pf for hypoelementary
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\BOOKMARK [1][-]{section.1}{\376\377\0001\000.\000\040\000I\000n\000t\000r\000o\000d\000u\000c\000t\000i\000o\000n}{}% 1
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\BOOKMARK [1][-]{section.2}{\376\377\0002\000.\000\040\000C\000y\000c\000l\000i\000c\000\040\000c\000o\000v\000e\000r\000s}{}% 2
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\BOOKMARK [1][-]{section.3}{\376\377\0003\000.\000\040\000H\000y\000p\000o\000e\000l\000e\000m\000e\000n\000t\000a\000r\000y\000\040\000c\000o\000v\000e\000r\000s}{}% 3
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\BOOKMARK [1][-]{section.4}{\376\377\0004\000.\000\040\000P\000r\000o\000o\000f\000\040\000o\000f\000\040\000M\000a\000i\000n\000\040\000T\000h\000e\000o\000r\000e\000m}{}% 4
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@ -143,14 +143,14 @@ hyperref, bbm, mathtools, mathrsfs}
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Let for any $\ZZ/p^n$-cover $X \to Y$
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%
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\begin{align*}
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u_{X/Y, P}^{(t)} &:= \min \{ t \ge 0 : G_P^{(t)} \cong \ZZ/p^{n-t???} \},\\
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l_{X/Y, P}^{(t)} &:= \min \{ t \ge 0 : G_{P, t} \cong \ZZ/p^{n-t???} \}.
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u_{X/Y, P}^{(t)} &:= \min \{ t \ge 0 : G_P^{(t)} \cong \ZZ/p^{n-t} \},\\
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l_{X/Y, P}^{(t)} &:= \min \{ t \ge 0 : G_{P, t} \cong \ZZ/p^{n-t} \}.
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\end{align*}
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%
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Note that if $G_P = \ZZ/p^n$, this coincides with the standard definition of
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the $t$th upper (resp. lower) ramification jump of $X \to Y$ at $P$.
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%
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\begin{Theorem}
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\begin{Theorem} \label{thm:cyclic_de_rham}
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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:
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%
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\[
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@ -187,7 +187,7 @@ Note also that for $j \ge 1$:
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\end{itemize}
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\begin{Lemma}
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\begin{Lemma} \label{lem:G_invariants_etale}
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If the $G$-cover $X \to Y$ is \'{e}tale, then the natural map
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%
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\[
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@ -200,7 +200,7 @@ Note also that for $j \ge 1$:
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????
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\end{proof}
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%
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\begin{Lemma}
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\begin{Lemma} \label{lem:trace_surjective}
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If the $G$-cover $X \to Y$ is totally ramified, then the map
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%
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\[
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@ -213,7 +213,7 @@ Note also that for $j \ge 1$:
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????
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\end{proof}
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%
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\begin{Lemma}
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\begin{Lemma} \label{lem:TiM_isomorphism}
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For any $i \le p^n - 1$:
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%
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\[
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@ -224,14 +224,27 @@ Note also that for $j \ge 1$:
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\end{proof}
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%
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\begin{proof}[Proof of Theorem ????]
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\begin{Lemma}
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Let $M$ be a $k[H]$-module. Let $T^i M$ be as above and
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$\mc T^i M := T^i_{H'} M$ for $H' \le H$, $H' \cong \ZZ/p^{n-1}$.
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If $\mc T^i M \cong \mc T^{i+1} M$ for some $i$ then:
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%
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\[
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T^{pi + p} M \cong T^{pi + p - 1} M \cong \ldots \cong T^{pi - p + 1} M.
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\]
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\end{Lemma}
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\begin{proof}
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??
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\end{proof}
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\begin{proof}[Proof of Theorem~\ref{thm:cyclic_de_rham}]
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We use the following notation: $H' := \langle \sigma^p \rangle \cong \ZZ/p^{n-1}$,
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$H'' := H/\langle \sigma^{p^{n-1}} \rangle \cong \ZZ/p^{n-1}$, $Y' := X/H'$, $X'' := X/H''$.
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Write also $M := H^1_{dR}(X)$.
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Write also $\mc M := H^1_{dR}(X)$.
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By induction hypothesis for $H'$ acting on $X$, we have the following isomorphism of $k[H']$-modules:
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%
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\[
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M \cong \mc J_{p^{n-1}}^{2 (g_{Y'} - 1)} \oplus \mc J_{p^{n-1} - p^{n - 1 -m'} + 1}^2 \oplus \bigoplus_{P \neq P_0} \mc J_{p^n - \frac{p^{n-1}}{e_{X/Y', P}}}^2
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\mc M \cong \mc J_{p^{n-1}}^{2 (g_{Y'} - 1)} \oplus \mc J_{p^{n-1} - p^{n - 1 -m'} + 1}^2 \oplus \bigoplus_{P \neq P_0} \mc J_{p^n - \frac{p^{n-1}}{e_{X/Y', P}}}^2
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\oplus \bigoplus_P \bigoplus_{t = 0}^{n-1} \mc J_{p^n - p^t}^{u_{X/Y', P}^{(t+1)} - u_{X/Y', P}^{(t)}}
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\]
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%
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@ -245,34 +258,40 @@ Note also that for $j \ge 1$:
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\end{cases}
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\]
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Therefore, for $???$
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Therefore, for $i \le p^n - p^{n-1}$
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%
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\begin{align*}
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\dim_k \mc T^i M =
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\dim_k \mc T^i \mc M =
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\begin{cases}
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???,
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\end{cases}
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\end{align*}
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%
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In particular, $\dim_k \mc T^1 M = \ldots = \dim_k \mc T^{p^{n-1} - p^{n-2}} M$.
