As often happens, after I do a large computation, a lot more theory makes sense. So, in this post, I’ll talk about the parts that I can explain without getting dirty with formulae. After that, I’ll think there willl be one more post on functions and divisors.

Suppose we have an abelian variety of dimension and we have a cohomology class in . Suppose that is the Chern class of a line bundle. We will be interested in the question of to what extent we can canonically choose a line bundle with Chern class . (Later, we will ask the same questions regarding whether there is a hypersurface with class Poincare dual to .) The application to our original question is when is the Jacobian of a Riemann surface and is the class of the disivor in . (The ambiguity in identifying with is topologically trivial.)

By the way, when is could be the Chern class of a line bundle? If and only if it is a Hodge class. This is the main case in which the Hodge conjecture has been proved. This is a fascinating subject, but I’ll be ignoring it.

We have a map which maps a holomorphic line bundle to its first Chern class. The kernel of this map is called and the image is called the Neron-Severi group, . So we have a short exact sequence:

The question we are asking is whether we can find a natural way of splitting this sequence. Let be the endomorphism of which sends every point to its inverse. Then acts on all three terms of . The interesting thing is that acts by negation on the left term and acts trivially on the right. If we were dealing with a short exact sequence of real vector spaces, or more generally of modules, this would let us split the sequence immediately. We would simply split into its eigenspace, which would be , and its eigenspace, which would be .

We can still try to mimic this, but life will be trickier because we can’t divide by . Choose an arbitrary lift of to . Then is independent of the choice of . There are bundles such that , each of which has class . So, there isn’t one canonical choice of , but we can get down to a discrete choices, as opposed to the entire continuum of choices that we would naively see.

Now, this post has been about line bundles. What happens with divisors? If a line bundle has Chern class , then is the class (Poincare dual to) the zero locus of any section of . In general, will have many sections. Precisely, assuming that is ample, induces a skew symmetric form on , and the dimension of is the Pfaffian of that form. In particular, as will be the case in our intended application to -functions, if induces a perfect pairing on , then has a unique nonzero section (up to scaling). So, in that case, we can use the methods above to find a canonical divisor , modulo translation by -torsion, whose cohomology class is .

What is this divisor explicitly? It is determined by the condition that . (Note that translating by a -torsion point preserves the truth of this condition.) It’s a good exercise to trace through the previous definitions and see that this works.

Now, all of this was in the context of a general abelian variety with a specified Hodge class that gives an ample line bundle and induces a perfect pairing on . (The terminology for this is a *principally polarized abelian variety*.) In the particular case of the Jacobian, it is a nice exercise (using Riemman-Roch) to check that, if is any point of with then . Thus, the previously described construction does yield the translate of the divisor by some square root of the canonical bundle. (As several commenters told me.)

That is the end of what can be done nicely, and without reference to explicit functions (or indeed, any analysis). What functions let you do, given a choice of basis for , is to make a choice of one of the possible line bundles on . More on this in the forthcoming (and final post) “Theta functions: the funky part”!

Nice! That clears up some issues I hadn’t been understanding…

I haven’t really worked out the details, but shouldn’t most of this story (all except paragraph -4) go through even if the variety is not principally polarized? The Prym isn’t principally polarized…

Can you say something about the picture in the non-principally-polarized case? (although this has nothing to do with your original question).

Yeah, I think that the only importance of principal polarization is that it gives you a unique section of the line bundle. In the general case, everything works the same way on the level of line bundles.

We can then ask, if we are given an isomorphism , how does act on ? To find a divisor of class , we’d want to find a section in which is an eigenvalue for . That divisor will vanish to an even order at the identity (possibly zero) if it comes from a -eigenvector and to odd order if it comes from a -eignevector. In the principally polarized case, is one dimensional and the zero locus of its section is called an even or odd divisor according to which of these cases holds.

The above seems to depend on our choice of isomorphism but we can remove the dependency. Any two such isomorphisms differ by a nonvanishing holomorphic function on , which must be a constant. We can normalize that constant by requiring our isomorphism to act trivially on the fiber over the identity. So don’t worry about that.

As you’ll see in the next post, we can write down a basis for using functions and that basis is an eigenbasis for . So it would be easy to work out how many and how many eigenvalues occur. But I haven’t found a nice way to formulate the result.

Small typo: “When is beta could be…”

I’ve been trying to learn this stuff myself, so I’ll reread this when I’m less jet-lagged and sleepy! Thanks for writing it.