%~Mouliné par MaN_auto v.0.27.3 2022-11-07 09:58:50
\documentclass[AHL,Unicode,longabstracts,published]{cedram}

\usepackage{amssymb}






%\NumberTheoremsIn{subsection}

\DeclareMathOperator{\Gal}{Gal}
\DeclareMathOperator{\Aut}{Aut}
\DeclareMathOperator{\Mat}{Mat}
\DeclareMathOperator{\Nm}{N}
\DeclareMathOperator{\sh}{sh}
\DeclareMathOperator{\orb}{orb}
\DeclareMathOperator{\cont}{cont}
\DeclareMathOperator{\cyc}{cyc}
\DeclareMathOperator{\fin}{fin}
\DeclareMathOperator{\val}{val}
\newcommand{\hyphen}{\!\operatorname{-}\!}
\DeclareMathOperator{\bigO}{O}
\DeclareMathOperator{\M}{M}
\DeclareMathOperator{\GL}{GL}
\DeclareMathOperator{\Id}{Id}
\newcommand{\tco}{{}^t\!\operatorname{co}}

\newcommand{\lal}{\mathrm{la}}


\newcommand{\Qp}{\mathbf{Q}_p}
\newcommand{\Zp}{\mathbf{Z}_p}
\newcommand{\Fp}{\mathbf{F}_p}
\newcommand{\ZZ}{\mathbf{Z}}
\newcommand{\RR}{\mathbf{R}}
\newcommand{\efont}{\mathbf{E}}
\newcommand{\et}{\widetilde{\mathbf{E}}}
\newcommand{\dfont}{\mathbf{D}}

\newcommand{\Gm}{\mathbf{G}_\mathrm{m}}
\newcommand{\BB}{\mathrm{B}_2(\Qp)}

\newcommand{\eps}{\varepsilon}

\newcommand{\HH}{\mathcal{H}}
\newcommand{\WW}{\mathcal{W}}

\newcommand{\vp}{\val_p}
\newcommand{\vm}{\val_M}
\newcommand{\vx}{\val_X}
\newcommand{\dcroc}[1]{[\![#1]\!]}
\newcommand{\dpar}[1]{(\!(#1)\!)}
\newcommand{\kf}{E}
\renewcommand{\geq}{\geqslant}
\renewcommand{\leq}{\leqslant}
\renewcommand{\phi}{\varphi}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\graphicspath{{./figures/}}

\newcommand*{\mk}{\mkern -1mu}
\newcommand*{\Mk}{\mkern -2mu}
\newcommand*{\mK}{\mkern 1mu}
\newcommand*{\MK}{\mkern 2mu}

%\hypersetup{urlcolor=purple, linkcolor=blue, citecolor=red}

\newcommand*{\romanenumi}{\renewcommand*{\theenumi}{\roman{enumi}}}
\newcommand*{\Romanenumi}{\renewcommand*{\theenumi}{\Roman{enumi}}}
\newcommand*{\alphenumi}{\renewcommand*{\theenumi}{\alph{enumi}}}
\newcommand*{\Alphenumi}{\renewcommand*{\theenumi}{\Alph{enumi}}}
\let\oldtilde\tilde
\renewcommand*{\tilde}[1]{\mathchoice{\widetilde{#1}}{\widetilde{#1}}{\oldtilde{#1}}{\oldtilde{#1}}}
\let\oldhat\hat
\renewcommand*{\hat}[1]{\mathchoice{\widehat{#1}}{\widehat{#1}}{\oldhat{#1}}{\oldhat{#1}}}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\title[Decompletion of cyclotomic perfectoid fields in positive characteristic]{Decompletion of cyclotomic perfectoid fields in positive characteristic}
\alttitle{Décomplétion de corps perfecto\"ides cyclotomiques en caractéristique positive}

\subjclass{11S80, 12J25, 13J05, 22E35}


\author[\initial{L.} \lastname{Berger}]{\firstname{Laurent} \lastname{Berger}}
\address{UMPA de l'{\'E}NS de Lyon \\
UMR 5669 du CNRS (France)}
\email{laurent.berger@ens-lyon.fr}
\urladdr{http://perso.ens-lyon.fr/laurent.berger/}

\author[\initial{S.} \lastname{Rozensztajn}]{\firstname{Sandra} \lastname{Rozensztajn}}
\address{UMPA de l'{\'E}NS de Lyon \\
UMR 5669 du CNRS (France)}
\email{sandra.rozensztajn@ens-lyon.fr}
\urladdr{http://perso.ens-lyon.fr/sandra.rozensztajn/}

\begin{abstract}
Let $\kf$ be a field of characteristic $p$. The group $\Zp^\times$ acts on $\kf \dpar{X}$ by $a \cdot f(X) = f((1+X)^a-1)$. This action extends to the $X$-adic completion $\et$ of $\cup_{n\,\geq\,0} \kf \dpar{X^{1/p^n}}$. We show how to recover $\kf \dpar{X}$ from the valued $E$-vector space $\et$ endowed with its action of $\Zp^\times$. To do this, we introduce the notion of super-H\"older vector in certain $E$-linear representations of $\Zp$. This is a characteristic $p$ analogue of the notion of locally analytic vector in $p$-adic Banach representations of $p$-adic Lie groups.
\end{abstract}

\begin{altabstract}
Soit $\kf$ un corps de caractéristique $p$. Le groupe $\Zp^\times$ agit sur $\kf \dpar{X}$ par $a \cdot f(X) = f((1+X)^a-1)$. Cette action s'étend à la complétion $X$-adique $\et$ de $\cup_{n\,\geq\,0} \kf \dpar{X^{1/p^n}}$. Nous montrons comment récupérer $\kf \dpar{X}$ à partir du $E$-espace vectoriel valué $\et$ muni de son action de $\Zp^\times$. Pour faire cela, nous introduisons la notion de vecteur super-H\"older dans certaines représentations $E$-linéaires de $\Zp$. Ceci est un analogue en caractéristique $p$ de la notion de vecteur localement analytique dans les représentations de groupes de Lie $p$-adiques sur des Banach $p$-adiques.
\end{altabstract}


\datereceived{2022-03-17}
\daterevised{2022-07-03}
\dateaccepted{2022-08-03}

\editor{X. Caruso}

\begin{DefTralics}
\newcommand{\kf}{E}
\newcommand{\Zp}{\mathbf{Z}_p}
\newcommand{\dpar}[1]{(\!(#1)\!)}
\newcommand{\et}{\widetilde{\mathbf{E}}}
\end{DefTralics}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\dateposted{2022-11-24}
\begin{document}
\maketitle

%\setcounter{tocdepth}{2}
%\tableofcontents

\section*{Introduction}

Let $p$ be a prime number, and let $\kf$ be a field of characteristic $p$. Let $\efont = \kf \dpar{X}$, and let $\et$ be the $X$-adic completion of $\cup_{n\,\geq\,0} \kf \dpar{X^{1/p^n}}$. Note that if $\kf$ is perfect, the field $\et$ is perfectoid. The group $\Zp^\times$ acts on $\efont$ by $\left(a \cdot f\right)(X) = f((1+X)^a-1)$. This action extends to $\cup_{n \,\geq\,0} \kf \dpar{X^{1/p^n}}$ by $(a \cdot f)(X^{1/p^n}) = f((1+X^{1/p^n})^a-1)$, and by continuity to $\et$. The question that motivated this paper is the following.

\begin{enonce*}{Question}
Can we recover $\cup_{n\,\geq\,0} \kf \dpar{X^{1/p^n}}$ or even $\kf \dpar{X}$ from the data of the valued $E$-vector space $\et$ endowed with the action of $\Zp^\times$?
\end{enonce*}

In characteristic zero, it is possible to answer an analogous question by using Schneider and Teitelbaum's theory of locally analytic vectors in $p$-adic Banach representations of $p$-adic Lie groups. For characteristic $p$ representations, there is no such theory. One of the main contributions of this article is to introduce a characteristic $p$ analogue of locally analytic functions and vectors.

Let $M$ be an $\kf$-vector space, endowed with a valuation $\vm$ such that $\vm(xm) = \vm(m)$ if $x \in \kf^\times$. We assume that $M$ is separated and complete for the $\vm$-adic topology. For example, we will consider $M= \efont$ or $\et$ with the $X$-adic valuation. We say that a function $f :\Zp \to M$ is super-H\"older if there exist constants $\lambda,\mu \in \RR$ such that $\vm(f(x)-f(y)) \geq p^\lambda \cdot p^i +\mu$ whenever $\vp(x-y) \geq i$, for all $x,y \in \Zp$ and $i \geq 0$. These super-H\"older functions are the characteristic $p$ analogue of locally analytic functions $\Zp \to \Qp$. We prove an analogue of Mahler's theorem for continuous functions $f : \Zp \to M$, and give a characterization of super-H\"older functions in terms of their Mahler expansions. This is a characteristic $p$ analogue of a theorem of Amice.

Assume now that $\Gamma$ is a group that is isomorphic to $\Zp$ via a coordinate map $c$, and that $M$ is endowed with an $\kf$-linear action of $\Gamma$ by isometries. We say that $m \in M$ is a super-H\"older vector if the orbit map $z \mapsto c^{-1}(z) \cdot m$ is a super-H\"older function $\Zp \to M$. This definition is a characteristic $p$ analogue of the notion of locally analytic vector of a $p$-adic Banach representation of a $p$-adic Lie group. We let $M^{\Gamma\hyphen\sh,\lambda}$ denote the space of super-H\"older vectors for a given constant $\lambda$ as in the definition above. We also let $M^{\sh}$ denote the set of super-H\"older vectors in $M$. Our main result is a complete answer to the question above. Consider $M=\et$, endowed with the action of $\Gamma=1+p^k \Zp$ for $k \geq 1$ (or $k \geq 2$ if $p=2$).

%\begin{enonce*}{Theorem}
\begin{theo*}
For all $n \geq 0$, we have $\et^{(1+p^k \Zp) \hyphen\sh, k-n} = \kf \dpar{X^{1/p^n}}$.

In particular, $\et^{\sh} = \cup_{n\,\geq\,0} \kf \dpar{X^{1/p^n}}$.
\end{theo*}

The main ingredients of the proof of this theorem are some simple computations in $\kf \dcroc{X}$, as well as Colmez' analogue of Tate traces for $\et$.

We give several applications of our main result. First, we compute the perfectoid commutant of $\Aut(\Gm)$, namely the set of $u \in \et^{\vx\,>\,0}$ such that $u \circ \gamma_a = \gamma_a \circ u$ for all $a \in \Zp^\times$, where $\gamma_a(X) = (1+X)^a-1$. Using our main theorem, and a result of Lubin-Sarkis on the classical commutant of $\Aut(\Gm)$, we prove that such a $u$ is of the form $\gamma_b(X^{p^n})$ for some $b \in \Zp^\times$ and $n \in \ZZ$. Next we study $(\phi,\Gamma)$-modules over $\efont$. We prove that the action of $\Gamma$ on a $(\phi,\Gamma)$-module $\dfont$ is always super-H\"older, and deduce that $(\et \otimes_{\efont} \dfont)^{\sh} = (\cup_{n\,\geq\,0} \kf \dpar{X^{1/p^n}}) \otimes_{\efont} \dfont$. This allows us to extend our computation of super-H\"older vectors to the finite extensions of $\Fp \dpar{X}$ provided by Fontaine and Wintenberger's theory of the field of norms. We finish this article with a computation that suggests that the theory of super-H\"older vectors could have some applications to the $p$-adic local Langlands correspondence.

\subsection*{Acknowledgements}
We thank Juan Esteban Rodríguez Camargo for asking LB the question that motivated this paper, as well as Christophe Breuil, Daniel Gulotta, Gal Porat and the referee for their comments and questions.

\section{Super-H\"older functions and vectors}\label{secshf}

In this section, we define super-H\"older functions $\Zp \to M$ and super-H\"older vectors in $M$ when $M$ is a representation of a group isomorphic to $\Zp$. We prove an analogue of Mahler's theorem for continuous functions $\Zp \to M$, and give a characterization of super-H\"older functions in terms of their Mahler expansions.

\subsection{Super-H\"older functions}\label{subshf}

We keep the notation of the introduction. Let $M$ be an $\kf$-vector space, endowed with a valuation $\vm$ such that $\vm(xm) = \vm(m)$ if $x \in \kf^\times$. We assume that $M$ is separated and complete for the $\vm$-adic topology. For example, we will consider $M= \kf \dcroc{X}$ with the $X$-adic valuation.

Let $C^0(\Zp,M)$ denote the space of continuous functions $f : \Zp \to M$.

\begin{defi}\label{defsh}
We say that $f : \Zp \to M$ is super-H\"older if there exist constants $\lambda,\mu \in \RR$ such that $\vm(f(x)-f(y)) \geq p^\lambda \cdot p^i +\mu$ whenever $\vp(x-y) \geq i$, for all $x,y \in \Zp$ and $i \geq 0$.
\end{defi}

We let $\HH^{\lambda,\mu}(\Zp,M)$ denote the space of functions such that $\vm(f(x)-f(y)) \geq p^\lambda \cdot p^i +\mu$ whenever $\vp(x-y) \geq i$, for all $x,y \in \Zp$ and $i \geq 0$, and $\HH^\lambda(\Zp,M) = \cup_{\mu\,\in\,\RR} \HH^{\lambda,\mu}(\Zp,M)$ and $\HH(\Zp,M) = \cup_{\lambda\,\in\,\RR} \HH^\lambda(\Zp,M)$.

For example, if $M = \kf \dcroc{X}$ with $\vm=\vx$, then $[a \mapsto (1+X)^a] \in\HH^{0,0}(\Zp,M)$. Indeed, $(1+X)^a - (1+X)^{a + p^i b} = (1+X)^a (1-(1+X^{p^i})^b) \in X^{p^i} \kf \dcroc{X}$ if $i \geq 0$.


\begin{rema}\label{hcdcl}
The space $\HH^{\lambda,\mu}(\Zp,M)$ is closed in $C^0(\Zp,M)$.
\end{rema}

\begin{rema}\label{isom}
If $\alpha : \Zp \to \Zp$ is an isometry, then $f : \Zp \to M$ belongs to $\HH^{\lambda,\mu}(\Zp,M)$ if and only if $f \circ \alpha \in \HH^{\lambda,\mu}(\Zp,M)$
\end{rema}

\begin{prop}\label{hrng}
Suppose that $M$ is a ring, and that $\vm(mm') \geq \vm(m) + \vm(m')$ for all $m,m' \in M$. If $c \in \RR$, let $M_c= M^{\vm\,\geq\,c}$.
\begin{enumerate}
\item\label{prop1.4.1} If $f \in \HH^{\lambda,\mu}(\Zp,M_c)$ and $g \in \HH^{\lambda,\nu}(\Zp,M_d)$, and $\xi=\min(\mu+d,\nu+c)$, then $fg \in \HH^{\lambda,\xi}(\Zp,M_{c+d})$.
\item\label{prop1.4.2} If $\lambda,\mu \in \RR$, then $\HH^{\lambda,\mu}(\Zp,M_0)$ is a subring of $C^0(\Zp,M)$.
\item\label{prop1.4.3} If $\lambda \in \RR$, then $\HH^{\lambda}(\Zp,M)$ is a subring of $C^0(\Zp,M)$
\item\label{prop1.4.4} If $d \geq 1$, we see $\GL_d(M)$ as a subset of the valued $E$-vector space $\M_d(M)$. If $\lambda,\nu \in \RR$ and $Q \in \HH^{\lambda}(\Zp,\GL_d(M))$ are such that $\vm(\det Q(x)) \leq \nu$ for all $x \in \Zp$, then $Q^{-1} \in \HH^{\lambda}(\Zp,\GL_d(M))$.
\end{enumerate}
\end{prop}

\begin{proof}
Items~\eqref{prop1.4.2} and~\eqref{prop1.4.3} follow from item~\eqref{prop1.4.1}, which we now prove. If $x,y \in \Zp$, then
\[
(fg)(x) - (fg)(y) = \left(f(x)-f(y)\right)g(x) + \left(g(x)-g(y)\right)f(y),
\]
which implies the claim. We now prove (4). If $d=1$, then
\[
Q^{-1}(y) - Q^{-1}(x) = \frac{Q(x)-Q(y)}{Q(x)Q(y)},
\]
which implies the claim. If $d \geq 1$, we can write $Q^{-1} = \tco(Q) \cdot \det(Q)^{-1}$, and the claim results from~\eqref{prop1.4.3}, and~\eqref{prop1.4.4} applied to $d=1$.
\end{proof}

\begin{rema}\label{notting}
Take $u \in X + X^2 \kf \dcroc{X}$, and let $u^{\circ n}$ be $u$ composed with itself $n$ times. Sen's theorem (\cite[Theorem~1]{S69}) implies that $\vx(u^{\circ p^k}(X)-X) \geq p^k$ if $k \geq 0$, so that $\vx(u^{\circ x}-u^{\circ y}) \geq p^i$ if $\vp(x-y) \geq i$. This implies that the map $\ZZ_{\geq\,0} \to X + X^2 \kf \dcroc{X}$, given by $n \mapsto u^{\circ n}$, extends to a super-H\"older function on $\Zp$.
\end{rema}

\subsection{Super-H\"older vectors}\label{secvec}

We now assume that $M$ is endowed with an $\kf$-linear action by isometries of a group $\Gamma$, where $\Gamma$ is isomorphic to $\Zp$, via a coordinate map $c$. If $m \in M$, let $\orb_m : \Gamma \to M$ denote the function defined by $\orb_m(a) = a \cdot m$, so that $\orb_m \circ c^{-1}$ is a function $\Zp \to M$.

