## A tale of two arithmetic lattices

For almost 50 years, Paul Sally was a towering figure in mathematics education at the University of Chicago. Although he was 80 years old, and had two prosthetic legs and an eyepatch (associated with the Type 1 diabetes he had his whole life), it was nevertheless a complete shock to our department when he passed away last December, and we struggled just to cover his undergraduate teaching load this winter and spring. As my contribution, I have been teaching an upper-division undergraduate class on “topics in geometry”, which I have appropriated and repurposed as an introduction to the classical geometry and topology of surfaces.

I have tried to include at least one problem in each homework assignment which builds a connection between classical geometry and some other part of mathematics, frequently elementary number theory. For last week’s assignment I thought I would include a problem on the well-known connection between Pythagorean triples and the modular group, perhaps touching on the Euclidean algorithm, continued fractions, etc. But I have introduced the hyperbolic plane in my class mainly in the hyperboloid model, in order to stress an analogy with spherical geometry, and in order to make it easy to derive the identities for hyperbolic triangles (i.e. hyperbolic laws of sines and cosines) from linear algebra, so it made sense to try to set up the problem in the language of the orthogonal group $O(2,1)$, and the subgroup preserving the integral lattice in $\mathbb{R}^3$.

First, let’s recall the definition of the hyperboloid model of the hyperbolic plane. In $\mathbb{R}^3$ we consider the quadratic form $Q:= x^2+y^2-z^2$, and let $O(2,1)$ denote the group of real $3\times 3$ matrices preserving this form. The vectors with $Q(v,v)=-1$ are those lying on a 2-sheeted hyperboloid; the positive sheet H is the one consisting of vectors whose z coefficient is positive, and $O^+(2,1)$ is the subgroup preserving this sheet. For each vector v in H, the tangent space $T_vH$ is naturally isomorphic to the set of vectors $w$ with $Q(v,w)=0$; i.e. the subspace of vectors “perpendicular” to v with respect to the form. The restriction of the quadratic form to the tangent space is positive definite, so it makes H into a Riemannian manifold, in such a way that $O^+(2,1)$ acts by isometries. This group acts transitively, and the stabilizer of a point is conjugate to $O(2)$; thus H with this metric is homogeneous and isotropic, and is a model for the hyperbolic plane.

Another model is the upper half-space model of the hyperbolic plane. In this model, we define H to be the subspace of complex numbers with positive imaginary part, and let $SL(2,\mathbb{R})$ denote the group of real $2\times 2$ matrices, which acts on H by fractional linear transformations:

$\begin{pmatrix} a & b \\ c & d \end{pmatrix}: z \to (az+b)/(cz+d)$

This action is not faithful; the subgroup $\pm \text{Id}$ acts trivially, so the action descends to the quotient $PSL(2,\mathbb{R})$. The group acts transitively, and the stabilizer of a point is conjugate to $PSO(2,\mathbb{R})$; thus (again) H is homogeneous and isotropic, and is a model for the hyperbolic plane. This reflects the exceptional isomorphism of groups $PSL(2,\mathbb{R}) \cong SO^+(2,1)$.

The subgroup $PSL(2,\mathbb{Z})$ acts discretely with finite covolume (i.e. it is a lattice in the Lie group $PSL(2,\mathbb{R})$); the quotient is the modular surface — an orbifold with underlying surface a sphere with one puncture, and two cone points with order 2 and 3 respectively; one sometimes calls this the $(2,3,\infty)$-triangle orbifold, since it is made from two semi-ideal hyperbolic triangles with angles $\pi/2,\pi/3,0$ at the vertices (the third “ideal” vertex is at infinity, and corresponds to the puncture). There is an associated tessellation of the hyperbolic plane by such triangles whose symmetry group is $PSL(2,\mathbb{Z})$ in which the ideal vertices lie exactly at the rational numbers (plus infinity) on the boundary of hyperbolic space. Thus $PSL(2,\mathbb{Z})$ acts in a natural way on the set of rational numbers union infinity, which can be thought of as the projective line over $\mathbb{Q}$. As an abstract group, $PSL(2,\mathbb{Z})$ is the free product of two cyclic groups of order 2 and 3 respectively, corresponding to the matrices

$\begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}$    and     $\begin{pmatrix} 1 & 1 \\ -1 & 0 \end{pmatrix}$

and all torsion elements in $PSL(2,\mathbb{Z})$ are conjugate to these elements or their inverse (note that these matrices have orders 4 and 6 respectively in $SL(2,\mathbb{Z})$; it is only in $PSL(2,\mathbb{Z})$ that they have orders 2 and 3).

