\input{preamble.tex}
\unittitle{Frobenius antecedents and Frobenius descendants}

In this unit, we introduce Dwork's technique of descent along
Frobenius (or more exactly, descent along the $p$-th power map
on an affine or projective line) 
to analyze the generic radius of convergence and subsidiary radii of a
differential module.

We retain notation as in the previous unit. In particular,
$K$ is a complete nonarchimedean field, and $F_\rho$ is the
completion of $K(t)$ for the $\rho$-Gauss norm for some $\rho>0$.

\section{Why Frobenius?}

It may be helpful to review the current state of
affairs, to clarify why we need to descend along Frobenius.

Let $V$ be a finite differential module over $F_\rho$.
Then the allowable values of the truncated spectral norm
$|D|_{\tspect,V}$ are the real numbers greater than or equal to
$|d|_{\spect,F_\rho} = p^{1/(p-1)} \rho^{-1}$, corresponding to 
generic radii of convergence less than or equal to $\rho$.

However, if we want to calculate the truncated spectral norm using
the Newton polygon of a twisted polynomial, we cannot distinguish among
values less than or equal to the operator norm $|d|_{F_\rho} = \rho^{-1}$.
In particular, we cannot use this technique to prove a decomposition
theorem for differential modules that separates components of spectral
norm between $p^{1/(p-1)} \rho^{-1}$ and $\rho^{-1}$.

One way one might want to get around this is to consider not $d$ but
a high power of $d$, particularly a $p^n$-th power. The trouble with this is
that iterating a derivation does not give another derivation, but something
much more complicated.

Instead, we will try to differentiate with respect to $t^{p^n}$ instead of
with respect to $t$. This will have the effect of increasing the spectral
norm, so that we can push it into the range where Newton polygons
become useful.

\section{$p$-th roots}

But first, we must make some calculations in answer to the following question:
if two $p$-adic numbers are close together, how close are their $p$-th
powers, or their $p$-th roots?

We observed in the
previous unit that when $m$ is a positive integer coprime to $p$,
\[
|t-\eta| < \lambda |\eta| \Leftrightarrow
|t^m - \eta^m| < \lambda |\eta|^m \qquad (\lambda \in (0,1)).
\]
This breaks down for $m = p$, because a primitive $p$-th root of unity
$\zeta_p$ satisfies $|1-\zeta_p | < 1$. The quantities
$1-\zeta_p^m$ for $m=1,\dots,p-1$ are Galois conjugates, so
\[
|1-\zeta_p| = \left| \prod_{m=1}^{p-1} (1 - \zeta_p^m) \right|^{1/(p-1)}
= |p|^{1/(p-1)} = p^{-1/(p-1)}
\]
since the product is the derivative of $T^p - 1$ evaluated at $T=1$.

\begin{lemma} \label{L:p-th root}
Pick $t, \eta \in K$.
\begin{enumerate}
\item[(a)]
For $\lambda \in (0,1)$, if $|t - \eta| \leq \lambda |\eta|$,
then
\[
|t^p - \eta^p| \leq 
\max\{\lambda^p, p^{-1} \lambda \} |\eta^p| = 
\begin{cases} \lambda^p |\eta^p| & \lambda \geq p^{-1/(p-1)} \\
p^{-1} \lambda  |\eta^p| & \lambda \leq p^{-1/(p-1)}. \end{cases}
\]
\item[(b)]
Suppose $\zeta_p \in K$.
If $|t^p - \eta^p| \leq \lambda |\eta^p|$, then
there exists $m \in \{0,\dots,p-1\}$
 such  that
\[
|t-\zeta_p^m \eta| \leq \min\{\lambda^{1/p}, p\lambda\} |\eta|
= \begin{cases} \lambda^{1/p} |\eta| & \lambda \geq p^{-p/(p-1)} \\
p \lambda |\eta| & \lambda \leq p^{-p/(p-1)}. \end{cases}
\]
Moreover, if $\lambda \geq p^{-p/(p-1)}$, we may always take $m=0$.
\end{enumerate}
\end{lemma}
We will use repeatedly, and without comment, the fact that
\[
\lambda \mapsto \max\{\lambda^p, p^{-1} \lambda\}, \qquad
\lambda \mapsto \min\{\lambda^{1/p}, p \lambda\}
\]
are strictly increasing functions from $[0,1]$ to itself that are
inverse to each other.
\begin{proof}
There is no harm in assuming $\zeta_p \in K$ for both parts.
For (a), factor $t^p-\eta^p$ as $t-\eta$ times $t - \eta \zeta_p^m$ for
$m=1,\dots,p-1$, and write
\[
t-\eta \zeta_p^m = (t-\eta) + \eta(1 - \zeta_p^m).
\]
If $|t-\eta| \geq p^{-1/(p-1)} |\eta|$,
then $t-\eta$ is the dominant term, otherwise
$\eta (1-\zeta_p^m)$ dominates. This gives the claimed bounds.