\begin{defi}\label{defshvec}
Let $M^{\Gamma\hyphen\sh,\lambda,\mu}$ denote the set of $m \in M$ such that $\orb_m \circ c^{-1} \in \HH^{\lambda,\mu}(\Zp,M)$, and let $M^{\Gamma\hyphen\sh,\lambda}$ and $M^{\Gamma\hyphen\sh}$ be the corresponding sub-$\kf$-vector spaces of $M$.
\end{defi}

This definition should be seen as a characteristic $p$ analogue of the locally analytic vectors of a Banach representation of a $p$-adic Lie group, as defined in~\cite[\S~7]{ST03}. The requirement that $\Gamma$ acts by isometries is the analogue of the condition that the norm be invariant.

\begin{rema}\label{mcont}
We assume that $\Gamma$ acts by isometries on $M$, but not that $\Gamma$ acts continuously on $M$, namely that $\Gamma \times M \to M$ is continuous. However, let $M^{\cont}$ denote the set of $m \in M$ such that $\orb_m \circ c^{-1} : \Zp \to M$ is continuous. It is easy to see that $M^{\cont}$ is a closed sub-$\kf$-vector space of $M$, and that $\Gamma \times M^{\cont} \to M^{\cont}$ is continuous (compare with~\cite[\S~3]{E17}). We then have $M^{\sh} \subset M^{\cont}$.
\end{rema}

\begin{lemm}\label{mshalt}
We have $m \in M^{\Gamma\hyphen\sh,\lambda,\mu}$ if and only if $\vm(g \cdot m - m) \geq p^\lambda \cdot p^i +\mu$ for all $g \in \Gamma$ such that $c(g) \in p^i \Zp$.
\end{lemm}

\begin{proof}
Since $\Gamma$ acts by isometries, we have $\vm(hg \cdot m - h \cdot m)= \vm (g \cdot m - m)$ for all $g,h \in \Gamma$.
\end{proof}

\begin{lemm}\label{cplmsh}
The space $M^{\Gamma \hyphen \sh,\lambda,\mu}$ is a closed sub-$\kf$-vector space of $M$.
\end{lemm}

\begin{lemm}\label{shsbgr}
If $k \geq 0$ and $\Gamma' = c^{-1}(p^k \Zp)$, then $g \mapsto c(g)/p^k$ is a coordinate on $\Gamma'$, and $M^{\Gamma \hyphen \sh,\lambda} = M^{\Gamma'\hyphen \sh,\lambda+k}$.
\end{lemm}

\begin{proof}
It is clear that $M^{\Gamma \hyphen \sh,\lambda} \subset M^{\Gamma' \hyphen \sh,\lambda+k}$. Conversely, let $C= \{1,\,\hdots,\,p^k-1\}$. If $m \in M^{\Gamma' \hyphen \sh,\lambda+k,\mu}$, let $\nu = \min_{c(h)\,\in\,C} \vm(h \cdot m - m)$. If $g \in \Gamma \setminus \Gamma'$, we can write $g=g_k h$ with $c(h) \in C$ and $g_k \in \Gamma'$. We then have $g \cdot m - m = (g_k \cdot h \cdot m - g_k \cdot m) + (g_k \cdot m - m)$ so that $\vm(g \cdot m-m) \geq \min(\mu,\nu)$.

This implies that $m \in M^{\Gamma \hyphen \sh,\lambda,\mu'}$ with $\mu' = \min(\mu,\nu)-p^{k+\lambda}$.
\end{proof}

In particular, the space $M^{\Gamma'\hyphen\sh}$ does not depend on the choice of open subgroup $\Gamma' \subset \Gamma$, and we denote it by $M^{\sh}$.

\begin{prop}\label{propmsh}
Suppose that $M$ is a ring, and that $g(mm') = g(m)g(m')$ and $\vm(mm') \geq \vm(m) + \vm(m')$ for all $m,m' \in M$ and $g \in \Gamma$.
\begin{enumerate}
\item\label{prop1.11.1} If $v \in \RR$ and $m,m' \in M^{\Gamma \hyphen \sh,\lambda,\mu} \cap M^{\vm\,\geq\,v}$, then $m \cdot m' \in M^{\Gamma \hyphen \sh,\lambda,\mu+v}$;
\item\label{prop1.11.2} If $m \in M^{\Gamma \hyphen \sh,\lambda,\mu} \cap M^\times$, then $1/m \in M^{\Gamma \hyphen \sh,\lambda,\mu-2 \vm(m)}$.
\end{enumerate}
\end{prop}

\begin{proof}
Item~\eqref{prop1.11.1} follows from Proposition~\ref{hrng} and Lemma~\ref{mshalt}. Item~\eqref{prop1.11.2} follows from
\[
g\left(\frac 1m\right) - \frac 1m = \frac{m-g(m)}{g(m) m}.\qedhere
\]
\end{proof}

\begin{rema}\label{jerclocan}
One can extend the definition of super-H\"older vectors to the setting of a $p$-adic Lie group $G$ acting by isometries on a valued $\kf$-vector space $M$ as follows (the details are in our paper \emph{Super-H\"older vectors and the field of norms}). Let $P$ be a nice enough open pro-$p$ subgroup of $G$. We say that $m \in M$ is super-H\"older if and only if there exists $\lambda,\mu \in \RR$ and $e>0$ such that $\vm(g\cdot m - m) \geq p^{\lambda + ei} + \mu$ whenever $g \in P^{p^i}$, for all $i \geq 0$. Juan Esteban Rodríguez Camargo pointed out to us that there is a similar purely metric characterization of locally analytic vectors for a $p$-adic Lie group acting on a Banach space.
\end{rema}

\subsection{Mahler's theorem}\label{secmahl}

In this section, we prove a characteristic $p$ analogue of Mahler's theorem for continuous functions $\Zp \to \Qp$. We then give a characterization of super-H\"older functions in terms of their Mahler expansions. If $z \in \Zp$ and $n \geq 0$, then $\binom{z}{n} \in \Zp$ and we still denote by $\binom{z}{n}$ its image in $\Fp$.

\begin{theo}\label{mahlthm}
If $\{m_n\}_{n\,\geq\,0}$ is a sequence of $M$ such that $m_n \to 0$, the function $f : \Zp \to M$ given by $f(z) = \sum_{n\,\geq\,0} \binom{z}{n} m_n$ belongs to $C^0(\Zp,M)$. We have 
\[
m_n = (-1)^n \sum_{i=0}^n(-1)^i\binom{n}{i} f(i)\quad\text{and}\quad\inf_{z\,\in\, \Zp} \vm(f(z)) =\inf_{n\,\geq\,0} \vm(m_n).
\]

Conversely, if $f \in C^0(\Zp,M)$, there exists a unique sequence $\{m_n(f)\}_{n\,\geq\,0}$ such that $m_n(f) \to 0$ and such that $f(z) = \sum_{n\, \geq\,0} \binom{z}{n} m_n(f)$.
\end{theo}

\begin{proof}
Our proof follows Bojanic's proof (cf~\cite{B74}) of Mahler's theorem. The first part of the theorem is easy: $f$ is continuous since it is a uniform limit of continuous functions, and if $f(z) = \sum_{n\,\geq\,0} \binom{z}{n} m_n$, then $\vm(f(z)) \geq \inf_{n\,\geq\,0} \vm(m_n)$. The fact that 
\[
m_n = (-1)^n \sum_{i=0}^n(-1)^i\binom{n}{i} f(i)
\]
is a classical exercise, given that $f(k) = \sum_{j=0}^k \binom{k}{j} m_j$ for all $k \geq 0$, and it implies that $\vm(m_n) \geq \inf_{z\,\in\,\Zp} \vm(f(z))$ for all $n$. In order to show the converse, it is enough to show that if $f$ is continuous and 
\[
m_n(f) = (-1)^n \sum_{i=0}^n(-1)^i\binom{n}{i} f(i),
\]
then $m_n(f) \to 0$. Indeed, the functions $f$ and $z \mapsto \sum_{n \geq 0} \binom{z}{n} m_n(f)$ are then two continuous functions on $\Zp$ with the same values on $\ZZ_{\geq 0}$, so that they are equal.

