This is a naive question but I hope that the answers will be educational. When is it the case that a finitely presented group $G$ admits a faithful $2$-dimensional complex representation, e.g. an embedding into $\text{GL}_2(\mathbb{C})$? (I am mostly interested in sufficient conditions.)

I think I can figure out the finite groups with this property (they can be conjugated into $\text{U}(2)$ and taking determinants reduces to the classification of finite subgroups of $\text{SU}(2)$ and an extension problem) as well as the f.g. abelian groups with this property (there can't be too much torsion). But already I don't know what finitely presented groups appear as, say, congruence subgroups of $\text{GL}_2(\mathcal{O}_K)$ for $K$ a number field.

What can be said if you are given, say, a nice space $X$ with fundamental group $G$? I hear that in this case linear representations of $G$ are related to vector bundles on $X$ with flat connection.

  • $\begingroup$ You can always embed $O_K$ into the complex numbers, so that will do for $GL(2,O_K)$, or not? $\endgroup$
    – Marc Palm
    Commented Jul 23, 2012 at 15:04
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    $\begingroup$ A minimum necessary condition is residual finiteness. But it should be even stronger. The group must live in GL_2 of a finitely generated ring. Such groups somehow are close to being free. For example GL_2(Z) is virtually free. Experts will say more. $\endgroup$ Commented Jul 23, 2012 at 15:07
  • $\begingroup$ @Mrc: yes, I am aware of that. My issue is that I already don't know what finitely presented groups appear as nice subgroups of $\text{GL}_2(\mathcal{O}_K)$. $\endgroup$ Commented Jul 23, 2012 at 15:52
  • $\begingroup$ The easiest thing is to consider fundamental groups of hyperbolic 3-manifolds, and analyse when their holonomy lifts from $PSL_2(\mathbb C)$ to $GL_2(\mathbb C)$. The latter involves obstruction theory considerations. On the other hand it is clear that many (most?) fp groups do not embed into $GL_2(\mathbb C)$. It would help if you could provide some motivation and explain what groups you care about. $\endgroup$ Commented Jul 23, 2012 at 20:10
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    $\begingroup$ There's this book (which is essentially about computing $SL_2$ character varieties of groups): books.google.com/… $\endgroup$
    – Ian Agol
    Commented Jul 31, 2012 at 17:19

5 Answers 5


If the group $G$ does not have property FA, then a necessary and sufficient condition is that the group embeds in $\operatorname{GL}_2(\mathcal{O}_K)$, for some number field $K$ (although there are such subgroups which do not have property FA). This follows from Bass-Serre theory. Of course, this begs the question of classifying finitely presented subgroups of $\operatorname{GL}_2(\mathcal{O}_K)$ with property FA.

More generally, Bass-Serre theory implies that a general finitely generated subgroup of $\operatorname{GL}_2(K)$ will have a graph of groups decomposition into subgroups of $\operatorname{GL}_2(\mathcal{O}_K)$ for some number fields $K$.

The geometrization theorem and ending lamination theorems classify discrete subgroups of $\operatorname{PSL}_2(\mathbb{C})$ (which up to finite-index embed in $\operatorname{SL}_2(\mathbb{C})$), by their topological type as a 3-orbifold and the ending lamination data.

You ask about congruence subgroups of $\operatorname{GL}_2(\mathcal{O}_K)$. If $K=\mathbb{Q}$ or $K=\mathbb{Q}(\sqrt{-D})$, for some $D>0$, then the group is a discrete non-uniform lattice in $\operatorname{PSL}_2(\mathbb{R})$ or $\operatorname{PSL}_2(\mathbb{C})$, and one may classify the congruence subgroups of $\operatorname{SL}_2(\mathbb{Z})$ by a result of Tim Hsu (more generally, I think there exists and algorithm to determine if a finite-index subgroup of $\operatorname{GL}_2(\mathcal{O}_{\mathbb{Q}(\sqrt{-D})})$ is congruence, but I don't know if it's written down - I could describe it for you though if you're interested). More generally, one can determine if a discrete non-uniform arithmetic lattice in $\operatorname{PSL}_2(\mathbb{R})$ or $\operatorname{PSL}_2(\mathbb{C})$ is congruence.

Otherwise, Serre essentially showed that a finite-index subgroup of $\operatorname{GL}_2(\mathcal{O}_K)$ will have the congruence subgroup property (and thus, non-uniform lattices in a product $(\mathbb{H}^2)^k\times (\mathbb{H}^3)^l$ will have this property if $k+l>1$).

For examples of groups which don't have property FA, there's a paper of Calegari and Dunfield which constructs an ascending HNN extension subgroup of $\operatorname{SL}_2(\mathbb{C})$.

