Theta divisors: the funky part

A note about the title. My last year in Berkeley, I lived just off of Telegraph Avenue, a few blocks from People’s Park. It was a neighborhood filled with street artists and musicians, great bookstores and interesting people — as well as a good supply of pot dealers, homeless folks and people in various altered states of reality. Whenever people asked me if I lived in a good neighborhood, I told them I lived in a funky neighborhood. I think the same thing can happen with math. Grothendieck and Serre taught us all how to work in a very austere and beautiful way, where every construction is completely natural. When we see math that doesn’t work that way, often we think that it is bad. At its best, though, all of those changes of basis and transition formulas can be, well, kind of funky.

Let’s review where we were. Let A be an abelian variety and \beta \in H^2(A, \mathbb{Z}) be the Chern class of an ample line bundle. As we discussed last time, the space of all line bundles with Chern class \beta is a principal homogeneous space for Pic^0(A). However, a little cleverness can pick out a subcollection of 2^{2g} such line bundles which is a principal homogeneous space for Pic^0(A)[2]. The use of \theta functions is to give us a way to label individual line bundles in this 2^{2g} element set, and write down sections of them.

We’ll write \Lambda for H_1(A, \mathbb{Z}). Recall also that \beta naturally induces a skew symmetric, integer valued form V on \Lambda.

Here is my main claim: The 2^{2g} special bundles of class \beta correspond naturally to the 2^{2g} functions r from \Lambda \to (1/2) \mathbb{Z}/\mathbb{Z} such that

r(\lambda+\mu) = r(\lambda) + r(\mu) + (1/2) V(\lambda, \mu).

A few notes before going on. The reader will notice that I have abandoned talking about curves altogether; as I discussed in the last post, I think that the only role of the curve is to define a class \beta in the Jacobian. It would probably be more elegant to work with \mathbb{Z}/2 \mathbb{Z} instead of (1/2) \mathbb{Z}/\mathbb{Z}, but I spent a lot of time getting the twos right in the previous posts so I don’t want to change conventions now. Finally, based on my comments, 50 percent of my readers want me not to restrict my attention to the principally polarized case. Oh vast readership, your wish is granted!

Let H be the universal cover of A, which has the structure of a \mathbb{C}-vector space and let \Omega : \Lambda \to H be the obvious map. Any line bundle \mathcal{L} on A pulls back to a line bundle on H. All line bundles on H are trivial. It turns out that we can choose a trivialization of this pullback such that sections of \mathcal{L} correspond to \Theta-functions on H, for some (L,J). However, there is no natural choice of (L,J).

As we saw before, L must be a linear map \Lambda \to H^{\vee}. Moreover, if \beta is the Chern class of \mathcal{L}, then

\langle L(\lambda), \Omega(\mu) \rangle - \langle L(\mu), \Omega(\lambda) \rangle = V(\lambda, \mu).

Incidentally, this last relation seems to be hard to verify using any algebraic definition of Chern classes. I suspect that it is easier using connections but haven’t worked it out. If anyone finds a simple argument, leave a note below.

There are a lot of interesting things to say about the existence and nonuniqueness of L. I don’t want to get sidetracked, so I’ll just say that L may or may not exist for a general map \Omega — the condition that it does is equivalent to there being a line bundle with Chern class \beta on H/\Omega(\Lambda). (The condition that this line bundle be ample turns out to be a certain positivity condition, which I won’t be getting into at all.) Also, given one L_0, all other L‘s are of the form L_0 + S \circ \Omega for some self-adjoint map S: H \to H^{\vee}.

Exercise (post solutions below): Choose coordinates so that V=\left( \begin{smallmatrix}0 & D \\ -D^T & 0 \end{smallmatrix} \right) and \Omega = \left( \begin{smallmatrix} \omega_1 & \omega_2 \end{smallmatrix}\right). (Here D, \omega_1 and \omega_2 are g \times g matrices, which you may assume to be invertible.) Give a necessary and sufficient condition, written as an equality of matrices, for L to exist. Harder: do this without assuming that V has the given special form.

As we saw in the one variable case, (L,J) are not uniquely determined by the line bundle \mathcal{L}, because of the possibility of multiplying by trivial \Theta functions. However, writing

J(\lambda) = (1/2) \langle L(\lambda), \Omega(\lambda) \rangle + r(\lambda) \mod \mathbb{Z},

we saw that the function r from \Lambda to \mathbb{C}/\mathbb{Z}, is well defined up to adding linear functions of \Omega(\lambda). Furthermore, we have

r(\lambda+\mu) = r(\lambda) + r(\mu) + (1/2) V(\lambda, \mu) \mod \mathbb{Z}.

Let’s look at some special cases. If \lambda = \mu =0, we deduce that r(0)=0 \mod \mathbb{Z}. If \mu=-\lambda, we deduce that r(-\lambda) = -r(\lambda) \mod \mathbb{Z}. (Since V is skew symmetric.)

