Martin's Blog

Polarisable complex tori are projective

Posted by Martin Orr on Saturday, 26 March 2011 at 16:07

Last time, we defined polarisations on H_1 Hodge structures and saw that if A is a complex abelian variety, then H_1(A) has a polarisation. This time we will prove the converse: if X is a complex torus such that H_1(X) has a polarisation, then X is an abelian variety (in other words, X can be embedded in projective space). The proof is based on studying invertible sheaves on X.

This is long, even though I have left out all the messy calculations. For full details, see Mumford's Abelian Varieties or Birkenhake-Lange's Complex Abelian Varieties. For the next post, you will only need to know the two statements labelled as theorems.

This theorem is a special case of the Kodaira Embedding Theorem, which tells you that any compact complex manifold is projective if it has a polarisation, but that is somewhat more difficult.

Appell-Humbert Theorem

Let X = V/\Lambda be a complex torus, and \pi : V \to X the quotient map. The Appell-Humbert Theorem gives us a classification of all invertible sheaves on X.

We can construct invertible sheaves on the quotient V/\Lambda as follows: Take a collection of holomorphic functions e_\lambda : V \to \mathbb{C}^\times for \lambda \in \Lambda. For open subsets U \subseteq X, define L(U) to be the set of holomorphic functions on \pi^{-1}(U) satisfying f(z+\lambda) = e_\lambda(z) f(z) for z \in \pi^{-1}(U) and \lambda \in \Lambda. This gives an invertible sheaf on X if the e_\lambda satisfy a certain cocycle condition.

Let H be a Hermitian form on V such that the symplectic form E = \operatorname{Im} H takes integer values on \Lambda (in the language of previous posts, E is a Hodge symplectic form). Let \alpha be a function \alpha : \Lambda \to S^1 (here S^1 is the unit circle in \mathbb{C}) satisfying  \alpha(\lambda_1 + \lambda_2) = (-1)^{E(\lambda_1, \lambda_2)} \alpha(\lambda_1) \alpha(\lambda_2). Such an \alpha is called a semi-character for H. Exercise: prove that there exists a semi-character for every H.

Then set  e_\lambda(z) = \alpha(\lambda) \exp \left( \pi H(z, \lambda) + \tfrac12 \pi H(\lambda, \lambda) \right). The associated invertible sheaf is denoted L(H, \alpha).

Its global sections are holomorphic functions on all of V satisfying f(z+\lambda) = e_\lambda(z) f(z); such functions are called theta functions for (H, \alpha). We shall write \Gamma(\Lambda, H, \alpha) for the vector space of theta functions.

By some boring calculations, we can prove the following:

Appell-Humbert Theorem. Every invertible sheaf on a complex torus is isomorphic to L(H, \alpha) for exactly one pair (H, \alpha) of a Hermitian form H and a semi-character \alpha.

Furthermore, there is a group operation on the pairs (H, \alpha) by adding the Hs and multiplying the \alphas, and this corresponds to tensor product of invertible sheaves.

Counting theta functions

We can count exactly the dimension of the space of theta functions for (H, \alpha). For our purposes it will suffice to do this whenever H is positive definite (it is only a little more complicated when H is positive semidefinite, and the dimension is 0 if H is not positive semidefinite).

Let \Lambda^\perp = \{ z \in V \mid E(z, \Lambda) \subseteq \mathbb{Z} \}. If E is nondegenerate (for example if H is positive definite), then \Lambda^\perp/\Lambda is finite, and its order is always a square. The Pfaffian of E is defined to be \sqrt{\lvert \Lambda^\perp/\Lambda \rvert}.

Proposition. If H is positive definite, then \dim \Gamma(\Lambda, H, \alpha) = \operatorname{Pf}(E).

Sketch proof. We define classical theta functions to be theta functions as defined above, multiplied by a certain holomorphic function. The benefit of classical theta functions is that they are periodic modulo a certain rank-g sublattice of \Lambda, so can be expanded as Fourier series. It is possible to write down a basis for the space of Fourier series of classical theta functions.

