# Is the moduli space of curves arising from wild ramification smooth?

Fix a natural number $g$, a prime $p$, and a $p$-group $P$.

Let $C$ be a smooth projective curve of genus $g$ with a faithful action of $P$ and an isomorphism $C / P \cong \mathbb P^1$ such that $P$ acts without stabilizers on points lying over $\mathbb A^1 \subset \mathbb P^1$. Then I will call $C$ a $(P,\infty)$-curve. In other words $(P,\infty)$-curves are finite Galois covers of $\mathbb P^1$ with Galois group $P$, etale away from $\infty$.

I think by general nonsense a moduli space for $(P,\infty)$-curves can be constructed as a Deligne-Mumford stack from the moduli space of smooth curves of genus $g$. There is some subtlety in that the condition that $P$ acts simply away from $\infty$ defines a closed subset and not a closed subscheme/substack, but one can give it a closed subscheme/substack structure by writing down a polynomial that vanishes on the fixed points (the discriminant of the cover) and setting it equal to a polynomial vanishing only at $\infty$. Because $(P,\infty)$-curves can only exist in characteristic $p$, it is most natural to view this as a space over $\mathbb F_p$.

Is this moduli space smooth?

My motivation is that $(P,\infty)$-curves over a perfect field $k$ are in bijection with surjective homomorphisms $\pi_1(\mathbb A^1_k) \to P$, and hence in bijection with surjective homomorphisms $\operatorname{Gal} ( k((x))) \to P$, because the maximal pro-$p$ quotients of those groups are isomorphic. I want to understand quotients of that group, so I want to understand this space.

• There is a Deligne-Mumford stack $\mathcal{N}$ of all finite, flat, generically etale morphisms $f:C\to \mathbb{P}^1$ such that $\Omega_f$ has length $b$. There is a branch morphism $\text{br}$ from $\mathcal{N}$ to the Hilbert scheme $\text{Hilb}^b_{\mathbb{P}^1_k/k} = \mathbb{P}^b_k$ that associates to $[f]$ the divisor (defined via "det-div" as in Knudsen-Mumford / Fogarty) of $f_*\Omega_f$. You ask about smoothness of an (open subset) of a fiber of $\text{br}$. This may hold. However, $\text{br}$ is not a smooth morphism. This is why the Oort conjecture is nontrivial. – Jason Starr Mar 5 '15 at 20:23
• @JasonStarr Cool, that makes sense. Do you know a reference where this stack and/or this morphism is discussed? – Will Sawin Mar 5 '15 at 21:20
• The branch morphism is discussed in a paper by Fantechi-Pandharipande. They probably only define it in characteristic 0 (to use for Gromov-Witten theory). However, the same construction works in all characteristics (or over Z). – Jason Starr Mar 5 '15 at 22:47