Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I'm having some trouble proving the following statement: let $g:S^{2n-1} \to S^{2n-1}, f: S^{2n-1} \to S^n$. Then $H(f\circ g) = \deg g H(f)$ where $H(f)$ is the Hopf invariant. The definition I am using for Hopf invariant is as follows: let $C_f = D^{2n} \sqcup_f S^{n}$ where $D^{2n}$ is the $2n$ disk which we attach to $S^{n}$ via $f$. Let $\alpha, \beta$ be generators for $H^n(C_f)$ and $H^{2n}(C_f)$ respectively. Then $H(f)$ is defined by $\alpha \cup \alpha = H(f) \beta$. I can prove the statement using the integral formula for $H(f)$ but would like to prove it cohomologically.

I think I should consider the map $G: C_{f\circ g} \to C_f$ that takes $x\in \partial D^{2n}$ to $g(x)$ and acts as the identity on everything else. But then I don't know how to compute what $G^* \alpha$ is.

Thanks!

share|improve this question
    
In your definition of $C_f$, shouldn't the sphere be $n$ dimensional, not $2n$ dimensional? –  Jason DeVito Dec 19 '10 at 5:03
    
Youre right, thanks Jason. –  Eric O. Korman Dec 19 '10 at 14:46

1 Answer 1

up vote 3 down vote accepted

So, if for $x\in \partial D^{2n}$ you take $G(x)=g(x)$, then probably you want to do "the same" map over the whole cone. It follows from the definition of the union topology (or whatever it's called) that this extends to a continuous map. Then, look at the induced map on the pairs $(K,K-D^{2n})$ (for either complex $K$), and here the map is just (up to confusing maps on pairs with maps on their quotients) the suspension of $g$. Note that for both complexes, the inclusion $(K,\emptyset)\rightarrow (K,K-D^{2n})$ is a cohomology isomorphism in degree $2n$, and use the fact that suspension induces an isomorphism $\pi_n(S^n)\rightarrow \pi_{n+1}(S^{n+1})$.

Edit

A diagram might help:

    (Cfg,o)    --->     (Cf,o)
       |                  |
       |                  |
       V                  V
(Cfg,Cfg-D^2n) ---> (Cf, Cf-D^2n)

Here, the o's stand for $\emptyset$. The vertical maps are inclusions of pairs, and the horizontal lines are really all just $G$ (as I defined it). The top arrow induces $\alpha_{Cf} \mapsto \alpha_{Cfg}$, and the vertical maps are cohomology isomorphisms in degree $2n$. On the bottom row, since $H^*(X,A)=\tilde{H}^*(X/A)$, this is basically a map $S^{2n}\rightarrow S^{2n}$, which by construction is exactly the suspension $Sg$ of $g$. Then, $\deg(Sg)=\deg(g)$; that is, $S:\pi_{2n-1}(S^{2n-1})\rightarrow \pi_{2n}(S^{2n})$ is an isomorphism. So this induces $\beta_{Cf}\mapsto \deg(g)\cdot \beta_{Cfg}$. By naturality,

$$ H(fg)\cdot \beta_{Cfg} = \alpha_{Cfg}^2=G^*(\alpha_{Cf}^2)=G^*(H(f)\cdot \beta_{Cf})=H(f)\cdot (\deg(g)\cdot \beta_{Cfg}).$$

share|improve this answer
    
(Sorry this might be a little off, I'm kind of rushed but I saw nobody had answered so I wanted to say something) –  Aaron Mazel-Gee Dec 19 '10 at 20:55
    
Thanks for the response. It may take me some time to digest it...from what you said does it follow that $G^*$ takes a generator $\beta$ of cohomology in degree $2n$ to $\deg g \beta$? –  Eric O. Korman Dec 19 '10 at 22:35
    
Yes, it's because you can talk about the degree of a map of spheres either as the induced map on $Z$-(co)homology or as an element of $\pi_n(S^n)=Z$. Let me know if anything else doesn't make sense, I'm realizing this was pretty terse. –  Aaron Mazel-Gee Dec 20 '10 at 4:08
    
Awesome, thanks! –  Eric O. Korman Dec 20 '10 at 13:03

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.