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On the other hand, by Lemma ??:
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In particular, $\dim_k \mc T^1 \mc M = \ldots = \dim_k \mc T^{p^{n-1} - p^{n-2}} \mc M$.
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On the other hand, by Lemma~\ref{lem:TiM_isomorphism}:
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%
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\begin{align*}
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\dim_k \mc T^1 M &= \dim_k T^1 M + \ldots + \dim_k T^p M\\
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&\ge \dim_k T^{p^n - p^{n-1}} M + \ldots + \dim_k T^{p^n - p^{n-1}} M
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= \dim_k \mc T^{p^{n-1} - p^{n-2}} M.
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\dim_k \mc T^1 \mc M &= \dim_k T^1 \mc M + \ldots + \dim_k T^p \mc M\\
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&\ge \dim_k T^{p^n - p^{n-1}} \mc M + \ldots + \dim_k T^{p^n - p^{n-1}} \mc M
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= \dim_k \mc T^{p^{n-1} - p^{n-2}} \mc M.
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\end{align*}
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%
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Since the left-hand side and right hand side are equal, we conclude by Lemma ???
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Since the left-hand side and right hand side are equal, we conclude by Lemma~\ref{lem:TiM_isomorphism}
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that
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%
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\[
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\dim_k T^1 M = \ldots = \dim_k T^{p^n - p^{n-1}} M = \frac{1}{p} \dim_k \mc T^1 M.
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\dim_k T^1 \mc M = \ldots = \dim_k T^{p^n - p^{n-1}} \mc M = \frac{1}{p} \dim_k \mc T^1 \mc M.
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\]
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%
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If the cover $X \to X''$ is \'{e}tale, then the cover $X \to Y$ must be also \'{e}tale.
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We consider now two cases. If the cover $X \to Y$ is \'{e}tale, then by Lemma~\ref{lem:G_invariants_etale} we have
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%
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\[
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\dim_k T^1 \mc M = 2 g_{X''}
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\]
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then the cover $X \to Y$ must be also \'{e}tale.
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Thus the proof follows in this case by~\cite{Nakajima??Inventiones}. Suppose now that
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$X \to X''$ 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:
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$X \to X''$ is not \'{e}tale. Then, by Lemma~\ref{lem:trace_surjective}, 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:
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%
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\[
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\tr_{X/X''} = \sum_{j = 0}^{p-1} (\sigma^{p^{n-1}})^j = (\sigma^{p^{n-1}} - 1)^{p-1} =
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@ -282,13 +301,13 @@ Note also that for $j \ge 1$:
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This implies that:
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%
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\[
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\ker(\tr_{X/X''} : M \to M'') = M^{(p^n - p^{n-1})}
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\ker(\tr_{X/X''} : \mc M \to \mc M'') = \mc M^{(p^n - p^{n-1})}
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\]
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%
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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:
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%
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\[
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\dim_k T^{i + p^n - p^{n-1}} M = \dim_k \mc T^i M'' = ....
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\dim_k T^{i + p^n - p^{n-1}} \mc M = \dim_k \mc T^i \mc M'' = ....
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\]
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%
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This ends the proof.
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@ -296,7 +315,117 @@ Note also that for $j \ge 1$:
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\section{Hypoelementary covers}
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%
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Assume now that $G = H \rtimes_{\chi} \ZZ/??n$.
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Assume now that $G = H \rtimes_{\chi} \ZZ/c$.
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%
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\begin{Proposition} \label{prop:main_thm_for_hypoelementary}
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Main Theorem holds for a hypoelementary $G$ as above and $k = \ol k$.
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\end{Proposition}
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%
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\begin{Lemma}
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Let $M$ be a $k[G]$-module of finite dimension. The $k[G]$-structure of $M$
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is uniquely determined by the $k[C]$-structure of $T^1 M, \ldots, T^{p^n} M$.
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\end{Lemma}
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\begin{proof}
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???
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\end{proof}
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%
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\begin{Lemma} \label{lem:N+Nchi+...}
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Let $N_1$, $N_2$ be $k[G]$-modules. Assume that for some $j$
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%
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\[
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N_1 \oplus N_1^{\chi} \oplus \ldots \oplus N_1^{\chi^j}
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\cong N_2 \oplus N_2^{\chi} \oplus \ldots \oplus N_2^{\chi^j}.
