The homotopy fibre of an map $f \colon S^{2n-1} \to S^n$

Question

I am trying to compute the cohomology of the homotopy fibre $F$ of a continuous map $f \colon S^{2n-1} \to S^n$ ($n$ even and non-zero Hopf invariant). It is easy to see that through the Serre spectral sequence the non zero groups are (with integer coefficients) $$ H^{n-1}(F) = \mathbb Z,\quad H^k(F) = \mathbb Z_m $$ for $k = 2n-1 + j(n-1)$ and $j=0,1,2,\dots$ ($m=1$ means here that all that the latter groups are all zero). Someone mentioned to me that $m$ is just the Hopf-invariant of $f$. He didnt had a proof and I can not show or see this.

Answer 1

In the following, any identity may actually be up to a sign. I don't want to keep track of that.

We are thinking about the fibration sequence $$ \Omega S^n \to F \to S^{2n-1} \xrightarrow{f} S^n.$$ It may be little easier to think about the homology Serre spectral sequence of $\Omega S^n\to F\to S^{2n-1}$ rather than the cohomology one: they are basically dual to each other. We are interested in the first non-trivial differential: $$ d\colon E^{2n-1,0}=H_{2n-1}S^{2n-1} \to E^{0,2n-2}= H_{2n-2}\Omega S^n.$$ A "standard" argument shows that $d$ sends the generator (which is the Hurewicz image of the identity map of $S^{2n-1}$) to the Hurewicz image of $$\tilde{f}\colon S^{2n-2} \to \Omega S^n,$$ the adjoint to $f$.

We have that $H_*\Omega S^n$ is a polynomial ring on one generator $a$ in (odd) degree $n-1$; the product is the Pontryagin product associated to the H-space structure on $\Omega S^n$ given by composition of loops. Note that although this is a polynomial ring, it is not commutative in the graded sense: for instance, since $a$ is in odd degree, the graded commutator gives $[a,a]= a.a - (-1) a.a = 2a^2$.

So $H_{2n-2}\Omega S^n \approx \mathbb Z$, generated by $a^2$. Given $f\colon S^{2n-1}\to S^n$, we obtain an invariant $H'(f)\in \mathbb Z$ by $h(\tilde f)= H'(f) a^2$, where $h\colon \pi_* \Omega S^n\to H_*\Omega S^n$ is the Hurewicz map. I claim that $H'(f)$ is the same as the Hopf invariant of $f$ (up to a sign).

There's surely a slick proof of that, but I don't remember it. Here's a sketch of a dirty proof:

1. Show that $H'$ is a group homomorphism (this basically amounts to the fact that the pinch map $S^{2n-1}\to S^{2n-1}\vee S^{2n-1}$ is a suspension). (Remember that the Hopf invariant $H$ is also a homomorphism).

2. By Serre, we know that $\pi_{2n-1}S^n\approx \mathbb Z\oplus \text{torsion}$. Because $H'$ is a homomorphism, it factors through the torsion-free quotient. Thus it suffices to show $H'(f)=H(f)$ for one element $f$ of infinite order.

3. Let $f=[\iota,\iota]$, the Whitehead product of the identity map of $S^n$ with itself. The adjoint $\tilde f$ is (up to sign) the "Samelson product" $\{ \tilde\iota, \tilde\iota \}$. Here $\tilde\iota\colon S^{n-1}\to \Omega S^n$ is the adjoint of identity: the Samelson product on $\pi_p \Omega X\times \pi_q \Omega X\to \pi_{p+q}\Omega X$ is the bilinear map on homotopy groups induced by the commutator operation on the "group" $\Omega X$.

The Hurewicz map $h\colon \pi_*\Omega X\to H_*\Omega X$ takes Samelson products to commutators in the Pontryagin ring.

It is standard that $H(f)=2$. On the other hand, $$h(\tilde f)= h(\{\tilde \iota, \tilde \iota\})= [h(\tilde \iota), h(\tilde\iota)]=[a,a]=2a^2,$$ so $H'(f)=2$.