Homology of the Torelli group and negative-dimensional vector spaces

OK, not really.  You know and I know there’s no such thing as a negative-dimensional vector space.

And yet…

The Torelli group T_g is a subject of hot interest to mapping class groups people — it’s the kernel of the natural surjection from the mapping class group Γ_g to Sp_{2g}(Z).  You can think of it as “the part of the mapping class group that arithmetic lattices can’t see,” or at least can’t see very well, and as such it is somewhat intimidating.  We know very little about it, even in small genera.  One thing we do know is that for g at least 3 the Torelli group is finitely generated; this is a theorem of Johnson, and a recent paper by Andy Putman provides a small generating set.  So H_1(T_g,Q) is finite-dimensional.  (From now on all cohomology groups will be silently assigned rational coefficients.)

But a charming argument of Akita shows that, in general, T_g has some infinite-dimensional homology groups.  How do we know?  Because if it didn’t, you would be able to compute the integer χ(T_g) from the formula

χ(T_g) =  χ( Γ_g)/ χ(Sp_{2g}(Z)).

But both the numerator and denominator of the right-hand-side are known, and their quotient is not an integer once g is at least 7.  Done!

At the Park City Mathematics Institute session I visited this summer, there was a lot of discussion of what these infinite-dimensional homology groups of Torelli might look like.  We should remember that the outer action of Sp_2g(Z) on Torelli yields an action of Sp_2g(Z) on the homology of Torelli — so one should certainly think of these spaces as representations of Sp_2g(Z), not as naked vector spaces.  In the few cases these groups have been described explicitly, they are induced from finite-dimensional representations of infinite-index subgroups H of Sp_2g(Z).

I just wanted to record the small observation that in cases like this, there’s a reasonably good way to assign a “dimension” to the homology group!  Namely:  suppose G is a discrete group and H a a subgroup, and suppose that both BG and BH are homotopic to finite complexes.  (This is not quite true for G = Sp_2g(Z), but surely you’re willing to spot me a little finite level structure wherever I need it.)  Let W be a finite-dimensional representation of H and let V be the induction of W up to G.

Now if H were finite-index in G you’d have

dim V = [G:H] dim W

or, what’s the same,

dim V = χ(BH)/χ(BG) dim W

But note that the latter formula makes sense even if H is infinite-index in G!  And this allows you to assign a “dimension” to some infinite-dimensional homology groups.

For instance, consider T_2, which is not finitely generated.  By a theorem of Mess, it’s a free group on a countable set of generators; these generators are naturally in bijection with cosets in Sp_4(Z) of a subgroup H containing SL_2(Z) x SL_2(Z) with index 2.  Compute the Euler characteristics of H and Sp_4 and you find that the “dimension” of H_1(T_2) is -5.

And when you ask Akita’s argument about this case, you find that the purported Euler characteristic of T_2 is 6; a perfectly good integer, but not such a great Euler characteristic for a free group to have.  Unless, of course, it’s a free group on -5 generators.

If you want to see this stuff written up a bit (but only a bit) more carefully, here’s a short .pdf version, which also includes a discussion of the hyperelliptic Torelli group in genus 3.

The braid group, analytic number theory, and Weil’s three columns

This post is about a new paper of mine with Akshay Venkatesh and Craig Westerland; but I’m not going to mention that paper in the post. Instead, I want to explain why topological theorems about the stable homology of moduli spaces are relevant to analytic number theory.  If you’ve seen me give a talk about this stuff, you’ve probably heard this spiel before.

“The mathematician who studies these problems has the impression of deciphering a trilingual inscription. In the first column one finds the classical Riemannian theory of algebraic functions. The third column is the arithmetic theory of algebraic numbers.  The column in the middle is the most recently discovered one; it consists of the theory of algebraic functions over finite fields. These texts are the only source of knowledge about the languages in which they are written; in each column, we understand only fragments.”

Let’s see how a classical question of analytic number theory works in Weil’s three languages.  Start with the integers, and ask:  how many of the integers between X and 2X are squarefree?  This is easy:  we have an asymptotic answer of the form

$\frac{6}{\pi^2}X + O(X^{1/2}) = \zeta(2)^{-1} X + O(X^{1/2}).$

(In fact, the best known error term is on order X^{17/54}, and the correct error term is conjectured to be X^{1/4}; see Pappalardi’s “Survey on k-freeness” for more on such questions.)

