ldots \mapsto [\bar{\eta_0}\gamma_1\eta_1][\bar{\eta_1}\gamma_2\eta_2]\ldots&lt;/math&gt;<br /> <br /> where each section is entirely in &lt;math&gt;G_{b_i}&lt;/math&gt;. This above mapping is &lt;math&gt;\Psi&lt;/math&gt;<br /> <br /> We need to show that &lt;math&gt;\Psi&lt;/math&gt; is well defined. I.e., <br /> <br /> 1) &lt;math&gt;\Psi&lt;/math&gt; is independent of the subdivision<br /> <br /> 2) &lt;math&gt;\Psi&lt;/math&gt; is independent of choice of the &lt;math&gt;\eta_i&lt;/math&gt;'s<br /> <br /> 3) If &lt;math&gt;\gamma_1\sim\gamma_2&lt;/math&gt; then &lt;math&gt;\Psi(\gamma_1) = \Psi(\gamma_2)&lt;/math&gt;<br /> <br /> <br /> For 1) it is enough to show that one can add or remove a single subdivision point. Suppose you add a new subdivision point c in between two others a and b. Traveling between a and b is the same as going from a to c then to the basepoint and back and then continuing on to b. But this is precisely what happens when you add a subdivision point. Likewise for removing points. <br /> <br /> <br /> For 2), Suppose &lt;math&gt;\eta_i&lt;/math&gt; goes from the basepoint to &lt;math&gt;x_i&lt;/math&gt;. Add a subdivision point y right beside &lt;math&gt;x_i&lt;/math&gt; and remove the one at &lt;math&gt;x_i&lt;/math&gt;. Then add again the original point &lt;math&gt;x_i&lt;/math&gt;. The trick is that the new &lt;math&gt;\eta_i'&lt;/math&gt; is the one used when adding y. The map from the new &lt;math&gt;x_i&lt;/math&gt; to the basepoint is this &lt;math&gt;\eta_i'&lt;/math&gt; with the infinitesimal connection between y and &lt;math&gt;x_i&lt;/math&gt; added. The subdivision point y is then removed leaving our original configuration of basepoints, only with the path from &lt;math&gt;x_i&lt;/math&gt; to b now being &lt;math&gt;\eta_i'&lt;/math&gt; instead of the original &lt;math&gt;\eta_i&lt;/math&gt;<br /> <br /> <br /> For 3) we consider the homotopy between &lt;math&gt;\gamma_1&lt;/math&gt; and &lt;math&gt;\gamma_2&lt;/math&gt; thought of as a square with &lt;math&gt;\gamma_1&lt;/math&gt; on the bottom and &lt;math&gt;\gamma_2&lt;/math&gt; on the top. We use the Lesbegue Lemma to subdivide the square into many subsquares such that each one is entirely in &lt;math&gt;U_{b_i}&lt;/math&gt;<br /> <br /> We further modify the homotopy H to a new homotopy &lt;math&gt;\tilde{H}&lt;/math&gt; such that each grid point gets mapped to b. <br /> <br /> This can be thought of as &quot;pinching&quot; each gird point and pulling it to b or, alternatively, as tossing a &quot;handkerchief on a bed of nails&quot; <br /> <br /> If we let &lt;math&gt;\gamma_1&lt;/math&gt; be broken in to section &lt;math&gt;\alpha_1,\ldots, \alpha_n&lt;/math&gt; along the bottom then &lt;math&gt;\tilde{H}&lt;/math&gt; first lifts &lt;math&gt;\alpha_1&lt;/math&gt; to the top and right sides of the subsquare such that the gridpoint goes to b. &lt;math&gt;\tilde{H}&lt;/math&gt; then moves the next square up in an analogous manner until we are at the top with &lt;math&gt;\gamma_2&lt;/math&gt;.</div> Trefor http://drorbn.net/index.php?title=0708-1300/Class_notes_for_Thursday,_January_24 0708-1300/Class notes for Thursday, January 24 2008-02-01T20:21:31Z <p>Trefor: /* Typed Notes */</p> <hr /> <div>INCOMPLETE AND UNEDITED: Completion coming soon. <br /> <br /> <br /> {{0708-1300/Navigation}}<br /> <br /> ==Typed Notes==<br /> <br /> &lt;span style=&quot;color: red;&quot;&gt;The notes below are by the students and for the students. Hopefully they are useful, but they come with no guarantee of any kind.