http://drorbn.net/api.php?action=feedcontributions&user=Franklin&feedformat=atom Drorbn - User contributions [en] 2022-07-06T15:54:17Z User contributions MediaWiki 1.21.1 http://drorbn.net/index.php?title=0708-1300/Homework_Assignment_10 0708-1300/Homework Assignment 10 2008-02-28T20:59:52Z <p>Franklin: /* Just for Fun */</p> <hr /> <div>{{0708-1300/Navigation}}<br /> <br /> ==Reading==<br /> Read, reread and rereread your notes to this point, and make sure that you really, really really, really really really understand everything in them. Do the same every week!<br /> <br /> Also, read section 1 through 6 (to page 183) of chapter IV of Bredon's book three times:<br /> * First time as if you were reading a novel - quickly and without too much attention to detail, just to learn what the main keywords and concepts and goals are.<br /> * Second time like you were studying for an exam on the subject - slowly and not skipping anything, verifying every little detail.<br /> * And then a third time, again at a quicker pace, to remind yourself of the bigger picture all those little details are there to paint.<br /> <br /> ==Doing==<br /> Solve the following problems from Bredon's book, but submit only the solutions of the problems marked with an &quot;S&quot; - problems 1, 2, S3 on pages 176-177, problems S2, 3 on page 182, and problem S5 on page 190.<br /> Also, solve and submit the following question:<br /> <br /> '''Problem 6.''' &quot;Homotopies between maps&quot; define an &quot;ideal&quot; within the category of topological spaces and continuous maps between them: the homotopy relation is an equivalence relation, and if &lt;math&gt;f_1\sim f_2&lt;/math&gt;, then &lt;math&gt;f_1\circ g\sim f_2\circ g&lt;/math&gt; and &lt;math&gt;g\circ f_1\sim g\circ f_2&lt;/math&gt; whenever these compositions make sense. Show that the same is true for the notion &quot;homotopy of chain maps between chain complexes&quot;, within the category of chain complexes and chain maps between them.<br /> <br /> ==Due Date==<br /> This assignment is due in class on Thursday March 13, 2008.<br /> <br /> ==Just for Fun==<br /> [[Image:0708-1300-MaximalTree.png|frame|left|A maximal tree in the 9x9 integer lattice.]]<br /> # A maximal tree is chosen within the edges of the &lt;math&gt;n\times n&lt;/math&gt; integer lattice. Show that there are two neigboring points on the lattice whose distance from each other, measured only along the tree, is at least &lt;math&gt;n&lt;/math&gt;.<br /> # In class we have shown that &lt;math&gt;I\times\Delta_p&lt;/math&gt;, that is, &lt;math&gt;\Delta_1\times\Delta_p&lt;/math&gt;, is a union of &lt;math&gt;\Delta_{p+1}&lt;/math&gt;'s in a nice and natural way. Can you show that &lt;math&gt;\Delta_q\times\Delta_p&lt;/math&gt; is a union of &lt;math&gt;\Delta_{p+q}&lt;/math&gt;'s in a nice and natural way, for general &lt;math&gt;p&lt;/math&gt; and &lt;math&gt;q&lt;/math&gt;?<br /> # Can you show that the &lt;math&gt;p&lt;/math&gt;-cube &lt;math&gt;I^p&lt;/math&gt; is a union of &lt;math&gt;\Delta_p&lt;/math&gt;'s in a nice and natural way?<br /> <br /> <br /> <br /> [[0708-1300/Just_for_Fun10|Here]] is a idea for problem 1 but don't look at it until you have think on it for a while.</div> Franklin http://drorbn.net/index.php?title=0708-1300/Just_for_Fun10 0708-1300/Just for Fun10 2008-02-28T20:59:23Z <p>Franklin: </p> <hr /> <div>First label the edges of the tree with ones. Then, for any edge not belonging to the tree, not already labeled and such that a contiguous square has the rest of the edges labeled, label it with the sum of those edges. Observe that in this process the labeling of all the edges will be well defined and every edge will be labeled since we started from a tree that was maximal. The non selected edges form like passages on a labyrinth through which you can move along. Observe that there are at least one edge in the boundary of the square that is not part of the tree. Consider the numbers on those edges (the ones in the boundary of the square and not on the tree). Let's assume that all of these edges has label strictly less than &lt;math&gt;n&lt;/math&gt;. Observe that when you move one step along the labyrinth then the number in the edge you are on can be reduced by at least two units. Then none of the passages starting outside te square reach the center of it. Remove from the figure all those passages and the edges that are common to two passages. Then you get a non-empty (contains some squares) part of the square that is completely surrounded by the tree. But this is a contradiction because this would be a cycle in the tree.<br /> <br /> If this solution is correct and you already read it it doesn't means that the fun is over because this problem is certainly here because it can be solved with the aid of homology theory. So, we still need to find that proof.</div> Franklin http://drorbn.net/index.php?title=0708-1300/Homework_Assignment_10 0708-1300/Homework Assignment 10 2008-02-28T20:44:42Z <p>Franklin: /* Just for Fun */</p> <hr /> <div>{{0708-1300/Navigation}}<br /> <br /> ==Reading==<br /> Read, reread and rereread your notes to this point, and make sure that you really, really really, really really really understand everything in them. Do the same every week!<br /> <br /> Also, read section 1 through 6 (to page 183) of chapter IV of Bredon's book three times:<br /> * First time as if you were reading a novel - quickly and without too much attention to detail, just to learn what the main keywords and concepts and goals are.<br /> * Second time like you were studying for an exam on the subject - slowly and not skipping anything, verifying every little detail.<br /> * And then a third time, again at a quicker pace, to remind yourself of the bigger picture all those little details are there to paint.<br /> <br /> ==Doing==<br /> Solve the following problems from Bredon's book, but submit only the solutions of the problems marked with an &quot;S&quot; - problems 1, 2, S3 on pages 176-177, problems S2, 3 on page 182, and problem S5 on page 190.<br /> Also, solve and submit the following question:<br /> <br /> '''Problem 6.''' &quot;Homotopies between maps&quot; define an &quot;ideal&quot; within the category of topological spaces and continuous maps between them: the homotopy relation is an equivalence relation, and if &lt;math&gt;f_1\sim f_2&lt;/math&gt;, then &lt;math&gt;f_1\circ g\sim f_2\circ g&lt;/math&gt; and &lt;math&gt;g\circ f_1\sim g\circ f_2&lt;/math&gt; whenever these compositions make sense. Show that the same is true for the notion &quot;homotopy of chain maps between chain complexes&quot;, within the category of chain complexes and chain maps between them.<br /> <br /> ==Due Date==<br /> This assignment is due in class on Thursday March 13, 2008.<br /> <br /> ==Just for Fun==<br /> [[Image:0708-1300-MaximalTree.png|frame|left|A maximal tree in the 9x9 integer lattice.]]<br /> # A maximal tree is chosen within the edges of the &lt;math&gt;n\times n&lt;/math&gt; integer lattice. Show that there are two neigboring points on the lattice whose distance from each other, measured only along the tree, is at least &lt;math&gt;n&lt;/math&gt;.<br /> # In class we have shown that &lt;math&gt;I\times\Delta_p&lt;/math&gt;, that is, &lt;math&gt;\Delta_1\times\Delta_p&lt;/math&gt;, is a union of &lt;math&gt;\Delta_{p+1}&lt;/math&gt;'s in a nice and natural way. Can you show that &lt;math&gt;\Delta_q\times\Delta_p&lt;/math&gt; is a union of &lt;math&gt;\Delta_{p+q}&lt;/math&gt;'s in a nice and natural way, for general &lt;math&gt;p&lt;/math&gt; and &lt;math&gt;q&lt;/math&gt;?<br /> # Can you show that the &lt;math&gt;p&lt;/math&gt;-cube &lt;math&gt;I^p&lt;/math&gt; is a union of &lt;math&gt;\Delta_p&lt;/math&gt;'s in a nice and natural way?<br /> <br /> [[0708-1300/Just_for_Fun10|Here]] is a idea for problem 1 but don't look at it until you have think on it for a while.</div> Franklin http://drorbn.net/index.php?title=0708-1300/Homework_Assignment_10 0708-1300/Homework Assignment 10 2008-02-28T20:44:17Z <p>Franklin: /* Just for Fun */</p> <hr /> <div>{{0708-1300/Navigation}}<br /> <br /> ==Reading==<br /> Read, reread and rereread your notes to this point, and make sure that you really, really really, really really really understand everything in them. Do the same every week!<br /> <br /> Also, read section 1 through 6 (to page 183) of chapter IV of Bredon's book three times:<br /> * First time as if you were reading a novel - quickly and without too much attention to detail, just to learn what the main keywords and concepts and goals are.<br /> * Second time like you were studying for an exam on the subject - slowly and not skipping anything, verifying every little detail.<br /> * And then a third time, again at a quicker pace, to remind yourself of the bigger picture all those little details are there to paint.<br /> <br /> ==Doing==<br /> Solve the following problems from Bredon's book, but submit only the solutions of the problems marked with an &quot;S&quot; - problems 1, 2, S3 on pages 176-177, problems S2, 3 on page 182, and problem S5 on page 190.<br /> Also, solve and submit the following question:<br /> <br /> '''Problem 6.''' &quot;Homotopies between maps&quot; define an &quot;ideal&quot; within the category of topological spaces and continuous maps between them: the homotopy relation is an equivalence relation, and if &lt;math&gt;f_1\sim f_2&lt;/math&gt;, then &lt;math&gt;f_1\circ g\sim f_2\circ g&lt;/math&gt; and &lt;math&gt;g\circ f_1\sim g\circ f_2&lt;/math&gt; whenever these compositions make sense. Show that the same is true for the notion &quot;homotopy of chain maps between chain complexes&quot;, within the category of chain complexes and chain maps between them.<br /> <br /> ==Due Date==<br /> This assignment is due in class on Thursday March 13, 2008.<br /> <br /> ==Just for Fun==<br /> [[Image:0708-1300-MaximalTree.png|frame|left|A maximal tree in the 9x9 integer lattice.]]<br /> # A maximal tree is chosen within the edges of the &lt;math&gt;n\times n&lt;/math&gt; integer lattice. Show that there are two neigboring points on the lattice whose distance from each other, measured only along the tree, is at least &lt;math&gt;n&lt;/math&gt;.<br /> # In class we have shown that &lt;math&gt;I\times\Delta_p&lt;/math&gt;, that is, &lt;math&gt;\Delta_1\times\Delta_p&lt;/math&gt;, is a union of &lt;math&gt;\Delta_{p+1}&lt;/math&gt;'s in a nice and natural way. Can you show that &lt;math&gt;\Delta_q\times\Delta_p&lt;/math&gt; is a union of &lt;math&gt;\Delta_{p+q}&lt;/math&gt;'s in a nice and natural way, for general &lt;math&gt;p&lt;/math&gt; and &lt;math&gt;q&lt;/math&gt;?<br /> # Can you show that the &lt;math&gt;p&lt;/math&gt;-cube &lt;math&gt;I^p&lt;/math&gt; is a union of &lt;math&gt;\Delta_p&lt;/math&gt;'s in a nice and natural way?<br /> <br /> [[Here|0708-1300/Just_for_Fun10]] is a idea for problem 1 but don't look at it until you have think on it for a while.</div> Franklin http://drorbn.net/index.php?title=0708-1300/Covering_Product 0708-1300/Covering Product 2008-02-18T21:16:57Z <p>Franklin: </p> <hr /> <div>Think about an infinite product of copies of &lt;math&gt;S^1&lt;/math&gt;. Even more, think about the possibility of this space having a universal covering space. At some point it is probably needed that &lt;math&gt;\mathbb{Z}^n\not\cong\mathbb{Z}^m&lt;/math&gt; for &lt;math&gt;n\neq m&lt;/math&gt;. It is interesting to produce a proof of this using the less knowledge you can. A possible idea is [[0708-1300/fact|here]].</div> Franklin http://drorbn.net/index.php?title=0708-1300/Covering_Product 0708-1300/Covering Product 2008-02-18T21:16:12Z <p>Franklin: </p> <hr /> <div>Think about an infinite product of copies of &lt;math&gt;S^1&lt;/math&gt;. Even more, think about the possibility of this space having a universal covering space. At some point it is probably needed that &lt;math&gt;\mathbb{Z}^n\not\cong\mathbb{Z}^m&lt;/math&gt;. It is interesting to produce a proof of this using the less knowledge you can. A possible idea is [[0708-1300/fact|here]].</div> Franklin http://drorbn.net/index.php?title=0708-1300/Covering_Product 0708-1300/Covering Product 2008-02-18T21:15:55Z <p>Franklin: </p> <hr /> <div>Think about an infinite product of copies of &lt;math&gt;S^1&lt;/math&gt;. Even more, think about the possibility of this space having a universal covering space. At some point it is probably needed that &lt;math&gt;\mathbb{Z}^n\not\equiv\mathbb{Z}^m&lt;/math&gt;. It is interesting to produce a proof of this using the less knowledge you can. A possible idea is [[0708-1300/fact|here]].