MathOverflow will be down for maintenance for approximately 3 hours, starting Monday evening (06/24/2013) at approximately 9:00 PM Eastern time (UTC-4).

## 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.

-
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 2010 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 2010 at 3:53
Sorry: not in $G(\mathbb Q)$, but in $H(\mathbb Q)$. – Emerton Dec 14 2010 at 3:59
Dear Brian, I think you are correct, but "secretly" I am of course asking a question about the corresponding Galois representations (which I imagine to be harder). Feel free to edit the question with your MO superpowers. – Lavender Honey Dec 14 2010 at 5:55
Dear Matt, Thanks for the remark, I've never really groked the Mumford-Tate group - although for conjugacy purposes it's presumably better to think of $H(\mathbf{C})$. – Lavender Honey Dec 14 2010 at 5:56
show 2 more comments