\input{preamble.tex}
\unittitle{Decomposition by subsidiary radii}

In this unit, we show that one can sometimes decompose a differential
module on a disc or annulus so as to separate the small and large
subsidiary radii into different components.

\section{Unit coefficients in Newton polygons}

For any complete extension $L$ of $K$ and any $\mu \in \gotho_L$,
we write $f_\mu: K \langle t \rangle \to L \langle t \rangle$
for the map $t \mapsto t + \mu$.

\begin{lemma}
For $x \in K \langle t \rangle$,
$x$ is a unit if and only if for any $L$ and $\mu$,
$|f_\mu(x)|_{e^{-r}}$ is independent of $r$.
(By concavity, it suffices to check this for $r$ in a neighborhood of $0$.)
\end{lemma}
\begin{proof}
Write $x = \sum_i x_i t^i$. Then $x$ is a unit if and only if
$|x_0| > |x_i|$ for all $i > 0$. If this holds, then clearly
$|f_\mu(x)|_{e^{-r}} = |x_0|$ for all $\mu,r$. Otherwise,
we can reduce $x_0^{-1} x$ modulo $\gothm_K$ to get a nonconstant element of
$\kappa_K[t]$; by making $L$ large enough, we can choose $\mu$ to reduce
to a root of this polynomial, and then
$|f_\mu(x)|_{e^{-r}}$ will not be constant.
\end{proof}


This immediately yields the following.
\begin{theorem} \label{T:split1}
Let $P = T^n + \sum_{i=0}^{n-1} P_i T^i
\in K \langle t \rangle\{T\}$ be a monic twisted polynomial. 
Choose $j \in \{1,\dots,n-1\}$ such that
there is a vertex of the Newton polygon of $P$, measured with respect to
$|\cdot|_1$, with $x$-coordinate
$(-n+j)$, and that the first $j$ slopes are all less than $-\log |d|_{F_1}$.
Then $P_j$ is a unit in $K \langle t \rangle$ if and only if 
for any $L$ and $\mu$, the sum of the first $j$ slopes of
the Newton polygon of $f_\mu(P) = T^n + \sum_{i=0}^{n-1} f_\mu(P_i) T^i$,
measured with respect to $|\cdot|_{e^{-r}}$, 
is independent of $r$. (By concavity,
it suffices to check this for $r$ in a neighborhood of $0$.)
Moreover, in this case (by the master factorization theorem), there
is a factorization of $P$ separating the first $j$ slopes.
\end{theorem}

For annuli, it is a bit more complicated to formulate the analogous
results; here is a weak but adequate formulation.
\begin{lemma}
For $x \in \cup_{\alpha \in (0,1)} K \langle \alpha/t, t \rangle$,
$x$ is a unit if and only if for any $L$ and $\mu$,
$|f_\mu(x)|_{e^{-r}}$ is independent of $r$ for
$r$ in a neighborhood of $0$.
\end{lemma}

\begin{theorem} \label{T:split2}
Let $P = T^n + \sum_{i=0}^{n-1} P_i T^i
\in (\cup_{\alpha \in (0,1)} K \langle \alpha/t, t \rangle)\{T\}$ 
be a monic twisted polynomial. 
Choose $j \in \{1,\dots,n-1\}$ such that
there is a vertex of the Newton polygon of $P$,
measured with respect to $|\cdot|_1$, with $x$-coordinate
$(-n+j)$,
and that the first $j$ slopes are all less than $-\log |d|_{F_1}$.
Then $P_j$ is a unit in $\cup_{\alpha \in (0,1)}
K \langle \alpha/t, t \rangle$ if and only if 
for any $L$ and $\mu$, the sum of the first $j$ slopes of
the Newton polygon of $f_\mu(P) = T^n + \sum_{i=0}^{n-1} f_\mu(P_i) T^i$,
measured with respect to $|\cdot|_{e^{-r}}$, 
is independent of $r$ for $r$ in a neighborhood of $0$.
Moreover, in this case (by the master factorization theorem), there
is a factorization of $P$ separating the first $j$ slopes.
\end{theorem}