</div> Franklin http://drorbn.net/index.php?title=0708-1300/Homework_Assignment_9 0708-1300/Homework Assignment 9 2008-02-18T21:09:12Z <p>Franklin: /* Just for Fun */</p> <hr /> <div>{{0708-1300/Navigation}}<br /> <br /> ==Reading==<br /> Read, reread and rereread your notes to this point, and make sure that you really, really really, really really really understand everything in them. Do the same every week!<br /> <br /> ==Doing==<br /> (Problems 1,2,4,5 below are taken with slight modifications from Hatcher's book, pages 79-80).<br /> <br /> # Show that if &lt;math&gt;p_1\colon X_1\to B_1&lt;/math&gt; and &lt;math&gt;p_2\colon X_2 \to B_2&lt;/math&gt; are covering spaces, then so is their product &lt;math&gt;p_1\times p_2\colon X_1\times X_2\to B_1\times B_2&lt;/math&gt;.<br /> # Construct (i.e., describe in explicit terms) a simply-connected covering space of the space &lt;math&gt;X\subset\mathbb{R}^3&lt;/math&gt; that is the union of a sphere and a diameter. Do the same when &lt;math&gt;X&lt;/math&gt; is the union of a sphere and a circle intersecting it in two points.<br /> # Do the same to the space &lt;math&gt;Y&lt;/math&gt; of the term test: &lt;math&gt;Y=\{z\in{\mathbb C}\colon|z|\leq 1\}/(z\sim e^{2\pi i/3}z\mbox{ whenever }|z|=1)&lt;/math&gt;.<br /> # Find all the connected 2-sheeted and 3-sheeted covering spaces of the &quot;figure eight space&quot; &lt;math&gt;S^1\vee S^1&lt;/math&gt; (two circles joined at a point), up to isomorphism of covering spaces without base points.<br /> # Let &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; be the generators of &lt;math&gt;\pi_1(S^1\vee S^1)&lt;/math&gt; corresponding to the two &lt;math&gt;S^1&lt;/math&gt; summands. Draw a picture of the covering space of &lt;math&gt;S^1\vee S^1&lt;/math&gt; corresponding to the normal subgroup generated by &lt;math&gt;a^2&lt;/math&gt;, &lt;math&gt;b^2&lt;/math&gt;, and &lt;math&gt;(ab)^4&lt;/math&gt;, and prove that this covering space is indeed the correct one.<br /> <br /> ==Due Date==<br /> This assignment is due in class on Thursday February 28, 2008.<br /> <br /> ==Just for Fun==<br /> * What happens if in problem 1 we consider infinitely many covering spaces. That is, is the product of an infinite family of covering spaces a covering space? [[0708-1300/Covering_Product|Here]] is an idea but don't look at it until you have think on the problem for a while.<br /> * This raises another question. A &quot;pathwise totally disconnected space&quot; is a space in which every path is a constant path. How much of the theory of covering spaces can be generalized to &quot;coverings&quot; in which the fibers are pathwise totally disconnected, instead of discrete?</div> Franklin http://drorbn.net/index.php?title=0708-1300/fact 0708-1300/fact 2008-02-18T21:03:39Z <p>Franklin: </p> <hr /> <div>If &lt;math&gt;n\neq m&lt;/math&gt; then &lt;math&gt;\mathbb{Z}^n\not\cong\mathbb{Z}^m&lt;/math&gt;.<br /> <br /> Proof<br /> <br /> Assume that &lt;math&gt;f:\mathbb{Z}^n\rightarrow\mathbb{Z}^m&lt;/math&gt; is an isomorphism. Let &lt;math&gt;A&lt;/math&gt; be the matrix of &lt;math&gt;f&lt;/math&gt; in the canonical basis and &lt;math&gt;M&lt;/math&gt; the maximum of the absolute values of the entries of &lt;math&gt;A&lt;/math&gt;. If we evaluate &lt;math&gt;f&lt;/math&gt; in all the vectors of &lt;math&gt;\mathbb{Z}^n&lt;/math&gt; who's entries have absolute values less than or equal to &lt;math&gt;r&lt;/math&gt; (there are &lt;math&gt;(2r)^n&lt;/math&gt; of such elements) then we get elements who's entries have absolute value less than or equal to &lt;math&gt;nrM&lt;/math&gt; (there are &lt;math&gt;(2rnM)^m&lt;/math&gt; of such elements in &lt;math&gt;\mathbb{Z}^m&lt;/math&gt;). Since &lt;math&gt;f&lt;/math&gt; is injective we must have &lt;math&gt;(2r)^n\leq(2rnM)^m&lt;/math&gt; for every &lt;math&gt;r&lt;/math&gt;. Replacing &lt;math&gt;f&lt;/math&gt; by its inverse if necessary we can assume that &lt;math&gt;n&gt;m&lt;/math&gt; but if this is the case the inequality above can not be true for arbitrarily large values of &lt;math&gt;r&lt;/math&gt;.</div> Franklin http://drorbn.net/index.php?title=0708-1300/Class_notes_for_Tuesday,_January_22 0708-1300/Class notes for Tuesday, January 22 2008-02-18T20:59:01Z <p>Franklin: /* First 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 [[0708-1300/fact|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> Franklin http://drorbn.net/index.php?title=0708-1300/Class_notes_for_Tuesday,_January_22 0708-1300/Class notes for Tuesday, January 22 2008-02-18T20:58:48Z <p>Franklin: /* First 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 [[0708-1300/fact 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> Franklin http://drorbn.net/index.php?title=0708-1300/Class_notes_for_Tuesday,_January_22 0708-1300/Class notes for Tuesday, January 22 2008-02-18T20:58:21Z <p>Franklin: /* First 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|0708-1300/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> Franklin http://drorbn.net/index.php?title=0708-1300/Class_notes_for_Tuesday,_January_22 0708-1300/Class notes for Tuesday, January 22 2008-02-18T20:57:25Z <p>Franklin: /* First 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> Franklin http://drorbn.net/index.php?title=0708-1300/Homework_Assignment_9 0708-1300/Homework Assignment 9 2008-02-18T16:20:25Z <p>Franklin: </p> <hr /> <div>{{0708-1300/Navigation}}<br /> <br /> ==Reading==<br /> Read, reread and rereread your notes to this point, and make sure that you really, really really, really really really understand everything in them. Do the same every week!<br /> <br /> ==Doing==<br /> (Problems 1,2,4,5 below are taken with slight modifications from Hatcher's book, pages 79-80).<br /> <br /> # Show that if &lt;math&gt;p_1\colon X_1\to B_1&lt;/math&gt; and &lt;math&gt;p_2\colon X_2 \to B_2&lt;/math&gt; are covering spaces, then so is their product &lt;math&gt;p_1\times p_2\colon X_1\times X_2\to B_1\times B_2&lt;/math&gt;.<br /> # Construct (i.e., describe in explicit terms) a simply-connected covering space of the space &lt;math&gt;X\subset\mathbb{R}^3&lt;/math&gt; that is the union of a sphere and a diameter. Do the same when &lt;math&gt;X&lt;/math&gt; is the union of a sphere and a circle intersecting it in two points.<br /> # Do the same to the space &lt;math&gt;Y&lt;/math&gt; of the term test: &lt;math&gt;Y=\{z\in{\mathbb C}\colon|z|\leq 1\}/(z\sim e^{2\pi i/3}z\mbox{ whenever }|z|=1)&lt;/math&gt;.<br /> # Find all the connected 2-sheeted and 3-sheeted covering spaces of the &quot;figure eight space&quot; &lt;math&gt;S^1\vee S^1&lt;/math&gt; (two circles joined at a point), up to isomorphism of covering spaces without base points.<br /> # Let &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; be the generators of &lt;math&gt;\pi_1(S^1\vee S^1)&lt;/math&gt; corresponding to the two &lt;math&gt;S^1&lt;/math&gt; summands. Draw a picture of the covering space of &lt;math&gt;S^1\vee S^1&lt;/math&gt; corresponding to the normal subgroup generated by &lt;math&gt;a^2&lt;/math&gt;, &lt;math&gt;b^2&lt;/math&gt;, and &lt;math&gt;(ab)^4&lt;/math&gt;, and prove that this covering space is indeed the correct one.<br /> <br /> ==Due Date==<br /> This assignment is due in class on Thursday February 28, 2008.<br /> <br /> ==Just for Fun==<br /> What happens if in problem &lt;math&gt;1&lt;/math&gt; we consider infinitely many covering spaces. <br /> This is, is the product of an infinite family of covering spaces a covering space?</div> Franklin http://drorbn.net/index.php?title=0708-1300/Hawaiian_Earring 0708-1300/Hawaiian Earring 2008-01-10T22:24:32Z <p>Franklin: </p> <hr /> <div>Assume &lt;math&gt;r\in[0,1)&lt;/math&gt; and &lt;math&gt;r=0.a_1a_2...&lt;/math&gt; is a decimal expansion without an infinite tail of nines.<br /> Let &lt;math&gt;\gamma&lt;/math&gt; be a path that goes around the first (the largest) circle &lt;math&gt;a_1&lt;/math&gt; times, then &lt;math&gt;a_2&lt;/math&gt; times around the second and so on. Observe that thanks to the shrinking of the circles it is possible the continuity of the curve at time &lt;math&gt;1&lt;/math&gt; for the irrational numbers. Proving that two such curves belongs to different homotopy classes tell us that the number of homotopy classes is uncountable.<br /> <br /> Now, assume that two such curves, corresponding to two different real numbers, are homotopic. Let &lt;math&gt;k&lt;/math&gt; be the first decimal digit in which the numbers differ and &lt;math&gt;P_k&lt;/math&gt; the quotient map of identifying all the circles but the &lt;math&gt;k&lt;/math&gt;-th one to the joint point. if &lt;math&gt;H&lt;/math&gt; is a homotopy of the curves then &lt;math&gt;P_k\circ H&lt;/math&gt; is a homotopy of the projection of the curves which winds a different number of times. But this contradict the computation of the fundamental group o the circle.</div> Franklin http://drorbn.net/index.php?title=0708-1300/Homework_Assignment_7 0708-1300/Homework Assignment 7 2008-01-10T22:06:07Z <p>Franklin: /* Just for fun */</p> <hr /> <div>{{0708-1300/Navigation}}<br /> ==Reading==<br /> Quickly read section 1 of Chapter III of Bredon's book, and then <br /> '''read''' sections 2-4 three times:<br /> * First time as if you were reading a novel - quickly and without too much attention to detail, just to learn what the main keywords and concepts and goals are.<br /> * Second time like you were studying for an exam on the subject - slowly and not skipping anything, verifying every little detail.<br /> * And then a third time, again at a quicker pace, to remind yourself of the bigger picture all those little details are there to paint.<br /> Also, '''pre-read''' (just once), the rest of chapter III.<br /> <br /> ==Doing==<br /> Solve the following problems from Bredon's book, but submit only the solutions of the problems marked with an &quot;S&quot;:<br /> {|align=center border=1 cellspacing=0 cellpadding=5<br /> |- align=center<br /> !Problems<br /> !On page(s)<br /> !Comments<br /> |- align=center<br /> |S1, S2, 3<br /> |138<br /> |&quot;Topological groups&quot; are defined in page 51. You need nothing beyond definition 15.1.<br /> |- align=center<br /> |1, S2, S3, 4<br /> |143<br /> |<br /> |- align=center<br /> |1, S2, 3<br /> |145-146<br /> |<br /> |}<br /> <br /> ==Due Date==<br /> This assignment is due in class on Thursday January 24, 2008.<br /> <br /> ==Just for fun==<br /> Prove that the fundamental group of the Hawaiian Earring, shown below, is uncountable. (The Hawaiian Earing is a countable union of circles shrinking to a point; the figure only shows the first few).<br /> [[Image:0708-1300-HawaiianEarings.png|center]]<br /> <br /> <br /> [[0708-1300/Hawaiian Earring|Here]] is an idea but don't look at it until you think on the problem for a while.<br /> <br /> ==Just to be Sure==<br /> Visit [[0708-1300/Register of Good Deeds]] and complain if I didn't notice something.</div> Franklin http://drorbn.net/index.php?title=0708-1300/Homework_Assignment_7 0708-1300/Homework Assignment 7 2008-01-10T22:05:33Z <p>Franklin: /* Just for fun */</p> <hr /> <div>{{0708-1300/Navigation}}<br /> ==Reading==<br /> Quickly read section 1 of Chapter III of Bredon's book, and then <br /> '''read''' sections 2-4 three times:<br /> * First time as if you were reading a novel - quickly and without too much attention to detail, just to learn what the main keywords and concepts and goals are.<br /> * Second time like you were studying for an exam on the subject - slowly and not skipping anything, verifying every little detail.<br /> * And then a third time, again at a quicker pace, to remind yourself of the bigger picture all those little details are there to paint.<br /> Also, '''pre-read''' (just once), the rest of chapter III.<br /> <br /> ==Doing==<br /> Solve the following problems from Bredon's book, but submit only the solutions of the problems marked with an &quot;S&quot;:<br /> {|align=center border=1 cellspacing=0 cellpadding=5<br /> |- align=center<br /> !Problems<br /> !On page(s)<br /> !Comments<br /> |- align=center<br /> |S1, S2, 3<br /> |138<br /> |&quot;Topological groups&quot; are defined in page 51. You need nothing beyond definition 15.1.<br /> |- align=center<br /> |1, S2, S3, 4<br /> |143<br /> |<br /> |- align=center<br /> |1, S2, 3<br /> |145-146<br /> |<br /> |}<br /> <br /> ==Due Date==<br /> This assignment is due in class on Thursday January 24, 2008.<br /> <br /> ==Just for fun==<br /> Prove that the fundamental group of the Hawaiian Earring, shown below, is uncountable. (The Hawaiian Earing is a countable union of circles shrinking to a point; the figure only shows the first few).