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NOTE: This page is used as a placeholder for incomplete typed class notes. |
NOTE: This page is used as a placeholder for incomplete typed class notes. |
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{{0708-1300/Navigation}} |
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==A Homology Theory is a Monster== |
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[[Image:0708-1300-AxiomsForHomology.png|thumb|center|540px|Page 183 of Bredon's book]] |
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'''Bredon's Plan of Attack:''' State all, apply all, prove all. |
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'''Our Route:''' Axiom by axiom - state, apply, prove. Thus everything we will do will be, or should be, labeled either "'''S'''tate" or "'''P'''rove" or "'''A'''pply". |
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==Typed Notes== |
==Typed Notes== |
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<span style="color: red;">The notes below are by the students and for the students. Hopefully they are useful, but they come with no guarantee of any kind.</span> |
<span style="color: red;">The notes below are by the students and for the students. Hopefully they are useful, but they come with no guarantee of any kind.</span> |
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===First Hour=== |
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We start with a one further comment on Van-Kampen: |
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Let <math>X = \bigcup_{\alpha\in A}U_{\alpha}</math>, with <math>U_{\alpha}</math> open and with <math>\forall \alpha,\beta</math> then <math>U_{\alpha}\cap U_{\beta}</math> connected. |
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Recall we had defined for a chain complex the associated homology groups: <math>H_p(C_*) :=\ ker\partial_p/im\partial_{p+1}</math> |
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Then, <math>\pi_1(X) = *\pi_1(U_{\alpha})/(\gamma\in\pi_1(U_{\alpha}\cap U_{\beta}),\ and\ j_{\alpha\beta*}(\gamma) = j_{\beta\alpha*}(\gamma))</math> |
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From this we get the pth homology for a topological space <math>H_p(X)</math> |
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We actually also need <math>\forall\alpha,\beta,\gamma,\delta</math> then <math>U_{\alpha}\cap U_{\beta}\cap U_{\gamma}\cap U_{\delta}</math> to be connected. Recall we previously had a square grid in our homotopy, so needed the four sections around a grid point to intersect in a connected way. |
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We have previously shown that |
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However, we can employ a clever trick to reduce this to only needing triples of <math>U_{\alpha_i}</math> to be connected. That is, instead of covering it was squares on top of each other, cover it with rectangles arranged in the way bricks are laid in a wall. |
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''1) <math>H_p(\cup X) = \oplus H_p(X_i)</math> for disjoint unions of spaces <math>X_i</math> |
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2) <math>H_p(pt) = \mathbb{Z}\delta_{p0}</math> |
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'''Definition''' |
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<math> |
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p:X\rightarrow B</math> is a "covering" if, for some fixed discrete set F, every <math>b\in B</math> has a neighborhood U such that <math>p^{-1}(U)\cong F\times U</math>. We have the natural map <math>\pi:F\times U\rightarrow U</math> |
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3) <math>H_0(connected) = \mathbb{Z}</math> |
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4) <math>H_1(connected) \cong \pi_1^{ab}(X)</math> via the map |
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Now a naive dream would be to classify all such covering spaces. |
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<math>\phi:[\gamma]_{\pi_1^{ab}}\mapsto[\gamma]_{H_1}</math> |
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<math>\psi:\sigma\in C_1\mapsto[\gamma_{\sigma(0)}\sigma\bar{\gamma_{\sigma(1)}}]</math> where <math>\sigma_x</math> is a path connecting <math>x_0</math> to x. |
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'''Example 1''' |
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The double annulus depicted above is a covering space for the single annulus |
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We need to check the maps are in fact inverses of each other. |
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Lets consider <math>\psi\circ\phi</math>. We start with a closed path starting at <math>x_0</math> (thought of as in the fundamental group). <math>\phi</math> means we now think of it as a simplex in X with a point at <math>x_0</math>. <math>\Psi</math> now takes this to the path that parks at <math>x_0</math> for a third of the time, goes around the loop and then parks for the remaining third of the time. Clearly this is homotopic this composition is homotopic to the identity. |