\section{Decomposition over a disc}

Retain notation as in the subharmonicity theorem from the previous unit.
\begin{theorem} \label{T:decomposition disc}
Let $M$ be a finite differential module over $K \langle t/\beta \rangle$
of rank $n$.
Suppose that the following conditions hold for some $i \in \{1,\dots,n-1\}$.
\begin{enumerate}
\item[(a)]
We have $f_i(-\log \beta) > f_{i+1}(-\log \beta)$.
\item[(b)]
The function $f_1+\cdots+f_i$ is constant in a neighborhood of $r = -\log \beta$.
\item[(c)]
The $i$-th discrepancy of $M$ is zero at $r = -\log \beta$.
\end{enumerate}
Then
the decomposition of $M \otimes F_\beta$ separating the first $i$ subsidiary
radii lifts to a decomposition of $M$ itself.
\end{theorem}

Before proving Theorem~\ref{T:decomposition disc}, we record a trivial
but useful observation.
\begin{lemma} \label{L:intersection}
Let $R,S,T$ be subrings of a common ring $U$ with $S \cap T = R$.
Let $M$ be a finite free $R$-module. Then the intersection
$(M \otimes S) \cap (M \otimes T)$ inside $M \otimes U$ is equal
to $M$ itself.
\end{lemma}
This also holds when $M$ is only locally free; see exercises. The immediate
application is to replace $K$ by a complete extension $L$ in 
Theorem~\ref{T:decomposition disc}; inside the completion
of $L(t)$ for the 1-Gauss norm, we have
\[
F_1 \cap L \langle t \rangle = K \langle t \rangle.
\]
Thus obtaining matching decompositions of $M \otimes F_1$ and 
$M \otimes L \langle t \rangle$ gives a corresponding decomposition
of $M$ itself.

\begin{lemma} \label{L:decomp disc}
Theorem~\ref{T:decomposition disc} holds if $f_i(-\log \beta) > 1/(p-1) \log p
- \log \beta$.
\end{lemma}
\begin{proof}
As in the previous unit, we may reduce to the case $\beta=1$.
(This requires enlarging $K$, but as noted above,  Lemma~\ref{L:intersection}
renders this enlargement harmless.)
Set notation as in Lemma~6 from the previous unit.
Then for $r$ near 0, the sum of the first $i$ slopes of
characteristic polynomial of $N$ equals the sum of the first $i$ slopes
of the Newton polygon of the twisted polynomial $P$, which is constant,
and the $i$-th slope on either side is strictly less than the
$(i+1)$-st slope.
Moreover, this all remains true after replacing $t$ by $t+c$ for any
$c$ of norm 1. Consequently, by Theorem~\ref{T:split1},
the coefficient of $T^{n-i}$ in the 
characteristic polynomial of $N$ is a unit in $K \langle t \rangle$,
and serves as a vertex of the Newton polygon for all $r$.
We may thus perform a slope decomposition
in $K \langle t \rangle [T]$ to separate the first $i$ slopes.

By applying the lattice lemma from the previous unit
(again with $K$ suitably large)
to both factors of the decomposition, we obtain
a matrix $U_1 \in \GL_n(K \langle t \rangle)$ 
with $|U_1|, |U_1|^{-1} \leq c$ such that
$U_1^{-1} N U_1$ is a block diagonal matrix, with the first $i$
Newton slopes in the first block and the rest in the second block.

