1
$\begingroup$

This is my question:

Given a projective noetherian scheme $X$, the structural sheaf $\mathcal{O}_X$ is a coherent sheaf, so every locally free sheaf is coherent. This means that the family of stable locally free sheaves (with fixed Hilbert polynomial) is parametrized by a subset $S$ of the moduli space $M$ of stable sheaves on $X$. Is it an open subset? i.e. is the locally free condition an open condition?

I use the definition that a stable sheaves is a pure sheaves with strictly decreasing reduced Hilbert polynomial on proper subsheaves, so ,in particular, every sheaf in $M$ is torsion free. But torsion free sheaves are not locally free in general, so $S$ is a proper subset.

I'm wondering also if there exists an example of a torsion free sheaf and a locally free sheaf with the same Mukai vector. If this is not the case, I can obtain the required subset by the union of the connected components, relative to Mukai vectors corresponding to locally free sheaves.

Thank you!

P.S.: In general which are the minimal conditions on $ X $ to have that the locally free condition is an open condition?

$\endgroup$
  • $\begingroup$ Is your scheme proper? How are you defining stable? For coherent sheaves on a non-proper scheme, "locally free" is not always an open condition. $\endgroup$ – Jason Starr Aug 18 '14 at 18:04
  • $\begingroup$ No, my scheme is not proper in general, but a positive answer in the proper case is very useful. For the definition of stable sheaves, I am using the one given in Huybrechts-Lehn "The geometry of moduli spaces", that a stable sheaf is a pure sheaf such that the reduced Hilbert polynomial of every proper subsheaf is strictly minor than its. Do you have a reference for that? Thanks! $\endgroup$ – Cla Aug 18 '14 at 18:14
  • 1
    $\begingroup$ How do you define the Hilbert polynomial for a sheaf with non-proper support? If you scheme is non-proper, how can any (nonzero) locally free sheaf have proper support? $\endgroup$ – Jason Starr Aug 18 '14 at 18:47
  • $\begingroup$ @JasonStarr: if I may say so, I think the OP wouldn't mind assuming X to be proper (while in the meantime he/she tries to make sense of the broader question). $\endgroup$ – bananastack Aug 19 '14 at 2:34
  • 1
    $\begingroup$ If your schemes are proper, then local freeness should be open, and I am sure it is written down somewhere in EGA. First of all, using "openness of flatness", the locus in the total space of your family where the universal coherent sheaf is flat is open. Thus its complement is closed. Since the family is proper over the base, the image in the base of this closed subset is also a closed subset. The open complement is the open subset that you want. $\endgroup$ – Jason Starr Aug 19 '14 at 11:38
1
$\begingroup$

I will answer the second question.

Let $X$ be any smooth projective surface and $P \in X$ a point. Let $F = O_X \oplus I_P$. Then $F$ is torsion free but not locally free. But its general deformation is locally free. The reason is that $\dim Ext^1(F,F) = 3$, so a generic deformation of $F$ should erase the singularity of $F$ (of course to turn it into a rigorous proof you should work a bit).

In case $X = \mathbb{P}^2$ the generic deformation of $F$ can be written as the kernel of a generic map $O_X \oplus O_X(1) \oplus O_X(1) \to O_X(2)$.

$\endgroup$

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.