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)


\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<d\mid n} 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}}.