# Every locally free sheaf is Cohen-Macaulay on complex variety with at canonical singualities?

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!

• Please confer the following MO question: mathoverflow.net/questions/69436/… Oct 19, 2022 at 11:26
• This holds if and only if $X$ is Cohen-Macaulay.
– abx
Oct 19, 2022 at 11:59
• Sorry, I hadn't paid attention to your title. If $X$ has canonical singularities, it is Cohen-Macaulay by Elkik's theorem.
– abx
Oct 19, 2022 at 12:19
• 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. Oct 19, 2022 at 18:46