&lt;/span&gt;<br /> <br /> <br /> '''Proof of Van Kampen'''<br /> <br /> Let &lt;math&gt;G_i = \pi_1(U_i)&lt;/math&gt;<br /> <br /> &lt;math&gt;H = \pi_1(Y_1\cap U_2)&lt;/math&gt;<br /> <br /> &lt;math&gt;G=\pi_1(U_1\cup U_2)&lt;/math&gt;<br /> <br /> We aim to show that &lt;math&gt;G=G_1*_H G_2&lt;/math&gt;<br /> <br /> <br /> Hence, we want to define two maps:<br /> <br /> &lt;math&gt;\Phi:G_1*_H G_2\rightarrow G &lt;/math&gt; and<br /> <br /> <br /> &lt;math&gt;\Psi:G\rightarrow G_1*_H G_2&lt;/math&gt;<br /> <br /> such that they are inverses of each other. <br /> <br /> <br /> Now, recall the commuting diagram: <br /> <br /> &lt;math&gt;\begin{matrix}<br /> &amp;\ \ \ \ U_1&amp;&amp;\\<br /> &amp;\nearrow^{i_1}&amp;\searrow^{j_1}&amp;\\<br /> U_1\cap U_2&amp;&amp;&amp;U_1\cup U_2 = X\\<br /> &amp;\searrow_{i_2}&amp;\nearrow^{j_2}&amp;\\<br /> &amp;\ \ \ \ U_2&amp;&amp;\\<br /> \end{matrix}&lt;/math&gt;<br /> <br /> <br /> Further, let &lt;math&gt;b_i&lt;/math&gt; alternate between 1 and 2 for successive i's. <br /> <br /> Hence we define &lt;math&gt;\Phi&lt;/math&gt; via for &lt;math&gt;\alpha_i\in G_{b_i}&lt;/math&gt;, <br /> <br /> [&lt;math&gt;\alpha_1][\alpha_2]\ldots[\alpha_n]\rightarrow [j_{b_1 *}\alpha_1\cdot\ldots\cdot j_{b_n *}\alpha_n]&lt;/math&gt;<br /> <br /> Clearly this is well defined. We need to check the relations in <br /> &lt;math&gt;G_1*_H G_2&lt;/math&gt; indeed hold. Well, the identity element corresponds to the identity path so the relation that removes identities holds. Furthermore, the concatenation of paths is a sum after &lt;math&gt;\Phi&lt;/math&gt; and the definition necessitates the third relation holds. <br /> <br /> <br /> Now for &lt;math&gt;\Psi&lt;/math&gt;:<br /> <br /> Elements in G correspond with paths &lt;math&gt;\gamma&lt;/math&gt; in &lt;math&gt;U_1\cup U_2&lt;/math&gt;<br /> <br /> <br /> On consider such a &lt;math&gt;\gamma&lt;/math&gt;. The Lesbegue Lemma let us break &lt;math&gt;\gamma&lt;/math&gt; up so &lt;math&gt;\gamma = \gamma_1\ldots\gamma_N&lt;/math&gt; such that each &lt;math&gt;\gamma_i&lt;/math&gt; is in just &lt;math&gt;G_{b_i}&lt;/math&gt;<br /> <br /> Now, &lt;math&gt;\gamma_i&lt;/math&gt; does not go from base point to base point, so we can't consider it as a loop itself. Let &lt;math&gt;x_i&lt;/math&gt; denote the endpoint of &lt;math&gt;\gamma_i&lt;/math&gt; (and hence &lt;math&gt;x_{i-1}&lt;/math&gt; is the beginning point) and further, &lt;math&gt;x_0 = b&lt;/math&gt; the base point. <br /> <br /> Choose paths &lt;math&gt;\eta_i&lt;/math&gt; connecting &lt;math&gt;x_i&lt;/math&gt; to b such that if &lt;math&gt;x_i\in U_{b_i}&lt;/math&gt; then &lt;math&gt;\eta_i\sub U_{b_i}&lt;/math&gt;<br /> <br /> Hence, &lt;math&gt;\gamma\sim\bar{\eta_0}\gamma_1\eta_1\bar{\eta_1}\gamma_2\eta_2\ldots \mapsto [\bar{\eta_0}\gamma_1\eta_1][\bar{\eta_1}\gamma_2\eta_2]\ldots&lt;/math&gt;<br /> <br /> where each section is entirely in &lt;math&gt;G_{b_i}&lt;/math&gt;. This above mapping is &lt;math&gt;\Psi&lt;/math&gt;<br /> <br /> We need to show that &lt;math&gt;\Psi&lt;/math&gt; is well defined. I.e., <br /> <br /> 1) &lt;math&gt;\Psi&lt;/math&gt; is independent of the subdivision<br /> <br /> 2) &lt;math&gt;\Psi&lt;/math&gt; is independent of choice of the &lt;math&gt;\eta_i&lt;/math&gt;'s<br /> <br /> 3) If &lt;math&gt;\gamma_1\sim\gamma_2&lt;/math&gt; then &lt;math&gt;\Psi(\gamma_1) = \Psi(\gamma_2)&lt;/math&gt;<br /> <br /> <br /> For 1) it is enough to show that one can add or remove a single subdivision point. Suppose you add a new subdivision point c in between two others a and b. Traveling between a and b is the same as going from a to c then to the basepoint and back and then continuing on to b. But this is precisely what happens when you add a subdivision point. Likewise for removing points. <br /> <br /> <br /> For 2), Suppose &lt;math&gt;\eta_i&lt;/math&gt; goes from the basepoint to &lt;math&gt;x_i&lt;/math&gt;. Add a subdivision point y right beside &lt;math&gt;x_i&lt;/math&gt; and remove the one at &lt;math&gt;x_i&lt;/math&gt;. Then add again the original point &lt;math&gt;x_i&lt;/math&gt;. The trick is that the new &lt;math&gt;\eta_i'&lt;/math&gt; is the one used when adding y. The map from the new &lt;math&gt;x_i&lt;/math&gt; to the basepoint is this &lt;math&gt;\eta_i'&lt;/math&gt; with the infinitesimal connection between y and &lt;math&gt;x_i&lt;/math&gt; added. The subdivision point y is then removed leaving our original configuration of basepoints, only with the path from &lt;math&gt;x_i&lt;/math&gt; to b now being &lt;math&gt;\eta_i'&lt;/math&gt; instead of the original &lt;math&gt;\eta_i&lt;/math&gt;<br /> <br /> <br /> For 3) we consider the homotopy between &lt;math&gt;\gamma_1&lt;/math&gt; and &lt;math&gt;\gamma_2&lt;/math&gt; thought of as a square with &lt;math&gt;\gamma_1&lt;/math&gt; on the bottom and &lt;math&gt;\gamma_2&lt;/math&gt; on the top. We use the Lesbegue Lemma to subdivide the square into many subsquares such that each one is entirely in &lt;math&gt;U_{b_i}&lt;/math&gt;<br /> <br /> We further modify the homotopy H to a new homotopy &lt;math&gt;\tilde{H}&lt;/math&gt; such that each grid point gets mapped to b. <br /> <br /> This can be thought of as &quot;pinching&quot; each gird point and pulling it to b or, alternatively, as tossing a &quot;handkerchief on a bed of nails&quot; <br /> <br /> If we let &lt;math&gt;\gamma_1&lt;/math&gt; be broken in to section &lt;math&gt;\alpha_1,\ldots, \alpha_n&lt;/math&gt; along the bottom then &lt;math&gt;\tilde{H}&lt;/math&gt; first lifts &lt;math&gt;\alpha_1&lt;/math&gt; to the top and right sides of the subsquare such that the gridpoint goes to b. &lt;math&gt;\tilde{H}&lt;/math&gt; then moves the next square up in an analogous manner until we are at the top with &lt;math&gt;\gamma_2&lt;/math&gt;.</div> Trefor http://drorbn.net/index.php?title=0708-1300/Class_notes_for_Thursday,_January_24 0708-1300/Class notes for Thursday, January 24 2008-01-28T18:03:25Z <p>Trefor: </p> <hr /> <div>INCOMPLETE AND UNEDITED: Completion coming soon. <br /> <br /> <br /> {{0708-1300/Navigation}}<br /> <br /> ==Typed Notes==<br /> <br /> &lt;span style=&quot;color: red;&quot;&gt;The notes below are by the students and for the students. Hopefully they are useful, but they come with no guarantee of any kind.&lt;/span&gt;<br /> <br /> <br /> '''Proof of Van Kampen'''<br /> <br /> Let G_i = \pi_1(U_i)<br /> H = \pi_1(Y_1\cap U_2)<br /> G=\pi_1(U_1\cup U_2)<br /> <br /> We aim to show that G=G_1*_H G_2<br /> <br /> <br /> Hence, we want to define two maps:<br /> <br /> \Phi:G_1*_H G_2\rightarrow G and <br /> <br /> \Psi:G\rightarrow G_1*_H G_2<br /> <br /> such that they are inverses of each other. <br /> <br /> <br /> Now, recall the commuting diagram: <br /> <br /> &lt;math&gt;\begin{matrix}<br /> &amp;\ \ \ \ U_1&amp;&amp;\\<br /> &amp;\nearrow^{i_1}&amp;\searrow^{j_1}&amp;\\<br /> U_1\cap U_2&amp;&amp;&amp;U_1\cup U_2 = X\\<br /> &amp;\searrow_{i_2}&amp;\nearrow^{j_2}&amp;\\<br /> &amp;\ \ \ \ U_2&amp;&amp;\\<br /> \end{matrix}&lt;/math&gt;<br /> <br /> <br /> Further, let b_i alternate between 1 and 2 for successive i's. <br /> <br /> Hence we define \Phi via for \alpha_i\in G_{b_i}, <br /> <br /> [\alpha_1][\alpha_2]\ldots[\alpha_n]\rightarrow [j_{b_1 *}\alpha_1\cdot\ldots\cdot j_{b_n *}\alpha_n]<br /> <br /> Clearly this is well defined. We need to check the relations in <br /> G_1*_H G_2 indeed hold. Well, the identity element corresponds to the identity path so the relation that removes identities holds.</div> Trefor