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Maths > Abelian varieties > Absolute Hodge classes

Absolute Hodge classes

Posted by Martin Orr on Thursday, 20 November 2014 at 18:55

Let A be an abelian variety over a field k of characteristic zero. For each embedding \sigma \colon k \hookrightarrow \mathbb{C}, we get a complex abelian variety A^\sigma by applying \sigma to the coefficients of equations defining A.

Whenever an object attached to A is defined algebraically, we will get closely related objects for each A^\sigma. On the other hand, whenever we use complex analysis to define an object attached to A^\sigma, we should expect to get completely unrelated things for different \sigma (if k = \mathbb{C} then most field embeddings k \hookrightarrow \mathbb{C} are horribly discontinuous so will mess up anything analytic).

Hodge classes provide a special case: the definition of Hodge classes on A^\sigma as H^{2p}(A^\sigma, \mathbb{Z}) \cap H^{p,p}(A^\sigma) is analytic so we expect no relation between Hodge classes on different A^\sigma. But the Hodge conjecture says that every Hodge class in H^{2p}(A^\sigma, \mathbb{C}) is an algebraic cycle class, and this implies the associated cohomology class in H^{2p}(A^{\sigma'}, \mathbb{C}) is also a Hodge class for every \sigma' \colon k \hookrightarrow \mathbb{C}. (We will explain in the post why there is a natural semilinear isomorphism H^{2p}(A^\sigma, \mathbb{C}) \to H^{2p}(A^{\sigma'}, \mathbb{C}).)

Deligne had the idea that we could pick this out as a partial step on the way to the Hodge conjecture: he defined an absolute Hodge class to be a cohomology class such that its associated class on A^\sigma is a Hodge class for every \sigma and proved that every Hodge class on an abelian variety is an absolute Hodge class. It turns out that this is sufficient to obtain some of the consequences which would follow from the Hodge conjecture. In this post we will explain the definition of absolute Hodge classes.

De Rham cohomology

So far in this blog, we have only seen an analytic construction of the cohomology with complex coefficients H^n(A^\sigma, \mathbb{C}) of a complex abelian variety. Before we can talk about absolute Hodge classes, we need to construct this algebraically, so that we get semilinear isomorphisms between H^n(A^\sigma, \mathbb{C}) for different \sigma, and hence know what it means to talk about a cohomology class being a Hodge class with respect to all \sigma.

Recall that a few posts ago, we constructed the tangent space of the universal vector extension E_A. This is a vector space over k. We proved that, looking at the complex abelian variety A^\sigma, there is a canonical isomorphism  T_0(E_{A^\sigma})  \cong  H_1(A^\sigma, \mathbb{C}).

Using the fact that the construction of the universal vector extension is compatible with extensions of the base field, for each \sigma \colon k \hookrightarrow \mathbb{C} we get a canonical isomorphism  T_0(E_A) \otimes_{k,\sigma} \mathbb{C}  \cong  H_1(A^{\sigma}, \mathbb{C}).

Following the fact that the cohomology groups of A are exterior powers of the dual of H_1(A), we define the n-th de Rham cohomology of A to be the k-vector space  H^n_{dR}(A/k) = \bigwedge^n T_0(E_A)^\vee (usually one would define de Rham cohomology in a more general way, for all algebraic varieties, and then it is a theorem that the above formula holds for abelian varieties). We get canonical isomorphisms  \sigma^* \colon H^n_{dR}(A/k) \otimes_{k,\sigma} \mathbb{C}  \to  H^n(A^{\sigma}, \mathbb{C}).

De Rham cohomology and the Hodge filtration

As we discussed previously for H_1, the Hodge decomposition for A^\sigma cannot defined algebraically (it cannot come from a decomposition of the k-vector space H_{1,dR}(A/k) = T_0(E_A)) but the Hodge filtration  H^{0,-1}(A^\sigma) \subset H_1(A^\sigma, \mathbb{C}) corresponds to the first part of the exact sequence  0 \to T_0(A^\vee)^\vee \to T_0(E_A) \to T_0(A) \to 0. Dualising, we get that \sigma^* maps the filtration  T_0(A)^\vee \otimes_{k,\sigma} \mathbb{C}  \subset  T_0(E_A)^\vee \otimes_{k,\sigma} \mathbb{C} to  H^{1,0}(A^\sigma) \subset H^1(A^\sigma, \mathbb{C}).

