X ---> X/C

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jgarnek 2024-12-18 12:52:42 +01:00
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\def\cprime{$'$} \def\cprime{$'$}
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@ -834,25 +834,24 @@ Let $X$ be a curve with an action of $G$ and write $Y := X/H$. For any $k[C]$-mo
is determined by higher ramification data as well. is determined by higher ramification data as well.
\end{proof} \end{proof}
% %
\section{Examples} The method of proof of Main Theorem allows to obtain explicit formulas in the style of the result of Chevalley--Weil for
% particular group. Assume that $G$ is a group with a normal $p$-Sylow subgroup $H$ of order~$p$. Let $C := G/H$. Then $G = H \rtimes_{\chi} C$
Assume that $G$ is a group with a normal $p$-Sylow subgroup $H$ of order~$p$. Let $C := G/H$. Then $G = H \rtimes_{\chi} C$
for a homomorphism $\chi : C \to \FF_p^{\times}$. for a homomorphism $\chi : C \to \FF_p^{\times}$.
% %
\begin{Proposition} \begin{Proposition}
Keep the above notation. {\color{red} Assume that $k$ is algebraically closed.} If $G$ acts on a curve $X$ and the cover $X \to X/H$ is not \'{e}tale, then: Keep the above notation. {\color{red} Assume that $k$ is algebraically closed.} If $G$ acts on a curve $X$ and the cover $X \to X/H$ is not \'{e}tale, then:
% %
\[ \[
H^1_{dR}(X) \cong \bigoplus_{W \in \Indec(C)} \mc V(W, p)^{\oplus a^{dR}_{Y, C}(W)} \oplus \mc V(W, p-1)^{\oplus b_W}, H^1_{dR}(X) \cong \bigoplus_{W \in \Indec(C)} \mc V(W, p)^{\oplus a^{dR}_{Y, C}(W)} \oplus \mc V(W, p-1)^{\oplus b_W},
\] \]
% %
where for any $W \in \Indec(k[C])$ the number $a_W$ is as in the equality~\eqref{eqn:cw} for the action of $C$ on $X$, where for any $W \in \Indec(k[C])$ the number $a_W$ is as in the equality~\eqref{eqn:cw} for the action of $C$ on $X$,
$a_W'$ is as in the equality~\eqref{eqn:cw} for the action of $C$ on $Y := X/H$ and $a_W'$ is as in the equality~\eqref{eqn:cw} for the action of $C$ on $Y := X/H$ and
% %
\begin{align*} \begin{align*}
b_W &:= a^{dR}_{X, C}(W) - \frac 1p \sum_{i = 0}^{p-2} a^{dR}_{X, C}(W \otimes \chi^i) - a^{dR}_{Y, C}(W \otimes \chi). b_W &:= a^{dR}_{X, C}(W) - \frac 1p \sum_{i = 0}^{p-2} a^{dR}_{X, C}(W \otimes \chi^i) - a^{dR}_{Y, C}(W \otimes \chi).
\end{align*} \end{align*}
% %
\end{Proposition} \end{Proposition}
\begin{proof} \begin{proof}
Theorem~\ref{thm:cyclic_de_rham} easily implies that Theorem~\ref{thm:cyclic_de_rham} easily implies that
@ -864,9 +863,14 @@ $a_W'$ is as in the equality~\eqref{eqn:cw} for the action of $C$ on $Y := X/H$
for some $A_W, B_W \in \ZZ$. ?? for some $A_W, B_W \in \ZZ$. ??
\end{proof} \end{proof}
% %
Let $p > 2$ be a prime and $p \nmid m$ an natural number. Fix a primitive root of unity $\zeta \in \ol{\FF}_p^{\times}$ of order $m \cdot (p-1)$.
\section{An example -- a superelliptic curve with a metacyclic action}
%
Let $p > 2$ be a prime and $p \nmid m$ an natural number. Let $k$ be an algebraically closed field of characteristic~ $p$.
Fix a primitive root of unity $\zeta \in \ol{\FF}_p^{\times}$ of order $m \cdot (p-1)$.
Note that $\zeta^m \in \FF_p$. Note that $\zeta^m \in \FF_p$.
We compute now the equivariant structure of the de Rham cohomology for the superelliptic curve $X$ with the affine part given by: In this section we compute the equivariant structure of the de Rham cohomology for the superelliptic curve $X$ with the affine part given by:
% %
\begin{equation*} \begin{equation*}
y^m = x^{p^n} - x. y^m = x^{p^n} - x.