Put $N_1 = U_1^{-1} N_0 U_1 + U_1^{-1} d(U_1)$. 
Let $s_{H,1},\dots, s_{H,n}$
be the Hodge slopes of $N_1$; they differ from the
corresponding Newton slopes by at most $4 \log c$.
We now repeat the following
operation to produce $N_2, N_3, \dots$. Suppose we have
\[
N_l = \begin{pmatrix} A_l & B_l \\ C_l & D_l \end{pmatrix}
\]
with $A_l$ having Hodge slopes $s_{H,1},\dots,s_{H,i}$,
and $|B_l|, |C_l|, |D_l| \leq e^{-s_{H,i}} \delta$ for some
$\delta < 1$. (These hold for $l=1$ if we take $c$ small enough.)
Put
\[
U_{l+1} = \begin{pmatrix} I_i & 0 \\ C_l A_l^{-1} & I_{n-i}
\end{pmatrix}.
\]
Note that $|A_l^{-1}| \leq e^{s_{H,i}}$ by Cramer's rule,
so $|U_{l+1} - I_n| \leq \delta < 1$. In particular, $U_{l+1}$ is invertible.
Set
\[
N_{l+1} = U_{l+1}^{-1} N_l U_{l+1} + U_{l+1}^{-1} d(U_{l+1}).
\]
In block form,
\[
N_{l+1} = \begin{pmatrix} A_l - B_l C_l A_l^{-1} & B_l \\
-C_l A_l^{-1} B_l C_l A_l^{-1} + D_l C_l A_l^{-1} + d(C_l A_l^{-1})
& -C_l A_l^{-1} B_l + D_l 
\end{pmatrix},
\]
so
\begin{align*}
\max\{|B_{l+1}|, |D_{l+1}|\}
&\leq e^{-s_{H,i}} \delta  \\
|C_{l+1}| &\leq \max\{e^{s_{H,i}}, \delta\} |C_l|.
\end{align*}
Since $\max\{e^{s_{H,i}}, \delta\} < 1$, 
$C_l \to 0$ and so $U_l \to I_n$. 

In the limit, we get
a basis of $M$ on which $D$ acts by a block upper triangular matrix;
this shows that the first component of the 
decomposition of $M \otimes F_1$ separating the first $i$ subsidiary
radii from the others lifts to $M$.
Repeating the argument with $M$ replaced by
$M^\dual$ shows that the second component
also lifts (because its annihilator in $M^\dual \otimes F_1$ lifts).
\end{proof}

To prove Theorem~\ref{T:decomposition disc} in general, we must reduce
the problem for $M$ to the problem for $\varphi_* M$. If $M'_1 \oplus
M'_2$ is the corresponding decomposition of $\varphi_* M$, then
this might not be induced by a decomposition of $M_1$, because
some factors of subsidiary radius $p^{-p/(p-1)}$ that are needed
in $M'_2$ are instead grouped into $M'_1$. To fix this,
consider instead the decomposition
\[
((M'_1 \otimes W_0) \cap \cdots \cap (M'_1 \otimes W_{p-1}))
\oplus
((M'_2 \otimes W_0) + \cdots + (M'_2 \otimes W_{p-1}));
\]
this is induced by the desired decomposition of $M_1$.

\section{Decomposition over a closed annulus}

A similar argument works for annuli, to prove the following.

\begin{theorem} \label{T:decomposition annulus}
Let $M$ be a finite differential module over $K \langle \alpha/t, t/\beta 
\rangle$ of rank $n$.
Suppose that the following conditions hold for some $i \in \{1,\dots,n-1\}$.
\begin{enumerate}
\item[(a)]
We have $f_i(r) > f_{i+1}(r)$ for $-\log \beta \leq r \leq -\log \alpha$.
\item[(b)]
The function 
$f_1 + \cdots + f_i$ is affine on $[-\log \beta, -\log \alpha]$.
\item[(c)]
The $i$-th discrepancy of $M$ is zero at both $r = -\log \beta$ and
$r=-\log \alpha$.
\end{enumerate}
Then there is a decomposition of $M$ inducing, for each $\rho \in
[\alpha, \beta]$, the
decomposition of $M \otimes F_\rho$ separating the first $i$ subsidiary
radii from the others.
\end{theorem}
In the following proof, we will skip some details which proceed as in the
disc case.
\begin{proof}
By the patching theorem for finite
free modules over $K \langle \alpha/t,t/\beta \rangle$, 
it suffices to show that we can cover the interval $[\alpha, \beta]$
with finitely many closed subintervals $[\gamma, \delta]$ of positive
length, such that the desired conclusion holds for
$M \otimes K \langle \gamma/t, t/\delta \rangle$.
Since the interval $[\alpha, \beta]$ is compact, it suffices to check
that each $\rho \in [\alpha, \beta]$ occurs as the endpoint of such
a good interval.

Note that condition (b) implies that the $i$-th discrepancy of $M$ is zero
for all $r \in (-\log \beta, -\log \alpha)$. That means that we may relabel
things so that $\rho$ becomes an endpoint of the interval, without losing
our hypotheses. That is,
it suffices to check that there exists $\gamma \in 
[\alpha,\beta)$ such that the desired decomposition exists for
$M \otimes \langle \gamma/t, t/\beta \rangle$. 

