Suppose we have a surface $S$ mapping to a curve $C$ and a finite cover $Y/S$ that is ramified at a divisor $D$. For each point $x \in C$ we get a ramified cover of the curve $S_x$, and we can study how the local monodromy of the cover at different points of $D_x$ varies as we change $x$.
In characteristic zero, or for tame ramification, this is not very interesting, except at the singularities of $D$, because for smooth $D$ the local monodromy must remain constant.
In characteristic $p$, on the other hand, even when $D$ is smooth, we can have a degeneration from one Galois representation at the generic point to a very different one at a special point. (For instance, the Artin-Schreier cover $t^p-t = x/y$ is ramified at $y=0$ for most $x$, but unramified when $x=0$.) By Deligne's semicontinuity theorem, we at least know that Galois representations can only degenerate to simpler Galois representations, as measured by the Swan conductor.
Still, I think this leaves a lot of questions about degeneration of Galois representations unanswered. I'm going to ask about one of the simplest cases, which is local monodromy representations into $A_5$ in characteristic $2$.
First let's describe local monodromy representations into $A_5$. Let $k$ be an algebraically closed field of characteristic $2$, and consider a representation $\operatorname{Gal} ( k((x))) \to A_5$. The image must be a group with normal $2$-Sylow subgroup, so if the order is even it must be contained in $A_4$. By taking the third root of $x$, we may assume that the the image is contained in the Klein four subgroup.
This subgroup has three characters of order $2$. The most important invariants of the local monodromy representation are the Swan conductors of the three characters. Using $sw( \chi_1 \otimes \chi_2) \leq \max (sw(\chi_1), sw(\chi_2))$, we can see that the two largest Swan conductors must be the same. So we can write the Swan conductors as $a,b,b$ with $a,b$ two natural numbers such that $a \leq b$. In fact $a$ and $b$ must be odd numbers because the Swan conductor of every character of order $p$ in characteristic $p$ is prime to $p$.
Given a discrete valuation ring $R$ with fraction field $K$ and residue field $k$ and a map $\pi_1 ( \operatorname{Spec} R((x)) \to A_5)$, we obtain by pullback a map $\operatorname{Gal} \left( \overline{K}((x))\right) \to A_5$ and a map $\operatorname{Gal} \left( \overline{k}((x))\right) \to A_5$. Let $(a,b)$ be the invariants of the first representation and $(c,d)$ be the invariants of the second representation.
What quadruples $(a,b,c,d)$ are possible?
Here the best upper bound I know comes from Deligne's semicontinuity theorem. The Swan conductor of the standard representation is $a+2b$, so we have the inequality $c+2d \leq a+ 2b$. Also, by tensoring the standard representation with other representations and applying semicontinuity, we get the inequality $b \leq d$.
The best lower bound I know comes from Artin-Schreier extensions, which shows that all quadruples with $c \leq a$ and $d \leq b$ are possible.
I'm interested in knowing, if not the exact answer, at least the general shape of the space of solutions - is it closer to the "rectangle" $c\leq a, d\leq b$ or the "triangle" $c+2d \leq a+2b$?
The case where the difference is most extreme is something like $a=1$, $b=2n+1$, where one inequality is $c+2d \leq 4n+3$ and the other is $c \leq 1$, $d \leq 2n+1$. So the lower bound for the number of possibilities is linear in $n$, and the upper bound is quadratic. Can we give a better-than-linear lower bound or a better-than-quadratic upper bound on the number of possibilities?
