# How does complex conjugation act on the Hodge decomposition?

Let $A$ be a principally polarized abelian variety over $\mathbf{Q}$. Let $G$ be the Mumford--Tate group of $A$.

The action of complex conjugation on $A(\mathbf{C})$ induces an involution on the de Rham cohomology (with coefficients in $\mathbf{C}$), and thus defines an involution $\iota$ of $G(\mathbf{C})$. Because $A$ is principally polarized, and since complex conjugation preserves the polarization (up to scalar), $\iota$ defines an element of the group $$H:= \mathrm{normalizer}(G(\mathbf{C}) \subseteq \mathrm{GSp}_{2g}(\mathbf{C})).$$ My question, which I will try to make slightly more precise below, is: to what extent is the image of $\iota$ in $H$ independent of $A$ (given $G$)?

The de Rham cohomology of $A$ admits a Hodge decomposition: $$H^1_{dR}(A(\mathbf{C}),\mathbf{C}) = H^{1,0} \oplus H^{0,1}$$ on which complex conjugation sends $H^{1,0}$ to $H^{0,1}$. If $H = \mathrm{GSp}_{2g}$, then this is enough to determine $\iota \in H$ up to conjugacy. Specifically, it is the conjugacy class of involutions whose action on the adjoint representation has the smallest ($=$ most negative) trace.

The only other example I can compute: If $E$ has CM by $K$, then $G = \mathrm{Res}_{K/\mathbf{Q}}(\mathbf{G}_m)$, and so $G(\mathbf{C}) = \mathbf{C}^{\times} \oplus \mathbf{C}^{\times}$. If $H \subset \mathrm{GL}_2(\mathbf{C})$ is the normalizer of $(\mathbf C^{\times})^2$, then $\iota$ also defines a unique conjugacy class of $H$, which can be specified uniquely by saying that it generates the Weyl group.

More generally, one is tempted to ask whether $\iota$ defines a specific conjugacy class of involutions in $H$, except that perhaps one should ask for the image of $\iota$ in the slightly larger class of involutions in $H$ modulo conjugation by $G(\mathbf{C})$. Or perhaps there is another formulation, any suggestions welcome.

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Why not formulate the question for abelian varieties over $\mathbf{R}$? The $\mathbf{Q}$-structure on $A$ seems to be a red herring. (I am just making sure I am not misunderstanding what is relevant to the question.) – BCnrd Dec 14 '10 at 2:29
If I have everything straight, then the $\mathbb Q$-structure on the MT group comes from its action on the $\mathbb Q$-cohom. of $A(\mathbb C)$. So complex conjugation (on $A(\mathbb C)$) induces an automorphism of this cohom. In other words, complex conjugation in fact lies in $G(\mathbb Q)$. I don't claim this has any significance for the question at hand, but just want to point it out. – Emerton Dec 14 '10 at 3:53
Sorry: not in $G(\mathbb Q)$, but in $H(\mathbb Q)$. – Emerton Dec 14 '10 at 3:59
Dear algeBraic, I'm not sure whether the $\mathbb Q$ structure can help with anything. I'm currently trying to think about the role of the parabolic subgroup (if that's what it is) that preserves $H^{0,1}$. (I want to say something like there is a $P$ preserving $H^{0,1}$, and a $\overline{P}$ preserving $H^{1,0}$, which are a fairly integral part of the story, and since complex conjugation interchanges them, it is forced to be something like the longest element of the Weyl group. Whether this is at all correct is another question ... .) Cheers, Matt – Emerton Dec 14 '10 at 6:59
(comment removed, since it was based on a fundamental misunderstanding of the question) – Keerthi Madapusi Pera Dec 15 '10 at 1:31