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The moduli space of sheaves over a smooth variety is in general not separated. That is, there exists a flat family of coherent sheaves over a punctured disk which extends to a flat family of coherent sheaves over a disk in several ways. Could anyone give me easy examples?

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Take sheaves of ideals of the origin with different multiplicities. –  Alex Degtyarev Mar 1 at 9:17
It is better to write "the moduli STACK of sheaves" here, because the moduli space implicitly means the moduli space of semistable sheaves, and the latter is separated. –  Sasha Mar 1 at 10:20
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1 Answer 1

up vote 6 down vote accepted

There is a vector bundle $E$ on $\mathbb{P}^1 \times \mathbb{A}^1$ whose restriction to $\mathbb{P}^1 \times (\mathbb{A}^1 - \{ 0\})$ is isomorphic to $\mathcal{O}^2$, but whose restriction to $\mathbb{P}^1 \times 0$ is $\mathcal{O}(-1)\oplus \mathcal{O}(1)$.

One can take $E$ to be the cokernel of the map $$ (x, y, t) : \mathcal{O}(-1) \to \mathcal{O}\oplus \mathcal{O} \oplus\mathcal{O}(-1),$$ where $x$ and $y$ are the coordinates on $\mathbb{P}^1$ and $t$ is the coordinate on $\mathbb{A}^1$. Indeed, for nonzero $t$ the map has a section, and for $t=0$ we get $\mathcal{O}(-1)\oplus\mathcal{O}(1)$.

The whole point of course is that $\mathcal{O}(-1)\oplus \mathcal{O}(1)$ is not semistable.

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In the moduli space of semistable sheaves one has to identify those semistable sheaves which have filtrations with isomorphic associated graded quotients. So the stable part is separated but already on the semistable locus one has non-separatedness of the stack. –  Vladimir Baranovsky Mar 1 at 21:51
It seems to me that graded quotients (with obvious filtration) of the two bundles are different. One consists of two $\mathcal{O}$ and the other consists of $\mathcal{O}(-1)$ and $\mathcal{O}(1)$. –  user2013 Mar 1 at 23:43
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