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Let $G\subset \mathrm{SL}_2(\mathbb R)$ be a subgroup such that $\mathrm{SL}_2(\mathbb Z)\subset G$.

What are the possible groups such that $\mathrm{SL}_2(\mathbb Z)\subset G$ is of finite index? Is $G=\mathrm{SL}_2(\mathbb Z)$ the only possibility?

What if we replace $\mathrm{SL}_2$ by $\mathrm{Sp}_{2n}$?

Question. Suppose that $\mathrm{Sp}_{2n}(\mathbb Z)\subset G$ is of finite index and $G\subset \mathrm{Sp}_{2n}(\mathbb{R})$. Is $G= \mathrm{Sp}_{2n}(\mathbb{Z})$?

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Even for $Sp_{2g}$ the only possibility is $Sp_{2g}(\mathbb{Z})$. To see this, suppose $\Gamma \subset Sp_{2g}(\mathbb{R})$ is a subgroup containing $Sp_{2g}(\mathbb{Z})$ as a finite index subgroup. Suppose we prove that $\Gamma $ consists of rational symplectic matrices. The fact that $Sp_{2g}(\mathbb{Z}_p)$ is a $maximal$ compact subgroup of $Sp_{2g}(\mathbb{Q}_p)$ for all primes $p$ (and strong approximation) implies that $Sp_{2g}(\mathbb{Z})$ is a maximal arithmetic subgroup of $Sp_{2g}(\mathbb{Q})$ (edited: if $\Gamma \subset Sp_{2g}(\mathbb{Q})$ contains $Sp_{2g}(\mathbb{Z})$ as a finite index subgroup, then $\Gamma $ is the integral symplectic group ) and hence this implies that $\Gamma =Sp_{2g}(\mathbb{Z})$.

Thus it remains to prove that $\Gamma $ consists of rational matrices. This is essentially a Galois cohomology argument. Since $\Gamma$ contains $Sp_{2g}(\mathbb{Z})$ as a finite index subgroup, in particular, there is a finite index subgroup $N\subset Sp_{2g}(\mathbb{Z})$ $normalised$ by $\Gamma$. $N$ acts irreducibly on $\mathbb{C}^{2g}$ (for example, snce $N$ is Zariski dense in the complex symplectic group). Thus, by Burnside's theorem, there is a basis $e_1,\cdots,e_N$ of $2g\times 2g$ rational matrices consisting of elements of $N$.

Suppose $x\in \Gamma \subset Sp_{2g}(\mathbb{R})\subset Sp_{2g}(\mathbb{C})$. Under conjugation by $x$, the basis $e_i$ goes into elements of $N \subset Sp_{2g}(\mathbb{Z})$ and hence $xe_ix^{-1}$ is a rational linear combination of the matrices $e_i$. If $\sigma \in Aut(\mathbb{C})$ then $\sigma (x)e_i\sigma (x)^{-1}=xe_ix^{-1}$. The irreducibility of $N$ now implies that the cocycle $x^{-1}\sigma (x)$ is a scalar matrix $\lambda$. Since $x$ preserves the symplectic form, this implies that $\lambda ^2=1$, and hence $\lambda =\pm 1$. If $\lambda =1$ for all $\sigma$ then $x$ is rational, and that is what we set out to prove.

If the Galois cocycle $\lambda =\lambda _{\sigma}= -1$, then we may write $\lambda =i^{-1}\sigma (i)$ (where $i^2=-1$) and hence $h=i^{-1}x$ is a rational matrix: $h^{-1}\sigma (h)=1$ for all $\sigma$. But, $\Gamma $ consists of real matrices, and hence $x$ is a real matrix, and hence, the equation $x=ih$ for some rational matrix $h$ is impossible. Thus, $\lambda =-1$ is ruled out.

So essentially, the fact that over the $p$-adic field $\mathbb{Q}_p$, the group $Sp_{2g}(\mathbb{Z}_p)$ is maximal compact implies that $Sp_{2g}(\mathbb{Z})$ is a subgroup of $Sp_{2g}(\mathbb{R}$ such that every intermediate subgroup contains $Sp_{2g}(\mathbb{Z})$ as an infinite index subgroup.

