## De-mystify me about nonstandard finite fields

In their paper, “Definable subgroups of algebraic groups over finite fields,” Hrushovski and Pillay write

… if V is an absolutely irreducible variety of dimension d over the finite field F_q, then the cardinality of V(F) is “roughly” q^d.  So for nonstandard q, the cardinality of V(F) is exactly q^d.

Do I have any nonstandard readers who can explain what is meant by this provocative statement?

## Do all curves over finite fields have covers with a sqrt(q) eigenvalue?

On my recent visit to Illinois, my colleage Nathan Dunfield (now blogging!) explained to me the following interesting open question, whose answer is supposed to be “yes”:

Q1: Let f be a pseudo-Anosov mapping class on a Riemann surface Sigma of genus at least 2, and M_f the mapping cylinder obtained by gluing the two ends of Sigma x interval together by means of f.  Then M_f is a hyperbolic 3-manifold with first Betti number 1.  Is there a finite cover M of M_f with b_1(M) > 1?

You might think of this as (a special case of) a sort of “relative virtual positive Betti number conjecture.”  The usual vpBnC says that a 3-manifold has a finite cover with positive Betti number; this says that when your manifold starts life with Betti number 1, you can get “extra” first homology by passing to a cover.

Of course, when I see “3-manifold fibered over the circle” I whip out a time-worn analogy and think “algebraic curve over a finite field.”  So here’s the number theorist’s version of the above question:

Q2: Let X/F_q be an algebraic curve of genus at least 2 over a finite field.  Does X have a finite etale cover Y/F_{q^d} such that the action of Frobenius on H^1(Y,Z_ell) has an eigenvalue equal to q^{d/2}?

## 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.

## Positive motivic measures are counting measures

A new, very short paper with Michael Larsen, “Positive motivic measures are counting measures” is up on the arXiv today.  I thought I’d say a bit here about where the problem came from, since we don’t do so in the paper.

In the project with Akshay that I talked about at the recent Columbia conference on rational curves on varieties, one thing you do is compute estimates for |M_n(F_q)|, where M_n is the moduli space of algebraic maps of degree n from P^1 to some fixed target variety X, and F_q is a finite field.  These inequalities turn out to be very nicely uniform in q, which leads one naturally to ask; do the proofs actually give “motivic estimates” for the class [M_n] in the Grothendieck ring K_0(Var_K), for various non-finite fields K?

Well, what does it mean for one element r of a ring R to “estimate” another element s?  It might mean that r-s is rather deep in some natural filtration on R.  Those don’t seem to be the kind of estimates our methods provide; rather, they say something more like

(r-s)^2 <= B

where B is some fixed element of K_0(Var_K).  But what does “<=” mean?  Well, it means that B – (r-s)^2 is nonnegative.  And what does “nonnegative” mean?  That’s the question.  What the proof really gives is that B – (r-s)^2 lies in a certain semiring N of “nonnegative motives” in K_0(Var_K).  Let’s not be too precise about what N is; let’s just say that it includes [V] for every variety V, and it has the property that |n(F_q)| >=0 for all q, whenever n lies in N.  In particular, that means that

(|r(F_q)| – |s(F_q)|)^2 <= B(F_q)

so that, on the level of counting points, s(F_q) really is a good estimate for r(F_q).

So one might ask:  are there other interesting positive motivic measures — that is, homomorphisms

f: K_0(Var_K) -> reals

which take N to nonnegative reals?  If so, f(s) would be a good estimate for f(r).

And the point of this note with Larsen is to say, with some regret, no — any motivic measure which assigns nonnegative values to the classes of varieties is in fact just counting points over some finite field.  Which sort of kills in its crib the initial hope of some exciting world of “motivic inequalities.”

Of course, the reals are not the only ordered ring!  As Bjorn Poonen pointed out to me, for a general field K you can find an ordered ring A and a measure

g: K_0(Var_K) -> A

which is positive in the sense that g sends every variety to an element of A greater than or equal to 0; these come from big ultraproducts of counting measures of different finite fields.  Whether these measures are “interesting” I’m not sure.

## How many points does the average curve have?

Felipe Voloch complained that I didn’t list Ruby’s BBQ in my last post as one of the charms of visiting UT. I’ll make it up to him by observing that one of the charms of visiting UT is talking math with Felipe! He asked me an interesting question, about which we had different intuitions — I’ll present the question here and those readers who have an opinion are encouraged to voice it. (Math below the fold to avoid shocking the modesty of non-mathy readers.)