@ -887,24 +891,80 @@ This action is given by:
\rho(x, y) &= (\zeta^m \cdot x, \zeta \cdot y). \rho(x, y) &= (\zeta^m \cdot x, \zeta \cdot y).
\end{align*} \end{align*}
% %
\begin{Proposition}
\[
H^1_{dR}(X) \cong ????.
\]
\end{Proposition}
%
Note that $X/G \cong \PP^1$ and the quotient map is given by $(x, y) \mapsto (x^p - x)^{p-1}$. Indeed, ????. Note that $X/G \cong \PP^1$ and the quotient map is given by $(x, y) \mapsto (x^p - x)^{p-1}$. Indeed, ????.
We claim that the set of branch points is given by $B := \{ 0, \infty \} \cup B'$, where We claim that the set of branch points is given by $B := \{ Q_{\infty}, Q_0, Q_1, \ldots, Q_N \}$, where
$N := \frac{p^{n-1} - 1}{p - 1}$, $Q_0 = 0$, $Q_{\infty} = \infty$ and $Q_1, \ldots, Q_N$ are
the elements of the set
% %
\[ \[
B' := \{ (\alpha^p - \alpha)^{p-1} : \alpha \in \FF_{p^n} \setminus \FF_p \}. \{ (\alpha^p - \alpha)^{p-1} : \alpha \in \FF_{p^n} \setminus \FF_p \}.
\] \]
% %
The set $B'$ has $\frac{p^{n-1} - 1}{p - 1}$ elements. We claim that: Write $C' := \langle \rho^{p-1} \rangle \cong \ZZ/m$ and note that $C'$ is in the center of $G$. We claim that:
% %
\begin{itemize} \begin{itemize}
\item $G_{Q_0} = C$, \item $G_{Q_0}$ is the conjugacy class of the subgroup $C$,
\item $G_Q = \langle \rho^{p-1} \rangle \cong \ZZ/m$ for $Q \in B'$, \item $G_{Q_{\infty}} = G$ and the lower ramification jump at $Q_{\infty}$ equals $m$,
\item $G_{Q_{\infty}} = G$ and the lower ramification jump at $Q_{\infty}$ equals $m$.
\item $G_{Q_i} = C'$ for $i = 1, \ldots, N$.
\end{itemize} \end{itemize}
% %
Indeed, ????. Indeed, ????. The ramification points of $\pi : X \to X/G$ are as follows:
%
\begin{itemize}
\item points $P_0^{(1)}, \ldots, P_0^{(p)}$ above $Q_0$
\item[] (their stabilizers are subgroups $C_1 = C$, $\ldots$, $C_p$
conjugated to $C$),
\item point $P_{\infty}$ above $Q_{\infty}$ (its stabilizer is $G$),
\item points $P_i^{(1)}, \ldots, P_i^{(p \cdot (p-1))}$ above $Q_i$ for $i = 1, \ldots, N$
\item[] (their stabilizers equal $C'$).
\end{itemize}
%
The same points are in the ramification locus of the morphism $X \to X/C$ with the following
ramification groups:
%
\begin{align*}
C_{P_0^{(1)}} &= C\\
C_{P_0^{(i)}} &= C' \qquad \textrm{ for } i > 1,\\
C_{P_{\infty}} &= C\\
C_{P_i^{(j)}} &= C' \qquad \textrm{ for } i = 1, \ldots, N, \, j = 1, \ldots, p \cdot (p-1).
\end{align*}
Note that $Y := X/H$ is given by the equation:
%
\[
y^m = z^{p^{n-1}} + \ldots + z^p + z.
\]
%
Let $\psi : C \to k^{\times}$ be a primitive character. We claim that:
%
\begin{align*}
a^{dR}_{X, C}(\psi^i) &=
\begin{cases}
p \cdot N, & \textrm{ if } m \nmid i,\\
0, & \textrm{ otherwise. }
\end{cases}\\
a^{dR}_{Y, C}(\psi^i) &=
\begin{cases}
\frac{p^{n-1} - 1}{p - 1}, & \textrm{ if } m \nmid i,\\
0, & \textrm{ otherwise. }
\end{cases}
\end{align*}
%
% %