<br /> [[Image:0708-1300-HawaiianEarings.png|center]]<br /> <br /> <br /> [[Hawaiian Earring|Here]] is an idea but don't look at it until you think on the problem for a while.<br /> <br /> ==Just to be Sure==<br /> Visit [[0708-1300/Register of Good Deeds]] and complain if I didn't notice something.</div> Franklin http://drorbn.net/index.php?title=0708-1300/justforfun6 0708-1300/justforfun6 2007-12-13T21:43:27Z <p>Franklin: </p> <hr /> <div>We can see that the linking number of any pair of components in &lt;math&gt;\gamma_3&lt;/math&gt; is non zero while there is a pair of components of &lt;math&gt;\gamma'_3&lt;/math&gt; having linking number zero. For this we need to find a practical way to compute the linking number but this comes with the solution of problem 4.3 of this homework.<br /> <br /> On the other hand, to see that the complements of these links are diffeomorphic look at the following figure.<br /> <br /> [[Image:0708-1300linking.jpeg|thumb|center|500px|]]<br /> <br /> In the first part of the figure we can see the complement of &lt;math&gt;\gamma'_3&lt;/math&gt;. Cut this along the disc which boundary is the horizontal circle and you will get something similar to the middle figure. The space that you see between the two discs is not space it is complete emptiness. Rotate the upper disc as indicated by the arrow and you will get the right part of the figure. Finally glue again the two discs and you will get the link &lt;math&gt;\gamma_3&lt;/math&gt;.<br /> <br /> This cut-and-pasting gives us a (maybe non-smooth) homeomorphism but, using the trick of immersing &lt;math&gt;\mathbb{R}^3&lt;/math&gt; in &lt;math&gt;S^3&lt;/math&gt;, using that this is compact to find a deformation of our map that is smooth we will get the desired diffeomorphism. Or even more simple, if a good drawer paints this picture again it will certainly be smooth enough.</div> Franklin http://drorbn.net/index.php?title=0708-1300/justforfun6 0708-1300/justforfun6 2007-12-13T21:43:16Z <p>Franklin: </p> <hr /> <div>We can see that the linking number of any pair of components in &lt;math&gt;\gamma_3&lt;/math&gt; is non zero while there is a pair of components of &lt;math&gt;\gamma'_3&lt;/math&gt; having linking number zero. For this we need to find a practical way to compute the linking number but this comes with the solution of problem 4.3 of this homework.<br /> <br /> On the other hand, to see that the complements of these links are diffeomorphic look at the following figure.<br /> <br /> [[Image:0708-1300linking.jpeg|thumb|center|300px|]]<br /> <br /> In the first part of the figure we can see the complement of &lt;math&gt;\gamma'_3&lt;/math&gt;. Cut this along the disc which boundary is the horizontal circle and you will get something similar to the middle figure. The space that you see between the two discs is not space it is complete emptiness. Rotate the upper disc as indicated by the arrow and you will get the right part of the figure. Finally glue again the two discs and you will get the link &lt;math&gt;\gamma_3&lt;/math&gt;.<br /> <br /> This cut-and-pasting gives us a (maybe non-smooth) homeomorphism but, using the trick of immersing &lt;math&gt;\mathbb{R}^3&lt;/math&gt; in &lt;math&gt;S^3&lt;/math&gt;, using that this is compact to find a deformation of our map that is smooth we will get the desired diffeomorphism. Or even more simple, if a good drawer paints this picture again it will certainly be smooth enough.</div> Franklin http://drorbn.net/index.php?title=0708-1300/justforfun6 0708-1300/justforfun6 2007-12-13T21:42:35Z <p>Franklin: </p> <hr /> <div>We can see that the linking number of any pair of components in &lt;math&gt;\gamma_3&lt;/math&gt; is non zero while there is a pair of components of &lt;math&gt;\gamma'_3&lt;/math&gt; having linking number zero. For this we need to find a practical way to compute the linking number but this comes with the solution of problem 4.3 of this homework.<br /> <br /> On the other hand, to see that the complements of these links are diffeomorphic look at the following figure.<br /> <br /> [[Image:0708-1300linking.jpeg|thumb|center|240px|]]<br /> <br /> In the first part of the figure we can see the complement of &lt;math&gt;\gamma'_3&lt;/math&gt;. Cut this along the disc which boundary is the horizontal circle and you will get something similar to the middle figure. The space that you see between the two discs is not space it is complete emptiness. Rotate the upper disc as indicated by the arrow and you will get the right part of the figure. Finally glue again the two discs and you will get the link &lt;math&gt;\gamma_3&lt;/math&gt;.<br /> <br /> This cut-and-pasting gives us a (maybe non-smooth) homeomorphism but, using the trick of immersing &lt;math&gt;\mathbb{R}^3&lt;/math&gt; in &lt;math&gt;S^3&lt;/math&gt;, using that this is compact to find a deformation of our map that is smooth we will get the desired diffeomorphism. Or even more simple, if a good drawer paints this picture again it will certainly be smooth enough.</div> Franklin http://drorbn.net/index.php?title=0708-1300/justforfun6 0708-1300/justforfun6 2007-12-13T21:42:03Z <p>Franklin: </p> <hr /> <div>We can see that the linking number of any pair of components in &lt;math&gt;\gamma_3&lt;/math&gt; is non zero while there is a pair of components of &lt;math&gt;\gamma'_3&lt;/math&gt; having linking number zero. For this we need to find a practical way to compute the linking number but this comes with the solution of problem 4.3 of this homework.<br /> <br /> On the other hand, to see that the complements of these links are diffeomorphic look at the following figure.<br /> <br /> [[Image:0708-1300linking.jpeg|thumb|center|200px|]]<br /> <br /> In the first part of the figure we can see the complement of &lt;math&gt;\gamma'_3&lt;/math&gt;. Cut this along the disc which boundary is the horizontal circle and you will get something similar to the middle figure. The space that you see between the two discs is not space it is complete emptiness. Rotate the upper disc as indicated by the arrow and you will get the right part of the figure. Finally glue again the two discs and you will get the link &lt;math&gt;\gamma_3&lt;/math&gt;.<br /> <br /> This cut-and-pasting gives us a (maybe non-smooth) homeomorphism but, using the trick of immersing &lt;math&gt;\mathb{R}^3&lt;/math&gt; in &lt;math&gt;S^3&lt;/math&gt;, using that this is compact to find a deformation of our map that is smooth we will get the desired diffeomorphism. Or even more simple, if a good drawer paints this picture again it will certainly be smooth enough.</div> Franklin http://drorbn.net/index.php?title=0708-1300/justforfun6 0708-1300/justforfun6 2007-12-13T21:40:34Z <p>Franklin: </p> <hr /> <div><br /> We can see that the linking number of any pair of components in &lt;math&gt;\gamma_3&lt;/math&gt; is non zero while there is a pair of components of &lt;math&gt;\gamma'_3&lt;/math&gt; having linking number zero. For this we need to find a practical way to compute the linking number but this comes with the solution of problem 4.3 of this homework.<br /> <br /> On the other hand, to see that the complements of these links are diffeomorphic look at the following figure.<br /> <br /> [[Image:0708-1300linking.jpeg]]<br /> <br /> In the first part of the figure we can see the complement of &lt;math&gt;\gamma'_3&lt;/math&gt;. Cut this along the disc which boundary is the horizontal circle and you will get something similar to the middle figure. The space that you see between the two discs is not space it is complete emptiness. Rotate the upper disc as indicated by the arrow and you will get the right part of the figure. Finally glue again the two discs and you will get the link &lt;math&gt;\gamma_3&lt;/math&gt;.<br /> <br /> This cut-and-pasting gives us a (maybe non-smooth) homeomorphism but, using the trick of immersing &lt;math&gt;\mathb{R}^3&lt;/math&gt; in &lt;math&gt;S^3&lt;/math&gt;, using that this is compact to find a deformation of our map that is smooth we will get the desired diffeomorphism. Or even more simple, if a good drawer paints this picture again it will certainly be smooth enough.</div> Franklin http://drorbn.net/index.php?title=File:0708-1300linking.jpeg File:0708-1300linking.jpeg 2007-12-13T21:26:19Z <p>Franklin: </p> <hr /> <div></div> Franklin http://drorbn.net/index.php?title=0708-1300/Homework_Assignment_6 0708-1300/Homework Assignment 6 2007-12-13T21:22:03Z <p>Franklin: /* Just for Fun */</p> <hr /> <div>{{0708-1300/Navigation}}<br /> ==Reading==<br /> At your leisure, read your class notes over the break, and especially at some point right before classes resume next semester. Here are a few questions you can ask yourself while reading:<br /> * Do you understand pullbacks of differential forms?<br /> * Do you think you could in practice integrate any differential form on any manifold (at least when the formulas involved are not too messy)?<br /> * Do you understand orientations and boundaries and how they interact?<br /> * Why is Stokes' theorem true? Both in terms of the local meaning of &lt;math&gt;d&lt;/math&gt;, and in terms of a formal proof.<br /> * Do you understand the two and three dimensional cases of Stokes' theorem?<br /> * Do you understand the Hodge star operator &lt;math&gt;\star&lt;/math&gt;?<br /> * How did we get &lt;math&gt;d\star dA=J&lt;/math&gt; from the least action principle?<br /> * Do you understand how Poincare's lemma entered the derivation of Maxewell's equations?<br /> * Do you understand the operator &lt;math&gt;P&lt;/math&gt;? (How was it used, formally derived, and what is the intuitive picture behind it?)<br /> * What was &lt;math&gt;H_{dR}&lt;/math&gt; and how did it relate to pullbacks and homotopy.<br /> <br /> ==Doing==<br /> Solve the following problems and submit your solutions of problems 2, 3 and 4. This is a very challenging collection of problems; I expect most of you to do problem 1 with no difficulty (it is a repeat of an older problem), problem 2 with some effort, and I hope each of you will be able to do at least one further problem. It will be great if some of you will do all problems!<br /> <br /> '''Problem 1.''' The standard volume form on &lt;math&gt;S^2&lt;/math&gt; is the form &lt;math&gt;\omega&lt;/math&gt; given by &lt;math&gt;\omega=\frac{1}{4\pi}\left(xdy\wedge dz+ydz\wedge dx+zdx\wedge dy\right)&lt;/math&gt;. Show that &lt;math&gt;\int_{S^2}\omega=1&lt;/math&gt;.<br /> <br /> '''Problem 2.''' If &lt;math&gt;M&lt;/math&gt; is a compact orientable &lt;math&gt;n&lt;/math&gt;-manifold with no boundary, show that &lt;math&gt;H^n_{dR}(M)\neq 0&lt;/math&gt;.<br /> <br /> '''Problem 3.''' Show that if &lt;math&gt;\omega\in\Omega^2(S^2)&lt;/math&gt; satisfies &lt;math&gt;\int_{S_2}\omega=0&lt;/math&gt;, then &lt;math&gt;\omega&lt;/math&gt; is exact. Deduce that if &lt;math&gt;w_1\in\Omega^2(S^2)&lt;/math&gt; and &lt;math&gt;w_2\in\Omega^2(S^2)&lt;/math&gt; satisfy &lt;math&gt;\int_{S_2}\omega_1=\int_{S_2}\omega_2&lt;/math&gt;, then &lt;math&gt;[\omega_1]=[\omega_2]&lt;/math&gt; as elements of &lt;math&gt;H^2_{dR}(S^2)&lt;/math&gt;. Deduce further that &lt;math&gt;\dim H^2_{dR}(S^2)=1&lt;/math&gt;.<br /> <br /> '''Problem 4.''' A &quot;link&quot; in &lt;math&gt;{\mathbb R}^3&lt;/math&gt; is an ordered pair &lt;math&gt;\gamma=(\gamma_1, \gamma_2)&lt;/math&gt;, in which &lt;math&gt;\gamma_1&lt;/math&gt; and &lt;math&gt;\gamma_2&lt;/math&gt; are smooth embeddings of the circle &lt;math&gt;S^1&lt;/math&gt; into &lt;math&gt;{\mathbb R}^3&lt;/math&gt;, whose images (called &quot;the components of &lt;math&gt;\gamma&lt;/math&gt;&quot;) are disjoint. Two such links are called &quot;isotopic&quot;, if one can be deformed to the other via a smooth homotopy along which the components remain embeddings and remain disjoint. Given a link &lt;math&gt;\gamma&lt;/math&gt;, define a map &lt;math&gt;\Phi_\gamma:S^1\times S^1\to S^2&lt;/math&gt; by &lt;math&gt;\Phi_\gamma(t_1,t_2):=\frac{\gamma_2(t_2)-\gamma_1(t_1)}{||\gamma_2(t_2)-\gamma_1(t_1)||}&lt;/math&gt;. Finally, let &lt;math&gt;\omega&lt;/math&gt; be the standard volume form of &lt;math&gt;S^2&lt;/math&gt;, and define &quot;the linking number of &lt;math&gt;\gamma=(\gamma_1, \gamma_2)&lt;/math&gt;&quot; to be &lt;math&gt;l(\gamma)=l(\gamma_1,\gamma_2):=\int_{S^1\times S^1}\Phi_\gamma^\star\omega&lt;/math&gt;. Show<br /> # If two links &lt;math&gt;\gamma&lt;/math&gt; and &lt;math&gt;\gamma'&lt;/math&gt; are isotopic, then their linking numbers are the same: &lt;math&gt;l(\gamma)=l(\gamma')&lt;/math&gt;.