Let $G$ be a simple linear group over a non-archimedean local field $F$.
Assume that the split-rank over $F$ is at least 2.
Let $\Gamma$ be a lattice in $G$. Then $\Gamma$ is a finitely generated Kazhdan group.


My question is this:
Does $\Gamma$ admit a finite-dimensional unitary representation $\rho$ such that the image $\rho(\Gamma)$ is infinite?



Unitary representations of lattices
The answer is yes, always, if $\operatorname{char}(F)=0$ and never if $\operatorname{char}(F)\neq 0$. It is easy to see that it is enough to prove this when $G$ is simply connected as an algebraic group; we then assume $G$ is simply connected.


Suppose $\Gamma \subset G=G(F)$ is a lattice and $F$ is a non-archimedean local field with $char (F)=0$. By Margulis's arithmeticity theorem, $\Gamma $ is commensurate to $G\_0(O\_S)$ where $G\_0$ is an absolutely simple algebraic group defined over a number field $K$, $S$ is a finite set of places of $K$ including all the Archimedean ones, and for some place $v$ of $K$, we have $F\simeq K\_v$ with an isomorphism $G(F)\simeq G\_o(K\_v)$. Since the projection of $\Gamma $ to $G\_0(K\_v)$ is to be discrete, we have $G\_0 (K\otimes \_{\mathbb Q} {\mathbb R})$ is compact. But, the map of $\Gamma $ in $G\_0(K \otimes \_{\mathbb Q} {\mathbb R})$ is faithful, and the latter compact real algebraic group has a faithful unitary finite dimensional representation (this is essentially the argument of @YCor in the comments). 


Suppose $\operatorname{char}(F)>0$. Since the rank of $G$ over $F$ is at least two, by super-rigidity in finite characteristic (due to myself and also proved in Margulis' book later), every finite dimensional representation of $\Gamma $ into $\mathrm{GL}\_n(\mathbb C)$ has finite image. Hence $\Gamma $ cannot have infinite image finite dimensional (unitary) representations. 