We now show that $m_n(f) \to 0$. If $s \geq 0$, there exists $t$ such that if $\vp(x-y) \geq t$ then $\vm(f(x)-f(y)) \geq s$, as $f$ is uniformly continuous. Take $n \geq p^t$ and write $n = qp^t+r$ with $0 \leq r < p^t$ and $q \geq 1$. Writing $i=a+jp^t$, we get
\[
m_n(f) = \sum_{a=0}^{p^t-1}\sum_{j = 0}^q (-1)^{n+a+jp^t}\binom{n}{a+jp^t}f\left(a+jp^t\right).
\]
As we are in characteristic $p$, Lucas' theorem implies that $\binom{n}{a+jp^t} =
\binom{r}{a}\binom{q}{j}$, so that:
\[
m_n(f) =
\sum_{a=0}^{p^t-1}(-1)^{n+a}\binom{r}{a}
\left(\sum_{j=0}^q (-1)^j\binom{q}{j}f\left(a+jp^t\right)\right).
\]
As $(\sum_{j=0}^q (-1)^j\binom{q}{j}) \cdot f(a)=0$, and $\vm(f(a+jp^t)-f(a)) \geq s$ for all $j$, we get that $\vm(m_n(f)) \geq s$ if $n \geq p^t$.
\end{proof}

We now give a characterization of super-H\"older functions in terms of their Mahler expansions.

\begin{prop}\label{shmahl}
If $f \in C^0(\Zp,M)$, then $f \in \HH^{\lambda,\mu}(\Zp,M)$ if and only if for all $i \geq 0$, we have $\vm(m_n(f)) \geq p^\lambda \cdot p^i +\mu$ whenever $n \geq p^i$.
\end{prop}

\begin{proof}
Take $f \in C^0(\Zp,M)$ such that $\vm(m_n(f)) \geq p^\lambda \cdot p^i +\mu$ whenever $n \geq p^i$. Recall that if $a \in \Zp$ and $i \geq 1$, then for all $j < p^i$ we have $\binom{a}{j} = \binom{a+p^i}{j}$ in $\Fp$. If $z \in \Zp$ and $i \geq 1$, then
\begin{align*}
f\left(z+p^i\right)-f(z) &=
\sum_{n\,\geq\,0} m_n(f) \left(\binom{z+p^i}{n}-\binom{z}{n}\right)\\
&=
\sum_{n\,\geq\,p^i}m_n(f)\left(\binom{z+p^i}{n}-\binom{z}{n}\right).
\end{align*}
Since $\vm(m_n(f)) \geq p^\lambda \cdot p^i +\mu$ whenever $n \geq p^i$, the formula above implies that $\vm(f(x+p^i)-f(x)) \geq p^\lambda \cdot p^i +\mu$. Iterating this, we get that $\vm(f(x+kp^i)-f(x))\linebreak \geq p^\lambda \cdot p^i +\mu$ for all $k \in \ZZ_{\geq\,0}$. By continuity, this implies that $\vm(f(y)-f(x)) \geq p^\lambda \cdot p^i +\mu$ for all $x,y \in \Zp$ such that $\vp(y-x) \geq i$.

Assume now that $f \in \HH^{\lambda,\mu}(\Zp,M)$. We prove that for all $i \geq 0$ and $n \geq p^i$, we have $\vm(m_n(f)) \geq p^\lambda \cdot p^i +\mu$. Fix $i \geq 0$ and take $a \in \{0,\dots,p^i-1\}$. Define a function $g$ on $\Zp$ by $g(z) = f(a+p^i z)-f(a)$. By hypothesis, we have $\vm(g(z)) \geq p^\lambda \cdot p^i +\mu$ for all $z$. This implies that $\vm(m_n(g)) \geq p^\lambda \cdot p^i +\mu$ for all $n$. We now compute $m_n(g)$. We have
\begin{align*}
g(z) &=
\sum_{n\,\geq\,0}\left(\binom{a+p^iz}{n}-\binom{a}{n}\right)m_n(f) \\
&=
\sum_{n\,\geq\,p^i}\left(\binom{a+p^iz}{n}-\binom{a}{n}\right)m_n(f) =
\sum_{n\,\geq\,p^i}\binom{a+p^iz}{n}m_n(f),
\end{align*}
since $a \leq p^i-1$. If we write $n = t+p^i \ell$, with $0 \leq t \leq p^i-1$ and $\ell \geq 1$, then $\binom{a+p^i z}{n} = \binom{a}{t}\binom{z}{\ell}$. This implies that
\[
g(z) =\sum_{t=0}^{p^i-1}\sum_{\ell\,\geq\,1}\binom{a}{t}\binom{z}{\ell}m_{t+p^i\ell}(f),
\]
which gives $m_\ell(g) =
\sum_{t=0}^{p^i-1}\binom{a}{t}m_{t+p^i\ell}(f)$ for all $\ell \geq 1$. This now implies that
\[
\vm\left(\sum_{t=0}^{p^i-1}\binom{a}{t}m_{t+p^i\ell}(f)\right) \geq p^\lambda \cdot p^i +\mu
\]
for all $\ell \geq 1$ and $a \in \{0,\,\dots,\,p^i-1\}$. The matrix $(\binom{a}{t})_{0\,\leq\,a,t\,\leq\,p^i-1}$ is unipotent with integral coefficients. Hence for a given $\ell \geq 1$, the above inequality implies that $\vm(m_{a+p^i\ell}(f)) \geq p^\lambda \cdot p^i +\mu$ for all $a \in \{0,\,\dots,\,p^i-1\}$. Writing $n \geq p^i$ as $n=a+p^i \ell$, we get $\vm(m_n(f)) \geq p^\lambda \cdot p^i +\mu$ for all $n \geq p^i$.
\end{proof}

\begin{rema}\label{mahlvssh}
Let $\WW^{\lambda,\mu}(\Zp,M)$ denote the set of $f \in C^0(\Zp,M)$ such that $\vm(m_n(f)) \geq p^\lambda n +\mu$ for all $n \geq 0$.

Proposition~\ref{shmahl} implies that $\WW^{\lambda,\mu}(\Zp,M) \subset \HH^{\lambda,\mu}(\Zp,M) \subset \WW^{\lambda-1,\mu}(\Zp,M)$.
\end{rema}

Proposition~\ref{shmahl} and Remark~\ref{mahlvssh} strengthen the analogy between our definition of super-H\"older functions and the classical theory of locally analytic functions. Indeed, if $f : \Zp \to
\Qp$ is a continuous function, and if $f(z) = \sum_{n\,\geq\,0}
\binom{z}{n} m_n(f)$ is its Mahler expansion, then by a result of Amice (\cite{A64}, see~\cite[Corollary~I.4.8]{C10}), $f$ is locally analytic if and only if there exists $\lambda,\mu \in \RR$ such that $\vp(m_n(f))
\geq p^\lambda \cdot n + \mu$ for all $n \geq 0$.

\begin{rema}\label{gulotta}
Daniel Gulotta pointed out to us that Gulotta (in~\cite[\S~3]{G19}), as well as Johansson and Newton (in~\cite[\S~3.2]{JN19}), had defined a generalization of locally analytic functions, for functions valued in certain general Tate $\Zp$-algebra. When $p=0$ in the algebra, their definition is equivalent to our definition of super-H\"older functions.
\end{rema}

\section{Decompletion of cyclotomic perfectoid fields}\label{secdecomp}

Let $\efont^+ = \kf \dcroc{X}$. For $n \geq 0$, let $\efont^+_n = \kf
\dcroc{X^{1/p^n}}$, so that $\efont^+ = \efont^+_0$. Let $\efont^+_\infty = \cup_{n\,\geq\,0} \,\efont^+_n$ and let $\et^+$ be the $X$-adic completion of $\efont^+_\infty$. We denote by $\efont$, $\efont_n$, $\efont_\infty$, $\et$ the fields $\efont^+[1/X]$, $\efont_n^+[1/X]$, $\efont_\infty^+[1/X]$, $\et^+[1/X]$ respectively. The ring $\et^+$ is the ring of integers of the field $\et = \et^+[1/X]$. If $\kf$ is perfect, then $\et$ is perfectoid.

