the hopf invariant of the hopf construction

Question

I'm having some trouble with a problem about the Hopf construction, in the exercises for Ch. 4 of Mosher & Tangora. Given a map $g : S^{n-1} \times S^{n-1} \rightarrow S^{n-1}$, we get a map $h(g) : S^{2n-1} \rightarrow S^n$ by considering

$S^{2n-1} = S^{n-1} * S^{n-1} = S^{n-1} \times I \times S^{n-1} / \sim$

$S^n = S(S^{n-1}) = S^{n-1} \times I / \sim$

and by putting $h(g)(a,t,b) = (g(a,b),t)$. There's an easy homotopy invariant $(a,b)$ of the map $g$, the degree of the map when restricted to either factor (times any point of the other $S^{n-1}$). The problem asks me to show that the Hopf invariant of the map $h(g)$ is $H(h(g))=ab$. This is defined by $s^2=H(h(g))t$ for generators $s\in H^n(X)$ and $t\in H^{2n}(X)$, where $X=S^n \cup_{h(g)} S^{2n}$ (here $h(g)$ is the attaching map).

I'm trying to mimic a proof in the chapter, which constructs a map with Hopf invariant 2 for even $n$ which is really just the Whitehead square of the identity map. They use the following diagram ($f$ is the folding map, $F$ is induced from $f$, $g$ is the attaching map to get from $S^n \vee S^n$ to $S^n \times S^n$, the vertical maps are inclusions, and $K=S^n \cup_{fg} e^{2n}$ is the complex in which we need to calculate cup products):

          g                     f
S^{2n-1}  ---->   S^n V S^n ------------> S^n
                      |                    |
                      |                    |
                      |                    | i
                      |                    |
                      V              F     V
S^n x S^n = ( S^n V S^n ) U_g e^{2n} ----->  K

[Edit: Uh oh, how do I get it to put stuff in a uniformly spaced font? I'm too illiterate to understand the "formatting reference" -- any help??]

Note that $F$ exists because the composition $ifg:S^{2n-1} \rightarrow K$ is nullhomotopic. We know that $f^*s=s_1 + s_2$, where $s$ and $s_i$ represent the obvious cohomology generators. Denoting 2n-dim. cohomology generators of $K$ and $S^n \times S^n$ by $t$ and $r$ respectively, since $F$ induces an isomorphism in $H^{2n}$, (we can choose orientations so that) $F^* t=r$. And by the Kunneth formula, in $H^*(S^n \times S^n)$ we have $s_1 s_2=r$ and $s_1^2=s_2^2=0$. So now we calculate: $F^*(s^2) = (F^*s)^2 = (f^*s)^2 = (s_1 + s_2)^2 = s_1^2 + s_1 s_2 + s_2 s_1 + s_2^2 = 2r$ (assuming $n$ is even; otherwise $F^*(s^2)=0$).

So somehow they're using things they know about the cohomology of $S^n \times S^n$ to compute cup products in $H^*(K)$. This seems like it's possible because the attaching map $fg$ factors through $S^n \vee S^n$. In the problem I'm stuck on, I can't figure out what should be the analogous factorization. Alternatively, I was thinking that I could possibly use Poincare duality and just try to find the self-intersection of the copy of $S^n$ inside of $K$, but that seem too silly/gimmicky. Another fact I know is that $A * B \simeq \Sigma(A \wedge B)$. But suspension doesn't preserve cup products, so I don't think this could help either...

Answer 1

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 $$S^n= \{ (c,t):c\in S^{n-1},t\in I \} /\sim$$ by $X=\{(c,t):t\leq 1/2\}$, $Y=\{(c,t):t\geq 1/2\}$, and $Z=\{(c,1/2)\}$. We define subsets of $$S^{2n-1}=\{ (a,t,b):a,b\in S^{n-1},t\in I\}$$ by $A=\{(a,t,b)\in S^{2n-1}:t\leq 1/2\}$ and $B=\{(a,t,b)\in S^{2n-1}:t\geq 1/2\}$. We define subsets of $A\cap B=\{(a,1/2,b)\}$ by $S^{n-1}_A = \{(a,1/2,b_0)\}$ and $S^{n-1}_B = \{(a_0,1/2,b)\}$ for fixed $a_0,b_0\in S^{n-1}$. 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 $K=S^n\cup_{h(g)}e^{2n}$ 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 $x_Y\cup x_X$ 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 $j^\ast(x_Y\cup x_X)$ is $\alpha\beta$ times a generator of $H^{2n}(e^{2n},S^{2n-1})$. Thus $x_Y\cup x_X$ is $\alpha\beta$ times a generator of $H^{2n}(K,S^n)$. So by the first diagram, $H(h(g))=\alpha\beta$.

Answer 2

This can be seen as follows in terms of the older geometric definitions of degree and Hopf invariant.

The degree of a smooth map $S^{n-1}\to S^{n-1}$ is the number of points mapping to a given point. More precisely, for a generic point in the codomain the map is transverse to it so it has finitely many points in its preimage and these can be assigned signs according to orientation. The degree is the sum of the signs.

The Hopf invariant of a smooth map $S^{2n-1}\to S^n$ is the linking number of the preimages of two points. More precisely, for a generic point in the codomain the map is transverse to it, making the preimage a closed $(n-1)$-manifold with a trivialized normal bundle and therefore an orientation. The Hopf invariant is the linking number of any two of these.

In the case at hand, you can think of $S^{2n-1}$ as the union of two pieces, $S^{n-1}\times D^n$ and $D^n\times S^{n-1}$, glued along $S^{n-1}\times S^{n-1}$. These two doughnuts are linked once, and I believe you can arrange for the point preimages to be a manifold in $S^{n-1}\times D^n$ that is homologically $a$ times $S^{n-1}\times point$ and a manifold in $D^n \times S^{n-1}$ that is homologically $b$ times $point \times S^{n-1}$.

Added: To connect this with the definition of Hopf invariant of $f:S^{2n-1}\to S^n$ in terms of cup product in $S^n\cup_fD^{2n}$: As intermediary use the following cochain-level description: Let $\alpha$ be a cocycle of $S^n$ whose value on the fundamental class is $1$. Then $f*\alpha$ is $\delta \beta$ for some $(n-1)$-cochain of $S^{2n-1}$. Evaluate the cocycle $\beta\cup f*\alpha$ on the fundamental class of $S^{2n-1}$. The result is independent of choices and equal to the Hopf invariant of $f$. It's also OK to choose $\beta$ so that $\delta\beta$ is a different cocycle in the class of $\alpha$. Now to get from here to the linking number let the $n$-cocycle $\alpha$ be given by a $0$-cycle, a point in $S^n$, so that $f*\alpha$ is given by an $(n-1)$-cycle, the preimage of the point, and the cochain $\beta$ corresponds to an $n$-chain with boundary is that preimage. Cup product corresponds to intersecting one point's preimage with something whose boundary is the other point's preimage, and that's the linking number.