I wonder if there exists a compact closed smooth aspherical manifold $M$ of dimension at least $4$, so that for any covering space $\tilde{M}$ over $M,$ we always have $H_2(\tilde{M},\mathbb{Z})=0$ and $H_2(\tilde{M},\mathbb{Z}/v\mathbb{Z})=0$ for all integers $v\ge 2$.
Of course, when $H_2$ is replaced with $H_1,$ then the answer is negative since we can always choose a covering with homotopy type of $S^1$. If we drop the closedness assumption, the answer is true by considering a regular neighborhood of a bouquet of circles in $\mathbb{R}^4.$
The motivation comes from the calculus of variations. I'm sorry if this problem has trivial answers since my field is somewhat far away from algebraic topology. Many thanks!
Best Answer
I think that the answer to this question is unknown in general. If one had a closed aspherical manifold with this property, then it could not contain a Baumslag-Solitar subgroup since such a group has a subgroup with non-trivial $H_2$ (possibly with finite field coefficients, eg $BS(1,2)$ contains a $BS(1,4)$ subgroup with $H_2(BS(1,4);\mathbb{F}_3)\neq 0$). Examples of groups that do not contain Baumslag-Solitar subgroups are hyperbolic groups. However, Gromov has conjectured that hyperbolic groups contain a surface subgroup, and hence would have a subgroup with non-trivial $H_2$ if that conjecture is true.
Moreover, until recently a group of finite type (such as the fundamental group of an aspherical manifold) which was non hyperbolic and contained no Baumslag-Solitar subgroup was not known. Thus I would argue that there is probably no example of this sort in the literature, since it would have resolved one of two possible well-known open questions.
On the other hand, maybe it is known that any aspherical n-manifold admits a cover with non-trivial $H_2$? I think a theorem of this sort in the literature for $n>3$ would be well-known if it existed (and at least @MoisheKohan and myself are not aware of such a theorem, if that carries any weight).