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 ?


2 Answers 2



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)

see section 6 of A survey around the Hodge, Tate and Mumford-Tate conjectures for abelian varieties

  • $\begingroup$ Isn't the leftward arrow special to the case of abelian varieties? $\endgroup$
    – Will Sawin
    Apr 11, 2021 at 17:03
  • $\begingroup$ indeed, that is the case treated in the cited paper. $\endgroup$ Apr 11, 2021 at 17:07
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    $\begingroup$ 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. $\endgroup$
    – Will Sawin
    Apr 11, 2021 at 17:07
  • 1
    $\begingroup$ 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 $\endgroup$ Apr 12, 2021 at 7:16

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