Let $X$ be a scheme and $\bar{v}$ a geometric point of $X$. Say that a sheaf $F$ on the étale site is locally constant around $\bar{v}$ if its restriction to an étale neighbourhood of $\bar{v}$ is constant. It is clear that if $F$ is locally constant around $\bar{v}$ then $F_{|\mathrm{Spec}(\mathcal{O}_{\bar{v}})}$ is constant, where $\mathcal{O}_{\bar{v}}$ is the strictly local ring at $\bar{v}$.
When is the converse true ?
Let's consider a special case. Let $F$ be a constructible sheaf on a 1-dimensional normal scheme $X$ (or something along those lines). Denote $\eta=\mathrm{Spec}(K)$ the generic point, $v$ a point, $G_K$ the Galois group of $K$, $I_v\subset G_K$ an inertia subgroup for $v$, $(-)_\eta$ the pullback to $\eta$. Identifying a sheaf on $\eta$ with a continuous $G_K$-module. In that situation, by Artin gluing $F_{|\mathrm{Spec}(\mathcal{O}_{\bar{v}})}$ is constant if and only if $I_v$ acts trivially on $F_\eta$, so my question is the following:
Is $F$ is locally constant around $\bar{v}$ if and only if $I_v$ acts trivially on $F_\eta$ ?
Best Answer
Suppose that $F$ is locally constructible, and let's show the converse holds. Replacing $X$ by an étale neighbourhood of $\overline v$, we may assume that $X$ admits a finite partition by locally closed subschemes such that $F$ is constant on each piece. As long as there exists a piece that does not meet $\operatorname{Spec}(\mathcal O_{\overline v})$, we can shrink $X$ in the Zariski topology to make sure this piece is closed and then take it out. This makes the number of pieces strictly decrease, so by induction we can assume that each piece meets $\operatorname{Spec}(\mathcal O_{\overline v})$. Then, the restriction of $F$ to any piece is canonically isomorphic to the constant sheaf with value $F(\operatorname{Spec}(\mathcal O_{\overline v}))$, so $F$ is constant as well.
The minimal hypotheses are most likely weaker than local constructibility of $F$, but this seems like a natural sufficient condition.