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'''Example 2''' |
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We now consider <math>\phi\circ\psi</math>. Start with just a path <math>\sigma</math>. Then <math>\psi</math> makes a loop adding two paths from the <math>x_0</math> to the start and finish of <math>\sigma</math> forming a triangular like closed loop. We think of this loop as <math>\sigma'\in C_1</math> |
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Now, we start from c being <math>c = \sum a_i\sigma_i</math> with <math>\partial c = 0</math>. So get <math>\sum a_i(\partial \sigma_1) = \sum a_i(\sigma_i(1)-\sigma_i(0))</math> |
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So <math>\Psi(c) = [\gamma_{\sigma_i(0)}\sigma_i\bar{\gamma_{\sigma_i(1)}}]_{\pi_1}</math> which maps to, under <math>\phi</math>, <math>\sum a_i(\gamma_{\sigma_i(0)} + \sigma_i - \gamma_{\sigma_i(1)}) = \sum a_i\sigma_i = c</math> ( in the homology gamma's cancel as <math>\partial c = 0</math>) |
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''Axiomized Homology'' |
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We now will move to an approach where we prove that our defined homology satisfies a series of established homology axioms that will allow us to apply the machinery of general homology to our specific "singular" homology defined via simplices. |
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Axiom 0) Homology if a function |
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'''Definition''' The "category of chain complexes" is a category whose objects are chain complexes (of abelian groups) and morphisms which is a homomorphism between each abelian group in one chain and the corresponding group in the other chain such that the resulting diagram commutes. I.e, <math>Mor((C_p)_{p=0}^{\infty}, (D_p)^{\infty}_{p=0}) = \{(f_p:C_p\rightarrow D_p)_{p=0}^{\infty}\ |\ f_{p-1}\partial_p^C=\partial_p^D f_p\}</math> |
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Now, in our case, the chain complexes do in fact commute because <math>\partial</math> is defined by pre-composition but f is defined by post-composition. Hence, associativity of composition yields commutativity. |
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'''Claim''' |
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Homology of chain complexes is a functor in the natural way. That is, if <math>f_p:C_p\rightarrow D_p</math> for each p induces the functor <math>f_*:H_p(C_*)\rightarrow H_p(D_*)</math> |
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The proof is by "diagram chasing". Well, let <math>c\in C_p</math>, <math>\partial c =0</math>. |
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Let <math>f*[c] = [fc]</math>. Now <math>\partial fc = f\partial c = 0</math>. Furthermore, suppose <math>c = \partial b</math>. Then, <math>f_*c = fc = \partial fb</math> so therefore <math>fc = \partial\beta</math> some <math>\beta</math>. This shows <math>f_*</math> is well defined. |
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Thus, for <math>c\in H_p(C_*)</math> get <math>fc\in H_p(D_*)</math> via the well defined functor <math>f_*</math>. |
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===Second Hour=== |
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'''1) Homotopy Axioms''' |
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If f,g:X\rightarrow Y are homotopic then g_* = g_*: H_p(X)\rightarrow H_p(Y) |
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Applications: If X and Y are homotopy equivalent then H_*(X) \cong H_*(Y) |
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''Proof:'' |
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let f:X\rightarrow Y, g:Y\rightarrow X such that f\circ g\sim id_y and g\circ f \sim id_x. Well f_*\circ g_* = id_{H(Y)} and g_*\circ f_* = id_{H(X)} |
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Hence, f_* and g_* are invertible maps of each other. ''Q.E.D'' |
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'''Definition''' |
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Two morphisms f,g:C_*\rightarrow D_* between chain complexes are ''homotopic'' if you can find maps h_p:C_p\rightarrow D_{p+1} such that f_p - g_p = \partial_{p+1} h_p = h_{p-1}\partial_p |
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Consider a series of rectangular planes stacked on top of each other. Make a small cut in each plane, and then glue one side of the cut to the opposite side in the plane above. Finally glue the remaining unglued sides in the very top and bottom planes to each other. |
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Analogously this can be done in higher dimensions. |
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'''Claim 1''' |
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Given H a homotopy connecting f,g:X\rightarrow Y we can construct a chain homotopy between f_*,g_*:C_*(X)\rightarrow C_*(Y) |
