I apologize in advance if this is well-known, but I can't seem to find the answer in the literature. Let me be precise about my question. I am looking for concrete examples of locally compact Hausdorff groups $G$ such that that there exists $a, b \in G$ with $a \ne b$, but for any continuous representation $\pi : G \to \operatorname{GL}_n(k)$ with $k$ a valuated field and $n$ a positive integer, we have $\pi(a) = \pi(b)$.

I am also interested in a weaker form: the case where $k$ is the field $\mathbb{C}$ of complex numbers.

If it turns out that there are no such groups, please indicate a reference or sketch a proof.

Also a relevant statement which my gut tells me should be true, but I don't know enough Lie theory yet to give a complete proof: for every real Lie group $G$, there are enough finite dimensional, continuous complex representations of $G$ to distinguish points of $G$. I was thinking maybe this can be proved by exploiting the close relationship between representations of a Lie group and representations of its Lie algebra. It would be nice if someone could give a sketch if this can indeed be achieved or describe why this "hand-waving" actually does not work.

Edit: Sorry for the poor choice of terminology. To truly get the "weaker form" as YCor pointed out, I should replace "valuated field" by "a field $k$ with an absolute value" $|\cdot|: k \to \mathbb{R}_{\ge 0}$, such that (1) $|ab| = |a| |b|$; (2) $|a| = 0$ if and only if $a=0$; (3) $|1 + a| \le 2$ for all $a$ with $|a| \le 1$ (or equivalently by a well-known argument, the triangle inequality). To exclude the discrete topology induced by this absolute value, we exclude the trivial absolute value that $|a| = 1$ for all $a \in k^\times$.

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    $\begingroup$ I didn't check carefully but I suspect Higman's group will serve as a (discrete) counterexample in the $k={\bf C}$ case. en.wikipedia.org/wiki/Higman_group $\endgroup$
    – Terry Tao
    Dec 7, 2021 at 1:55
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    $\begingroup$ $SL_2\mathbb C$ is simply connected, but $SL_2\mathbb R$ has infinite fundamental group. Its double cover is a counterexample. It is connected, so it doesn't have any $p$-adic representations. Its finite dimensional real reprentations yield representations of its Lie algebra, which complexify to complex representations of $SL_2\mathbb C$, which doesn't have a double cover so can't detect the kernel of the extension. $\endgroup$ Dec 7, 2021 at 1:57
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    $\begingroup$ The object to understand, for a group $G$, is the intersection $K_G$ of all kernels of all finite-dimensional continuous representations over all valued fields (and similarly $K'_G$ defined considering finite-dimensional representations). You're asking when $K_G=\{1\}$ or $K'_G=\{1\}$. For instance for compact groups $G$, Peter-Weyl ensures $K'_G=\{1\}$. On the other hand, if $G$ is a finitely generated group then its "finite residual" (the intersection of all its finite index subgroups) is contained in $K_G\cap K'_G$. In particular if $G$ has no nontrivial finite quotient, $K_G=K'_G=G$. $\endgroup$
    – YCor
    Dec 7, 2021 at 8:28
  • $\begingroup$ As regards the difference between $K_G$ and $K'_G$: it seems to me that $K_G=G$ for every connected group $G$, since every valued field is totally disconnected (if I understand what you mean by valued field). While $K_G=\{1\}$ for many connected Lie groups. At the opposite, for $\mathrm{SL}_n(\mathbf{Q}_p)$ we have $K_G=G$ and $K'_G=\{1\}$. $\endgroup$
    – YCor
    Dec 7, 2021 at 8:32
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    $\begingroup$ You've ignored my previous comment while editing: I explained that the "weaker form" is not weaker. Unless you mean by "valued field" something other that what I imagine (which is in particular totally disconnected). $\endgroup$
    – YCor
    Dec 7, 2021 at 17:26

3 Answers 3


There is an example which satisfies something much stronger: there exist nontrivial groups $G$ such that any homomorphism (not even necessarily continuous) $\pi:G\to GL_n(k)$ for any field $k$ (and, in fact, any commutative integral domain) is trivial, so in particular for any $a,b\in G$ we have $\pi(a)=\pi(b)$. Since any group can be given a locally compact Hausdorff topology, namely the discrete topology, these will in particular answer your question.

As for examples of such groups $G$, we can take any finitely generated which has no finite quotients, e.g. the Higman group mentioned in the comment by Terry Tao.

