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The question is in the title, do you have simple examples of reflexive sheaves on a complex manifold $X$ (it can be Kähler if you prefer, or even smooth algebraic) that are not locally free (i.e. vector bundles) ?

I know that such a sheaf $\mathcal{F}$ is such that $\mathrm{S}(\mathcal{F}) = \{x \in X|\mathcal{F}_x$ is not free$\}$ has codimension greater or equal than $3$ because $\mathcal{F}$ is a $2$-syzygy sheaf hence we must look for $\dim(X) \geqslant 3$. I also know that if $\mathcal{F}$ has rank $1$, $\det(\mathcal{F}) = \mathcal{F}^{**}$ is a line bundle so $\mathcal{F}$ reflexive $\Leftrightarrow \mathcal{F}$ locally free in this case hence we must look for $\mathrm{rk}(\mathcal{F}) \geqslant 2$.

It is easy to find torsion-free sheaves that are not locally free, by taking sheaves of ideals generating non-empty codimension $\geqslant 2$ analytic/algebraic subsets but it won't make a refexive sheaf because it has rank $1$. Do you have a simple example, like on $\mathbb{C}^3$ or $\mathbb{P}^3(\mathbb{C})$ for example ? And do you have example of reflexive stable coherent sheaves that are not locally free in the Kähler case ?

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Take $R$ be the co-ordinate ring of algebraic functions on $\mathbb{C}^3$, so $R$ is just the polynomial ring in three variables $x,y,z$. Let $M= Re_1\oplus Re_2\oplus Re_3/xe_1+ye_2+ze_3$. Then, $M$ is reflexive, but not locally free.

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  • $\begingroup$ It seems to work, thank you ! $\endgroup$
    – Cactus
    Jul 11 at 20:12

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