Projective embeddings

Recall that we can get a projective embedding of X by taking a basis of the global sections of a very ample sheaf. We shall show that if the Hermitian form H is positive definite (in other words E is a polarisation), then 3L(H, \alpha) is very ample for any semicharacter \alpha.

First we will need a lemma. If \sigma \in V, then I shall write t_\sigma for the map "translate by \sigma"; this could be a map V \to V or X \to X.

Lemma. If H is positive definite, then there is a divisor D \in |L(H, \alpha)| such that t_\sigma^* D \neq D for all \sigma \in V - \Lambda.

Sketch proof. Suppose that \theta \in \Gamma(\Lambda, H, \alpha), and let D be the divisor of zeroes of \theta. Some calculations show that if t_\sigma^* D = D for some \sigma \in V - \Lambda, then E takes integer values on the lattice \Lambda' = \Lambda + \mathbb{Z}\sigma (which is strictly larger than \Lambda) and that \theta is a theta function with respect to \Lambda', H and some semi-character \alpha' on \Lambda' extending \alpha.

But the Pfaffian of E with respect to \Lambda' is smaller than its Pfaffian with respect to \Lambda, so \dim \Gamma(\Lambda', H, \alpha') < \dim \Gamma(\Lambda, H, \alpha). Furthermore there are only finitely many possible \Lambda' and \alpha', so the union of these \Gamma(\Lambda', H, \alpha') is not all of \Gamma(\Lambda, H, \alpha).

Now for the main theorem. Recall that to proof that a sheaf is very ample, we must prove that its linear system

  1. is base point free
  2. separates points
  3. separates tangents.

Theorem. If H is positive definite, then 3L(H, \alpha) is very ample.

Sketch proof. Let L = L(H, \alpha). Choose a divisor D as in the lemma: D \in |L| and t_\sigma^* D \neq D for all \sigma \in V - \Lambda. With a little more work we may also suppose that D is reduced i.e. D is a codimension-1 submanifold of X (with no multiplicities).

The reason for using 3L is that if \theta \in \Gamma(\Lambda, H, \alpha), then \theta(z + a) \theta(z + b) \theta(z - a - b) \in \Gamma(\Lambda, 3H, \alpha^3) for every a, b \in V. In terms of divisors, this says that t_a^* D + t_b^* D + t_{-a-b}^* D is in |3L| for every a, b \in X. By varying a and b, this gives us a lot of theta functions to play with.

  1. Suppose that x is a base point for |3L|. Choose a such that x-a \not\in D. Then for every b \in X, the fact that x is a base point gives  x \in t_a^* D + t_b^*D + t_{-a-b}^* D. So at least one of x+b and x-a-b is in D i.e.  b \in t_x^*D + (-1)^* t_{x-a}^* D. But this cannot hold for every b since D has codimension 1 in X.

  2. Suppose that |3L| does not separate x, y \in X. By a more advanced version of the argument in 1, we conclude that t_{x-y}^* D = D so by the lemma, x = y.

  3. Suppose that |3L| does not separate the tangent vector t at x \in X. We may suppose wlog that x = 0. More of the same style of argument tells us that D is everywhere tangent to the invariant vector field on X generated by t. It follows that D is fixed by the 1-parameter subgroup generated by t, contradicting the lemma.

Tags abelian-varieties, alg-geom, hodge, maths


  1. Dual abelian varieties over the complex numbers From Martin's Blog

    In this post I will define dual abelian varieties over the complex numbers. The motivation is that polarisations can be interpreted as isogenies from an abelian variety to its dual. For the moment, all this is tied to Hodge structures so only work...

  2. Dual abelian varieties and line bundles From Martin's Blog

    The definition I gave last time of dual abelian varieties was very much dependent on complex analytic methods. In this post I will explain how dual varieties can be interpreted geometrically: the points of correspond to a certain group of line bun...

  3. Weil pairings: the skew-symmetric pairing From Martin's Blog

    Last time, we defined a pairing By composing this with a polarisation, we get a pairing of with itself. This pairing is symplectic; the proof of this will occupy most of the post. We will also see that the action of the Galois group on this pairin...


