Background. Let $\mathbb H= \lbrace z \in \mathbb C \; | \; Im(z)>0 \rbrace$ be the upper half-plane, and let $\Gamma \subset PSL(2,\mathbb R)\times PSL(2,\mathbb R)$ be an irreducible lattice. More concretely $\Gamma$ is a discrete subgroup non-commensurable with a product $\Gamma_1 \times \Gamma_2$ and such that the quotient of $\mathbb H \times \mathbb H$ by $\Gamma$ has finite volume.

Examples of such irreducible lattices come from the natural embedding of $PSL(2,\mathcal O_K) \to PSL(2,\mathbb R) \times PSL(2,\mathbb R)$ where $K$ is any totally real quadratic number field and the embedding is given by $A \mapsto (\sigma_1(A),\sigma_2(A))$ with $\sigma_1,\sigma_2$ being the two distinct embeddings of $K$ into $\mathbb R$.

Let's call the quotient of $\mathbb H \times \mathbb H$ by an irreducible lattice a Hilbert modular surface when the quotient is a quasi-projective surface, even if this is not the standard terminology. Usually the term Hilbert modular surface is reserved to the quotient of $\mathbb H \times \mathbb H$ by certain sublattices of $PSL(2,\mathcal O_K)$.

Back in the seventies and eighties there was some activity around the determination of totally real number fields $K$ for which $X(K)$, a desingularization of a compactification of $\mathbb H\times \mathbb H/PSL(2,\mathcal O_K)$, has Kodaira dimension smaller than two. Much of the research on the subject is summarized on van der Geer's book Hilbert Modular Surfaces. There, if I remember correctly, one can find an exhaustive (and very short) list of number fields for which $X(K)$ is rational.

Question. Let $\mathcal S$ be the set of conjugacy classes of irreducible lattices $\Gamma \subset PSL(2,\mathbb R)\times PSL(2,\mathbb R)$ for which the quotient $\mathbb H\times \mathbb H/ \Gamma$ is rational. Is $\mathcal S$ finite? In other words, is the number of rational Hilbert modular surfaces finite ?

  • 1
    $\begingroup$ "A nice fact about Hilbert modular surfaces, even in this larger sense, is that the spaces of modular forms induce embeddings of them on projetive spaces as quasi-projective surfaces." Just out of curiosity, what makes it obvious that there are enough "modular forms" for a completely general lattice to make this the case? $\endgroup$
    – JSE
    Commented Jan 30, 2010 at 3:25
  • $\begingroup$ It is certainly not obvious, and may be even wrong. I remember reading somewhere that irreducible lattices on $SL(2,\mathbb R)^2$ are commensurable to arithmetic lattices, but perhaps my memory is betraying me. Anyway, I am interested in the lattices with quasi-projective quotients. I will edit the question accordingly. Thanks. $\endgroup$ Commented Jan 30, 2010 at 4:22

2 Answers 2


The answer is (probably) yes.

A theorem of Margulis et al. shows that an irreducible lattice in a Lie group is arithmetic unless the group is isogenous to $SO(1,n)\times(\text{compact})$ or $SU(1,n)\times(\text{compact})$.

A theorem of Mumford shows that the quotient of a hermitian symmetric domain by a neat arithmetic subgroup is of logarithmic general type in the sense of Iitaka (hence not rational). (An arithmetic subgroup is neat if the subgroup generated by the eigenvalues of any element of the subgroup is torsion-free. Every sufficiently small subgroup is neat).

For a discussion of the first theorem, see Section 5B of Witte Morris' book, Introduction to Arithmetic Groups, https://arxiv.org/abs/math/0106063.

For the second theorem, see:

  • Mumford, D. Hirzebruch's proportionality theorem in the noncompact case, Invent. Math. 42 (1977), 239--272. MR0471627, DASH version

Added: As @moonface points out, this doesn't prove it.

The congruence subgroup problem is known for $SL_{2}$ over totally real fields $F$ other than $Q$. Let $\Gamma$ be an irreducible lattice in $SL_{2}(\mathbb{R})\times SL_{2}(\mathbb{R})$. By Margulis, it is arithmetic, hence congruence. Moreover, after conjugating we may suppose that $\Gamma\subset\Gamma(1)=SL_{2}(O)$. Now $\Gamma\backslash\mathbb{H}\times\mathbb{H}$ covers $\Gamma(1)\backslash \mathbb{H}\times\mathbb{H}$, and so if the first is rational, so is the second (Zariski 1958; requires dimension 2). Thus, we have a finite list of the possible totally real quadratic extensions $K$ to work with. For some $N$, $\Gamma\supset\Gamma(N)$, and we may suppose that $\Gamma(N)$ is neat. Since $\Gamma(1)/\Gamma(N)$ is finite, for any integer $N$ we get (in sum) only finitely many rational Hilbert modular surfaces of level $N$.

The problem remains to show that, for each totally real quadratic field on our list, every rational Hilbert modular surface is of level $N$ for some fixed $N$. Apart from looking case by case, I don't see how to do this (but it is surely true).

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    $\begingroup$ I do not follow: surely there are infinitely many non-neat arithmetic subgroups? $\endgroup$
    – moonface
    Commented Jan 30, 2010 at 18:05

I don't know the answer to this question, but here's a possible strategy to prove finiteness, following the approach of Zograf, as extended by Long and Reid for congruence arithmetic fuchsian groups of genus 0 (i.e. birational to $\mathbb{CP}^1$).

Lattices in $SL_2(\mathbb{R})^2$ are congruence arithmetic groups, by results of Margulis. There should be a uniform lower bound on the first eigenvalue of the Laplacian for arithmetic lattices, but I do not know a reference (I only know this in certain other contexts, such as rank one arithmetic lattices). Then by results of Bourguignon, Li, and Yau, the first eigenvalue times the volume of a rational surface should be universally bounded. Thus, the volume should be bounded, and therefore there should be only finitely many. I'm not quite sure about the eigenvalue bound though, and I might be misunderstanding some algebraic geometric issues for orbifolds.

Addendum: Actually, I think this argument can only work for Hilbert modular surfaces which embed in $\mathbb{CP}^2$ as a quasi-projective variety, which is probably much stronger than assuming it is a rational surface. The lattice can have torsion, but the point stabilizers must have quotient locally $\mathbb{C}^2$, which is quite restrictive on the torsion.

  • $\begingroup$ +1: The result about all arithmetic subgroups being of congruence type is the key difference from the one-dimensional case. By the way, could you give a specific reference to a paper of Margulis (or somewhere else) where this congruence-subgroup property is proven? $\endgroup$ Commented Jan 30, 2010 at 9:18
  • $\begingroup$ MR1978430 (2005g:20082) Sury, B. The congruence subgroup problem. An elementary approach aimed at applications. Texts and Readings in Mathematics, 24. Hindustan Book Agency, New Delhi, 2003. xvi+301 pp. ISBN: 81-85931-38-0 $\endgroup$ Commented Jan 30, 2010 at 10:14
  • $\begingroup$ The original reference would be Serre: ams.org/mathscinet-getitem?mr=272790 $\endgroup$
    – Ian Agol
    Commented Jan 31, 2010 at 1:15

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