# How does complex conjugation act on the Hodge filtration?

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?

• "each $\mathcal{F}^{i}$ is the complexification of a $\mathbb{R}$-sub vector space of $H^{i}_{\operatorname{dR} }(X)$": no, this is false.
– abx
Commented May 1 at 5:32
• Isn't that a corollary of the degeneration of the Hodge to de Rham spectral sequence in the $E^1$-page for smooth projective schemes over a field? What am I missing here? Why is it true that $\mathcal{F^1H^1}$ is even rationally defined for an $\mathbb{R}$-defined elliptic curve? Commented May 1 at 14:48
• I think this is well explained in the two answers below, in particular that of Donu Arapura.
– abx
Commented May 1 at 16:51

## 2 Answers

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 $$p+q>i$$.

• Thank you very much for explaining the definition. Where would be a good place to read about the motivation behind it? I guess that in the complex analytic case, there is an interesting interaction between the Hodge decomposition and complex conjugation which is somehow what we are trying to capture with this algebraic description. Commented May 1 at 15:04

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 $$F_B(\alpha) = \bar\lambda d\bar z\not=\alpha$$