\subsection{The action of \texorpdfstring{$\Zp^\times$}{Zp x }}\label{secgam}

The group $\Zp^\times$ acts continuously by isometries on each $\efont^+_n$ by the formula $a \cdot X^{1/p^n} = (1+X^{1/p^n})^a-1$. This action is compatible when $n$ varies, extends to the fields $\efont_n$, and extends by continuity to $\et^+$ and $\et$.

\begin{rema}\label{perfcyc}
If $\kf=\Fp$, then $\et$ is the tilt of $\widehat{\Qp(\mu_{p^\infty})}$ (see \S~\ref{secfon} for more details). The group $\Gamma=\Gal(\Qp(\mu_{p^\infty})/\Qp)$ is isomorphic to $\Zp^\times$ via the cyclotomic character $\chi_{\cyc}$, and acts on $\et$ by $g(f) = \chi_{\cyc}(g) \cdot f$.
\end{rema}

If $k \geq 1$ (or $k \geq 2$ if $p=2$), let $\Gamma_k = 1+p^k \Zp$. The natural coordinate on $\Gamma_k$ is given by $1+p^k a \mapsto \log_p(1+p^k a) / p^k$. It differs from the coordinate $1+p^k a \mapsto a$ (which is not a group homomorphism) by an isometry. By Remark~\ref{isom}, the definition of $(\et^+)^{\Gamma_k \hyphen \sh,\lambda,\mu}$ does not depend on the choice of one of those coordinates, and we use $1+p^k a \mapsto a$.

\begin{prop}\label{etnsh}
We have $\efont^+_n = (\efont^+_n)^{\Gamma_k \hyphen \sh,k-n,0}$.
\end{prop}

\begin{proof}
We have $(1+X^{1/p^n})^{1+p^{k+i}b} = (1+X^{1/p^n}) \cdot (1+X^{p^{k+i-n}})^b$, so that
\[
\vx\left(\left(1+X^{1/p^n}\right)^{1+p^{k+i}b} - \left(1+X^{1/p^n}\right)\right) \geq p^{k-n} \cdot p^i.
\]
This implies that $X^{1/p^n} \in (\efont^+_n)^{\Gamma_k \hyphen \sh,k-n,0}$. The claim now follows from Proposition~\ref{propmsh} and Lemma~\ref{cplmsh}.
\end{proof}

Taking $n=0$ in Proposition~\ref{etnsh}, we find that $\kf \dcroc{X} = \kf \dcroc{X}^{\Gamma_k \hyphen \sh,k}$. Let $\efont=\efont^+[1/X]$.

\begin{coro}\label{egsk}
We have $\efont = \efont^{\Gamma_k \hyphen \sh,k}$.
\end{coro}

\begin{proof}
This follows from Propositions~\ref{etnsh} and~\ref{propmsh}.
\end{proof}

\begin{prop}\label{levelzer}
If $\eps>0$, then $\kf \dcroc{X}^{\Gamma_k \hyphen \sh,k + \eps} \subset \kf \dcroc{X^p}$.
\end{prop}

\begin{proof}
Take $f(X) \in \kf \dcroc{X}$. There is a power series $h(Y,Z) \in \kf \dcroc{Y,Z}$ such that
\[
f(Y+Z) = f(Y) + Z \cdot f'(Y) + Z^2 \cdot h(Y,Z).
\]
If $m \geq 0$, this implies that
\begin{align*}
f\left((1+X)^{1+p^m}-1\right) & = f \left(X+X^{p^m}(1+X)\right) \\
& = f(X) + X^{p^m}(1+X) \cdot f'(X) + \bigO\left(X^{2p^m}\right).
\end{align*}
If $f(X) \notin \kf \dcroc{X^p}$, then $f'(X) \neq 0$. Let $\mu=\vx(f'(X))$. The above computations imply that $\vx((1+p^{i+k}) \cdot f(X) - f(X)) = p^{i+k}+\mu$ for $i \gg 0$. This implies the claim.
\end{proof}

\begin{coro}\label{etshlevel}
We have $(\efont^+_\infty)^{\Gamma_k \hyphen \sh, k-n} = \efont^+_n$.
\end{coro}

\begin{proof}
Take $f(X^{1/p^m}) \in (\efont^+_\infty)^{\Gamma_k \hyphen \sh, k-n}$ where $f(X) \in \kf \dcroc{X}$. Since $\vx(h^p) = p \cdot \vx(h)$ for all $h \in \et^+$, we have $f^{p^m}(X) \in (\efont^+_\infty)^{\Gamma_k\hyphen \sh, k+m-n}$, where $f^{p^m}(X) \in \kf \dcroc{X}$ is $f^{p^m}(X) = f(X^{1/p^m})^{p^m}$. If $m \geq n+1$, then Proposition~\ref{levelzer} implies that $f^{p^m}(X) \in \kf \dcroc{X^p}$, so that $f(X) = g(X^p)$, and $f(X^{1/p^m}) = g(X^{1/p^{m-1}})$. This implies the claim.
\end{proof}

\subsection{Tate traces}\label{sectate}

We recall some constructions of Colmez (see~\cite[\S~8.2]{C08}). For $m \geq 0$ let $I_m = p^{-m}\ZZ\cap [0,1)$, and let $I = \cup_m I_m$. Note that if $i
\in I_m$, then $(1+X)^i \in \efont^+_m$.

\begin{lemm}
The elements $(1+X)^i$, $i\in I_m$, form a basis of $\efont^+_m$ over $\efont^+_0$.
\end{lemm}

\begin{proof}
See~\cite[Lemma~8.2]{C08}. Colmez works with $\kf = \Fp$, but the proofs are the same with arbitrary coefficients.
\end{proof}

\begin{prop}\label{colmet+}
Any $f \in \et^+$ can be written uniquely as $\sum_{i\,\in\,I}(1+X)^ia_i(f)$, with $a_i(f) \in \efont^+_0$, and $a_i(f) \to 0$. Moreover, $\vx(f)-1 < \inf_{i\,\in\,I} \vx(a_i(f))\linebreak \leq \vx(f)$.
\end{prop}

\begin{proof}
See~\cite[Props 4.10 and 8.3]{C08}.
\end{proof}

In particular, for all $i \in I$, the map $\et^+ \to \efont^+_0$, given by $f \mapsto a_i(f)$ is continuous.

\begin{prop}\label{colmtn}
There exists a family $\{ T_n \}_{n\,\geq\,0}$ of continuous maps $T_n : \et^+ \to \efont^+_n$ satisfying the following properties:
\begin{enumerate}
\item\label{prop2.8.1} The restriction of $T_n$ to $\efont^+_n$ is the identity map.
\item We have $T_n(f) \to f$ as $n \to +\infty$.
\item\label{prop2.8.2} We have $\vx(T_n(f)) \geq \vx(f)-1$ for all $n$.
\item\label{prop2.8.3} Each $T_n$ is $\Zp^\times$-equivariant.
\end{enumerate}
\end{prop}

\begin{proof}
If $f = \sum_{i\,\in\,I}(1+X)^ia_i(f)$, let $T_n(f) = \sum_{i\,\in\,I_n}(1+X)^ia_i(f)$. With this definition, the first property is immediate. The second and third one follow from Proposition~\ref{colmet+}.

For the last one, observe that if $i \in I$ and $g \in \Zp^\times$, then $g
\cdot (1+X)^i =(1+X)^{ig}$ so $g \cdot (1+X)^i$ can be written uniquely as $(1+X)^{\sigma_g(i)}u_{i,g}(X)$ with $\sigma_g(i) \in I$ and $u_{i,g}(X) \in \efont^+_0$. The map $\sigma_g$ induces a bijection from $I_m$ to itself for all $m$. Take $f \in \et^+$, and write $f = \sum_{i\,\in\,I}(1+X)^ia_i(f)$. We have 
\begin{align*}
g \cdot f &= \sum_{i\,\in\,I}(1+X)^{\sigma_g(i)}u_{i,g}(X)(g\cdot a_i(f)),\\
\intertext{so that}
T_n(g \cdot f) &= \sum_{i\,\in\,I_n}(1+X)^{\sigma_g(i)}u_{i,g}(X)(g\cdot a_i(f)) = g \cdot T_n(f).\qedhere
\end{align*}
\end{proof}

\subsection{Decompletion of \texorpdfstring{$\et$}{et}}\label{subdec}

We now prove that $\et^{\sh} = \efont_\infty$. More precisely, we have the following result.

\begin{theo}\label{shdecet}
We have $\et^{\Gamma_k\hyphen\sh, k-m} = \efont_m$ for all $m \geq 0$, and $\et^{\sh} = \efont_\infty$.
\end{theo}

\begin{prop}\label{tnsh}
If $f \in (\et^+)^{\Gamma_k\hyphen \sh, \lambda,\mu}$, then $T_n(f) \in (\efont^+_n)^{\Gamma_k \hyphen \sh, \lambda,\mu-1}$.
\end{prop}

\begin{proof}
If $g \in \Gamma_k$, then $g(T_n(f)) - T_n(f) = T_n(g(f)-f)$ so that
\[
\vx\left(g(T_n(f)) - T_n(f)\right) = \vx \left(T_n(g(f)-f)\right) \geq \vx(g(f)-f)-1
\]
by Proposition~\ref{colmtn}. This implies the claim.
\end{proof}

\begin{proof}[Proof of Theorem~\ref{shdecet}]
Take $f \in (\et^+)^{\Gamma_k \hyphen \sh, k-m}$. By Proposition~\ref{tnsh}, we have $T_n(f) \in (\efont^+_n)^{\Gamma_k \hyphen \sh, k-m}$ for all $n \geq 0$. By Corollary~\ref{etshlevel}, $T_n(f) \in \efont^+_m$ for all $n$. Since $T_n(f) \to f$ as $n \to +\infty$, we have $f \in \efont^+_m$.

Hence $(\et^+)^{\Gamma_k\hyphen\sh, k-m} = \efont^+_m$, and this implies the theorem by Proposition~\ref{propmsh}.
\end{proof}

\section{Applications}\label{secappli}

We now give several applications of the fact that $\et^{\sh} = \efont_\infty$.

