Pullback of a connection

Question

let $X,Y$ be smooth schemes (or rigid spaces etc..) over a base $S$, let $f:Y \rightarrow X$ be a $S$-morphisn and let $\mathcal{F}$ be a locally free $\mathcal{O}_X$-module with connection $\nabla$. How do we define the "pull-back connection" along $f$? We get a connection $$f^{-1}\mathcal{F} \overset{f^{-1}\nabla}{\rightarrow} f^{-1}\mathcal{F} \otimes_{f^{-1}\mathcal{O}_X} f^{-1} \Omega^1_X,$$ do we extend this to $f^{\ast}\mathcal{F}$ by taking $f^{\ast}\nabla= f^{-1}\nabla \otimes d_Y : f^{-1}\mathcal{F} \otimes_{f^{-1}\mathcal{O}_X} \mathcal{O}_Y$ (as tensor product of connections) or how?

Sorry for the stupid question but it is hard to find references on connections which are not written in the language of differential geometry.

Answer 1

Recall that you have a map $f^* \colon f^{-1} \Omega_X \to \Omega_Y$ (pull-back of differentials). Consider the composition $$ \nabla' \colon f^{-1} E \xrightarrow{f^{-1} \nabla} f^{-1} E\otimes_{f^{-1} \mathcal{O}_X} f^{-1}\Omega_X \xrightarrow{{\rm id}\otimes f^*} f^{-1} E\otimes_{f^{-1}\mathcal{O}_X} \Omega_Y = f^* E \otimes_{\mathcal{O}_Y} \Omega_Y. $$ We want to extend $\nabla'$ to $f^* E$; i.e. we want to check that $$ (f^*\nabla)(e\otimes y) = \nabla'(e)\cdot y + (e\otimes 1)\otimes dy, \quad y\in\mathcal{O}_Y, ', e\in f^{-1} E$$ gives a well-defined map $f^* E\to f^* E \otimes_{\mathcal{O}_Y} \Omega_Y$. This amounts to checking that $$ \nabla'(e)\cdot xy + (e\otimes 1)\otimes dxy = \nabla'(ex)y + (ex\otimes 1)\otimes dy, \quad x\in f^{-1}\mathcal{O}_X,\, y\in \mathcal{O}_Y,\, e\in f^{-1} E. $$ For this, use the fact that the Leibniz rule for $E$ implies $\nabla'(ex) = e\otimes dx + \nabla'(e)\cdot x$.

One also needs to check that $f^*\nabla$ satisfies the Leibniz rule.

Answer 2

Another option is to proceed as follows : show that there exists a unique connection $f^*\nabla$ on $f^*\mathcal F$ verifying :

$$ (f^*\nabla)(f^*s) = f^*(\nabla(s))$$

where on the right-hand side you use the canonical morphism $f^* (\mathcal F\otimes \Omega^1_X)\to f^* \mathcal F\otimes \Omega^1_Y$.

The uniqueness follows from Leibniz rule, as the $f^*s$ generate $f^* \mathcal F$ (locally).

To show the existence locally, you can trivialize $\mathcal F$. Connections on the trivial bundle are of the form $d+\Omega$, where $\Omega$ is a matrix of $1$-forms. It is enough to take the matrix obtained by pulling back each form individually.

Finally, the uniqueness ensures that you can glue these local connections together to get a global solution.