The group $PSL(2,\mathbb{Z})$ is an example of what is known as an arithmetic lattice; roughly speaking, the arithmetic lattices in semisimple Lie groups G are those with “integer entries”, in a suitable sense.  Arithmetic lattices are characterized by the existence of many hidden symmetries — i.e. their finite index subgroups have surprisingly large normalizers in G. More formally, for a subgroup $\Gamma$ in G, we define the commensurator of $\Gamma$ to be the subgroup of G consisting of elements g such that the conjugate of $\Gamma$ by g intersects $\Gamma$ in a finite index subgroup. With this definition, Margulis famously proved that the arithmetic lattices are precisely those whose commensurators are dense, and that all other lattices (i.e. the non-arithmetic ones) have a commensurator which is discrete (and hence contains the lattice itself with finite index). In $PSL(2,\mathbb{R})$, all the arithmetic lattices are derived from quaternion algebras over totally real number fields. Roughly speaking, if $K$ is a totally real number field — i.e. a finite extension of $\mathbb{Q}$ obtained by adjoining some root of an integer polynomial with all real roots — and if $A$ is a quaternion algebra over $K$, then we can find a group $\Gamma$ consisting of “integer” elements of $A$ of norm 1. Each real embedding of $K$ embeds $A$ in a quaternion algebra over $\mathbb{R}$; this is either the Hamiltonian quaternions (which is a division algebra), or the algebra of $2\times 2$ real matrices (which has zero divisors). Then $\Gamma$ embeds as a lattice in a product of copies of $SU(2)$ and $SL(2,\mathbb{R})$, one for each real embedding in the Hamiltonian quaternions and in $M_2(\mathbb{R})$ respectively. The $SU(2)$ factors are compact, so if there is exactly one $SL(2,\mathbb{R})$ factor, $\Gamma$ embeds as a lattice in it, and projects to a lattice in $PSL(2,\mathbb{R})$; these are exactly the arithmetic lattices.

It is a theorem of Borel that the only way to get an arithmetic lattice in $PSL(2,\mathbb{R})$ which is not cocompact is to take $K=\mathbb{Q}$ — in other words, $PSL(2,\mathbb{Z})$.

OK, now — how to reproduce this picture in the hyperboloid model? The most natural guess is to look at $O^+(2,1;\mathbb{Z})$ — the group of $3\times 3$ matrices with integer entries preserving the quadratic form $Q$ and the positive sheet of the hyperboloid. So, what exactly is this group? Let’s let A be a matrix in this group, and denote its column vectors by u,v,w. One obvious matrix to take is the identity matrix; for that matrix, the vector w is $(0,0,1)$ which lies on the hyperboloid H, whereas the vectors u and v are orthonormal vectors in $T_wH = w^\perp$. But this property of a triple of vectors is preserved by the action of any element of $O^+(2,1)$, and therefore in general there is a bijection between such matrices and triples u,v,w where w lies on H, and u,v are orthonormal vectors in $T_wH$.

Now consider the condition that the entries of the matrix be integers. Let’s abstract the discussion slightly. Suppose V is a real vector space of dimension n, with a symmetric nondegenerate quadratic form Q. Let L be a lattice in V; this is a slightly different use of the word “lattice” than above (at least in flavor) — it means a discrete cocompact additive subgroup, isomorphic as a group to $\mathbb{Z}^n$. We suppose that the lattice L is integral and unimodular; the first condition means that $Q(v_1,v_2)$ is an integer for all $v_1,v_2$ in L, and the second means that the $n\times n$ matrix with entries $Q(e_i,e_j)$ has determinant 1 or -1 for any basis $e_1,e_2,\cdots,e_n$ of L. Now, for any nonzero vector $v$ the linear function $Q(\cdot,v):L \to \mathbb{Z}$ has image of finite index (because Q is nondegenerate and L has full rank) and therefore the kernel $L \cap v^\perp$ has rank (n-1). If $v$ has norm 1 or -1, then $L \cap v^\perp$ is itself an integral unimodular lattice in the vector space $v^\perp$ with respect to the quadratic form which is the restriction of Q.

