[This was first posted on MSE but did not get any answer. I apologize if it is not suited for MO.]

Let $C$ be a curve, i.e. a purely one-dimensional scheme, embedded in a smooth projective threefold $X$. For a coherent sheaf $E$ of codimension $c$ on $X$, let $E^D=\mathscr Ext_X^c(E,\omega_X)$ be the Grothendieck dual of $E$. It is a reflexive sheaf of codimension $c$ on $X$. Reflexive means that the natural map to the double dual is an isomorphism. Let $\mathscr O_C$ be the structure sheaf of $C$, viewed as a torsion sheaf on $X$.

Question. For which curves $C\subset X$ is $\mathscr O_C$ reflexive?

Certainly, when $C$ is smooth. Let us assume $C$ is singular. I was thinking that maybe if $C$ is a local complete intersection then the first dual $\mathscr O_C^D$ could already be isomorphic to $\mathscr O_C$. Is this true? If so, then in particular dualizing twice does not do anything. But probably someone knows a more convincing (or even more true) statement. Thank you for any help!


2 Answers 2


Indeed, this condition is equivalent to $C$ being CM (Regardless of $X$ as long as it is also CM). This "dual" is called the $\omega$-dual in Kol13. You might be interested in reading section 2.5, or just the part directly dealing with this on pp.80-83. In particular, the statement you need is Cor. 2.70. Furthermore, if $X$ is not necessarily CM, then you get the same result using the dualizing complex instead of the dualizing sheaf.

As Mohan points out, the first dual will only be isomorphic to $\mathscr O$ if its own dualizing sheaf is trivial. Then again, why would you want that? That's a freak accident and doesn't give you too much mileage. Reflexivity is a (or "the") natural condition.


If $C$ is Cohen-Macaulay (in particular, if it is local complete intersection), then $\mathcal{O}_C$ is reflexive with your definition. Even when $C$ is smooth, $\mathcal{O}_C^D$ is not necessarily isomorphic to $\mathcal{O}_C$, since the dual is just $K_C$, which may or may not be isomorphic to $\mathcal{O}_C$.


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