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'''Theorem''' |
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For a "decent" B (a topological condition to be discussed later) there is a bijection between connected coverings of B and subgroups of <math>\pi_1(B)</math> |
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'''Claim 2''' |
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given by <math>x\mapsto p_*\pi_1(X)</math> included in <math>\pi_1(B)</math> |
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If f,g:C_*\rightarrow C_* are chain homotopic then they induce equal maps on homology. |
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'''Example 3''' |
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Consider <math>\mathbb{R}P^3</math> |
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''Proof of 2'' |
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Now, <math>\pi_1(\mathbb{R}P^3) \cong\pi_1(\mathbb{R}P^3 - \{pt\})\cong \pi_1</math>(equator with identification of antipodes)<math>\cong\mathbb{Z}_2</math> as we have previously computed. |
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Assume [c]\in H_p(C_*), that is, \partial c =0 |
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There are only two subgroups of <math>\mathbb{Z}_2</math>, namely {e} and <math>\mathbb{Z}_2</math> |
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[f_* c] - [g_* c] = [(f_*-g_*)] =[(\partial h + h\partial)c] = 0 (as \partial c = 0 and homology ignores exact forms) |
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Hence, at the level of homology they are the same. |
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Lets consider the case where the subgroup is <math>\mathbb{Z}_2</math>. Then <math>I:\mathbb{R}P^3\rightarrow\mathbb{Z}P^3</math> the identity covering with F merely a point. So <math>p_*\pi_1(X) = \pi(X)</math> |
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''Proof of 1'' |
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For {e}, <math>p:S^3\rightarrow\mathbb{R}P^3</math> is the other covering. |
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Consider a simplex in X. Now consider its image, a simplex, in Y under g and f respectively. Because of the homotopy we can construct a triangular based cylinder in Y with the image under f at the top and the image under y at the bottom. |
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Define h\sigma = the above prism formed by \sigma and the homotopy H. |
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''Aside:'' |
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<math> |
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\mathbb{R}P^3 = SO(3)</math> |
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(f_*-g_*)\sigma = h\partial\sigma = \partial h\sigma |
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Consider a belt. A point on it can be associated with three vectors. One vector is tangent to the belt in the direction of one end. The other vector is normal to the belt. And the third is normal to both of these. |
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This, pictorially is correct but we need to be able to break up the prism into a union of images of simplexes. |
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Now fix the orientation of the end points. Hence, we can think of the belt as a path in SO(3) as a homotopy. Pulling the belt tight is the identity homotopy. |
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Suppose |
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We note that if one twists the orientation of the end 360 degree, then this is non trivial and there is no way to return this to the identity while holding the ends fixed in orientation. However, if you twists the orientation of an end point 720 degrees then in fact you CAN "untwist" the belt without changing the orientation of the endpoints! |
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Revision as of 19:46, 7 March 2008
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NOTE: This page is used as a placeholder for incomplete typed class notes.
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A Homology Theory is a Monster
Bredon's Plan of Attack: State all, apply all, prove all.
Our Route: Axiom by axiom - state, apply, prove. Thus everything we will do will be, or should be, labeled either "State" or "Prove" or "Apply".
Typed Notes
The notes below are by the students and for the students. Hopefully they are useful, but they come with no guarantee of any kind.
First Hour
Recall we had defined for a chain complex the associated homology groups:
From this we get the pth homology for a topological space
We have previously shown that
1) for disjoint unions of spaces
2)
3)
4) via the map
![{\displaystyle \phi :[\gamma ]_{\pi _{1}^{ab}}\mapsto [\gamma ]_{H_{1}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/007f783b8f177daa4eba4fcddacb2d0b4db34426)
where is a path connecting to x.
