9
$\begingroup$

Is the condition that a module is reflexive an open condition?

That is, if $X$ is a smooth projective complex variety, $T$ a quasi-projective variety, and $F$ a finitely presented module on $X \times T$ that is $T$-flat, then we can form the locus $T' \subset T$ of points $t$ such that the restriction of $F$ to $X \times t$ is reflexive.

Is $T' \subset T$ open?

If not, is it locally closed?

Recall that a coherent sheaf $F$ is said to be reflexive if the natural map $F \to (F^{\vee})^{\vee}$ to the double dual is an isomorphism.

Improve this question
$\endgroup$
0

3 Answers 3

Reset to default
9
$\begingroup$

This locus is indeed open. I will explain why using Kollar's "Hulls and Husks" (arXiv:0805.0576). More generally, this article studies in great detail when taking the double dual commutes with base change.

First, we may restrict to the open locus of $T$ where $F_t$ is torsion-free (because reflexive sheaves are torsion-free). Then, we choose an ample line bundle $H$ on $X$, and we compute the Hilbert polynomials relatively to $H$. The Hilbert polynomial $P(F_t)$ of $F_t$ is constant by flatness. From the exact sequence $0\to F_t\to F^{\vee\vee}_t\to F^{\vee\vee}_t/F_t\to 0$, we see that $F_t$ is reflexive exactly when $P(F^{\vee\vee}_t)=P(F_t)$, i.e. exactly when $P(F^{\vee\vee}_t)$ takes its minimal value. But the polynomial $P(F^{\vee\vee}_t)$ is constructible and upper semicontinuous by Proposition 28 (3) of Hulls ans Husks. This proves that this locus is open.

Improve this answer
$\endgroup$
2
  • $\begingroup$ Olivier, how do you know that the polynomial $P(F_t^{\vee\vee})$ is constructible? The problem is that a priori $F_t^{\vee\vee}$ is just a collection of sheaves on the fibers, it is NOT given by restrictions of one coherent sheaf to the fibers... $\endgroup$
    – Sasha
    Commented Apr 3, 2012 at 2:46
  • $\begingroup$ @Sasha : It is part of the statement of Kollar's article (arXiv:0805.0576 Proposition 28 (3)), and it is proven there. By noetherian induction, you only need to show that it is constant on an open subset of $T$, hence, you will only have to show that it is "given by restrictions of one coherent sheaf to the fibers" on an open subset of $T$. $\endgroup$
    – Olivier Benoist
    Commented Apr 3, 2012 at 6:54
3
$\begingroup$

[EDIT: as Sasha points out, this does not answer the question. Please see it as a comment explaining why $X$ should be proper!]

The answer is no in general if $X$ is not proper: take $X=\mathrm{Spec}\,\mathbb{C}[x]$, $T=\mathrm{Spec}\,\mathbb{C}[t]$, and $F=$ the structure sheaf of $Z=\mathrm{Spec}\,(\mathbb{C}[t,x]/(1-tx))$. Then $T'$ is just the origin.

Variant: if instead you take $T=\mathrm{Spec}\,\mathbb{C}[t,u]$ and $Z=\mathrm{Spec}\,(\mathbb{C}[t,u,x]/(u,1-tx))$ (i.e. the same $Z$ as before, but embedded in 3-space), then $T'$ is the union of the origin and the complement of the $t$-axis, hence not locally closed (but still constructible).

Of course the point here is that $Z$ "goes to infinity" at the origin. I don't have a counterexample where $X$ is proper, but the main problem then is "taking the dual in the fibers", as in Sasha's comment above.

Improve this answer
$\endgroup$
2
  • 2
    $\begingroup$ In original question $X$ was proper (even projective). $\endgroup$
    – Sasha
    Commented Apr 2, 2012 at 15:23
  • $\begingroup$ @Sasha: oops, sorry! I ha read "quasiprojective" but that was about $T$. $\endgroup$
    – Laurent Moret-Bailly
    Commented Apr 2, 2012 at 17:02
2
$\begingroup$

I'll assume you meant to say that $F$ was locally finitely presented or coherent in your second sentence. The locus where $F$ is reflexive is the complement of the union of the supports of the kernel and cokernel of $F\to (F^\vee)^\vee$. This will be open.

Postscript As Sasha points out the argument is incomplete because it is not clear that duals commute with restriction to the fibres.

PPS Now Laurent has a counter example, so I guess that finishes it.

Improve this answer
$\endgroup$
4
  • 3
    $\begingroup$ Donu, the problem is that operation of taking the dual sheaf does not commute with restriction to the fiber over $T$. So, it is not clear why $((F^\vee)^\vee)_{|X\times\{t\}} \cong ((F_{|X\times\{t\}})^\vee)^\vee$. $\endgroup$
    – Sasha
    Commented Apr 2, 2012 at 12:00
  • $\begingroup$ what happens if you take the derived dual? $\endgroup$
    – Yosemite Sam
    Commented Apr 2, 2012 at 15:28
  • 1
    $\begingroup$ Well, in the bounded derived category of a smooth variety, everything is reflexive in the sense that $$F\cong RHom(RHom(F, O_X), O_X)$$ $\endgroup$
    – Donu Arapura
    Commented Apr 2, 2012 at 16:08
  • $\begingroup$ ah, of course, silly me. it's an open condition then! :) $\endgroup$
    – Yosemite Sam
    Commented Apr 3, 2012 at 23:54

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .