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Let $X$ be a complex manifold (for example $\mathbb CP^n$), let $v$ be a holomorphic vector field on $X$, and let $F$ be a coherent sheaf (for example a vector bundle or a structure sheaf of a point). Then $v$ defines an element in $Ext^1(F,F)$. Indeed $v$ generates an action of $\mathbb C$ on $X$ and taking pull-backs of $F$ under this action we get a deformation of $F$, hence an element of $Ext^1(F,F)$.

Question. Is there any fancy (or not fancy) way to express the corresponding element of $Ext^1(F,F)$ in terms of $v$ and $F$? Maybe there is some construction with jets? (I understand, this is a bit vague)

Added. Two equally nice and far leading answers were given to this question. I would like summarise here what I understood from David's answer in down to earth terms. So, we want to associate an extension $F\to E\to F$ to a vector field $v$. Suppose (just for the sake of been very much down to earth), that $F$ is a vector bundle, and $v$ has no zeros. Then we can consider $1$-jets of sections of $F$ in the direction of $v$ (or if you want, along trajectories of $v$). It is not hard to see that this is a bundle of rank $2rank(F)$, and this is exactly $E$, that we are looking for.

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up vote 6 down vote accepted

Holomorphic vector fields give elements of the first Hochschild cohomology group of the structure sheaf $$HH^1(O_X,O_X)\simeq H^1(X,O_X)\oplus H^0(X,T_X)$$ (via an easy part of the Hochschild-Kostant-Rosenberg theorem).

On the other hand the Hochschild cohomology may be identified (or indeed defined as) self-Ext of the identity functor on the category of coherent sheaves. Hence, by definition of self-ext of the identity, there's a functorial map $$HH^1(O_X,O_X)=Ext^1(Id_{O_X-mod})\longrightarrow Ext^1(F,F)$$ for any coherent sheaf $F$. This is one fancy way to express your construction. Likewise the second Hochschild cohomology gives deformations of the (derived) category of sheaves, and the corresponding element in $Ext^2(F,F)$ is the obstruction for $F$ to deform.

EDIT: Here's a little more to explain the compatibility with Torsten's answer. Any sheaf $F$ has a canonical extension by $F\otimes \Omega^1$, given by 1-jets of sections of $F$ $$0\to F\otimes \Omega^1 \to J^1(F) \to F\to 0.$$ We can contract this extension against a vector field to get an $Ext^1$ of $F$ by itself as desired. The functor taking $F$ to its extension is represented (as an integral transform) by the first-order neighborhood of the diagonal, i.e. by the 1-jet of the identity functor (which again can be contracted against any vector field). This is explicitly the map from vector fields to $Ext^1$ of the identity functor, i.e. Hochschild $HH^1$.

This universal extension can be rewritten as an element $$Hom(T_X[-1]\otimes F,F),$$ which in fact defines a homotopy Lie algebra structure on the shifted tangent sheaf $T_X[-1]$ (the Atiyah bracket). The sheaf Hochschild cohomology can then be identified as the universal enveloping algebra of this homotopy Lie algebra. So in some precise sense this universal action of vector fields (as Torsten describes) IS Hochschild cohomology, acting as the derived center of the category.

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Dear David, very interesting, thanks a lot! I will think about what you have said. – Dmitri May 28 '10 at 22:35

Here is a not so fancy description.

There is a general principle (in algebraic geometry but applicable to some neighbouring disciplines) that says that anything that is functorial and commutes with base change admits an action by vector fields. The reason is that a vector field on $X$ gives an automorphism on $X[\delta]$, the base extension of $X$ to the ring of dual numbers, that is the identity modulo $\delta$.

As an example that is unrelated to the question but illustrative take the example of a vector bundle $E$ and its symmetric algebra $S^*E$. The fact that the symmetric algebra is defined by a universal property and commutes with base extension immediately gives an action of derivations of $\mathcal O_X$ on $S^*E$ (the usual proof of this uses these properties implicitly but is less transparent as it doesn't use them explicitly).

In any case turning to the problem at hand, given a vector field $v$ we get an automorphism $\varphi$ of $X[\delta]$ and hence an $\mathcal O_{X[\delta]}$-module which is the pullback by $\varphi$ of the constant extension $F[\delta]$ of $F$ to $X[\delta]$. Concretely it means that we consider $F[\delta]=F\bigoplus F\delta$ with action of $f\in\mathcal O_X$ given by $f\cdot e=fe+v(f)e\delta$ (and $\delta$ acting in the obvious way). The identification of first order deformations of $F$ with $\mathrm{Ext}^1(F,F)$ is the done by considering this as an extension of $F=F\delta$ by $F[\delta]/F\delta=F$.

This can be elaborated as follows (which should give a closer relation with the answer provided by David): Instead of taking the pullback of $F[\delta]$ by $\varphi$ we can instead consider the vector field as a morphism $\varphi\colon X[\delta]\rightarrow X$ and pullback $F$ along it. We of course also have the constant morphism $X[\delta]\rightarrow X$ and the coincide on the closed immersion $X\hookrightarrow X[\delta]$. Hence the induced map $X[\delta]\rightarrow X\times X$ factors through the first order neighbourhood of the diagonal $X^{(2)}:=\mathrm{Spec}\mathcal O_{X\times X}/\mathcal I^2_\Delta \hookrightarrow X\times X$, i.e., a map $X[\delta]X^{(2)}$ such that the composite of the first projection $X^{(2)}\rightarrow X$ is $\varphi$ and with the second is the constant map. (Concretely, we have $\mathcal I_\Delta/\mathcal I^2_\Delta=\Omega^1_X$ and $v$ is a map $\Omega^1_X\rightarrow\mathcal O_X$ and the map is just the pushout along this map.) Hence the pullback of $F$ along $\varphi\colon X[\delta]\rightarrow X$ is obtained by first taking the pullback $F'$ of $F$ along the first projection $X^{(2)}\rightarrow X$ and the pullback further on to $X[\delta]$. However, $F'$ as an extension $F\bigotimes\Omega^1_X$ by $F$ is the Atiyah extension and we the extension of $F$ by $F$ associated to $v$ is just the pushout of it along the map $F\bigotimes\Omega^1_X \rightarrow F$ induced by $v$. Hence, the extension is the pushout of the universal Atiyah extension.

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Dear Torsten, huge thanks! I feel your answer is what I wanted. – Dmitri May 29 '10 at 11:13

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