# Examples of locally compact groups that do not admit enough finite dimensional representations

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 $$a \in k^\times$$.

• 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 Dec 7 '21 at 1:55
• $SL_2\mathbb C$ 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. Dec 7 '21 at 1:57
• 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, $K_G=K'_G=G$.
– YCor
Dec 7 '21 at 8:28
• 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 $K'_G=\{1\}$.
– YCor
Dec 7 '21 at 8:32
• 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).
– YCor
Dec 7 '21 at 17:26

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 $$GL_n(R/m)$$.

• 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. Dec 10 '21 at 13:36
• @RickSternbach You can find the details for instance in section 10 of these notes Dec 10 '21 at 13:46
• 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. Dec 12 '21 at 8:57
• @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. Dec 12 '21 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 $$p$$.

Proof.

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.

• "infinite simple" should be "finitely generated infinite simple". Otherwise $\mathrm{SL}_3(\mathbf{Q})$ is a counterexample to the first assertion.
– YCor
Dec 7 '21 at 8:29
• 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 Dec 7 '21 at 19:06
• 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).
– YCor
Dec 8 '21 at 8:28
• (...) 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.
– YCor
Dec 8 '21 at 8:31
• Just pointing out that you mean Thompson's group $T$ or $V$, not $F$ ($F$ is not simple). Dec 12 '21 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.

• 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).
– YCor
Dec 7 '21 at 12:26