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    $\begingroup$ $Sp_{2g}(\mathbf{Z})$ is not maximal in $Sp_{2g}(\mathbf{Q})$ (an intermediate subgroup is $Sp_{2g}(\mathbf{Z}[1/2])$). Actually, a finitely generated group cannot be maximal in an infinitely generated group since any group is generated by any of its maximal subgroups and one further element. $\endgroup$ – YCor Oct 13 '16 at 4:56
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    $\begingroup$ That is not the claim: every intermediate subgroup contains the integral symp group as an infinite index subgroup. Pls read carefully $\endgroup$ – Venkataramana Oct 13 '16 at 6:30
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    $\begingroup$ Oops, you are right. I meant to write every intermediate subgroup contains the integral subgroup of infinite index $\endgroup$ – Venkataramana Oct 13 '16 at 6:34
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    $\begingroup$ @Venkataramana Thank you for your answer. Is it a general fact of life that, if $G$ is a connected reductive group over $\mathbb Z$, then $G(\mathbb Z)$ is a maximal arithmetic subgroup of $G(\mathbb Q)$? $\endgroup$ – Honing Oct 13 '16 at 8:13
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    $\begingroup$ @Honig: not quite. If $G$ is a Chevalley group which is simply connected, then yes. Otherwise, there are conditions. There are results by Adler about these matters (there is an article in the Algebraic Groups and Discontinuous subgroups edited by Borel, where these issues are discussed). $\endgroup$ – Venkataramana Oct 13 '16 at 9:02
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$G = SL_2(\mathbb{Z})$ is indeed the only possibility. This follows from the structure of the quotient orbifold $\mathcal{O} = \mathbb{H}^2 / SL_2(\mathbb{Z})$, which is a once-punctured sphere with a $\mathbb{Z}/2\mathbb{Z}$ cone point and a $\mathbb{Z}/3\mathbb{Z}$ cone point. The key fact is that the orbifold $\mathcal{O}$ is not a proper orbifold-covering space of any other oriented 2-orbifold. The proof of this fact is essentially an Euler characteristic calculation: given that $$\chi(\mathcal{O}) = 2 - 1 - (1 - \frac{1}{2}) - (1 - \frac{1}{3}) = -\frac{1}{6} $$ one then verifies that no fraction of the form $-\frac{1}{6n}$ with $n \ge 2$ can be the Euler characteristic of any finite type oriented 2-orbifold having at least one puncture.

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    $\begingroup$ Thank you for this nice argument. To show that there is no finite type oriented 2-orbifold with Euler characteristic $-\frac{1}{6n}$, I would argue using Riemann-Hurwitz (applied to the coarse spaces). This would get a bit messy I presume. Is there a more direct way of proving that the Euler characteristic is never $-\frac{1}{6n}$? $\endgroup$ – Honing Oct 13 '16 at 8:07
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    $\begingroup$ It's a messy proof by cases. To start, an orbifold $\mathcal{P}$ obtained from a genus $g$ surface $S_g$ satisfies $$\chi(\mathcal{P}) = 2-2g-\#(\text{punctures}) - \sum_{n \ge 2} \#(\text{$\mathbb{Z}/n\mathbb{Z}$ cone points})(1-\frac{1}{n})$$ Assume $\chi(\mathcal{P}) = -\frac{1}{6n}$ with $n \ge 2$, and that $\mathcal{P}$ has 1 puncture (since $\mathcal{O}$ has 1). It follows that $g = 0$. Also, $\mathcal{P}$ has between 1 and 2 cone points (since $\mathcal{O}$ has 2), so it has 2 (by hyperbolicity). Enumerate the assignments of cone angle for which $\chi \ge 1/12n$, and eliminate each. $\endgroup$ – Lee Mosher Oct 13 '16 at 13:42
  • $\begingroup$ I think the answer as written isn't quite right because of counterexamples like the $(2,3,7)$ triangle group that have no cusps. To fix it, add the hypothesis of at least one puncture (because $\cal O$ has one). $\endgroup$ – Noam D. Elkies Jan 20 '17 at 16:28
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I answer here the case of $\mathrm{SL}_2$, with actually an argument carrying over $\mathrm{SL}_n$. Namely, $\mathrm{SL}_n(\mathbf{Z})$ is maximal among lattices in $\mathrm{SL}_n(\mathbf{R})$. Indeed, let $\Lambda$ be an overgroup of finite index.

First assume that $\Lambda\subset\mathrm{SL}_n(\mathbf{Q})$. Then there is a bound on denominators in $\Lambda\subset\mathrm{M}_n(\mathbf{Q})$. It follows that the subgroup of $\mathbf{Q}^n$ generated by the $g\mathbf{Z}^n$ when $g$ ranges over $\Lambda$, is a lattice. So $\Lambda$ preserves this lattice and hence is conjugate to a subgroup of $\mathrm{SL}_n(\mathbf{Z})$. But since the automorphism group of $\mathrm{SL}$ preserves the volume, this implies that $\Lambda=\mathrm{SL}_n(\mathbf{Z})$.

