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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.

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    $\begingroup$ Take a look at a book on D-modules. $\endgroup$ – user2520938 Nov 3 '20 at 9:05
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    $\begingroup$ Let $\Delta \hookrightarrow X\times X$ be the diagonal, $\Delta '$ its first infinitesimal neighborhood (defined by $\mathscr{I}_{\Delta }^2$), $p_1,p_2$ the two projections from $\Delta '$ to $X$. You can view a connection on a vector bundle $E$ as an isomorphism $p_1^*E\rightarrow p_2^*E$ on $\Delta '$ inducing the identity on $\Delta $. Then you just have to pull back this isomorphism by the morphism $\Delta '_Y\rightarrow \Delta '_X$ induced by $f$. $\endgroup$ – abx Nov 3 '20 at 11:36
  • $\begingroup$ Thank you both, I am aware of both notions (D-modules and stratifications) but I am looking for an explicit statement in terms of connections. Shouldn't there be a simple answer here? $\endgroup$ – John Nov 3 '20 at 12:24
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    $\begingroup$ @user2520938 the connection is not assumed to be integrable $\endgroup$ – Piotr Achinger Nov 3 '20 at 13:12
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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.

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  • $\begingroup$ Great answer, thank you! $\endgroup$ – John Nov 3 '20 at 14:24
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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.

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