# Structure of second homotopy group of a compact CW complex

I am interested in the second (and higher as well) homotopy groups of compact CW complexes. I know these groups don't need to be finitely generated (e.g. for $$S^1 \vee S^2$$ they are not), but I'd like to know more about the structure of these groups. Below $$X$$ always denotes a compact CW complex.

For example, it seems unlikely to me that one could have something like $$\pi_2(X) \cong \mathbb Q$$ or $$\pi_2(X) \cong \mathbb Q / \mathbb Z$$, but then again I really don't know much about the topic. Here is are some specific questions:

(1) Does $$\pi_2$$ decompose into torsion-free and torsion parts as follows: $$\pi_2(X) \cong \bigoplus_I \mathbb Z \oplus \bigoplus_i \bigoplus_{J_i} \mathbb Z/p_i^{r_i}$$, where the $$I$$ and $$J_i$$ are index sets, $$p_i$$ are some primes? In all the examples I have ever seen, this happens. Of course we can't use the structure theorem for finitely generated Abelian groups to conclude something like this, but maybe it is still true?

(2) If the above is false, is the following weaker thing true? If $$\pi_2(X) \otimes k = 0$$ for every field $$k$$, then is $$\pi_2(X) = 0$$?

(3) Do only finitely many different primes occur as the orders elements in $$\pi_2(X)$$?

I'm aware that via Hurewicz we have $$\pi_2(X) \cong H_2(\widetilde X)$$, I don't know if this helps.

Let $$X=S^1 \vee S^2$$ have $$\pi_1(X) = \mathbb{Z} = \langle t \rangle$$, and so $$\pi_2(X) = \mathbb{Z}[t, t^{-1}]$$, generated by the inclusion $$\iota : S^2 \to X$$. Attach a 3-cell to $$X$$ along $$t\iota - 2 \iota : S^2 \to X$$ to form $$Y$$, which is a finite cell complex having $$\pi_2(Y) = \mathbb{Z}[t, t^{-1}]/(t- 2) = \mathbb{Z}[\tfrac{1}{2}].$$ This shows that (1) is not true.
I expect that starting with a more interesting fundamental group than $$\mathbb{Z} = \langle t \rangle$$ one could make an example with $$\pi_2 = \mathbb{Q}$$, but I don't know how offhand.
• I think it is unlikely that you could get $\pi_2(Y)\cong\mathbf Q$ this way, for ring theoretic reasons. In this construction, $\pi_2(Y)$ is finitely generated over $\mathbf Z[\pi_1(Y)]$, which in turn is a finitely presented $\mathbf Z$-algebra. This is impossible at least when $\pi_1(Y)$ is abelian (if $\pi_2(Y)\cong\mathbf Q$, then it is a cyclic $\mathbf Z[\pi_1(Y)]$-module, hence isomorphic to $\mathbf Z[\pi_1(Y)]/I$ for an ideal $I$, so $\mathbf Q$ would be a finitely generated $\mathbf Z$-algebra). I think this is also impossible if $\pi_1(Y)$ is non-abelian, but I don't have a proof. Commented Dec 27, 2023 at 17:46
• @R.vanDobbendeBruyn : $\mathbf Q$ is also not finitely generated as an associative $\mathbf Z$-algebra, simply because it is a filtered colimit of the $\mathbf Z[1/p, p \in F]$. I'm confused as to how you deduce that $\pi_2(Y)$ has to be a cyclic module if it is $\mathbf Q$ ? Commented Dec 27, 2023 at 19:40
• @MaximeRamzi the difficulty in the non-commutative setting is that one-sided ideals need not be two-sided, so getting $\mathbf Q$ as a cyclic module does not mean it becomes an algebra. (To show it is cyclic if it is finitely generated as $\mathbf Z[\pi_1(X)]$-module (say by $x_1,\ldots,x_k \in \mathbf Q$) take a generator of the subgroup of $\mathbf Q$ generated by $x_1,\ldots,x_k$.) Commented Dec 27, 2023 at 23:38
• @SFeesh for now it is referring to the construction of this answer. If I knew how to rule out $\mathbf Q$ in general, I would probably post it as a separate answer. But even in the case described here, my argument has some gaps in it. Commented Dec 27, 2023 at 23:41