A reflexive sheaf (i.e its double dual is equal itself) is locally free (i.e., a holomorphic vector bundle) outside a subvariety of codimension greater than or equal to two. Let $\mathcal F$ be a coherent subsheaf of holomorphic vector bundle $E$, then there is an analytic subset $S \subset M$ of codimension bigger than two and a holomorphic vector bundle $F$ on $X \setminus S$ such that $$\mathcal F|_{X\setminus S}=\mathcal O(F)$$

We know that a reflexive sheaf $\mathcal F$, on Kahler variety $X$ outside of codimension at least 3 is a holomorphic vector bundle,

Under which condition such singular variety $S$ is an analytic sub-variety?

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    $\begingroup$ Always, right? Coherence implies that one may locally write $\mathscr{F}$ as the cokernel of a map of trivial holomorphic vector bundles; $S$ is the locus where the rank of this map drops, so it's cut out by some matrix minors, which are analytic... $\endgroup$ – Daniel Litt May 11 '17 at 0:33
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    $\begingroup$ How did you know that statement about a reflexive sheaf being a vector bundle if it is .....? $\endgroup$ – Mohan May 11 '17 at 4:23
  • $\begingroup$ Mohan, This is a theorem in the book of Kobayashi $\endgroup$ – user21574 May 11 '17 at 9:49
  • $\begingroup$ The title of the question seems to be misleading. An "almost vector bundle" might be a vector bundle in the context of "almost mathematics". Besides that, the question is not even about the sheaf but about the variety S. $\endgroup$ – Helene Sigloch May 11 '17 at 10:16
  • $\begingroup$ @HeleneSigloch ,ALMOST VECTOR BUNDLE in the following sense: If you take $Sing(\mathcal F) = \{x \in X | \mathcal F_x \text{is not free}\}$ where $\mathcal F_x$ is the stalk of $\mathcal F$ over $x$ , Then any torsion free sheaf $\mathcal F$ out side of $Sing (\mathcal F)$ is vector bundle , since it is locally free. We say $\mathcal F$ is torsion free if $\mathcal F\to \mathcal F^{**}$ be injective and if it be isomorphism then it is called reflexive $\endgroup$ – user21574 May 11 '17 at 10:57

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