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\]
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%
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If $\GCD(j, p-1) = 1$, then $N_1 \cong N_2$. If $p-1 | j$, then
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$N_1 \cong N_2^{\chi^i}$ for some $i$.
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\end{Lemma}
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\begin{proof}
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\end{proof}
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%
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\begin{Lemma} \label{lem:chevalley_weil_for_Z/p}
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If $X$ has a $G$-action and $Y := X/H$,
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then as $k[C]$-modules:
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%
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\[
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H^1_{dR}(X) \cong H^1_{dR}(Y) \oplus N^{p-1}
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\]
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%
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for a $k[C]$-module $N$ such that $N^{\chi} \cong N$.
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\end{Lemma}
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\begin{proof}
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??Chevalley--Weil??
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??is it really needed ??
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\end{proof}
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%
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\begin{Lemma} \label{lem:TiM_isomorphism_hypoelementary}
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For any $i \le p^n - 1$:
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%
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\[
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(\sigma - 1) : T^{i+1} M \hookrightarrow (T^i M)^{\chi^{-1}}.
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\]
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\end{Lemma}
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\begin{proof}
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\end{proof}
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\begin{proof}[Proof of Proposition~\ref{prop:main_thm_for_hypoelementary}]
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We prove this by induction on $n$. If $n = 0$, then it follows by Chevalley--Weil theorem.
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Consider now two cases. Firstly, we assume that $X \to Y$ is \'{e}tale.
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Recall that by proof of Theorem~\ref{thm:cyclic_de_rham}, the map $(\sigma - 1)$
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is an isomorphism of $k$-vector spaces between $T^{i+1} \mc M$ and $T^i \mc M$ for
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$i = 2, \ldots, p^n$. This yields an isomorphism of $k[C]$-modules for $i \ge 2$:
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%
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\begin{equation} \label{eqn:TiM=T1M_chi_etale}
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T^i \mc M \cong (T^2 \mc M)^{\chi^{-i+2}}
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\end{equation}
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%
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Observe that $\mc T^i \mc M$ has the filtration $\mc M^{(pi)} \supset \mc M^{(pi - 1)} \supset \ldots \supset \mc M^{(pi - p)}$ with subquotients $T^{pi} \mc M, \ldots, T^{pi - p} \mc M$.
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Thus, since the category of $k[C]$-modules is semisimple, for $i \le p^n - p^{n-1}$:
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%
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\begin{align*}
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\mc T^i \mc M &\cong
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\begin{cases}
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T^1 \mc M \oplus T^2 \mc M \oplus \ldots \oplus (T^2 \mc M)^{\chi^{-p + 1}}, & i = 1\\
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T^2 \mc M \oplus \ldots \oplus (T^2 \mc M)^{\chi^{-p}}, & i > 1.
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\end{cases}
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\end{align*}
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%
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Thus, since by induction hypothesis $\mc T^i \mc M$ is determined by ramification data,
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we have by Lemma~\ref{lem:N+Nchi+...} that $T^2 \mc M$ is determined by ramification data.
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Moreover, by Lemma~\ref{lem:G_invariants_etale}, $T^1 \mc M \cong H^1_{dR}(X'')$
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is also determined by ramification data (???).
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Assume now that $X \to Y$ is not etale. Analogously as in the previous case, Lemma~\ref{lem:TiM_isomorphism_hypoelementary} and proof of Theorem~\ref{thm:cyclic_de_rham}
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yield an isomorphism of $k[C]$-modules:
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%
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\begin{equation} \label{eqn:TiM=T1M_chi}
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T^{i+1} \mc M \cong (T^1 \mc M)^{\chi^{-i}}
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\end{equation}
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%
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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 $T^{pi} \mc M, \ldots, T^{pi - p + 1} \mc M$.
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Thus, since the category of $k[C]$-modules is semisimple, for $i \le p^n - p^{n-1}$:
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%
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\begin{align*}
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\mc T^i \mc M &\cong T^{pi - p + 1} \mc M \oplus \ldots \oplus T^{pi} \mc M\\
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&\cong T^1 \mc M \oplus (T^1 \mc M)^{\chi^{-1}} \oplus \ldots \oplus
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(T^1 \mc M)^{\chi^{-p}}.
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\end{align*}
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%
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By induction assumption, the $k[C]$-module structure of $\mc T^i \mc M$ is uniquely determined by the 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 ramification data for $i \le p^n - p^{n-1}$.
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By similar reasoning, $\tr_{X/X'}$ yields an isomorphism:
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%
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\[
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T^{i + p^n - p^{n-1}} \mc M \cong (\mc T^i \mc M'')^{\chi^{-1??}}.
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\]
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%
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Thus, by induction hypothesis for $\mc M''$, the $k[C]$-structure of $T^{i + p^n - p^{n-1}} \mc M$
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is determined by ramification data as well.
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\end{proof}
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\section{Proof of Main Theorem}
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%
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(Conlon induction ???) (algebraic closure ???)
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\bibliography{bibliografia}
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\end{document}
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