Say $A_0$ is an ordinary abelian variety over ${\mathbf{F}}_q$. Call $\mathcal{A}$ the canonical lift of $A_0$ over $R := W({\mathbf{F}}_q)$. It carries a lift of the $q$-th power map on $A_0$. We call $\phi : \mathcal{A}\to\mathcal{A}$ this lift. It exists by functoriality of the canonical lift.

Call $K = \text{Frac}(R)$ and choose a field embedding $K\subset\mathbf{C}$.

Call $A$ the complex torus $(\mathcal{A}\times_K\mathbf{C})(\mathbf{C})$ and $F$ the endomorphism of $A$ induced by the Frobenius lift on $\mathcal{A}$, i.e. $$F = (\phi \times_{\text{Spec}(R)}\text{id}_{\mathbf{\text{Spec}(C)}})^{\rm an}:A\to A.$$

$F$ acts on the Betti cohomology $H^*(A,\mathbf{C})$. By the Weil conjectures (or by an argument of Serre, in this case), its eigenvalues are of the form $q^{*/2}\zeta$ for algebraic numbers $\zeta$ of complex absolute value $1$.

Suppose $v \in H^{2m}(A,\mathbf{C})$ is an eigenvector of $F$ whose eigenvalue is of the form $q^m\zeta$ for $\zeta$ a root of unity. Is $v$ a class of type $(m,m)$?

  • $\begingroup$ You have not said what $q$ is. (Presumably $A_0$ is defined over $\mathbf{F}_q$ and the Frobenius is the relative Frobenius corresponding to this field.) $\endgroup$
    – naf
    Aug 2, 2021 at 7:41
  • $\begingroup$ @naf Yes, that is correct. I will clarify, just in case $\endgroup$
    – user322153
    Aug 2, 2021 at 9:23

1 Answer 1


Let's examine how $\phi$ acts on the algebraic Dolbeaut cohomology $$H^1(\mathcal A_K , \mathcal O_{\mathcal A})+ H^0 ( \mathcal A_K, \Omega^1_{\mathcal A}).$$

I claim its eigenvalues on $H^1(\mathcal A_K , \mathcal O_{\mathcal A})$ are units and its eigenvalues on $H^0 ( \mathcal A_K, \Omega^1_{\mathcal A})$ are $q$ times units.

For the first claim, we can use the Artin-Schreier exact sequence $$H^1 ( A_{0, \overline{\mathbb F_q}}, \mathbb Z/p) \to H^1 ( A_{0, \overline{\mathbb F_q}} , \mathcal O_{ A_0} )\to H^1 ( A_{0, \overline{\mathbb F_q}} , \mathcal O_{ A_0} ) ,$$ the image of whose first arrow is a basis for $H^1 ( A_{0, \overline{\mathbb F_q}} , \mathcal O_{ A_0} )$ on which Frobenius acts invertibly, showing that Frobenius acts invertibly on $H^1 ( A_0, \mathcal O_{A_0})$ and thus invertibly on $H^1 (\mathcal A, \mathcal O_{\mathcal A})$.

For the second claim, we can use the fact that the pullback of a polarization along $\phi$ is $q$ times that polarization, so Frobenius is a symplectic similitude with similitude character $q$ for the form on $H^1(\mathcal A_K)$ induced by that polarization, hence for each eigenvalue $\lambda$ that is a $p$-adic unit, $q/\lambda$ must also be an eigenvalue.

Using that claim and the isomorphism $$\wedge ^a H^1(\mathcal A_K , \mathcal O_{\mathcal A})\otimes \wedge^b H^0 ( \mathcal A_K, \Omega^1_{\mathcal A}) \to H^a ( \mathcal A_K, \Omega^b_{\mathcal A})$$ we conclude that all eigenvalues of $\Phi$ on $$H^a ( \mathcal A_K, \Omega^b_{\mathcal A})$$ are $q^b$ times a $p$-adic unit.

So any eigenvalues in degree $2m$ of the form $q^m$ times a unit must occur in $H^{m,m}$, as desired.

  • 1
    $\begingroup$ A few quick questions. (1) Do you mean $H^1$ instead of $H^0$ when you use the Artin-Schreier sequence? (2) Also I guess Frobenius acts invertibly on $H^1(A_{0,\overline{\mathbf{F}}_q},\mathbf{Z}/p)$ because the arithmetic Frobenius is its inverse? (3) I assume your $K$ is $\text{Frac}(W(\overline{\mathbf{F}}_q))$, am I right? (4) And finally, why does the image of the first map in the Artin-Schreier sequence generate $H^1(A_{0,\overline{\mathbf{F}}_q},\mathcal{O}_{A_0})$? Thank you very much! Great answer $\endgroup$
    – user322153
    Aug 6, 2021 at 1:56
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    $\begingroup$ @Arty (1) Yes, fixed. (2) Exactly. (3) No, I mean the same as you. When taking algebraic de Rham cohomology there is no need to pass to the unramified extension. (4) The key thing is to observe that the next arrow looks like the identity minus a semilinear Frobenius map semilinear under $x \mapsto x^p$. One can check that any relation between vectors in the kernel of such a map must be defined over $\mathbb F_p$: $\endgroup$
    – Will Sawin
    Aug 6, 2021 at 2:21
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    $\begingroup$ @Arty It suffices to check this for the smallest relation among a given set of vectors, and then its $p$th power must also be a relation. Some linear combination of this relation and the original must have vanishing coefficient of the first vector, so by assumptions all coefficients vanish, forcing the original relation to be defined over $\mathbb F_p$ (up to scalar multiplication). Using this, because the kernel has rank $g$ as the abelian variety is ordinary, we get that the kernel forms a basis. Maybe there is a better way... $\endgroup$
    – Will Sawin
    Aug 6, 2021 at 2:22
  • $\begingroup$ I have a basic question about the denotation 'algebraic Dolbeaut cohomology' in this context. Do you really mean aDc $H^1(\mathcal A_K , \mathcal O_{\mathcal A})+ H^0 ( \mathcal A_K, \Omega^1_{\mathcal A})$ or should the $\mathcal A_K$ replaced by $\mathcal{A}\times_K\mathbf{C}$? Because the only 'algebraic Dolbeaut cohomology' I know (up to now) deals with smooth complex varieties (as analoga of complex manifolds with usual Dolbeaut cohomology). $\endgroup$
    – user267839
    Aug 7, 2021 at 7:26
  • $\begingroup$ Or is the terminology'algebraic Dolbeaut cohomology' you are using here a proper generalization to Abelian varieties over other fields than $\mathbf{C}$? $\endgroup$
    – user267839
    Aug 7, 2021 at 7:27

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