Let $X$ be an algebraic scheme over an algebraically closed field of characteristic $p$, let $U$ be an open dense subset and let $\mathcal F$ be a lisse $\overline{\mathbb Q_l}$ sheaf on $U$.
It is known that $\mathcal F$ corresponds to a finite dimensional $\overline{\mathbb Q_l}$ vector space $V$ together with a continuous $\pi_1(U, \bar x)$-action (for an $x \in |U|$).
Here I wonder what is $(j_*\mathcal F)_{\bar s}$, where $s \in X-U$ and $j : U \rightarrow X$ is the open immersion.
The textbook I read says it is $V^{I_s}$, where $I_s \subseteq \pi_1(U, \bar x)$ is the "ramification group" at $s$, which seems to be the absolute Galois group of the quotient field of the strict henselization of the local ring at $s$ in $\bar s$.
I don't see why the absolute Galois group is contained in $\pi_1(U, \bar x)$ and $(j_*\mathcal F)_{\bar s} = V^{I_s}$.
Thanks in advance.
Someone brought this to my attention so hopefully I can resolve this satisfactorily.
Let $X$ be a noetherian normal $1$-dimensional scheme over a field $k$. Let $S$ be a finite set of closed points of $X$ and let $U=X-S$. Let $\mathcal{F}$ be a local system on $U$ and so choosing the geometric point $\bar{\eta}$ lying above the generic point $\eta\in U$ we can equivalently think of $\mathcal{F}$ as the $\pi_1(U,\bar{\eta})$ representation $\mathcal{F}_{\bar{\eta}}$.
Now let $s\in S$ and consider the strict localisation $\mathcal{O}_{X,\bar{s}}^{sh}$ at a geometric point $\bar{s}$ lying above $s$. This is a dvr and note that its spectrum $Z=\operatorname{Spec}(\mathcal{O}_{X,\bar{s}}^{sh})$ consists of precisely two points, the closed point $\bar{s}$ and the generic point which we again denote by $\eta$ by abuse of notation. We let $K_{\bar{s}}^{sh}$ denote $\operatorname{Frac}(\mathcal{O}_{X,\bar{s}}^{sh})$. Notice that $Z^0:=\operatorname{Spec}(\mathcal{O}_{X,\bar{s}})-\{\bar{s}\}=\operatorname{Spec}(K_{\bar{s}}^{sh}).$ So in fact $\pi_1(Z^0,\bar{\eta})=\operatorname{Gal}(k(\bar{\eta})/K_{\bar{s}}^{sh})$ maps into $\pi_1(U,\bar{\eta})$. The image of this morphism is the inertia group at $s$ denoted $I_s$.
Let $j\colon U\to X$ be the canonical immersion; we want to compute $(j_*\mathcal{F})_{\bar{s}}.$ To do this we compute that $$(j_*\mathcal{F})_{\bar{s}}=H^0(\operatorname{Spec}(\mathcal{O}^{sh}_{\bar{s}})\times_X U, p^*\mathcal{F})$$ where $p\colon \mathcal{O}^{sh}_{\bar{s}}\times_X U\to U$ is the projection.
So it is enough to understand the scheme $\operatorname{Spec}(\mathcal{O}^{sh}_{\bar{s}})\times_X U$. This is the fiber of the canonical morphism $j\colon U\to X$ above the image of $\operatorname{Spec}(\mathcal{O}^{sh}_{\bar{s}})$ in $X$. The fiber above the closed point $\bar{s}$ is empty since $s\in S=X-U$ and the fiber above the generic point is precisely the generic point of $U$. So the fiber is canonically isomorphic to $\operatorname{Spec}(K^{sh}_{\bar{s}})\times_U X$. By replacing $U$ by an affine cover containing $\eta$ and similarly for $X$ we see that $\operatorname{Spec}(K^{sh}_{\bar{s}})\times_U X=\operatorname{Spec}(K^{sh}_{\bar{s}}).$
Thus we have $$(j_*\mathcal{F})_{\bar{s}}=H^0(\operatorname{Spec}(K^{sh}_{\bar{s}}), \mathcal{F}_{\bar{\eta}}).$$
Since étale cohomology over a point is just Galois cohomology we have the right side is equal to
$$H^0(\operatorname{Gal}(k(\bar{\eta})/K_{\bar{s}}^{sh}),\mathcal{F}_{\bar{\eta}})=\mathcal{F}_{\bar{\eta}}^{\operatorname{Gal}(k(\bar{\eta})/K_{\bar{s}}^{sh})}=\mathcal{F}_{\bar{\eta}}^{I_s}.$$