<br /> # If &lt;math&gt;\omega'&lt;/math&gt; is a second 2-form on &lt;math&gt;S^2&lt;/math&gt; for which &lt;math&gt;\int_{S^2}\omega'=1&lt;/math&gt; and if &lt;math&gt;l'(\gamma)&lt;/math&gt; is defined in the same manner as &lt;math&gt;l(\gamma)&lt;/math&gt; except replacing &lt;math&gt;\omega&lt;/math&gt; with &lt;math&gt;\omega'&lt;/math&gt;, then &lt;math&gt;l(\gamma)=l'(\gamma)&lt;/math&gt;. (In particular this is true if &lt;math&gt;\omega'&lt;/math&gt; is very close to a &lt;math&gt;\delta&lt;/math&gt;-function form at the north pole of &lt;math&gt;S^2&lt;/math&gt;).<br /> # Compute (but just up to an overall sign) the linking number of the link {{KAT Link|L11a193|L11a193}}, displayed below:<br /> {| align=center<br /> |-<br /> |[[Image:L11a193.png]]<br /> |[[Image:0708-1300-LinkComplementExample1.png]]<br /> |[[Image:0708-1300-LinkComplementExample2.png]]<br /> |-<br /> |colspan=3 align=center|The links L11a193, &lt;math&gt;\gamma_3&lt;/math&gt; and &lt;math&gt;\gamma'_3&lt;/math&gt;.<br /> |}<br /> <br /> ==Just for Fun==<br /> Prove that the two (3-component) links &lt;math&gt;\gamma_3&lt;/math&gt; and &lt;math&gt;\gamma'_3&lt;/math&gt; shown above are not isotopic, yet their complements are diffeomorphic. (See more at {{Home Link|classes/0405/Topology/HW5/HW.html|Classes: 2004-05: Math 1300Y - Topology: Homework Assignment 5}})<br /> <br /> [[0708-1300/justforfun6|Here]] is one idea. Don't look at it before thinking the problem for a while.<br /> <br /> ==Due Date==<br /> This assignment is due in class on Thursday January 10, 2007.</div> Franklin http://drorbn.net/index.php?title=0708-1300/Homework_Assignment_5 0708-1300/Homework Assignment 5 2007-11-23T16:29:40Z <p>Franklin: /* Doing */ Correcting a minor typo.</p> <hr /> <div>{{0708-1300/Navigation}}<br /> ==Reading==<br /> '''Read''' sections 1-3 of chapter V of Bredon's book three times:<br /> * First time as if you were reading a novel - quickly and without too much attention to detail, just to learn what the main keywords and concepts and goals are.<br /> * Second time like you were studying for an exam on the subject - slowly and not skipping anything, verifying every little detail.<br /> * And then a third time, again at a quicker pace, to remind yourself of the bigger picture all those little details are there to paint.<br /> <br /> '''Also,''' do the same with your own class notes - much of what we do for this part of the class is '''not''' in the textbook!<br /> <br /> ==Doing==<br /> Solve all of the following problems, but submit only your solutions of problems 1, 3, 8, 9 and 10:<br /> <br /> '''Problem 1.''' Let &lt;math&gt;M^n&lt;/math&gt; be a manifold. Show that the following definitions for the orientability of &lt;math&gt;M&lt;/math&gt; are equivalent:<br /> # There exists a nowhere vanishing &lt;math&gt;n&lt;/math&gt;-form on &lt;math&gt;M&lt;/math&gt;.<br /> # There exists an atlas &lt;math&gt;\{(U_\alpha,\phi_\alpha:U_\alpha\to{\mathbb R}^n)\}&lt;/math&gt; for &lt;math&gt;M&lt;/math&gt;, so that &lt;math&gt;\det(\phi_\alpha\phi^{-1}_\beta)&gt;0&lt;/math&gt; wherever that makes sense.<br /> <br /> '''Problem 2.''' Show that the tangent space &lt;math&gt;TM&lt;/math&gt; of any manifold &lt;math&gt;M&lt;/math&gt; is orientable.<br /> <br /> '''Problem 3.'''<br /> # Show that if &lt;math&gt;M&lt;/math&gt; and &lt;math&gt;N&lt;/math&gt; are orientable then so is &lt;math&gt;M\times N&lt;/math&gt;.<br /> # Show that if &lt;math&gt;M&lt;/math&gt; and &lt;math&gt;M\times N&lt;/math&gt; are orientable then so is &lt;math&gt;N&lt;/math&gt;.<br /> <br /> '''Problem 4.''' Show that &lt;math&gt;S^n&lt;/math&gt; is always orientable.<br /> <br /> '''Problem 5.''' Recall that a form is called closed if it is in the kernel of &lt;math&gt;d&lt;/math&gt; and exact if it is in the image of &lt;math&gt;d&lt;/math&gt;. Show that every exact form is closed.<br /> <br /> '''Problem 6.''' Let &lt;math&gt;f:{\mathbb R}_t\to S^1\subset{\mathbb C}&lt;/math&gt; be given by &lt;math&gt;f(t)=e^{it}&lt;/math&gt;.<br /> # Show that there exists a unique &lt;math&gt;\omega\in\Omega^1(S^1)&lt;/math&gt; such that &lt;math&gt;f^\star\omega=dt&lt;/math&gt;.<br /> # Show that &lt;math&gt;\omega&lt;/math&gt; is closed but not exact.<br /> <br /> '''Problem 7.''' Show, directly from the definitions, that every closed 1-form on &lt;math&gt;{\mathbb R}^2&lt;/math&gt; is exact.<br /> <br /> '''Problem 8.''' Compute the integral &lt;math&gt;\int_{S^2}zdx\wedge dy&lt;/math&gt; twice:<br /> # Using Stokes' theorem.<br /> # Directly from the definition, by using a two- or three-chart atlas for &lt;math&gt;S^2&lt;/math&gt; (or for &lt;math&gt;S^2&lt;/math&gt; minus a single point).<br /> (Repeat 1 and 2 until they stop giving different answers).<br /> <br /> '''Problem 9.''' Show that the form &lt;math&gt;\omega_0=xdy\wedge dz+ydz\wedge dx+zdx\wedge dy\in\Omega^2({\mathbb R}^3_{x,y,z})&lt;/math&gt; is invariant under rigid orientation-preserving rotations of &lt;math&gt;{\mathbb R}^3&lt;/math&gt;. That is, if &lt;math&gt;A&lt;/math&gt; is such a rotation matrix (&lt;math&gt;AA^T=I&lt;/math&gt; and &lt;math&gt;\det A=1&lt;/math&gt;) considered also as a linear transformation &lt;math&gt;A:{\mathbb R}^3\to{\mathbb R}^3&lt;/math&gt;, then &lt;math&gt;A^\star\omega_0=\omega_0&lt;/math&gt;.<br /> <br /> '''Problem 10.''' Let &lt;math&gt;\omega_\alpha&lt;/math&gt; be the form &lt;math&gt;(x^2+y^2+z^2)^{\alpha/2}\omega_0&lt;/math&gt; defined on &lt;math&gt;{\mathbb R}^3\backslash\{0\}&lt;/math&gt;, where &lt;math&gt;\alpha&lt;/math&gt; is a real number and &lt;math&gt;\omega_0&lt;/math&gt; is the form from the previous question.<br /> # Find a number &lt;math&gt;\alpha&lt;/math&gt; so that &lt;math&gt;\omega_\alpha&lt;/math&gt; would be closed.<br /> # For that value of &lt;math&gt;\alpha&lt;/math&gt;, compute &lt;math&gt;\int_S\omega_\alpha&lt;/math&gt;, where &lt;math&gt;S&lt;/math&gt; is the sphere of radius &lt;math&gt;r&lt;/math&gt; around a point &lt;math&gt;p\in{\mathbb R}^3&lt;/math&gt;. (If you end up writing complicated formulas, you missed the point).<br /> # For the same &lt;math&gt;\alpha&lt;/math&gt;, is &lt;math&gt;\omega_\alpha&lt;/math&gt; exact?<br /> <br /> ==Due Date==<br /> This assignment is due in class on Thursday December 6, 2007.<br /> <br /> <br /> [http://www.brainyquote.com/quotes/authors/d/douglas_adams.html Douglas Adams] - &quot;I love deadlines. I like the whooshing sound they make as they fly by.&quot;<br /> <br /> ==Just for Fun==<br /> A planimeter is a (mechanical!) measuring instrument used to measure the area of an arbitrary two-dimensional shape by tracing along its boundary; if you have a map of Ontario, for example, you can roll the measuring end of a planimeter counterclockwise around the borders of the province and the gadget will tell you its area. Figure out how a planimeter works, both on the mechanical level and on the mathematical level. Wikipedia has an [http://en.wikipedia.org/wiki/Planimeter explanation] which is is right in the spirit but wrong on the details. Fix it, if you feel inspired (and if you do right and let me know about it, it'll count as a good deed).</div> Franklin http://drorbn.net/index.php?title=0708-1300/Homework_Assignment_5 0708-1300/Homework Assignment 5 2007-11-22T21:33:25Z <p>Franklin: /* Due Date */</p> <hr /> <div>{{0708-1300/Navigation}}<br /> ==Reading==<br /> '''Read''' sections 1-3 of chapter V of Bredon's book three times:<br /> * First time as if you were reading a novel - quickly and without too much attention to detail, just to learn what the main keywords and concepts and goals are.<br /> * Second time like you were studying for an exam on the subject - slowly and not skipping anything, verifying every little detail.<br /> * And then a third time, again at a quicker pace, to remind yourself of the bigger picture all those little details are there to paint.<br /> <br /> '''Also,''' do the same with your own class notes - much of what we do for this part of the class is '''not''' in the textbook!<br /> <br /> ==Doing==<br /> Solve all of the following problems, but submit only your solutions of problems 1, 3, 8, 9 and 10:<br /> <br /> '''Problem 1.''' Let &lt;math&gt;M^n&lt;/math&gt; be a manifold. Show that the following definitions for the orientability of &lt;math&gt;M&lt;/math&gt; are equivalent:<br /> # There exists a nowhere vanishing &lt;math&gt;n&lt;/math&gt;-form on &lt;math&gt;M&lt;/math&gt;.<br /> # There exists an atlas &lt;math&gt;\{(U_\alpha,\phi_\alpha:U_\alpha\to{\mathbb R}^n)\}&lt;/math&gt; for &lt;math&gt;M&lt;/math&gt;, so that &lt;math&gt;\det(\phi_\alpha\phi^{-1}_\beta)&gt;0&lt;/math&gt; wherever that makes sense.<br /> <br /> '''Problem 2.''' Show that the tangent space &lt;math&gt;TM&lt;/math&gt; of any manifold &lt;math&gt;M&lt;/math&gt; is orientable.<br /> <br /> '''Problem 3.'''<br /> # Show that if &lt;math&gt;M&lt;/math&gt; and &lt;math&gt;N&lt;/math&gt; are orientable then so is &lt;math&gt;M\times N&lt;/math&gt;.<br /> # Show that if &lt;math&gt;M&lt;/math&gt; and &lt;math&gt;M\times N&lt;/math&gt; are orientable then so is &lt;math&gt;N&lt;/math&gt;.<br /> <br /> '''Problem 4.''' Show that &lt;math&gt;S^n&lt;/math&gt; is always orientable.<br /> <br /> '''Problem 5.''' Recall that a form is called closed if it is in the kernel of &lt;math&gt;d&lt;/math&gt; and exact if it is in the image of &lt;math&gt;d&lt;/math&gt;. Show that every exact form is closed.<br /> <br /> '''Problem 6.''' Let &lt;math&gt;f:{\mathbb R}_t\to S^1\subset{\mathbb C}&lt;/math&gt; be given by &lt;math&gt;f(t)=e^{it}&lt;/math&gt;.<br /> # Show that there exists a unique &lt;math&gt;\omega\in\Omega^1(S^1)&lt;/math&gt; such that &lt;math&gt;f^\star\omega=dt&lt;/math&gt;.<br /> # Show that &lt;math&gt;\omega&lt;/math&gt; is closed but not exact.<br /> <br /> '''Problem 7.''' Show, directly from the definitions, that every closed 1-form on &lt;math&gt;{\mathbb R}^2&lt;/math&gt; is exact.<br /> <br /> '''Problem 8.''' Compute the integral &lt;math&gt;\int_{S^2}zdx\wedge dy&lt;/math&gt; twice:<br /> # Using Stokes' theorem.<br /> # Directly from the definition, by using a one- or two-chart atlas for &lt;math&gt;S^2&lt;/math&gt; (or for &lt;math&gt;S^2&lt;/math&gt; minus a single point).<br /> (Repeat 1 and 2 until they stop giving different answers).<br /> <br /> '''Problem 9.''' Show that the form &lt;math&gt;\omega_0=xdy\wedge dz+ydz\wedge dx+zdx\wedge dy\in\Omega^2({\mathbb R}^3_{x,y,z})&lt;/math&gt; is invariant under rigid orientation-preserving rotations of &lt;math&gt;{\mathbb R}^3&lt;/math&gt;. That is, if &lt;math&gt;A&lt;/math&gt; is such a rotation matrix (&lt;math&gt;AA^T=I&lt;/math&gt; and &lt;math&gt;\det A=1&lt;/math&gt;) considered also as a linear transformation &lt;math&gt;A:{\mathbb R}^3\to{\mathbb R}^3&lt;/math&gt;, then &lt;math&gt;A^\star\omega_0=\omega_0&lt;/math&gt;.<br /> <br /> '''Problem 10.''' Let &lt;math&gt;\omega_\alpha&lt;/math&gt; be the form &lt;math&gt;(x^2+y^2+z^2)^{\alpha/2}\omega_0&lt;/math&gt; defined on &lt;math&gt;{\mathbb R}^3\backslash\{0\}&lt;/math&gt;, where &lt;math&gt;\alpha&lt;/math&gt; is a real number and &lt;math&gt;\omega_0&lt;/math&gt; is the form from the previous question.<br /> # Find a number &lt;math&gt;\alpha&lt;/math&gt; so that &lt;math&gt;\omega_\alpha&lt;/math&gt; would be closed.<br /> # For that value of &lt;math&gt;\alpha&lt;/math&gt;, compute &lt;math&gt;\int_S\omega_\alpha&lt;/math&gt;, where &lt;math&gt;S&lt;/math&gt; is the sphere of radius &lt;math&gt;r&lt;/math&gt; around a point &lt;math&gt;p\in{\mathbb R}^3&lt;/math&gt;. (If you end up writing complicated formulas, you missed the point).<br /> # For the same &lt;math&gt;\alpha&lt;/math&gt;, is &lt;math&gt;\omega_\alpha&lt;/math&gt; exact?<br /> <br /> ==Due Date==<br /> This assignment is due in class on Thursday December 6, 2007.<br /> <br /> <br /> [http://www.brainyquote.com/quotes/authors/d/douglas_adams.html Douglas Adams] - &quot;I love deadlines. I like the whooshing sound they make as they fly by.&quot;<br /> <br /> ==Just for Fun==<br /> A planimeter is a (mechanical!) measuring instrument used to measure the area of an arbitrary two-dimensional shape by tracing along its boundary; if you have a map of Ontario, for example, you can roll the measuring end of a planimeter counterclockwise around the borders of the province and the gadget will tell you its area. Figure out how a planimeter works, both on the mechanical level and on the mathematical level. Wikipedia has an [http://en.wikipedia.org/wiki/Planimeter explanation] which is is right in the spirit but wrong on the details. Fix it, if you feel inspired (and if you do right and let me know about it, it'll count as a good deed).