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usual definition of ramification jumps

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\title{Galois Action on Homology of the Heisenberg Curve.}
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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}
}
\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?}
% 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 $e'_P := e_{X/Y', P}$.
%
\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}