# Pushforwards of Line Bundles and Stability

I recently finished reading this paper, and was wondering about a couple of things relating to theorem 1, which says that for any curve X there is a curve Y and f:Y->X such that pushforward is a dominant rational map from the Jacobian of Y to the moduli of semistable vector bundles on X (with numerical invariants fixed to make things more definite.) So I had two questions:

1) Given a morphism of curves f:Y->X, is there a good characterization of the line bundles L on Y with f_*(L) semistable (or not semistable, equivalently)?

2) Given a morphism of curves f:Y->X, is there a good characterization of which semistable bundles are in/not in the image of f_*?

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Re: first question.

For semistability, we need a homomorphism from a line bundle U -> X of a certain degree to the pushdown of a line bundle L, which is the same thing as having a section on Y of f ^ * U ^ * \otimes L.

This can be expressed in terms of special spaces of divisors, and you can find details worked out explicitly for an example in rank 2 (i.e. a holomorphic double covering f: Y -> X) in pages 103-105 of [NJH,87].

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Just got a chance to look at it, and this is pretty much what I was after. Thanks – Charles Siegel Nov 6 '09 at 0:36

A later paper of Beauville (here, preprint from 2000) seems to address your first question, but more generally by taking direct images of vector bundles. He makes the following conjecture:

If f: X'-> X is a finite morphism of smooth projective curves and E is a generic vector bundle on X', then f_*E is stable if g(X) \geq 2 and is semistable if g(x) = 1.

The problem is actually equivalent to that of a pushing down a line bundle L (from a different cover).

In the paper, he shows that the conjecture holds with some restrictions on L (e.g. when \chi(X) is small), although obviously one wants it to hold more generally. He shows, as well, that the conjecture holds as worded above whenever f is an etale covering.

Does this help?

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Eh, not really. I'm looking for more "what does generic mean in this case?" type of answer, and I don't really see anything in there addressing this. – Charles Siegel Oct 28 '09 at 5:03
Yeah, I realise now that you were looking for more, and that I should have read the questions better. Please see my next attempt below... (Revision note for those following this thread: originally I had directly replaced my old answer with the new one, but I think this made things confusing, since Charles' comment above was directed at my original attempt. So now my new answer is a separate post below. Sorry for the confusion.) – 1-- Nov 1 '09 at 23:32