![{\displaystyle \psi :\sigma \in C_{1}\mapsto [\gamma _{\sigma (0)}\sigma {\bar {\gamma _{\sigma (1)}}}]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/848546805d00e89f0d90fda2d1cadd85c606722c)
We need to check the maps are in fact inverses of each other.
Lets consider . We start with a closed path starting at (thought of as in the fundamental group). means we now think of it as a simplex in X with a point at . now takes this to the path that parks at for a third of the time, goes around the loop and then parks for the remaining third of the time. Clearly this is homotopic this composition is homotopic to the identity.
We now consider . Start with just a path . Then makes a loop adding two paths from the to the start and finish of forming a triangular like closed loop. We think of this loop as
Now, we start from c being with . So get
So which maps to, under , ( in the homology gamma's cancel as )
![{\displaystyle \Psi (c)=[\gamma _{\sigma _{i}(0)}\sigma _{i}{\bar {\gamma _{\sigma _{i}(1)}}}]_{\pi _{1}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/339085df669a0a1f378ef9dcee5797adbf862dab)
Axiomized Homology
We now will move to an approach where we prove that our defined homology satisfies a series of established homology axioms that will allow us to apply the machinery of general homology to our specific "singular" homology defined via simplices.
Axiom 0) Homology if a function
Definition The "category of chain complexes" is a category whose objects are chain complexes (of abelian groups) and morphisms which is a homomorphism between each abelian group in one chain and the corresponding group in the other chain such that the resulting diagram commutes. I.e,
Now, in our case, the chain complexes do in fact commute because is defined by pre-composition but f is defined by post-composition. Hence, associativity of composition yields commutativity.
Claim
Homology of chain complexes is a functor in the natural way. That is, if for each p induces the functor
The proof is by "diagram chasing". Well, let , .
Let . Now . Furthermore, suppose . Then, so therefore some . This shows is well defined.
![{\displaystyle f*[c]=[fc]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/05478b17f5a435754f25ed17f1d883a88b3b37f3)
Thus, for get via the well defined functor .
Second Hour
1) Homotopy Axioms
If f,g:X\rightarrow Y are homotopic then g_* = g_*: H_p(X)\rightarrow H_p(Y)
Applications: If X and Y are homotopy equivalent then H_*(X) \cong H_*(Y)
Proof:
let f:X\rightarrow Y, g:Y\rightarrow X such that f\circ g\sim id_y and g\circ f \sim id_x. Well f_*\circ g_* = id_{H(Y)} and g_*\circ f_* = id_{H(X)}
Hence, f_* and g_* are invertible maps of each other. Q.E.D
Definition
Two morphisms f,g:C_*\rightarrow D_* between chain complexes are homotopic if you can find maps h_p:C_p\rightarrow D_{p+1} such that f_p - g_p = \partial_{p+1} h_p = h_{p-1}\partial_p
Claim 1
Given H a homotopy connecting f,g:X\rightarrow Y we can construct a chain homotopy between f_*,g_*:C_*(X)\rightarrow C_*(Y)
Claim 2
If f,g:C_*\rightarrow C_* are chain homotopic then they induce equal maps on homology.
Proof of 2
Assume [c]\in H_p(C_*), that is, \partial c =0
[f_* c] - [g_* c] = [(f_*-g_*)] =[(\partial h + h\partial)c] = 0 (as \partial c = 0 and homology ignores exact forms)
Hence, at the level of homology they are the same.
Proof of 1
Consider a simplex in X. Now consider its image, a simplex, in Y under g and f respectively. Because of the homotopy we can construct a triangular based cylinder in Y with the image under f at the top and the image under y at the bottom.
Define h\sigma = the above prism formed by \sigma and the homotopy H.
(f_*-g_*)\sigma = h\partial\sigma = \partial h\sigma
This, pictorially is correct but we need to be able to break up the prism into a union of images of simplexes.
Suppose
