Equivalence between connections on $\mathcal{E}$ and $\mathcal{P}^1$-linear isomorphisms that induce the identity modulo $\Omega^1_{X|S}$

Question

In Berthelot and Ogus' book "notes on crystalline cohomology", I don't understand the proof of proposition 2.9:
Given an $O_X$-module $\mathcal{E}$ on an $S$-scheme $X,$ a connection $\nabla$ on $\mathcal{E}$ is equivalent to a $\mathcal{P}^1$-linear isomorphism $\varepsilon:\mathcal{P}^1\otimes_{O_X}\mathcal{E}\rightarrow \mathcal{E}\otimes_{O_X}\mathcal{P}^1$ that induces the identity when reduced modulo $\Omega^1_{X|S}$ (where $\mathcal{P}^1=O_x\otimes_{O_S}O_X/\mathcal{I}^2$ and $\mathcal{I}$ is the kernel of $O_X\otimes_{O_S}O_X\rightarrow O_X,\ x\otimes y\mapsto xy$)
In the proof, given a connection $\nabla,$ they define the morphism $\theta:\mathcal{E}\rightarrow \mathcal{E}\otimes_{O_X}\mathcal{P}^1$ by $\theta(x)=\nabla(x)+x\otimes 1$ and they say that it is $O_X$-linear. I don't see how this is $O_X$-linear, isn't $x\mapsto x\otimes 1$ already $O_X$-linear ? If $\theta$ is also $O_X$-linear then that would imply that $\nabla$ is also $O_X$-linear which is not true in general ? I think that I don't understand well the structures of $O_X$-modules considered here and that is probably the source of my confusion.

Answer 1

You are right: your confusion arises from the $O_X$-module structure you have to consider.

In the notes the authors denote by left $O_X$-module structure of $\mathcal{P}^1$ the structure in which multiplication by a scalar is given by $a \cdot (x \otimes y) := (ax) \otimes y$, whereas the right $O_X$ structure is given by $a \cdot (x \otimes y) := x \otimes (ay)$. (Here and in the rest $a \in O_X$ and $x$ is in the $O_X$-module.) The two structures are not the same since the tensor product is taken over $O_S$.

The statement you quote of proposition 2.9 says that $\theta$ is linear for the right $O_X$-module structure of $E \otimes \mathcal{P}^1$, where multiplication by scalar works as follows: $a \cdot (x \otimes b \otimes_{O_S} c) = x \otimes b \otimes_{O_S} ac$.

In particular, with respect to the right $O_X$-module structure of $\mathcal{P}^1$, the map $x \mapsto x \otimes \mathbf{1}$ (here recall that $\mathbf{1} = 1 \otimes_{O_S} 1$) is not linear: given $x \in E$ and $a \in O_X$, then $\theta(ax) = (ax) \otimes \mathbf{1}$, whereas $a \cdot (x \otimes \mathbf{1}) = x \otimes 1 \otimes_{O_S} a$.

For linearity of $\theta$ it boils down to recalling that $da = 1\otimes a - a \otimes 1 \in \mathcal{I}/\mathcal{I}^2$. Recall that $\theta(x) = \nabla(x) + x \otimes \mathbf{1}$ and $\nabla(ax) = a\nabla(x) + x\otimes da$. Then $$\theta(ax) = a \nabla(x) + x \otimes (1 \otimes a - a \otimes 1) + ax \otimes \mathbf{1} = a \nabla(x) + x \otimes 1 \otimes a,$$ where the last equality holds since $a x \otimes \mathbf{1} = x \otimes_{O_X} a \otimes_{O_S} 1$. But this is exactly $a\theta(x)$ if we give to $E \otimes \mathcal{P}^1$ the right $O_X$-module structure.