## Monday, 25 November 2013

### How many lattices are there?

The basic question is the one in the title: How many lattices are there? I intend to answer two quite different interpretations of this question.
1. How many sublattices of the standard lattice $\mathbf{Z}^n$ are there? Here sublattice just means subgroup. There are infinitely many of course. Less trivially, how many sublattices of $\mathbf{Z}^n$ are there of index at most $m$?
2. How many lattices are there in $\mathbf{R}^n$ of covolume $1$? Here a lattice means a discrete subgroup of rank $n$, and covolume refers to the volume of a fundamental domain. Again, the answer is infinitely many. Less trivially, what (in an appropriate sense) is the volume of the set of covolume-$1$ lattices?
Question number 2 obviously needs some explanation. Covolume, as can easily be proved, is equal to the determinant of any complete set of spanning vectors, from which it follows that covolume-$1$ lattices are in some way parameterised by $\text{SL}_n(\mathbf{R})$. Actually, we are counting each lattice several times here, because each lattice has a stabliser isomorphic to $\text{SL}_n(\mathbf{Z})$. Thus, the space of covolume-$1$ lattices can be identified with the quotient space $\text{SL}_n(\mathbf{R})/\text{SL}_n(\mathbf{Z})$.

This subgroup $\text{SL}_n(\mathbf{Z})$ itself is also considered a lattice, in the group $\text{SL}_n(\mathbf{R})$. Generally, let $G$ be a locally compact group, and recall that $G$ possesses a unique (up to scale) left-invariant regular Borel measure, its Haar measure $\mu$. Suppose moreover that $G$ is unimodular, i.e., that $\mu$ is also right-invariant. If $H\leq G$ is a unimodular subgroup, and we fix a Haar measure on $H$ as well, then $\mu$ descends to a unique $G$-invariant regular Borel measure on the homogeneous space $G/H$ which we also denote by $\mu$. If $\mu(G/H)<\infty$, we say that $H$ has finite covolume. For example, if $H$ is a discrete subgroup then we can fix the counting measure as a bi-invariant Haar measure. If in this case $H$ has finite covolume then we say that $H$ is a lattice.

It is a classical theorem due to Minkowski, which we prove in a moment, that $\text{SL}_n(\mathbf{Z})$ is a lattice in the unimodular group $\text{SL}_n(\mathbf{R})$. We thus consider the volume of $\text{SL}_n(\mathbf{R})/\text{SL}_n(\mathbf{Z})$, for a natural choice of Haar measure, to be the answer to question 2 above, at least in some sense. The actual computation of this volume, I understand, is originally due to Siegel, but I learnt it from this MO answer.

Let $\lambda$ be the usual Lebesgue measure on $\mathbf{R}^{n^2}$, normalized (as usual) so that $\lambda(\mathbf{R}^{n^2}/\mathbf{Z}^{n^2}) = 1$, and note that $\lambda$ is invariant under the left multiplication action of $\text{SL}_n(\mathbf{R})$. Given closed $E\subset\text{SL}_n(\mathbf{R})$, we denote $\text{cone}(E)$ the union of all the line segments which start at $0\in\mathbf{R}^{n^2}$ and end in $E$, and we define
$$\mu(E) = \lambda(\text{cone}(E)).$$
This function $\mu$ extends to a nontrivial Borel measure in the usual way, and inherits regularity and both left- and right-invariance from $\lambda$. Thus $\text{SL}_n(\mathbf{R})$ is unimodular, and $\mu$ is a Haar measure. We consider this $\mu$ to be the natural choice of Haar measure.

Now let $E\subset\text{SL}_n(\mathbf{R})$ be a measurable fundamental domain for $\text{SL}_n(\mathbf{Z})$. Then, for $R>0$,
$\mu(\text{SL}_n(\mathbf{R})/\text{SL}_n(\mathbf{Z})) = \mu(E) = \lambda(\text{cone}(E)) = \frac{\lambda(R\,\text{cone}(E))}{R^{n^2}}.$
But $\lambda(R\,\text{cone}(E))$ is asymptotic, as $m\rightarrow\infty$, to $|R\,\text{cone}(E)\cap M_n(\mathbf{Z})|$, which is just the number of sublattices of $\mathbf{Z}^n$ of index at most $R^n$. Thus we have reduced question 2 to question 1.

To finish the proof of Minkowski's theorem, it suffices to show that there are $O(R^{n^2})$ sublattices of $\mathbf{Z}^n$ of index at most $R^n$, and this much can be accomplished by a simple inductive argument. If $\Gamma\leq\mathbf{Z}^n$ has index at most $R^n$, then by Minkowski's convex bodies theorem there exists some nonzero $\gamma\in\Gamma$ of $\|\gamma\|_\infty\leq R$. Without loss of generality, $\Gamma\cap\mathbf{R}\gamma = \mathbf{Z}\gamma$. But then $\Gamma^\prime = \Gamma/(\mathbf{Z}\gamma)$ is a lattice in $\mathbf{Z}^n/(\mathbf{Z}\gamma)\cong\mathbf{Z}^{n-1}$ of index at most $R^n$. Since there are no more than $(2R+1)^n$ choices for $\gamma$ and, by induction, at most $O(R^{n(n-1)})$ choices for $\Gamma^\prime$, there are at most $O(R^{n^2})$ choices for $\Gamma$.

But with a more careful analysis, this calculation can actually be carried out explicitly. An elementary(-ish) calculation shows that the number $a_m(\mathbf{Z}^n)$ of subgroups of $\mathbf{Z}^n$ of index exactly $m$ is the coefficient of $m^{-s}$ in the Dirichlet series of
$\zeta_{\mathbf{Z}^n}(s) = \zeta(s)\zeta(s-1)\cdots\zeta(s-n+1),$
where $\zeta(s) = \sum_{n\geq1} n^{-s}$ is the Riemann zeta function. Thus, by Perron's formula, if $x\notin\mathbf{Z}$,
$\frac{1}{x^n}\sum_{1\leq m< x} a_m(\mathbf{Z}^n) = \frac{1}{i2\pi} \int_{n+1/2-i\infty}^{n+1/2+i\infty} \zeta_{\mathbf{Z}^n}(z) \frac{x^{z-n}}{z}\,dz.$
But, by Cauchy's residue theorem, this integral differs from
$\text{Res}_{z=n}\left(\frac{\zeta_{\mathbf{Z}^n}(z) x^{z-n}}{z}\right) = \frac{1}{n}\zeta(2)\zeta(3)\cdots\zeta(n)$
by
$\frac{1}{i2\pi} \int_{n-1/2-i\infty}^{n-1/2+i\infty} \zeta_{\mathbf{Z}^n}(z) \frac{x^{z-n}}{z}\,dz,$
which tends to $0$ as $x\to\infty$. Thus there are about
$\frac{1}{n}\zeta(2)\zeta(3)\cdots\zeta(n) m^n$
sublattices of $\mathbf{Z}^n$ of index at most $m$, and
$\mu(\text{SL}_n(\mathbf{R})/\text{SL}_n(\mathbf{Z})) = \frac{1}{n}\zeta(2)\zeta(3)\cdots\zeta(n).$

#### 1 comment:

1. Sean,

I can't follow all of the proof, but you have a very clear way of presenting your findings.