at.algebraic topology - the hopf invariant of the hopf construction

Tuesday, 19 August 2008

at.algebraic topology - the hopf invariant of the hopf construction

This is my attempt at an answer. I think I might be off in justifying the second diagram. It's based on the proof in Bredon, but I think it might be a little simpler. Corrections are welcome, of course!

For convenience, we introduce some notation and conventions. We define subsets of

$S^n= { (c,t):cin S^{n-1},tin I } /sim$ by

$X={(c,t):tleq 1/2}$ ,

$Y={(c,t):tgeq 1/2}$ , and

$Z={(c,1/2)}$ . We define subsets of

$S^{2n-1}={ (a,t,b):a,bin S^{n-1},tin I}$ by

$A={(a,t,b)in S^{2n-1}:tleq 1/2}$ and

$B={(a,t,b)in S^{2n-1}:tgeq 1/2}$ . We define subsets of

$Acap B={(a,1/2,b)}$ by

$S^{n-1}_A = {(a,1/2,b_0)}$ and

$S^{n-1}_B = {(a_0,1/2,b)}$ for fixed

$a_0,b_0in S^{n-1}$ . We consider

$S^{2n-1}$ as the boundary of

$e^{2n}$ ; in this cell,

$S^{n-1}_A$ is the boundary of a cell

$e^n_A$ and

$S^{n-1}_B$ is the boundary of a cell

$e^n_B$ . We write

$K=S^ncup_{h(g)}e^{2n}$ for the complex in which we must compute

$H(h(g))$ . Throughout, all cohomology is integral.

To compute cup products in $K$ , we use the commutative diagram

H^n(K) x H^n(K) ---------> H^{2n}(K)
       ^             cup       ^
       |                       |
       |             cup       |
H^n(K,X) x H^n(K,Y) -----> H^{2n}(K,S^n)

which follows from the map $(K,emptyset,emptyset)rightarrow (K,X,Y)$ and naturality; the vertical maps are isomorphisms by the long exact sequence in cohomology for pairs. Let the generator $x$ of $H^n(K)$ correspond to the generator $x_Y$ of $H^n(K,X)$ and the generator $x_X$ of $H^n(K,Y)$ , so that $x_Ycup x_X$ in $H^{2n}(K,S^n)$ corresponds to $x^2$ in $H^{2n}(K)$ .

To understand the image of the cup product, we use the evident map $j:(e^{2n},A,B)rightarrow (K,X,Y)$ . We have the diagram

H^n(K,X)  =  H^n(S^n,X)  =  H^n(Y,Z)  =  H^{n-1}(Z)
   |                                       |
   | j^*                                   | j^*
   |                                       |
   V                                       V
H^n(e^{2n},A) = H^n(e^n_A,S^{n-1}_A) = H^{n-1}(S^{n-1}_A)

which commutes because of the naturality of the isomorphisms involved (namely, along the top: cellular cohomology, excision and homotopy invariance, lexseq for pairs; along the bottom: homotopy invariance, lexseq for pairs). By definition of degree, the map $j^ast:H^{n-1}(Z)rightarrow H^{n-1}(S^{n-1}_A)$ is multiplication by $alpha$ . Hence, so is the map $j^ast : H^n(K,X)rightarrow H^n(e^{2n},A)$ . Similarly, the map $j^ast:H^n(K,Y)rightarrow H^n(e^{2n},B)$ is multiplication by $beta$ . Now, $j^ast:H^{2n}(K,S^n)rightarrow H^{2n}(e^{2n},S^{2n-1})$ is an isomorphism, and $j^ast(x_Ycup x_X)$ is $alphabeta$ times a generator of $H^{2n}(e^{2n},S^{2n-1})$ . Thus $x_Ycup x_X$ is $alphabeta$ times a generator of $H^{2n}(K,S^n)$ . So by the first diagram, $H(h(g))=alphabeta$ .

AnswerDesk

Tuesday, 19 August 2008