For (b), consider the Newton polygon of 
\[
t^p - \eta^p - c = \sum_{i=0}^{p-1} \binom{p}{i} \eta^i (t-\eta)^{p-i}
- c
\]
viewed as a polynomial in $t-\eta$. Suppose $|c| = \lambda |\eta^p|$.
If $\lambda \geq p^{-p/(p-1)}$,
then the terms $(t-\eta)^p$ and $c$ dominate, and all roots have norm
$\lambda^{1/p} |\eta|$. Otherwise, the terms
$(t-\eta)^p$, $p (t-\eta) \eta^{p-1}$, and $c$ dominate,
so one root has norm $p \lambda |\eta|$ and the others are larger;
repeating with $\eta$ replaced by $\zeta_p^m \eta$ for $m=0,\dots,p-1$
gives $p$ distinct roots, which accounts for all of them.
\end{proof}

\begin{cor} \label{C:p-th root}
Let $T: K \llbracket t^p - \eta^p \rrbracket \to K \llbracket t - \eta
\rrbracket$ be the substitution $t^p - \eta^p \mapsto
((t-\eta) + \eta)^p - \eta^p$.
\begin{enumerate}
\item[(a)]
If $f \in K \langle (t^p-\eta^p)/(\lambda |\eta^p|) \rangle$ for some
$\lambda \in (0,1)$, then 
$T(f) \in K \langle (t-\eta)/(\lambda'|\eta|) \rangle$ for
$\lambda' = \min\{\lambda^{1/p}, p\lambda\}$.
\item[(b)]
If $T(f) \in K \langle (t-\eta)/(\lambda |\eta|) \rangle$ for some
$\lambda \in (p^{-1/(p-1)},1)$, then 
$f \in K \langle (t^p-\eta^p)/(\lambda' |\eta^p|) \rangle$
for $\lambda' = \lambda^p$.
\item[(c)]
Suppose $K$ contains a primitive $p$-th root of unity $\zeta_p$.
For $m = 0,\dots, p-1$, let
$T_m: K \llbracket t^p - \eta^p \rrbracket \to K \llbracket t - \zeta_p^m \eta
\rrbracket$ be the substitution $t^p - \eta^p \mapsto
((t-\zeta_p^m \eta) + \zeta_p^m \eta)^p - \eta^p$.
If for some $\lambda  \in (0, p^{-1/(p-1)}]$ one has
$T_m(f) \in K \langle (t-\zeta_p^m \eta)/(\lambda |\eta|) \rangle$
for $m=0,\dots,p-1$, then
$f \in K \langle (t^p-\eta^p)/(\lambda' |\eta^p|) \rangle$
for $\lambda' = p^{-1} \lambda$.
\end{enumerate}
\end{cor}

