# Do deformations of a vector bundle form a vector space?

I wonder whether deformations of a vector bundle $F$ over a local Artinian ring $(A, m)$ form a vector space $\operatorname{Ext}^1(F, F) \otimes m$, or what happens in a sequence (*) below? I read two different discussions, one in Huybrechts, Lehn Geometry of moduli space of sheaves, another in Hartshorne Lectures on deformation theory, but it still looks bizarre.

Definitions. Let $F$ be a vector bundle on a nice variety $X$ over a field $k$. Let $A$ be a local Artinian ring over $k$, for example $A=k[t]/(t^n)$, and recall that a deformation of $F$ on $A$ is a vector bundle $F'$ on $X \times \operatorname{Spec} A$ with a fixed isomorphism $F'|_X \to F$. Denote by $D_F(A)$ the set of all deformations of $F$ over $A$, so that $D_F$ is a covariant functor from local Artinian rings to sets. A small extension is a sequence $$0 \to K \to A' \to A \to 0$$ such that the maximal ideal $m'$ of $A'$ goes to the maximal ideal $m$ of $A$, and such that $Km'=0$ holds. A typical example is $$0 \to k \xrightarrow{t^n} k[t]/(t^{n+1}) \to k[t]/(t^n) \to 0.$$

Theorem. Every small extension gives a sequence $$\operatorname{Ext}^1(F, F) \otimes K \to D_F(A') \to D_F(A) \to \operatorname{Ext}^2(F, F) \otimes K,\qquad(*)$$ which is exact in a sense of sets with a chosen element, and also functorial.

But $D_F(k)=0$, so $D_F(A)=\operatorname{Ext}^1(F, F) \otimes m$ holds at least for $A=k[t]/(t^2)$. So does this equality holds for a general $A$? And why a sequence $$\operatorname{Ext}^1(F, F) \otimes K \to \operatorname{Ext}^1(F, F) \otimes m' \to \operatorname{Ext}^1(F, F) \otimes m \to \operatorname{Ext}^2(F, F) \otimes K$$ is so similar to a long exact sequence for $\operatorname{Hom}$-functor? At last, what is a correct way to think about the sequence (*): Huybrechts, Lehn choose an injective resolution, while Hartshorne uses Čech cocycles, but both look like just ways to calculate a cohomology by hand.

• Aren't the deformations a torsor? So, they shouldn't even naturally be a group. Specifically, you say that $D_F(A)$ is a pointed set--what is the distinguished element? – Alex Youcis Nov 28 '16 at 21:43
• @AlexYoucis, a trivial deformation $F'=F \otimes A$ (here, deformations are on $X \times \operatorname{Spec} A$, not some general $X' \supset X$). – evgeny Nov 28 '16 at 21:46
• Ah, thank you. I misread! Another silly question--is it even true that $D_F(A)$ is a group if the rank of $F$ is greater than $1$? Even passing to isomorphism classes this seems unlikely. Sorry if I'm missing something silly! – Alex Youcis Nov 28 '16 at 21:48
• @AlexYoucis, a tensor product of deformations over $F$ is a deformation over $F \otimes F$, not $F$. About whether it is a group/vector space I do not understand, probably not) – evgeny Nov 28 '16 at 21:53
• Right, precisely--the only case where this seems doable is if $F$ is the trivial bundle. Anyways, good luck with your question! – Alex Youcis Nov 28 '16 at 21:54

The deformations over $k[t]/(t^2)$ do form a vector space: this is a general property, namely that the Zariski tangent space at a point of a scheme is a vector space. In the case of deformations of a vector bundle, it admits the description as an $\mathrm{Ext}^1$ that you recall.