Hodge structure versus Weight structure

This is a naive question.

One is told that, somehow, Hodge theory for varieties over complex numbers, is an analog of weight theory for varities over finite fields. In weight theory, one considers eigenvalues of Frobenius and so on. Hodge theory should capture symmetries of the Galois group of complex numbers over real numbers.

My question is: It is confusing; For a variety over a finite field, it makes sense that the Galois group will act on different things and give some information etc.; But in Hodge theory, we begin with a variety over the complex numbers, not over the real numbers; So what is the logic by which the Galois group of complex numbers over real numbers enters the picture? What is the logic behind the words "hidden symmetries" in the book by Gelfand and Manin?

i.e., when we have a variety over the algebraic closure of a finite field, we do not claim to have "weight theory"; We need structures to be defines over the finite field..

I hope my naive question is clear,

Thank you, Sasha

• The Hodge structure does not come from the field of definition (so the Galois group of $\mathbb{C}/\mathbb{R}$ does not really enter the picture). The structure comes from the field of coefficients: You take cohomology with coefficients in $\mathbb{Q}$ (or $\mathbb{Z}$) and then extend scalars to $\mathbb{C}$ (remembering the $\mathbb{Q}$-structure). If you would start with $\mathbb{C}$-coefficients, there is no way to get the Hodge structure. Hope this small comment clarifies the confusion a bit. – jmc Nov 12 '13 at 9:31
• @jmc and all others: So, if we have a variety over an algebraically closed field, and we consider l-adic cohomology, there should be some weight theory attached (related to Galois group of Q_l bar over Q_l)? How is it called? – Sasha Nov 20 '13 at 8:47
• Dear Sasha, I am not sure I understand you. The fact that there is a Hodge structure on singular cohomology is not just a consequence of abstract nonsense involving some action of Galois. To prove it you need lots of “analytic” stuff, such as harmonic forms. I am not an expert, but I don't think there is a sensible analogue at the $\ell$-adic side. There are things such as algebraic de Rham cohomology (which is comparable with singular cohomology and its Hodge structure), and $p$-adic Hodge theory (involving the $p$-adic cohomology, where $p$ is the residue characteristic). – jmc Nov 20 '13 at 10:22
• Dear jmc, thank you. Of course I did not mean that it is something trivial, just wondered if there is such a theory. – Sasha Nov 21 '13 at 12:28

I'm not sure how helpful this is, but anyway: in the étale cohomology of a variety over a finite field there is an action of $\mathrm{Gal}(\overline{\mathbf F}_q/\mathbf F_q)$ on $$H^\bullet_{\text{ét}}(X \times_{\mathbf F_q} \overline{\mathbf F}_q,\mathbf Q_\ell)$$ and in the Betti cohomology of a complex variety there is a Galois action of $\mathrm{Gal}(\mathbf C/\mathbf R)$ on $$H^\bullet(X(\mathbf C),\mathbf C).$$ In the first case the Galois action is on the first factor, i.e. on the variety. In the second case the Galois action is on the second factor, i.e. on the coefficients. This is why there is no need for $X$ to be defined over the real numbers in the second case.
I think your misconception is that since the weight filtration in étale cohomology encodes the absolute values of Frobenius eigenvalues, that there should be a similarly strong relation between weight filtration and Galois action in Betti cohomology. The action of $\mathrm{Gal}(\mathbf C/\mathbf R)$ is important in Hodge theory because it's what allows you to talk about the $(p,q)$-decomposition. But it can't be what gives rise to weights, since the weight filtration exists already with $\mathbf Q$-coefficients.
I don't know how helpful this will be either. But let me modify your question a bit to ask what is the correct analogue of the Galois group (of say a finite field or number field) in Hodge theory? It is certainly true that when $X$ is a smooth complex projective variety, $Gal(\mathbb{C}/\mathbb{R})=\mathbb{Z}/2\mathbb{Z}$ acts on $H=H^i(X,\mathbb{C})$, by virtue of the fact that it is obtained by complexifying real de Rham cohomology. However, this is rather weak information; certainly too weak to recover the Hodge structure. Fortunately, one can do better. Deligne's torus $\mathbb{S}$ is the real algebraic group whose real points $\mathbb{S}(\mathbb{R})=\mathbb{C}^\ast$. So that $\mathbb{S}(\mathbb{C})=\mathbb{C}^\ast\times \mathbb{C}^\ast$. Note that we have natural cocharacter $\mathbb{G}_m\to \mathbb{S}$ corresponding to the inclusion $\mathbb{R}^\ast\subset\mathbb{C}^\ast$ on real points. $H$ becomes a real representation of this group, by letting $(z_1, z_2)\in \mathbb{S}(\mathbb{C})$ act on $H^{pq}\subset H$, by $z_1^pz_2^q$. This is a better answer to the question, because this action completely determines the real Hodge structure on $H$. The weight $i$ is gotten by restricting the action to $\mathbb{G}_m$, and observing that the characters of $H$ are just $i$th twists of the standard character.
If you look at the rational Hodge structure, then there is a bigger group called the Mumford-Tate group $MT(H)$, that I won't attempt to define, that acts on $H_{\mathbb{Q}}= H^i(X,\mathbb{Q})$ and competely determines the Hodge structure on it. When $X$ is defined over a number field $K$, then conjecturally $MT(\mathbb{Q}_\ell)$ should be (essentially) the same as the image of the action of $Gal(\bar K/K)$ on $\ell$-adic cohomology $H^i(\bar X_{et},\mathbb{Q}_\ell)$.
• Dear Donu Arapura, so if I understand you correctly the MT-group is the analogue of the (Zariski closure) of the image of the Galois group. Would the analogue of the Galois group be the (pro-)group corresponding to the Tannakian category of $\mathbb{Q}$-Hodge structures? I guess that is also the categorical limit of the MT-groups, right? – jmc Nov 21 '13 at 14:51