There are many necessary conditions which show that various groups cannot embed in $\operatorname{GL}_2(\mathbb{C})$, some of which you describe. But I think a general classification is beyond reach at this point.

As you say, if a space $X$ has a $\mathbb{C}^2$ bundle with flat connection, then you get a representation of $G=\pi_1(X)$ into $\operatorname{GL}_2(\mathbb{C})$. The space of such flat bundles is computable if $G$ is finitely presented, it amounts to computing the character variety of $G$ into $\operatorname{GL}_2(\mathbb{C})$. However, it is difficult to tell if there is a faithful representation. If you can solve the word problem in $G$, then in principle one can determine if a representation is not faithful. Also, it seems difficult to certify that a representation is faithful, except if it is discrete. The difficulty is to find a nice fundamental domain for the action on a product of symmetric spaces on which the group acts discretely (in fact, it might not exist).

  • $\begingroup$ @Ian: word problem is solvable for any finitely generated group of matrices with entries in a commutative ring. See e.g. Miller's survey, after theorem 5.1: ms.unimelb.edu.au/~cfm/papers/paperpdfs/msri_survey.all.pdf $\endgroup$ Commented Jul 23, 2012 at 17:33
  • $\begingroup$ @ Igor: good point! I guess what I mean is whether given the presentation, there is a procedure which finds the solution to the word problem. $\endgroup$
    – Ian Agol
    Commented Jul 23, 2012 at 17:44
  • $\begingroup$ I've deleted the part about unsolvable word problem. $\endgroup$
    – Ian Agol
    Commented Jul 23, 2012 at 17:48
  • $\begingroup$ I am puzzled how given a group with solvable word problem one could check whether there is a faithful representation into $GL_2(\mathbb C)$? Are there specific groups for which the method works other than discrete subgroups of $PSL_2(\mathbb C)$? Say how would it work for the mapping class group? $\endgroup$ Commented Jul 23, 2012 at 20:25
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    $\begingroup$ Oh, I see that Henry's answer above sheds light on my question. Maybe this is what you meant. $\endgroup$ Commented Jul 23, 2012 at 20:36

For a finitely generated group $G$, a necessary and sufficient condition to be embeddable into $\text{GL}_2(\mathbf{C})$ is the following strengthening of being residually finite: for every finite subset $F$ of $G$ and any $m$ there exists a finite field $K$ of characteristic $>m$ and a homomorphism from $G$ to $\text{GL}_2(K)$ that is injective on $F$.

Indeed, if this condition is satisfied, write $G=\bigcup F_n$ (increasing union) and consider such homomorphisms $G\to\text{GL}_2(K_n)$ with $K_n$ of characteristic $>n$. If $K$ is an ultraproduct of the $K_n$ then it is isomorphic to a subfield of $\mathbf{C}$ and $G\to\text{GL}_2(K)$ is an injective homomorphism. For the converse, we have to know that for every finitely generated domain of characteristic zero $R$ the intersection of maximal ideals is reduced to 0 (it's indeed a Jacobson ring). It follows that $\text{GL}_2(R)$ satisfies the above property. Indeed let $F$ be a finite subset. For every non-identity element $b$ in $FF^{-1}$ pick a nonzero entry of $b-1$ and let $x$ be the product of all these nonzero entries. Take a maximal ideal $M$ not containing $m!x$. Then $R/M$ is a finite field of characteristic $>n$ in which $x$ is nonzero, so the homomorphism $\text{GL}_2(R)\to\text{GL}_2(R/M)$ is injective on $F$.

If you remove the condition "characteristic $>m$", you similarly characterize being isomorphic to a subgroup of $\text{GL}$ over an arbitrary field $K$. Also this works the same with $\text{GL}_k$ and $\text{SL}_k$, etc.


This is a very nice question which, as Agol says, is probably out of reach at the moment. To say that there is a classification of the fp subgroups of $GL_2(\mathbb{C})$ would be to say that the set of presentations of those subgroups is recursively enumerable and has solvable Isomorphism Problem. It's hard to guess which way to jump for either of these properties: the Isomorphism Problem is known to be unsolvable for finitely presented subgroups of $GL_N(\mathbb{Z})$ for some large $N$, by work of Bridson--Miller and Haglund--Wise.

There is one nice, tangential but positive, result that I know of. Oddly enough, it says that we have some sort of limited decision-theoretic understanding of which groups are not subgroups of $GL_2(\mathbb{C})$. Apparently it was known to Mal'cev; Daniel Groves and I re-discovered it for ourselves, but then found it in Lubotzky and Segal's book.

It's well known that most nice (more precisely, Markov) classes of finitely presented groups aren't recursively recognizable, by the Adian--Rabin Theorem. This applies to subgroups of $GL_2(\mathbb{C})$, essentially because the word problem is solvable, as Igor Belegradek pointed out above. However, in nice cases, it turns out that the word problem is the only obstruction.

