Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question

1 Answer 1

up vote 4 down vote accepted

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 $\mathcal G=\mathcal O_{X_1}/\mathcal I$).

share|improve this answer
    
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.