Now, we want to restrict to the case that \mathcal{L} = [-1]^* \mathcal{L}. In this case it is plausible, and in fact true, that we can choose our trivialization symmetrically, so that we are interested in \Theta functions with \theta(z) = \pm \theta(-z).

So

\theta(z+\Omega(\lambda)) = \pm \theta(-z-\Omega(\lambda))

and

e^{2 \pi i (\langle L(\lambda), z \rangle + J(\lambda))} \theta(z) = \pm  e^{2 \pi i (\langle L(-\lambda), -z \rangle + J(-\lambda))}.

Using \theta(z) = \theta(-z), we have J(\lambda) = J(- \lambda) \mod \mathbb{Z}. Since \langle L(\lambda), \Omega(\lambda) \rangle = \langle L(-\lambda), \Omega(-\lambda) \rangle , this translates into the condition that r(\lambda) = r(-\lambda) \mod \mathbb{Z}. On the other hand, r(-\lambda) = -r(\lambda) \mod \mathbb{Z} for any r. So we see that \theta(z) = \pm \theta(-z) if and only if r is half-integer valued.

We have now proven the claim, up to believing everything I said about choosing trivializations. Given a half-integer valued r, such that
r(\lambda+\mu) = r(\lambda) + r(\mu) + (1/2) V(\lambda, \mu) \mod \mathbb{Z},
the sections of the corresponding line bundle are the quotients of \Theta-functions with monodromy (L, (1/2) \langle L(\bullet), \Omega(\bullet) \rangle + r(\bullet)).

We’ve reached the main result. This would be a good time to stretch your legs and get a cup of coffee.

First, an exercise: We said before that we would consider a set of line bundles which would be a principal homogeneous space for (Pic^0(A))[2]. We also noticed that the difference of any two r-functions is linear, so our r-functions are a principal homogenous space for Hom(\Lambda, (1/2) \mathbb{Z}/\mathbb{Z}). Why are these two pictures consistent?

Next, how does this relate to the classical presentation? Classically, one begins by choosing a decomposition \Lambda = M \oplus N where M and N are two lattices of rank n on which V is zero. Then there is a unique map D: N \to M^{\vee} so that V(n,m) = \langle D(n), m \rangle for m \in M and n \in N. In general, V is given by

V(m_1+n_1, m_2+n_2) = \langle D(n_1), m_2 \rangle - \langle D(n_2), m_1 \rangle.

In the principally polarized case, D is invertible (in GL_n(\mathbb{Z})) and one generally chooses coordinates so that it is the identity. In general, my references like to diagonalize D but I see no reason to do so.

One then makes the following choice of L: L is zero on M and, for n \in N and m \in M, we have \langle L(n), \Omega(m) \rangle = \langle D(n), m \rangle. Note that the requirement that

\langle L(n_1), \Omega(n_2) \rangle - \langle L(n_2), \Omega(n_1) \rangle=V(n_1,n_2)=0

for n_1 and n_2 \in N gives a nontrivial condition on (\Omega, D): is it the same condition you got in the exercise above?

Finally, one sets r(m+n) = (1/2) \langle D(n), m \rangle for m \in M and n \in N.

Then \Theta-functions with monodromy (L,J) are \Omega(M) periodic so we can write them as Fourier series. Identify H with M \otimes \mathbb{C}, so \langle D(n), \bullet \rangle can be thought of as a \mathbb{C}-linear function on H which take integer values on \Omega(M).

Write

\theta(z) = \sum_{\mu \in M^{\vee}} a(\mu) e^{(2 \pi i) \langle \mu, z \rangle }.

It is required that, for n \in N,

\theta(z+\Omega(n)) = e^{(2 \pi i) \left( \langle D(n), z \rangle + (1/2) \langle D(n), \Omega(n) \rangle \right)} \theta(z)

so

a(\mu) e^{(2 \pi i) \langle \mu, \Omega(n) \rangle} = a(\mu - D(n)) e^{(\pi i)  \langle D(n), \Omega(n) \rangle}.

In particular, a is determined by its values on a collection of coset representatives for M^{\vee}/D(N). Ignoring convergence issues, this shows that the dimension of the space of \theta functions with monodromy (L,J) is revealed to be \# (M^{\vee}/D(N)) = \det D.

One last thought. I’m not going to work this out myself, but I’d gladly link to someone who did. Here is a possible way to avoid analysis altogether. Given a class \beta on an abelian variety A as before, let \gamma be one of the 2^{2g} lifts of \beta to Pic(A). Then {[2]}^* \gamma is a line bundle L whose isomorphism class is independent of the choice of \gamma. So we can describe the 2^{2g} various choices of \gamma by giving \beta, which will determine L, and giving descent data (link is to PostScript file) for the etale map {[2]}: A \to A. I suspect that this lets you do all of the above over any field of characteristic not {2}, without ever mentioning an infinite sum.