6
$\begingroup$

In P. Deligne. Extensions centrales non résiduellement finies de groupes arithmétiques. CR Acad. Sci. Paris, série A-B, 287, 203–208, 1978. Deligne proves the existence of a certain central extension of a residually finite group.

See the section Deligne's central extension in Cornulier-Guyot-Pitsch's paper for a quick discussion of the group.

A countable, discrete group $\Gamma$ is sofic if for every $\epsilon>0$ and finite subset $F$ of $\Gamma$ there exists an $(\epsilon,F)$-almost action of $\Gamma$. See, for example Theorem 3.5 of the nice survey of Pestov. Gromov asked whether all countable discrete groups are sofic. It is now widely believed that there should be a counterexample to this.

Is Deligne's central extension sofic?

This question is related to the one here, but is not sharpened enough to be an answer. (In fact, in its original form not even to be a question! Thank you Henry and Andreas.)

Improve this question
$\endgroup$
3
  • 2
    $\begingroup$ A few points: you haven't said anything about $H$, for example it could be the trivial subgroup. In general, the largest such $H$ I thought was called the finite residuum of $G$ (although that doesn't seem very popular with Google), and is the intersection of all finite index subgroups of $G$. The distinction you make betewen "every finite index subgroup" and "every finite index normal subgroup" turns out not to matter - you can always drop by a finite index to gain normality (this doesn't depend on finite generation). $\endgroup$
    – ndkrempel
    Dec 8, 2010 at 18:12
  • 4
    $\begingroup$ The interest of Deligne's example is not that it isn't residually finite. The interest is that it's a non-residually finite group which is a central extension of a residually finite (indeed, linear) group. $\endgroup$
    – HJRW
    Dec 8, 2010 at 18:15
  • 2
    $\begingroup$ (The arxiv link is broken. Here a working one: arxiv.org/abs/0804.3968 ) $\endgroup$
    – Greg Graviton
    Dec 8, 2010 at 18:26

2 Answers 2

Reset to default
5
$\begingroup$

If I understand your question correctly then I think you've already answered it! The 'above property' is precisely 'not being residually finite'. To see this, just consider the intersection of all finite-index subgroups: if it's trivial, your group is residually finite; if not, then that intersection is your subgroup $H$.

As we found in a previous question of yours, Baumslag's group

$B=\langle a,b\mid (a^b)^{-1}a(a^b)a^{-2}\rangle$

which is certainly not residually finite, is sofic.

Improve this answer
$\endgroup$
4
  • 2
    $\begingroup$ Incidentally, I think the question in the title is still an interesting one. $\endgroup$
    – HJRW
    Dec 8, 2010 at 18:15
  • $\begingroup$ Hi Henry. Sorry about the nonsense! Please have another look. $\endgroup$
    – Jon Bannon
    Dec 8, 2010 at 20:35
  • 1
    $\begingroup$ Gladly! Unfortunately, I think it's pretty unlikely that I can say anything useful about the modified question. $\endgroup$
    – HJRW
    Dec 8, 2010 at 21:30
  • 1
    $\begingroup$ For reference: It's also known as "Baumslag-Gersten group". Its soficity follows from stability of soficity under HNN over amenable groups. Cf Remark 2.2 in J. Bannon, A non-residually solvable hyperlinear one-relator group, Proc. Amer. Math. Soc. 139 (2011), 1409-1410, freely accessible: ams.org/journals/proc/2011-139-04/S0002-9939-2010-10668-8 $\endgroup$
    – YCor
    Jun 7, 2017 at 23:48
5
$\begingroup$

Deligne's central extension is certainly an interesting group to consider. I can say something about the question whether this group is hyperlinear. For hyperlinearity, you ask for approximation of the group law by unitary matrices instead of permutations. This is (as you of course know) weaker than being sofic, but being sofic implies that the group is hyperlinear. Hence, it is a neccesary condition and natural generalization.

In the paper Examples of hyperlinear groups without factorization property, Groups Geom. Dyn. 4 (2010), no. 1, 195–208 I obverved that a central extension of a group $G$ by an abelian group $A$ is hyperlinear if and only if the twisted group von Neumann algebra $L_{\phi \circ \alpha} G$ is embeddable, where $$\alpha \colon G \times G \to A$$ is the 2-cocycle which classifies the central extension, and $\phi$ belongs to a dense set in the Pontrjagin dual of $A$. This is the same as asking for a approximation of the group laws of $G$ with unitaries on a finite-dimensional Hilbert space, twisting the multiplication with the cocycle $\phi \circ \alpha$. The twisted setup is natural anyway and I propose to call a $S^1$-valued 2-cocycle on a group $G$ hyperlinear if you can find such an approximation by unitaries. It seems natural to consider the possibility that there are $S^1$-valued 2-cocycles even on residually finite groups which are not hyperlinear. However, I do not have any examples.

On the other side, if $G$ is residually finite and you can at the same time approximate $\phi \circ \alpha$ for sufficiently many $\phi$'s by suitable almost 2-cocycles defined on the finite quotients, you are in business. This would show that the central extension is at least hyperlinear.

I do not know about sofic in place of hyperlinear, since in the combinatorial setup, it is not possible to disintegrate the central extension over the Pontrjagin dual of $A$.

Improve this answer
$\endgroup$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.