7
$\begingroup$

Let we consider the Quot scheme $Q(X,N,P)$ of coherent sheaves on an algebraic variety with given Hilbert polinomial $P$ and quotient of $\mathcal{O}_X^N$, i.e. all $\mathcal{F}$ of the form $\mathcal{O}_X^N\longrightarrow\mathcal{F}\longrightarrow 0$. Recall that a simple sheaves $\mathcal{F}$ is a coherent sheaves such that $End(\mathcal{F})$ is trivial. Then, why is the Quot scheme smooth in the point represented by $\mathcal{F}$, where $\mathcal{F}$ is simple?

I know that the tangent space of the Quot scheme in the point $$q=[\mathcal{O}_X^N\stackrel{q}{\longrightarrow}\mathcal{F}\longrightarrow 0]$$ is $$T_qQuot\cong Hom(\mathcal{G},\mathcal{F}),$$ where $\mathcal{G}:=\ker q$, but I am not able to answer the above question only from this.

Improve this question
$\endgroup$
6
  • $\begingroup$ what's the tangent space? $\endgroup$
    – bananastack
    Commented May 19, 2014 at 15:09
  • $\begingroup$ maybe I should point out that my comment from a month ago wasn't implying the answer was obvious once you knew what the tangent space was. I don't know the answer myself, I was hoping to be able to use some long exact sequences to relate simpleness to smoothness (ie constancy of dim Tangent space). Why do you believe the result to be true? $\endgroup$
    – bananastack
    Commented Jun 20, 2014 at 3:07
  • $\begingroup$ I am sure that simple sheaves are smooth point in the Quot scheme because it is used in some contexts, for example a way to create a Kuranishi family of simple sheaves is the following. Take the Hilbert scheme and consider the action (that is linearly reductive) of the group $GL(n)$ as (with the above notation) $$ g\cdot q:=g\circ q,$$ where we see $g\in GL(n)$ as a morfism $g:\,\mathcal{O}_X^N\to\mathcal{O}_X^N$. $\endgroup$
    – Lal
    Commented Jun 20, 2014 at 13:40
  • $\begingroup$ If we take a point $q$ in the Quot represented by a simple sheaves than the Slice étale, whose existence is ensured by the Luna's theorem, is the required Kuranishi family (the universal property is inherited by that of Quot). This is possible only if q is smooth, so I decided to try to prove it but I was not able. (This is the rest of Claudio's comment -ed.) $\endgroup$
    – Ben Webster ♦
    Commented Jun 20, 2014 at 14:38
  • $\begingroup$ Have you looked at Proposition 2.2.8 in Huybrechts&Lehn - Geometry of the moduli spaces of sheaves? It says that $hom(G,F) \geq dim_q Quot \geq hom(G,F)-ext^1(G,F)$ and if $ext^1(G,F)=0$ then $Quot$ is smooth at $q$. But i was not able to show that this $Ext$ vanishes. Are there any more assumptions on $F$ that could help to simplify this situation? $\endgroup$
    – DonD
    Commented Jun 21, 2014 at 17:33

1 Answer 1

Reset to default
2
$\begingroup$

The claim is not true in general! The obstructions for a coherent (semistable) sheaves to be a smooth point in the Quot scheme are contained in the second Ext group $Ext^2(\mathcal{F},\mathcal{F})$. Even if $\mathcal{F}$ is simple, this group can be very big and there is no matter because the obstructions must vanish. Anyway if $X$ is a surface (with trivial canonical bundle) we can use Serre Duality to claim that $Ext^2(\mathcal{F},\mathcal{F})\cong Hom(\mathcal{F},\mathcal{F})$ that is trivial, by the assumption that $\mathcal{F}$ is simple.

References for this can be: 1) Huybrechts, Lehn, "The geometry of moduli spaces of sheaves" section 4.5. 2) Mukai, "Moduli of vector bundles on K3 surfaces and symplectic manifolds" Sugaku Expositions, vol.1, n.2, 1988, pag. 151-153.

Improve this answer
$\endgroup$

You must log in to answer this question.

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