This result is apparently due to Mal'cev, but I don't have a reference at hand so here is a sketch of an argument. It is enough to show that if you have a finitely generated subgroup $\Gamma$ of a group $GL_n(R)$ for some integral domain $R$ (e.g. the image of $G$ under a representation), then $\Gamma$ is residually finite (hence trivial, if $\Gamma$ is a quotient of $G$ which has no finite quotients). The main idea of the proof is to replace $R$ by some ring which is finitely generated over $\mathbb Z$, meaning it is a quotient of a polynomial algebra over $\mathbb Z$. Then $R$ is a Jacobson ring, so in particular the intersection of all maximal ideals of $R$ is trivial, and moreover for all maximal ideals $m$ of $R$ we have $R/m$ finite. Now if $\gamma\in\Gamma$ is any nontrivial element, then there is a maximal ideal $m$ not containing all coefficients of $\gamma$, so $\gamma$ has nontrivial image in the finite group $GL_n(R/m)$.

  • $\begingroup$ I marked this as accepted answer though I couldn't completely check all the details. It would be more helpful if you could provide a reference when you have the time. $\endgroup$ Dec 10, 2021 at 13:36
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    $\begingroup$ @RickSternbach You can find the details for instance in section 10 of these notes $\endgroup$
    – Wojowu
    Dec 10, 2021 at 13:46
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    $\begingroup$ In fact, there are groups which have no non-trivial homomorphism into $\text{G}(R)$ for any commutative unital ring $R$. For this, take an infinite simple group with property (T), see appendix A in arxiv.org/abs/1608.06265. There exists finitely presented such groups by the work of Caprace-Remy, see arxiv.org/abs/math/0607664. $\endgroup$
    – Uri Bader
    Dec 12, 2021 at 8:57
  • $\begingroup$ @UriBader Thank you for the references! I believed this more general statement to be the case too, but I couldn't find it in the literature. $\endgroup$
    – Wojowu
    Dec 12, 2021 at 10:08

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 $p$.


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

  1. 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.

  2. 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.

  • $\begingroup$ "infinite simple" should be "finitely generated infinite simple". Otherwise $\mathrm{SL}_3(\mathbf{Q})$ is a counterexample to the first assertion. $\endgroup$
    – YCor
    Dec 7, 2021 at 8:29
  • $\begingroup$ By the first assertion do you mean “an infinite simple group G has no non-trivial representations in finite fields”? This is true because the image is a finite quotient of G of which there is only the trivial group. I further imposed the group should be finitely presented in the second paragraph $\endgroup$ Dec 7, 2021 at 19:06
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    $\begingroup$ Sorry, indeed I didn't see you assumed to be finite and thought you were implicitly using Malcev's theorem that f.g. subgroups of $GL_n$ over any field, are residually finite. So my point is that a finitely generated group with no nontrivial finite quotient has no nontrivial (finite-dim) linear representation over any field. By the way, Malcev's result is classical and finite generation is enough (no finite presentability is needed). $\endgroup$
    – YCor
    Dec 8, 2021 at 8:28
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    $\begingroup$ (...) Malcev's result is proved by showing that every f.g. domain is residually a finite field. Once this is known, it follows that every f.g. group having a non-trivial $n$-dimensional representation over some field, also has a non-trivial $n$-dimensional representation over some finite field. $\endgroup$
    – YCor
    Dec 8, 2021 at 8:31
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    $\begingroup$ Just pointing out that you mean Thompson's group $T$ or $V$, not $F$ ($F$ is not simple). $\endgroup$ Dec 12, 2021 at 0:55

Towards the weaker form of the question, i.e., when $k = \mathbb{C}$, one can also give the following example.

First, recall the following fact. Any finite dimensional complex representation of a totally disconnected locally compact group, which is continuous with respect to the complex topology factors through a discrete quotient, see the Uri Bader's answer to this question. In particular it will be smooth (i.e., the stabilizer of any vector will be open).

Example. Now, let $p$ be a prime. The $p$-adic group $G = {\rm GL}_2(\mathbb{Q}_p)$ is locally profinite. By the above fact, any finite-dimensional continuous complex representation of $G$ will be smooth. Now, it is a well-known fact that only finite-dimensional smooth representations of $G$ are one-dimensional. (E.g., one checks that the kernel of any such representation contains all upper and lower triangular matrices, hence contains ${\rm SL}_2(\mathbb{Q}_p)$.) Then, for any $a,b \in {\rm SL}_2(\mathbb{Q}_p)$ one has $\pi(a) = \pi(b)$ for all $\pi$ as above.

Remark. The fact above breaks down without the finite-dimensionality assumption: see, for example, this answer.

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    $\begingroup$ I don't think it's "weaker", it's a variant of the question. That continuous complex representations, vs continuous representation over valued fields, separate points, are independent conditions, as I explained in a comment (where I explicitly mentioned that $\mathrm{SL}_2(\mathbf{Q}_p)$ has no nontrivial continuous complex representation). $\endgroup$
    – YCor
    Dec 7, 2021 at 12:26

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