If you take an infinite simple group $G$ (say Thompson’s $T$) and put the discrete topology on it this will have your property when k is a finite field (since the image in $GL_n(k)$ will necessarily be trivial.

Say further that $G$ is finitely presented (as indeed for example Thompson's $T$ is). Then you can generalize this to $\mathbb{C}$ as follows by following the techniques in this paper. Basically, a non-trivial homomorphism into $GL_n(\mathbb{C})$ can be turned into a homomorphism into $GL_n(\mathbb{F}_p)$ for some prime $p$. See theorem 3.4. There they use it to turn a $\mathbb{C}$ representation with non-commutative image into a $\mathbb{F}_p$ representation with non-commutative image, but you can do the same thing for "non-trivial" instead of "non-commutative" $\mathbb{C}$ representation. I will spell it out below.

**The key result it relies on is a very cool and useful theorem that says "solutions to systems of equations over $\mathbb{C}$ imply over some finite field."** Specifically, if $f_1(x_1, \dots, x_N), \dots f_m(x_1, \dots, x_N)$ is a system of polynomials with integer coefficients that has a solution over $\mathbb{C}$, then it also has a solution over $\mathbb{F}_p$ for some prime $p$. **Edit:** *This was using a theorem cited in the attached paper. I replaced this with an easy to prove theorem that gives you a solution over a finite field (not necessarily prime order), proven below.*

**Next we want to construct a system of polynomials such that a solution to that system in a field $k$ will correspond to a $k$-representation.**

This we do as follows. Since our group $G$ is finitely presented (say with generators $a_1, \dots, a_s$ and relations $w_1, \dots, w_r$), for any $n$, we can write down matrices $A_1, \dots, A_s, B_1, \dots, B_s$ with variable entries (like $a_{i, j}$ and $b_{i, j}$) and consider the products of these matrices corresponding to the relations $w_i$, where we replace $a_i$ with $A_i$ and $a_i^{-1}$ with $B_i$. By equating the resulting polynomial-entried matrix with the identity matrix, we get $n^2$ integer-coefficient polynomial equations.

We then do this for all matrices simultaneously, plus the additional relations $A_i B_i - I_n = 0$, to get a big system of integer-coefficient polynomials for which a solution in $k$ is exactly the data of a $k$ representation of $G$.

Now, let's say we have a non-trivial $\pi \colon G \to GL_n(\mathbb{C})$. If it has a non-trivial image, there is some $x \in G$ with $\pi(x) \neq 1$. Write $x$ as a word $w$ in the alphabet $a_1^\pm, \dots a_s^\pm$. Denote by $X$ the product of $A_i, B_i$s corresponding to the word $w$ (and thus to the element $x \in G$). We want to augment the above system of equations to say that the image of $w$ in $GL_n(\mathbb{C})$ is not the identity matrix.

This is just a bit of hacking -- we want to enforce that for some $i, j$ the $i,j$ entry of $I_n$ differs from $X$. I.e., some $\delta_{i, j} - X_{i,j}$ is non-zero. For each $i, j$ introduce new variables $z_{i, j}$ and $r_{i, j}$ and add equations

- $z_{i, j} (\delta_{i, j} - X_{i, j}) - (1 - r_{i, j})$
- $z_{i, j} r_{i, j}$

You can check that these force $r_{i, j}$ to be 0 if $\delta_{i, j} - X_{i, j}$ is non-zero, and 1 if $\delta_{i, j} - X_{i, j}$ is 0.

Now we want to say that some $r_{i, j}$ is 0. This we can do by adding $\prod_{i, j} r_{i, j}$ to our system of equations.

**At this point, a solution to the system over $k$ is exactly a $k$ representation that sends $x$ to a non-trivial element of $GL_n(k)$.** We assumed such a representation exists for $\mathbb{C}$. Thus, the lemma proved below yields a representation for some finite field. But this is a contradiction, as we showed earlier that there are no representations over finite fields. Thus, there cannot be representations over $\mathbb{C}$ either.

**Edit:** After some reflection I realized we can do without the results of the paper in this context by proving an easier version of a similar theorem.

**Lemma.** Fix any prime $p$. If $f_1(x_1, \dots, x_N), \dots f_m(x_1, \dots, x_N) \in \mathbb{Z}[x_1, \dots, x_N]$ have a common zero over $\mathbb{C}$, then they have a common zero over some finite field of characteristic $p$.

*Proof.*

Consider the ideal $I_p = (f_1, \dots, f_m) \subseteq \mathbb{F}_p[x_1, \dots, x_N]$.

First suppose this ideal is trivial (i.e., equal to $\mathbb{F}_p[x_1, \dots, x_N]$). Then there exist some $g_1, \dots, g_m \in \mathbb{F}_p[x_1, \dots, x_N]$ with
$$
g_1 f_1 + \dots + g_m f_m = 1 \mod p
$$
Now let $G_i \in \mathbb{Z}[x_1, \dots, x_N]$ be any polynomial which when reduced mod $p$ is $g_i$. Then
$$
G_1 f_1 + \dots + G_m f_m = 1 + a p
$$
for some $a \in \mathbb{Z}$.
Then we have
$$
\frac{G_1}{1 + ap} f_1 + \dots + \frac{G_m}{1+ ap} f_m = 1
$$
which means the ideal $(f_1, \dots, f_m) \subseteq \mathbb{C}[x_1, \dots, x_N]$ is trivial. But that means there can be no solution to the $f_i$ over $\mathbb{C}$, which is a contradiction to our assumption.

So, it must be the case that $I_p$ is not trivial. Then, by the weak nullstellensatz, the $f_i$ have a common solution $\alpha_1, \dots, \alpha_m$ in $\overline{\mathbb{F}_p}$, the algebraic closure of $\mathbb{F}_p$.
Since each of the $\alpha_i$ is contained is some finite extension of $\mathbb{F}_p$, we can take a finite extension $k$ of $\mathbb{F}_p$ that contains all the $\alpha_i$. So we are done: $k$ is our finite field of characteristic $p$ which has a solution $(\alpha_1, \dots, \alpha_m)$ to our system of equations.

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