</div> Franklin http://drorbn.net/index.php?title=0708-1300/Homework_Assignment_4 0708-1300/Homework Assignment 4 2007-11-18T17:30:52Z <p>Franklin: /* Just for Fun */</p> <hr /> <div>__NOTOC__<br /> {{0708-1300/Navigation}}<br /> <br /> ==Reading==<br /> '''Read''' section 11 of chapter II and sections 1-3 of chapter V of Bredon's book three times:<br /> * First time as if you were reading a novel - quickly and without too much attention to detail, just to learn what the main keywords and concepts and goals are.<br /> * Second time like you were studying for an exam on the subject - slowly and not skipping anything, verifying every little detail.<br /> * And then a third time, again at a quicker pace, to remind yourself of the bigger picture all those little details are there to paint.<br /> <br /> ==Doing==<br /> Solve the following problems from Bredon's book, but submit only the solutions of the problems marked with an &quot;S&quot;:<br /> {|align=center border=1 cellspacing=0 cellpadding=5<br /> |- align=center<br /> !problems<br /> !on page(s)<br /> |- align=center<br /> |S1, S2<br /> |100-101<br /> |- align=center<br /> |S1, S2, 3<br /> |264<br /> |}<br /> <br /> Also, solve and submit the following question:<br /> <br /> '''Question 6.'''<br /> # Show that if &lt;math&gt;n\neq m&lt;/math&gt; then &lt;math&gt;{\mathbf R}^n&lt;/math&gt; is not diffeomorphic (homeomorphic via a smooth map with a smooth inverse) to &lt;math&gt;{\mathbf R}^m&lt;/math&gt;.<br /> # &lt;strike&gt;Show that if &lt;math&gt;n\neq m&lt;/math&gt; then &lt;math&gt;{\mathbf R}^n&lt;/math&gt; is not homeomorphic to &lt;math&gt;{\mathbf R}^m&lt;/math&gt;.&lt;/strike&gt;<br /> Note that a priori the second part of this question is an order of magnitude harder than the first. I am not sure how to do it with our current techniques, though later on it will become an easy consequence of &quot;homology theory&quot;.<br /> <br /> ==Due Date==<br /> This assignment is due in class on Tuesday November 20, 2007.<br /> <br /> ==Just for Fun==<br /> Find a ''geometric'' interpretation to the formula<br /> {{Equation*|&lt;math&gt;d\omega(X,Y)=X(\omega(Y))-Y(\omega(X))-\omega([X,Y])&lt;/math&gt;.}}<br /> (Of course, you have to first obtain a ''geometric'' understanding of &lt;math&gt;[X,Y]&lt;/math&gt;, and this in itself is significant and worthwhile).<br /> <br /> {{Dror/Students Divider}}<br /> <br /> Look at the story of [[0708-1300/Barnie the polar bear | Barnie the polar bear]].<br /> <br /> If &lt;math&gt;n\neq m&lt;/math&gt; then &lt;math&gt;{\mathbf R}^n&lt;/math&gt; is [[0708-1300/not homeomorphic|not homeomorphic]] to &lt;math&gt;{\mathbf R}^m&lt;/math&gt;.</div> Franklin http://drorbn.net/index.php?title=0708-1300/not_homeomorphic 0708-1300/not homeomorphic 2007-11-18T17:29:51Z <p>Franklin: </p> <hr /> <div>Please, read the following carefully. It can contain some mistake.<br /> <br /> Assume &lt;math&gt;f_0 : R^n \rightarrow R^m&lt;/math&gt; is a homeomorphism. Since &lt;math&gt;f_0&lt;/math&gt; is proper we can extend it to a continuous map &lt;math&gt;f : S^n \rightarrow S^m&lt;/math&gt; which in fact will be a homeomorphism. Taking inverse if necessary we may assume &lt;math&gt;n &lt; m&lt;/math&gt;. <br /> Let &lt;math&gt;F:S^n\times[0, 1] \rightarrow S^m&lt;/math&gt; be a homotopy of &lt;math&gt;f&lt;/math&gt; to a smooth map i.e. &lt;math&gt;F&lt;/math&gt; is continuous, &lt;math&gt;F(x, 0) = f(x)&lt;/math&gt; and &lt;math&gt;F(x, 1)&lt;/math&gt; is smooth. Since &lt;math&gt;F(x, 1)&lt;/math&gt; is smooth and &lt;math&gt;n &lt; m&lt;/math&gt; all of its image points are singular values and by Sard's theorem constitute a set of measure zero. Then there is a point in &lt;math&gt;S^m&lt;/math&gt; not in the image of &lt;math&gt;F(x, 1)&lt;/math&gt;, but the complement of that point is contractible. This means that we can extend &lt;math&gt;F&lt;/math&gt; to &lt;math&gt;F_0:S^n\times[0, 2]\rightarrow S^m&lt;/math&gt; to be a homotopy of &lt;math&gt;f&lt;/math&gt; to a constant map. But then &lt;math&gt;f^{-1}F_{0}&lt;/math&gt; is a contraction of &lt;math&gt;S^n&lt;/math&gt; which is a contradiction with the fact that no such contraction exists.</div> Franklin http://drorbn.net/index.php?title=0708-1300/not_homeomorphic 0708-1300/not homeomorphic 2007-11-18T17:29:04Z <p>Franklin: </p> <hr /> <div>Please, read the following carefully. It can contain some mistake.<br /> <br /> Assume &lt;math&gt;f_0 : R^n \rightarrow R^m&lt;/math&gt; is a homeomorphism. Since &lt;math&gt;f_0&lt;/math&gt; is proper we can extend it to a continuous map &lt;math&gt;f : S^n \rightarrow S^m&lt;/math&gt; which in fact will be a homeomorphism. Taking inverse if necessary we may assume &lt;math&gt;n &lt; m&lt;/math&gt;. <br /> Let &lt;math&gt;F:S^n\times[0, 1] \rightarrow S^m&lt;/math&gt; be a homotopy of &lt;math&gt;f&lt;/math&gt; to a smooth map i.e. &lt;math&gt;F&lt;/math&gt; is continuous, &lt;math&gt;F(x, 0) = f(x)&lt;/math&gt; and &lt;math&gt;F(x, 1)&lt;/math&gt; is smooth. Since &lt;math&gt;F(x, 1)&lt;/math&gt; is smooth and &lt;math&gt;n &lt; m&lt;/math&gt; all of its image points are singular values and by Sard's theorem constitute a set of measure zero. Then there is a point in &lt;math&gt;S^m&lt;/math&gt; not in the image of &lt;math&gt;F(x, 1)&lt;/math&gt;, but the complement of that point is contractible. This means that we can extend &lt;math&gt;F&lt;/math&gt; to &lt;math&gt;F_0:S^n\times[0, 2]\rightarrow S^m&lt;/math&gt; to be a homotopy of &lt;math&gt;f&lt;/math&gt; to a constant map. But then is a contraction of &lt;math&gt;S^n&lt;/math&gt; which is a contradiction with the fact that no such contraction exists.</div> Franklin http://drorbn.net/index.php?title=0708-1300/not_homeomorphic 0708-1300/not homeomorphic 2007-11-18T17:27:28Z <p>Franklin: </p> <hr /> <div>Please, read the following carefully. It can contain some mistake.<br /> <br /> Assume &lt;math&gt;f_0 : R^n \rightarrow R^m&lt;/math&gt; is a homeomorphism. Since &lt;math&gt;f_0&lt;/math&gt; is proper we can extend it to a continuous map &lt;math&gt;f : S^n \rightarrow S^m&lt;/math&gt; which in fact will be a homeomorphism. Taking inverse if necessary we may assume &lt;math&gt;n &lt; m&lt;/math&gt;. <br /> Let &lt;math&gt;F:S^n\times[0, 1] \rightarrow S^m&lt;/math&gt; be a homotopy of &lt;math&gt;f&lt;/math&gt; to a smooth map i.e. &lt;math&gt;F&lt;/math&gt; is continuous, &lt;math&gt;F(x, 0) = f(x)&lt;/math&gt; and &lt;math&gt;F(x, 1)&lt;/math&gt; is smooth. Since &lt;math&gt;F(x, 1)&lt;/math&gt; is smooth and &lt;math&gt;n &lt; m&lt;/math&gt; all of its image points are singular values and by Sard's theorem constitute a set of measure zero. Then there is a point in &lt;math&gt;S^m&lt;/math&gt; not in the image of &lt;math&gt;F(x, 1)&lt;/math&gt;, but the complement of that point is contractible. This means that we can extend &lt;math&gt;F&lt;/math&gt; to &lt;math&gt;F_0:S^n\times[0, 2]\rightarrow S^m&lt;/math&gt; to be a homotopy of &lt;math&gt;f&lt;/math&gt; to a constant map. But then &lt;math&gt; f^{－1}F_{0} &lt;/math&gt; is a contraction of &lt;math&gt;S^n&lt;/math&gt; which is a contradiction with the fact that no such contraction exists.</div> Franklin http://drorbn.net/index.php?title=0708-1300/not_homeomorphic 0708-1300/not homeomorphic 2007-11-18T17:27:09Z <p>Franklin: </p> <hr /> <div>Please, read the following carefully. It can contain some mistake.<br /> <br /> Assume &lt;math&gt;f_0 : R^n \rightarrow R^m&lt;/math&gt; is a homeomorphism. Since &lt;math&gt;f_0&lt;/math&gt; is proper we can extend it to a continuous map &lt;math&gt;f : S^n \rightarrow S^m&lt;/math&gt; which in fact will be a homeomorphism. Taking inverse if necessary we may assume &lt;math&gt;n &lt; m&lt;/math&gt;. <br /> Let &lt;math&gt;F:S^n\times[0, 1] \rightarrow S^m&lt;/math&gt; be a homotopy of &lt;math&gt;f&lt;/math&gt; to a smooth map i.e. &lt;math&gt;F&lt;/math&gt; is continuous, &lt;math&gt;F(x, 0) = f(x)&lt;/math&gt; and &lt;math&gt;F(x, 1)&lt;/math&gt; is smooth. Since &lt;math&gt;F(x, 1)&lt;/math&gt; is smooth and &lt;math&gt;n &lt; m&lt;/math&gt; all of its image points are singular values and by Sard's theorem constitute a set of measure zero. Then there is a point in &lt;math&gt;S^m&lt;/math&gt; not in the image of &lt;math&gt;F(x, 1)&lt;/math&gt;, but the complement of that point is contractible. This means that we can extend &lt;math&gt;F&lt;/math&gt; to &lt;math&gt;F_0:S^n\times[0, 2]\rightarrow S^m&lt;/math&gt; to be a homotopy of &lt;math&gt;f&lt;/math&gt; to a constant map. But then &lt;math&gt; f^{－1}F_0 &lt;/math&gt; is a contraction of &lt;math&gt;S^n&lt;/math&gt; which is a contradiction with the fact that no such contraction exists.</div> Franklin http://drorbn.net/index.php?title=0708-1300/not_homeomorphic 0708-1300/not homeomorphic 2007-11-18T17:26:06Z <p>Franklin: </p> <hr /> <div>Please, read the following carefully. It can contain some mistake.<br /> <br /> Assume &lt;math&gt;f_0 : R^n \rightarrow R^m&lt;/math&gt; is a homeomorphism. Since &lt;math&gt;f_0&lt;/math&gt; is proper we can extend it to a continuous map &lt;math&gt;f : S^n \rightarrow S^m&lt;/math&gt; which in fact will be a homeomorphism. Taking inverse if necessary we may assume &lt;math&gt;n &lt; m&lt;/math&gt;. <br /> Let &lt;math&gt;F:S^n\times[0, 1] \rightarrow S^m&lt;/math&gt; be a homotopy of &lt;math&gt;f&lt;/math&gt; to a smooth map i.e. &lt;math&gt;F&lt;/math&gt; is continuous, &lt;math&gt;F(x, 0) = f(x)&lt;/math&gt; and &lt;math&gt;F(x, 1)&lt;/math&gt; is smooth. Since &lt;math&gt;F(x, 1)&lt;/math&gt; is smooth and &lt;math&gt;n &lt; m&lt;/math&gt; all of its image points are singular values and by Sard's theorem constitute a set of measure zero. Then there is a point in &lt;math&gt;S^m&lt;/math&gt; not in the image of &lt;math&gt;F(x, 1)&lt;/math&gt;, but the complement of that point is contractible. This means that we can extend &lt;math&gt;F&lt;/math&gt; to &lt;math&gt;F_0:S^n\times[0, 2]\rightarrow S^m&lt;/math&gt; to be a homotopy of &lt;math&gt;f&lt;/math&gt; to a constant map. But then &lt;math&gt;f^{－1}F_0&lt;/math&gt; is a contraction of &lt;math&gt;S^n&lt;/math&gt; which is a contradiction with the fact that no such contraction exists.</div> Franklin http://drorbn.net/index.php?title=0708-1300/not_homeomorphic 0708-1300/not homeomorphic 2007-11-18T17:25:21Z <p>Franklin: </p> <hr /> <div>Please, read the following carefully. It can contain some mistake.<br /> <br /> Assume &lt;math&gt;f_0 : R^n \rightarrow R^m&lt;/math&gt; is a homeomorphism. Since &lt;math&gt;f_0&lt;/math&gt; is proper we can extend it to a continuous map &lt;math&gt;f : S^n \rightarrow S^m&lt;/math&gt; which in fact will be a homeomorphism. Taking inverse if necessary we may assume &lt;math&gt;n &lt; m&lt;/math&gt;. <br /> Let &lt;math&gt;F:S^n\times[0, 1] \rightarrow S^m&lt;/math&gt; be a homotopy of &lt;math&gt;f&lt;/math&gt; to a smooth map i.e. &lt;math&gt;F&lt;/math&gt; is continuous, &lt;math&gt;F(x, 0) = f(x)&lt;/math&gt; and &lt;math&gt;F(x, 1)&lt;/math&gt; is smooth. Since &lt;math&gt;F(x, 1)&lt;/math&gt; is smooth and &lt;math&gt;n &lt; m&lt;/math&gt; all of its image points are singular values and by Sard's theorem constitute a set of measure zero. Then there is a point in &lt;math&gt;S^m&lt;/math&gt; not in the image of &lt;math&gt;F(x, 1)&lt;/math&gt;, but the complement of that point is contractible. This means that we can extend &lt;math&gt;F&lt;/math&gt; to &lt;math&gt;F_0:S^n\times[0, 2] \rightarrow S^m&lt;/math&gt; to be a homotopy of &lt;math&gt;f&lt;/math&gt; to a constant map. But then &lt;math&gt;f^{－1}oF_0&lt;/math&gt; is a contraction of &lt;math&gt;S^n&lt;/math&gt; which is a contradiction with the fact that no such contraction exists.</div> Franklin http://drorbn.net/index.php?title=0708-1300/not_homeomorphic 0708-1300/not homeomorphic 2007-11-18T17:24:40Z <p>Franklin: </p> <hr /> <div>Please, read the following carefully. It can contain some mistake.<br /> <br /> Assume &lt;math&gt;f_0 : R^n \rightarrow R^m&lt;/math&gt; is a homeomorphism. Since &lt;math&gt;f_0&lt;/math&gt; is proper we can extend it to a continuous map &lt;math&gt;f : S^n \rightarrow S^m&lt;/math&gt; which in fact will be a homeomorphism. Taking inverse if necessary we may assume &lt;math&gt;n &lt; m&lt;/math&gt;. <br /> Let &lt;math&gt;F:S^n\times[0, 1] \rightarrow S^m&lt;/math&gt; be a homotopy of &lt;math&gt;f&lt;/math&gt; to a smooth map i.e. &lt;math&gt;F&lt;/math&gt; is continuous, &lt;math&gt;F(x, 0) = f(x)&lt;/math&gt; and &lt;math&gt;F(x, 1)&lt;/math&gt; is smooth. Since &lt;math&gt;F(x, 1)&lt;/math&gt; is smooth and &lt;math&gt;n &lt; m&lt;/math&gt; all of its image points are singular values and by Sard's theorem constitute a set of measure zero. Then there is a point in &lt;math&gt;S^m&lt;/math&gt; not in the image of &lt;math&gt;F(x, 1)&lt;/math&gt;, but the complement of that point is contractible. This means that we can extend &lt;math&gt;F&lt;/math&gt; to &lt;math&gt;F_0:S^n\times[0, 2] \rightarrow S^m&lt;/math&gt; to be a homotopy of &lt;math&gt;f&lt;/math&gt; to a constant map. But then &lt;math&gt;f^{－1}\circ F_0&lt;/math&gt; is a contraction of &lt;math&gt;S^n&lt;/math&gt; which is a contradiction with the fact that no such contraction exists.