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A subgroup of $SL_2(\mathbb{R})$ is called arithmetic if it is commensurable with $SL_2(\mathbb{Z})$.

An arithmetic subgroup is called congruence if it contains a subgroup of type $\Gamma(N)$ for some $N\in \mathbb{N}$.

Question: What are concrete examples of subgroups of $SL_2(\mathbb{R})$, which are arithmetic, but not congruence?

I have heard the Belyi theorem produces some examples, but I have never seen a concrete one.

Further Question: Can such things exist in higher rank Lie groups (real rank $\geq 2$)?

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This is my motivation: mathoverflow.net/questions/129637/… –  plusepsilon.de May 8 '13 at 7:37
    
Dear Marc, If $\Gamma$ is a congruence subgroup of $\mathrm{SL}(2,\mathbb Z)$, not containing $-I$ for simplicity, then there is an absolute bound on the size of an abelian quotient of $\Gamma$ whose kernel is again a congruence subgroup. The reason is that for all but finitely many primes $p$, the image of $\Gamma$ in $\mathrm{SL}(2,\mathbb Z/p^r)/{\pm I}$ is the whole group (here $p^r$ is any power of $p$), and for $p \geq 5$ (if I've not blundering) this group has no abelian quotients. So any abelian image of $\Gamma$ with congruence kernel is a quotient of the image of $\Gamma$ in ... –  Emerton May 10 '13 at 5:01
    
$\mathrm{SL}(2,\mathbb Z/q^r)$ for some finite list of primes $q$ (those that divide the level of $\Gamma$), and these images again admit only finitely many possible abelian quotients. So if $\Gamma$ is any torsion free congruence subgroup for which $\mathbb H/\Gamma$ has non-trivial $H^1$, then (identifying $\Gamma$ with $\pi_1$ of the quotient) we get a surjection $\Gamma \to \mathbb Z$, and if we choose $N$ huge, then the kernel of $\Gamma \to \mathbb Z/N$ will be a non-congruence subgroup. Now actually $\mathbb H/\Gamma$ always has non-trivial $H^1$ if it has more than one cusp, so ... –  Emerton May 10 '13 at 5:05
    
this gives many examples. (JSE's answer below is a particular case, applied to a (torsion free modification of) $\Gamma(2)$.) Regards, Matthew –  Emerton May 10 '13 at 5:06
    
Thanks Matthew for both your comments:-) –  plusepsilon.de May 10 '13 at 9:08
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4 Answers

up vote 17 down vote accepted

I would quibble with your definition of "arithmetic". There are discrete subgroups of $SL(2, \mathbf{R})$ coming from quaternion algebras, which are not commensurable with $SL(2, \mathbf{Z})$, but are commensurable with $G(\mathbf{Z})$ for some twisted form $G$ of $SL(2)$ which becomes isomorphic to the usual form over $\mathbf{R}$. These are arithmetic subgroups of $G$, but not of the usual $SL(2)$. So the notion of "arithmetic subgroup" depends on a choice of model over a number field.

With that quibble out of the way, examples of noncongruence arithmetic subgroups of $SL_2 / \mathbf{Q}$ are not too hard to come by. The easiest construction I know goes like this. We know $G = PSL(2, \mathbf{Z})$ has a rather simple presentation; it's the free product of $C_2$ and $C_3$. So if we have any other group $H$ containing elements $a,b$ of orders 2 and 3, there is a unique homomorphism $G \to H$ sending the generators to these elements.

In particular, $S_7$ is generated by two such elements, so there is a normal subgroup of $G$ with quotient isomorphic to $S_7$. But $S_7$ is not a quotient of $SL(2, \mathbf{Z} / m)$ for any $m$ (easy check once you know that $PSL(2, \mathbf{Z}/p)$ is simple for $p > 2$) hence this subgroup cannot be congruence.

(If you look not at the kernel of this homomorphism but the preimage of the stabilizer of 1, then you get a noncongruence subgroup of index 7, which is, if I remember correctly, the smallest possible.)

