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).

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