\subsection{The perfectoid commutant of \texorpdfstring{$\Aut(\Gm)$}{Aut(Gm)}}\label{secperfcom}

In this section, we assume that $\kf=\Fp$. If $a \in \Zp^\times$, let $\gamma_a(X) = (1+X)^a-1 \in \Fp \dcroc{X}$. Note that if $f \in \et$, then $a \cdot f = f \circ \gamma_a$. If $u \in \et^+$ is such that $\vx(u)>0$, the series $\gamma_a \circ u$ converges in $\et^+$. If $u = \gamma_b(X^{p^n})$ for some $b \in \Zp^\times$ and $n \in \ZZ$, then $u \circ \gamma_a = \gamma_a \circ u$ for all $a \in \Zp^\times$.

\begin{theo}\label{gmcom}
If $u \in \et^+$ is such that $\vx(u) > 0$ and $u \circ \gamma_a = \gamma_a \circ u$ for all $a \in \Zp^\times$, then there exists $b \in \Zp^\times$ and $n \in \ZZ$ such that $u(X) = \gamma_b(X^{p^n})$.
\end{theo}

Recall that a power series $f(X) \in \Fp\dcroc{X}$ is separable if $f'(X)
\neq 0$. If $f(X) \in X \cdot \Fp\dcroc{X}$, we say that $f$ is invertible if $f'(0) \in \Fp^\times$, which is equivalent to $f$ being invertible for composition (denoted by $\circ$). We say that $w(X) \in X\cdot \Fp\dcroc{X}$ is nontorsion if $w^{\circ n}(X) \neq X$ for all $n
\geq 1$. The following is a reformulation of~\cite[Lemma~6.2]{L94}.

\begin{lemm}\label{lubnarch}
Let $w(X) \in X + X^2\cdot \Fp\dcroc{X}$ be an invertible nontorsion series, and let $f(X) \in X \cdot \Fp\dcroc{X}$ be a separable power series. If $w \circ f = f \circ w$, then $f$ is invertible.
\end{lemm}

\begin{lemm}\label{locansh}
If $u \in \et^+$ is such that $\vx(u) > 0$ and $u \circ \gamma_a = \gamma_a \circ u$ for all $a \in \Zp^\times$, then $u \in (\et^+)^{\sh}$.
\end{lemm}

\begin{proof}
The group $\Zp^\times$ acts on $\et^+$ by $a \cdot u = u \circ \gamma_a$, so we need to check that the function $a \mapsto \gamma_a \circ u$ is super-H\"older. This is clear since 
\[
\gamma_a(u) = \sum_{n\,\geq\,1} \binom{a}{n} u^n\quad \text{and}\quad \vx(u)>0.\qedhere
\]
\end{proof}

\begin{proof}[Proof of Theorem~\ref{gmcom}]
Take $u \in \et^+$ such that $\vx(u) > 0$ and $u \circ \gamma_a = \gamma_a \circ u$ for all $a \in \Zp^\times$. By Lemma~\ref{locansh} and Theorem~\ref{shdecet}, there exists $m \geq 0$ such that $u \in \efont^+_m$. Hence there is an $n \in \ZZ$ such that $f(X) = u(X^{1/p^n})$ belongs to $X \cdot \Fp\dcroc{X}$ and is separable. Take $g \in 1+p \Zp$ such that $g$ is nontorsion, and let $w(X) = \gamma_g(X)$ so that $u \circ w = w \circ u$. We also have $f \circ w = w \circ f$. By Lemma~\ref{lubnarch}, $f$ is invertible. Since $f \circ \gamma_a = \gamma_a \circ f$ for all $a \in \Zp^\times$, \cite[Theorem~6]{LS07} implies that $f \in \Aut(\Gm)$. Hence there exists $b \in \Zp^\times$ such that $f(X)=\gamma_b(X)$. This implies the theorem.
\end{proof}

\subsection{Decompletion of \texorpdfstring{$(\phi,\Gamma)$}{(phi, Gamma)}-modules}\label{secpgm}

Let $\Gamma_k = 1+p^k \Zp$ with $k \geq 1$, as in \S~\ref{secgam}. Let $M$ be a finite-dimensional $\efont$-vector space with a continuous semi-linear action of $\Gamma_k$.

\begin{prop}\label{gamlat}
There is an $\efont^+$-lattice in $M$ that is stable under $\Gamma_k$.
\end{prop}

\begin{proof}
Choose any lattice $M_0^+$ of $M$. The map $\pi : \Gamma_k \times M \to M$ is continuous, so there is an open subgroup $H$ of $\Gamma_k$ and an $n \geq 0$ such that $\pi^{-1}(M_0^+)$ contains $H \times X^n M_0^+$. In particular, $h(m) \in X^{-n} M_0^+$ for all $h \in H$ and $m \in M_0^+$. Since $H$ is open in the compact group $\Gamma_k$, it is of finite index, and there exists $d \geq n$ such that $g(m) \subset X^{-d} M_0^+$ for all $g \in \Gamma_k$ and $m \in M_0^+$. The space $M^+ = \sum_{g\,\in\,\Gamma_k} g(M_0^+)$ is an $\efont^+$-module such that $M_0^+ \subset M^+ \subset X^{-d} M_0^+$, so that $M^+$ is a lattice of $M$. It is clearly stable under $\Gamma_k$.
\end{proof}

Choosing such an $\efont^+$-lattice in $M$ defines a valuation $\vm$ on $M$, such that $\Gamma_k$ acts on $M$ by isometries. We make such a choice, and we can therefore define $M^{\sh}$ and $M^{\Gamma_k\hyphen\sh,\lambda}$ as in Definition~\ref{defshvec}. We say that the action of $\Gamma_k$ on $M$ is super-H\"older if $M=M^{\sh}$.

\begin{lemm}\label{mshsub}
The space $M^{\Gamma_k\hyphen\sh,\lambda}$ does not depend on the choice of $\Gamma_k$-stable lattice of $M$. If $\lambda \leq k$ then $M^{\Gamma_k\hyphen\sh,\lambda}$ is sub-$\efont$-vector space of $M$.
\end{lemm}

\begin{proof}
The first assertion results from the fact that if we choose two $\efont^+$-lattices $M_1^+$ and $M_2^+$ in $M$, then there exists a constant $C$ such that $|{\val_1-\val_2} | \leq C$.

Next, recall that by Corollary~\ref{egsk}, $\efont =
\efont^{\Gamma_k\hyphen\sh,k}$. If $m \in M^{\sh,\lambda}$, $f \in\efont$, and $g \in \Gamma_k$, then $g(fm)-fm = g(f)(g(m)-m) + (g(f)-f)m$, so that $fm \in M^{\sh,\lambda}$ by Lemma~\ref{mshalt}.
\end{proof}

Lemma~\ref{mshsub} implies that $M^{\sh}$ is a sub-$\efont$-vector space of $M$. We say that a basis of $M$ is good if it generates a lattice that is stable under $\Gamma_k$.

\begin{prop}\label{mshbas}
Take $\lambda \leq k$ and fix a good basis of $M$. We have $M=M^{\Gamma_k\hyphen\sh,\lambda}$ if and only if the map $\Gamma_k \to \M_n(\efont^+)$, given by $g \mapsto \Mat(g)$, is in $\HH^\lambda(\Gamma_k,\M_n(\efont^+))$.
\end{prop}

\begin{proof}
We fix a good basis $(m_1,\,\dots,\,m_n)$ of $M$, and work with the corresponding valuation $\vm$ on $M$. By Lemma~\ref{mshsub}, we have $M=M^{\Gamma_k\hyphen\sh,\lambda}$ if and only if $m_j \in M^{\Gamma_k\hyphen\sh,\lambda}$ for all $j$. We have $g \cdot m_j = \sum_{i=1}^n\Mat(g)_{i,j} m_i$ by definition of $\Mat(g)$. Hence if $g,h \in \Gamma_k$, then $g \cdot m_j - h \cdot m_j = \sum_{i=1}^n (\Mat(g)_{i,j} - \Mat(h)_{i,j}) m_i$. This implies that if $\ell \geq 0$ and $\mu \in \RR$, then $\vm(g \cdot m_j - h \cdot m_j) \geq p^{\lambda+\ell} + \mu$ if and only if $\vx(\Mat(g)-\Mat(h)) \geq p^{\lambda+\ell} + \mu$. This implies the claim.
\end{proof}

If $M$ is a finite-dimensional $\efont$-vector space with a semi-linear action of $\Gamma_k$, then $\et \otimes_{\efont} M$ is a finite-dimensional $\et$-vector space with a semi-linear action of $\Gamma_k$. If $M$ is super-H\"older, there exists $m_0 = m_0(M) \geq 0$ such that $M=M^{\Gamma_k\hyphen\sh,k-m_0}$

\begin{prop}\label{vecbash}
If $M$ is super-H\"older and $m \geq m_0(M)$, then we have $(\et \otimes_{\efont} M)^{\Gamma_k\hyphen\sh,k-m} = \efont_m \otimes_{\efont} M$.
\end{prop}

\begin{proof}
By the same argument as in the proof of Lemma~\ref{mshsub}, we see that for $m \geq m_0$, $(\et \otimes_{\efont} M)^{\Gamma_k\hyphen\sh,k-m}$ is a sub-$\efont_m$-vector space of $\et \otimes_{\efont} M$. The space $(\et \otimes_{\efont} M)^{\Gamma_k\hyphen\sh,k-m}$ contains $M$, and therefore also $\efont_m \otimes_{\efont} M$. This proves one inclusion.

We now prove that $(\et \otimes_{\efont} M)^{\Gamma_k\hyphen\sh,k-m} \subset \efont_m \otimes_{\efont} M$. Fix a good basis $(m_1,\,\dots,\,m_n)$ of $M$, the corresponding valuation $\vm$ on $\et \otimes_{\efont} M$, and $m \geq m_0$. Take $x = \sum_{i=1}^n x_i m_i \in \et \otimes_{\efont} M$ and write $g(x) = \sum_{i=1}^n f_i(g) m_i$. We have $x \in (\et \otimes_{\efont} M)^{\Gamma_k\hyphen\sh,k-m}$ if and only if $f_i \in \HH^{k-m}(\Gamma_k,\et)$ for all $i$. In addition, $g(x) = \sum_{i,j} g(x_i) \Mat(g)_{j,i} m_j$. Hence $f_j : g \mapsto \sum_{i=1}^n g(x_i) \Mat(g)_{j,i}$ belongs to $\HH^{k-m}(\Gamma_k,\et)$ for all $j$. We have $g(x_\ell) = \sum_{j=1}^n f_j(g) (\Mat(g)^{-1})_{\ell,j}$. By Propositions~\ref{mshbas} and~\ref{hrng}, $[g \mapsto g(x_\ell)] \in \HH^{k-m}(\Gamma_k,\et)$ and therefore $x_\ell \in \et^{\Gamma_k\hyphen\sh,k-m} = \efont_m$ for all $\ell$.
\end{proof}

\begin{coro}\label{vecbashcor}
If $M$ is super-H\"older, then $(\et \otimes_{\efont} M)^{\sh} =
\efont_\infty \otimes_{\efont} M$.
\end{coro}

The field $\efont = \kf \dpar{X}$ is equipped with its action of $\Zp^\times$ and with the $\kf$-linear Frobenius map $\phi$ given by $\phi(f)(X) = f(X^p)$. Let $\Gamma = \Gamma_k$ with $k \geq 1$. A $(\phi,\Gamma)$-module $\dfont$ over $\efont$ is a finite-dimensional $\efont$-vector space, endowed with commuting, semi-linear actions of $\phi$ and $\Gamma$, such that the action of $\Gamma$ is continuous and such that $\Mat(\phi)$ is invertible (in any basis of $\dfont$).

\begin{prop}\label{phigsh}
If $\dfont$ is a $(\phi,\Gamma)$-module over $\efont$, then $\dfont=\dfont^{\Gamma_k\hyphen\sh,k}$.
\end{prop}

\begin{lemm}\label{hmnep}
If $\ell \geq 1$ and $\lambda,\mu \in \RR$, then $\HH^{\lambda,\mu}(\Gamma_\ell,\M_n(\efont^+))$ is a ring, that is stable under $\phi$.
\end{lemm}

\begin{proof}
The first claim follows from Proposition~\ref{hrng}. The second one follows from the fact that if $M \in \M_n(\efont^+)$, then $\vx(\phi(M)) \geq \vx(M)$.
\end{proof}

\begin{proof}[Proof of Proposition~\ref{phigsh}]
Choose a good basis $(d_1,\,\dots,\,d_n)$ of $\dfont$. We can replace $(d_1,\,\dots,\,d_n)$ by $(X^s d_1,\,\dots,\,X^s d_n)$ for some $s \geq 0$, and assume that $P = \Mat(\phi) \in \M_n(\efont^+)$. Take $r \geq 1$ such that $X^r P^{-1} \in X \M_n(\efont^+)$. Let $G_g$ be the matrix of $g \in \Gamma$. By continuity of the map $\Gamma \to \GL_n(\efont^+)$, $g \mapsto G_g$, there exists $\ell \geq k$ such that for all $g \in \Gamma_\ell$, we have $\vx(G_g-\Id) \geq r$. Write $G_g = \Id + X^r H_g$ with $H_g \in \M_n(\efont^+)$.

By definition of $r$, we have $X^r g(P)^{-1} \in X \M_n(\efont^+)$, so that if $Q_g = X^{r(p-1)}g(P)^{-1}$, then $Q_g \in X\M_n(\efont^+)$. The commutation relation between $\phi$ and $\Gamma_\ell$ gives $P\phi(G_g)\linebreak = G_g g(P)$ for all $g \in \Gamma_\ell$. Therefore, $P\phi(\Id+X^rH_g) = (\Id+X^rH_g)g(P)$, so that
\[
Pg(P)^{-1}-\Id = X^r\left(H_g-P\phi(H_g)Q_g\right).
\]
This implies that $Pg(P)^{-1}-\Id \in X^r\M_n(\efont^+)$. Let
\[
f(g) = H_g-P\phi(H_g)Q_g = X^{-r}\left(Pg(P)^{-1}-\Id\right).
\]
Recall that $Q_g, f(g) \in \M_n(\efont^+)$ for all $g \in \Gamma_\ell$, and that (compare with Proposition~\ref{hrng}$\MK$\eqref{prop1.4.4})
\begin{align*}
Q_g &= X^{r(p-1)} g(P)^{-1} = X^{r(p-1)} g(\tco(P)) g\left(\det(P)^{-1}\right)
\\
\intertext{and}
f(g) &= X^{-r}\left(P g(\tco(P)) g\left(\det(P)^{-1}\right)-\Id\right).
\end{align*}
By Propositions~\ref{propmsh} and~\ref{etnsh}, and Lemma~\ref{hmnep}, there exists $\mu \in \RR$ such that $g \mapsto Q_g$ and $g \mapsto f(g)$ belong to $\HH^{\ell,\mu}(\Gamma_\ell,\M_n(\efont^+))$.

Let $f_0 = f$ and for $i \geq 1$, let $f_i : \Gamma_\ell \to
\M_n(\efont^+)$ be the function
\[
g \mapsto P\phi(P)\cdots\phi^{i-1}(P) \cdot \phi^i(f(g)) \cdot
\phi^{i-1}(Q_g)\cdots\phi(Q_g)Q_g.
\]

Since $P \in \M_n(\efont^+)$, Lemma~\ref{hmnep} implies that $f_i \in
\HH^{\ell,\mu}(\Gamma_\ell,\M_n(\efont^+))$. In addition, $\vx(Q_g) \geq 1$, so that $\vx(\phi^{i-1}(Q_g)\cdots\phi(Q_g)Q_g) \geq (p^i-1)/(p-1)$. Hence $\sum_{i\,\geq\,0} f_i$ converges in $\HH^{\ell,\mu}(\Gamma_\ell,\M_n(\efont^+))$, and we let $T(f)$ be its limit.

We have $T(f)(g) = H_g$. This implies that $g \mapsto H_g$ belongs to $\HH^{\ell,\mu}(\Gamma_\ell,\M_n(\efont^+))$, and hence so does $g
\mapsto G_g = \Id + X^r H_g$.

We therefore have $\dfont=\dfont^{\Gamma_\ell\hyphen\sh,\ell}$, so that $\dfont=\dfont^{\Gamma_k\hyphen\sh,k}$ by Lemma~\ref{shsbgr}.
\end{proof}

\begin{coro}\label{decompgm}
If $\dfont$ is a $(\phi,\Gamma)$-module over $\efont$, then $(\et \otimes_{\efont} \dfont)^{\Gamma_k\hyphen\sh,k-m} = \efont_m \otimes_{\efont} \dfont$ for $m \geq 0$.
\end{coro}

We now prove the following result, which generalizes Proposition~\ref{phigsh}. Note that the underlying constants are not as good as in the case of a $(\phi,\Gamma)$-module.

\begin{prop}\label{gamodsh}
If $M$ is a finite-dimensional $\efont$-vector space with a continuous semi-linear action of $\Gamma_k$, then $M=M^{\sh}$.
\end{prop}

\begin{proof}
Choose a good basis of $M$. Let $f(g)$ denote the matrix of $g \in \Gamma$ in this basis. If $\ell \geq 1$, there exists $k \geq \ell+1$ such that $f(g) \in \Id + X^{p^\ell} \M_n(\efont^+)$ for all $g \in 1+p^k \Zp$. Write $f(g) = \Id + X^{p^\ell} H$. The cocycle formula gives
\[
f(g^p) = \left(\Id + X^{p^\ell} H\right) \left(\Id + g\left(X^{p^\ell} H\right)\right) \cdots \left(\Id + g^{p-1}\left(X^{p^\ell} H\right)\right).
\]
Proposition~\ref{etnsh}, with $n=0$, implies that $g^m(X^{p^\ell} H) \equiv X^{p^\ell} H \bmod{X^{p^k}}$ for all $0 \leq m \leq p-1$. Hence $f(g^p) \equiv (\Id + X^{p^\ell} H)^p \bmod{X^{p^k}}$. This implies that $f(g^p) \equiv \Id + X^{p^{\ell+1}} H^p \bmod{X^{p^k}}$ so that $f(g^p) = \Id \bmod{X^{p^{\ell+1}}}$ since $k \geq \ell+1$.

