Let $X,S$ be $k-$schemes with $k$ a field of characteristic $p$ and let $\omega$ be a line bundle on $X$.
Let $\nabla$ be the canonical connection on $\omega$ (with $p$-curvature $0$).
Let $z$ be a local coordinate near a geometric point $x \in X$.
- How can we choose trivializations of $\omega$ so that $\nabla$ is equal to the derivation $\frac{d}{dz}$?
There is some identification that I don't understand as $\nabla: \omega \to \omega \otimes \Omega^1_{X|S}$ but $\frac{d}{dz}\in \operatorname{Der}(\mathscr O_X, \mathscr O_X)$. How can we see a derivation like a connection?
Thank you for your help.
Since we are talking about line bundles I will restrict myself to this case. Note that to give a trivialization of a line bundle is the same as to give a local section without zeros. Indeed a trivialization gives a commutative diagram
where $\pi_1$ is the projection on the first component. Thus $x \mapsto \psi^{-1}(x,1)$ gives a local section that does not vanish on $U$. On the other hand given a section $s$ on $U$ that does not vanish we may construct a trivialization. Since the rank of $L$ is one, every $v \in L_x$ is a scalar multiple of $s(x)$, say $v= \lambda s(x)$ then we define $v \mapsto (x, \lambda)$.
Now returning to our case we have a local section $s$ of $\omega$ and $dz$ a local section for $\Omega^1$ then $$\nabla(gs) = g\nabla(s) + s\otimes dg = \left(Ag+ \frac{dg}{dz}\right)s\otimes dz$$ i.e $\nabla$ corresponds under this trivialization to the map $g\mapsto Ag+\frac{dg}{dz}$.
Therefore the question is reduced to whether we can give $s$ such that $A=0$ or equivalently $\nabla(s) =0$.
This does not happen for every connection but you have a special one. I don't know this cannonical connection but I found this discussion that may be interesting to you: Canonical connection on the Frobenius pull-back