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

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.

• Care to explain the downvote? I'd be happy to improve the question. – Bruno Stonek Jan 13 '15 at 20:01

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.

• Thanks a lot for your answer. I will check the details tomorrow. I think I see it for the first question. As for the second one, it's odd that you use those results, because in Davis & Kirk's book it comes after the computation that $\pi_4(S^3)=\mathbb Z/2$... – Bruno Stonek Jan 13 '15 at 22:03
• Ok, it's clear for the first question. For the second one, I think I got it assuming that the homology groups of $X$ are finitely generated. Which is true if its homotopy groups are f.g. (Serre classes). But its homotopy groups are those of spheres, which are f.g. because their homology groups are f.g. (Serre classes). – Bruno Stonek Jan 15 '15 at 16:56

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 $H_4(X)=\mathbb Z/2$.

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).