The congruence subgroup problem for more general reductive groups is a difficult problem and one that's far from solved. As I mentioned above, it depends on a choice of model over a number field. The group $SL(n) / \mathbf{Q}$ has no non-congruence arithmetic subgroups for $n \ge 3$ (Bass-Milnor-Serre), while $SL(2)$ over an imaginary quadratic field has lots of them (it behaves like $SL(2) / \mathbf{Q}$). There are many more recent results; googling brings up a survey by Raghunathan.

(EDIT: I suddenly realized I wasn't sure if the bit about $S_7$ being a quotient of $PSL(2, \mathbf{Z})$ was true or not. So I wrote a computer program and it is indeed true that the elements (2,3,4)(5,6,7) and (1,2)(3,5)(4,6) generate $S_7$).

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Excellent. Many thanks! –  plusepsilon.de May 8 '13 at 8:24
    
Another useful reference are K. Conrad's notes : math.uconn.edu/~kconrad/blurbs/grouptheory/SL(2,Z).pdf –  François Brunault May 8 '13 at 20:05
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To construct non-congruence subgroups of $PSL(2,\mathbf Z)$, it is easier to start from the congruence subgroup $\Gamma(2)$ — matrices congruent to identity modulo $2$, for this group is free on two generators. Since the symmetric group $S_k$ is generated by a transposition and a $k$-cycle, it is a quotient of $\Gamma(2)$. –  ACL May 8 '13 at 20:18
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@David: A helpful answer, including the clarification about passage to an algebraic group and arithmetic subgroups in the formulation. As Raghunathan's 2004 bibliography shows, the CSP has been solved in the bulk of natural cases where it comes up (especially in his work with Prasad); so it seems to be an overstatement to call the problem "far from solved". Maybe I'm misunderstanding what are the "more general reductive groups" you mention. –  Jim Humphreys May 8 '13 at 20:20
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See also mathoverflow.net/questions/118376 for more on low index example and why index 7 is the least possible. –  Misha May 8 '13 at 20:50
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Gamma(2) is a free group on two generators, so it surjects onto (Z/pZ)^2. For all odd p, the kernel is a non-congruence subgroup, corresponding to a non-congruence cover of the modular curve X(2), and this cover is none other than the Fermat curve x^p + y^p + z^p = 0. So in some sense the "Frey trick" used in the proof of Fermat can be thought of as a passage between noncongruence modular curves and classical modular curves!

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If you want to go really concrete, take any connected finite $3$-valent graph in the $2$-sphere with disks as components of the complement.

This yields a conjugacy class of genus $0$ finite index torsion free subgroups of $\Gamma(1)=PSL(2,\mathbb{Z})$. In fact the index is $6n$ if the graph has $2n$ vertices (hence $3n$ edges and $n+2$ faces). ADDED: Indeed, since $\Gamma(1)\simeq Z/2*Z/3$, there is a transitive action of $\Gamma(1)$ on oriented edges of the graph, flipping orientation for $Z/2$, and rotating around source vertex for $Z/3$. The conjugacy class of stabilizers subgroups determines the graph.

But there are clearly infinitely many such graphs (up to oriented homeomorphism), whereas there are only finitely many congruence subgroups of genus $0$ in $\Gamma(1)$ (about $33$ conjugacy classes of torsion free ones, with maximum index $60$, according to this table).

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Grant Lakeland's paper "Dirichlet-Ford Domains and Arithmetic Reflection Groups (arxiv version)" also addresses the OP's first question. In light of David Loeffler's thoughtful quibble, Lakeland's paper employs a definition of arithmetic consistent with the OP's question. Using that definition for arithmeticity, this paper exhibits maximal arithmetic reflection groups that are not congruence. In either version of that paper, Figure 2 together with section six's Claims 1,2, and 3 gives an explicit fundamental domain of a non-congruence arithmetic group. This group is an index 2 subgroup of a reflection group. (Seven other examples are provided in Lakeland's thesis.)

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