Deligne's Principle A and the Mumford-Tate conjecture

Maths > Abelian varieties > Absolute Hodge classes

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 of 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 is in the Mumford-Tate group if and only if every Hodge class in (for every $p$ ) is an eigenvector of $g$ . However this turns out to be false.

To recognise exactly which are in the Mumford-Tate group, we need more Hodge classes. We can find these by looking at Cartesian powers of $A$ . Observe that is the direct sum of $r$ copies of , so there is a natural action of on , and hence on .

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

Theorem. An element 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 , 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

Of course we will apply this with 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

By the Künneth 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 is in the $\ell$ -adic algebraic monodromy group of $A$ if and only if every Tate class (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 $v \in V^{\otimes p} \otimes V^{\vee \otimes p}$ (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 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 and . Applying the same dualising trick as in the Hodge case, we get such that

We can regard $v'$ as an element of . Then $v'$ is a Tate class, and we can apply the same argument as before to conclude that, if then .

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 of the $\ell$ -adic algebraic monodromy group is equal to (as subgroups of ).

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 is equal to .

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 -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, .

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

Trackbacks

Absolute Hodge classes in l-adic cohomology From Martin's Blog

We can define absolute Hodge classes in l-adic cohomology in the same way as absolute Hodge classes in de Rham cohomology. We can then prove Deligne's theorem, that Hodge classes on an abelian variety are absolute Hodge, for l-adic cohomology. ...
Motivic Galois groups and periods From Martin's Blog

In my last post, I discussed ... an upper bound for the transcendence degree of the extended period matrix of an abelian variety, namely the dimension of the general symplectic group. ... In this post, I will discuss how this can be generalised ...

Comments

No comments.

Name (required)
Email (required)
Website
Comment
Enter any number with three digits
	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.

Martin Orr's Blog