Computing $\pi_4(S^3)$ using Serre spectral sequence

Question

I'm following Davis & Kirk's computation that $\pi_4(S^3)=\mathbb Z/2$ using the Serre spectral sequence but I'm having problems at the very end.

We consider a homotopy fibration $X\to S^3 \to K(\mathbb Z,3)$ where the second map induces an iso on $\pi_3$. Looking at the long exact sequence we get that the homotopy groups of $X$ are those of $S^3$ except at 3 where it is 0. In particular, by Hurewicz we get that $H_4(X)=\pi_4(S^3)$ and this is how we will deduce $\pi_4(S^3)$ using $X$.

Taking the homotopy fiber of the above fibration yields a homotopy fibration $K(\mathbb Z,2)\to X \to S^3$. We work with the cohomological Serre spectral sequence of this homotopy fibration.

First: We get that $\mathbb Z/2=E_4^{3,2}=E_\infty^{3,2}$ and $0=E_4^{0,4}=E_\infty^{0,4}$. I agree with this. However, they conclude that $\mathbb Z/2=H^5(X)$ and $0=H^4(X)$. Why is that? The $E_\infty$ terms give in general some quotients of two consecutive subgroups in the filtrations of $H^5(X)$ and of $H^4(X)$ respectively, why do they give all the group in this case?

Second: They conclude that $H_4(X)=\mathbb Z/2$ by the universal coefficient theorem. I don't see how this goes. Using the universal coefficient theorem for $H^4$ gives $\hom(H_4(X),\mathbb Z)=0$ and using it for $H^5$ gives something involving $Ext_1(H_4(X),\mathbb Z)$ that I don't see how it could be useful.

Answer 1

First: Are there any other terms $E_2^{p,q}$ with $p+q=5$ which survives to $E_\infty?$

No, in fact, the only non-trivial $E_2^{p,q}$ with $p+q=5$ is $E_2^{3,2}$. To see this, it may be useful to observe that $H^p(S^3;\mathbb{Z})$ is non-trivial only at $p=0,3$, and that $\mathbb{C}P^\infty$ is a model for $K(\mathbb{Z},2)$, so $E_2^{0,5}$ is also trivial. If there is only one non-trivial factor in your filtration...

Similar considerations also answers the second part of this question.

Second: You'll probably have to use the fact that $\pi_i(S^n)$ is finite for $i \not=n $, $n$ odd, which should help you calculate the Ext group. This fact follows from Serre's mod $\mathcal{C}$ theory, which you would undoubtedly have seen before if you're calculating homotopy groups of spheres.

Answer 2

Ok, I'll elaborate on fixedp's answer which was sort of cryptic for me.

1. You only have to observe the general fact that if there is only one $E_\infty^{p,q}$ not zero for $p+q=n$ given, then that one gives $H^{p+q}(X)$. This follows from the definition of convergence.

2. This is a bit more tricky. I don't know how Davis & Kirk might have concluded the computation without having Serre classes at hand.

Theorem (Serre): Let $X$ be a simply connected space. Then $X$ has all its homotopy groups finitely generated iff it has all its homology groups finitely generated.

We apply it twice, getting:

1) homotopy groups of spheres are finitely generated, and thus

2) $X$ has finitely generated homology groups.

Remark: Let $A$ be an abelian group.

a) If $A$ is finite, then $\hom(A,\mathbb Z)=0$.

b) If $A$ is finitely generated, then $\hom(A,\mathbb Z)$ is free. If moreover $\hom(A,\mathbb Z)=0$ then $A$ is finite.

Proof: for a), a homomorphism sends finite order elements in finite order elements. For b): structure theorem. $\square$

Now, the universal coefficients theorem for cohomology gives:

$\mathbb Z/2=H^5(X)=\hom(H_5(X),\mathbb Z)\oplus Ext(H_4(X),\mathbb Z)$

$0=H^4(X)=\hom(H_4(X),\mathbb Z)\oplus Ext(H_3(X),\mathbb Z)$.

Since $H_3(X)=0$, using the second equation and the remark above, we get that $H_4(X)$ is a finite abelian group.

On the other hand, $H_5(X)$ is finitely generated, so by the first equation we get that $\hom(H_5(X),\mathbb Z)$ is $\mathbb Z/2$ or $0$. But it must be $0$, since it is free by the remark above.

So $\mathbb Z/2=Ext(H_4(X),\mathbb Z)$. Since $H_4(X)$ is finite, we decomopose it and use basic Ext properties & computations to get that $H_4(X)=\mathbb Z/2$.

Answer 3

For the first part, there is only one $E_\infty$ term element in the diagonal corresponding to $p+q=3+2$. The same thing is true for $p+q=0+4$.

$\newcommand{\Z}{\mathbb{Z}}$ For the second part, you can use that there is an exact sequence $Ext(H^{n+1}(Y),\Z) \to Hom(H^{n}(Y,\Z),G) \to H_n(Y,G)$. This implies that $H_n(Y)=free H^n(Y) \oplus torsion H^{n+1}(Y)$. This version of UCT is an immediate consequence of the fact that $C*(Y,G)=C^*(X,\Z) \otimes G$ whenever $C^*(Y)$ is finitely generated over $\Z$(assume this holds).