More precisely, Groves, Manning and I call a class $\mathcal{C}$ of fp groups recursive modulo the word problem if $\mathcal{C}\cap\mathcal{D}$ can be recursively recognized in any class of groups $\mathcal{D}$ in which the word problem is uniformly solvable. This holds for the trivial group, abelian groups, nilpotent groups of class at most $k$ for fixed $k$, free groups, Sela's limit groups, surface groups, 3-manifold groups...

There is some hope that this could hold for subgroups of $GL_2(\mathbb{C})$. As I said above, it's unknown whether the class of these subgroups is recursively enumerable. However, it is true that the complement of this class is recursively enumerable modulo the word problem. That is, the following holds:

Theorem: Let $\mathcal{L}_n$ be the set of all finite presentations of subgroups of $GL_n(\mathbb{C})$. There is a Turing machine that determines in finite time if a presentation $P\notin\mathcal{L}_n$, using a solution to the word problem in $P$.

The key point is that a group is in $\mathcal{L}_n$ if and only if it is fully residually $\mathcal{L}_n$, and this latter property can be reduced to the decidability of the elementary theory of $\mathbb{C}$ (Tarski).


Added details:

As far as I know, the theorem above isn't written down anywhere. The proof I had in mind is essentially an application of the following theorem.

Theorem (Mal'cev): A finitely generated group $G$ is a subgroup of a group $GL_n(\mathbb{C})$ if and only if $G$ is fully residually $GL_n(\mathbb{C})$; that is, for any finite subset $X\subseteq G\smallsetminus 1$ there is a homomorphism $f:G\to GL_n(\mathbb{C})$ with $1\notin f(X)$.

This is Theorem 16.4.1 in Lubotsky and Segal's book Subgroup growth.

Proof of the first theorem. Let $G$ be the group presented by $P$. Using the word problem, enumerate balls $B(n)$ in $G$. If $G$ is not embeddable in $GL_n(\mathbb{C})$ then, by Mal'cev's theorem, for some $n$, every homomorphism $f:G\to GL_n(\mathbb{C})$ kills a non-trivial element of $B(n)$. This last statement can be rephrased as a system of equations and inequations over $\mathbb{C}$ with integral coefficients, and hence solved. QED

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    $\begingroup$ HW: nice theorem; where is this written? $\endgroup$ Commented Jul 23, 2012 at 20:39
  • $\begingroup$ Igor - I don't know that it's actually written down anywhere, but I tried to explain how to deduce it from a theorem in Lubotzky and Segal's book. $\endgroup$
    – HJRW
    Commented Jul 24, 2012 at 13:19

First, instead of subgroups of $GL(2,C)$ it suffices to consider subgroups of $SL(2, C)$. I will also restrict to subgroups which are not virtually solvable (since these should be easy to classify). Lastly, I will consider characterization of subgroups of $SL(2, C)$ "up to abstract commensuration" since the criterion is much cleaner in this setting. (Recall that two abstract groups $G_1, G_2$ are called "commensurable" if they contain isomorphic finite-index subgroups.)

Definition. Let $p$ be a prime and $c$ an integer. A $p$-congruence structure of degree $c$ for a group $\Gamma$, is a descending chain of finite-index normal subgroups of $\Gamma$: $\Gamma=N_0\supset N_1 \supset ... \supset N_k ... $, such that:

(i) $\bigcap_{k=0}^\infty N_k= 1$;

(ii) $N_1/N_k$ is a finite $p$-group for every $k\ge 2$;

(iii) $d(N_i/N_j)\le c$ for all $j\ge i\ge 1$, where $d(H)$ denotes the minimal number of generators of a group $H$.

Theorem. A finitely-generated (non-virtually solvable) group $\Gamma$ is commensurable to a subgroup of $SL(2,C)$ if and only if $\Gamma$ admits (for some prime $p$) a $p$-congruence structure of degree $c=3$.

This theorem is an application of "A Group Theoretic Characterization of Linear Groups" by Alex Lubotzky (Journal of Algebra, 113 (1988), 207-214), combined with the classification of Lie algebras of dimension $\le 3$ over fields. Note that Lubotzky does not get a group-theoretic characterization of subgroups of $SL(n,C)$ for particular values of $n$ (even up to commensurability), however, for $n=2$ the situation is better than for the general $n$ since dimension of the Lie group is so small in this case.


I'm not sure whether it is worth remarking on, but if $G$ is any irreducible subgroup of ${\rm GL}(2,\mathbb{C}),$ its non-central Abelian subgroups have some very restrictive properties: If $A$ is a non-central Abelian subgroup of $G$, then $C_{G}(A)$ is also Abelian and $[N_{G}(A):C_{G}(A)] \leq 2.$


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