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}?

One nice thing about Q2 is that I know some cases where the answer is yes:  namely, the case where X is a hyperelliptic curve and the characteristic of F_q is congruent to 2 mod 3.  When X is hyperelliptic, a theorem of Bogomolov and Tschinkel guarantees that X has an etale cover which dominates an elliptic curve E_0 with CM by Q(sqrt(-3)).  Now for every p congruent to 2 mod 3, E_0 is supersingular mod p, and in particular has a Frobenius eigenvalue of q^{1/2} whenever q is a sufficiently divisible power of p.

When I mentioned this fact to Nathan, he remarked that Q1 also has a positive answer when X is hyperelliptic!  This is a theorem of Joe Masters.

What about the general case?  Let’s jot down a careless heuristic argument and see what happens.

First of all, what’s the probability that a genus g curve Y over F_q has a Frobenius eigenvalue of q^{1/2}?  A theorem of DiPippo and Howe suggests that the number of Weyl polynomials of degree 2g is on order q^{cg^2}.  Of these, about q^{c(g-1)^2} can be expressed as (X – q^{1/2})^2 P(x) for a Weyl polynomial P of degree 2g-2.  If the characteristic polynomial of Frobenius on H^1(Y) is a randomly chosen Weyl polynomial, then the chance that it has q^{1/2} as a root should thus be exponential in -g.


Guess 1: The probability that a random genus-g curve over F_{q^d} has a positive real Frobenius eigenvalue (i.e. an eigenvalue of q^{d/2}) is on order of q^{-dg}.

Now we need to think about the fields of definition of etale covers of X.  Let C_n be the set of all degree-n covers of X/F_qbar.  Then Frobenius acts on C.  But this action is clearly not transitive; for instance, if Y -> X is a degree-n etale cover, the image of pi_1(X) in the permutation group of the n sheets is a subgroup of S_n defined up to conjugacy, the monodromy group of the cover.  (We might restrict to connected covers, i.e. to transitive monodromy groups, but I don’t think this makes a big difference.)

Guess 2: The action of Frobenius is well-modeled by a permutation which permutes randomly the covers with each monodromy group.

Suppose this is the case.  Let G in S_n be a monodromy group and let C_{n,G} be the set of covers with monodromy group G.  Then a random permutation on C_{n,G} will have one fixed point, 1/2 cycles of length 2, 1/3 cycles of length 3 and so on.  Thus, there should on average be 1/k covers in C_{n,G} whose field of definition is F_{q^k}.  These covers should have genus on order ng, where g is the genus of X.   And so, according to Guess 1, the expected number of positive real eigenvalues among the covers with monodromy group G is

\sum_{k=1}^\infty (1/k)q^{-kng}

which is dominated by its main term, q^{-ng}.

But there are a lot of subgroups of S_n; at least on order of c^{n^2}, according to a theorem of Pyber.  For large n this swamps the negative exponential in q^{-ng}, suggesting that there are indeed lots of covers of X with a positive real Frobenius eigenvalue.  In fact, it seems to suggest that you could find a cover with a ton of positive real eigenvalues; maybe even a cover Y whose Jacobian has constant * g(Y) supersingular elliptic curves as isogeny factors.  That last seems unreasonably strong, but I don’t see that it’s immediately ruled out.

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6 thoughts on “Do all curves over finite fields have covers with a sqrt(q) eigenvalue?

  1. JSE says:

    Two comments on my own post:

    1. The prediction that there exist infinitely many covers Y/F_{q^d}, some constant proportion of whose Frobenius eigenvalues are equal to q^{d/2}, is perhaps not so unreasonable; the analogue back on the 3-manifold side would be the prediction that M_f has a sequence of covers whose first Betti number grows linearly with degree. But people believe that 3-manifolds have finite covers which are large, i.e. whose fundamental group surjects onto a free group of rank 2, and this implies exactly such a statement on growth of Betti numbers.

    2. Guess 2 is more precise than necessary — if Frobenius is thought of as drawn from some group much smaller than the full group of permutations, it only increases the expected number of short cycles, and thus the expected number of degree-n covers of X defined over small extensions of F_q.

  2. Jim says:

    Wait, can you explain why you consider q1 and q2 to be analogous? q1 is about dimensions of cohomology groups of the total space, and q2 is about eigenvalues of the monodromy on the cohomology of the geometric fiber (or at least that’s what I think you mean by the notation H^1(Y,Z_l)).

  3. JSE says:

    So: if M_f is the mapping torus of some diffeo f of a Sigma_g, then passage around the circle induces a symplectic automorphism of H_1(Sigma_g), which is the analogue of the Frobenius action on H_1(Y). The Betti number of M_f is bigger than 1 just when this automorphism has an eigenvalue of 1. Of course, Frobenius _can’t_ have an eigenvalue of 1, but (and you’re welcome to reject this analogy) I take q^{1/2} to be analogous to 1 here.

  4. Shouldn’t one count the subgroups of S_n up to conjugacy? Pyber’s result (according to the link to OEIS) is about subgroups themselves, and the corresponding sequence for conjugacy classes (here) seems to grow much slower (though still quite fast).

  5. JSE says:

    But this only diminishes by a factor of n! at most, right? So it should still be exponential in n^2.

  6. Yes, of course… Certain growth rates dissolve intuition…

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