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.



Measuring how far from being cocompact a lattice is 
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".


