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Let $D$ be a reduced projective scheme over $\mathbb{C}$ such that $H^1(D,\mathcal{O}_D) = 0$ and $D$ is Gorenstein. There is a map

\begin{equation} r:= \frac{d \log}{2 \pi i }: H^1(D, \mathcal{O}_D^{\ast}) \otimes \mathbb{C} \rightarrow H^1(D,\Omega_D^1) \end{equation} locally defined by $f \mapsto \frac{df}{f}$.

Question Is this homomorphism injective?

If $D$ is smooth or a V-manifold, it is known that this is injective. If $D$ is general, I considered the following argument. If there is a mistake, please let me know about it. ;

Let $(\underline{\Omega}_D^{\bullet},F)$ be the Du bois complex on $D$. Then there is a homomorphism $H^1(D, \Omega_D^1) \rightarrow \mathbb{H}^1(D, \underline{\Omega}^1_D) $ where $\underline{\Omega}^1_D := Gr_F^1 \underline{\Omega}^{\bullet}_D[1]$.

Consider homomorphisms $ t : H^1(D, \mathcal{O}_D^{\ast}) \rightarrow H^2(D, \mathbb{C})$ which is induced by the exponential exact sequence $0 \rightarrow \mathbb{Z} \rightarrow \mathcal{O}_D \rightarrow \mathcal{O}_D^{\ast} \rightarrow 0$ and the homomorphism which is composition of the form $H^1(D, \mathcal{O}_D^{\ast}) \stackrel{r}{\rightarrow} H^1(D, \Omega^1_D) \stackrel{s}{\rightarrow} \mathbb{H}^1(D, \underline{\Omega}^1_D) \stackrel{u}{\rightarrow} H^2(D, \mathbb{C})$.

I don't know how to define $u$ in a natural way. Take a hyperresolution $f_{\bullet}: D_{\bullet} \rightarrow D $. I think that $\underline{\Omega}_D^1$

can be expressed by using the terms come from $\Omega^1_{D_{\bullet}}$ on each $D_{\bullet}$.

and there is a complex homomorphism $\underline{\Omega}^1_D \rightarrow \underline{\Omega}_D^{\bullet}$ induced by the expression above and define $u$ by this complex homomorphism.
If $t = u \circ s \circ r$, then $r$ is injective. Is there a mistake in this argument?

Moreover, I'm not familiar with arguments using Du Bois complex. I don't know the above arguments make sense. If there are useful literatures, please let me know about it.

(add) I had a mistake in the definition of $\underline{\Omega}_D^1$. I forgot a shift by 1.
I'm readin p.174 of the book "Mixed Hodge Structures" by Peters and Steenbrink. I thought that I can define $\underline{\Omega}_D^1 \rightarrow \underline{\Omega}_D^{\bullet}$ whose homomorphisms on $k$-th term are defined by using the direct summand inclusions $ (f_k)_{*} \Omega_{D_k}^1 \rightarrow \oplus_{p+q = k+1} ( f_q )_{*} \Omega_{D_q}^p$.

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  • $\begingroup$ What is the homomorphism $\underline{\Omega}^1_D \rightarrow \underline{\Omega}_D^{\bullet}$ you mention? $\endgroup$ Commented Jul 6, 2011 at 16:26
  • $\begingroup$ I added comments in the last. I hope it makes sense. $\endgroup$
    – tarosano
    Commented Jul 6, 2011 at 17:40
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    $\begingroup$ The problem with your definition is that this is not going to be a map of complexes. $\endgroup$ Commented Jul 6, 2011 at 21:15
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    $\begingroup$ If I may be honest, your argument doesn't look quite right. But the end result should be. Here's how to fix it: The image of $t=c_1$ should lie in $$ker[H^1(D,\mathbb{C})\to H^1(D,\mathcal{O}_D)]\subseteq F^1H^1(D,\mathbb{C})$$ But $im(c_1)$ is real, so it also lies in $\bar F^1$. You can identify $F^1\cap \bar F^1=H^1(D,\underline{\Omega}_D^\dt)$. Now proceed as above. (See Barbieri-Viale/Srinivas Crelle 1994, and my paper with Kang in Comm. Algebra, 2011 for some related stuff.) $\endgroup$ Commented Jul 6, 2011 at 21:21
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    $\begingroup$ Thank you for great comments! I wish I had a right to vote. $\endgroup$
    – tarosano
    Commented Jul 7, 2011 at 7:28

1 Answer 1

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Let me convert my obscure comment into an (obscure?) answer, with some corrections.

The problem with your argument as it stands is that a map is not well defined: $H^1(D,\underline{\Omega}_D^1)$ is only a subquotient of $H^2(D,\mathbb{C})$. But the problem is minor. To fix things observe that $H^2(D)$ carries a mixed Hodge structure. Also the image $im(c_1)$ of the Chern class map (what you call $t$) lies in $F^1\cap \bar F^1$ because it lies in $$ker[H^2(D,\mathbb{C})\to H^2(D,\mathcal{O}_D)]\subseteq F^1$$ and is invariant under conjugation. The space $F^1\cap \bar F^1$ maps injectively to $H^1(D,\underline{\Omega}_D^1)=Gr_F^1H^2(D,\mathbb{C})$. Now by your assumption $c_1$ is injective, so the map to $H^1(D,\underline{\Omega}_D^1)$ is also injective, and as (the normalized) $dlog$ factors through it, it is also injective.

References: I'm using the standard facts from the papers of Deligne and Du Bois (which should be in Peters-Steenbrink). For more information about $c_1$ in this setting, see Barbieri Viale and Srinivas "The Neron Severi groups and the mixed Hodge structure on $H^2$" Crelles (1994); for higher Chern classes, see my paper with Kang "Kaehler-de Rham cohomology and Chern classes" Commun. Alg. 2011.

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  • $\begingroup$ Thank you very much for the details. Can I ask a silly question? $F^1 \cap \overline{F}^2 = 0$ because they are filtration associated to the mixed hodge structure on $H^2(D, \mathbb{C})$, is it right? They don't always satisfy that $F^1 + \overline{F}^2 = H^2(D, \mathbb{C})$, right? $\endgroup$
    – tarosano
    Commented Jul 7, 2011 at 14:00
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    $\begingroup$ Yes, that's right. It helps to visualize the Hodge numbers of $H^2(D)$ as lying in the triangle $p+q\le 2$ in the $pq$-plane. $F^1$ is the intersection with the half plane $p\ge 1$, and $\bar F^2$ is $q\ge 2$. $\endgroup$ Commented Jul 7, 2011 at 14:54

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