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I am reading the paper Frobenius splitting of Hilbert schemes of points on surfaces by Kumar and Thomsen. At the end of Lemma 11, they seem to imply that the dualizing sheaf on a Cohen-Macaulay scheme is reflexive. I am not familiar with reflexive sheaves, so I would like to ask whether this statement is true or not.

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2 Answers 2

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The answer is yes. Let me briefly explain why.

There are several ways to define a canonical sheaf on a projective variety $X$. One of them, that works for any normal $X$ is the following. Let $U$ be the smooth locus of $X$ and $i \colon U \hookrightarrow X$ be the natural inclusion. Then define $$\omega_X:=i_*(\Omega^n_U),$$ where $n:=\dim X$. As the sections of $\Omega^n_U$ do not depend on a subset of codimension $2$, neither do the sections of $\omega_X$. In other words, $\omega_X$ is a so-called normal sheaf. Moreover, by construction $\omega_X$ is also torsion-free, so it is reflexive by the second characterization in [1, Proposition 1.6].

When $X$ is Cohen-Macaulay, this sheaf coincides with the dualizing sheaf in the sense of Serre duality. A different way to define a canonical sheaf for embedded varieties $X \subset \mathbb{P}^{N}$ is to consider $$\omega_X' := h^{-n}(\omega_X^{\bullet}),$$
where $\omega_X^{\bullet}$ is the so-called dualizing complex. Again, when $X$ is Cohen-Macaulay this construction provides a reflexive sheaf, that coincides with $\omega_X$ outside a subset of codimension $\geq 2$. Then reflexivity implies $$\omega_X'\simeq \omega_X,$$ see [2, Section 5].

References.

[1]$\,$ Robin Hartshorne, Stable reflexive sheaves, Math. Ann. 254 (1980), no. 2, 121--176.

[2] $\,$ Sándor J. Kovács, Singularities of stable varieties, Handbook of moduli. Vol. II 159--203.

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  • $\begingroup$ This seems to settle the question when $X$ is Cohen–Macaulay and normal (I think that [2] has this as a running assumption, at least when they mention reflexivity). $\endgroup$ Jan 30, 2017 at 17:20
  • $\begingroup$ @Francesco, thanks for your explanation. I don't see immediately that your construction of canonical sheaf coincides with dualizing sheaf in the sense of Serre duality. I will write my partial proof to this statement in another question "dualizing sheaf on cohen-macaulay scheme". It would be great if you can take a look at it and point out what I missed. $\endgroup$
    – helloWorld
    Jan 30, 2017 at 22:02
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Just for the record, this does not need CM. The following is true:

Let $Z$ be an excellent scheme that admits a dualizing complex. Then $ω_Z$ is torsion-free and $S_2$ on $Z$. If in addition Z is normal, then $ω_Z$ is a reflexive $\mathscr O_Z$ -module.

Also note that torsion-free and $S_2$ seems to be the "right" notion to replace reflexivity on non-normal schemes.

If there is interest in this, I will add a proof.


As requested, here is a proof: (The following is Lemma 3.7.5 in this paper.)

Lemma Let $Z$ be an excellent scheme that admits a dualizing complex. Then $\omega_Z$ is torsion-free and $S_2$ on $Z$. If in addition $Z$ is normal, then $\omega_Z$ is a reflexive $\mathscr O_Z$-module.

Proof. The statement is local, so we may assume that $Z$ is a noetherian affine local scheme. Then, since it admits a dualizing complex, it can be embedded into a finite dimensional Gorenstein affine local scheme W as a closed subscheme by [Kaw02, Cor. 1.4]. Being Gorenstein and local, $W$ must be pure dimensional. Let $r = \mathrm{codim}(Z, W)$. Then by [Mat80, (16.B) Theorem 31(i)] there exists a length $r$ regular sequence in the ideal of $Z$ in $W$ and let $W'$ be the common zero locus. Then $W'$ is also Gorenstein, $Z\subseteq W$ is a closed subscheme and $\dim Z = \dim W$ .

It follows that $\omega_ {W'}^\bullet \simeq \omega_{W'}[m]$, where $\omega_{W'}$ is a line bundle on ${W'}$. By Grothendieck duality $\omega^\bullet_ Z\simeq RHom_{W'}(\mathscr O_Z, \omega^\bullet_{W'})$, and since $\dim Z=\dim W'$, $\omega_Z\simeq Hom_{W'}(\mathscr O_Z, \omega_{W'})$ hence it is indeed torsion-free on $Z$.

By [Stacks Project, Tag 0AWE] $\omega_Z$ is $S_2$. In particular, if $Z$ is normal, then it is reflexive by [Stacks Project,Tag 0AVB].


As far as non-trivial examples of schemes admitting dualizing complexes are concerned, try anything that can be embedded (locally) into a Gorenstein scheme. In fact, those are exactly the ones you are looking for.

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  • $\begingroup$ Certainly! In fact, I would be much more interested in seeing some natural examples in algebraic geometry which is not CM but is excellent and admits a dualizing complex. $\endgroup$
    – helloWorld
    Feb 1, 2017 at 13:10
  • $\begingroup$ Any finite type scheme over a field (or over an affine regular scheme) admits a dualizing complex and, of course they needn't be CM. $\endgroup$
    – Leo Alonso
    Nov 27, 2017 at 18:16

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