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\date{\today}


\title{Galois Action on Homology of the Heisenberg Curve.}






% \author[A. Kontogeorgis]{Aristides Kontogeorgis}
% \address{Department of Mathematics, National and Kapodistrian  University of Athens
% Pane\-pist\-imioupolis, 15784 Athens, Greece}
% \email{kontogar@math.uoa.gr}

% \author[D. Noulas]{Dimitrios Noulas}
% \address{Department of Mathematics, National and Kapodistrian University of Athens\\
% Panepistimioupolis, 15784 Athens, Greece}
% \email{dnoulas@math.uoa.gr}


% \author[I. Tsouknidas]{Ioannis Tsouknidas }
% \address{Department of Mathematics, National and Kapodistrian University of Athens\\
% Panepistimioupolis, 15784 Athens, Greece}
% \email{iotsouknidas@math.uoa.gr}


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\begin{document}
\section{Cyclic Ramification}

Serre Local fields p. 77 Hasse- Arf for cyclic groups. 

For a cyclic group $G= \mathbb{Z}/p^n$ and $G(i)=\mathbb{Z}/p^{n-i}$,    there are integers $i_0,i_1, \ldots , i_{n-1}>0$ such that
\begin{align}
    \label{eq:serreI}
    G(0) & =G_0= \cdots = G_{i_0} &&=G^0 = \cdots = G^{i_0} \\
    G(1) & = G_{i_{0}+1 }= \cdots = G_{i_{0}+ p i_1 } &&= G^{i_0+1} = \cdots = G^{i_0+i_1}
    \nonumber \\
    G(2) &= G_{i_0+p i_1 +1} = \cdots = G_{i_0 +p i_1 + p^2 i_2} & &= G^{i_0+i_1+1} = \cdots = G^{i_0+ i_1 +i_2}  \nonumber
\end{align}
We also set $i_{-1}=-1$.
This means that the lower jumps occur at the integers 
\[
    i_0, i_0+i_1 p, i_0+ i_1 p + i_2 p^2, \ldots , i_0 + i_1 p + i_2 p^2 + \cdots i_{n-1} p^{n-1}
\]
while the upper jumps occur at 
\[
    i_0, i_0+i_1, i_0+ i_1  + i_2, \ldots , i_0 + i_1  + i_2  + \cdots i_{n-1} 
\]
%
\begin{definition}
Let for any $\ZZ/p^n$-cover $X \to Y$
%
{\color{blue}
\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*}
}

{\color{red}
\begin{align*}
	u_{X/Y, P}^{(t)} &:= \min \{ \nu \ge 0 : G_P^{(\nu)} \cong \ZZ/p^{n-t} \},\\
	l_{X/Y, P}^{(t)} &:= \min \{ \nu \ge 0 : G_{P, \nu} \cong \ZZ/p^{n-t} \}.
\end{align*}
but maybe you mean
\begin{align}
	u_{X/Y, P}^{(t)} &:= \min \{ \nu \ge 0 : G_P^{(\nu)} \cong \ZZ/p^{n-t} \}-1,\\
	l_{X/Y, P}^{(t)} &:= \min \{ \nu \ge 0 : G_{P, \nu} \cong \ZZ/p^{n-t} \}-1.
\end{align}
}
{\color{green}
Adding one to usual jumps was unitentional. It doesn't change any thing in the formula for $H^1_{dR}(X)$ (we have differences there),
but let's return to the usual definition of ramification jumps.	
}
\end{definition}
%
{\color{blue}
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 $P$.
If $G_P = \ZZ/p^m$, then (??relation with usual jumps??). By Hasse--Arf theorem (cf. ???),
the numbers $u_{X/Y, P}^{(t)}$ are integers.
%
}
Observe that if $G_P= \ZZ/ p^{n}$ with corresponding integers $i_0=i_0(P), \ldots , i_{n-1}=i_{n-1}(P)$ at $P$ then eq. (\ref{eq:serreI}) gives us 
\begin{align*}
    l^{(t)}_{X/Y,P} &=
    \begin{cases}
        0 , &\text{ if } t=0 \\
        i_0 + i_{1} p+ \cdots+  i_{t-1} p^{t-1} 
        % {\color{blue} +1}, &\text{ if } t>0 \\
    \end{cases}  
\\
    u^{(t)}_{X/Y,P} &=
    \begin{cases}
        0 , &\text{ if } t=0 \\
        i_0 + i_{1} + \cdots+  i_{t-1} 
        % {\color{blue} +1}, &\text{ if } t>0 \\
    \end{cases}  
\end{align*}
% that is not the upper jump but the next number.
We then have:
\[
    i_{j-1} =u_{X/Y,P}^{(j)}-u_{X /Y,P}^{(j-1)}= \frac{1}{p^{j-1}} (l_{X/ Y,P}^{j} - l_{X/ Y,P}^{j-1}) = \frac{1}{p^{j-1}} p^{j-1} i_{j-1}  
\]
{\color{red} you have written it in the other way out, do you agree?}
{\color{green} Yes, it was the other way around!}

% Now the ramification jumps for a subgroup I thing are a little bit different from what you write. 

The lower ramification jumps for the subgroup $ \mathbb{Z}/p^{n-N} =G(N)$ are given by 
\begin{align*}
    I_0(N) &=i_0 + i_1 p + \cdots + i_{N} p^{t},\\
    I_0(N) + p^{N+1} i_{i+1} &=I_0(N)+ p I_1(N),\\ 
     I_0(N) + p I_1(N) + p^{N+2} i_{N+2} &= I_0(N) + p I_1(N) + p^2 I_2(N),\\ 
     \ldots    
\end{align*}
that is 
\begin{align*}
    I_0(N)&= i_0 + i_1 p + \cdots + i_{N} p^{N},\\ 
    I_1(N)&= p^N i_{N+1}, 
    \\
    I_2(N)& = p^N i_{N+2},\\
     \ldots 
\end{align*}
This proves that if $\Gal(X/ X^{\prime} )= G(N)$ then 
\begin{align*}
    l_{X/X^{\prime},P}^{(t)} &=I_0 + I_1 p + \cdots + I_{t-1}p^{t-1}+1\\
      &= 
    i_0 + i_1 p + \cdots + i_{t+N-1} p^{t+N-1}+1\\
     &= l_{X/Y,P}^{(t+N)}
\end{align*}
and 
\begin{align*}
    u_{X/X^{\prime},P}^{(t)} &=I_0 + I_1  + \cdots + I_{t-1}+1\\
      &= 
    (i_0 + i_1 p + \cdots + i_{N} p^{N})+  i_{N+1} p^N + \cdots +  i_{N+t} p^N + 1\\
     &= 
\end{align*}
\vskip 2cm
Assume now that $X \to Y$ is not \'{e}tale. Therefore $X \to X''$ is also not \'{e}tale.
$\Gal(X / X'')= \ZZ/p$ and  $\Gal(X /Y^{\prime} ) = \ZZ/p^{n-1}$. 
\[
    \xymatrix{ 
       & X \ar[dl]_{ \ZZ / p\cong \langle \sigma^{p^{n-1} } \rangle } \ar[dr]^{H^{\prime} =\langle \sigma^p \rangle \cong \ZZ / p^{n-1} =G(1) } \\ 
    X '' \ar[dr]& & Y^{\prime}  \ar[dl] \\ 
    & Y 
    }
\]


	 Note that for any $P \in X(k)$:
    {\color{blue}
	%
	\begin{equation}
		\label{eq:pul}
		p \cdot u^{(n)}_{X/Y, P} = u^{(n-1)}_{X/Y', P} + (p-1) \cdot l^{(1)}_{Y'/Y, Q},
    \end{equation}
	%
    }
    Indeed, {\color{blue} blue color means that it is going to be erased}
    {\color{red}
    \begin{align}
         u^{(n)}_{X/ Y, P } &= (i_0 + i_1 + \cdots + i_{n-1} 
         {\color{blue} +1 }  ) \\ 
         u^{(n-1)}_{X/Y', P} &= (i_0 + i_1 p) + i_2 p + \cdots + i_{1+n-1} p
         {\color{blue}  +1} 
		 \\ \label{eq:l1}
         l^{(1)}_{Y'/Y, Q} &= u^{(1)}_{Y^{\prime} /Y,Q}= u^{(1)}_{(X/ Y,Q)} = l^{(1)}_{(X/ Y,Q)} = i_0
         {\color{blue} + 1}.
    \end{align}
    }
	where $Q$ denotes the image of~$P$ in~$Y'$. 

