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I am trying to understand the correspondence between isomorphism classes of infinitessimal extensions of a k-scheme $X$ by a coherent sheaf $\mathcal{F}$ and the cohomology group $H^1(X,\mathcal{F}\otimes \mathcal{T}_X)$ where $\mathcal{F}_X$ is teh tangent sheaf of $X$.

Let me recall some definitions and properties in the first place:

  • Infinitessimal lifting property: given an algebraically closed field $k$, a finitely generated $k$-algebra $A$ with $X=Spec A$ non-singular, and an exact sequence $0\rightarrow \mathcal{I} \rightarrow B' \rightarrow B \rightarrow 0$, where $B,B'$ are k-agebras and $I\subset B'$ is an ideal such that $\mathcal{I}^2=0$, any k-algebra homomorphism $A\rightarrow B$ lifts to a h-algebra homomorphism $A\rightarrow B'$, and two such homomorphism differ by a k-derivation of A into $\mathcal{I}$, namely an element in $Hom_A(\Omega_{A/k},\mathcal{I})$.

  • An infinitessimal extension of a k-scheme $X$ by a coherent sheaf $\mathcal{F}$ is a pair $(X',\mathcal{I})$ where $X'$ is a k-scheme and $\mathcal{I}$ is a sheaf of ideals on $X'$ with $\mathcal{I}^2=0$ and such that we have isomorphisms $(X',\mathcal{O}_{X'}/\mathcal{I})\cong (X,\mathcal{O}_X)$ (as k-schemes) and $\mathcal{I}\cong \mathcal{F}$ (as $\mathcal{O}_X$-modules). For instance, the trivial extension of $X$ by $\mathcal{F}$ is given by the pair $(X,\mathcal{F})$, where the $X$ has structure sheaf $\mathcal{O}_X'=\mathcal{O}_X\oplus \mathcal{F}$ with product $(a\oplus f)\cdot (a'\oplus f')=aa'\oplus (af'+a'f)$, so that $\mathcal{F}$ becomes an ideal sheaf in $X$.

  • If $X=Spec A$ is affine and $\mathcal{F}$ is a coherent sheaf, then any extension is isomorphic to the trivial one: we just use the previous lifting property to construct a splitting of an appropriate short exact sequence.

Now regarding the correspondence I wrote down at the offset: let $X',\mathcal{I})$ be an infinitessimal extension of $X$ by $\mathcal{F}$ and let $\{U_i\}$ be an affine open cover of $X$ (so that sheaf cohomology is isomorphic to Cech cohomology). By the previous remark, we know that on every open affine set the extension is trivial, namely of the form $(U_i,\mathcal{I}_{|U_i}=\mathcal{O}_{U_i}\oplus \mathcal{F}_{|U_i})$. From the way we constructed the trivialisation (choosing a lift, and noting that the difference of two lifts gives an element of $Hom_A(\Omega_{A/k},\mathcal{I})$, we see that this gives a cocyle in $\check{H}^1(X,\mathcal{F}\otimes \mathcal{T}_X)$. I am confused about the converse:

  1. How do we obtain an infinitesimmal extension of $X$ by $\mathcal{F}$ from a cohomology class in $H^1(X,\mathcal{F}\otimes \mathcal{T}_X)$?
  2. For instace, if $X=P_k^2$ and $\omega_X$ is the sheaf of differential two forms, then $H^1(X,\omega_X^1)\cong H^1(X,\omega_X\otimes \mathcal{T}_X)$. A nontrivial cocycle is given by $\xi \in H^1(X,\omega_X^1)$ given over $U_{ij}=U_i\cap U_j$ (where the $\{U_i\}$ are the standard open subsets covering $P_k^2$) by $\xi_{ij}=\frac{x_j}{x_i}d\left(\frac{x_i}{x_j}\right)$. What infinitessimal extension $X'$ does it yield?

Thanks in advance for any insight.

share|improve this question
    
Without getting technical, just reverse your construction. Start with an open affine cover $\{U_i\}$ of $X$ and a Cech cocycle $\eta=(f_{ij}) \in \check{H}^1(X,\mathcal F \otimes \mathcal T_X)$. Now begin with the trivial extension of $U_i$ by $\mathcal F_{|U_i}$, and use $\eta$ to glue these extensions together to get an infinitesimal extension of $X$ by $\mathcal F$. –  Parsa Apr 14 '12 at 10:23

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