# Differential construction of mixed Hodge structure on smooth open varieties

Let $$\bar{X}$$ be a complete smooth variety over $$\mathbb{C}$$ and $$D$$ be a simple normal crossing divisor. Denote $$X:=\bar{X}\backslash D$$. Then it is known that $$H^\ast(X,\mathbb{C})$$ admits a canonical mixed Hodge structure.

Denote $$j:X\to\bar{X}$$ be the open immersion and $$A_X^\ast$$ be the complex of sheaves of differential forms on $$X$$. Let $$F^p$$ be the subcomplex of $$A_X^\ast$$ consists of $$(i,j)$$-forms such that $$i\geq p$$. My question is

Dose $$(j_\ast A_X^\ast,j_\ast F^\bullet,\tau^{\leq\bullet} j_\ast A_X^\ast)$$ induces the canonical mixed Hodge structure on $$H^\ast(X)\simeq H^\ast(\bar{X},j_\ast A_X^\ast)$$?

• Does your definition of $A^*_X$ include any hypothesis that forms are $\overline{\partial}$-closed? The usual holomorphic de Rham complex, whose filtration $\tau^{\bullet}$ does induce the Hodge filtration, is for $\overline{\partial}$-closed forms, not all forms. Nov 13, 2018 at 7:45
• @ Starr. There is no hypothesis on the form except smoothness. By Deligne's Hodge II Prop. 3.1.8, the filtration $\tau$ contributes to the weight filtration on $H^\ast(X)$.
– stjc
Nov 13, 2018 at 11:56
• It is not clear to me if this induces the correct Hodge filtration, as it does not use holomorphic forms with logarithmic poles to define the filtration $F^{\cdot}$. Nov 13, 2018 at 17:21

Here is a counterexample. Let $$\bar X$$ be an elliptic curve in Legendre form $$y^2= x(x-1)(x-\lambda)$$ Let $$\omega_i$$ be the meromorphic differential forms $$\omega_1 = \frac{dx}{y}$$ $$\omega_2= \frac{x(x-1)dx}{y^3}$$ The first form is actually holomorphic on $$\bar X$$. The second form does have poles. Let $$X$$ be the complement of these. The form $$\omega_2$$ is classically what is called a differential of the second kind, which means that all it's residues are zero. The last property implies that $$\omega_2$$, which defines an element of de Rham cohomology $$H^1(X)$$, lifts to a class $$\tilde \omega_2$$ in $$H^1(\bar X)$$. Note that $$\tilde \omega_2$$ is not in $$F^1H^1(\bar X)$$, which is spanned by $$\omega_1$$. By the strictness, the image $$\omega_2$$ cannot be in Deligne's $$F^1H^1(X)$$. However, obviously it does lie in your $$F^1$$.
I have found a reason why the "Hodge filtration" $$j_\ast F^\bullet$$ may not induce the right one. By the Grothedieck-Dolbeault lemma, $$(A_X^\bullet, F^{\geq p})$$ is filtered quasi-isomorphic to $$(\Omega_{X^{an}}^\bullet,\Omega_{X^{an}}^{\geq p})$$. Since $$j$$ is a stein morphism, one has $$(j_\ast A_X^\bullet, j_\ast F^{\geq p})\simeq_{\rm qis}(j_\ast\Omega_{X^{an}}^\bullet,j_\ast\Omega_{X^{an}}^{\geq p}).$$ As a consequence, the subspace of $$H^i(X)\simeq H^i(X,\Omega_{X^{an}}^{\bullet})\simeq H^i(\bar{X},j_\ast\Omega_{X^{an}}^{\bullet})$$ induced by $$j_\ast F^{\geq p}$$ is $${\rm im}(H^i(\Omega_{X^{an}}^{\geq p})\to H^i(\Omega_{X^{an}}^{\bullet})).$$ Now let us assume that $$X$$ is affine (=stein), then this subspace is $${\rm im}(H^i(\Gamma(X,\Omega_{X^{an}}^{\geq p}))\to H^i(\Gamma(X,\Omega_{X^{an}}^{\bullet})))=\begin{cases} 0, & p>i \\ H^i(\Gamma(X,\Omega_{X^{an}}^{\bullet})), & p\leq i \end{cases}.$$ Therefore when $$X$$ is affine, the filtration $$j_\ast F^\bullet$$ induces on $$H^i(X)$$ the trivial filtration. This also explain the interesting counterexample given by Donu.