Riemann Hurwitz formula for the cover $Y^{\prime} / Y$, together with eq. (\ref{eq:l1}), implies that 
\begin{equation}
	\label{eq:RH}
	2(g_{Y^{\prime} }-1) = 
	2p(g_Y- 1) + \sum_{P\in Y^{\prime} (k)} (p-1)(l^{(1)}_{Y^{\prime} / Y}+1)
\end{equation}
By induction hypothesis for $H'$ acting on $X$, we have the following isomorphism of $k[H']$-modules:
%
\[
	\mc M \cong \mc J_{p^{n-1}}^{2 (g_{Y'} - 1)} \oplus \mc J_{p^{n-1} - p^{n - m} + 1}^2 \oplus \bigoplus_{\substack{P \in X(k)\\
	P \neq P_0}} \mc J_{p^{n-1} - p^{n-1}/e'_P}^2
	\oplus \bigoplus_{P \in X(k)} \bigoplus_{t = 0}^{n-2} \mc J_{p^n - p^t}^{u_{X/Y', P}^{(t+1)} - u_{X/Y', P}^{(t)}}
\]
%
where $e'_P := e_{X/Y', P}$.
{\color{green} The formula above needs a correction -- I want to sum over branch locus in $Y(k)$! This matters if the cover is not completely ramified.}
	%
	\begin{align*}
		\dim_k \mc T^i \mc M &=
		2(g_{Y'} - 1) + 2 + 2(\# R - 1) + \sum_{P \in X(k)} (u_{X/Y', P}^{(n-1)} - 1)\\
		&
		\stackrel{(\ref{eq:RH})}{=} 2 p (g_Y - 1) + \sum_{Q \in Y'(k)} (p-1) \cdot (l_{Y'/Y, Q}^{(1)} + 1)\\
		 &+ 2 + 2(\# R - 1) + \sum_{P \in X(k)} (u_{X/Y', P}^{(n-1)} - 1)\\
		&
		\stackrel{(\ref{eq:pul})}{=} p \cdot \left( 2(g_Y - 1)  + 2\# R 
		{\color{red}/p} 
		{\color{red}+
		\frac{1}{p}\sum_{Q \in Y'(k)} (p-1)
		}
		+ \sum_{P \in X(k)} (u_{X/Y, P}^{(n)} - 1
		{\color{red}/p}
		) \right)
		\\
		&= 
		{\color{red}
		p \cdot \left( 2(g_Y - 1)  + \# R 
		+ \sum_{P \in R} u_{X/Y, P}^{(n)} 
		\right)
		}
	\end{align*}
	{\color{red}
	I guess that we want to combine 
$2\# R/ p + \frac{1}{p}\sum_{Q \in Y'(k)} (p-1)$ together. This depends on the ramification of all ramified points in $H^{\prime}$...
	}
	%
	where
	%
	\[ R := \{ P \in X(k) : e_P > 1 \} = \{ P \in X(k) : e'_P > 1 \}. \]
	%
	In particular, $\dim_k \mc T^1 \mc M = \ldots = \dim_k \mc T^{p^{n-1} - p^{n-2}} \mc M$.
	Thus by Lemma~\ref{lem:lemma_mcT_and_T}
	%
	\begin{align*}
		\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\\
		&= 2(g_Y - 1) + 2 + 2(\# R - 1) + \sum_{P \in X(k)} (u_{X/Y, P}^{(n)} - 1).
	\end{align*}
	%
	By Lemma~\ref{lem:trace_surjective} since $X \to X''$ is not \'{e}tale, the map $\tr_{X/X''} : H^1_{dR}(X) \to H^1_{dR}(X'')$ is surjective. Recall that
	in $\FF_p[x]$ we have the identity:
	%
	\[
		1 + x + \ldots + x^{p-1} = (x - 1)^{p-1}.
	\]
	%
	Therefore 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:
	%
	\[
		\ker(\tr_{X/X''} : \mc M \to \mc M'') = \mc M^{(p^n - p^{n-1})}
	\]
	%
	and that $\tr_{X/X''}$ induces a $k$-linear isomorphism $T^{i + p^n - p^{n-1}} \mc M \to \mc T^i \mc M''$ for any $i \ge 1$. Thus:
	%
	\[
		\dim_k T^{i + p^n - p^{n-1}} \mc M = \dim_k \mc T^i \mc M'' = ....
	\]
	%
	This ends the proof.


\end{document}