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?



profinite completion and linear representations of finitely presented groups
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.




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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 $x$).


