How does complex conjugation act on the Hodge filtration?

Question

Let $X$ be a $\mathbb{R}$-defined smooth proper scheme, and let $H^i_{\text{dR}}(X)$, denote its algebraic de Rham cohomology. The Hodge filtration gives an $\mathbb{R}$-defined pure Hodge structure on $H^i_{\text{dR}}(X)$, which yields in practice a (Hodge) filtration on this $\mathbb{R}$-vector space.

It is said that, as a part of the definition of a Hodge structure, this filtration must satisfy a certain property, namely that for every pair of non-negative integers, $p+q = i + 1$, $\mathcal{F}^pH^i_{\text{dR}}(X)\cap \overline{\mathcal{F}^qH^i_{\text{dR}}(X)} = 0$.

My question is: since the filtration is $\mathbb{R}$-defined, in the sense that each $\mathcal{F}^i$ is the complexification of an $\mathbb{R}$-sub vector space of $H^i_{\text{dR}}(X)$, doesn't this mean that complex conjugation acts trivially on $\mathcal{F}^qH^i_{\text{dR}}(X)$, in the sense that $\mathcal{F}^qH^i_{\text{dR}}(X) = \overline{\mathcal{F}^qH^i_{\text{dR}}(X)}$? In other words, is the complex conjugation redundant whenever my base scheme is stable under complex conjugation?

Answer 1

Let us first consider the singular cohomology group $H^i(X(\mathbb{C}),\mathbb{C})$. There are three ways to let the group $G:=\mathrm{Gal}(\mathbb{C}/\mathbb{R})\simeq\mathbb{Z}/2$ act on it : by letting complex conjugation act on the coefficients $\mathbb{C}$, or on the space $X(\mathbb{C})$, or on both. Denote by $F_{\mathrm{B}}$, $F_{\infty}$ and $F_{\mathrm{dR}}$ these three involutions of $H^i(X(\mathbb{C}),\mathbb{C})$. They commute and $F_{\mathrm{dR}}=F_{\infty}\circ F_{\mathrm{B}}$. The involution $F_{\infty}$ is $\mathbb{C}$-linear, and $F_B$ and $F_{\mathrm{dR}}$ are $\mathbb{C}$-antilinear. The involution $F_{\mathrm{B}}$ exists for all complex varieties (and is the antiholomorphic involution associated with the real structure $H^i(X(\mathbb{C}),\mathbb{R})$ of $H^i(X(\mathbb{C}),\mathbb{C})$), but defining $F_{\infty}$ and $F_{\mathrm{dR}}$ requires $X$ to be defined over $\mathbb{R}$. The involution $F_{\mathrm{dR}}$ preserves the Hodge decomposition (as the next paragraph shows), but $F_{\mathrm{B}}$ and hence $F_{\infty}$ reverse it.

Now consider the Grothendieck comparison theorem $H^i(X(\mathbb{C}),\mathbb{C})=H^i_{\mathrm{dR}}(X_{\mathbb{C}}/\mathbb{C})$ between singular and algebraic de Rham cohomology. Inspecting the proof of this theorem shows that the antiholomorphic involution of this space associated with the real structure $H^i_{\mathrm{dR}}(X/\mathbb{R})$ of $H^i_{\mathrm{dR}}(X_{\mathbb{C}}/\mathbb{C})$ is $F_{\mathrm{dR}}$. Consequently, $F_{\mathrm{B}}$ and $F_{\infty}$ descend to the same $\mathbb{R}$-linear involution on $H^i_{\mathrm{dR}}(X/\mathbb{R})$ (let us still call it $F_{\infty}$). It is with respect to this involution, which is very much nontrivial and difficult to compute on the de Rham side, that $F^pH^i_{\mathrm{dR}}(X/\mathbb{R})\cap F_{\infty}(F^qH^i_{\mathrm{dR}}(X/\mathbb{R}))=0$ when $p+q>i$.

Answer 2

I thought it would be useful to give an explicit example to supplement Olivier Benoist's answer; I will use the same notation as in his answer. Let $f(x)\in \mathbb{R}[x]$ be cubic with distinct roots. Then $y^2=f(x)$ defines an elliptic curve $X$ over $\mathbb{R}$. The differential $\alpha = dx/y$ spans $ H^0(X,\Omega_X^1)= F^1 H^1_{dR}(X/\mathbb{R})$. The de Rham cohomology of the complex curve $X(\mathbb{C})$ carries a conjugation $F_{dR}$ which leaves $\alpha$ invariant. However, if we uniformize this curve as $\mathbb{C}/\Lambda$, and let $z$ denote the coordinate on $\mathbb{C}$. Then $\alpha = \lambda dz$, for some nonzero constant $\lambda$. With respect to the other conjugation $F_B$, relevant to Hodge theory, we have $$F_B(\alpha) = \bar\lambda d\bar z\not=\alpha$$