So far so good.  Now let’s apply the popular analogy between number fields and function fields, going over to Weil’s column 3, and ask: what’s the analogous statement when Z is replaced by F_q[T]?

Anabelian puzzle 4: What is the probability that a set of n points has no 3 collinear?

OK, this isn’t really an anabelian puzzle, but it was presented to me at the anabelian conference by Alexei Skorobogatov.

Let X_n be the moduli space of n-tuples of points in A^2 such that no three are collinear.  The comment section of this blog computed the number of components of X_n(R) back in January.  Skorobogatov asked what I could say about the cohomology of X_n(C).  Well, not a lot!  But if I were going to make a good guess, I’d start by trying to estimate the number of points on X_n over a finite field F_q.

So here’s a question:  can you estimate the number of degree-n 0-dimensional subschemes S of A^2/F_q which have no three points collinear?  It seems very likely to me that the answer is of the form

$P(1/q) q^{2n} + o(q^{2n})$

for some power series P.

One way to start, based on the strategy in Poonen’s Bertini paper:  given a line L, work out the probability P_L that S doesn’t have three points on L.  Now your first instinct might be to take the product of P_L over all lines in A^2; this will be some version of a special value of the zeta function of the dual P^2.  But it’s not totally clear to me that “having three points on L_1” and “having three points on L_2” are independent.

Lieblich’s counter counterexample example

Max Lieblich gave a great talk at WAGS yesterday about something that looks like a counterexample to the Hasse principle, but secretly isn’t!  All mistakes in this summary are my own.  For a more authoritative take on the material below, see Max’s recent arXiv preprint.

The counterexample Max countered is the central simple algebra A over the field Q(t) obtained as the tensor product of the two generalized quaternion algebras $(17,t)$ and $(13,6(t-1)(t-11))$.  This algebra has index 4, which is to say (if I understand correctly) that its Brauer class isn’t the cup product of two elements in Q(t)^*.  On the other hand, it turns out that $A \otimes \mathbf{Q}_v$ has index 1 or 2 — that is, it’s either trivial or a cup product — for all places v of Q.

This seems like an example of a situation where there’s no Hasse principle; A fails to be a cup product despite the fact that A is a cup product over every completion of Q.  But the truth, as Max explained, is more complicated.

We know (and if we don’t, we read A Course in Arithmetic to remember) that the quaternion algebra (a,b) is trivial (i.e., has index 1) over a field k precisely when the conic $x^2 - ay^2 - bz^2$ has a k-rational point.  You might ask whether there’s a similar criterion for the tensor product of two quaternion algebras to have index 2.  It turns out that (at least over Q(t)) there is indeed a variety that does this trick — it’s a coarse moduli space M parametrizing certain twisted vector bundles on — well, not quite P^1/Q, but a certain orbifold version of P^1 with a bunch of stacky points with inertia Z/2Z.  And the assertion that A has index 4 is equivalent to the assertion that M(Q) is empty.

To really talk about what M is would take us too far afield; I just want to record Max’s observation that the definition of M is truly global, in the sense that the scheme $M_{\mathbf{Q}_v}$ is not determined by $A \otimes \mathbf{Q}_v$.  In particular, the fact that $A \otimes \mathbf{Q}_v$ has index 2 doesn’t imply that M has a $\mathbf{Q}_v$-point.  And indeed, in the case at hand, M has no points over $\mathbf{Q}_{17}$.  So there is, after all, a local obstruction to A having index 2; but it’s a local obstruction which, it seems, can’t be seen except in this rather intricate geometric way.

It makes you wonder what should actually be meant by “Hasse principle.”  Suppose, for instance, you had some class C of varieties X/Q, and suppose you had some construction which attached to each X in C a variety Y/Q such that X(Q) is empty if and only if Y(Q) is empty.   Now one way to prove X(Q) empty would be to prove that Y(Q) was empty, which you could in turn prove if you knew that Y(Q_v) was empty for some place v.  Do you consider this a local obstruction to rational points on X?