(the title got out of hand)


Say I have a surface $X$, then I also have M, the Hilbert scheme of curves and points on X.
This can be seen as a moduli space of quotients $O\_X \to O\_Z$.


If $I\_Z$ is the kernel of that map, I would like to impose the condition that $Ext^2(I\_Z,O\_Z) = 0$. I imagine this is badly behaved, but let me ask anyway:


`Is this condition open, or closed, or something on M?`


I wouldn't expect to be, but I lack the knowledge to cook up some convincing example.



is there some condition I can impose on families of curves on a surface such that the second Ext between the ideal sheaf and the structure sheaf is zero?
If I am not mistaken, we *always* have $\mathrm{Ext}^2(\mathcal I\_Z,\mathcal O\_Z)=0$ for any closed subscheme $Z\subset X$, so the question is somewhat vacuous.


We have an exact sequence 


$$0\to \mathcal I\_Z \to \mathcal O\_X \to \mathcal O\_Z\to 0,$$ 


and applying $\mathrm{Hom}( -,\mathcal O\_Z)$ gives a surjection $$\mathrm{Ext}^2(\mathcal O\_X,\mathcal O\_Z)\to \mathrm{Ext}^2(\mathcal I\_Z,\mathcal O\_Z)\to 0.$$ But $\mathrm{Ext}^2(\mathcal O\_X,\mathcal O\_Z) = H^2(\mathcal O\_Z)$, which vanishes since the support of $\mathcal O\_Z$ has dimension at most $1$.


(Edit addressing your comment below:


If $X$ is a threefold and $Z$ has dimension at most $1$, it is still true that $\mathrm{Ext}^i(\mathcal O\_X,\mathcal O\_Z)=0$ for $i=2,3$, so $\mathrm{Ext}^2(\mathcal I\_Z,\mathcal O\_Z) = \mathrm{Ext}^3( \mathcal O\_Z,\mathcal O\_Z)$. But $$\mathrm{Ext}^3(\mathcal O\_Z,\mathcal O\_Z) = \mathrm{Hom}(\mathcal O\_Z , \mathcal O\_Z \otimes K\_X) = H^0(K\_X|\_Z),$$ which is typically nonzero.)