  1. Barinder Banwait said on Tuesday, 29 March 2011 at 18:54 :

    It seems that tori can be defined over any base scheme S, and I guess there is a similar "Polarisable tori are projective" theorem there. How different in spirit is the proof of that case (assuming it is true) to the one sketched above? I am particularly interested in the case where S = Spec(\mathcal{O}_K), for K a number field.

    Also, can X be projective without having a very ample sheaf on it?

  2. Martin Orr said on Saturday, 02 April 2011 at 23:01 :

    “Complex tori” (as I talk about here) are completely different from “tori over \operatorname{Spec} \mathbb{C}”.

    Specifically, complex tori are \mathbb{C}^n mod a lattice, which is topologically an n-torus (product of n circles). This is an analytic thing so I don’t think that it makes sense to generalise it to other base schemes (maybe p-adic fields). It is only once you have embedded your complex torus in projective space that it becomes an algebraic-geometry object (if it is polarisable of course).

    On the other hand, I guess that when you talk about “tori over S”, then you probably mean algebraic groups over S which become isomorphic to \mathbb{G}_m^n over the algebraic closure (at least for S a field). These are completely different. The relation with topological tori comes from real points rather than complex points. They are affine schemes, so never projective.

    A projective scheme always has a very ample sheaf on it - this is built into the definition of very ample sheaf. Very ample sheaves are covered in II.5 and II.7 of Hartshorne and in any other similar book on algebraic geometry.

  3. Giovanni Rosso said on Tuesday, 05 April 2011 at 12:39 :

    Martin, nice posts! I was wondering if you can tell me where the Galois representation associated to the Tate module of an abelian varieties of dimension g takes values.

    I suppose all the elements of the Galois group should preserve the alternating form giving the polarization, which over \mathbb{Q} becomes the classical sympletic form, so I would guess it should takes values in a group closely related to GSp_{2g}.

    I think that the image of the representation right-translated by \sqrt{A^{-1}} (where A is the matrix of the alternating form over \mathbb{Z}) should be in GSp_{2g} but I could not find any precise reference about this.

    As it is not so close to my interest, I don't want to spend too much time on it, but I still have this curiosity! Do you know anything about it?

  4. Martin Orr said on Thursday, 07 April 2011 at 14:41 :

    Over \mathbb{Q}, I would just say that the Galois representation takes values in \operatorname{GSp}_{2g} since all non-degenerate symplectic forms are equivalent. (The Galois group does not quite preserve the symplectic form - it twists it by the cyclotomic character, hence why we get \operatorname{GSp}_{2g} and not \operatorname{Sp}_{2g}.)

    Over \mathbb{Z}, I think the correct statement is that there is a matrix A \in \operatorname{GL}_{2g}(\mathbb{Q}) such that, if J is the standard alternating form and J' the one coming from the polarization, then

     J' = A J A^t.

    Then the image of the representation will be contained in \operatorname{GSp}_{2g}(\mathbb{Q}) conjugated by A.

  5. Giovanni Rosso said on Friday, 08 April 2011 at 13:34 :

    Thank you very much, Martin!

    Sorry to bother you again, but now I have other questions: is it easy to see that the Galois representation twists the symplectic form by the cyclotomic character? and which is the determinant of this representation?

    Thank you again.

  6. Martin Orr said on Saturday, 09 April 2011 at 16:00 :

    I am not sure about how to calculate the action of the Galois representation on the symplectic form, but it is something I intend to cover later in this series of blog posts. At my current rate, it might be a few months before I get there.

    The determinant is easier: if \omega \in \bigwedge^2 V is the symplectic form, then \omega^g generates \bigwedge^{2g} V so the determinant of the representation is the g-th power of the character by which it acts on the symplectic form.

Post a comment

Markdown syntax with embedded LaTeX.
Type LaTeX between dollar signs, and enclose them between backticks to protect it from Markdown.
All comments are subject to moderation before they appear on the blog.