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Polarisations on Hodge structures

Posted by Martin Orr on Saturday, 26 February 2011 at 18:27

In the last post, I discussed Hodge symplectic forms. Now I shall show that the H_1 of an abelian variety has a polarisation, which is defined to be a Hodge symplectic form satisfying a positivity condition. The importance of polarisations is that they give a way of recognising which H_1 Hodge structures come from abelian varieties - I shall discuss this application next time.

Recap: Hodge symplectic forms

It's been a long time since the last post, so let us recall what it said. Let L be an H_1 \mathbb{Z}-Hodge structure. A symplectic form on L is a bilinear map \psi : L_\mathbb{Z} \times L_\mathbb{Z} \to \mathbb{Z} satisfying \psi(v, v) = 0 for all v. A Hodge symplectic form is a symplectic form \psi satisfying any of the following equivalent properties:

  1. \psi vanishes on L^{-1,0} \times L^{-1,0} and on L^{0,-1} \times L^{0,-1}.
  2. If h : \mathbb{C} \to \operatorname{End}(L_\mathbb{R}) is the complex structure defining the H_1 Hodge structure, then \psi(h(z)u, h(z)v) = z \bar{z} \, \psi(u, v).
  3. \psi lies in the (1,1)-part of the H^2 Hodge structure \wedge^2 L^*.

If A is a complex abelian variety, then H_1(A, \mathbb{Z}) has a Hodge symplectic from, induced by an embedding of A in projective space (and in general there are many Hodge symplectic forms coming from different embeddings into projective space).

Polarisations and Hermitian forms

A polarisation on an H_1 Hodge structure L is a Hodge symplectic form \psi satisfying  \psi(h(i)v, v) > 0 for all v \in L_\mathbb{R} - \{ 0 \}. Note that it would not make sense to ask for \psi(v, v) > 0 because \psi is a symplectic form so by definition \psi(v, v) = 0.

At first sight this definition may be a little odd, but it can be interpreted as positive definiteness for a certain Hermitian form as follows:

Let V be L_\mathbb{R} viewed as a complex vector space via the complex structure h (this is not at all the same thing as L_\mathbb{C} = L_\mathbb{R} \otimes_\mathbb{R} \mathbb{C}!) Then a Hodge symplectic form \psi extends to an \mathbb{R}-bilinear symplectic form \tilde\psi : V \times V \to \mathbb{R}. It is not \mathbb{C}-bilinear, but by definition 2 of a Hodge symplectic form, it satisfies  \tilde\psi(u, v) = \tilde\psi(iu, iv).

This corresponds to a Hermitian form H on V via the following lemma.

Lemma. Let V be a \mathbb{C}-vector space. There is a bijection between Hermitian forms H on V and symplectic \mathbb{R}-bilinear forms \tilde\psi : V \times V \to \mathbb{R} satisfying \tilde\psi(iu, iv) = \tilde\psi(u, v) given by  \tilde\psi(u, v) = \operatorname{Im} H(u, v),  H(u, v) = \tilde\psi(iu, v) + i\tilde\psi(u, v).

Now the Hermitian form H is positive definite if and only if \psi is a polarisation. (Exercise: prove this fact and the lemma.)

Polarisable Hodge structures and abelian varieties

The following (and its converse) is the key theorem to explain the interest in polarisations.

Theorem. If A is an abelian variety, then the Hodge structure H_1(A, \mathbb{Z}) has a polarisation.

Sketch of proof. Recall that to construct a Hodge symplectic form, we took an embedding of A in \mathbb{P}^N and let H be a hyperplane in \mathbb{P}^N. We looked at the homology class of A \cap H in H_{2g-2}(A, \mathbb{Z}) and took its Poincaré dual, which was the desired Hodge symplectic form.

We could do this in a slightly different order: Look at the homology class of H in H_{2N-2}(\mathbb{P}^N, \mathbb{Z}). Take its Poincaré dual to get \omega \in H^2(\mathbb{P}^N, \mathbb{Z}). Then restrict \omega to A.

By working on \mathbb{P}^N, we can do explicit calculations in homogeneous coordinates. We can write down a differential form, something like the following, and check that its de Rham cohomology class is \omega:  \frac{i}{2\pi} \partial \bar\partial \log \lvert X_0^2 + \dotsb + X_N^2 \rvert. We can also calculate the corresponding Hermitian form on the tangent space of \mathbb{P}^N and check that it is positive definite, and so \omega restricts to a polarisation of H_1(A, \mathbb{Z}).

The converse says that any H_1 \mathbb{Z}-Hodge structure with a polarisation is the H_1 of some abelian variety, and I will discuss it next time.