\section{Moving along Frobenius}

Let $F'_\rho$ be the completion of $K(t^p)$ for the $\rho^p$-Gauss norm,
viewed as a subfield of $F_\rho$, and equipped with the derivation
$d' = \frac{d}{dt^p}$. We then have 
\[
d = \frac{dt^p}{dt} d' = pt^{p-1} d'.
\]
Given a finite differential module $(V',D')$ over $F'_\rho$, we may view
$\varphi^* V' = V' \otimes F_\rho$ as a differential 
module over $F_\rho$ for the derivation $D =
pt^{p-1}D' \otimes d$ as a differential 
\[
D(v \otimes f) = pt^{p-1}D'(v) \otimes f + v \otimes d(f).
\]

\begin{lemma} \label{L:pullback radius}
Let $(V',D')$ be a finite differential module over $F'_\rho$. Then
\[
IR(\varphi^* V') \geq \min\{IR(V')^{1/p}, p IR(V') \}.
\]
\end{lemma}
\begin{proof}
For any $\lambda < IR(\varphi^* V')$,
any complete extension $L$ of $K$,
and any generic point $t_\rho \in L$ relative to $K$
of norm $\rho$,
$(\varphi^* V') \otimes L \langle (t^p-t_\rho^p)/(\lambda \rho^p) \rangle$
admits a basis of horizontal sections.
By Corollary~\ref{C:p-th root}(a), $V' \otimes L \langle (t-t_\rho)/
(\min\{\lambda^{1/p}, p \lambda\} \rho) \rangle$ does likewise.
\end{proof}

For $V$ a differential module over $F_\rho$,
define the \emph{Frobenius descendant} of $V$ as 
the module $\varphi_* V$ obtained from $V$ by restriction along
$F'_{\rho} \to F_\rho$, viewed
as a differential module over $F'_\rho$ with differential
$D' = p^{-1} t^{-p+1} D$. Note that this operation commutes with duals.

For $m=0,\dots,p-1$,
let $W_m$ be the differential module over $F'_\rho$
with one generator $v$, such that
\[
D(v) = \frac{m}{p} t^{-p} v.
\] 
From the Newton polynomial associated to $v$, we read off
$IR(W_m) = p^{-p/(p-1)}$ for $m \neq 0$.
(You may think of the generator $v$
as a proxy for $t^m$.)
\begin{lemma} \label{L:push pull}
\begin{enumerate}
\item[(a)]
For $V$ a differential module over $F_\rho$, there are
canonical isomorphisms $\iota_m: (\varphi_* V) 
\otimes W_m \cong \varphi_* V$ for
$m=0,\dots,p-1$.
\item[(b)]
For $V$ a differential module over $F_\rho$, 
a submodule $U$ of $\varphi_* V$ is itself the Frobenius descendant
of a submodule of $V$ if and only if $\iota_m(U \otimes W_m) = U$
for $m=0,\dots,p-1$.
\item[(c)]
For $V'$ a differential module over $F'_\rho$, there is a canonical
isomorphism
\[
\varphi_* \varphi^* V' \cong \bigoplus_{m=0}^{p-1} (V' \otimes W_m).
\]
\item[(d)]
For $V$ a differential module over $F_\rho$, there is a canonical
isomorphism
$\varphi^* \varphi_* V
\cong V^{\oplus p}$.
\item[(e)]
For $V$ a differential module over $F_\rho$,
there are canonical bijections $H^i(V) \cong H^i(\varphi_* V)$ for $i=0,1$.
\item[(f)]
For $V_1, V_2$ differential modules over $F_\rho$,
there is a canonical isomorphism
\[
\varphi_* V_1 \otimes \varphi_* V_2 \cong
\bigoplus_{m=0}^{p-1} W_m \otimes \varphi_* (V_1 \otimes V_2).
\]
\end{enumerate}
\end{lemma}
\begin{proof}
Exercise.
\end{proof}

\section{Frobenius antecedents and descendants}

Unlike Frobenius descendants, Frobenius antecedents can only be constructed
in some cases, namely when the intrinsic radius is sufficiently large.

\begin{theorem}[after Christol-Dwork] \label{T:Frobenius antecedent}
Let $(V,D)$ be a finite differential module over $F_\rho$ such that
$IR(V) > p^{-1/(p-1)}$. 
Then there exists a unique differential
module $(V',D')$ over $F'_\rho$ such that 
$V \cong \varphi^* V'$ and $IR(V') > p^{-p/(p-1)}$.
For this $V'$, one has in fact $IR(V') = IR(V)^p$.
\end{theorem}
The module $V'$ in the theorem is called the \emph{Frobenius antecedent}
of $V$.
\begin{proof}[Proof of Theorem~\ref{T:Frobenius antecedent}]
We may assume $\zeta_p \in K$, as otherwise we may check everything by
adjoining $\zeta_p$ and then performing a Galois descent at the end.

