The following is bugging the heck out of me.

Let be a complete algebraic curve of genus defined over the complex numbers. Then is the abelian group gotten by taking formal sums of points on and modding out by the relation that, for any nonzero meromorphic function , the sum of the zeroes of equals the sum of the poles. Such a formal (finite) sum of points of is called a “divisor”. If is a divisor then the degree of is . Since any meromorphic function has equally many zeroes and poles, we can define the degree of a point in ; let be the component of coming from divisors of degree . Then is an abelian variety of dimension and is a principal homogeneous space for . Obviously, for any nonnegative integer , there is a map called “take the sum”. When , the image of this map is called the Theta divisor and it is a complex subvariety of codimension one in .

That’s one description of the Theta divisor. Here is another. Let be the vector space of global homolomorphic 1-forms on and let be the dual vector space. There is an injection of into . Namely, if is a 1-cycle in and is a global holomorphic 1-form, set . Then there is an isomorphism defined as follows: for any degree zero divisor , choose a 1-chain with and map to the functional .

One then writes down a holomorphic function on , whose definition I have not quite fully internalized. is not quite -periodic, but it has the property that, for any , the ratio for some functions and on . In particular, the zero locus of is -periodic. The Theta divisor is the image of this zero locus in .

So, in the first perspective I gave, the Theta divisor is a hypersurface in , whereas, in the second, it is a hypersurface in . There is no canonical isomorphism between these two, so what gives?

Now that I have written all this out, I see sort of what the answer must be. To define , one does something like choose a basis of , and this choice must somehow give me an isomorphism between and . But I’d still appreciate hearing from anyone who has thought through the details and has an elegant way to present them.

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I’ve been doing a lot of reading about this for my orals. The Theta Divisor and the divisor of the function are only really defined up to translation by an element of , and by choosing a point , you get an isomorphism by , and if you change base point, this only translates things. If you want more details, I can provide them (and I’m working towards talking about Jacobians and Theta divisors on my blog anyway)

But the theta divisor, in , is defined without any choices at all. And, as far as I can tell, the only choice in defining is a discrete choice — a basis for the first homology. There isn’t a whole continuous of options here.

I will begin by apologizing for not really being somebody who has thought through the details or an elegant way to present them, and not being an expert in the field in any way, shape, or form (or even an algebraic geometer)… but since this is a blog, I suppose one may make irresponsible (trivial or wrong) comments…

The way I understand it, it isn’t that you’re choosing a basis for $\Lambda$ (if it is, I don’t understand how), but rather that you’re choosing a basepoint for $Pic^{g-1}$. You see, $Pic^0$ does have a canonical choice of basepoint- the trivial divisor. Both are abelian varieties of the same dimension, and the isomorphism is given by choosing some divisor in $Pic^{g-1}$ and tensoring everything with its inverse.

So what happens when you change basepoints? You act on the coordinate of the theta function, and this corresponds to a translation of the line bundle- i.e. to a change in theta characteristic. But if you just want the divisor then you don’t care, because it’s only defined up to translation…

Now, the only setting I’ve thought this through is the setting of Prym varieties (the Hitchin fibration setting), where you have the Jacobian of a double-cover of $C$ branched over some even-degree divisor. This divisor has a square root (sketch-proof: take it to $Pic^0$, take the square root, and bring it back), and in fact has $2^{3g-3}$ square roots- a discrete choice. Here you can actually calculate how the theta divisor changes and get explicit formulae…

So I’m guessing that for $C$ there’s also some discrete choice of basepoints (probably $2^g$ choices)… not sure how that works.

I believe that you have to choose a square root of the canonical

bundle. Such a square root has degree g-1, as desired, and

there are 2^{2g} choices.

I should maybe give more detail of that sketch-proof, because I was really confused about this for some time, and it’s really simple:

If $deg(D)=2deg(E)$ then $\frac{-1}{2}(2E-D)+E$ is a square root of $D$. If it’s over a torus of degree $n$, there are $2^n$ such choices.

Thanks Matthew! I think I can now possibly see the picture- you have to tell me if I’m right though. It’s the Jacobian not the Prym, so the number of choices is $2^{2g}$ not $2^g$, as you pointed out… anyway.

You take the canonical bundle, which you express on $Pic^0$ as a sum of differences of points, and choose a square root. This square root is given in terms of the Abel-Jacobi map as taking “$a-c$” instead of $a-b$ where $\gamma$ is the path from $a$ to $b$ and $c$ is a midpoint such that the integral via $\gamma$ for $a$ to $c$ gives half of the integral of $gamma$ from $a$ to $b$. The issue is that I have two choices for $c$, corresponding to the fact that $gamma$ is half of a cycle- half of a basis element of $\Lambda=H_1(C;Z)$. If I choose an orientation for the basis of $\Lambda$— i.e. if I choose signs for these generators— then $a-c$ is given uniquely as a divisor. There are $2^{2g}$ such choices because $\Lambda$ has $2g$ generators, and I can choose a sign for each generator…

Does this make any sense at all?

This makes some sense, but let me take out the arbitrary choices.

Regarding square roots of the canonical bundle: the canonical bundle lives in , not . So its square roots live in as wanted. There are indeed such square roots, because the ratio of any two such roots is a square root of the identity; these are described by . The square roots of the canonical bundle are a principal homogeneous space for .

So it is plausible that, in constructing the function, we make some discrete choice which allows us to choose a square root of the canonical bundle. What I would like to know is how that process works.

