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\begin{document}

\title[The de Rham...]{The de Rham cohomology of covers\\ with cyclic $p$-Sylow subgroup}
\author[A. Kontogeorgis and J. Garnek]{Aristides Kontogeorgis and J\k{e}drzej Garnek}
\address{???}
\email{jgarnek@amu.edu.pl}
\subjclass[2020]{Primary 14G17, Secondary 14H30, 20C20} 
\keywords{de~Rham cohomology, algebraic curves, group actions,
	characteristic~$p$}
\urladdr{http://jgarnek.faculty.wmi.amu.edu.pl/}
\date{}  

\begin{abstract}
	????
\end{abstract}

\maketitle
\bibliographystyle{plain}
%
\section{Introduction}
%
The classical Chevalley--Weil formula 
(cf. \cite{Chevalley_Weil_Uber_verhalten}, 
{\color{red} 
\cite{Ellingsrud_Lonsted_Equivariant_Lefschetz})
}
 gives an explicit description
of the equivariant structure of the cohomology of a curve $X$ with a group action over a field of characteristic~$0$. Their formula depends on the so-called \emph{fundamental characters} of points $x \in X$ that are ramified in the cover $X \to X/G$. ????

It is hard to expect such a formula over fields of characteristic~$p$.
Indeed, if $G$ is a finite group with a non-cyclic $p$-Sylow subgroup, the set of indecomposable $k[G]$-modules is infinite. If, moreover, $p > 2$ then the indecomposable $k[G]$-modules are considered impossible to classify (cf. \cite{Prest}). There are many results concerning equivariant structure of cohomologies  for particular groups
(see e.g.~\cite{Valentini_Madan_Automorphisms} for the case of cyclic groups, \cite{WardMarques_HoloDiffs} for abelian groups, \cite{Bleher_Chinburg_Kontogeorgis_Galois_structure} for groups with a cyclic Sylow subgroup, or \cite{Bleher_Camacho_Holomorphic_differentials} for the Klein group) or curves (cf. \cite{Lusztig_Coxeter_orbits}, \cite{Dummigan_99}, \cite{Gross_Rigid_local_systems_Gm}, \cite{laurent_kock_drinfeld}). Also, one may expect that that (at least in the case of $p$-groups) determining cohomologies comes down to Harbater--Katz--Gabber covers (cf. \cite{Garnek_p_gp_covers}, \cite{Garnek_p_gp_covers_ii}). However, there is no hope of obtaining a result similar to the one of Chevalley and Weil.\\

This brings attention to groups with cyclic $p$-Sylow subgroup. For those, the set of 
{\color{red} equivalence classes of }
indecomposable modules is finite (cf. \cite{Higman}, \cite{Borevic_Faddeev}, \cite{Heller_Reiner_Reps_in_integers_I}). While their representation theory still
seems a bit too complicated to derive a general formula for the cohomologies, 
the article~\cite{Bleher_Chinburg_Kontogeorgis_Galois_structure} proved that 
the $k[G]$-module structure of $H^0(X, \Omega_X)$ is determined by the higher ramification data (i.e. higher ramification groups and the fundamental characters of the ramification locus). The main result of this article is a similar statement for the de Rham cohomology.
%
\begin{mainthm}
	Suppose that $G$ is a group with a $p$-cyclic Sylow subgroup.
	Let $X$ be a curve with an action of~$G$ over a field $k$ of characteristic $p$.
	The $k[G]$-module structure of $H^1_{dR}(X)$ is uniquely determined by the
	higher ramification data of the cover $X \to X/G$ and the genus of $X$.
\end{mainthm}
%
Note that if $p > 2$ and the $p$-Sylow subgroup of $G$ is not cyclic, the structure
of $H^1_{dR}(X)$ isn't determined uniquely by the higher ramification data. Indeed, see \cite{garnek_indecomposables} for a construction of $G$-covers with the same higher ramification data, but varying $k[G]$-module structure of $H^0(X, \Omega_X)$ and~$H^1_{dR}(X)$.\\

We elaborate now on the proof of Main Theorem. The first step is to prove Main Theorem for $G = \ZZ/p^n$. In this case we give an explicit formula for the structure of $H^1_{dR}(X)$, depending only on the ramification indices, ramification jumps and genus of the quotient
curve (see Theorem~\ref{thm:cyclic_de_rham}). This formula is proven inductively, by applying induction twice: once for the curve $X$ with an action of $\ZZ/p^{n-1}$ and once for the curve $X'' := X/(\ZZ/p)$.
In the second step, we use similar methods to show the result for a group of the form
$\ZZ/p^n \rtimes_{\chi} \ZZ/c$.  Finally, we use Conlon induction theorem to deduce Main Theorem
for an arbitrary group with a cyclic $p$-Sylow subgroup.
%
\section{Notation and preliminaries}
%
Assume that $\pi : X \to Y$ is a $G$-cover of smooth projective curves over an field $k$
of characteristic $p$.
Throughout the paper we will use the following notation for any $P \in X(\ol k)$:
\begin{itemize}
	\item $e_{X/Y, P}$ is the ramification index at $P$,
	
	\item $m_{X/Y, P} := \ord_p(e_{X/Y, P})$ is the maximal power of~$p$
	dividing the ramification index,
	
	\item $m_{X/Y} := \max \{ m_{X/Y, P} : P \in X(k) \}$,
	
	\item $u_{X/Y, P}^{(t)}$ (resp. $l_{X/Y, P}^{(t)}$) is the $t$th upper (resp. lower) ramification jump
	at $P$ for $t \ge 1$,
	
	\item $u^{(0)}_{X/Y, P} := 1$ for any ramified point $P \in X(\ol k)$
	(note that this is not a standard convention),
	
	\item $u_{X/Y, P} := u_{X/Y, P}^{(m_{X/Y, P})}$ is the last ramification jump.
\end{itemize}
%
By Hasse--Arf theorem (cf. ???), if the $p$-Sylow subgroup of $G$ is abelian, the numbers $u_{X/Y, P}^{(t)}$ are integers.
For any $Q \in Y(\ol k)$ we denote also by abuse of notation $G_Q := G_P$, $e_{X/Y, Q} := e_{X/Y, P}$,
$u_{X/Y, Q}^{(t)} := u_{X/Y, P}^{(t)}$ etc. for arbitrary $P \in \pi^{-1}(Q)$.
Let
%
\[
	B_{X/Y} := \{ Q \in Y(\ol k) : e_{X/Y, Q} > 1 \}
\]
%
be the branch locus of $\pi$. In the article we often use the Iverson bracket notation:
%
\[
\llbracket P \rrbracket = 
\begin{cases}
	1, & \textrm{ if $P$ is true,}\\
	0, & \textrm{ if $P$ is false.}
\end{cases}
\]
%
We review now some facts from representation theory of finite groups.
Recall that $\ZZ/p^n$ has $p^n$ indecomposable representations over a field of characteristic~$p$.
We denote them by $J_1, \ldots, J_{p^n}$. Observe that $J_i$ is given by the Jordan block of size $i$ and eigenvalue $1$. Assume now that $G$ is a finite group with a normal cyclic $p$-Sylow subgroup $H = \langle \sigma \rangle \cong \ZZ/p^n$. Let $C := G/H$.
Recall that if $U$ is an indecomposable $k[G]$-module
then $U^{\sigma} := \ker(\sigma - 1)$ (the socle of $U$) is an indecomposable
$k[C]$-module. It turns out that the map
%
\begin{align*}
	\Indec(k[G]) \to \Indec(k[C]) \times \{ 1, \ldots, p^n \}\\
	U \mapsto \left(U^{\sigma}, \frac{\dim_k U}{\dim_k U^{\sigma}} \right)
\end{align*}
%
is a bijection (cf. \cite[p. 35--37, 42 -- 43]{Alperin_local_rep}). We write
$\mc V(M, i)$ for the $k[G]$-module corresponding to a pair $(M, i) \in \Indec(k[C]) \times \{ 1, \ldots, p^n \}$.

Finally, we recall the classical Chevalley-Weil formula. Keep the above notation and assume that $p \nmid \# G$. For any $Q \in Y(k)$ let $\chi_Q : G_Q \to k^{\times}$ be the fundamental character of $G_Q$ acting on the tangent space of $Q$. Then:
%
\begin{equation}
	H^0(X, \Omega_X) \cong \bigoplus_{W \in \Indec(k[G])} M^{\oplus a_M},
\end{equation}
%
where:
%
\begin{align*}
	a_M := - \dim_k M + \sum_{Q \in Y(k)} \sum_{i = 0}^{e_{X/Y, Q} - 1} \left\langle \frac{-i}{e_{X/Y, Q}} \right\rangle \cdot N_{P, i}(M),
\end{align*}
%
and $N_{P, i}(M) := ???$.

\section{Cyclic covers}
%
\begin{Theorem} \label{thm:cyclic_de_rham}
	Let $k$ be an algebraically closed field of characteristic~$p$.
	Suppose that $\pi : X \to Y$ is a $\ZZ/p^n$-cover. Pick arbitrary $Q_0 \in Y(k)$
	with $m_{X/Y, Q_0} = m_{X/Y}$. Then, as a $k[\ZZ/p^n]$-module
	$H^1_{dR}(X)$ is isomorphic to:
	%
	\begin{equation} \label{eqn:HdR_formula}
		J_{p^n}^{2 (g_Y - 1)} \oplus J_{p^n - p^{n-m} + 1}^2 \oplus \bigoplus_{\substack{Q \in B\\ Q \neq Q_0}} J_{p^n - p^n/e_{Q}}^2
		\oplus \bigoplus_{Q \in B} \bigoplus_{t = 0}^{m_{Q}} J_{p^n - p^{n+t}/e_Q}^{u_Q^{(t+1)} - u_Q^{(t)}},
	\end{equation}
	%
	where $B := B_{X/Y}$, $e_Q := e_{X/Y, Q}$ and $u_Q^{(t)} := u_{X/Y, Q}^{(t)}$, $m := m_{X/Y, Q}$, $m_Q := m_{X/Y, Q}$.
\end{Theorem}
%
\begin{Remark}
	Note that for $g_Y = 0$ the first exponent is negative. However, since $m_{X/Y} = n$ (as $\PP^1$ doesn't have any \'{e}tale covers), the first two summands in~\eqref{eqn:HdR_formula} cancel out. Thus also in this case the module~\eqref{eqn:HdR_formula} is well-defined.
\end{Remark}

