# Definition of Grothendieck Connection

I suppose this is not hard, but I thought quite a time about it and don't get what the point is.

I'm reading Deligne's "Equations differentielles...", where he defines what a Grothendieck Connection is:

you consider a smooth projective variety $X$ over the complex numbers and a coherent sheaf $F$ on it. Let $X_1$ be the first infinitesimal neighborhood of the diagonal of $X$ and $p_1,p_2$ the two projections of $X_1$ to $X$.

Then one defines a connection as a homomorphism $p_1^*F \rightarrow p_2^*F$, which restricts to the identiy on $X$.

Deligne says in a bracket that such a homomorphism is always an isomorphism. What's the reason for this?

Addition: can one formulate a general principle, which roughly says that if I have a homomorphism on the first oder neighborhood of the diagonal, which is an iso on $X$, then it is already an iso? How general does something like this hold?

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You can write everything in an affine charte and suppose $X$ is affine. Let $A$ be the algebra of $X$ and let $B$ be the algebra of $X_1$, let $M$ be a finitely generated $A$-module. You have a $B$-linear map $$\varphi : N:=M\otimes_A B \to N$$ such that $\varphi(v)=v + \psi(v)$ with $\psi(v)\in IN$ where $I$ is the ideal of $B$ definning $A$ (so $I^2=0$). As $\varphi$ is surjective modulo a nilpotent ideal, it is surjective by Nakayama lemma.
Let's prove the injectivity. Notice that if $v\in IN$, then by the $B$-linearity of $\psi$, $\psi(v)\in I(IN)=0$. Now let $v\in N$ be in the kernel of $\varphi$. Then $v=-\psi(v)\in IN$, hence $\psi(v)=0$, thus $v=0$.
In general, if $\varphi : \mathcal F\to \mathcal G$ is a homomorphism of coherents sheaves on $X_1$ which restricts to an isomorphism to $X$, then Nakayama implies that $\varphi$ is surjective. Let $\mathcal I$ be the sheaf ideal on $X_1$ definning $X$, let $\mathcal L$ be the kernel of $\varphi$. If $\mathcal G$ is flat over $X_1$, then tensoring the exact sequence $$0 \to \mathcal L \to \mathcal F\to \mathcal G\to 0$$ by $\mathcal O_{X_1}/\mathcal I$ implies that $\mathcal L\subseteq \mathcal I\mathcal L\subseteq \mathcal I^2\mathcal L=0$. Therefore $\varphi$ is an isomorphism. Without flatness of $\mathcal G$, the result doesn't hold in general (take $\mathcal F=\mathcal O_{X_1}$ and $\mathcal G=\mathcal O_{X_1}/\mathcal I$).
A very fine comment, thanks! Just one tiny question: how do you get your decomposition of $\phi$ in the fourth line? –  Veen Oct 27 '11 at 7:24
This is because mod $IN$, $\varphi$ is identity by hypothesis. –  user18119 Oct 27 '11 at 10:00