Isocrystal with no $F$-structure

Question

$\DeclareMathOperator\Isoc{Isoc}$Let $X_k$ be a quasiprojective $k$ scheme, with $k$ finite, and let $X_K$ be the rigid analytic space lifting it to the fraction field of its Witt ring, which I denote by $K$.

Let $\Isoc(X_k/K)$ denote the category of convergent isocrystals on $X_K$.

If $F$ is a Frobenius lift of the absolute Frobenius on $X_k$ to $X_K$, then pulling back an integrable connection along $F$ induces a tensor functor from $\Isoc(X_k/K)$ to itself.

An isocrystal is called an $F$ isocrystal if it is isomorphic to its pullback along Frobenius. Is there an explicit example of an isocrystal with no $F$-structure? For example, perhaps of a line bundle on the affine line or $\mathbb{G}_m$?

Answer 1

The connection $D - \dfrac{a dx}{x}$ on the trivial line bundle on $\mathbb G_m$ has formal solutions proportional to $x^a$ which is convergent for $a \in \mathbb Z_p$. Hence the connection is integrable for $a\in \mathbb Z_p$.

Frobenius multiplies $a$ by $|k|$ since $x \mapsto x^{|k|}$ is an explicit Frobenius lift, and two connections are equivalent if and only if their difference is an integer, so the $F$-isocrystals correspond to $a \in \frac{1}{ |k|-1} \mathbb Z$.

So any irrational $a$ in $\mathbb Z_p$ will do the trick.