# 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 by Alex Eskin.

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

$\displaystyle \mu(E) = \lambda(\text{cone}(E)).$

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 ${R>0}$,

$\displaystyle \mu(\text{SL}_n(\mathbf{R})/\text{SL}_n(\mathbf{Z})) = \mu(E) = \lambda(\text{cone}(E)) = \frac{\lambda(R\,\text{cone}(E))}{R^{n^2}}.$

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

$\displaystyle \zeta_{\mathbf{Z}^n}(s) = \zeta(s)\zeta(s-1)\cdots\zeta(s-n+1),$

where ${\zeta(s) = \sum_{n\geq1} n^{-s}}$ is the Riemann zeta function. Thus, by Perron’s formula, if ${x\notin\mathbf{Z}}$,

$\displaystyle \frac{1}{x^n}\sum_{1\leq m< x} a_m(\mathbf{Z}^n) = \frac{1}{i2\pi} \int_{n+1/2-i\infty}^{n+1/2+i\infty} \zeta_{\mathbf{Z}^n}(z) \frac{x^{z-n}}{z}\,dz.$

But, by Cauchy’s residue theorem, this integral differs from

$\displaystyle \text{Res}_{z=n}\left(\frac{\zeta_{\mathbf{Z}^n}(z) x^{z-n}}{z}\right) = \frac{1}{n}\zeta(2)\zeta(3)\cdots\zeta(n)$

by

$\displaystyle \frac{1}{i2\pi} \int_{n-1/2-i\infty}^{n-1/2+i\infty} \zeta_{\mathbf{Z}^n}(z) \frac{x^{z-n}}{z}\,dz,$

which tends to ${0}$ as ${x\rightarrow\infty}$. Thus there are about

$\displaystyle \frac{1}{n}\zeta(2)\zeta(3)\cdots\zeta(n) m^n$

sublattices of ${\mathbf{Z}^n}$ of index at most ${m}$, and

$\displaystyle \mu(\text{SL}_n(\mathbf{R})/\text{SL}_n(\mathbf{Z})) = \frac{1}{n}\zeta(2)\zeta(3)\cdots\zeta(n).$

## One thought on “How many lattices are there?”

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

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