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Deligne's Principle A and the Mumford-Tate conjecture

Posted by Martin Orr on Wednesday, 10 September 2014 at 11:20

In this post I will fill in a missing detail from two weeks ago, where I mentioned that the Mumford-Tate group is determined by the Hodge classes. More precisely, I will show that an element g of \operatorname{GL}(H_1(A, \mathbb{Z})) is in the Mumford-Tate group if and only if every Hodge class on every Cartesian power A^r is an eigenvector of g. In the context of Deligne's theorem on absolute Hodge classes, this is known as Principle A.

We will also see that a version of this statement holds for the \ell-adic monodromy group and Tate classes. This implies a link between the Hodge, Tate and Mumford-Tate conjectures.

Characterisation of the Mumford-Tate group in terms of Hodge classes

Let A be an abelian variety over \mathbb{C}.

We saw previously that an element of H^{2p}(A, \mathbb{Q}) is a Hodge class if and only if it is an eigenvector for the Mumford-Tate group of A. The obvious guess at a converse to this would be that an element g \in \operatorname{GL}(H_1(A, \mathbb{Z})) is in the Mumford-Tate group if and only if every Hodge class in H^{2p}(A, \mathbb{Q}) (for every p) is an eigenvector of g. However this turns out to be false.

To recognise exactly which g are in the Mumford-Tate group, we need more Hodge classes. We can find these by looking at Cartesian powers of A. Observe that H_1(A^r, \mathbb{Q}) is the direct sum of r copies of H_1(A, \mathbb{Q}), so there is a natural action of \operatorname{GL}(H_1(A, \mathbb{Q})) on H_1(A^r, \mathbb{Q}), and hence on H^n(A^r, \mathbb{Q}) \cong \bigwedge^n H_1(A^r, \mathbb{Q})^\vee.

Using these actions, we can state and prove the following.

Theorem. An element g \in \operatorname{GL}(H_1(A, \mathbb{Q})) is in the Mumford-Tate group of A if and only if every Hodge class v \in H^{2p}(A^r, \mathbb{Q}) (for every p and r) is an eigenvector of g.

Proof. The part we have to prove is that if every Hodge class v \in H^{2p}(A^r, \mathbb{Q}) (for every p and r) is an eigenvector of g_0 \in \operatorname{GL}(H_1(A, \mathbb{Q})), then g_0 \in MT(A).

We will prove this using Chevalley's theorem on algebraic subgroups:

Theorem. Let G \subset \operatorname{GL}(V) be an algebraic subgroup. Then there exist nonnegative integers m, n and a vector v \in V^{\otimes m} \otimes V^{\vee \otimes n} such that  G = \{ g \in \operatorname{GL}(V) \mid v \text{ is an eigenvector for } g \}.

Of course we will apply this with V = H_1(A, \mathbb{Q}) and G = MT(A).

But first we recall that MT(A) \subset \operatorname{GSp}(V, \psi) where \psi is a polarisation. As representations of \operatorname{GSp}(V, \psi), V is isomorphic to V^\vee \otimes \chi where \chi is the character telling us how \operatorname{GSp}(V, \psi) changes \psi.

Tensoring with the character \chi does not change which vectors are eigenvectors (only the characters associated with those eigenvectors) so applying the isomorphism of \operatorname{GSp}(V, \psi)-representations V^{\otimes m} \otimes V^{\vee \otimes n} \cong V^{\vee \otimes (m+n)} \otimes \chi^m, we can get v' \in V^{\vee \otimes (m+n)} such that  MT(A) = \{ g \in \operatorname{GSp}(V, \psi) \mid v' \text{ is an eigenvector for } g \}. \tag{*}

By the K√ľnneth formula, V^{\vee \otimes (m+n)} is a sub-\operatorname{GSp}(V, \psi)-representation of  H^{m+n}(A^{m+n}, \mathbb{Q}).

Thus we can regard v' as a class in H^{m+n}(A^{m+n}, \mathbb{Q}). By construction v' is an eigenvector for MT(A), so it is a Hodge class. Thus, by main hypothesis of the theorem, v' is an eigenvector for g_0.

We also know that the polarisation \psi is a Hodge class of weight 1 and hence, by hypothesis, an eigenvector for g_0. In other words g_0 \in \operatorname{GSp}(V, \psi). We conclude that g_0 satisfies the condition of (*) to be in MT(A), as required.

