# 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

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

# Total ergodicity

A measure-preserving transformation ${T}$ of a (finite) measure space ${(X,\Sigma,\mu)}$ is called ergodic if whenever ${A\subset X}$ satisfies ${T^{-1}A=A}$ we have ${\mu(A)=0}$ or ${\mu(A^c)=0}$. To aid intuition, it is useful to keep in mind the ergodic theorem, which roughly states that ergodicity equals equidistribution.

Following a discussion today, I’m concerned here with the ergodicity of ${T^n}$ for ${n\in\mathbf{N}}$. This is an entertaining topic, and I thank Rudi Mrazovic for bringing it to my attention. Here’s an amusing example: While ${T}$ being ergodic does not generally imply that ${T^2}$ is ergodic, ${T^2}$ being ergodic does imply that ${T^4}$ is ergodic.

Let’s start with a quite simple observation.

Lemma 1 If ${n\in\mathbf{N}}$ and ${T^n}$ is ergodic then ${T}$ is ergodic.

Proof: Suppose ${T^{-1}A=A}$. Then ${T^{-n}A=A}$, so ${\mu(A)=0}$ or ${\mu(A^c)=0}$. $\Box$

What is the picture of ${T}$ being ergodic and ${T^n}$ not being ergodic? Certainly one picture to imagine is a partition ${X = E\cup T^{-1}E\cup\cdots\cup T^{-(d-1)}E}$ with ${1 and ${T^{-d}E=E}$, or in other words a factor ${X\rightarrow \mathbf{Z}/d\mathbf{Z}}$. Possibly surprisingly, this is the whole story.

Theorem 2 If ${T}$ is ergodic and ${T^n}$ is not ergodic then for some divisor ${d>1}$ of ${n}$ there exists a partition of ${X}$ as ${E\cup T^{-1}E\cup\cdots\cup T^{-(d-1)}E \cup N}$ where ${T^{-d}E=E}$ and ${\mu(N)=0}$.

Proof: Since ${T^n}$ is not ergodic there exists a set ${A}$ with ${T^{-n}A=A}$ and ${\mu(A)>0}$ and ${\mu(A^c)>0}$. For any subset ${S\subset \mathbf{Z}/n\mathbf{Z}}$ let ${X_S}$ consist of those ${x\in X}$ such that

$\displaystyle \{j\in\mathbf{Z}/n\mathbf{Z} : T^j x \in A\} = t + S$

for some ${t\in\mathbf{Z}/n\mathbf{Z}}$. (Note that because ${x\in A}$ iff ${T^n x\in A}$, the sentence “${T^j x \in A}$” makes sense for ${j\in \mathbf{Z}/n\mathbf{Z}}$.) Note that ${T^{-1}X_S = X_S}$, so ${\mu(X_S)=0}$ or ${\mu(X_S^c)=0}$. Since ${X = \bigcup_S X_S}$, we must have ${\mu(X_S^c)=0}$ for some ${S}$.

Let ${d}$ be the smallest positive period of ${S}$. Note that ${d=1}$ iff ${S=\emptyset}$, contradicting ${\mu(A)>0}$, or ${S=\mathbf{Z}/n\mathbf{Z}}$, contradicting ${\mu(A^c)>0}$, so ${d>1}$. Let

$\displaystyle E = \bigcap_{j\in S} T^{-j} A.$

(Again note that ${T^{-j}A}$ makes sense for ${j\in\mathbf{Z}/n\mathbf{Z}}$.) Then ${\mu(T^{-i}E \cap T^{-j} E) = 0}$ unless ${i+S = j+S}$, i.e., iff ${i-j}$ is divisible by ${d}$, in which case ${T^{-i}E=T^{-j}E}$. Thus we have the desired partition

$\displaystyle X_S = E \cup T^{-1}E \cup \cdots \cup T^{-(d-1)}E.$

This finishes the proof. $\Box$

Note as a corollary that “only the primes matter” insofar as which ${n\in N}$ make ${T^n}$ ergodic: if ${n>1}$ and ${T^n}$ is not ergodic then for some prime ${p|n}$, ${T^p}$ is not ergodic. Combining this with the initial lemma, which shows that if ${T^n}$ is ergodic then ${T^p}$ is ergodic for every ${p|n}$, we have thus proved one half of the following theorem.

Theorem 3 Given a set ${S\subset\mathbf{N}}$, there exists a measure-preserving system ${(X,\Sigma,\mu,T)}$ such that ${S = \{n\in\mathbf{N} : T^n \text{ is ergodic}\}}$ if and only if ${S}$ is empty or the set of ${n\in\mathbf{N}}$ not divisible by any ${p\in\mathcal{P}}$, for some set of primes ${\mathcal{P}}$.

It remains only to construct a measure-preserving system for a given set ${\mathcal{P}}$ of primes. For ${\mathcal{P}}$ finite this can be achieved even with finite spaces. In general it suffices to look at ${\prod_{p\in\mathcal{P}} \mathbf{Z}/p\mathbf{Z}}$.