In $\mathbb{R}^3$ with the quadratic form Q as above, suppose we can find an integer vector w on the hyperboloid H. Then the intersection of $T_wH$ with the lattice of integer vectors has rank 2, and since the form Q is positive definite there, we can find an orthonormal basis u,v of integer vectors for $T_wH$. Hence there is a matrix A in $O^+(2,1;\mathbb{Z})$ taking $(0,0,1)$ to w, and $O^+(2,1;\mathbb{Z})$ acts transitively on such vectors, with stabilizer isomorphic to $O^+(2;\mathbb{Z})$, the group of symmetries of the square. If we want to restrict attention to orientation-preserving symmetries, then $SO^+(2;\mathbb{Z})$ is cyclic of order 4, generated by

$R:=\begin{pmatrix} 0 & 1 & 0 \\ -1 & 0 & 0 \\ 0 & 0 & 1 \end{pmatrix}$

Let’s find another matrix. An integral vector w on the hyperbolic H is a triple of integers x,y,z so that $x^2+y^2-z^2 = -1$; one simple example is $(2,2,3)$, and then it is straightforward to find vectors $(2,1,2)$ and $(1,2,2)$ for u and v. This gives the matrix

$T:=\begin{pmatrix} 2 & 1 & 2 \\ 1 & 2 & 2 \\ 2 & 2 & 3 \end{pmatrix}$

Actually, it is pretty easy to see that no other integral vector on H is closer to $(0,0,1)$ than $(2,2,3)$, since $2^2-1=3$ is not a sum of two squares. Let’s let $\Gamma$ be the group generated by R and T. Some experimentation with fundamental domains confirms that this group is a lattice, and that the quotient is a sphere with one puncture and two orbifold points of orders 2 and 4; in particular, this is the entire group $SO^+(2,1;\mathbb{Z})$, and its quotient is the $(2,4,\infty)$ triangle orbifold.

So, this group is certainly not $PSL(2,\mathbb{Z})$. In fact, a rotation of order 4 realized as an element of $PSL(2,\mathbb{R})$ necessarily has a trace of $\sqrt{2}$, so it can’t even have rational entries. But wait — this is surely an arithmetic lattice (for any conceivable definition of arithmetic), and therefore corresponds to some lattice derived from a quaternion algebra over a totally real number field. Since it is not cocompact, the only possibility is that the number field is $\mathbb{Q}$, so that this lattice is commensurable with $PSL(2,\mathbb{Z})$. At this point I vaguely recall something from a course on arithmetic lattices I took from Walter Neumann over 20 years ago in Melbourne, in which he stressed that the trace field of an arithmetic lattice (i.e. the field generated by the traces of the elements, thought of as a subgroup of $PSL(2,\mathbb{R})$) is not by itself a commensurability invariant — rather the trace field generated by the squares of the elements is invariant; and the squares of the elements in this group all have integer trace after conjugating into $PSL(2,\mathbb{R})$. So mathematics is consistent after all, and I learn the surprising (to me) fact that the $(2,3,\infty)$ and $(2,4,\infty)$ triangle orbifolds are commensurable. Hard to believe I have been working with Kleinian groups for 20 years without noticing that before . . .

Here’s a picture of the tiling of the hyperbolic plane whose symmetry group is $O^+(2,1;\mathbb{Z})$:

The center is the projection of $(0,0,1)$ and the adjacent 8-valent vertices are the projection of $(\pm 2, \pm 2,3)$.

(Update May 20, 2014): As galoisrepresentations points out, the fact that the field generated by traces of squares of elements is a commensurability invariant is a theorem of Alan Reid.

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### 2 Responses to A tale of two arithmetic lattices

1. A nice example! I also remember that course (assuming you mean the one that ran during summer of 93-94). It was my first job; I got \$250 a week to learn math, watch the cricket while drinking milkshakes at the student union, and have two hour lunches with Matt which usually involved a stop for gelati on Lygon street.

I think the invariant trace field was first defined in a paper by Walter with Alan (Reid). There a further 7 non-compact arithmetic triangle groups (classified by Takeuchi), some of which are more transparently commensurable to the modular group than others. For example, if you take a fundamental domain for $(\infty,\infty,\infty)$ to consist of the ideal triangle with vertices $(-1/2,\infty,1/2)$ then the corresponding group is the normalizer of the congruence subgroup $\Gamma_0(4)$, if I computed correctly.