Characterisation of the \ell-adic monodromy group in terms of Tate classes

Now let A be an abelian variety defined over a finitely generated field. Similarly to above, we can prove:

Theorem. An element g \in \operatorname{GSp}(T_\ell A \otimes \mathbb{Q_\ell}) is in the \ell-adic algebraic monodromy group of A if and only if every Tate class v \in H^{2p}_{et}(A_{\bar{k}}^r, \mathbb{Q_\ell})(p) (for every p and r) is fixed by g.

This requires a little more work to prove than the statement about Hodge classes, because not every eigenvector of the \ell-adic monodromy group is a Tate class but only those eigenvectors with eigenvalue 1. They key point is that, if G \subset \operatorname{GL}(V) is reductive, then we can extend Chevalley's theorem to say that there is some p and some v \in V^{\otimes p} \otimes V^{\vee \otimes p}  G = \{ g \in \operatorname{GL}(V) \mid gv = v \}. (We have altered Chevalley's theorem in two ways: elements of g fix v rather than just having v as an eigenvector, and V and V^\vee are raised to the same tensor power.)

The fact that the \ell-adic algebraic monodromy group G_\ell is reductive was proved by Faltings as part of his work on the Mordell conjecture. So we can apply the strengthened Chevalley's theorem to V = T_\ell A \otimes \mathbb{Q}_\ell and G = G_\ell. Applying the same dualising trick as in the Hodge case, we get v' \in V^{\vee \otimes 2p}(p) such that  G_\ell = \{ g \in \operatorname{GSp}(V, \psi) \mid gv' = v \}.

We can regard v' as an element of H^{2p}_{et}(A_{\bar{k}}^{2p}, \mathbb{Q}_\ell)(p). Then v' is a Tate class, and we can apply the same argument as before to conclude that, if  g_0 \in \{ g \in \operatorname{GSp}(T_\ell A \otimes \mathbb{Q_\ell}) \mid gv' = v' \} then g_0 \in G_\ell.

We can also modify the above theorem, replacing the \ell-adic algebraic monodromy group by its identity component and Tate classes by potentially Tate classes.

The Mumford-Tate conjecture

Now we focus on abelian varieties defined over a number field k (with a chosen embedding into \mathbb{C}). Recall the statement of the Mumford-Tate conjecture.

Mumford-Tate conjecture. The identity component G_\ell^\circ of the \ell-adic algebraic monodromy group is equal to MT(A) \times_\mathbb{Q} \mathbb{Q}_\ell (as subgroups of \operatorname{GL}(T_\ell A \otimes \mathbb{Q}_\ell) = \operatorname{GL}(H_1(A, \mathbb{Q})) \times_\mathbb{Q} \mathbb{Q}_\ell).

Using the above characterisations of MT(A) and G_\ell^\circ, this is equivalent to the following.

Conjecture. For every r and p, the subspace of potentially Tate classes in H^{2p}(A_{bar{k}}^r, \mathbb{Q}_\ell)(p) is equal to \operatorname{Hdg}^p(A^r)(p) \otimes_{\mathbb{Q}} \mathbb{Q}_\ell.

The Hodge conjecture says that the space of Hodge classes is equal to the \mathbb{Q}-span of algebraic cycle classes, and the Tate conjecture says that the space of potentially Tate classes is equal to the \mathbb{Q}_\ell-span of algebraic cycle classes. Hence we can deduce that if any two out of the Hodge, Tate and Mumford-Tate conjectures are true (for all Cartesian powers of a given variety), the third is also true.

Note that the last paragraph requires one additional fact we have not mentioned before. The Hodge conjecture concerns classes of algebraic cycles defined over \mathbb{C} while the Tate conjecture (in its potentially Tate classes version) concerns classes of algebraic cycles defined over \bar{k}. Fortunately this does not matter because if K \subset L are algebraically closed fields, X is a variety defined over K, and Z is a subvariety of X defined over L, then there exists a subvariety Z' \subset X defined over K which is algebraically equivalent to Z. Algebraically equivalent varieties are mapped to the same cohomology class by the cycle class map (into any Weil cohomology theory).

Deligne's theory of absolute Hodge classes, which I intend to discuss in my next posts, proves that every Hodge class on an abelian variety is a potentially Tate class. This is half of the Mumford-Tate conjecture and allows us to conclude that, for an abelian variety, G_\ell^\circ \subset MT(A) \times_{\mathbb{Q}} \mathbb{Q}_\ell.

Tags abelian-varieties, alg-geom, alg-groups, hodge, maths, number-theory

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