I am reading "Ample divisors on fine moduli spaces on the Projective plane" by Stromme. In the proof of Proposition 2.4, he seems to claim that if $E$ is a torsion free sheaf of projective dimension at most 1 on a smooth variety and the singular locus of $E$ has codimension at least $3$, then $E$ is reflexive. Of course the converse is true without the projective dimension assumption. 

Does anyone know an explanation or reference for this fact?

  • 1
    $\begingroup$ I would guess that the condition $pd(F) \leq 1$ is important. Without it you can get a counterexample by considering the ideal sheaf of a point on a variety of dimension $\geq 3$. $\endgroup$
    – naf
    Dec 20, 2013 at 11:33
  • $\begingroup$ Ah yes, of course. Edited. $\endgroup$ Dec 20, 2013 at 11:49

1 Answer 1


In general the converse is not true --- take for example the ideal of a point on 3-fold (its reflexive hull is the structure sheaf). On the other hand, if you know that $pd(F) \le 1$ then there is an exact sequence $$ 0 \to E_1 \stackrel{f}\to E_2 \to F \to 0 $$ with $E_1$ and $E_2$ locally free. Taking the dual one gets $$ 0 \to F^* \to E_2^* \stackrel{f^*}\to E_1^* \to F_1 \to 0, $$ where the last sheaf $F_1 := \mathcal{Ext}^1(F,O)$ is supported in codimension 3. In particular, $\mathcal{Ext}^i(F_1,O) = 0$ for $i \le 2$. It follows that $\mathcal{Ext}^i(F_1,O)$ do not contribute into $F^{**}$ and so there is an exact sequence $$ 0 \to E_1^{**} \stackrel{f^{**}}\to E_2^{**} \to F^{**} \to 0, $$ and so since $E_i$ are locally free, it follows that $F^{**} = F$.

  • $\begingroup$ Great, thanks! This seems to do the trick. $\endgroup$ Dec 20, 2013 at 16:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.