We first check existence.
Since $|D|_{\tspect,V} < \rho^{-1}$, for any
$x \in V$, we may define an action of $\ZZ/p\ZZ$ on $V$ using Taylor
series:
\[
\zeta_p^m(x) = \sum_{i=0}^\infty \frac{(\zeta^m_p t - t)^i}{i!} D^i(x).
\]
Take $V'$ to be the fixed space for this action; then $V'$ is
an $F'_\rho$-subspace of $V$, and the map
$\phi^* V' \to V$ is an isomorphism by Hilbert's Theorem 90.
(You can also show this explicitly by writing down projectors onto the eigenspaces
of $V$ for the $\ZZ/p\ZZ$-action.)
By applying the $\ZZ/p\ZZ$-action to a basis of horizontal sections of $V$ 
in a generic disc $|t-t_\rho| \leq \lambda \rho$, 
and invoking Corollary~\ref{C:p-th root}(b),
we may construct
horizontal sections of $V'$ in a generic disc
$|t^p - t_\rho^p| \leq \lambda^p \rho^p$.
Hence $IR(V') \geq IR(V)^p > p^{-p/(p-1)}$.

To check uniqueness, suppose $V \cong \varphi^* V' \cong \varphi^* V''$
with $IR(V'), IR(V'') > p^{-p/(p-1)}$.
By Lemma~\ref{L:push pull}, we have
\[
\varphi_* V \cong \oplus_{m=0}^{p-1} (V' \otimes W_m)
\cong \oplus_{m=0}^{p-1} (V'' \otimes W_m).
\]
For $m=1,\dots,p-1$, we have $IR(W_m) = p^{-p/(p-1)}$; 
since $IR(V') > IR(W_m)$, we have $IR(V' \otimes W_m) = p^{-p/(p-1)}$.
Since $IR(V'') > p^{-p/(p-1)}$,
the factor $V'' \otimes W_0$ must be contained in $V' \otimes W_0$
and vice versa.

For the last assertion, note that the proof of existence gives
$IR(V') \geq IR(V)^p$, whereas Lemma~\ref{L:pullback radius}
gives the reverse inequality.
\end{proof}
\begin{cor} \label{C:antecedent}
Let $V'$ be a differential module over $F'_\rho$ such that
$IR(V') > p^{-p/(p-1)}$. Then $V'$ is the Frobenius antecedent of
$\varphi^* V'$, so $IR(V') = IR(\varphi^* V')^p$.
\end{cor}

The construction of Frobenius antecedents carries over to discs and
annuli as follows.
\begin{theorem} \label{T:Frobenius antecedent2}
Let $M$ be a finite 
differential module over $K \langle \alpha/t, t/\beta \rangle$
(we may allow $\alpha = 0$),
such that $IR(M \otimes F_\rho) > p^{-1/(p-1)}$ for
$\rho \in [\alpha,\beta]$ (or equivalently, for $\rho =\alpha$
and $\rho=\beta$).
Then there exists a unique differential module $M'$ over
$K \langle \alpha^p/t^p, t^p/\beta^p \rangle$ such that $M = M' \otimes
K \langle \alpha/t,t/\beta \rangle$ and
$IR(M' \otimes F'_\rho) > p^{-p/(p-1)}$ for 
$\rho \in [\alpha,\beta]$; this $M'$ also satisfies
$IR(M' \otimes F'_\rho) = IR(M \otimes F_\rho)^p$ for
$\rho \in [\alpha,\beta]$.
\end{theorem}
\begin{proof}
For existence and the last assertion, 
use the $\ZZ/p\ZZ$-action as in the proof of
Theorem~\ref{T:Frobenius antecedent}.
(Note that the proof does not apply directly when $\alpha = 0$; we must
make a separate calculation on a disc around the origin on which $M$
is trivial.)
For uniqueness, apply Theorem~\ref{T:Frobenius antecedent} for any single
$\rho \in [\alpha, \beta]$.
\end{proof}

