# Pullback of a connection

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.

• Take a look at a book on D-modules. – user2520938 Nov 3 '20 at 9:05
• 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 $f$. – abx Nov 3 '20 at 11:36
• 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? – John Nov 3 '20 at 12:24
• @user2520938 the connection is not assumed to be integrable – Piotr Achinger Nov 3 '20 at 13:12

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.

• Great answer, thank you! – John Nov 3 '20 at 14:24

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.