A splitting of an exact sequence is a connection

110 Views Asked by At

Let $X$ be a smooth projective curve over a scheme $S$ and let $\mathcal E$ be a vector bundle on $X$.

For $\mathcal P^1_{X/S}$ the sheaf of principal parts on $X$, Let $1$ be the global section on $\mathcal P^1_{X/S}$ with constant value $1$.

There is an exact sequence:

$0\to \Omega^1_{X/S}\otimes \mathcal E\to \mathcal P^1_{X/S}\otimes \mathcal E \xrightarrow[]{p} \mathcal E \to 0 $.

Let $s\in Hom(\mathcal E, \mathcal P^1_{X/S}\otimes \mathcal E)$ be a split.

Is it true that the map $\nabla: e\to 1.e - s(e)$ is a connection?

Suppose it is true, then we shoud check the "Leibniz rule":

$\nabla(fe) = f\nabla(e) +df\otimes e \Leftrightarrow s(fe) = f.s(e) -df\otimes e$.

But I see no evidence of this last equation being verified by the split $s$ so I have a doubt regarding this question.

Thank you for your lights.

edit

Denote by $ \mathcal I $ the quasi-coherent ideal sheaf of $ \mathcal O_X $-modules corresponding to $ \Delta: X\hookrightarrow X\otimes_S X$, we have $\mathcal{P}_{X/S}^n:=\Delta^{*}( \mathcal{O}_{X\times_S X}) / \mathcal{I}^{n+1}$

And there is a fundamental exact sequence (that we can get by the 3rd isomorphism theorem): $0\to \mathcal I^j/\mathcal I^{j+1} \to \mathcal P^j_{X/S}\to \mathcal P^{j-1}_{X/S} \to 0$ for all $j\ge 0$.

Taking $j=1$, we have $ 0\to \omega_{X/S} \to \mathcal P^1_{X/S}\xrightarrow[]{} \mathcal O_x \to 0$ and tensoring with $\mathcal E$ gives the exact sequence.