We can define a descending filtration on H^n_{dR}(A/k) = \bigwedge^n T_0(E_A)^\vee by letting F^p H^n_{dR}(A/k) be the subspace generated by elements of the form v_1 \wedge \cdots \wedge v_n where v_1, \dotsc, v_p \in T_0(A)^\vee and v_{p+1}, \dotsc, v_n \in T_0(E_A)^\vee. Then the above gives us  \sigma^*(F^p H^n_{dR}(A/k) \otimes_{k,\sigma} \mathbb{C}) = \bigoplus_{s \geq p} H^{s,n-s}(A^\sigma), or in other words "the Hodge filtration is defined algebraically."

Hodge classes and de Rham cohomology

We say that a de Rham cohomology class v \in H^{2p}_{dR}(A/k) is a Hodge class relative to \sigma if \sigma^*(v) is a Hodge class on A^\sigma i.e. if  \sigma^*(v) \in H^{p,p}(A^\sigma) \cap H^{2p}(A^\sigma, \mathbb{Z}). We can equivalently say  v \in F^p H^{2p}_{dR}(A/k) \text{ and } \sigma^*(v) \in H^{2p}(A^\sigma, \mathbb{Z}) because if  \sigma^*(v) \in \bigl( \bigoplus_{s \geq p} H^{s,2p-s}(A^\sigma) \bigr) \cap H^{2p}(A^\sigma, \mathbb{Z}) then \sigma^*(v) is its own complex conjugate (because it is in H^{2p}(A^\sigma, \mathbb{Z})) and so the complex conjugation condition on the Hodge decomposition forces \sigma^*(v) \in H^{p,p}(A^\sigma).

Observe that the condition v \in F^p H^{2p}_{dR}(A/k) is independent of \sigma, while the condition v \in H^{2p}(A^\sigma, \mathbb{Z}) depends on \sigma and because v \in H^{2p}(A^\sigma, \mathbb{Z}) is an analytic object, even if the second condition holds for one \sigma, it probably does not hold for most \sigma.

Enter the Hodge conjecture

Suppose that we have a cohomology class v \in H^{2p}_{dR}(A/k) which is a Hodge class relative to one embedding \sigma \colon k \hookrightarrow \mathbb{C}. The Hodge conjecture says that v is in the \mathbb{Q}-span of the classes cl(Z) of algebraic subvarieties Z \subset A^\sigma.

Now suppose we are given another embedding \sigma' \colon k \hookrightarrow \mathbb{C}, and suppose that there is an automorphism \tau of \mathbb{C} such that \tau\sigma = \sigma'. (Such a \tau does not always exist when k = \mathbb{C} because there are non-surjective embeddings \mathbb{C} \to \mathbb{C}. There is a way of dealing with \sigma' for which there is no \tau, but we will ignore them for now.)

Then applying \tau to the coefficients of equations defining the subvariety Z \subset A^\sigma, we get a subvariety Z^\tau \subset A^{\sigma'}. Furthermore, there is an algebraic recipe for defining cycle classes in de Rham cohomology, and this implies that  cl(Z^\tau) = \tau^*(cl(Z)) \in H^{2p}(A^{\sigma'}, \mathbb{C}). Hence \tau^*(cl(Z)) is an algebraic cycle class, and hence a Hodge class.

Thus the Hodge conjecture implies that \sigma'^*(v) is a Hodge class for every \sigma' \colon k \hookrightarrow \mathbb{C}.

We define an absolute Hodge class to be a de Rham cohomology class v \in H^{2p}_{dR}(A/k) which is a Hodge class relative to every \sigma \colon k \hookrightarrow \mathbb{C}. Deligne proved:

Theorem. If A is an abelian variety over k and v \in H^{2p}_{dR}(A/k) is a Hodge class relative to one embedding \sigma \colon k \hookrightarrow \mathbb{C}, then v is an absolute Hodge class.

By the above argument, the Hodge conjecture would imply that the theorem is true for all algebraic varieties. However it is only known for abelian varieties, K3 surfaces and a few other special cases.

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


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