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On p. 205 of Katz's paper entitled "p-adic L-functions for CM fields" Katz says that

"Shimura's algebraicity theorem, in our context, is an easy consequence of the fact that Hodge decomposition of the $H_{dR}^1$ of our CM abelian varieties is purely algebraic (i.e. $H^{0,1}$ is defined over $\overline{\mathbb{Q}}$), and is valid for any CM-type"

Since Katz makes reference to $H^{0,1}$, $H_{dR}^1$ probably means the classical $C^{\infty}$ de Rham cohomology with coefficients in $\mathbf{C}$. So if $A$ is an abelian variety of dimension $g$ defined over $k:=\overline{\mathbb{Q}}$, then, looking at the algebraic De Rham cohomology, which we denote by $H_{alg-dR}^{1}(A/k)$, one has a basis of $H_{alg-dR}^{1}(A/k)$ which consists of $g$ regular algebraic differentials and $g$ algebraic differentials of the second kind, all defined over $k$. Using the comparison isomorphism (after base change to $\mathbf{C}$) between $H_{alg-dR}^{1}(A/k)\otimes \mathbf{C}$ and $H_{dR}^1(A,\mathbf{C})$ we see that it is always possible to find a basis of $H^{1,0}$ with regular differentials defined over $k$ and a basis of $H^{0,1}$ corresponding to (via the comparison isomorphism) to algebraic differentials of the second kind defined over $k$. But this is always true regardless of the assumption that $A$ is CM. This seems to use only the fact that $A$ is defined over $k$.

Q: So what does Katz really mean by the "Hodge decomposition is purely algebraic in the case where $A$ is CM"?

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Suppose, to simplify, that $A$ is defined over $\mathbb{Q}$. Then $H^1(A_{\mathbb{C}},\mathbb{C})$ has two $\mathbb{Q}$-structures, one coming from singular cohomology, the other one from the algebraic de Rham cohomology; they are different. For instance, for an elliptic curve $A=\mathbb{C}/(\mathbb{Z}+\mathbb{Z}\tau )$, if $(e_1,e_\tau)$ is the basis of $H_1(A_{\mathbb{C}},\mathbb{Z})$ corresponding to $(1,\tau )$, the subspace $H^{1,0}$ of $H^1(A_{\mathbb{C}},\mathbb{C})=\mathrm{Hom}(H_1(A_{\mathbb{C}},\mathbb{Z}),\mathbb{C}$ is $e_1^*+\tau e_\tau^*$; it is defined over $\bar{\mathbb{Q}}$ iff $\tau \in \bar{\mathbb{Q}}$, which is of course the case if $A$ is CM. In higher dimension it is easy to see that $H^{1,0}(A)$ is defined over $\bar{\mathbb{Q}}$ if $A$ is CM -- in fact it is defined over the CM-field of $A$.

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  • $\begingroup$ I see, so hee simply meant algebraic over the base field of the abelian variety. Thanks. $\endgroup$ Feb 19, 2014 at 22:40

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