Differential construction of mixed Hodge structure on smooth open varieties

Question

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 $H^\ast(X)\simeq H^\ast(\bar{X},j_\ast A_X^\ast)$?

Answer 1

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 $F^1$.

Answer 2

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.