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.

- How many sublattices of the standard lattice are there? Here sublattice just means subgroup. There are infinitely many of course. Less trivially, how many sublattices of are there of index at most ?
- How many lattices are there in of covolume ? Here a lattice means a discrete subgroup of rank , 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- 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- lattices are in some way parameterised by . Actually, we are counting each lattice several times here, because each lattice has a stabliser isomorphic to . Thus, the space of covolume- lattices can be identified with the quotient space .

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

It is a classical theorem due to Minkowski, which we prove in a moment, that is a lattice in the unimodular group . We thus consider the volume of , 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 be the usual Lebesgue measure on , normalized (as usual) so that , and note that is invariant under the left multiplication action of . Given closed , we denote the union of all the line segments which start at and end in , and we define

This function extends to a nontrivial Borel measure in the usual way, and inherits regularity and both left- and right-invariance from . Thus is unimodular, and is a Haar measure. We consider this to be the natural choice of Haar measure.

Now let be a measurable fundamental domain for . Then, for ,

But is asymptotic, as , to , which is just the number of sublattices of of index at most . Thus we have reduced question 2 to question 1.

To finish the proof of Minkowski’s theorem, it suffices to show that there are sublattices of of index at most , and this much can be accomplished by a simple inductive argument. If has index at most , then by Minkowski’s convex bodies theorem there exists some nonzero of . Without loss of generality, . But then is a lattice in of index at most . Since there are no more than choices for and, by induction, at most choices for , there are at most choices for .

But with a more careful analysis, this calculation can actually be carried out explicitly. An elementary(-ish) calculation shows that the number of subgroups of of index exactly is the coefficient of in the Dirichlet series of

where is the Riemann zeta function. Thus, by Perron’s formula, if ,

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

by

which tends to as . Thus there are about

sublattices of of index at most , and

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

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