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Suppose $X$ is a normal complex space with at most singularities, can we say any locally free sheaf on it is Cohen-Macauly? Recall that a coherent sheaf $\mathcal{F}$ over $X$ is called (maximal) Cohen-Macaulay if $depth_{\mathcal{O}_x}(\mathcal{F}_x) = \dim\mathcal{O}_x$ for any $x\in X$, where $\mathcal{O}_x$ is the local ring of $X$ at $x$.

How badly the singularities can the complex space $X$ possess to remain the above property?

Thanks!

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    $\begingroup$ Please confer the following MO question: mathoverflow.net/questions/69436/… $\endgroup$
    – Jason Starr
    Commented Oct 19, 2022 at 11:26
  • $\begingroup$ This holds if and only if $X$ is Cohen-Macaulay. $\endgroup$
    – abx
    Commented Oct 19, 2022 at 11:59
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    $\begingroup$ Sorry, I hadn't paid attention to your title. If $X$ has canonical singularities, it is Cohen-Macaulay by Elkik's theorem. $\endgroup$
    – abx
    Commented Oct 19, 2022 at 12:19
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    $\begingroup$ There is also a variant of canonical in the non-Q-Gorenstein case due to de Fernex-Hacon which doesn't imply Cohen-Macaulay if I recall correctly. $\endgroup$
    – Karl Schwede
    Commented Oct 19, 2022 at 18:46

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