1
$\begingroup$

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

$\endgroup$
3
  • $\begingroup$ 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. $\endgroup$ Nov 13, 2018 at 7:45
  • $\begingroup$ @ 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 $H^\ast(X)$. $\endgroup$
    – stjc
    Nov 13, 2018 at 11:56
  • $\begingroup$ 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 $F^{\cdot}$. $\endgroup$ Nov 13, 2018 at 17:21

2 Answers 2

5
$\begingroup$

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 $F^1$.

$\endgroup$
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.