</div> Franklin http://drorbn.net/index.php?title=0708-1300/not_homeomorphic 0708-1300/not homeomorphic 2007-11-18T17:21:39Z <p>Franklin: </p> <hr /> <div>Please, read the following carefully. It can contain some mistake.<br /> <br /> Assume &lt;math&gt;f_0 : R^n --&gt; R^m&lt;/math&gt; is a homeomorphism. Since &lt;math&gt;f_0&lt;/math&gt; is proper we can extend it to a continuous map &lt;math&gt;f : S^n --&gt; S^m&lt;/math&gt; which in fact will be<br /> a homeomorphism. Taking inverse if necessary we may assume &lt;math&gt;n &lt; m&lt;/math&gt;. Let<br /> &lt;math&gt;F : Sn × [0, 1] --&gt; Sm&lt;/math&gt; be a homotopy of &lt;math&gt;f&lt;/math&gt; to a smooth map i.e. &lt;math&gt;F&lt;/math&gt; is continuous, &lt;math&gt;F(x, 0) = f(x)&lt;/math&gt; and &lt;math&gt;F(x, 1)&lt;/math&gt; is smooth. Since &lt;math&gt;F(x, 1)&lt;/math&gt; is smooth and &lt;math&gt;n &lt; m&lt;/math&gt; all of its image points are singular values and by Sard's theorem constitute a set of measure zero. Then there is a point in &lt;math&gt;S^m&lt;/math&gt; not in the image of &lt;math&gt;F(x, 1)&lt;/math&gt;, but the complement of that point is contractible. This means that we can extend &lt;math&gt;F&lt;/math&gt; to &lt;math&gt;F_0 : Sn × [0, 2] --&gt; S^m&lt;/math&gt; to be a homotopy of &lt;math&gt;f&lt;/math&gt; to a constant map. But then &lt;math&gt;f^{－1}\circ F_0&lt;/math&gt; is a contraction of &lt;math&gt;S^n&lt;/math&gt; which is a contradiction with the fact that no such contraction exists.</div> Franklin http://drorbn.net/index.php?title=0708-1300/not_homeomorphic 0708-1300/not homeomorphic 2007-11-18T17:20:18Z <p>Franklin: </p> <hr /> <div>Please, read the following carefully. It can contain some mistake.<br /> <br /> Assume &lt;math&gt;\overline{f} : R^n --&gt; R^m&lt;/math&gt; is a homeomorphism. Since &lt;math&gt;\overline{f}&lt;/math&gt; is proper we can extend it to a continuous map &lt;math&gt;f : S^n --&gt; S^m&lt;/math&gt; which in fact will be<br /> a homeomorphism. Taking inverse if necessary we may assume &lt;math&gt;n &lt; m&lt;/math&gt;. Let<br /> &lt;math&gt;F : Sn × [0, 1] --&gt; Sm&lt;/math&gt; be a homotopy of &lt;math&gt;f&lt;/math&gt; to a smooth map i.e. &lt;math&gt;F&lt;/math&gt; is continuous, &lt;math&gt;F(x, 0) = f(x)&lt;/math&gt; and &lt;math&gt;F(x, 1)&lt;/math&gt; is smooth. Since &lt;math&gt;F(x, 1)&lt;/math&gt; is smooth and &lt;math&gt;n &lt; m&lt;/math&gt; all of its image points are singular values and by Sard's theorem constitute a set of measure zero. Then there is a point in &lt;math&gt;S^m&lt;/math&gt; not in the image of &lt;math&gt;F(x, 1)&lt;/math&gt;, but the complement of that point is contractible. This means that we can extend &lt;math&gt;F&lt;/math&gt; to &lt;math&gt;\overline{F} : Sn × [0, 2] --&gt; S^m&lt;/math&gt; to be a homotopy of &lt;math&gt;f&lt;/math&gt; to a constant map. But then &lt;math&gt;f^{－1}\circ\overline{F}&lt;/math&gt; is a contraction of &lt;math&gt;S^n&lt;/math&gt; which is a contradiction with the fact that no such contraction exists.</div> Franklin http://drorbn.net/index.php?title=0708-1300/not_homeomorphic 0708-1300/not homeomorphic 2007-11-18T17:18:33Z <p>Franklin: </p> <hr /> <div>Please, read the following carefully. It can contain some mistake.<br /> <br /> Assume &lt;math&gt;\~{f} : R^n --&gt; R^m&lt;/math&gt; is a homeomorphism. Since &lt;math&gt;\~{f}&lt;/math&gt; is proper<br /> we can extend it to a continuous map &lt;math&gt;f : S^n --&gt; S^m&lt;/math&gt; which in fact will be<br /> a homeomorphism. Taking inverse if necessary we may assume &lt;math&gt;n &lt; m&lt;/math&gt;. Let<br /> &lt;math&gt;F : Sn × [0, 1] --&gt; Sm&lt;/math&gt; be a homotopy of &lt;math&gt;f&lt;/math&gt; to a smooth map i.e. &lt;math&gt;F&lt;/math&gt; is continuous,<br /> &lt;math&gt;F(x, 0) = f(x)&lt;/math&gt; and &lt;math&gt;F(x, 1)&lt;/math&gt; is smooth. Since &lt;math&gt;F(x, 1)&lt;/math&gt; is smooth and &lt;math&gt;n &lt; m&lt;/math&gt; all<br /> of its image points are singular values and by Sard's theorem constitute a set<br /> of measure zero. Then there is a point in &lt;math&gt;S^m&lt;/math&gt; not in the image of &lt;math&gt;F(x, 1)&lt;/math&gt;, but<br /> the complement of that point is contractible. This means that we can extend<br /> &lt;math&gt;F&lt;/math&gt; to &lt;math&gt;\~{F} : Sn × [0, 2] --&gt; S^m&lt;/math&gt; to be a homotopy of &lt;math&gt;f&lt;/math&gt; to a constant map. But then<br /> &lt;math&gt;f^{－1}\circ\~{F}&lt;/math&gt; is a contraction of &lt;math&gt;S^n&lt;/math&gt; which is a contradiction with the fact that no<br /> such contraction exists.</div> Franklin http://drorbn.net/index.php?title=0708-1300/Homework_Assignment_4 0708-1300/Homework Assignment 4 2007-11-18T17:15:18Z <p>Franklin: /* Just for Fun */</p> <hr /> <div>__NOTOC__<br /> {{0708-1300/Navigation}}<br /> <br /> ==Reading==<br /> '''Read''' section 11 of chapter II and sections 1-3 of chapter V of Bredon's book three times:<br /> * First time as if you were reading a novel - quickly and without too much attention to detail, just to learn what the main keywords and concepts and goals are.<br /> * Second time like you were studying for an exam on the subject - slowly and not skipping anything, verifying every little detail.<br /> * And then a third time, again at a quicker pace, to remind yourself of the bigger picture all those little details are there to paint.<br /> <br /> ==Doing==<br /> Solve the following problems from Bredon's book, but submit only the solutions of the problems marked with an &quot;S&quot;:<br /> {|align=center border=1 cellspacing=0 cellpadding=5<br /> |- align=center<br /> !problems<br /> !on page(s)<br /> |- align=center<br /> |S1, S2<br /> |100-101<br /> |- align=center<br /> |S1, S2, 3<br /> |264<br /> |}<br /> <br /> Also, solve and submit the following question:<br /> <br /> '''Question 6.'''<br /> # Show that if &lt;math&gt;n\neq m&lt;/math&gt; then &lt;math&gt;{\mathbf R}^n&lt;/math&gt; is not diffeomorphic (homeomorphic via a smooth map with a smooth inverse) to &lt;math&gt;{\mathbf R}^m&lt;/math&gt;.<br /> # &lt;strike&gt;Show that if &lt;math&gt;n\neq m&lt;/math&gt; then &lt;math&gt;{\mathbf R}^n&lt;/math&gt; is not homeomorphic to &lt;math&gt;{\mathbf R}^m&lt;/math&gt;.&lt;/strike&gt;<br /> Note that a priori the second part of this question is an order of magnitude harder than the first. I am not sure how to do it with our current techniques, though later on it will become an easy consequence of &quot;homology theory&quot;.<br /> <br /> ==Due Date==<br /> This assignment is due in class on Tuesday November 20, 2007.<br /> <br /> ==Just for Fun==<br /> Find a ''geometric'' interpretation to the formula<br /> {{Equation*|&lt;math&gt;d\omega(X,Y)=X(\omega(Y))-Y(\omega(X))-\omega([X,Y])&lt;/math&gt;.}}<br /> (Of course, you have to first obtain a ''geometric'' understanding of &lt;math&gt;[X,Y]&lt;/math&gt;, and this in itself is significant and worthwhile).<br /> <br /> {{Dror/Students Divider}}<br /> <br /> Look at the story of [[0708-1300/Barnie the polar bear | Barnie the polar bear]].<br /> If &lt;math&gt;n\neq m&lt;/math&gt; then &lt;math&gt;{\mathbf R}^n&lt;/math&gt; is [[0708-1300/not homeomorphic|not homeomorphic]] to &lt;math&gt;{\mathbf R}^m&lt;/math&gt;.</div> Franklin http://drorbn.net/index.php?title=0708-1300/Homework_Assignment_4 0708-1300/Homework Assignment 4 2007-11-18T17:14:49Z <p>Franklin: </p> <hr /> <div>__NOTOC__<br /> {{0708-1300/Navigation}}<br /> <br /> ==Reading==<br /> '''Read''' section 11 of chapter II and sections 1-3 of chapter V of Bredon's book three times:<br /> * First time as if you were reading a novel - quickly and without too much attention to detail, just to learn what the main keywords and concepts and goals are.<br /> * Second time like you were studying for an exam on the subject - slowly and not skipping anything, verifying every little detail.<br /> * And then a third time, again at a quicker pace, to remind yourself of the bigger picture all those little details are there to paint.<br /> <br /> ==Doing==<br /> Solve the following problems from Bredon's book, but submit only the solutions of the problems marked with an &quot;S&quot;:<br /> {|align=center border=1 cellspacing=0 cellpadding=5<br /> |- align=center<br /> !problems<br /> !on page(s)<br /> |- align=center<br /> |S1, S2<br /> |100-101<br /> |- align=center<br /> |S1, S2, 3<br /> |264<br /> |}<br /> <br /> Also, solve and submit the following question:<br /> <br /> '''Question 6.'''<br /> # Show that if &lt;math&gt;n\neq m&lt;/math&gt; then &lt;math&gt;{\mathbf R}^n&lt;/math&gt; is not diffeomorphic (homeomorphic via a smooth map with a smooth inverse) to &lt;math&gt;{\mathbf R}^m&lt;/math&gt;.<br /> # &lt;strike&gt;Show that if &lt;math&gt;n\neq m&lt;/math&gt; then &lt;math&gt;{\mathbf R}^n&lt;/math&gt; is not homeomorphic to &lt;math&gt;{\mathbf R}^m&lt;/math&gt;.&lt;/strike&gt;<br /> Note that a priori the second part of this question is an order of magnitude harder than the first. I am not sure how to do it with our current techniques, though later on it will become an easy consequence of &quot;homology theory&quot;.<br /> <br /> ==Due Date==<br /> This assignment is due in class on Tuesday November 20, 2007.<br /> <br /> ==Just for Fun==<br /> Find a ''geometric'' interpretation to the formula<br /> {{Equation*|&lt;math&gt;d\omega(X,Y)=X(\omega(Y))-Y(\omega(X))-\omega([X,Y])&lt;/math&gt;.}}<br /> (Of course, you have to first obtain a ''geometric'' understanding of &lt;math&gt;[X,Y]&lt;/math&gt;, and this in itself is significant and worthwhile).<br /> <br /> {{Dror/Students Divider}}<br /> <br /> Look at the story of [[0708-1300/Barnie the polar bear | Barnie the polar bear]].<br /> If &lt;math&gt;n\neq m&lt;/math&gt; then &lt;math&gt;{\mathbf R}^n&lt;/math&gt; is [[0708-1300/not homeomorphic not homeomorphic]] to &lt;math&gt;{\mathbf R}^m&lt;/math&gt;.</div> Franklin http://drorbn.net/index.php?title=0708-1300/the_unit_sphere_in_a_Hilbert_space_is_contractible 0708-1300/the unit sphere in a Hilbert space is contractible 2007-11-07T14:44:47Z <p>Franklin: </p> <hr /> <div>Let &lt;math&gt;H=L^2[0,1]&lt;/math&gt; and define &lt;math&gt;S^{\infty}=\{x\in H| ||x||=1\}&lt;/math&gt;<br /> <br /> '''Claim'''<br /> <br /> &lt;math&gt;S^{\infty}&lt;/math&gt; is contractible <br /> <br /> '''Proof'''<br /> <br /> For any &lt;math&gt;t\in[0,1]&lt;/math&gt; and any &lt;math&gt;f\in H&lt;/math&gt; define &lt;math&gt;f_t(x)= f&lt;/math&gt; for &lt;math&gt;0\leq x \leq t&lt;/math&gt; and &lt;math&gt;f_t(x)=1&lt;/math&gt; for &lt;math&gt;t&lt;x\leq1&lt;/math&gt;.<br /> Observe that &lt;math&gt;t\rightarrow f_t/||f_t||&lt;/math&gt; is continuous and gives the desired retraction to the point &lt;math&gt;f=1&lt;/math&gt;.<br /> <br /> This proof only works in '''separable''' Hilbert spaces? Is the unit ball in a non-separable Hilbert space contractible?<br /> <br /> The answer seems to be '''YES''' see [http://www.jstor.org/view/00029939/di970909/97p01032/0?frame=noframe&amp;userID=80644483@utoronto.ca/01c0a80a6600501ced693&amp;dpi=3&amp;config=jstor Spheres in infinite-dimensional normed spaces are Lipschitz contractible]</div> Franklin http://drorbn.net/index.php?title=0708-1300/the_unit_sphere_in_a_Hilbert_space_is_contractible 0708-1300/the unit sphere in a Hilbert space is contractible 2007-11-07T14:22:01Z <p>Franklin: </p> <hr /> <div>Let &lt;math&gt;H=L^2[0,1]&lt;/math&gt; and define &lt;math&gt;S^{\infty}=\{x\in H| ||x||=1\}&lt;/math&gt;<br /> <br /> '''Claim'''<br /> <br /> &lt;math&gt;S^{\infty}&lt;/math&gt; is contractible <br /> <br /> '''Proof'''<br /> <br /> For any &lt;math&gt;t\in[0,1]&lt;/math&gt; and any &lt;math&gt;f\in H&lt;/math&gt; define &lt;math&gt;f_t(x)= f&lt;/math&gt; for &lt;math&gt;0\leq x \leq t&lt;/math&gt; and &lt;math&gt;f_t(x)=1&lt;/math&gt; for &lt;math&gt;t&lt;x\leq1&lt;/math&gt;.<br /> Observe that &lt;math&gt;t\rightarrow f_t/||f_t||&lt;/math&gt; is continuous and gives the desired retraction to the point &lt;math&gt;f=1&lt;/math&gt;.