lematy
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\begin{thebibliography}{}
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\begin{thebibliography}{1}
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\bibitem{Bleher_Chinburg_Kontogeorgis_Galois_structure}
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F.~M. Bleher, T.~Chinburg, and A.~Kontogeorgis.
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\newblock Galois structure of the holomorphic differentials of curves.
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\newblock {\em J. Number Theory}, 216:1--68, 2020.
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\bibitem{Valentini_Madan_Automorphisms}
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R.~C. Valentini and M.~L. Madan.
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\newblock Automorphisms and holomorphic differentials in characteristic~{$p$}.
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\newblock {\em J. Number Theory}, 13(1):106--115, 1981.
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\end{thebibliography}
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@ -202,6 +202,7 @@ Note also that for $j \ge 1$:
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\end{proof}
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%
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\begin{Lemma} \label{lem:trace_surjective}
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Suppose that $G$ is a $p$-group.
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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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@ -211,18 +212,58 @@ Note also that for $j \ge 1$:
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is an epimorphism.
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\end{Lemma}
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\begin{proof}
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????
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%
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By induction, it suffices to prove this in the case when $G = \ZZ/p$.
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Consider the following commutative diagram:
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%
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\begin{center}
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% https://tikzcd.yichuanshen.de/#N4Igdg9gJgpgziAXAbVABwnAlgFyxMJZABgBpiBdUkANwEMAbAVxiRGJAF9T1Nd9CKMgEYqtRizYduvbHgJFh5MfWatEIABIA9YgAoAGqQAEAHVMB5ALYwA5nQD6BgJRceIDHIGLSo6qskNHX0ATRNzaztHENcZDz55QWQAJmV-CXUtbWEHYCgAJU5DWPdPfgUUVL9xNTYdHLzCvRi3WXKkgGY0msCs4UNw0ysAY2MLJxK2xKIu6oDM+ubBkbGHFriy6ZQAFm75qVb4rwrkXbmMg84xGChbeCJQADMAJwgrJDIQHAgkZLiXt6-ajfJDbf6vd6IXZfH6IABs4MB8OBsIAHIjIQB2FFIACcGKQAFYcYhMQTEF0YUTyUoqRTyak6ZT9hpzDhnrkDAB6EKcQ4AyHQkGIYk9TJsjnAbm8-kQpBwknYsVsCWcnl8q6cIA
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\begin{tikzcd}
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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 \\
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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
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\end{tikzcd}
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\end{center}
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%
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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
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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
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of the image of
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%
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\begin{equation} \label{eqn:trace_H0_Omega}
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\tr_{X/Y} : H^0(X, \Omega_X) \to H^0(Y, \Omega_Y)
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\end{equation}
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%
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is $g_Y$. Therefore the map~\eqref{eqn:trace_H0_Omega} is surjective.
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Similarly, by Serre's duality, also $H^1(X, \mc O_X)$ contains $k[G]^{\oplus g_Y}$ as a direct summand
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and one shows similarly that the trace map
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%
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\begin{equation*} %\label{eqn:trace_H0_Omega}
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\tr_{X/Y} : H^1(X, \mc O_X) \to H^1(Y, \mc O_Y)
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\end{equation*}
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%
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is surjective. Therefore, since the outer trace maps in the diagram are surjective,
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the trace map on the de Rham cohomology must be surjective as well.
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%
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\end{proof}
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%
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\begin{Lemma} \label{lem:TiM_isomorphism}
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For any $i \le p^n - 1$ we have the following $k$-linear monomorphism:
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%
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\[
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(\sigma - 1) : T^{i+1} M \hookrightarrow T^i M.
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m_{\sigma - 1} : T^{i+1} M \hookrightarrow T^i M.
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\]
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\end{Lemma}
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\begin{proof}
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%
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We define $m_{\sigma - 1}$ as follows:
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%
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\[
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m_{\sigma - 1}(\ol x) := (\sigma - 1) \cdot x,
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\]
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%
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where for $\ol x \in T^i M$ we picked any representative $x \in M^{(i)}$.
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If $x \in M^{(i+1)} := \ker((\sigma - 1)^{i+1})$ then clearly $(\sigma - 1) x \in M^{(i)}$.
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Moreover $(\sigma - 1) \cdot x \in M^{(i-1)}$ holds if and only if $x \in M^{(i)}$. This
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shows that $m_{\sigma - 1}$ is well-defined and injective.
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\end{proof}
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%
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\begin{Lemma} \label{lem:lemma_mcT_and_T}
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@ -346,7 +387,7 @@ 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 $G = H \rtimes_{\chi} C = \langle \sigma \rangle \rtimes_{\chi} \langle \rho \rangle \cong \ZZ/p^n \rtimes_{\chi} \ZZ/c$.
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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 $M^{\psi} := M \otimes_{k[G]} \psi$.
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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 $M^{\psi} := M \otimes_{k[C]} \psi$.
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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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@ -357,33 +398,84 @@ Let $X$ be a curve with an action of $G$ and write $Y := X/H$. For any $k[C]$-mo
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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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See \cite[????]{Bleher_Chinburg_Kontogeorgis_Galois_structure} for a proof.