Using Lemma~\ref{L:intersection}, we may enlarge $K$ and then reduce
to the case $\beta=1$.
Moreover, using Frobenius descendants as in the
proof of Theorem~\ref{T:decomposition disc}, we may reduce to
the case where $f_i(0) > 1/(p-1) \log p$.

Again, set notation as in Lemma~6 from the previous unit.
This time, hypotheses (a) and (c) plus
Theorem~\ref{T:split2} imply that 
for $\gamma \in [\alpha, 1)$ sufficiently close to $1$,
the coefficient of $T^{n-i}$ in the characteristic polynomial of $N$
is a unit in $K \langle \gamma/t, t \rangle$,
and serves as a vertex of the Newton polygon for all $r$.
We may thus perform a slope decomposition
in $K \langle \gamma/t, t \rangle [T]$ to separate the first $i$ slopes,
and continue as in Lemma~\ref{L:decomp disc}.
\end{proof}

\section{Decomposition over an open disc or annulus}

If we are willing to lose information at the endpoints of our given
interval, we can drop condition (c) from 
Theorem~\ref{T:decomposition annulus}. 

\begin{theorem} \label{T:decomposition disc2}
Let $M$ be a finite differential module over $K \langle t/\beta \rangle$
of rank $n$.
Suppose that the following conditions hold for some $i \in \{1,\dots,n-1\}$.
\begin{enumerate}
\item[(a)]
For $r$ in a neighborhood of $-\log \beta$,
$f_1 + \cdots + f_i$ is constant.
\item[(b)]
For $r \in (-\log \beta, \infty)$ sufficiently small,
$f_i(r) > f_{i+1}(r)$. (We do not require the inequality for $r=-\log \beta$.)
\end{enumerate}
Then for any $\delta < \beta$, $M \otimes K \langle t/\delta \rangle$
admits a unique decomposition separating the first $i$ subsidiary
radii.
\end{theorem}
\begin{proof}
It suffices to check this for a sequence of values of $\delta$
approaching $\beta$. To do this,
note that by subharmonicity, the $i$th discrepancy of $M$ can only be 
nonzero at a value of $r$ where the function
$f_1 +\cdots +f_i$ has a change of slope. In particular, those values occur
discretely in any interval, so for $\delta< \beta$ sufficiently close to
$\beta$, $-\log \delta$ is not such a value. We may then apply
Theorem~\ref{T:decomposition disc} to conclude.
\end{proof}

We have a similar result for annuli; the proof is similar, so we omit it.
\begin{theorem} \label{T:decomposition annulus2}
Let $M$ be a finite differential module over $K \langle \alpha/t, t/\beta 
\rangle$ of rank $n$.
Suppose that the following conditions hold for some $i \in \{1,\dots,n-1\}$.
\begin{enumerate}
\item[(a)]
We have $f_i(r) > f_{i+1}(r)$ for $-\log \beta < r < -\log \alpha$.
(We do not require the inequality for $r = - \log \beta$ or 
$r = -\log \alpha$.)
\item[(b)]
The function 
$f_1 + \cdots + f_i$ is affine on $[-\log \beta, -\log \alpha]$.
\end{enumerate}
Then for any $\alpha < \gamma \leq \delta < \beta$, 
$M \otimes K \langle \gamma/t, t/\delta  \rangle$
admits a unique decomposition separating the first $i$ subsidiary
radii.
\end{theorem}

\section{Modules solvable at a boundary}

Let $M$ be a finite differential module on the half-open annulus
with closed inner radius $\alpha$ and open outer radius $\beta$.
We say $M$ is 
\emph{solvable at $\beta$} if $R(M \otimes F_\rho) \to \beta$ as $\rho \to
\beta^-$, or equivalently, if $IR(M \otimes F_\rho) \to 1$ as $\rho \to 
\beta^-$.

