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S Mar 30, 2023 at 10:21 history bounty ended Zhenhua Liu
S Mar 30, 2023 at 10:21 history notice removed Zhenhua Liu
Mar 28, 2023 at 18:52 vote accept Zhenhua Liu
Mar 28, 2023 at 17:32 answer added Ian Agol timeline score: 4
Mar 24, 2023 at 13:06 comment added HJRW @MoisheKohan: Of course, thanks for resolving my worries!
Mar 24, 2023 at 2:25 comment added Moishe Kohan It will be a wasted bounty.
Mar 24, 2023 at 0:39 comment added Moishe Kohan @HJRW: It contains $BS(1,4)$, thus, is not a problem.
S Mar 23, 2023 at 21:36 history bounty started Zhenhua Liu
S Mar 23, 2023 at 21:36 history notice added Zhenhua Liu Draw attention
Mar 22, 2023 at 18:22 comment added HJRW @IanAgol: Ah, OK. (There’s a typo in that part of the question, so it’s somewhat unclear.) I’m still a little worried about BS(1,2)…
Mar 22, 2023 at 17:35 comment added Ian Agol @hjrw: the question asks for trivial homology with finite coefficients as well.
Mar 22, 2023 at 17:08 history edited Moishe Kohan CC BY-SA 4.0
added 5 characters in body
Mar 22, 2023 at 17:08 comment added Moishe Kohan Please, check the correctness of my edit.
Mar 22, 2023 at 11:09 history edited Zhenhua Liu CC BY-SA 4.0
Fixing typo
Mar 22, 2023 at 7:08 comment added HJRW @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.)
Mar 22, 2023 at 6:11 comment added Fernando Muro What’s the condition for $v\geq 2$?
Mar 22, 2023 at 0:03 comment added Moishe Kohan I am quite sure that no such examples are known.
Mar 21, 2023 at 21:29 comment added Igor Belegradek 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.
Mar 21, 2023 at 21:05 comment added Ian Agol 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
Mar 21, 2023 at 20:01 history asked Zhenhua Liu CC BY-SA 4.0