In the other direction, we can control the intrinsic radius of a
Frobenius descendant.
\begin{prop} \label{P:Frobenius descendant1}
Let $V$ be a differential module over $F_\rho$. Then
\[
IR(\varphi_* V) = \min\{p^{-1} IR(V), p^{-p/(p-1)}\}.
\]
\end{prop}
\begin{proof}
First suppose $IR(V) > p^{-1/(p-1)}$. 
By Theorem~\ref{T:Frobenius antecedent}, we can write $V \cong \varphi^*
V'$ for $V'$ the Frobenius antecedent. By Lemma~\ref{L:push pull},
$\varphi_* V \cong \oplus_{m=0}^{p-1}
(V' \otimes W_m)$. 
In this direct sum, $IR(V' \otimes W_0) = IR(V') > p^{-p/(p-1)}$
and $IR(V' \otimes W_m) = IR(W_m) = p^{-p/(p-1)}$ for $m \neq 0$.
Hence $IR(\varphi_* V) = p^{-p/(p-1)}$.

Next suppose $IR(V) \leq p^{-1/(p-1)}$.
By Lemma~\ref{L:push pull}, $\varphi^* \varphi_* V \cong V^{\oplus p}$,
so by Lemma~\ref{L:pullback radius}, 
$IR(V) \geq \min\{IR(\varphi_* V)^{1/p}, p IR(\varphi_* V)\}$.
This forces $IR(\varphi_* V) \leq p^{-1} IR(V)$.

In the other direction, for $t_\rho$ a generic point of radius
$\rho$ and $\lambda \in (0, p^{-1/(p-1)})$, the module $\varphi_* V
\otimes L \langle (t^p - t_\rho^p)/(p^{-1} \lambda \rho^p) \rangle$
splits as the direct sum of
$V \otimes L \langle (t-\zeta_p^m t_\rho)/(\lambda \rho) \rangle$
over $m=0,\dots,p-1$. 
If $\lambda < IR(V)$, by applying Corollary~\ref{C:p-th root}(c),
we obtain $IR(\varphi_* V) \geq p^{-1} \lambda$.
\end{proof}

You might be tempted to think that one can run the last part of the
previous proof also
in the case $IR(V) > p^{-1/(p-1)}$ to prove that $IR(\varphi_* V)
\geq IR(V)^p$, which would contradict the first part of the proof. 
What breaks down in the
argument is that in this case,
pushing forward a basis of local horizontal sections of $V$
only gives you $(\dim V)$ local horizontal sections of $\varphi_* V$;
what they span is precisely the Frobenius antecedent of $V$.

\section{Subsidiary radii and Frobenius}

We now refine Proposition~\ref{P:Frobenius descendant1} to cover subsidiary
radii.
This will be tremendously important when we study variation of the
subsidiary radii in the next unit.

\begin{theorem} \label{T:Frobenius descendant2}
Let $V$ be a finite differential module over $F_\rho$ with
intrinsic subsidiary radii $s_1, \dots, s_n$. Then
the intrinsic subsidiary radii of $\varphi_* V$
comprise the multiset
\[
\bigcup_{i=1}^n \begin{cases} \{ s_i^p, \,p^{-p/(p-1)} \mbox{ ($p-1$ times)}\}
& s_i > p^{-1/(p-1)} \\ \{p^{-1} s_i \mbox{ ($p$ times)}\} & 
s_i \leq p^{-1/(p-1)}. 
\end{cases}
\]
\end{theorem}
\begin{proof}
It suffices to consider $V$ irreducible. First suppose
$IR(V) > p^{-1/(p-1)}$. Let $V'$ be the Frobenius antecedent
of $V$ (as per Theorem~\ref{T:Frobenius antecedent});
note that $V'$ is also irreducible.
By Lemma~\ref{L:push pull}, $\varphi_* V \cong
\oplus_{m=0}^{p-1} (V' \otimes W_m)$.
Since each $W_m$ has rank 1, $V' \otimes W_m$ is also irreducible.
Since $IR(V') = IR(V)^p$ and $IR(V' \otimes W_m) = p^{-p/(p-1)}$ for
$m \neq 0$, we have the claim.

