diff --git a/article_de_rham_cyclic.bbl b/article_de_rham_cyclic.bbl index 1d94378..e7e97c7 100644 --- a/article_de_rham_cyclic.bbl +++ b/article_de_rham_cyclic.bbl @@ -1,3 +1,13 @@ -\begin{thebibliography}{} +\begin{thebibliography}{1} + +\bibitem{Bleher_Chinburg_Kontogeorgis_Galois_structure} +F.~M. Bleher, T.~Chinburg, and A.~Kontogeorgis. +\newblock Galois structure of the holomorphic differentials of curves. +\newblock {\em J. Number Theory}, 216:1--68, 2020. + +\bibitem{Valentini_Madan_Automorphisms} +R.~C. Valentini and M.~L. Madan. +\newblock Automorphisms and holomorphic differentials in characteristic~{$p$}. +\newblock {\em J. Number Theory}, 13(1):106--115, 1981. \end{thebibliography} diff --git a/article_de_rham_cyclic.synctex.gz b/article_de_rham_cyclic.synctex.gz index 8d59263..77bb0e3 100644 Binary files a/article_de_rham_cyclic.synctex.gz and b/article_de_rham_cyclic.synctex.gz differ diff --git a/article_de_rham_cyclic.tex b/article_de_rham_cyclic.tex index 47ed993..204e378 100644 --- a/article_de_rham_cyclic.tex +++ b/article_de_rham_cyclic.tex @@ -202,6 +202,7 @@ Note also that for $j \ge 1$: \end{proof} % \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 % \[ @@ -211,18 +212,58 @@ Note also that for $j \ge 1$: 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 trace 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: % \[ - (\sigma - 1) : T^{i+1} M \hookrightarrow T^i M. + m_{\sigma - 1} : T^{i+1} M \hookrightarrow T^i M. \] \end{Lemma} \begin{proof} - +% +We define $m_{\sigma - 1}$ as follows: +% +\[ + m_{\sigma - 1}(\ol x) := (\sigma - 1) \cdot x, +\] +% +where for $\ol x \in T^i M$ we picked any representative $x \in M^{(i)}$. +If $x \in M^{(i+1)} := \ker((\sigma - 1)^{i+1})$ then clearly $(\sigma - 1) x \in M^{(i)}$. +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} @@ -346,7 +387,7 @@ Note also that for $j \ge 1$: \section{Hypoelementary covers} % Assume now that $G = H \rtimes_{\chi} C = \langle \sigma \rangle \rtimes_{\chi} \langle \rho \rangle \cong \ZZ/p^n \rtimes_{\chi} \ZZ/c$. -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 $M^{\psi} := M \otimes_{k[G]} \psi$. +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{Proposition} \label{prop:main_thm_for_hypoelementary} Main Theorem holds for a hypoelementary $G$ as above and $k = \ol k$. @@ -357,33 +398,84 @@ Let $X$ be a curve with an action of $G$ and write $Y := X/H$. For any $k[C]$-mo is uniquely determined by the $k[C]$-structure of $T^1 M, \ldots, T^{p^n} M$. \end{Lemma} \begin{proof} - ??? + See \cite[????]{Bleher_Chinburg_Kontogeorgis_Galois_structure} for a proof. \end{proof} % \begin{Lemma} \label{lem:N+Nchi+...} - Let $N_1$, $N_2$ be $k[G]$-modules. Assume that for some $j$ + Keep the above notation. Let $M$, $N$ be $k[C]$-modules. Assume that % \[ - N_1 \oplus N_1^{\chi} \oplus \ldots \oplus N_1^{\chi^j} - \cong N_2 \oplus N_2^{\chi} \oplus \ldots \oplus N_2^{\chi^j}. + M \cong N \oplus N^{\chi} \oplus \ldots \oplus N^{\chi^{p-1}}. \] % - If $\GCD(j, p-1) = 1$, then $N_1 \cong N_2$. + Then $N$ is uniquely determined by $M$. %If $p-1 | j$, then $N_1 \cong N_2^{\chi^i}$ for some $i$. \end{Lemma} \begin{proof} - + Note that + % + \[ + M \cong N^{\oplus 2} \oplus N^{\chi} \oplus N^{\chi^2} \oplus \ldots \oplus N^{\chi^{p-2}}. + \] + % + By tensoring this isomorphism by $\chi^i$ we obtain: + % + \begin{align*} + 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}} + \cong (N^{\chi^i})^{\oplus 2} \oplus \bigoplus_{\substack{j = 0\\j \neq i}}^{p-2} N^{\chi^j} + \end{align*} + % + for $i = 0, \ldots, p-2$. Therefore: + % + \begin{equation} \label{eqn:N+M=M} + N^{\oplus p} \oplus M^{\chi} \oplus M^{\chi^2} \oplus \ldots \oplus M^{\chi^{p-2}} \oplus + \cong M^{\oplus (p-1)}. + \end{equation} + % + Indeed, for the proof of~\eqref{eqn:N+M=M} note that + % + \begin{align*} + N^{\oplus p} &\oplus M^{\chi} \oplus M^{\chi^2} \oplus \ldots \oplus M^{\chi^{p-2}} + \cong N^{\oplus p} \oplus \bigoplus_{i = 1}^{p-2} \left((N^{\chi^i})^{\oplus 2} + \oplus \bigoplus_{\substack{j = 0\\j \neq i}}^{p-2} N^{\chi^j} \right)\\ + &\cong \left( N^{\oplus 2} \oplus N^{\chi} \oplus N^{\chi^2} \oplus \ldots \oplus N^{\chi^{p-2}} \right)^{\oplus (p-1)} + \cong M^{\oplus (p-1)}. + \end{align*} + % + The isomorphism~\eqref{eqn:N+M=M} clearly proves the thesis. \end{proof} % \begin{Lemma} \label{lem:TiM_isomorphism_hypoelementary} - For any $i \le p^n - 1$: + For any $i \le p^n - 1$ the map~$m_{\sigma - 1}$ from Lemma~\ref{lem:TiM_isomorphism} + yields a $k[C]$-equivariant monomorphism: % \[ - (\sigma - 1) : T^{i+1} M \hookrightarrow (T^i M)^{\chi^{-1}}. + m_{\sigma - 1} : T^{i+1} M \hookrightarrow (T^i M)^{\chi^{-1}}. \] \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 Proposition~\ref{prop:main_thm_for_hypoelementary}] @@ -403,15 +495,15 @@ Let $X$ be a curve with an action of $G$ and write $Y := X/H$. For any $k[C]$-mo \begin{align*} \mc T^i \mc M &\cong \begin{cases} - T^1 \mc M \oplus T^2 \mc M \oplus \ldots \oplus (T^2 \mc M)^{\chi^{-p + 1}}, & i = 1\\ - T^2 \mc M \oplus \ldots \oplus (T^2 \mc M)^{\chi^{-p}}, & i > 1. + 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\\ + T^2 \mc M \oplus \ldots \oplus (T^2 \mc M)^{\chi^{-(p-1)}}, & i > 1. \end{cases} \end{align*} % Thus, since by induction hypothesis $\mc T^2 \mc M$ is determined by ramification data, we have by Lemma~\ref{lem:N+Nchi+...} that $T^2 \mc M$ is determined by ramification data. Moreover, by Lemma~\ref{lem:G_invariants_\'{e}tale} and induction hypothesis, $T^1 \mc M \cong H^1_{dR}(X'')$ - is also determined by ramification data (???). + is also determined by ramification data. 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} yield an isomorphism of $k[C]$-modules: