# Singularities of reflexive sheaves

I am studying reflexive sheaves (on $\mathbb{P}^3$) by the Hartshorne's paper ''Stable reflexive sheaves''. As far I understood, reflexive sheaves fail to be locally free at a finite number of points (on $\mathbb{P}^3$, for higher dimensional varieties it fail to be locally free in a scheme of codimension 3). Therefore, if one restricts a reflexive sheaf in an open of $\mathbb{P}^3$ not containing the singular points, the restricted sheaf should isomorphic to copies of the trivial. I would like to understand what happens when one restrict the sheaf to an open containing the singular point. Is there a way of computing explicit what the restricted sheaf will look like?

Thank you in advance.

• The line ` the restricted sheaf should (be) isomorphic to copies of the trivial (line bundle) ' is false. – Mohan Mar 26 '18 at 1:46
• What do you mean by explicit? It won't be more explicit that the description of the sheaf on the whole $\Bbb{P}^3$. – abx Mar 26 '18 at 5:15
• A reflexive sheaf is the kernel of a morphism of locally free sheaves. Locally, you can assume the locally free sheaves to be trivial, so you can think of a reflexive sheaf as a subsheaf of $\mathcal{O}^{\oplus n}$ given by a finite number of fiberwise linear conditions. – Sasha Mar 26 '18 at 6:26
• Dear @Mohan, I mean the restricted sheaf in the open is the sum of the structural sheaf of the open. This should happens by the definition of locally free sheaf, don't it? Is there a counter example? – User43029 Mar 26 '18 at 11:41
• Dear @abx by explicit I mean explicit in that open, not in the whole $\mathbb{P}^3$, e.g., a not trivial locally free sheaf is not sum of the trivial on whole $\mathbb{P}^3$, but it is in some open subset. – User43029 Mar 26 '18 at 11:43