Since $(1+p^k \Zp)^p = 1+p^{k+1} \Zp$, the above computation implies by induction on $i$ that $f(1+p^{k+i} \Zp) \subset \Id + X^{p^{\ell+i}} \M_n(\efont^+)$ for all $i \geq 0$.

This implies that $M=M^{\Gamma_k\hyphen\sh,\ell,0}$ by Lemma~\ref{mshalt}.
\end{proof}

\begin{coro}\label{senfin}
Let $N$ be an $\efont$-vector space, with a compatible valuation and a semi-linear action of $\Gamma_k$ by isometries. Let $N^{\fin}$ denote the set of $x \in N$ that belong to a finite dimensional $\efont$-vector space stable under $\Gamma_k$, in analogy with classical Sen theory.

Proposition~\ref{gamodsh} implies that $N^{\fin} \subset N^{\sh}$. In particular, if $N=\et$, then $\et^{\fin} = \et^{\sh} = \efont_\infty$.
\end{coro}

\subsection{The field of norms}\label{secfon}

Let $K$ be a finite extension of $\Qp$. Let $K_n = K (\mu_{p^n})$ and let $K_\infty = \cup_{n\,\geq\,0} K_n$. The field of norms of the extension $K(\mu_{p^\infty})/K$ is defined and studied in~\cite{W83}. It is the set of sequences $\{x_n\}_{n \,\geq\,0}$ where $x_n \in K_n$ and $\Nm_{K_{n+1}/K_n}(x_{n+1}) = x_n$ for all $n \geq 0$. This set has a natural structure of a field of characteristic $p$ whose residue field is that of $K_\infty$ (\S~2.1 of ibid), which we denote by $\efont_K$. If $K=\Qp$, then $\efont_{\Qp} = \Fp \dpar{X}$, where $X=\{x_n\}_{n\,\geq\,0}$ with $x_n = 1-\zeta_{p^n}$ for $n \geq 1$. When $K$ is a finite extension of $\Qp$, $\efont_K$ is a finite separable extension of $\efont_{\Qp}$ of degree $[K_\infty:(\Qp)_\infty]$ (\S~3.1 of ibid).

Let $\Gamma_K = \Gal(K_\infty / K)$, so that $\Gamma_K$ is isomorphic to an open subgroup of $\Zp^\times$ via the cyclotomic character $\chi_{\cyc}$. The group $\Gamma_K$ acts naturally on $\efont_K$, and if $g \in \Gamma_K$, then $g(X) = (1+X)^{\chi_{\cyc}(g)}-1$. Let $\phi : \efont_K \to \efont_K$ denote the map $y \mapsto y^p$. Let $\et_K$ denote the $X$-adic completion of $\cup_{n \,\geq\,0} \phi^{-n}(\efont_K)$. In particular, $\et_{\Qp} = \et$ in the notation of \S~\ref{secdecomp}, and $\et_K$ is the tilt of $\widehat{K}_\infty$ (\S~4.3 of ibid and~\cite[\S~3]{S12}).

\begin{lemm}\label{ffpgm}
We have $\phi^{-n}(\efont_K) = \efont_n \otimes_{\efont} \efont_K$ for all $n$, and $\et_K = \et \otimes_{\efont} \efont_K$.
\end{lemm}

\begin{proof}
The extensions $\efont_n/\efont$ and $\efont_K/\efont$ are linearly disjoint since the first is purely inseparable and the second is separable. By comparing degrees, we get the first claim. It implies that $ \et \otimes_{\efont} \efont_K \to \et_K$ is surjective, and the second claim follows, since $[\et_K:\et] = [\efont_K:\efont] = [K_\infty:(\Qp)_\infty]$.
\end{proof}

\begin{coro}\label{ffsh}
We have $\et_K^{\sh} = \cup_{n\,\geq\,0} \phi^{-n}(\efont_K)$.
\end{coro}

\begin{proof}
This follows from Lemma~\ref{ffpgm} and Corollary~\ref{decompgm}, as $\efont_K$ is a $(\phi,\Gamma_K)$-module over $\efont$, and $\cup_{n\,\geq\,0} \phi^{-n}(\efont_K) = \efont_\infty \otimes_\efont \efont_K$.
\end{proof}

\begin{rema}\label{bcla}
In characteristic zero, $\hat{K}_\infty$ is a $p$-adic Banach representation of $\Gamma_K$, and by~\cite[Theorem~3.2]{BC16}, $K_\infty$ is the space $\hat{K}_\infty^{\lal}$ of locally analytic vectors in~$\hat{K}_\infty$.
\end{rema}

\subsection{The \texorpdfstring{$p$}{p}-adic local Langlands correspondence}\label{secllp}

We now prove a result that suggests that the theory of super-H\"older vectors could have some applications to the $p$-adic local Langlands correspondence. In order to avoid too many technicalities, we consider only the simplest example. Recall that if $f \in \efont^+$, there exist $f_0,\,\hdots,\,f_{p-1} \in \efont^+$ such that $f = \sum_{i=0}^{p-1}
\phi(f_i) (1+X)^i$. We define $\psi(f)=f_0$. The map $\psi : \efont^+ \to
\efont^+$ has the following properties: $\psi(f \phi(h)) = h \psi(f)$ if $f,h \in \efont^+$ and $\psi \circ g = g \circ \psi$ if $g \in
\Zp^\times$.

Let $M = \varprojlim_\psi \efont^+$ be the set of sequences $m = (m_0,m_1,\,\hdots)$ with $m_i \in \efont^+$ and $\psi(m_{i+1}) = m_i$ for all $i \geq 0$. The space $M$ is endowed with an action of $\Zp^\times$ given by $(g \cdot m)_i = g \cdot m_i$ and the structure of an $\efont^+$-module given by $(f(X) m)_i = \phi^i(f(X))m_i$. Following Colmez, we could extend these structures to an action of the Borel subgroup $\BB$ of $\GL_2(\Qp)$ on $M$, and this idea is an important step in the construction of the $p$-adic local Langlands correspondence. The representation $M$ is then the dual of most of the restriction to $\BB$ of a parabolic induction. However, we don't use this here.

Let $\vx$ be the $X$-adic valuation on $M$: $\vx(m)$ is the max of the $n \geq 0$ such that $m \in X^n M$. The space $M$ is separated and complete for the $X$-adic topology, although this is not the natural topology on $M$ (the natural topology is induced by the product topology $\varprojlim_\psi \efont^+ \subset \prod \efont^+$. The action of $\Zp^\times$ on $M$ is not continuous for the $X$-adic topology: $M \neq M^{\cont}$ in the notation of Remark~\ref{mcont}).

We have an injection~$i : \efont^+ \to M$, given by $i(f) = (f,\phi(f),\phi^2(f),\,\hdots)$.

\begin{prop}\label{llpsh}
We have $M^{\Gamma_k\hyphen\sh,k} = i(\efont^+)$.
\end{prop}

\begin{proof}
Recall that if $m \in M$ and $f(X) \in \efont$, then $(f(X) m)_j = \phi^j(f(X))m_j$ for all $j \geq 0$. We have $\vx(\phi^j(f(X))) = p^j \vx(f(X))$. In particular, if $m \in M^{\Gamma_k\hyphen\sh,k}$, then $m_j \in (\efont^+) ^{\Gamma_k\hyphen\sh,k+j}$. The results of \S~\ref{secgam} imply that $m_j \in \phi^j(\efont^+)$. If $m_j = \phi^j(f_j)$, the $\psi$-compatibility implies that $f_j=f_0$ for all $j \geq 0$. This implies the claim.
\end{proof}

A generalization of Proposition~\ref{llpsh} to representations of $\BB$ obtained from $(\phi,\Gamma)$-modules using Colmez' construction shows that using the theory of super-H\"older vectors, we can recover the $(\phi,\Gamma)$-module giving rise to such a representation of $\BB$. One of the main results of~\cite{BV} is that every infinite dimensional smooth irreducible $\kf$-linear representation of $\BB$ having a central character comes from a $(\phi,\Gamma)$-module by Colmez' construction. Is it possible to reprove this result using super-H\"older vectors?


\bibliography{Berger}
\end{document}
