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Suppose $M^4$ is a compact hyperbolic (i.e. curvature $-1$) $4$-manifold and $\Gamma\cong\pi_1(M^4)$.

Is there any expectation whether $\Gamma$ acts properly and co-compactly on a $\rm CAT(0)$ cube complex?

Note that by work of Bergeron Wise based on work of Kahn-Markovic the answer to the question is always positive for $\Gamma=\pi_1(M^3)$, where $M^3$ is any hyperbolic $3$-manifold. However, if instead $\Gamma$ is a co-compact lattice in $U(2,1)$, the answer is always negative (by Delzant Py (I guess)).

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    $\begingroup$ "properly discontinuously" is a single word (also called, more simply "properly"). It's not "properly and discontinuously". The property of acting properly on a CAT(0) cube complex is known as "Property PW". $\endgroup$
    – YCor
    Commented Jan 12, 2017 at 19:03
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    $\begingroup$ About Delzant-Py, you certainly refer to arxiv.org/abs/1609.08474; however they never claimed that cocompact subgroups of $U(2,1)$ don't have Property PW. They proved that they don't act properly cocompactly on a CAT(0) cube complex. Whether they have Property PW is probably open. $\endgroup$
    – YCor
    Commented Jan 12, 2017 at 19:07
  • $\begingroup$ Yves, really sorry for the confusion, indeed I wanted to write "properly and cocompactly"... Concerning Bergeron-Wise I was just citing Theorem 9.3 in arxiv.org/pdf/1204.2810.pdf $\endgroup$
    – aglearner
    Commented Jan 12, 2017 at 19:29

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This question is certainly open in general. I don't know if anyone has formally expressed an 'expectation' in print, but you might be interested in the following pieces of positive evidence.

I believe that all known examples of such 4-manifolds essentially come either from arithmetic constructions or from Coxeter groups. Haglund--Wise showed that all hyperbolic Coxeter groups are cocompactly cubulated (indeed virtually special) [3], and Bergeron--Haglund--Wise showed that 'standard' cocompact arithmetic lattices in $SO(n,1)$ are virtually special [1]. So most known examples are certainly cocompactly cubulated.

One might worry that closed hyperbolic 3-manifolds are too rigid to be constructed in large numbers, in the way that Kahn--Markovic did with surfaces. But Calegari and I were able to construct very many acylindrical hyperbolic 3-manifolds in random groups [2]. (They don't cubulate for other reasons, but it makes the point that a single group may contain many rigid subgroups.)

[1] Bergeron, Nicolas(F-PARIS6-IMJ); Haglund, Frédéric(F-PARIS11-M); Wise, Daniel T.(3-MGL) Hyperplane sections in arithmetic hyperbolic manifolds. (English summary) J. Lond. Math. Soc. (2) 83 (2011), no. 2, 431–448.

[2] Calegari, D; Wilton, H. 3-manifolds everywhere, https://arxiv.org/abs/1404.7043

[3] Haglund, Frédéric(F-PARIS11-M); Wise, Daniel T.(3-MGL) Coxeter groups are virtually special. (English summary) Adv. Math. 224 (2010), no. 5, 1890–1903

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  • $\begingroup$ Thank you Henri, this is very helpful. I am a bit confused with terminology, is "cocompactly cubulated" = "cubulated "= "admits a properly discontinuous cocompact action by isometries on some CAT(0) cube complex"? Also from what you say I deduce that at the present moment there is no known example of $\Gamma=\pi(M^4)$ with $M^4$ hyperbolic such that $\Gamma$ is not cubulated. Am I right? $\endgroup$
    – aglearner
    Commented Jan 12, 2017 at 23:12
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    $\begingroup$ @aglearner, yes, "cocompactly cubulated" means what you want, and when I just wrote "cubulated" I should have said "cocompactly". (I'll edit.) You are correct that there are no examples known not to be cubulated. I believe that the status of 'non-standard' arithmetic lattices (ie those constructed from quarternionic lattices) is unknown. $\endgroup$
    – HJRW
    Commented Jan 13, 2017 at 7:16
  • $\begingroup$ Sorry, when I wrote 'quarternionic lattices' I meant 'quaternion algebras'. $\endgroup$
    – HJRW
    Commented Jan 13, 2017 at 7:35
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    $\begingroup$ Note that these "non-standard" arithmetic lattices exist only in odd dimensions, so it seems there is no known example of a compact hyperbolic 4--manifold which is not known to be cubulable. $\endgroup$
    – Jean Raimbault
    Commented Jan 15, 2017 at 9:05
  • $\begingroup$ @JeanRaimbault, thanks for the addendum -- I did not know that! $\endgroup$
    – HJRW
    Commented Jan 15, 2017 at 19:42

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