# The Mumford-Tate conjecture

The Mumford-Tate conjecture asserts that, via the Betti-étale comparison isomorphism, and for any smooth projective variety $$X$$, over a number field $$K$$, the $$\mathbb{Q}_{ \ell }$$-linear combinations of Hodge cycles coincide with the $$\ell$$-adic Tate cycles.

Question. Would that mean that if the Hodge conjecture and the Tate conjecture hold, then the Mumford-Tate conjecture holds as well ?

Yes.

Under the Hodge conjecture, the Hodge cycles are the algebraic cycles, so the $$\mathbb Q_\ell$$-linear combinations of Hodge cycles are the $$\mathbb Q_\ell$$-linear combinations of algebraic cycles.

Under the Tate conjecture, the $$\ell$$-adic Tate classes are the $$\mathbb Q_\ell$$-linear combinations of algebraic cycles.

So under the Hodge and Tate conjectures, these are both equal.

This then implies that the identity component of the $$\ell$$-adic monodromy group is isomorphic over $$\mathbb Q_\ell$$ to the Mumford-Tate group, by a Tannakian argument.

One can deduce the Tate conjecture for every abelian variety which satisfies the Mumford-Tate and the Hodge conjecture, and vice versa: (MT) + (H) $$\Leftrightarrow$$ (T)

• Isn't the leftward arrow special to the case of abelian varieties? Apr 11 at 17:03
• indeed, that is the case treated in the cited paper. Apr 11 at 17:07
• Also, one step in the argument for that equivalence is "Moreover one can also prove that (H) + (T ) ⇒ (MT ) using motives." I think that step is in fact not restricted to abelian varieties. Apr 11 at 17:07
• Another discussion of the relationship between the conjectures, using André's notion of motivated cycles, is in Section 3.2 of link.springer.com/content/pdf/10.1007/s00032-017-0273-x.pdf Apr 12 at 7:16