Measuring how far from being cocompact a lattice is - MathOverflow

Measuring how far from being cocompact a lattice is

2012-10-24T11:47:02Z

Let $G$ be a locally compact group and $\Gamma$ a lattice (=discrete subgroup of $G$ such that $G/\Gamma$ carries a probability measure $\mu$ that is invariant under the action of $G$ by left-multiplication).

My vague question is: "How to measure the lack of cocompactness of $\Gamma$"?.

Edit: My question was indeed unclear. I am not looking for a criterion that says me whether a lattice is cocompact. Given a non-cocompact lattice, I am looking for a way of quantifying "how much non-cocompact" it is. I propose below such a measurement (and ask what can be said about it), but if somebody has other propositions I will be happy.

A natural such measurement in the case when $G$ has a compact generating set is the following. One can equip $G$ with the word length metric $d$ corresponding to some compact generating set, and consider the sequence $n \mapsto u_n=1 - \mu (B(0,n) \Gamma)$. It is easy to see that $\Gamma$ is cocompact if and only if $u_n=0$ for all sufficiently large $n$. The speed of convergence of $u_n$ to zero in some sense should measure how far we are from this ideal situation. This is related to the integrability properties of $w \mapsto d(1,w)$ on a suitably chosen fundamental domain $\Omega \in G$ for the action of $\Gamma$ on $G$ by right-multiplication.

Has this been studied? What is the typical behaviour of $u_n$ in the case of (arithmetic) lattices in Lie or algebraic groups? In a first time I would already be happy to have an answer for $\Gamma=SL(3,\mathbf Z)$. My guess is that number theorists may have already studied this.

Answer by Aakumadula for Measuring how far from being cocompact a lattice is

2012-10-24T12:12:22Z

If $G$ is a simple linear algebraic group defined over the rationals, then $G({\mathbb Z}\subset G({\mathbb R})$ is a lattice (a result of Borel and Harisch-Chandra). Moreover, $G({\mathbb R}/G({\mathbb Z})$ is non-compact if and only if $G({\mathbb Z})$ contains unipotent elements (this is called Godement's criterion).

In particular, in your case, $SL_3({\mathbb Z})$ is indeed a non-cocompact lattice in $SL_n({\mathbb R})$ although this special case was known probably to Hermite (if not earlier).

Actually, this is true for all lattices in simple Lie groups (I believe this is due to Kazhdan-Margulis).

There is aslo a result of Selberg which says that if $\Gamma $ is a lattice in a locally compact group $G$, and if $G/\Gamma $ is non-compact, then there exists a sequence of elements $\gamma _n \neq 1$ in the lattice, and a sequence of elements $g_n\in G$ such that $g_n\gamma _n g_n^{-1}$ tends to the identity.

Answer by Alex Eskin for Measuring how far from being cocompact a lattice is

2012-10-24T22:15:34Z

It seems to be that what you are asking for is roughly the measure of a neighborhood of the "cusp" of $G/\Gamma$.

For the case of $\Omega(n) = SL(n,\mathbb{R})/SL(n,\mathbb{Z})$ there is a classical calculation of a closely related quantity. Recall that $\Omega(n)$ is the moduli space of unimodular lattices in $\mathbb{R}^n$.

For $\epsilon > 0$, let $\Omega_\epsilon(n) \subset \Omega(n)$ denote the subset parametrizing lattices in which the shortest vector has length at least $\epsilon$. Then the complement of $\Omega_\epsilon(n)$ is compact (This is called Mahler compactness).

I will describe below how to estimate the measure $\mu(\Omega_\epsilon(n))$. (This is similar to the quantity you wanted to compute).

Let $f: \mathbb{R}^n \to \mathbb{R}$ be a compactly supported function. Define the "Siegel Transform" $\hat{f}: \Omega_n \to \mathbb{R}$ by $$\hat{f}(\Delta) = \sum_{v \in \Delta'} f(v),$$ where $\Delta'$ is the set of primitive vectors in the lattice $\Delta$. The Siegel integral formula (which you can prove by unfolding) says that $$\int_{\Omega_n} \hat{f} \,d\mu = \frac{1}{\zeta(n)} \int_{\mathbb{R}^n} f.$$ Now take $f$ to be the characterisric function of the ball of radius $\epsilon$ around the origin. Then the integral on the left is supported on $\Omega_\epsilon(n)$, and $\hat{f} \ge 1$ on $\Omega_\epsilon(n)$. It turns out that to compute the leading term in the asymptotics as $\epsilon \to 0$, you might as well assume that $\hat{f} = 1$ on $\Omega_{\epsilon}(n)$. You get, that as $\epsilon \to 0$, $$\mu(\Omega_\epsilon(n)) \sim \frac{1}{\zeta(n)} Vol(B(0,\epsilon)).$$

Similar calculations can be done for other non-uniform lattices but things get more technical. The terms to google for are "Eisenstein Series" and "precise reduction theory".