Riemann’s Bilinear Relations

Posted on April 19, 2010 by Charles Siegel

This post begins my series on some classical geometry of curves and abelian varieties. We’ll start with some talk of line bundles and polarizations on abelian varieties in general, and the first big theorem I’m really targeting is the Torelli theorem (going to go with Andreotti’s proof, though once some other stuff is covered, might reprove it another way or two) but I might get distracted by other things along the way. We’ll see how this goes. Posting will be sporadic at best, and most of this material is, in more detailed form, going to appear in my master’s thesis (hopefully).

So before we begin, pretty much everything basic is covered in Arbarello, Cornalba, Griffiths and Harris, Birkenhake and Lange, or Griffiths and Harris.  There are a few things that I can’t point out where in these they appear, but this is all pretty standard stuff, and can be done by direct computation, if need be.

First, a complex manifold {M} is called a complex torus if it is biholomorphic to {\mathbb{C}^g/\Lambda} for some {g} and some full rank lattice {\Lambda}.  We call complex torus an abelian variety if it is a projective variety.  Our first goal should be to determine when a torus is an abelian variety.

We define a polarization on an abelian variety is the class of a Hodge form {\omega=\sum \delta_\alpha dx_\alpha\wedge dx_{g+\alpha}} where {d_\alpha|d_{\alpha+1}} and the {x_i} are a dual basis to the {\lambda_i}, a basis for the lattice {\Lambda}.  A principal polarization is a polarization where {\delta_\alpha=1} for all {\alpha}.

First, we should note that a polarization is, an equivalence class of line bundles, which are ample.  So the existence of a polarization is enough, but we’ll look for a principal one, because these will have better properties (for instance, we’ll manage a moduli space of them!)

Theorem (Riemann’s Bilinear Relations): A {g\times g} matrix {\Omega} over {\mathbb{C}} determines a principally polarized abelian variety {\mathbb{C}^g/\mathbb{Z}^g\oplus \Omega\mathbb{Z}^g} if and only if

  1. {\Omega^t=\Omega}
  2. {\mathrm{Im}\Omega>0}

Proof: Let {X=\mathbb{C}^g/\Lambda}, with basis {e_1,\ldots e_g} of {\mathbb{C}^g} and {\lambda_1,\ldots,\lambda_{2g}} for {\Lambda}. Denote the period matrix by {\Pi}. We’ll first prove that our complex torus is an abelian variety if and only if there is a nondegenerate alternating matrix {A} over {\mathbb{Z}} such that {\Pi A^{-1} \Pi^t=0} and {\Pi A^{-1}\bar{\Pi}^t>0}. Given these, we use Gaussian elimination to write {\Pi=[I, \Omega]}, and the relations in our theorem follow.

Let {E} be any nondegenerate alternating form on {\Lambda}, and denote by {A} its matrix in the basis of {\lambda_i}‘s. We extend {E} to {\mathbb{C}^g} and set {H(u,v)=E(iu,v)+iE(u,v)}. The form {H} is Hermitian if and only if {\Pi A^{-1}\Pi^t=0}. To see this, first note that it is if and only if {E(iu,iv)=E(u,v)} for all {u,v\in\mathbb{C}^g}. Now, define {I=\left(\begin{array}{c}\Pi\\\bar{\Pi}\end{array}\right)^{-1}\left(\begin{array}{cc}i1 & 0\\ 0 &-i1\end{array}\right)\left(\begin{array}{c}\Pi\\\bar{\Pi}\end{array}\right)}. Then we have {i\Pi=\Pi I}, and as {E(\Pi x,\Pi y)=x^t Ay}, {H} is hermitian if and only if {I^t AI=A}, expanding, we get precisely the condition desired.

Now, given that {H} is hermitian, we can compute the matrix representation in terms of {e_1,\ldots, e_g}. Let {u=\Pi x} and {v=\Pi y}. Then {E(iu,v)=x^t I^t Ay=i u^t(\bar{\Pi}A^{-1}\Pi^t)^{-1}\bar{v}-i\bar{u}^t(\Pi A^{-1}\bar{\Pi}^t)^{-1}v} by direct computation. Similarly, {E(u,v)=u^t(\bar{\Pi}A^{-1}\Pi)^{-1}\bar{v}+\bar{u}^t(\Pi A^{-1}\bar{\Pi}^t)^{-1}v}. This implies that {H(u,v)=2i u^t(\bar{\Pi} A^{-1} \Pi^t)^{-1}\bar{v}}. Thus, we have the positivity condition for {H} to be a positive hermitian form.

So then {H} defines the polarization, giving us an abelian variety. \Box

So, that’s a big messy computation.  What’s it mean? Well, a nondegenerate alternating matrix is a symplectic form, so we’re assuming we have a symplectic pairing on the lattice.  We’re actually using that we have a principal polarization, because a nonprincipal one wouldn’t look like \left(\begin{array}{cc}0&I\\ -I & 0\end{array}\right), but instead would have some nonzero weights on the diagonal.  Given one of these, the intersection pairing is looking quite a bit like the one on a genus g curve’s first homology.  We’ll see later that there’s a natural abelian variety and a natural principal polarization on any curve of genus g, called the Jacobian.  But next up: line bundles!

About Charles Siegel

Charles Siegel is currently a postdoc at Kavli IPMU in Japan. He works on the geometry of the moduli space of curves.
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2 Responses to Riemann’s Bilinear Relations

  1. aelle says:
    April 12, 2011 at 9:31 am

    Two questions:

    “We extend E to C^g and set H(u,v) = E(iu,v) + i E(u,v)”. Surely if we extend E to C^g such that it becomes sesquilinear, it is clear that H(u,v)=iE(u,v) for H to be Hermitian? The linear combination you specify seems “too much” (in fact zero if E is sesquilinear).

    What’s the motivation for the definition of the matrix I (the underlying reason it has the property required)?

    Many thanks!

    Reply
    • Charles Siegel says:
      May 6, 2011 at 10:25 am

      We’re extending E really to a bilinear form on \Lambda\otimes\mathbb{R}, and then identifying that space with \mathbb{C}^g.

      As for the definition of I, now that I’m looking at it more carefully, I’m not seeing it immediately. Once my thesis is a bit more done, I’m going to get back to blogging at least a little, and I intend to do a series on abelian varieties, theta functions, and their moduli, and I’ll revisit the bilinear relations.

      Reply

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