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Let $\underline{\Omega}_X^{\bullet}$ denote the Deligne -- Du Bois complex of a normal variety $X$. What kind sigularities satisfy $gr^k\underline{\Omega}_X^{\bullet}[k] \simeq \Omega_X^{[k]}$ where $\Omega_X^{[k]} := j_*\Omega^k_{X^{\operatorname{reg}}}$ and $j\colon X^{\operatorname{reg}} \hookrightarrow X$ is the inclusion of the regular locus of $X$? ADDED LATER: Is there even a map like $\Omega_X^{[k]} \to gr^k\underline{\Omega}_X^{\bullet}[k]$ to begin with?

Recall that the case $k=0$ is by definition the Du Bois singularities and in order to define it, one does not need to assume that $X$ is normal. Log canonical singularities (normal) are known to be Du Bois (https://arxiv.org/abs/0902.0648).

EDIT: As Donu pointed out, when $X$ is smooth $h^{-k}\omega_X^{\bullet} = 0$ for $k\neq -\dim X$ but $gr^k\underline{\Omega}_X^{\bullet}[k] = \Omega_X^k$. So the following question does not make sense. One could perhaps ask more generally when is $gr^k\underline{\Omega}_X^{\bullet}[k] \simeq h^{-k}\omega_X^{\bullet}$ where $\omega_X^{\bullet}$ is the dualizing complex of $X$?

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    $\begingroup$ Re. question "Added later". There is the adjunction map, which goes the other way. $\endgroup$
    – Donu Arapura
    Jun 28, 2021 at 15:31
  • $\begingroup$ Oh ok, my impression was that for $k=0$ the natural map goes from $\mathcal{O}_X \to gr^0\underline{\Omega}_X^{\bullet}$. $\endgroup$
    – guest0803
    Jun 30, 2021 at 12:16

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Good question. Your first question has a positive answer when $X$ has finite quotient singularities, i.e. locally analytically $\mathbb{C}^n/G$, with $G\subset GL_n(\mathbb{C})$ finite. This was essentially proved in Du Bois' original paper "Complexe de de Rham filtré...", see also Steenbrink "Mixed Hodge structure on the vanishing cohomology". I have a hunch (see added comment) this true when $G$ is reductive. There are some other classes as well, like toric singularities. I can say more later, or perhaps Sándor will.

Added A better reference for examples would be the book by Navarro Aznar et. al. "Hyperrésolutions cubiques et descente cohomologique".

Also I overlooked your second question: Think about the case when $X$ is smooth. It's almost never true. Perhaps you meant something else?

2nd edit I suppose I may as well make my old notes public. See conjecture 2.8 of that for a more precise statement of what I'm expecting.

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    $\begingroup$ I believe that Sándor told me after one seminar that this does work for reductive group quotients. $\endgroup$
    – Jason Starr
    Jun 27, 2021 at 14:32
  • $\begingroup$ Dear Donu, thanks for the answer and for sharing your expectation. You are right, the second question does not make sense for smooth case, edited away. $\endgroup$
    – guest0803
    Jun 28, 2021 at 11:49
  • $\begingroup$ On today's news: arxiv.org/abs/2107.06619 $\endgroup$
    – guest0803
    Jul 15, 2021 at 8:11

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