## The congruence subgroup property for thin groups.

I finally had a chance to visit Yale, which (by various orderings) is the fanciest US university at which I had never given a talk (nor even visited). The town itself struck me, at first, as a cross between Oxford and New Jersey. That aside, my coffee research led me to Blue State Coffee, which was more than up to the task of preparing a decent 8 ounce latte. (As a comparison, it is significantly better than Small World Coffee in Princeton. Small World has all the correct hipster attitude without enough of the corresponding aptitude.) Mathematically, I had a great chat with Hee Oh and Gregg Zuckerman over several hours. At one point, I raised the following idle question about thin groups.

Question: Let $G = \mathrm{SL}_N(\mathbf{R})$ where $N > 2$. Let $\Gamma$ be an arithmetic lattice in $G$. Suppose that $\Phi \subset \Gamma$ is a subgroup such that the following two conditions are satisfied:

1. The Zariski closure of $\Phi$ in $G$ is $G$.
2. The induced map of profinite completions: $\widehat{\Phi} \rightarrow \widehat{\Gamma}$ is injective.

Then is $\Phi$ necessarily of finite index in $\Gamma$?

If $\Gamma = \mathrm{SL}_N(\mathbf{Z})$, then the first condition implies that the image of the induced map of profinite completions has finite index; I presume this is true more generally. Hence the question asked can be phrased as follows: “can congruence subgroups be determined by their pro-finite completions?” Alternatively, in the opposite direction, one can ask: “are there thin groups which satisfy the congruence subgroup property?” I have no particular reason to believe that the answer to the question above is positive, and I might even guess that one could write down a counter-example, but I don’t know how to write one down myself.

On the other hand, suppose that the answer to the question is positive. Then it might prove useful for determining whether, given a finitely presented group $H:=\langle G \ | \ R \rangle$ and an explicit homomorphism: $\phi: H \rightarrow \Gamma$

whether its image has finite index or (even more strongly) whether $\phi$ is an isomorphism onto a finite index subgroup. Namely, if the image of $\phi$ does not have finite index, then a positive answer to the question above would imply that $H$ must have a finite quotient which does not come from $\Gamma$, and (since finite quotients of $H$ may be enumerated) this leads to an algorithm which terminates if $\phi$ has infinite index. On the other hand, if $H$ does have such a quotient, then certainly $\phi$ will not be an isomorphism onto a finite index subgroup.

This problem explicitly came up in some work of Curt McMullen (see question 5.6 of this paper), who produced explicit maps of various finitely presented groups into lattices (not quite in $\mathrm{SL}_N(\mathbf{R})$, but one can of course ask the more general question for lattices in semi-simple groups of rank at least two) and asked whether these maps were isomorphisms onto finite index subgroups. So the hope is that (in the contexts in which one expected the answer to be negative) this could always be answered by considering the pro-finite completion of the finitely presented group in question. Alas, I believe that I explicitly tried to find non-congruence quotients of the associated explicitly presented groups (in contexts where one expected $\phi$ to have infinite index) and didn’t find any (not that I carried out this computation in anything approaching a sophisticated manner, of course).

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### 4 Responses to The congruence subgroup property for thin groups.

1. Kevin says:

Aren’t both conditions satisfied by the trivial subgroup?

• galoisrepresentations says:

The trivial subgroup is not Zariski dense in $\mathrm{SL}_N(\mathbf{R})$.

2. Pingback: Math blog roundup | Quomodocumque

3. Placi says:

For some related (mainly negative) results see Bridson-Grunewald:,Grothendieck’s problems concerning profinite completions and representations of groups ,Ann.Math.160(2004)359-373
(and of course references+citations).