Homological restrictions on certain $4$-manifolds

Question

I am not very familiar with the non-compact $4$-manifold theory. So I apologize if the following question is very silly.

Let $X$ be a non-compact, orientable $4$ manifold that is homotopic to an infinite wedge of $S^2$'s. Let $G$ be a finitely generated group acting freely on $X$ so that $X\to X/G$ is universal covering, here $X/G$ is also non-compact, orientable. Can we show that $H_3(X/G;\mathbb{Z})$ is a free group of finite rank? Or we put some condition on $G$ so that $H_3(X/G;\mathbb{Z})$ is a free group of finite rank?

Any suggestion will be really helpful. Thanks in advance!

Answer 1

It seems to me that one does get $H_3(X/G)$ finitely generated under some natural assumptions, namely:

1. the action of $G$ on $H_2(X)$ makes it into a free $G$-module;

2. the group homology $H_3(G)$ is finitely generated.

So let's make these two assumptions.

By the first assumption I can equivariantly glue a disjoint union of $3$-balls $B$ to $X$ to kill $H_2(X)=\pi_2(X)$. The result is a contractible space $Z=X\cup B$ equipped with a free $G$-action, so the homology of the quotient $Z/G$ is the group homology of $G$.

Since the action of $G$ just permutes the components of $B$, the quotient $Z/G$ is just constructed by gluing a set of 3-balls to $X/G$, so the third row of the Mayer--Vietoris sequence gives

$$ 0\to H_3(X/G) \to H_3(G)\to\ldots $$

Thus, $H_3(X/G)$ injects into $H_3(G)$, and in particular is finitely generated by the second assumption.

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Although I don't have a counterexample to the general version of your question, I think it's reasonable to believe that one exists at least wihout the second assumption.

Stallings famously exhibited a finitely presented group $G$ with $H_3(G)$ infinitely generated (AJM, 1963). In fact, Stallings' example is 3-dimensional, being a subgroup of a direct product of three free groups. Thus, there is an aspherical 3-complex $K$ with $\pi_1(K)\cong G$. Let $L$ be the result of deleting a single 3-cell from $K$.

Now, it seems entirely possible to me that $L$ can be thickened to an open 4-manifold $M$. (At least, I don't see any a priori reason why such a thickening shouldn't exist.)

If this were true then the universal cover $X$ of $M$ would be homotopy equivalent to the universal cover of $L$, which is in turn homotopy equivalent to a wedge of 2-spheres, since it is constructed by deleting a disjoint set of 3-cells from the (contractible) universal cover of $K$.

On the other hand, Mayer--Vietoris tells us that the rank of $H_3(L)$ is only one less than the rank of $H_3(K)=H_3(G)$, which is infinite by Stallings' construction.

So one is left wondering if $L$, or even $K$, or some similar example, can be thickened to a 4-manifold. This probably has something to do with the links of the vertices. It seems like an interesting question!