Hang Xue, genus 4 curves, “Franchetta in bounded degree”

Very interesting algebraic geometry seminar by Hang Xue today, about his paper “The height of a canonical point in the Jacobian of a genus 4 curve.”

How do you get a canonical point in the Jacobian of a genus 4 curve?

Well, that turns out to be kind of cool!

First of all, what does it mean for a point to be “canonical”?  It should mean something like “functorial,” which should in turn mean something like “defined over the whole moduli space.”   In other words:  let K be the function field of M_4/C, and let X/K be the (restriction to generic moduli) of the universal genus-4 curve.  Then we are asking for an element of the Mordell-Weil group Jac(X)(K).  Except this Mordell-Weil group is trivial!  This fact (or rather — this fact with 4 replaced by an arbitrary genus g) used to be called the Franchetta conjecture before it was proved by Beauville, Arbarello, and Cornalba.

So where does Xue’s point come from?  It’s defined, not over K, but over a quadratic extension of K.  The generic  genus-4 curve embeds canonically in a quadric surface; the two families of lines aren’t defined over K, but over a quadratic extension K’, and passing to K’ you have made the quadric surface isomorphic to P^1 x P^1.  Now there are two line bundles on P^1 x P^1 — pulling back each one to X gives you two degree-3 divisors on X, and their difference is a point on Jac(X)(K’).  By construction, it is negated by conjugation over K.

Here’s another way to construct this point.  Your generic genus-4 curve X has a degree-3 map to P^1 which is simply ramified at 12 points.  So on X you have a ramification divisor D of degree 12.  And then 2 K_X – D is your point on the Jacobian!  When Hang explained this to me I said, wait, what happened to the quadratic extension? and he pointed out to me that the generic genus-4 curve actually has two trigonal maps to P^1; and these are defined over K’.

Anyway, Hang shows that this point P generates Jac(X)(K’) up to torsion, and moreover gives a nice formula for the Neron-Tate height of the restriction of P to any family of genus-4 curves over a 1-dimensional base.

Here is an idle question.  Let d >= 1 be an integer.  For each g, let K_g be the function field of M_g and J_g the Jacobian of the generic genus-g curve.

Are there only finitely many g such that J_g has a non-torsion point over some extension of K_g of degree at most d?

When d=1, this is the Franchetta conjecture.

I have no direct need to know the answer, but it seems a nice structural question about “universal Mordell-Weil groups.”

9 thoughts on “Hang Xue, genus 4 curves, “Franchetta in bounded degree”

  1. Jason Starr says:

    Isn’t Franchetta a consequence of Harer’s theorem? Was this really proved by Beauville, Arbarello and Cornalba? Schroeer extended Franchetta to positive characteristic.

  2. Jason Starr says:

    Okay, I just checked. Arbarello and Cornalba showed that Harer’s theorem implies the Franchetta conjecture.

  3. jlk says:

    It is worth noting that there are actually 2 Franchetta Conjectures: the Weak Conjecture and the Strong Conjecture. In “Picard Groups of Moduli Spaces of Curves” (MR0895568), Arbarello and Cornalba prove that the only lines bundles on the generic universal curve C_{g} –> Spec(K) are the multiples of the dualizing sheaf. This is sometimes known as the Weak Franchetta Conjecture because one could ask for a stronger statement: That the only sections of the degree d Jacobian J^{d} –> \Spec(K) of C_{g} are the sections coming from a multiple of the dualizing sheaf.

    The Weak Franchetta Conjecture does not formally imply the Strong Franchetta Conjecture because a section of the Jacobian corresponds to a line bundle on $C_{g} \otimes L$ for $L/K$ a (possibly non-trivial extension) étale extension that satisfies a compatibility condition; More precisely, the Jacobian represents the étale sheaf associated to the functor parameterizing line bundles, not the functor itself. I think the first proof of the Strong Conjecture was given by Mestrano in her paper “Conjecture de Franchetta forte” (MR0870734).

    @Jason Starr: Do you know if anything is known about the Brauer group of the function field of $M_{g}$? (This Brauer group contains the obstruction to a section being a line bundle.)

  4. JSE says:

    This is a great point — so just checking, jlk, the assertion I made in the post is the strong one proved by Mestrano, not the weak one, is that right right?

  5. jlk says:

    Yes, the d=1 assertion in your post is the strong one; the Strong Franchetta Conjecture implies that the only K_g-valued point of the degree 0 Jacobian J_g is the identity (for g at least 3).

  6. Jason Starr says:

    @jlk: Of course the Brauer group of the function field of Mg is enormous, as is every function field of dimension at least 2. But that isn’t really the correct question. By the Weil extension property, every rational section of the relative Picard extends to a regular section at least over the open subset parameterizing curves of compact type. The Brauer group of this scheme is much smaller.

  7. Michael Thaddeus says:

    Am I to believe that, although there exists a Poincare line bundle over C x J_0(C) for any fixed smooth curve C, there exists no such line bundle over the family of such products parametrized by any nonempty open subset of M_g? Is that what the second paragraph of jlk’s remark amounts to?

  8. Jason Starr says:

    @Michael: Yes, that is precisely the issue. Of course as soon as you have a rational section, or even a “degree 1 zero-cycle”, i.e., a collection of multisections whose (relative) degrees are coprime, then you can use that to produce a Poincare bundle. So you have a Poincare bundle for a single curve over an algebraically closed field. But you don’t have a degree 1 zero-cycle for the universal curve over Mg; Franchetta tells us that the degrees of multi-sections are all divisible by 2g-2. Thus, it is a priori possible that some sections of the relative Picard variety do not correspond to line bundles.

    Of course the obstruction is a gerbe, i.e., an element of the Brauer group (so, because of Tsen’s theorem, you don’t even start to see this issue until you deal with curves over a base of “cohomological dimension 2” or higher). For Franchetta, it is the Brauer group of an open subscheme of Mgbar whose complement has codimension > 1 (I realized that the Neron model can be used in place of Weil extension to extend across delta_0). By “purity”, that is *probably* the same as the Brauer group of all of Mgbar (there are issues because of the finite quotient singularities, but I think this is okay). Using the exponential sequence and Harer’s theorem again, probably you can prove that the Brauer group of all of Mgbar is trivial: it is trapped between H^2(M,O_M), which is zero, and H^3(M,Z)_tor, which I believe is zero by Harer. I am pretty sure I looked at Mestrano’s article when Harris and Khosla were doing their work. So probably I am just repeating what I learned then (I will look again at Mestrano and check).

  9. Jason Starr says:

    I know that I should just check Mestrano, but the fog is starting to clear on its own. I believe I remember now that the key is to work with Mg,1 rather than Mg. For the universal curve of Mg,1, you do have a degree 1 zero-cycle, namely the diagonal. So the Brauer obstruction vanishes if we work over Mg,1. I believe that Harer’s theorem implies that the Cartier divisors on the universal curve over Mg,1 are all Z-linear combinations of pullbacks from Mg,1, the relative dualizing sheaf, and the diagonal divisor. Thus, the only sections of the relative Picard over Mg,1 that are pulled back from sections of the relative Picard of Mg are tensor powers of the relative dualizing sheaf.

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