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(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.

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another condition to impose which I think implies what I want is for $I_Z$ to be a vector bundle. This would seem to be an open condition, but it might too strong. – Jacob Bell Dec 1 '12 at 4:52
Regarding your comment above: assume $X$ is smooth. Then $I_Z$ a line bundle if and only if $Z$ is a divisor (consists only of curves), which I believe is both open and closed... – Piotr Achinger Dec 1 '12 at 5:58
ah, thanks Piotr, good point. But is it really a component? I didn't think it would be closed (I thought people tried to compactify such things). – Jacob Bell Dec 1 '12 at 6:48
If the surface is smooth, being a Cartier divisor is indeed an open and closed condition. This is explained in Mumford's "Curves on an algebraic surface." – Angelo Dec 1 '12 at 8:55
thank you Angelo, I'll have a look. I think my bias comes from the fact that I'm used to thinking about threefolds, where curves aren't divisors. – Jacob Bell Dec 1 '12 at 20:35
up vote 5 down vote accepted

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.)

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you're right, I haven't thought about that! This fails in higher dimensions right? I mean, if $X$ is a threefold say and $Z$ is still a curve. – Jacob Bell Dec 1 '12 at 20:34
In general yes, see the edit. – Jack Huizenga Dec 1 '12 at 21:16

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