<br /> <br /> This proof only works in '''separable''' Hilbert spaces? Is the unit ball in a non-separable Hilbert space contractible?</div> Franklin http://drorbn.net/index.php?title=0708-1300/the_unit_sphere_in_a_Hilbert_space_is_contractible 0708-1300/the unit sphere in a Hilbert space is contractible 2007-11-02T20:29:46Z <p>Franklin: </p> <hr /> <div>Let &lt;math&gt;H=\{(x_1,x_2,...)| \sum x_n^2&lt;\infty\}&lt;/math&gt; and define &lt;math&gt;S^{\infty}=\{x\in H| ||x||=1\}&lt;/math&gt;<br /> <br /> '''Claim'''<br /> <br /> &lt;math&gt;S^{\infty}&lt;/math&gt; is contractible <br /> <br /> '''Proof'''<br /> <br /> A way to see this is via the cellular structure of &lt;math&gt;S^{\infty}&lt;/math&gt;. If &lt;math&gt;S^{\infty}=C_0 C_1 ...&lt;/math&gt; you can always contract &lt;math&gt;C_k&lt;/math&gt; along &lt;math&gt;C_{k+1}&lt;/math&gt; like moving contracting the equator along the surface of the earth.<br /> <br /> Does this proof only works in '''separable''' Hilbert spaces? Is the unit ball in a non-separable Hilbert space contractible?</div> Franklin http://drorbn.net/index.php?title=0708-1300/Class_notes_for_Thursday,_November_1 0708-1300/Class notes for Thursday, November 1 2007-11-02T15:17:52Z <p>Franklin: </p> <hr /> <div>{{0708-1300/Navigation}}<br /> ==Today's Agenda==<br /> * [[0708-1300/Homework Assignment 4|HW4]] and [[0708-1300/Term Exam 1|TE1]].<br /> * Continue with [[0708-1300/Class notes for Tuesday, October 30|Tuesday's]] agenda:<br /> ** Debt on proper functions.<br /> ** Prove that &quot;the sphere is not contractible&quot;.<br /> ** Complete the proof of the &quot;tubular neighborhood theorem&quot;.<br /> ===Proper Implies Closed===<br /> '''Theorem.''' A proper function &lt;math&gt;f:X\to Y&lt;/math&gt; from a topological space &lt;math&gt;X&lt;/math&gt; to a locally compact (Hausdorff) topological space &lt;math&gt;Y&lt;/math&gt; is closed.<br /> <br /> '''Proof.''' Let &lt;math&gt;B&lt;/math&gt; be closed in &lt;math&gt;X&lt;/math&gt;, we need to show that &lt;math&gt;f(B)&lt;/math&gt; is closed in &lt;math&gt;Y&lt;/math&gt;. Since closedness is a local property, it is enough to show that every point &lt;math&gt;y\in Y&lt;/math&gt; has a neighbourhood &lt;math&gt;U&lt;/math&gt; such that &lt;math&gt;f(B)\cap U&lt;/math&gt; is closed in &lt;math&gt;U&lt;/math&gt;. Fix &lt;math&gt;y\in Y&lt;/math&gt;, and by local compactness, choose a neighbourhood &lt;math&gt;U&lt;/math&gt; of &lt;math&gt;y&lt;/math&gt; whose close &lt;math&gt;\bar U&lt;/math&gt; is compact. Then<br /> {{Equation*|&lt;math&gt;f(B)\cap U=f(B\cap f^{-1}(U))\cap U\subset f(B\cap f^{-1}(\bar U))\cap U\subset f(B)\cap U&lt;/math&gt;,}}<br /> so that &lt;math&gt;f(B)\cap U=f(B\cap f^{-1}(\bar U))\cap U&lt;/math&gt;. But &lt;math&gt;\bar U&lt;/math&gt; is compact by choice, so &lt;math&gt;f^{-1}(\bar U)&lt;/math&gt; is compact as &lt;math&gt;f&lt;/math&gt; is proper, so &lt;math&gt;B\cap f^{-1}(\bar U)&lt;/math&gt; is compact as &lt;math&gt;B&lt;/math&gt; is closed, so &lt;math&gt;f(B\cap f^{-1}(\bar U))&lt;/math&gt; is compact (and hence closed) as a continuous image of a compact set, so &lt;math&gt;f(B)\cap U&lt;/math&gt; is the intersection &lt;math&gt;f(B\cap f^{-1}(\bar U))\cap U&lt;/math&gt; of a closed set with &lt;math&gt;U&lt;/math&gt;, hence it is closed in &lt;math&gt;U&lt;/math&gt;.<br /> <br /> ===Note===<br /> The example of a non-contractible &quot;comb&quot; seen today is, in fact, &quot;Cantor's comb&quot;. See, for example, page 25 of www.karlin.mff.cuni.cz/~pyrih/e/e2000v0/c/ect.ps<br /> <br /> <br /> ===Typed Notes===<br /> <br /> '''Definition 1'''<br /> <br /> X is contractible to &lt;math&gt;x_0&lt;/math&gt; if there exists a continuous function &lt;math&gt;H:X\times I\rightarrow X&lt;/math&gt;<br /> where<br /> <br /> 1) &lt;math&gt;H(x,0) = x&lt;/math&gt; &lt;math&gt;\forall x&lt;/math&gt;<br /> <br /> 2) &lt;math&gt;H(x,1) = x_0&lt;/math&gt; &lt;math&gt;\forall x&lt;/math&gt;<br /> <br /> 3) &lt;math&gt;H(x_0,t)=x_0&lt;/math&gt; &lt;math&gt;\forall t&lt;/math&gt;<br /> <br /> <br /> ''Example 1''<br /> <br /> The singleton &lt;math&gt;\{x_0\}&lt;/math&gt; is contractible<br /> <br /> <br /> ''Example 2''<br /> <br /> &lt;math&gt;B^n&lt;/math&gt;, &lt;math&gt;D^n&lt;/math&gt; and &lt;math&gt;\mathbb{R}^n&lt;/math&gt; are all contractible. For instance, &lt;math&gt;\mathbb{R}^n&lt;/math&gt; is contractible via the function &lt;math&gt;H:\mathbb{R}^n\times I\rightarrow\mathbb{R}^n&lt;/math&gt; given by H(x,t) = (1-t)x. <br /> <br /> <br /> ''Example 3''<br /> <br /> &lt;math&gt;S^{\infty}&lt;/math&gt; [[0708-1300/the unit sphere in a Hilbert space is contractible|the unit sphere in a Hilbert space is contractible]]<br /> <br /> ''Example 4''<br /> <br /> &lt;math&gt;S^n&lt;/math&gt; is NOT contractible for all n (as we shall prove)<br /> <br /> <br /> ''Example 5'' <br /> <br /> Consider the Cantor Comb consisting of the subset of &lt;math&gt;\mathbb{R}^2&lt;/math&gt; consisting of the unit interval along the x axis with a spike of height 1 going up perpendicularly to the x axis at the location 1/n for n = 0,1,2,...<br /> <br /> This set is contractible to (0,0) via the mapping that shrinks all the spikes on the comb to the real axis in time t=1/2 and then shrinks the interval to the point (0,0) by time t=1. <br /> <br /> <br /> However, this set is NOT contrabible to (0,1). This is because should we try to flatten to the real axis points on the teeth of the comb within an epsilon neighborhood of (1,0) they would pull (1,0) down to the real axis as well otherwise continuity would be broken. Hence the third requirement of a contraction is broken. <br /> <br /> <br /> '''Proposition'''<br /> <br /> ''&lt;math&gt;S^n&lt;/math&gt; is not contractible''<br /> <br /> Assume not, thus it is contractible with ''contraction'' H(x,t). <br /> <br /> Consider &lt;math&gt;D^{n+1}&lt;/math&gt; and consider the spherical shell of diameter less than or equal to 1 inside &lt;math&gt;D^{n+1}&lt;/math&gt; where each shell comes to rest tangentially to a point &lt;math&gt;x_0&lt;/math&gt; on &lt;math&gt;S^n&lt;/math&gt;. This is like the idea of making a new spherical shell for each such diameter inside of &lt;math&gt;D^{n+1}&lt;/math&gt; and letting all the shells &quot;fall&quot; to the bottom. <br /> <br /> We now define a retract r on &lt;math&gt;D^{n+1}&lt;/math&gt; by associating H(-,t) with the spherical shell of diameter t. I.e. the outside shell of &lt;math&gt;D^{n+1}, S^{n}&lt;/math&gt; is associated with H(x,0). <br /> <br /> Hence &lt;math&gt;r|_{\partial D^{n+1}} = Id|_{S^n}&lt;/math&gt;. R is clearly continuous and is a retract. But we proved last class that such retracts are impossible and this establishes the contradiction. <br /> <br /> ''Q.E.D''</div> Franklin http://drorbn.net/index.php?title=0708-1300/Class_notes_for_Thursday,_November_1 0708-1300/Class notes for Thursday, November 1 2007-11-02T15:17:24Z <p>Franklin: </p> <hr /> <div>{{0708-1300/Navigation}}<br /> ==Today's Agenda==<br /> * [[0708-1300/Homework Assignment 4|HW4]] and [[0708-1300/Term Exam 1|TE1]].<br /> * Continue with [[0708-1300/Class notes for Tuesday, October 30|Tuesday's]] agenda:<br /> ** Debt on proper functions.<br /> ** Prove that &quot;the sphere is not contractible&quot;.<br /> ** Complete the proof of the &quot;tubular neighborhood theorem&quot;.<br /> ===Proper Implies Closed===<br /> '''Theorem.''' A proper function &lt;math&gt;f:X\to Y&lt;/math&gt; from a topological space &lt;math&gt;X&lt;/math&gt; to a locally compact (Hausdorff) topological space &lt;math&gt;Y&lt;/math&gt; is closed.<br /> <br /> '''Proof.''' Let &lt;math&gt;B&lt;/math&gt; be closed in &lt;math&gt;X&lt;/math&gt;, we need to show that &lt;math&gt;f(B)&lt;/math&gt; is closed in &lt;math&gt;Y&lt;/math&gt;. Since closedness is a local property, it is enough to show that every point &lt;math&gt;y\in Y&lt;/math&gt; has a neighbourhood &lt;math&gt;U&lt;/math&gt; such that &lt;math&gt;f(B)\cap U&lt;/math&gt; is closed in &lt;math&gt;U&lt;/math&gt;. Fix &lt;math&gt;y\in Y&lt;/math&gt;, and by local compactness, choose a neighbourhood &lt;math&gt;U&lt;/math&gt; of &lt;math&gt;y&lt;/math&gt; whose close &lt;math&gt;\bar U&lt;/math&gt; is compact. Then<br /> {{Equation*|&lt;math&gt;f(B)\cap U=f(B\cap f^{-1}(U))\cap U\subset f(B\cap f^{-1}(\bar U))\cap U\subset f(B)\cap U&lt;/math&gt;,}}<br /> so that &lt;math&gt;f(B)\cap U=f(B\cap f^{-1}(\bar U))\cap U&lt;/math&gt;. But &lt;math&gt;\bar U&lt;/math&gt; is compact by choice, so &lt;math&gt;f^{-1}(\bar U)&lt;/math&gt; is compact as &lt;math&gt;f&lt;/math&gt; is proper, so &lt;math&gt;B\cap f^{-1}(\bar U)&lt;/math&gt; is compact as &lt;math&gt;B&lt;/math&gt; is closed, so &lt;math&gt;f(B\cap f^{-1}(\bar U))&lt;/math&gt; is compact (and hence closed) as a continuous image of a compact set, so &lt;math&gt;f(B)\cap U&lt;/math&gt; is the intersection &lt;math&gt;f(B\cap f^{-1}(\bar U))\cap U&lt;/math&gt; of a closed set with &lt;math&gt;U&lt;/math&gt;, hence it is closed in &lt;math&gt;U&lt;/math&gt;.<br /> <br /> ===Note===<br /> The example of a non-contractible &quot;comb&quot; seen today is, in fact, &quot;Cantor's comb&quot;. See, for example, page 25 of www.karlin.mff.cuni.cz/~pyrih/e/e2000v0/c/ect.ps<br /> <br /> <br /> ===Typed Notes===<br /> <br /> '''Definition 1'''<br /> <br /> X is contractible to &lt;math&gt;x_0&lt;/math&gt; if there exists a continuous function &lt;math&gt;H:X\times I\rightarrow X&lt;/math&gt;<br /> where<br /> <br /> 1) &lt;math&gt;H(x,0) = x&lt;/math&gt; &lt;math&gt;\forall x&lt;/math&gt;<br /> <br /> 2) &lt;math&gt;H(x,1) = x_0&lt;/math&gt; &lt;math&gt;\forall x&lt;/math&gt;<br /> <br /> 3) &lt;math&gt;H(x_0,t)=x_0&lt;/math&gt; &lt;math&gt;\forall t&lt;/math&gt;<br /> <br /> <br /> ''Example 1''<br /> <br /> The singleton &lt;math&gt;\{x_0\}&lt;/math&gt; is contractible<br /> <br /> <br /> ''Example 2''<br /> <br /> &lt;math&gt;B^n&lt;/math&gt;, &lt;math&gt;D^n&lt;/math&gt; and &lt;math&gt;\mathbb{R}^n&lt;/math&gt; are all contractible. For instance, &lt;math&gt;\mathbb{R}^n&lt;/math&gt; is contractible via the function &lt;math&gt;H:\mathbb{R}^n\times I\rightarrow\mathbb{R}^n&lt;/math&gt; given by H(x,t) = (1-t)x. <br /> <br /> <br /> ''Example 3''<br /> <br /> &lt;math&gt;S^{\infty}&lt;/math&gt; [[0708-1300/the unit sphere in a Hilbert space is contractible<br /> the unit sphere in a Hilbert space is contractible]]<br /> <br /> ''Example 4''<br /> <br /> &lt;math&gt;S^n&lt;/math&gt; is NOT contractible for all n (as we shall prove)<br /> <br /> <br /> ''Example 5'' <br /> <br /> Consider the Cantor Comb consisting of the subset of &lt;math&gt;\mathbb{R}^2&lt;/math&gt; consisting of the unit interval along the x axis with a spike of height 1 going up perpendicularly to the x axis at the location 1/n for n = 0,1,2,...<br /> <br /> This set is contractible to (0,0) via the mapping that shrinks all the spikes on the comb to the real axis in time t=1/2 and then shrinks the interval to the point (0,0) by time t=1. <br /> <br /> <br /> However, this set is NOT contrabible to (0,1). This is because should we try to flatten to the real axis points on the teeth of the comb within an epsilon neighborhood of (1,0) they would pull (1,0) down to the real axis as well otherwise continuity would be broken. Hence the third requirement of a contraction is broken. <br /> <br /> <br /> '''Proposition'''<br /> <br /> ''&lt;math&gt;S^n&lt;/math&gt; is not contractible''<br /> <br /> Assume not, thus it is contractible with ''contraction'' H(x,t). <br /> <br /> Consider &lt;math&gt;D^{n+1}&lt;/math&gt; and consider the spherical shell of diameter less than or equal to 1 inside &lt;math&gt;D^{n+1}&lt;/math&gt; where each shell comes to rest tangentially to a point &lt;math&gt;x_0&lt;/math&gt; on &lt;math&gt;S^n&lt;/math&gt;. This is like the idea of making a new spherical shell for each such diameter inside of &lt;math&gt;D^{n+1}&lt;/math&gt; and letting all the shells &quot;fall&quot; to the bottom. <br /> <br /> We now define a retract r on &lt;math&gt;D^{n+1}&lt;/math&gt; by associating H(-,t) with the spherical shell of diameter t. I.e. the outside shell of &lt;math&gt;D^{n+1}, S^{n}&lt;/math&gt; is associated with H(x,0). <br /> <br /> Hence &lt;math&gt;r|_{\partial D^{n+1}} = Id|_{S^n}&lt;/math&gt;. R is clearly continuous and is a retract. But we proved last class that such retracts are impossible and this establishes the contradiction. <br /> <br /> ''Q.E.D''</div> Franklin http://drorbn.net/index.php?title=0708-1300/the_unit_sphere_in_a_Hilbert_space_is_contractible 0708-1300/the unit sphere in a Hilbert space is contractible 2007-11-02T15:13:33Z <p>Franklin: </p> <hr /> <div>Let &lt;math&gt;H=\{(x_1,x_2,...)