\begin{lemma} \label{L:linear}
Let $M$ be a finite differential module on the half-open annulus
with closed inner radius $\alpha$ and open outer radius $\beta$,
which is solvable at $\beta$.
There exist $b_1 \geq \dots \geq b_n \in [0,\infty)$
such that for $\rho \in [\alpha, \beta)$ sufficiently close to
$\beta$, the intrinsic subsidiary radii of $M \otimes F_\rho$
are $(\rho/\beta)^{b_1}, \dots, (\rho/\beta)^{b_n}$.
Moreover, if $i=n$ or $b_i > b_{i+1}$, then $b_1 + \cdots + b_i \in \ZZ$.
\end{lemma}
\begin{proof}
Define $f_1(r), \dots, f_n(r)$ as before. For $r \to (-\log \beta)^+$,
$f_1 + \cdots + f_i - ir$ is convex with slopes in a discrete subset of $\RR$
and has limit 0; this implies that the slopes are all positive.
However, the slopes lie in a discrete subgroup of $\RR$, so they must
eventually stabilize. We deduce that each $f_i$ is linear in a neighborhood
of $-\log \beta$, and we may infer the desired conclusions from the
known properties of the $f_i$.
\end{proof}
The quantities $b_1, \dots, b_n$ are called the \emph{(differential) slopes}
of $M$ at $\beta$. We will have a closer look at these numbers in a later 
unit.

We now recover a decomposition theorem of Christol-Mebkhout; see notes.
\begin{theorem}[Christol-Mebkhout] \label{T:decomposition boundary}
Let $M$ be a finite differential module on the half-open annulus
with closed inner radius $\alpha$ and open outer radius $\beta$,
which is solvable at $\beta$. 
Then for some $\gamma \in [\alpha, \beta)$,
the restriction of $M$ to 
the annulus with closed inner radius $\alpha$ and open outer radius $\beta$
splits as a direct 
sum $\oplus_{b \in [0, \infty)} M_b$, such that for each $b \in [0, \infty)$,
for all $\rho \in [\gamma, \beta)$,
the intrinsic subsidiary radii
of $M_b \otimes F_\rho$ are all equal to $(\rho/\beta)^b$.
\end{theorem}
\begin{proof}
By Lemma~\ref{L:linear}, we are in a case where
Theorem~\ref{T:decomposition annulus2} may be applied.
\end{proof}

The condition of solvability at a boundary is often established by exhibiting
a Frobenius structure for the differential equation. We will explain what
this means in a later unit.

\section{Notes}

Our results on modules solvable at a boundary are originally
due to Christol and Mebkhout [CM00], [CM01]. In particular,
Lemma~\ref{L:linear} for the generic radius is [CM00, Th\'eor\`eme~4.2.1],
and the decomposition theorem (which implies Lemma~\ref{L:linear}
in general) is [CM01, Corollaire~2.4--1].
The proof technique of Christol and Mebkhout is significantly different
from ours: they construct the decomposition
by exhibiting convergent sequences for a certain topology
on the ring of differential operators. 

Note that Christol and Mebkhout do not
obtain quantitative results; that is, they do not control the
range over which the decomposition occurs. We are not sure whether this
is really a limitation of their method, or whether they did not bother
to exert such control because they did not have an application in mind 
for it. (Remember that for the quantitative results here, it was
crucial to use Frobenius descendants, which do not
appear previously elsewhere.)

Note also that Christol and Mebkhout work directly with a differential
module on an open annulus as a ring-theoretic object; this requires
a freeness result of the following form.
If $K$ is spherically complete, any finite free module on the
half-open annulus with closed inner radius $\alpha$ and open outer radius
$\beta$ is induced by a finite free module over the ring
$\cap_{\rho \in [\alpha,\beta)} K \langle \alpha/t, t/\rho \rangle$.
(That is, any locally free
coherent sheaf on this annulus is freely generated by global sections.)
For a proof, see for instance [Ked05, Theorem~3.14]. 
A result of Lazard [Laz62] implies that this property, even when restricted
to modules of rank 1, is in fact equivalent to spherical
completeness of $K$.

\section{Exercises}

\begin{enumerate}
\item
Prove the analogue of Lemma~\ref{L:intersection} where $M$ is only
locally free. This is [Ked07b, Lemma~1.2.7].
\end{enumerate}

\end{document}