Remark on terminology

Polarisations on H_1 Hodge structures are also called Riemann forms. If you are being careful, a polarisation is defined on the Hodge structure L (\psi above) and a Riemann form is defined on the complex vector space L_\mathbb{R} (\tilde\psi above) but they are pretty much the same thing.

People also talk about polarisations on abelian varieties, which are algebro-geometrically defined, and make sense over any field. Fortunately if we work over \mathbb{C} then polarisations on the abelian variety (in the algebraic geometry sense) are in 1-1 correspondence with polarisations on the Hodge structure (as defined above) so again it doesn't matter.

A polarisable Hodge structure is defined to be a Hodge structure L for which there exists at least one polarisation (but we have not chosen any particular polarisation). On the other hand, a polarised Hodge structure is a pair (L, \psi) where L is a Hodge structure and \psi is a polarisation on L. This difference matters when we consider categories - a morphism of polarised Hodge structures is required to preserve the chosen polarisation, while a morphism of polarisable Hodge structures is any morphism of Hodge structures.

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

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Comments

  1. Barinder Banwait said on Monday, 07 March 2011 at 20:11 :

    B1. In the last paragraph of "Recap: Hodge Sympletic forms", "an embedding in A in projective space" should be replaced with "an embedding of A in projective space".

    A2. Let E be a complex elliptic curve. In your setup, you think of E as \mathbb{C}/\Lambda, where \Lambda \cong \mathbb{Z}^2, and \Lambda admits a given complex structure h; indeed, this complex structure is what tells apart two elliptic curves.

    Now I embed E into \mathbb{P}^2. You say that this induces a symplectic form \psi : \mathbb{Z}^2 \times \mathbb{Z}^2 \to \mathbb{Z}, and a Hodge one at that! What is this symplectic form explicitly? And why is it Hodge?

    C3. Is embedding E into \mathbb{P}^2 like choosing a Weierstraß equation for E? But how is varying the complex structure giving you different Weierstraß equations?

    B4. In the "Polarisations and Hermitian forms" section, right before the lemma, "...form on H on V..." should be replaced with "...form H on V...".

    C5. Trieste is a remarkable place. Venice is a very crowded place at Carnival time. The Library at the ICTP is the best I have ever been in.

  2. Martin Orr said on Thursday, 17 March 2011 at 15:51 :

    Thanks for the corrections. Sorry I have taken so long to reply, because I have been without a computer for a couple of weeks.

    A2. In the case of an elliptic curve, everything simplifies (I probably should have talked about this). That is why you don't encounter polarisations in the theory of elliptic curves, because they don't tell you much.

    The family of symplectic forms \mathbb{Z}^2 \times \mathbb{Z}^2 \to \mathbb{Z} is isomorphic to \mathbb{Z}: choose an ordered basis (a, b) of \mathbb{Z}^2. If C = \psi(a, b), then by linearity and anti-symmetry, for every p, q, r, s \in \mathbb{Z} we have  \psi(pa+qb, ra+sb) = (ps-qr)C.

    So \psi \mapsto \psi(a,b) is an isomorphism \wedge^2 \Lambda^* \to \mathbb{Z}. There are exactly two possible isomorphisms differing by a sign, which correspond to choosing a clockwise or an anti-clockwise ordered basis.

    This implies that all symplectic forms on \Lambda = \mathbb{Z}^2 are Hodge: the space V = \wedge^2 \Lambda_\mathbb{R}^* is one-dimensional, but we know that V^{2,0} = \overline{V^{0,2}}, so if either V^{0,2} or V^{2,0} is non-zero then both of them are and \dim V \geq 2. So V = V^{1,1} and all symplectic forms are automatically Hodge.

    Let us fix the sign of the isomorphism \wedge^2 \Lambda^* \to \mathbb{Z} by insisting that positive integers should correspond to symplectic forms where the corresponding Hermitian form is positive definite.

    Then the only question is: to which integer does a Weierstrass embedding E \hookrightarrow \mathbb{P}^2 correspond? The answer turns out to be 3 because a divisor on E induced by a line in \mathbb{P}^2 has degree 3.

    If we embed \Lambda in \mathbb{C} so that it has a basis \{1, \tau\} with \operatorname{Im} \tau > 0, then the associated Hermitian form \mathbb{C} \times \mathbb{C} \to \mathbb{C} is  H(z, w) = \frac{3 z \bar{w}}{\operatorname{Im} \tau}.

    C3. Embedding E in \mathbb{P}^2 is exactly choosing a Weierstrass equation for E. As I just said, so long as you stick to elliptic curves and Weierstrass equations then it doesn't matter which equation you pick: you always get the same Hodge form. This is not too surprising because any two Weierstrass equations are equivalent via an automorphism of \mathbb{P}^2.

    C5. Yes the library at the ICTP is nice, but I don't think you have been in the one at Orsay.

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