| \sum x_n^2&lt;\infty\}&lt;/math&gt; and define &lt;math&gt;S^{\infty}=\{x\in H| ||x||=1\}&lt;/math&gt;<br /> <br /> '''Claim'''<br /> <br /> &lt;math&gt;S^{\infty}&lt;/math&gt; is contractible <br /> <br /> '''Proof'''<br /> <br /> Suppose &lt;math&gt;x=(x_1,x_2,...)\in S^{\infty}&lt;/math&gt; then &lt;math&gt;\sum x_n^2=1&lt;/math&gt; <br /> <br /> Define &lt;math&gt;F_1:S^{\infty}\times I\rightarrow S^{\infty}&lt;/math&gt; by &lt;math&gt;F_1(x,t)=(sgn(x_1)((1-t)|x_1|+t\sqrt{x_1^2+x_2^2}),(1-t)x_2,x_3,x_4,...)/||(sgn(x_1)((1-t)|x_1|+t\sqrt{x_1^2+x_2^2}),(1-t)x_2,x_3,x_4,...)||&lt;/math&gt;<br /> <br /> &lt;math&gt;F_2:S^{\infty}\times I\rightarrow S^{\infty}&lt;/math&gt; by &lt;math&gt;F_2(x,t)=(sgn(x_1)((1-t)|x_1|+t\sqrt{x_1^2+x_3^2}),0,(1-t)x_3,x_4,x_5,...)/||(sgn(x_1)((1-t)|x_1|+t\sqrt{x_1^2+x_3^2}),0,(1-t)x_3,x_4,x_5,...)||&lt;/math&gt;<br /> <br /> &lt;math&gt;F_3:S^{\infty}\times I\rightarrow S^{\infty}&lt;/math&gt; by &lt;math&gt;F_3(x,t)=(sgn(x_1)((1-t)|x_1|+t\sqrt{x_1^2+x_4^2}),0,0,(1-t)x_4,x_5,x_6,...)/||(sgn(x_1)((1-t)|x_1|+t\sqrt{x_1^2+x_4^2}),0,0,(1-t)x_4,x_5,x_6,...)||&lt;/math&gt;<br /> <br /> and so on ...<br /> <br /> applying the homotopy &lt;math&gt;F_1&lt;/math&gt; in the time interval &lt;math&gt;[0,1/2]&lt;/math&gt;, &lt;math&gt;F_2&lt;/math&gt; in the interval &lt;math&gt;[1/2,3/4]&lt;/math&gt;, &lt;math&gt;F_3&lt;/math&gt; in &lt;math&gt;[3/4,5/6]&lt;/math&gt; etc...<br /> <br /> we get the desired contraction to the point &lt;math&gt;(1,0,0,...)&lt;/math&gt;.<br /> <br /> <br /> A different way to see this is via the cellular structure of &lt;math&gt;S^{\infty}&lt;/math&gt;. If &lt;math&gt;S^{\infty}=C_0 C_1 ...&lt;/math&gt; you can always contract &lt;math&gt;C_k&lt;/math&gt; along &lt;math&gt;C_{k+1}&lt;/math&gt; like moving contracting the equator along the surface of the earth.<br /> <br /> This proof only works in '''separable''' Hilbert spaces. Is the unit ball in a non-separable Hilbert space contractible?</div> Franklin http://drorbn.net/index.php?title=0708-1300/the_unit_sphere_in_a_Hilbert_space_is_contractible 0708-1300/the unit sphere in a Hilbert space is contractible 2007-11-02T15:05:50Z <p>Franklin: </p> <hr /> <div>Let &lt;math&gt;H=\{(x_1,x_2,...)| \sum x_n^2&lt;\infty\}&lt;/math&gt; and define &lt;math&gt;S^{\infty}=\{x\in H| ||x||=1\}&lt;/math&gt;<br /> <br /> '''Claim'''<br /> <br /> &lt;math&gt;S^{\infty}&lt;/math&gt; is contractible <br /> <br /> '''Proof'''<br /> <br /> Suppose &lt;math&gt;x=(x_1,x_2,...)\in S^{\infty}&lt;/math&gt; then &lt;math&gt;\sum x_n^2=1&lt;/math&gt; <br /> <br /> Define &lt;math&gt;F_1:S^{\infty}\times I\rightarrow S^{\infty}&lt;/math&gt; by &lt;math&gt;F_1(x,t)=(sgn(x_1)((1-t)|x_1|+t\sqrt{x_1^2+x_2^2}),(1-t)x_2,x_3,x_4,...)/||(sgn(x_1)((1-t)|x_1|+t\sqrt{x_1^2+x_2^2}),(1-t)x_2,x_3,x_4,...)||&lt;/math&gt;<br /> <br /> &lt;math&gt;F_2:S^{\infty}\times I\rightarrow S^{\infty}&lt;/math&gt; by &lt;math&gt;F_2(x,t)=(sgn(x_1)((1-t)|x_1|+t\sqrt{x_1^2+x_3^2}),0,(1-t)x_3,x_4,x_5,...)/||(sgn(x_1)((1-t)|x_1|+t\sqrt{x_1^2+x_3^2}),0,(1-t)x_3,x_4,x_5,...)||&lt;/math&gt;<br /> <br /> &lt;math&gt;F_3:S^{\infty}\times I\rightarrow S^{\infty}&lt;/math&gt; by &lt;math&gt;F_3(x,t)=(sgn(x_1)((1-t)|x_1|+t\sqrt{x_1^2+x_4^2}),0,0,(1-t)x_4,x_5,x_6,...)/||(sgn(x_1)((1-t)|x_1|+t\sqrt{x_1^2+x_4^2}),0,0,(1-t)x_4,x_5,x_6,...)||&lt;/math&gt;<br /> <br /> and so on ...<br /> <br /> applying the homotopy &lt;math&gt;F_1&lt;/math&gt; in the time interval &lt;math&gt;[0,1/2]&lt;/math&gt;, &lt;math&gt;F_2&lt;/math&gt; in the interval &lt;math&gt;[1/2,3/4]&lt;/math&gt;, &lt;math&gt;F_3&lt;/math&gt; in &lt;math&gt;[3/4,5/6]&lt;/math&gt; etc...<br /> <br /> we get the desired contraction to the point &lt;math&gt;(1,0,0,...)&lt;/math&gt;.<br /> <br /> <br /> A different way to see this is via the cellular structure of &lt;math&gt;S^{\infty}&lt;/math&gt;. If &lt;math&gt;S^{\infty}=C_0 C_1 ...&lt;/math&gt; you can always contract &lt;math&gt;C_k&lt;/math&gt; along &lt;math&gt;C_{k+1}&lt;/math&gt; like moving contracting the equator along the surface of the earth.</div> Franklin http://drorbn.net/index.php?title=0708-1300/the_unit_sphere_in_a_Hilbert_space_is_contractible 0708-1300/the unit sphere in a Hilbert space is contractible 2007-11-02T15:05:31Z <p>Franklin: </p> <hr /> <div>Let &lt;math&gt;H=\{(x_1,x_2,...)| \sum x_n^2&lt;\infty\}&lt;/math&gt; and define &lt;math&gt;S^{\infty}=\{x\in H| ||x||=1\}&lt;/math&gt;<br /> <br /> '''Claim'''<br /> <br /> &lt;math&gt;S^{\infty}&lt;/math&gt; is contractible <br /> <br /> '''Proof'''<br /> <br /> Suppose &lt;math&gt;x=(x_1,x_2,...)\in S^{\infty}&lt;/math&gt; then &lt;math&gt;\sum x_n^2=1&lt;/math&gt; <br /> <br /> Define &lt;math&gt;F_1:S^{\infty}\times I\rightarrow S^{\infty}&lt;/math&gt; by &lt;math&gt;F_1(x,t)=(sgn(x_1)((1-t)|x_1|+t\sqrt{x_1^2+x_2^2}),(1-t)x_2,x_3,x_4,...)/||(sgn(x_1)((1-t)|x_1|+t\sqrt{x_1^2+x_2^2}),(1-t)x_2,x_3,x_4,...)||&lt;/math&gt;<br /> <br /> &lt;math&gt;F_2:S^{\infty}\times I\rightarrow S^{\infty}&lt;/math&gt; by &lt;math&gt;F_2(x,t)=(sgn(x_1)((1-t)|x_1|+t\sqrt{x_1^2+x_3^2}),0,(1-t)x_3,x_4,x_5,...)/||(sgn(x_1)((1-t)|x_1|+t\sqrt{x_1^2+x_3^2}),0,(1-t)x_3,x_4,x_5,...)||&lt;/math&gt;<br /> <br /> &lt;math&gt;F_3:S^{\infty}\times I\rightarrow S^{\infty}&lt;/math&gt; by &lt;math&gt;F_3(x,t)=(sgn(x_1)((1-t)|x_1|+t\sqrt{x_1^2+x_4^2}),0,0,(1-t)x_4,x_5,x_6,...)/||(sgn(x_1)((1-t)|x_1|+t\sqrt{x_1^2+x_4^2}),0,0,(1-t)x_4,x_5,x_6,...)||&lt;/math&gt;<br /> <br /> and so on ...<br /> <br /> applying the homotopy &lt;math&gt;F_1&lt;/math&gt; in the time interval &lt;math&gt;[0,1/2]&lt;/math&gt;, &lt;math&gt;F_2&lt;/math&gt; in the interval &lt;math&gt;[1/2,3/4]&lt;/math&gt;, &lt;math&gt;F_3&lt;/math&gt; in &lt;math&gt;[3/4,5/6]&lt;/math&gt; etc...<br /> <br /> we get the desired contraction to the point &lt;math&gt;(1,0,0,...)&lt;/math&gt;.<br /> <br /> <br /> A different way to see this is via the cellular structure of &lt;math&gt;S^{\infty}&lt;/math&gt;. If &lt;math&gt;S^{\infy}=C_0 C_1 ...&lt;/math&gt; you can always contract &lt;math&gt;C_k&lt;/math&gt; along &lt;math&gt;C_{k+1}&lt;/math&gt; like moving contracting the equator along the surface of the earth.</div> Franklin http://drorbn.net/index.php?title=0708-1300/the_unit_sphere_in_a_Hilbert_space_is_contractible 0708-1300/the unit sphere in a Hilbert space is contractible 2007-11-02T15:04:49Z <p>Franklin: </p> <hr /> <div>Let &lt;math&gt;H=\{(x_1,x_2,...)| \sum x_n^2&lt;\infty\}&lt;/math&gt; and define &lt;math&gt;S^{\infty}=\{x\in H| ||x||=1\}&lt;/math&gt;<br /> <br /> '''Claim'''<br /> <br /> &lt;math&gt;S^{\infty}&lt;/math&gt; is contractible <br /> <br /> '''Proof'''<br /> <br /> Suppose &lt;math&gt;x=(x_1,x_2,...)\in S^{\infty}&lt;/math&gt; then &lt;math&gt;\sum x_n^2=1&lt;/math&gt; <br /> <br /> Define &lt;math&gt;F_1:S^{\infty}\times I\rightarrow S^{\infty}&lt;/math&gt; by &lt;math&gt;F_1(x,t)=(sgn(x_1)((1-t)|x_1|+t\sqrt{x_1^2+x_2^2}),(1-t)x_2,x_3,x_4,...)/||(sgn(x_1)((1-t)|x_1|+t\sqrt{x_1^2+x_2^2}),(1-t)x_2,x_3,x_4,...)||&lt;/math&gt;<br /> <br /> &lt;math&gt;F_2:S^{\infty}\times I\rightarrow S^{\infty}&lt;/math&gt; by &lt;math&gt;F_2(x,t)=(sgn(x_1)((1-t)|x_1|+t\sqrt{x_1^2+x_3^2}),0,(1-t)x_3,x_4,x_5,...)/||(sgn(x_1)((1-t)|x_1|+t\sqrt{x_1^2+x_3^2}),0,(1-t)x_3,x_4,x_5,...)||&lt;/math&gt;<br /> <br /> &lt;math&gt;F_3:S^{\infty}\times I\rightarrow S^{\infty}&lt;/math&gt; by &lt;math&gt;F_3(x,t)=(sgn(x_1)((1-t)|x_1|+t\sqrt{x_1^2+x_4^2}),0,0,(1-t)x_4,x_5,x_6,...)/||(sgn(x_1)((1-t)|x_1|+t\sqrt{x_1^2+x_4^2}),0,0,(1-t)x_4,x_5,x_6,...)||&lt;/math&gt;<br /> <br /> and so on ...<br /> <br /> applying the homotopy &lt;math&gt;F_1&lt;/math&gt; in the time interval &lt;math&gt;[0,1/2]&lt;/math&gt;, &lt;math&gt;F_2&lt;/math&gt; in the interval &lt;math&gt;[1/2,3/4]&lt;/math&gt;, &lt;math&gt;F_3&lt;/math&gt; in &lt;math&gt;[3/4,5/6]&lt;/math&gt; etc...<br /> <br /> we get the desired contraction to the point &lt;math&gt;(1,0,0,...)&lt;/math&gt;.<br /> <br /> <br /> A different way to see this is via the cellular structure of &lt;math&gt;S^{\infty}&lt;/math&gt;. If &lt;math&gt;S^{\infy}=C_0C_1...&lt;/math&gt; you can always contract &lt;math&gt;C_k&lt;/math&gt; along &lt;math&gt;C_{k+1}&lt;/math&gt; like moving contracting the equator along the surface of the earth.</div> Franklin http://drorbn.net/index.php?title=0708-1300/the_unit_sphere_in_a_Hilbert_space_is_contractible 0708-1300/the unit sphere in a Hilbert space is contractible 2007-11-02T14:50:58Z <p>Franklin: </p> <hr /> <div>Let &lt;math&gt;H=\{(x_1,x_2,...)| \sum x_n^2&lt;\infty\}&lt;/math&gt; and define &lt;math&gt;S^{\infty}=\{x\in H| ||x||=1\}&lt;/math&gt;<br /> <br /> '''Claim'''<br /> <br /> &lt;math&gt;S^{\infty}&lt;/math&gt; is contractible <br /> <br /> '''Proof'''<br /> <br /> Suppose &lt;math&gt;x=(x_1,x_2,...)\in S^{\infty}&lt;/math&gt; then &lt;math&gt;\sum x_n^2=1&lt;/math&gt; <br /> <br /> Define &lt;math&gt;F_1:S^{\infty}\times I\rightarrow S^{\infty}&lt;/math&gt; by &lt;math&gt;F_1(x,t)=(sgn(x_1)((1-t)|x_1|+t\sqrt{x_1^2+x_2^2}),(1-t)x_2,x_3,x_4,...)/||(sgn(x_1)((1-t)|x_1|+t\sqrt{x_1^2+x_2^2}),(1-t)x_2,x_3,x_4,...)||&lt;/math&gt;<br /> <br /> &lt;math&gt;F_2:S^{\infty}\times I\rightarrow S^{\infty}&lt;/math&gt; by &lt;math&gt;F_2(x,t)=(sgn(x_1)((1-t)|x_1|+t\sqrt{x_1^2+x_3^2}),0,(1-t)x_3,x_4,x_5,...)/||(sgn(x_1)((1-t)|x_1|+t\sqrt{x_1^2+x_3^2}),0,(1-t)x_3,x_4,x_5,...)||&lt;/math&gt;<br /> <br /> &lt;math&gt;F_3:S^{\infty}\times I\rightarrow S^{\infty}&lt;/math&gt; by &lt;math&gt;F_3(x,t)=(sgn(x_1)((1-t)|x_1|+t\sqrt{x_1^2+x_4^2}),0,0,(1-t)x_4,x_5,x_6,...)/||(sgn(x_1)((1-t)|x_1|+t\sqrt{x_1^2+x_4^2}),0,0,(1-t)x_4,x_5,x_6,...)||&lt;/math&gt;<br /> <br /> and so on ...<br /> <br /> applying the homotopy &lt;math&gt;F_1&lt;/math&gt; in the time interval &lt;math&gt;[0,1/2]&lt;/math&gt;, &lt;math&gt;F_2&lt;/math&gt; in the interval &lt;math&gt;[1/2,3/4]&lt;/math&gt;, &lt;math&gt;F_3&lt;/math&gt; in &lt;math&gt;[3/4,5/6]&lt;/math&gt; etc...<br /> <br /> we get the desired contraction to the point &lt;math&gt;(1,0,0,...)&lt;/math&gt;.</div> Franklin http://drorbn.net/index.php?title=0708-1300/the_unit_sphere_in_a_Hilbert_space_is_contractible 0708-1300/the unit sphere in a Hilbert space is contractible 2007-11-02T14:46:18Z <p>Franklin: </p> <hr /> <div>Let &lt;math&gt;H=\{(x_1,x_2,...)| \sum x_n^2&lt;\infty\}&lt;/math&gt; and define &lt;math&gt;S^{\infty}=\{x\in H| ||x||=1\}&lt;/math&gt;<br /> <br /> '''Claim'''<br /> <br /> &lt;math&gt;S^{\infty}&lt;/math&gt; is contractible <br /> <br /> '''Proof'''<br /> <br /> Suppose &lt;math&gt;x=(x_1,x_2,...)\in S^{\infty}&lt;/math&gt; then &lt;math&gt;\sum x_n^2=1&lt;/math&gt; <br /> <br /> Define &lt;math&gt;F_1:S^{\infty}\times I\rightarrow S^{\infty}&lt;/math&gt; by &lt;math&gt;F_1(x,t)=((1-t)x_1+t\sqrt{x_1^2+x_2^2},(1-t)x_2,x_3,x_4,...)/||((1-t)x_1+t\sqrt{x_1^2+x_2^2},(1-t)x_2,x_3,x_4,...)||&lt;/math&gt;<br /> <br /> &lt;math&gt;F_2:S^{\infty}\times I\rightarrow S^{\infty}&lt;/math&gt; by &lt;math&gt;F_2(x,t)=((1-t)x_1+t\sqrt{x_1^2+x_3^2},0,(1-t)x_3,x_4,x_5,...)/||((1-t)x_1+t\sqrt{x_1^2+x_3^2},0,(1-t)x_3,x_4,x_5,...)||&lt;/math&gt;<br /> <br /> &lt;math&gt;F_3:S^{\infty}\times I\rightarrow S^{\infty}&lt;/math&gt; by &lt;math&gt;F_3(x,t)=((1-t)x_1+t\sqrt{x_1^2+x_4^2},0,0,(1-t)x_4,x_5,x_6,...)/||((1-t)x_1+t\sqrt{x_1^2+x_4^2},0,0,(1-t)x_4,x_5,x_6,...)||&lt;/math&gt;<br /> <br /> and so on ...<br /> <br /> applying the homotopy &lt;math&gt;F_1&lt;/math&gt; in the time interval &lt;math&gt;[0,1/2]&lt;/math&gt;, &lt;math&gt;F_2&lt;/math&gt; in the interval &lt;math&gt;[1/2,3/4]&lt;/math&gt;, &lt;math&gt;F_3&lt;/math&gt; in &lt;math&gt;[3/4,5/6]&lt;/math&gt; etc...<br /> <br /> we get the desired contraction to the point &lt;math&gt;(1,0,0,...)&lt;/math&gt;.</div> Franklin http://drorbn.net/index.php?title=0708-1300/the_unit_sphere_in_a_Hilbert_space_is_contractible 0708-1300/the unit sphere in a Hilbert space is contractible 2007-11-02T14:40:02Z <p>Franklin: </p> <hr /> <div>Let &lt;math&gt;H=\{(x_1,x_2,...)| \sum x_n^2&lt;\infty\}&lt;/math&gt; and define &lt;math&gt;S^{\infty}=\{x\in H| ||x||=1\}&lt;/math&gt;<br /> <br /> '''Claim'''<br /> <br /> &lt;math&gt;S^{\infty}&lt;/math&gt; is contractible <br /> <br /> '''Proof'''<br /> <br /> Suppose &lt;math&gt;x=(x_1,x_2,...)\in S^{\infty}&lt;/math&gt; then &lt;math&gt;\sum x_n^2=1&lt;/math&gt; <br /> <br /> Define &lt;math&gt;F:S^{\infty}\times I\rightarrow S^{\infty}&lt;/math&gt; by &lt;math&gt;F(x,t)=((1-t)x_1+t\sqrt{x_1^2+x_2^2},(1-t)x_2,x_3,x_4,...)/||((1-t)x_1+t\sqrt{x_1^2+x_2^2},(1-t)x_2,x_3,x_4,...)||&lt;/math&gt;</div> Franklin