Write $H := \langle \sigma \rangle \cong \ZZ/p^n$.
For any $k[H]$-module $M$ denote:
%
\begin{align*}
	M^{(i)} &:= \ker ((\sigma - 1)^i : M \to M),\\
	T^i M &= T^i_H M := M^{(i)}/M^{(i-1)} \quad \textrm{ for } i = 1, \ldots, p^n.
\end{align*}
%
Recall that $\dim_k T^i M$, 
{\color{red} for $i=1, \ldots,p^n$}
determines the structure of $M$ completely (see \cite[p. 108]{Valentini_Madan_Automorphisms} -- they give the argument for $M := H^0(X, \Omega_X)$,
but it works for an arbitrary module).
Moreover, for $i > 0$:
%
\begin{equation} \label{eqn:dim_of_Ti_Jl}
	\dim_k T^i J_l = \llbracket i \le l \rrbracket.
\end{equation}
%
In the inductive step we use also the group $H' := \ZZ/p^{n-1}$. In this case
we denote the indecomposable $k[H']$-modules by $\mc J_1, \ldots, \mc J_{p^{n-1}}$
and $\mc T^i M := T^i_{H'} M$ for any $k[H']$-module $M$.
%
\begin{Lemma} \label{lem:G_invariants_\'{e}tale}
	If the $G$-cover $X \to Y$ is \'{e}tale, then
	%
	\[
		\dim_k H^1_{dR}(X)^G = 2g_Y - \dim_k H^1(G, k) + \dim_k H^2(G, k).
	\]
	%
	In particular, if $G \cong \ZZ/p^n$ then $\dim_k H^1_{dR}(X)^G = 2g_Y$.
\end{Lemma}
\begin{proof}
	Let $\HH^i(Y, \mc F^{\bullet})$ be the $i$th hypercohomology of a complex $\mc F^{\bullet}$.  
	Write also $\mc H^i(G, -)$ for the $i$th derived functor of the functor
	%
	\[
		\mc F \mapsto \mc F^G.
	\]
	%
	Since $X \to Y$ is \'{e}tale, $\mc H^i(G, \pi_* \mc F) = 0$ for any $i > 0$ and any coherent sheaf $\mc F$ on $X$ by \cite[Proposition~2.1]{Garnek_equivariant}.
	Therefore the spectral sequence~\cite[(3.4)]{Garnek_equivariant} applied for the complex $\mc F^{\bullet} := \pi_* \Omega_{X/k}^{\bullet}$ yields $\RR^i \Gamma^G(\pi_* \Omega_{X/k}^{\bullet}) = \HH^1(Y, \pi_*^G \Omega_{X/k}^{\bullet}) = H^1_{dR}(Y)$, since $\pi_*^G \Omega_X^{\bullet} \cong \Omega_Y$ (cf. ???).
	On the other hand, the seven-term exact sequence applied for the spectral sequence~\cite[(3.5)]{Garnek_equivariant} yields:
	%
	\begin{align*}
		0 \to H^1(G, H^0_{dR}(X)^G) \to H^1_{dR}(Y) \to H^1_{dR}(X)^G \to H^2(G, H^0_{dR}(X)^G) \to K,
	\end{align*}
	%
	where:
	%
	\[
		K := \ker(H^2_{dR}(Y) \to H^2_{dR}(X)^G) = \ker(k \stackrel{\id}{\rightarrow} k) = 0.
	\]
	%
	Therefore, since $H^0_{dR}(X)^G \cong k$:
	%
	\begin{align*}
		\dim_k H^1_{dR}(X)^G = \dim_k H^1_{dR}(Y) - \dim_k H^1(G, k) + \dim_k H^2(G, k)\\
		= 2g_Y - \dim_k H^1(G, k) + \dim_k H^2(G, k).
	\end{align*}
	%
	Finally, note that if $G$ is cyclic then $\dim_k H^1(G, k) = \dim_k H^2(G, k)$ by 
	{\color{red}
	\cite[th. 6.2.2]{Weibel}.
	}
\end{proof}
{\color{red}
\begin{Remark}
The equality $\dim_k H^1(G, k) = \dim_k H^2(G, k)$ does not hold for non-cyclic groups. For example it is known \cite[cor. II.4.3,th. II.4.4]{MR2035696} that the cohomological ring for the elementary abelian group $\mathbb{F}_p^s$ is given by 
\[
	H^* (G, \mathbb{F}_p)=
	\begin{cases}
		\mathbb{F}_2[x_1, \ldots,x_s] & \text{ if } p=2 \\
		\wedge(x_{1}, \ldots, x_s) \otimes \mathbb{F}_p[x_1, \ldots,x_s] & \text{ if } p>2
	\end{cases}
\]
Therefore, for $s>1$ the degree one and two parts of the cohomological ring, which correspond to the first and second cohomology groups, have different dimensions. 
\end{Remark}
}
%
\begin{Lemma} \label{lem:trace_surjective}
	Suppose that $G$ is a $p$-group.
	If the $G$-cover $X \to Y$ is totally ramified, then the map
	%
	\[
	\tr_{X/Y} : H^1_{dR}(X) \to H^1_{dR}(Y)
	\]
	%
	is an epimorphism.
\end{Lemma}
\begin{proof}
%
By induction, it suffices to prove this in the case when $G = \ZZ/p$.
Consider the following commutative diagram:
%
\begin{center}
	% https://tikzcd.yichuanshen.de/#N4Igdg9gJgpgziAXAbVABwnAlgFyxMJZABgBpiBdUkANwEMAbAVxiRGJAF9T1Nd9CKMgEYqtRizYduvbHgJFh5MfWatEIABIA9YgAoAGqQAEAHVMB5ALYwA5nQD6BgJRceIDHIGLSo6qskNHX0ATRNzaztHENcZDz55QWQAJmV-CXUtbWEHYCgAJU5DWPdPfgUUVL9xNTYdHLzCvRi3WXKkgGY0msCs4UNw0ysAY2MLJxK2xKIu6oDM+ubBkbGHFriy6ZQAFm75qVb4rwrkXbmMg84xGChbeCJQADMAJwgrJDIQHAgkZLiXt6-ajfJDbf6vd6IXZfH6IABs4MB8OBsIAHIjIQB2FFIACcGKQAFYcYhMQTEF0YUTyUoqRTyak6ZT9hpzDhnrkDAB6EKcQ4AyHQkGIYk9TJsjnAbm8-kQpBwknYsVsCWcnl8q6cIA
	\begin{tikzcd}
		0 \arrow[r] & {H^0(X, \Omega_X)} \arrow[r] \arrow[d, "\tr_{X/Y}"] & H^1_{dR}(X) \arrow[r] \arrow[d, "\tr_{X/Y}"] & {H^1(X, \mc O_X)} \arrow[r] \arrow[d, "\tr_{X/Y}"] & 0 \\
		0 \arrow[r] & {H^0(Y, \Omega_Y)} \arrow[r]                        & H^1_{dR}(Y) \arrow[r]                        & {H^1(Y, \mc O_Y)} \arrow[r]                        & 0
	\end{tikzcd}
\end{center}
%
where the rows are Hodge--de Rham exact sequences. Recall that by~\cite[Theorem~1]{Valentini_Madan_Automorphisms}, in this case $H^0(X, \Omega_X)$ contains
a copy of $k[G]^{\oplus g_Y}$ as a direct summand. Thus, since trace is injective on $k[G]^{\oplus g_Y}$, the dimension
of the image of
%
\begin{equation} \label{eqn:trace_H0_Omega}
	\tr_{X/Y} : H^0(X, \Omega_X) \to H^0(Y, \Omega_Y)
\end{equation}
%
is $g_Y$. Therefore the map~\eqref{eqn:trace_H0_Omega} is surjective. 
Similarly, by Serre's duality, also $H^1(X, \mc O_X)$ contains $k[G]^{\oplus g_Y}$ as a direct summand
and one shows similarly that the trace map
%
\begin{equation*} %\label{eqn:trace_H0_Omega}
	\tr_{X/Y} : H^1(X, \mc O_X) \to H^1(Y, \mc O_Y)
\end{equation*}
%
is surjective. Therefore, since the outer vertical maps in the diagram are surjective,
the trace map on the de Rham cohomology must be surjective as well.
%
\end{proof}
%
\begin{Lemma} \label{lem:TiM_isomorphism}
	For any $i \le p^n - 1$ we have the following $k$-linear monomorphism:
	%
	\[
		m_{\sigma - 1} : T^{i+1} M \hookrightarrow T^i M.
	\]
\end{Lemma}
\begin{proof}
%
We define $m_{\sigma - 1}$ as follows:
%
\[
	m_{\sigma - 1}(\ol x) := (\sigma - 1) \cdot x,
\]
%
where for $\ol x \in T^{i+1} M$ we picked any representative $x \in M^{(i+1)}$.
Indeed, if $x \in M^{(i+1)}$ then clearly $(\sigma - 1) \cdot x \in M^{(i)}$.
Moreover $(\sigma - 1) \cdot x \in M^{(i-1)}$ holds if and only if $x \in M^{(i)}$. This
shows that $m_{\sigma - 1}$ is well-defined and injective.
\end{proof}
%
\begin{Lemma} \label{lem:lemma_mcT_and_T}
	Let $M$ be a $k[H]$-module. Let $T^i M$ and $\mc T^i M$ be as above.
	If $\dim_k \mc T^i M = \dim_k \mc T^{i+1} M$ for some $i$ then:
	%
	\[
		\dim_k T^{pi + p} M = \dim_k T^{pi + p - 1} M = \ldots = \dim_k T^{pi - p + 1} M.
	\]
\end{Lemma}
\begin{proof}
	Note that $\mc T^i M = M^{(pi)}/M^{(pi - p)}$. This easily implies that:
	%
	\begin{align*}
		\dim_k \mc T^i M &= \dim_k T^{pi} M + \ldots + \dim_k T^{pi - p + 1} M\\
		&\ge \dim_k T^{pi+p} M + \ldots + \dim_k T^{pi+1} M
		= \dim_k \mc T^{i+1} M.
	\end{align*}
	%
	Since the left-hand side and right hand side are equal, we conclude by Lemma~\ref{lem:TiM_isomorphism}
\end{proof}
%
\begin{Lemma} \label{lem:u_equals_ul}
	Assume that $Y' \to Y$ is a $\ZZ/p$-subcover of $X \to Y$.
	Then:
	%
	\[
	p \cdot \sum_{Q \in B_{X/Y}} (u_{X/Y, Q} - 1) = \sum_{Q' \in B_{X/Y'}} (u_{X/Y', Q'} - 1)
	+ (p-1) \cdot \sum_{Q \in B_{Y'/Y}} (l^{(1)}_{Y'/Y, Q} + 1).
	\]
	%
\end{Lemma}
\begin{proof}
	%
	Pick a point $Q \in B_{X/Y}$. If $Q \not \in B_{Y'/Y}$ then 
	$u_{X/Y, Q} = u_{X/Y', Q'}$ for all $p$ points $Q' \in Y'(k)$ in the preimage of $Q$ and:
	%
	\begin{equation} \label{eqn:Q_not_in_B'}
		p \cdot (u_{X/Y, Q} - 1) = \sum_{Q'} (u_{X/Y', Q'} - 1).
	\end{equation}
	%
	Assume now that $Q \in B_{Y'/Y}$. Then there exists a unique point $Q' \in Y'(k)$
	in the preimage of $Q$ through $Y' \to Y$. Moreover, $m_{X/Y, Q} = n$, $m_{X/Y', Q'} = n-1$. 
	Recall also that by 
	{\color{red}
	\cite[Example p.76]{Serre1979}
	 }
	there exist integers $i_{X/Y, P}^{(0)}, i_{X/Y, P}^{(1)}, \ldots$ such that for every $t \ge 0$:
	%
	\begin{align*}
		u_{X/Y, P}^{(t)} &= i_{X/Y, P}^{(0)} + i_{X/Y, P}^{(1)} + \cdots + i_{X/Y, P}^{(t-1)}\\
		l_{X/Y, P}^{(t)} &= i_{X/Y, P}^{(0)} + i_{X/Y, P}^{(1)} \cdot p + \cdots + i_{X/Y, P}^{(t-1)} \cdot p^{t-1}.
	\end{align*}
	%
	Observe that:
	%
	\begin{align*}
		i_{X/X', P}^{(0)} &= i_{X/Y, P}^{(0)} + i_{X/Y, P}^{(1)} \cdot p,\\
		i_{X/X', P}^{(t)} &= p \cdot (i_{X/Y, P}^{(t + 1)} + \cdots + i_{X/Y, P}^{(n-1)}) \quad \textrm{ for } t > 0.
	\end{align*}
	%
	This implies that
	%
	\begin{equation} \label{eqn:Q_in_B'}
		p \cdot (u_{X/Y, Q} - 1) = (u_{X/Y', Q'} - 1) + (p-1) \cdot (l^{(1)}_{X/Y, Q} + 1).
	\end{equation}
	%
	Indeed, using the above formulas:
	%
	\begin{align*}
		p \cdot (u_{X/Y, Q} - 1) &=
		p \cdot (i^{(0)}_{X/Y, Q} + \cdots + i^{(m_Q - 1)}_{X/Y, Q} - 1)\\
		&= (p-1) \cdot (i^{(0)}_{X/Y, Q} + 1) + (i^{(0)}_{X/Y, Q} + p \cdot i^{(1)}) \\
		&+ p \cdot (i^{(2)}_{X/Y, Q} + i^{(3)}_{X/Y, Q} + \cdots) - 1 \\
		&= (p-1) \cdot (l^{(1)}_{X/Y, Q} + 1) + (i^{(0)}_{X/Y', Q'} + i^{(1)}_{X/Y', Q'} + \cdots - 1)\\
		&= (p-1) \cdot (l^{(1)}_{X/Y, Q} + 1) + (u_{X/Y', Q'} - 1).
	\end{align*}
	%
	The proof follows by summing~\eqref{eqn:Q_not_in_B'} and~\eqref{eqn:Q_in_B'} over all $Q \in B_{X/Y}$.
\end{proof}

\begin{proof}[Proof of Theorem~\ref{thm:cyclic_de_rham}]
	We use the following notation: $H' := \langle \sigma^p \rangle \cong \ZZ/p^{n-1}$,
	$H'' := H/\langle \sigma^{p^{n-1}} \rangle \cong \ZZ/p^{n-1}$, $Y' := X/H'$, $X'' := X/\langle \sigma^{p^{n-1}} \rangle$. Note that $H''$ naturally acts on $X''$.
	Let also $\mc M := H^1_{dR}(X)$ and write $\mc M_0$ for the module~\eqref{eqn:HdR_formula}.
	We consider now two cases. If the cover $X \to Y$ is \'{e}tale, then by induction assumption, since $2(g_{Y'} - 1) = p \cdot 2 \cdot (g_Y - 1)$:
	%
	\[
	\mc M \cong \mc J_{p^{n-1}}^{2 p \cdot (g_Y - 1)} \oplus k^{\oplus 2}.
	\]
	%
	Therefore $\dim_k \mc T^2 \mc M = \ldots = \dim_k \mc T^{p^{n-1}} \mc M = 2 p (g_Y - 1)$,
	which by Lemma~\ref{lem:lemma_mcT_and_T} implies that
	%
	\[
		\dim_k T^p \mc M = \ldots = \dim_k T^{p^n} \mc M = 2(g_Y - 1) = \dim_k T^p \mc M_0.
	\]
	%
	Thus, for $i = 2, \ldots, p$:
	%
	\[
		\dim_k T^i \mc M \ge 2(g_Y - 1) = \dim_k T^{p+1} \mc M.
	\]
	%
	On the other hand, by Lemma~\ref{lem:G_invariants_\'{e}tale} we have
	%
	$
	\dim_k T^1 \mc M = 2 g_Y = \dim_k T^1 \mc M_0
	$. Thus:
	%
	\begin{align*}
		\sum_{i = 2}^p \dim_k T^i \mc M = 2g_X - \dim_k T^1 \mc M - \sum_{i = p+1}^{p^n} \dim_k T^i \mc M = (p-1) \cdot 2(g_Y - 1).
	\end{align*}
	%
	Thus $\dim_k T^i \mc M = 2(g_Y - 1) = \dim_k T^i \mc M_0$ for every $i \ge 2$, which ends the proof in this case.
	
	Assume now that $X \to Y$ is not \'{e}tale. Therefore $X \to X''$ is also not \'{e}tale.
	By induction hypothesis for $H'$ acting on $X$, we have the following isomorphism of $k[H']$-modules:
	%
	\[
		\mc M \cong \mc J_{p^{n-1}}^{2 (g_{Y'} - 1)} \oplus \mc J_{p^{n-1} - p^{n - m} + 1}^2 \oplus \bigoplus_{\substack{Q \in Y'(k)\\Q \neq Q_1}} \mc J_{p^{n-1} - p^{n-1}/e'_Q}^2
		\oplus \bigoplus_{Q \in Y'(k)} \bigoplus_{t = 0}^{n-2} \mc J_{p^n - p^t}^{u_{X/Y', Q}^{(t+1)} - u_{X/Y', Q}^{(t)}}
	\]
	%
	where $e'_Q := e_{X/Y', Q}$ and $Q_1 \in \pi^{-1}(Q_0)$.  Therefore, for $i \le p^{n-1} - p^{n-2}$, using the Riemann--Hurwitz formula (cf. \cite[Corollary~IV.2.4]{Hartshorne1977}) and Lemma~\ref{lem:u_equals_ul}:
	%
	\begin{align*}
		\dim_k \mc T^i \mc M &=
		2(g_{Y'} - 1) + 2 + 2(\# B - 1) + \sum_{Q' \in Y'(k)} (u_{X/Y', Q'} - 1)\\
		&= 2 p (g_Y - 1) + \sum_{Q' \in Y'(k)} (p-1) \cdot (l_{Y'/Y, Q'}^{(1)} + 1)\\
		 &+ 2 + 2(\# B - 1) + \sum_{Q' \in Y'(k)} (u_{X/Y', Q'} - 1)\\
		&= p \cdot \left( 2(g_Y - 1) + 2 + 2(\# B - 1) + \sum_{Q' \in Y(k)} (u_{X/Y, Q'} - 1) \right).
	\end{align*}
	%
	In particular, $\dim_k \mc T^1 \mc M = \ldots = \dim_k \mc T^{p^{n-1} - p^{n-2}} \mc M$.
	Thus by Lemma~\ref{lem:lemma_mcT_and_T} for any $1 \le i \le p^n - p^{n-1}$:
	%
	\begin{align*}
		\dim_k T^i \mc M &= \frac{1}{p} \dim_k \mc T^1 \mc M\\
		&= 2(g_Y - 1) + 2 + 2(\# B - 1) + \sum_{Q \in Y(k)} (u_{X/Y, P} - 1)\\
		&= \dim_k T^i \mc M_0.
	\end{align*}
	%
	By Lemma~\ref{lem:trace_surjective} since $X \to X''$ is not \'{e}tale, the map $\tr_{X/X''} : H^1_{dR}(X) \to H^1_{dR}(X'')$ is surjective. Recall that
	in $\FF_p[x]$ we have the identity:
	%
	\[
		1 + x + \ldots + x^{p-1} = (x - 1)^{p-1}.
	\]
	%
	Therefore in the group ring $k[H]$ we have:
	%
	\[
		\tr_{X/X''} = \sum_{j = 0}^{p-1} (\sigma^{p^{n-1}})^j = (\sigma^{p^{n-1}} - 1)^{p-1} = 
		(\sigma - 1)^{p^n - p^{n-1}}.
	\]
	%
	This implies that:
	%
	\[
		\ker(\tr_{X/X''} : \mc M \to \mc M'') = \mc M^{(p^n - p^{n-1})}
	\]
	%
	and that $\tr_{X/X''}$ induces a $k$-linear isomorphism $T^{j + p^n - p^{n-1}} \mc M \to \mc T^j \mc M''$ for any $j \ge 1$. ?? Therefore, if $i \in (p^n - p^{N+1}, p^n - p^N]$:
	%
	\begin{align*}
		\dim_k T^i \mc M_0 &= 2 \cdot (g_Y - 1) + 2 \cdot \llbracket N < n - m \rrbracket\\
		&+ 2 \cdot \# \{ Q \in Y(k) \setminus \{Q_0\} : N \le n - m_Q \}\\
		&+ \sum_{Q \in Y(k)} \sum_{t = 0}^{m_Q - 1} \llbracket t \ge m_Q + N - n \rrbracket \cdot (u_{Q}^{(t+1)} - u_{Q}^{(t)}).
	\end{align*}
	%
	Suppose now that
	$i = p^n - p^{n-1} + j$, where $j \in (p^{n-1} - p^{N+1}, p^{n-1} - p^N]$. Then, by induction assumption:
	%
	\begin{align*}
		\dim_k T^i \mc M &= \dim_k \mc T^j \mc M'' = 2 \cdot (g_Y - 1) + 2 \cdot \llbracket N < (n - 1) - (m - 1) \rrbracket\\
		&+ 2 \cdot \# \{ Q \in Y(k) \setminus \{Q_0\} : N \le (n-1) - (m_{X''/Y, Q}) \}\\
		 &+ \sum_{Q \in Y(k)} \sum_{t = 0}^{m_{X''/Y, Q}} \llbracket t \ge m_{X''/Y, Q} + N - (n - 1) \rrbracket \cdot (u_{X''/Y, Q}^{(t+1)} - u_{X''/Y, Q}^{(t)})\\
		 &= 2 \cdot (g_Y - 1) + 2 \cdot \llbracket N < n - m \rrbracket\\
		 &+ 2 \cdot \# \{ Q \in Y(k) \setminus \{Q_0\} : N \le n - m_Q \}\\
		 &+ \sum_{Q \in Y(k)} \sum_{t = 0}^{m_Q - 1} \llbracket t \ge m_Q + N - n \rrbracket \cdot (u_{Q}^{(t+1)} - u_{Q}^{(t)})\\
		 &= \dim_k T^i \mc M_0.
	\end{align*}
	%
	This ends the proof.
\end{proof}

\section{Proof of Main Theorem}
%
\begin{Lemma} \label{lem:reductions}
	Suppose $M$ is a finitely generated $k[G]$-module. 	
	\begin{enumerate}[leftmargin=*]
		\item The $k[G]$-module structure of $M$
		is uniquely determined by the restrictions $M|_H$ as $H$ ranges over all $p$-hypo-elementary subgroups of $G$.
		
		\item The $k[G]$-module structure of $M$ is uniquely determined by the $\ol k[G]$-module structure of $M \otimes_k \ol k$.
	\end{enumerate}	
\end{Lemma}
\begin{proof}
	\begin{enumerate}[leftmargin=*]
		\item This follows easily from Conlon induction theorem (cf. \cite[Theorem~(80.51)]{Curtis_Reiner_Methods_II}), see e.g. \cite[Lemma~3.2]{Bleher_Chinburg_Kontogeorgis_Galois_structure}.
		
		\item This is \cite[Proposition~3.5. (iii)]{Bleher_Chinburg_Kontogeorgis_Galois_structure}
	\end{enumerate}
\end{proof}
%
By Lemma~\ref{lem:reductions} we may assume that $G = H \rtimes_{\chi} C = \langle \sigma \rangle \rtimes_{\chi} \langle \rho \rangle \cong \ZZ/p^n \rtimes_{\chi} \ZZ/c$ and that $k$ is algebraically closed.
Let $X$ be a curve with an action of $G$ and write $Y := X/H$. For any $k[C]$-module $M$ and any character $\psi$ of $H$ we write $M^{\psi} := M \otimes_{k[C]} \psi$.
%
\begin{Lemma} 
	Let $k$ and $G$ be as above. Assume that $M$ is a $k[G]$-module of finite dimension. The $k[G]$-structure of $M$
	is uniquely determined by the $k[C]$-structure of $T^1 M, \ldots, T^{p^n} M$.
\end{Lemma}
\begin{proof}
	This is basically \cite[proof of Theorem~1.1]{Bleher_Chinburg_Kontogeorgis_Galois_structure}. We sketch the proof for reader's convenience. Let $\psi : C \to k^{\times}$ be a primitive character. Write
	%
	\[
		M \cong \bigoplus_{a, b} \mc V(\psi^a, b)^{\oplus n(a, b)}.
	\]
\end{proof}
%
\begin{Lemma} \label{lem:N+Nchi+...}
	Keep the above notation. Let $M$, $N$ be $k[C]$-modules. 
	If
	%
	\[
		N \cong M \oplus M^{\chi} \oplus \ldots \oplus M^{\chi^{p-1}},
	\]
	then $N$ is uniquely determined by $M$.
\end{Lemma}
\begin{proof}
	Note that
	%
	\[
	N \cong M^{\oplus 2} \oplus M^{\chi} \oplus M^{\chi^2} \oplus \ldots \oplus M^{\chi^{p-2}}.
	\]
	%
	By tensoring this isomorphism by $\chi^i$ we obtain:
	%
	\begin{align*}
		N^{\chi^i} \cong (M^{\chi^i})^{\oplus 2} \oplus M^{\chi^{i+1}} \oplus M^{\chi^{i+2}} \oplus \ldots \oplus M^{\chi^{i + p-2}}
		\cong (M^{\chi^i})^{\oplus 2} \oplus \bigoplus_{\substack{j = 0\\j \neq i}}^{p-2} M^{\chi^j}
	\end{align*}
	%
	for $i = 0, \ldots, p-2$. Therefore:
	%
	\begin{equation} \label{eqn:M+N=N}
		M^{\oplus p} \oplus N^{\chi} \oplus N^{\chi^2} \oplus \ldots \oplus N^{\chi^{p-2}} 
		\cong N^{\oplus (p-1)}.
	\end{equation}
	%
	Indeed, for the proof of~\eqref{eqn:M+N=N} note that
	%
	\begin{align*}
		M^{\oplus p} &\oplus N^{\chi} \oplus N^{\chi^2} \oplus \ldots \oplus N^{\chi^{p-2}}
		\cong M^{\oplus p} \oplus \bigoplus_{i = 1}^{p-2} \left((M^{\chi^i})^{\oplus 2}
		\oplus \bigoplus_{\substack{j = 0\\j \neq i}}^{p-2} M^{\chi^j} \right)\\
		&\cong \left( M^{\oplus 2} \oplus M^{\chi} \oplus M^{\chi^2} \oplus \ldots \oplus M^{\chi^{p-2}} \right)^{\oplus (p-1)}
		\cong N^{\oplus (p-1)}.
	\end{align*}
	%
	The isomorphism~\eqref{eqn:M+N=N} clearly proves the thesis.
\end{proof}
%
\begin{Lemma} \label{lem:TiM_isomorphism_hypoelementary}
	For any $i \le p^n - 1$ the map~$m_{\sigma - 1}$ from Lemma~\ref{lem:TiM_isomorphism}
	yields a $k[C]$-equivariant monomorphism:
	%
	\[
	m_{\sigma - 1} : T^{i+1} M \hookrightarrow (T^i M)^{\chi^{-1}}.
	\]
\end{Lemma}
\begin{proof}
	By Lemma~\ref{lem:TiM_isomorphism} this map is injective. Thus it suffices to check that it is $k[C]$-equivariant.
	Note that we have the following identity in the ring~$k[C]$:
	%
	\[
		(\sigma - 1) \cdot \rho = \rho \cdot (\sigma^{\chi(\rho)^{-1}} - 1)
		= \rho \cdot (\sigma - 1) \cdot (1 + \sigma + \sigma^2 + \ldots + \sigma^{\chi(\rho)^{-1} - 1})
	\]
	%
	Note that $\sigma$ acts trivially on $T^i M$, so that for any $\ol x \in T^i M$:
	%
	\[
		(1 + \sigma + \sigma^2 + \ldots + \sigma^{\chi(\rho)^{-1} - 1}) \cdot \ol x = \chi(\rho)^{-1} \cdot \ol x.
	\]
	%
	This easily shows that
	%
	\[
		m_{\sigma - 1}(\rho \cdot \ol x) = \chi(\rho)^{-1} \cdot \rho \cdot m_{\sigma - 1}(\ol x),
	\]
	%
	which ends the proof.
	%
\end{proof}

\begin{proof}[Proof of Main Theorem]
	As explained at the beginning of this section, it suffices to show this in the case when $G = H \rtimes_{\chi} C = \langle \sigma \rangle \rtimes_{\chi} \langle \rho \rangle \cong \ZZ/p^n \rtimes_{\chi} \ZZ/c$ and $k = \ol k$ by Lemma~\ref{lem:reductions}.
	We prove this by induction on~$n$. If $n = 0$, then it follows by Chevalley--Weil theorem.
	Consider now two cases. Firstly, we assume that $X \to Y$ is \'{e}tale. Then by Lemma~\ref{lem:G_invariants_\'{e}tale} and \cite[Corollary~2.4]{Garnek_equivariant} we have $\dim_k H^1_{dR}(X)^H = 2g_Y = \dim_k H^0(X, \Omega_X)^H + \dim_k H^1(X, \mc O_X)^H$. Therefore the Hodge--de Rham exact sequence splits by \cite[Lemma~5.3]{Garnek_equivariant} and
	%
	\begin{align*}
		H^1_{dR}(X) &\cong H^0(X, \Omega_X) \oplus H^1(X, \mc O_X)\\
		&\cong H^0(X, \Omega_X) \oplus H^1(X, \mc O_X)^{\vee}
	\end{align*}
	%
	(the last isomorphism follows from Serre's duality, cf. ???).
	Now it suffices to note that by~\cite[Theorem~1.1]{Bleher_Chinburg_Kontogeorgis_Galois_structure}
	the $k[G]$-module structure of $H^0(X, \Omega_X)$ is determined by the higher ramification data. This ends the proof in this case.\\
	
	Assume now that $X \to Y$ is not \'{e}tale. Lemma~\ref{lem:TiM_isomorphism_hypoelementary} and proof of Theorem~\ref{thm:cyclic_de_rham}
	yield an isomorphism of $k[C]$-modules:
	%
	\begin{equation} \label{eqn:TiM=T1M_chi}
		T^{i+1} \mc M \cong (T^1 \mc M)^{\chi^{-i}}
	\end{equation}
	%
	for $i \le p^n - p^{n-1}$. Observe that $\mc T^i M$ has the filtration $\mc M^{(pi)} \supset \mc M^{(pi - 1)} \supset \ldots \supset \mc M^{(pi - p)}$ with subquotients $T^{pi} \mc M, \ldots, T^{pi - p + 1} \mc M$.
	Thus, since the category of $k[C]$-modules is semisimple, for $i \le p^n - p^{n-1}$:
	%
	\begin{align*}
		\mc T^i \mc M &\cong T^{pi - p + 1} \mc M \oplus \ldots \oplus T^{pi} \mc M\\
		&\cong T^1 \mc M \oplus (T^1 \mc M)^{\chi^{-1}} \oplus \ldots \oplus
		(T^1 \mc M)^{\chi^{-p}}.
	\end{align*}
	%
	By induction assumption, the $k[C]$-module structure of $\mc T^i \mc M$ is uniquely determined by the higher ramification data. Thus, by Lemma~\ref{lem:N+Nchi+...} for $N := T^1 \mc M$ and by~\eqref{eqn:TiM=T1M_chi} the $k[C]$-structure of the modules $T^i \mc M$ is uniquely determined by the higher ramification data for $i \le p^n - p^{n-1}$.
	By a similar reasoning, $\tr_{X/X'}$ yields an isomorphism:
	%
	\[
		T^{i + p^n - p^{n-1}} \mc M \cong (\mc T^i \mc M'')^{\chi^{-1??}}.
	\]
	%
	Thus, by induction hypothesis for $\mc M''$, the $k[C]$-structure of $T^{i + p^n - p^{n-1}} \mc M$
	is determined by higher ramification data as well.
\end{proof}
%
\section{Examples}
%
\noindent Let $p > 2$. Consider the Mumford curve
%
\[
	X : (x^p - x) \cdot (y^p - y) = 1.
\]
%
It is a curve of genus $(p-1)^2$ and an action of the group $(\ZZ/p \times \ZZ/p) \rtimes D_{2(p-1)}$ given by:
%
\begin{align*}
	\sigma_0(x, y) &= (x+1, y),\\
	\sigma_1(x, y) &= (x, y+1),\\
	s(x, y) &= (y, x),\\
	\theta(x, y) &= (\zeta \cdot x, \zeta^{-1} \cdot y) \quad \textrm{ for } \FF_p^{\times} = \langle \zeta \rangle.
\end{align*}
%
Recall representation theory of $D_{2(p - 1)}$ (cf. \cite[Example~8.2.3]{Steinberg_Representation_book}).
For $1 \le j \le p-2$ let $\chi_j$ be the character of the representation of $D_{2(p-1)}$
induced from
%
\[
	\ZZ/(p-1) = \langle \theta \rangle \to \FF_p^{\times}, \quad \theta \mapsto \zeta^j.
\]
%
One easily checks that $\chi_j$ is given by the matrices:
%
\begin{align*}
	\theta \mapsto
	\left(
	\begin{matrix}
		\zeta^j & 0\\
		0 & \zeta^{-j}
	\end{matrix}
	\right),
	\qquad 
	s \mapsto
	\left(
	\begin{matrix}
		0 & 1\\
		1 & 0
	\end{matrix}
	\right).
\end{align*}
%
Moreover, $\chi_j$ is irreducible and isomorphic to $\chi_{p - 1 - j}$.
Let also $\chi_0$ be the representation:
%
\[
	D_{2(p-1)} \to \FF_p^{\times}, \qquad \theta \mapsto 1, \qquad
	s \mapsto -1.
\]
%
We claim that as $k[C]$-modules: ??k or $\FF_p$??
%
\begin{equation}
	H^1_{dR}(X) \cong V_0^{\oplus (p-1)} \oplus \bigoplus_{j = 1}^{\frac{p-1}{2}} V_j^{\oplus 2(p-1)}.
\end{equation}



{\color{gray}
Basis of holomorphic differentials:
%
\[
	\omega_{a, b} = \frac{x^a \cdot y^b \, dx}{(x^p - x)} \qquad 0 \le a, b \le p-2.
\]		
}


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% 
% \bibliography{AKGeneral}
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