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$\DeclareMathOperator\Hom{Hom}\DeclareMathOperator\GL{GL}$Fix a group $\Gamma$ and a positive integer $n$. Let $c(q):=\lvert\Hom(\Gamma, \GL_n(\mathbb{F}_q)\rvert$ denote the number of homomorphisms from $\Gamma$ to $\GL_n(\mathbb{F}_q)$.

Question A: is $c_n(q)$ a polynomial in $q$? I.e., does there exist a polynomial $P=P_\Gamma$ such that $c_n(q)=P(q)$ for all prime powers $q$?

This is known to be true if $\Gamma$ is finite or Fuchsian (which includes surface groups). For finite groups, see for instance the 2018 article "Asymptotic bounds for the size of $\Hom(A, \GL_n(q))$" by Bate and Gullon. For Fuchsian groups, this can be extracted from the discussions in the 2005 paper "Fuchsian groups, finite simple groups and representation varieties" by Liebeck and Shalev.

As the latter reference makes clear, (if $\Gamma$ is finitely generated), then this is about counting points on the representation variety $\Hom(\Gamma, \GL_n)$. Thus, it can be considered as a question in arithmetic geomety.

Added in response to comments: the question should be rephrased as:

Question B: is $c_n(q)$ "essentially" a polynomial in $q$?

Here by "essentially" we could mean a number of things, e.g., excluding some primes, or allowing quasi-polynomials, etc.

Even a broader question:

Question C: What do we know about $c_n(q)$ for groups $\Gamma$ beyond the ones mentioned?

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    $\begingroup$ This is not a polynomial when $\Gamma$ is a cyclic group of order $k$ and $n=1$; the formula is $\text{GCD}(k, q-1)$. Perhaps you want to ask for a quasi-polynomial? en.wikipedia.org/wiki/Quasi-polynomial $\endgroup$
    – David E Speyer
    Commented Mar 9, 2022 at 9:29
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    $\begingroup$ It's strange. The Bate-Gullon article (linked in the question) claims that $c_n(q)$ (which they denote $X(n,q)$ is always a polynomial. But their reference for this is this article, specifically Proposition~4.1, and that article has the condition that (in OP's notation) $|\Gamma|$ is not a multiple of $p$, where $q = p^k$. The article also references this article, which deals with that excluded case by giving polynomial bounds on $c_n(q)$, but doesn't claim that it's polynomial. (cont...) $\endgroup$
    – Carl-Fredrik Nyberg Brodda
    Commented Mar 9, 2022 at 9:43
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    $\begingroup$ (...cont) But then Bate-Gullon claims to prove things about the leading term of $c_n(q)$, which is not defined in the excluded case at all (such as observed by David E Speyer)! Either I am, or Bate-Gullon are, somewhat confused here; maybe I am misreading something. $\endgroup$
    – Carl-Fredrik Nyberg Brodda
    Commented Mar 9, 2022 at 9:44
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    $\begingroup$ @Carl-FredrikNybergBrodda At the beginning of the "Preliminaries" section of Bate-Gullon, they say that their standing assumptions include that the order of the group is not divisible by the characteristic of the field. $\endgroup$
    – Jeremy Rickard
    Commented Mar 9, 2022 at 10:24
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    $\begingroup$ @Carl-FredrikNybergBrodda I think Bate-Gullon also include in their assumptions "Suppose that $\mathbb F_q$ is a splitting field for $A$." $\endgroup$
    – Will Sawin
    Commented Mar 10, 2022 at 0:15

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