The following is my best understanding of the definition of the function. The usual phrasing is that we choose cycles , …, in which are represented by disjoint, nonhomologous simple closed curves. (I suspect that this is more data than we need, and we actually just need to choose a rank sublattice of with a certain isotropy condition, but let’s ignore that for now.)

We denote the lattice spanned by the ‘s by . We call the dual lattice ; note that lives naturally in . Then we somehow build a quadratic form on and set

.

So, that is where we choose something like a basis. As far as I can tell, this construction doesn’t choose a base point on the curve.

Ok, so I failed in my first attempt at saying what I meant, let’s see if I do better this time.

You’re using the Abel-Jacobi map, which, as a map into requires a choice of basepoint and of a basis for . Call this basis . Then for any point , we can pick a path from the basepoint to , and the Abel-Jacobi map will take . We then just extend linearly to get the map .

The upshot of this view is that we want to think about the Theta divisor solely in terms of the theta function. Now, call the image of the th symmetric product of . Then use to denote the fundamental cohomology classes.

Then there’s a theorem, called Poincare’s Formula, which says that for , we have .

What this really tells you is that if you put in , you get that the cohomology classes of and are the same. Now, on an abelian variety, this means that the line bundles that come from these divisors differ by pullback along a translation map, and so the divisors are the same up to translation.

I think the confusion stems from there being a LOT of ways to view the Jacobian, the Abel-Jacobi map, and the like, and in this case, you want to think of the Jacobian just as the quotient, and the Abel-Jacobi map as integration, rather than as mapping divisors to the line bundle associated to the divisor. This help at all?

The divisor defined by the vanishing of the theta function is even, i.e., equal to its negative. The other divisor depends on identifying classes of degree g-1 with classes of degree 0. Pick a divisor D_0 of degree g-1. Then the other divisor is the locus of the classes of D-D_0 as D ranges over positive divisors of degree g-1. For such a thing to be even, we need that, for all D as above, there is a D’ (also positive of degree g-1) with D-D_0 ~ D_0 – D’, which I rewrite as 2D_0 ~ D+D’. As D is allowed to vary we conclude that 2D_0 defines a linear system of dimension g-1. But a divisor of degree 2g-2 defining a linear system of dimension g-1 must be a canonical divisor, by Riemann-Roch. So 2D_0 is canonical and D_0 is a theta characteristic.

David,

This question has been bugging me too.

I’m a little confused by what you wrote in the previous comment.

The lattice $H_{1}(X,\mathbb{Z})$ has a natural pairing coming from Poincare duality (no choices needed) and this induces a pairing on the dual lattice. Why isn’t the quadratic form $Q$ just equal to the form associated to this pairing?

I don’t have an answer to your question, but I can point out another point of confusion.

Without making an arbitrary choice, the variety J^{0} does not contain a theta divisor. However, I think that this variety does carry a distinguished line bundle.

I like to think of J^{0} as parametrizing degree 0 line bundles on C. The relation between this description and your description in terms of linear equivalence classes of divisors is simple. Given a divisor D the corresponding line bundle is O(D), the line bundle of rational functions with at worst a pole along D.

By general formalism, the product $C \times J^{0}$ carries a tautological line bundle(*) called P (for Poincare). The fiber of P at a point (x,L) of C \times J^{0} is the fiber L|_{x}. Consider the derived direct image:

$Rf_{*}(P)$

Here $f: C \times J^{0} –> J^{0}$ is the projection. By general formalism, I can represent this object in the derived category by a 2-term complex of vector bundles:

0—> E_0 –> E_1 –> 0

I get a distinguished line bundle on J^{0} by taking the top exterior powers of the E_i’s and then taking the difference. One can check that this line bundle does not change if I replace the complex E^{.} by a quasi-isomorphic complex. This line bundle is sometimes called the determinant of the cohomology of P.

If D is a _chosen_ theta divisor on J^{0}, then I would guess that O(D) should be related to this line bundle.

One can also carry out this construction for J^{g-1} too. If D is the image of the Abel map C^{g-1} –> J^{g-1}, then one can show that O(D) is equal to the determinant of the cohomology of the tautological line bundle.

(*) I am cheating here. The tautological line bundle P is defined only up to taking the tensor product with a line bundle obtained by pulling back via $f: C \times J^{0} —> J^{0}$. However, one can check that choices of Poincare line bundles all lead to isomorphic line bundles on J^{0}.

-jk

Matt beat me to it, and it’s been years since I thought about this business precisely, but the thing you want to read is the first chapter of volume I of Mumford’s “Tata Lectures on Theta” — there you will see that there are indeed 2^{2g} theta functions, which naturally differ from each other by 2-torsion points of the Jacobian (i.e. the set of these is a torsor for J[2], as you say.)

I’d like to know more…

I don’t have a copy of Tata lectures on Theta here, so it might be in there, but still I wonder whether anyone could post a full answer- in particular how and if choosing a square root to the canonical bundle chooses signs for generators of $\Lambda$ (it must be something fairly similar to what I wrote in post #6, but how to make it rigourous?). I think you need to do this on $Pic^0$ via the procedure in post 5- I can’t see another way to do it.

It’s really interesting how to do these general things concretely, although it’s never presented that way…

Well, I have Mumford’s lectures now. They are certainly relevant, but the answer isn’t just sitting there waiting to be picked up. The do confirm that, as people have been saying: (1) there are different functions associated to a Riemmann surface, differing by the two torsion points of and (2) if we translate the divisor back by one of the square roots of the canonical bundle, we get the zero section of one of those functions. I’ll post again if I understand more. Thanks for all the comments!