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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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Keep the above notation. Let $M$, $N$ be $k[C]$-modules. Assume that
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%
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\[
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N_1 \oplus N_1^{\chi} \oplus \ldots \oplus N_1^{\chi^j}
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\cong N_2 \oplus N_2^{\chi} \oplus \ldots \oplus N_2^{\chi^j}.
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M \cong N \oplus N^{\chi} \oplus \ldots \oplus N^{\chi^{p-1}}.
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\]
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%
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If $\GCD(j, p-1) = 1$, then $N_1 \cong N_2$.
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Then $N$ is uniquely determined by $M$.
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%If $p-1 | j$, then $N_1 \cong N_2^{\chi^i}$ for some $i$.
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\end{Lemma}
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\begin{proof}
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Note that
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%
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\[
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M \cong N^{\oplus 2} \oplus N^{\chi} \oplus N^{\chi^2} \oplus \ldots \oplus N^{\chi^{p-2}}.
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\]
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%
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By tensoring this isomorphism by $\chi^i$ we obtain:
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%
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\begin{align*}
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M^{\chi^i} \cong (N^{\chi^i})^{\oplus 2} \oplus N^{\chi^{i+1}} \oplus N^{\chi^{i+2}} \oplus \ldots \oplus N^{\chi^{i + p-2}}
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\cong (N^{\chi^i})^{\oplus 2} \oplus \bigoplus_{\substack{j = 0\\j \neq i}}^{p-2} N^{\chi^j}
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\end{align*}
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%
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for $i = 0, \ldots, p-2$. Therefore:
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%
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\begin{equation} \label{eqn:N+M=M}
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N^{\oplus p} \oplus M^{\chi} \oplus M^{\chi^2} \oplus \ldots \oplus M^{\chi^{p-2}} \oplus
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\cong M^{\oplus (p-1)}.
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\end{equation}
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%
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Indeed, for the proof of~\eqref{eqn:N+M=M} note that
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%
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\begin{align*}
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N^{\oplus p} &\oplus M^{\chi} \oplus M^{\chi^2} \oplus \ldots \oplus M^{\chi^{p-2}}
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\cong N^{\oplus p} \oplus \bigoplus_{i = 1}^{p-2} \left((N^{\chi^i})^{\oplus 2}
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\oplus \bigoplus_{\substack{j = 0\\j \neq i}}^{p-2} N^{\chi^j} \right)\\
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&\cong \left( N^{\oplus 2} \oplus N^{\chi} \oplus N^{\chi^2} \oplus \ldots \oplus N^{\chi^{p-2}} \right)^{\oplus (p-1)}
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\cong M^{\oplus (p-1)}.
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\end{align*}
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%
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The isomorphism~\eqref{eqn:N+M=M} clearly proves the thesis.
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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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For any $i \le p^n - 1$ the map~$m_{\sigma - 1}$ from Lemma~\ref{lem:TiM_isomorphism}
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yields a $k[C]$-equivariant monomorphism:
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%
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\[
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(\sigma - 1) : T^{i+1} M \hookrightarrow (T^i M)^{\chi^{-1}}.
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m_{\sigma - 1} : T^{i+1} M \hookrightarrow (T^i M)^{\chi^{-1}}.
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\]
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\end{Lemma}
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\begin{proof}
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By Lemma~\ref{lem:TiM_isomorphism} this map is injective. Thus it suffices to check that it is $k[C]$-equivariant.
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Note that we have the following identity in the ring~$k[C]$:
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%
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\[
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(\sigma - 1) \cdot \rho = \rho \cdot (\sigma^{\chi(\rho)^{-1}} - 1)
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= \rho \cdot (\sigma - 1) \cdot (1 + \sigma + \sigma^2 + \ldots + \sigma^{\chi(\rho)^{-1} - 1})
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\]
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%
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Note that $\sigma$ acts trivially on $T^i M$, so that for any $\ol x \in T^i M$:
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%
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\[
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(1 + \sigma + \sigma^2 + \ldots + \sigma^{\chi(\rho)^{-1} - 1}) \cdot \ol x = \chi(\rho)^{-1} \cdot \ol x.
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\]
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%
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This easily shows that
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%
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\[
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m_{\sigma - 1}(\rho \cdot \ol x) = \chi(\rho)^{-1} \cdot \rho \cdot m_{\sigma - 1}(\ol x),
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\]
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%
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which ends the proof.
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%
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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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@ -403,15 +495,15 @@ Let $X$ be a curve with an action of $G$ and write $Y := X/H$. For any $k[C]$-mo
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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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T^1 \mc M \oplus T^2 \mc M \oplus (T^2 \mc M)^{\chi^{-1}} \oplus \ldots \oplus (T^2 \mc M)^{\chi^{-(p - 2)}}, & i = 1\\
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T^2 \mc M \oplus \ldots \oplus (T^2 \mc M)^{\chi^{-(p-1)}}, & 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 $\mc T^2 \mc M$ 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_\'{e}tale} and induction hypothesis, $T^1 \mc M \cong H^1_{dR}(X'')$
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is also determined by ramification data (???).
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is also determined by ramification data.
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Assume now that $X \to Y$ is not \'{e}tale. 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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