6
$\begingroup$

Let $G$ be a finitely presented group. It is clear that if the profinite completion $\widehat{G} $ of $G$ is finite, then any finite dimensional complex linear representation $\rho: G\to \text{GL}(m, \mathbb{C})$ is finite, i.e., $\rho(G)$ is a finite group. For the converse, is there a counterexample for such $G$ such that any finite dim representation $\rho$ is finite, but $\widehat{G}$ is a infinite group?

$\endgroup$
1
  • 5
    $\begingroup$ "It is clear that" is rather "It is a clear consequence of Malcev's theorem that" $\endgroup$
    – YCor
    Commented Jan 7, 2021 at 21:03

2 Answers 2

10
$\begingroup$

Yes, there exists such a finitely presented group.

Let $\Gamma$ be a cocompact arithmetic lattice in a product of $\ge 2$ rank 1 groups simple groups (with trivial center) over locally compact fields of finite characteristic. So $\Gamma$ is finitely presented (it is even CAT($0$)). By Malcev, $\Gamma$ is residually finite. By Kazhdan-Margulis, $\Gamma$ is just-infinite: all normal subgroups in $\Gamma$ except $\{1\}$ have finite index. By Venkataramana's superrigidity, every finite-dimensional complex representation of $\Gamma$ has a finite image.


My initial answer was conditional answer:

Let $\Gamma$ be a cocompact lattice in $\mathrm{PSp}(n,1)$, $n\ge 2$. This is a non-elementary hyperbolic group. Hence it is known that for many $x\in\Gamma-\{1\}$, the quotient $\Lambda_x=\Gamma/\langle\!\langle x\rangle\!\rangle$ is non-elementary hyperbolic.

It is also known (superrigidity) that every linear proper quotient of $\Gamma$ is finite. Hence, every linear quotient of $\Lambda_x$ has a finite image. Accordingly

  1. either $\Lambda_x$ has an infinite profinite completion, hence answers your question
  2. or $\Lambda_x$ has finite profinite completion, and hence there exists a non-residually-finite hyperbolic group. The existence of such a group is a famous open problem.

I'm not sure that any well-accepted conjecture predicts which of these properties holds (this might even depend on $x$).

$\endgroup$
1
  • $\begingroup$ Thank you for the nice example! $\endgroup$
    – SDUer
    Commented Jan 8, 2021 at 9:32
7
$\begingroup$

Yes if you weaken to finitely generated. Take an infinite finitely generated torsion group that is residually finite like the Grigorchuk group. By a theorem is Schur any finitely generated torsion linear group is finite. I am not sure of the general case.

$\endgroup$
6
  • $\begingroup$ The question asks for a finitely presented example. :) $\endgroup$
    – HJRW
    Commented Jan 7, 2021 at 17:50
  • $\begingroup$ It is unknown if there are finitely presented infinite torsion groups. $\endgroup$ Commented Jan 7, 2021 at 17:50
  • $\begingroup$ Sorry. You are right $\endgroup$ Commented Jan 7, 2021 at 17:50
  • 3
    $\begingroup$ But I'll leave this up since it might be useful $\endgroup$ Commented Jan 7, 2021 at 17:50
  • $\begingroup$ I think your answer is worth keeping, though. Perhaps the OP can clarify how important the "finitely presented" hypothesis is to them. $\endgroup$
    – HJRW
    Commented Jan 7, 2021 at 17:51

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .