de_rham_cyclic/article_de_rham_cyclic.tex

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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}
%
\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 lower ramification groups and the fundamental characters of closed
	points $x$ of $X$ that are ramified in the cover $X \to X/G$.
\end{mainthm}

\section{Cyclic covers}
%
Let for any $\ZZ/p^n$-cover $X \to Y$
%
\begin{align*}
	u_{X/Y, P}^{(t)} &:= \min \{ t \ge 0 : G_P^{(t)} \cong \ZZ/p^{n-t???} \},\\
	l_{X/Y, P}^{(t)} &:= \min \{ t \ge 0 : G_{P, t} \cong \ZZ/p^{n-t???} \}.
\end{align*}
%
Note that if $G_P = \ZZ/p^n$, this coincides with the standard definition of
the $t$th upper (resp. lower) ramification jump of $X \to Y$ at $P$.
%
\begin{Theorem}
	Suppose that $\pi : X \to Y$ is a $\ZZ/p^n$-cover. Let $\langle G_P : P \in X(k) \rangle = \ZZ/p^m = G_{P_0}$ for $P_0 \in X(k)$. Then, as $k[\ZZ/p^n]$-modules:
	%
	\[
	H^1_{dR}(X) \cong J_{p^n}^{2 (g_Y - 1)} \oplus J_{p^n - p^{n-m} + 1}^2 \oplus \bigoplus_{P \neq P_0} J_{p^n - \frac{p^n}{e_{X/Y, P}}}^2
	\oplus \bigoplus_P \bigoplus_{t = 0}^{n-1} J_{p^n - p^t}^{u_{X/Y, P}^{(t+1)} - u_{X/Y, P}^{(t)}}.
	\]
\end{Theorem}
%
Write $H := \ZZ/p^n = \langle \sigma \rangle$.
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 $\dim_k T^i M$ determines the structure of $M$ completely (cf. ????).
In the inductive step we use also the group $\ZZ/p^{n-1}$. In this case
we denote the irreducible $k[\ZZ/p^{n-1}]$-modules by $\mc J_1, \ldots, \mc J_{p^{n-1}}$
and $\mc T^i M := T^i_{\ZZ/p^{n-1}} M$ for any $k[\ZZ/p^{n-1}]$-module $M$.

Note also that for $j \ge 1$:
%
\[
	l_{X/Y, P}^{(j)} - l_{X/Y, P}^{(j-1)} = \frac{1}{p^{j-1}} (u_{X/Y, P}^{(j)} - u^{(j-1)}_{X/Y, P})
\]
%
(in particular, $u_{X/Y, P}^{(1)} = l_{X/Y, P}^{(1)}$). Moreover, if $X' \to Y$ is the $\ZZ/p^N$-subcover of $X \to Y$ for $N \le n$ then:
%
\begin{itemize}
	\item $u_{X'/Y, P}^{(t)} = u_{X'/Y, P}^{(t)}$ for $t \le N$,
	
	\item $l_{X/X', P}^{(t)} = l_{X/X', P}^{(t + N)}$ for $t \le n-N$. 
\end{itemize}


\begin{Lemma}
	If the $G$-cover $X \to Y$ is \'{e}tale, then the natural map
	%
	\[
		H^1_{dR}(Y) \to H^1_{dR}(X)^G
	\]
	%
	is an isomorphism.
\end{Lemma}
\begin{proof}
	????
\end{proof}
%
\begin{Lemma}
	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}
	????
\end{proof}
%
\begin{Lemma}
	For any $i \le p^n - 1$:
	%
	\[
		(\sigma - 1) : T^{i+1} M \hookrightarrow T^i M.
	\]
\end{Lemma}
\begin{proof}
	
\end{proof}
%
\begin{proof}[Proof of Theorem ????]
	We use the following notation: $H' := \langle \sigma^p \rangle \cong \ZZ/p^{n-1}$,
	$H'' := H/\langle \sigma^{p^{n-1}} \rangle \cong \ZZ/p^{n-1}$, $Y' := X/H'$, $X'' := X/H''$.
	Write also $M := H^1_{dR}(X)$.
	By induction hypothesis for $H'$ acting on $X$, we have the following isomorphism of $k[H']$-modules:
	%
	\[
		M \cong \mc J_{p^{n-1}}^{2 (g_{Y'} - 1)} \oplus \mc J_{p^{n-1} - p^{n - 1 -m'} + 1}^2 \oplus \bigoplus_{P \neq P_0} \mc J_{p^n - \frac{p^{n-1}}{e_{X/Y', P}}}^2
		\oplus \bigoplus_P \bigoplus_{t = 0}^{n-1} \mc J_{p^n - p^t}^{u_{X/Y', P}^{(t+1)} - u_{X/Y', P}^{(t)}}
	\]
	%
	where
	%
	\[
	m' :=
		\begin{cases}
			n-1, & \textrm{ if } m = n,\\
			m, & \textrm{ otherwise.}
		\end{cases}
	\]
	
	Therefore, for $???$
	%
	\begin{align*}
		\dim_k \mc T^i M =
		\begin{cases}
			???, 
		\end{cases}
	\end{align*}
	%
	In particular, $\dim_k \mc T^1 M = \ldots = \dim_k \mc T^{p^{n-1} - p^{n-2}} M$.
	On the other hand, by Lemma ??:
	%
	\begin{align*}
		\dim_k \mc T^1 M &= \dim_k T^1 M + \ldots + \dim_k T^p M\\
		&\ge \dim_k T^{p^n - p^{n-1}} M + \ldots + \dim_k T^{p^n - p^{n-1}} M
		= \dim_k \mc T^{p^{n-1} - p^{n-2}} M.
	\end{align*}
	%
	Since the left-hand side and right hand side are equal, we conclude by Lemma ???
	that
	%
	\[
		\dim_k T^1 M = \ldots = \dim_k T^{p^n - p^{n-1}} M = \frac{1}{p} \dim_k \mc T^1 M.
	\]
	%
	If the cover $X \to X''$ is \'{e}tale, then the cover $X \to Y$ must be also \'{e}tale.
	Thus the proof follows in this case by~\cite{Nakajima??Inventiones}. Suppose now that 
	$X \to X''$ is not \'{e}tale. Then, by Lemma ???, the map $\tr_{X/X''} : H^1_{dR}(X) \to H^1_{dR}(X'')$ is surjective. Moreover, note that 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:
	%
	\[
		\ker(\tr_{X/X''} : M \to M'') = M^{(p^n - p^{n-1})}
	\]
	%
	and that $\tr_{X/X''}$ induces a $k$-linear isomorphism $T^{i + p^n - p^{n-1}} M \to \mc T^i M''$ for any $i \ge 1$. Thus:
	%
	\[
		\dim_k T^{i + p^n - p^{n-1}} M = \dim_k \mc T^i M'' = ....
	\]
	%
	This ends the proof.
\end{proof}

\section{Hypoelementary covers}
%
Assume now that $G = H \rtimes_{\chi} \ZZ/??n$.

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