Now we have to show that $\Gamma$ is contained in $\mathrm{SL}_n(\mathbf{Q})$. To show this, let us first understand orbits in $\mathbf{R}^n$ of finite index subgroups $\Xi$ of $\mathrm{SL}_n(\mathbf{Z})$. Start from a nonzero vector $v$, we wish to describe the closure $X$ of the additive subgroup generated by the orbit $\Lambda v$. Note that $X$ is the closure of $Av$, where $A$ is the $\mathbf{Z}$-subalgebra generated by $\Gamma$. For some $m\ge 1$, all elementary matrices $e_{ij}(m)=I+E_{ij}(m)$ belong to $\Xi$, and hence $E_{ij}(m)$ belongs to $A$. So, denoting by $(e_i)$ the canonical basis, $mv_je_i$ belongs to $X$ for all $i\neq j$. If the $v_j$ and $v\in X$ for $j\neq i$ generate a dense subgroup of $\mathbf{R}$, we deduce that the line $\mathbf{R}e_i$ is contained in $X$, and reiterating, we see that $\mathbf{R}e_j$ is contained in $X$ for all $j$ and hence that $X=\mathbf{R}$. We can do this as soon as $n\ge 3$ and $v$ is not scalar multiple of a rational vector. If $n=2$, assuming that $v$ is not multiple of a scalar vector, we can assume $v=(1,t)$ with $t$ irrational. Then $X$ contains the lattice $t\mathbf{Z}\times\mathbf{Z}$. Clearly $v$ has infinite order modulo this lattice, so the closure of $\mathbf{Z}v$ modulo this lattice is non-discrete, hence $X$ is not discrete, thus contains a line. Applying either $e_{12}(m)$ or $e_{21}(m)$ to this line yields another line and hence $X=\mathbf{R}^2$.

To summarize, we have shown that for every $v$ that is not scalar multiple of a rational vector, the subgroup generated by $\Xi v$ is dense. Thus, the lattices preserved by $\Xi$ are all scalar multiples of rational lattices (i.e., contained in $\mathbf{Q}^n$).

If $g\in\Lambda$, there exist two finite index subgroups $\Xi_1$ and $\Xi_2$ of $\mathrm{SL}_n(\mathbf{Z})$ such that $g\Xi_1=\Xi_2$. From the previous fact, we deduce that $g$ maps rational lattices to scalar multiples of rational lattices. Thus we can write $g=\lambda g'$ with $g'$ a rational matrix. The determinant condition implies that $1=\lambda^n\det(g)$, and thus $\lambda^n$ is rational. Also, $\lambda$ is unique in $\mathbf{R}^*/\mathbf{Q}^*$ and $g\mapsto \lambda=\lambda_g$ is a homomorphism into this group. The kernel is the intersection $\Lambda\cap\mathrm{SL}_n(\mathbf{Q})$, which has been shown to be $\mathrm{SL}_n(\mathbf{Z})$. So $\mathrm{SL}_n(\mathbf{Z})$ is normal in $\Lambda$.

So now we need to understand the normalizer of $\mathrm{SL}_n(\mathbf{Z})$. Using that $\mathrm{SL}_n(\mathbf{Z})$ is transitive on primitive elements in $\mathbf{Z}^n$, we see that the lattices preserved by $\mathrm{SL}_n(\mathbf{Z})$ are precisely the scalar multiples of $\mathrm{Z}^n$. So if an element normalizes it and preserves the volume, it has to preserve $\mathrm{Z}^n$. This shows that $\mathrm{SL}_n(\mathbf{Z})$ is self-normalized in $\mathrm{SL}_n(\mathbf{R})$, and finally $\Lambda=\mathrm{SL}_n(\mathbf{Z})$.

A similar argument shows that every lattice containing a finite index subgroup of $\mathrm{SL}_n(\mathbf{Z})$ is actually contained in a conjugate of $\mathrm{SL}_n(\mathbf{Z})$ by some rational matrix.

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    $\begingroup$ what is asked is $Sp_{2g}(\mathbb{Z})$ and not $SL_n(\mathbb{Z})$. $\endgroup$ – Venkataramana Oct 13 '16 at 3:59
  • $\begingroup$ It was not at all clear that you meant this to be a partial answer and that you were giving a more general argument for SL(n). I tried to remove the downvote (in the light of your remarks) but it is locked. Next time, I will wait till it is clear what is meant (not what is written). $\endgroup$ – Venkataramana Oct 13 '16 at 4:48
  • $\begingroup$ @YCor Thank you for your answer. I'm glad to see the result is also true for $SL_n(\mathbb Z)$. I was wondering about the following concerning your answer. Suppose that $G$ contains $Sp(2g,\mathbb Z)$ as a finite index subgroup and is contained in $Sp(2g,\mathbb Q)$. Does it follow from your argument in the second paragraph of your answer that $G = Sp(2g,\mathbb Z)$? $\endgroup$ – Honing Oct 13 '16 at 8:10
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    $\begingroup$ @Honing note that Venky proved this (granted the fact that $Sp_{2g}$ of $p$-adic integers is maximal compact); it does not follow directly from what I said but the argument can be completed: you get an invariant lattice, of finite index over $\mathbf{Z}^{2g}$. You have to show that this is a multiple of $\mathbf{Z}^{2g}$ and this can indeed be shown by hand, using that the action of $\mathrm{Sp}_{2g}(\mathbf{Z})$ is irreducible on $\mathbf{Z}/p\mathbf{Z}^{2g}$ for every $p$. $\endgroup$ – YCor Oct 13 '16 at 8:56
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See the paper: Ramanathan, K. G., 'Discontinuous Groups II', Nachr. Akad. Wiss. Gottingen, II. Math-Phys. Klasse, (1964), 145-164. for an elementary proof without using big tool as Venk's proof.

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