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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!

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    $\begingroup$ If one had a closed aspherical manifold with this property, then its fundamental group could not contain a Baumslag-Solitar subgroup; such a group has a subgroup with non-trivial $H_2$. 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 which was non hyperbolic and contained no Baumslag-Solitar subgroup was not known.arxiv.org/abs/2105.14795 $\endgroup$
    – Ian Agol
    Commented Mar 21, 2023 at 21:05
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    $\begingroup$ A weak version of what you ask for: There is a closed aspherical $16$-dimensional manifold $M$ such that every finitely-sheeted coverings $\bar M$ of $M$ has finite $H_i(\bar M)$ for $i=1,2,3$. Here $M$ is any closed Cayley hyperbolic manifold, see section 8 in "Nonarithmetic Superrigid Groups: Counterexamples to Platonov's Conjecture" by Bass and Lubotzky, arxiv.org/pdf/math/0005302.pdf. $\endgroup$
    – Igor Belegradek
    Commented Mar 21, 2023 at 21:29
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    $\begingroup$ I am quite sure that no such examples are known. $\endgroup$
    – Moishe Kohan
    Commented Mar 22, 2023 at 0:03
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    $\begingroup$ What’s the condition for $v\geq 2$? $\endgroup$
    – Fernando Muro
    Commented Mar 22, 2023 at 6:11
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    $\begingroup$ @IanAgol: which subgroup do you have in mind for the solvable Baumslag—Solitar groups? The finite-sheeted covers have trivial $H_2$, as does the infinite cyclic cover. (I mean with $\mathbb{Z}$ coefficients, of course.) $\endgroup$
    – HJRW
    Commented Mar 22, 2023 at 7:08

1 Answer 1

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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).

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