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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.

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3 Answers

up vote 8 down vote accepted

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.

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@Olivier Benoist: Thanks! –  jlk Apr 2 '12 at 21:10
    
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... –  Sasha Apr 3 '12 at 2:46
    
@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 $T$. –  Olivier Benoist Apr 3 '12 at 6:54
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[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.

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In original question $X$ was proper (even projective). –  Sasha Apr 2 '12 at 15:23
    
@Sasha: oops, sorry! I ha read "quasiprojective" but that was about $T$. –  Laurent Moret-Bailly Apr 2 '12 at 17:02
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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.

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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 $((F^\vee)^\vee)_{|X\times\{t\}} \cong ((F_{|X\times\{t\}})^\vee)^\vee$. –  Sasha Apr 2 '12 at 12:00
    
what happens if you take the derived dual? –  Yosemite Sam Apr 2 '12 at 15:28
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Well, in the bounded derived category of a smooth variety, everything is reflexive in the sense that $$F\cong RHom(RHom(F, O_X), O_X)$$ –  Donu Arapura Apr 2 '12 at 16:08
    
ah, of course, silly me. it's an open condition then! :) –  Yosemite Sam Apr 3 '12 at 23:54
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