Frobenius action on the trivial connection

Question

Let $F$ denote the absolute Frobenius acting on a smooth quasiprojective scheme $X$ over a finite field $k$.

Denote the trivial connection on $\mathcal{O}_X$ by $d$. Denote its pullback by Frobenius by $d_f$.

I am trying to understand how $F$ acts on $d$. Since pulling back connection induces a tensor functor between the category of integrable connections on $X$ to itself, implies that it must send a unit object to a unit object, hence we must have $d\cong d_f$.

An isomorphism of connections on a line bundle is the same as a global section of $\mathbb{G}_m$, denoted $\rho$, having the property that $\rho d_f(s) - d(\rho s) = 0$, which is equivalent to $\rho F^*ds - \rho ds = sd\rho$.

However, since $F^*ds = d(s^p) = 0$, we find that a horizontal morphism depends on $s$, so shouldn't exist in general. What am I missing?

Answer 1

Since both connections are defined on the same domain, I managed to confuse myself with the definition of the pullback connection.

The pullback connection extends via the Leibniz rule on a local base.

This means in particular that $d_f(s)$ is not defined as $F^*ds$, but rather '$s$' is actually $s\otimes_{F^{-1}\mathcal{O}_X(U)}1$, for $s\in \mathcal{O}_X(U)$, so that $d_f(s) = d(s)\cdot 1 + sd_f(1) = ds$. The formula I wrote now makes sense: it says $\rho$ needs to be chosen so that $sd\rho = 0$. That is, any constant $\rho$ would do.

Another point I notice here is that the automorphisms of a connection are not quite the same thing as an automorphism of the underlying line bundle, which are given by $\Gamma(X,\mathbb{G}_m)$, but rather we are supposed to quotient this space by $k^{\times}$, or more generally by global $\mathbb{G}_m$ sections on the base.