Next suppose $IR(V) \leq p^{-1/(p-1)}$. By
Proposition~\ref{P:Frobenius descendant1}, we have $IR(\varphi_* V) 
= p^{-1} IR(V)$.
Let $W'$ be any irreducible subquotient of $\varphi_* V$; then
$IR(W') \geq IR(\varphi_* V)$, so Lemma~\ref{L:pullback radius} gives
\begin{equation} \label{eq:sub radii}
IR(\varphi^* W') \geq 
\min\{IR(W')^{1/p}, p IR(W')\} \geq 
\min\{IR(\varphi_* V)^{1/p}, p IR(\varphi_* V)\} 
= IR(V).
\end{equation}
On the other hand, $\varphi^* W'$ is a subquotient of
$\varphi^* \varphi_* V$, which by
Lemma~\ref{L:push pull} is isomorphic to $V^{\oplus p}$. Since $V$
is irreducible,
each Jordan-H\"older constituent of $\varphi^* W'$ must be isomorphic
to $V$, yielding $IR(\varphi^* W') = IR(V)$.
That forces each inequality in \eqref{eq:sub radii} to be an equality;
in particular, $IR(W')$ and $IR(\varphi_* V)$ have the same image under
the injective map $s \mapsto \min\{s^{1/p}, ps\}$.
We conclude that $IR(W') = IR(\varphi_* V) = p^{-1} IR(V)$,
proving the claim.
\end{proof}

\begin{cor}
Let $s_1 \leq \dots \leq s_n$ be the intrinsic subsidiary radii of $V$.
\begin{enumerate}
\item[(a)]
For $i$ such that $s_i < p^{-1/(p-1)}$, 
the product of the $pi$ smallest intrinsic subsidiary
radii of $\varphi_* V$ is equal to $p^{-pi} s_1^p \cdots s_i^p$.
\item[(b)]
For $i$ such that either $i=n$ or $s_{i+1} \geq p^{-1/(p-1)}$,
the product of the $pi + (p-1)(n-i)$
smallest intrinsic subsidiary radii of $\varphi_* V$ is equal to
$p^{-ni} s_1^p \cdots s_i^p$.
\end{enumerate}
In particular, the
product of the intrinsic subsidiary radii of $\varphi_* V$
is $p^{-np} s_1^p \cdots s_n^p$.
\end{cor}
Note that both conditions apply 
when $s_i = p^{-1/(p-1)}$; this will be important later.

\section{Decomposition by spectral norm}

We now extend the decomposition by spectral
norm across the barrier $|d|_{F_\rho}$. This cannot be done 
using Frobenius antecedents alone: they give no information
in case $IR(V) = p^{-1/(p-1)}$. 

\begin{prop} \label{P:split irreducible seq}
Let $V_1, V_2$ be irreducible finite differential modules over $F_\rho$
with $IR(V_1) \neq IR(V_2)$. Then $H^1(V_1 \otimes V_2) = 0$.
\end{prop}
\begin{proof}
By dualizing if necessary, we can ensure that $IR(V_2) > IR(V_1)$.
If $IR(V_1) < p^{-1/(p-1)}$, then 
any short exact sequence $0 \to V_2 \to V \to V_1^\dual \to 0$
splits by the original decomposition theorem.

Suppose that $IR(V_1) = p^{-1/(p-1)}$. 
Let $V'_2$ be the Frobenius antecedent of $V_2$; it is also irreducible,
and $IR(V'_2) = IR(V_2)^p > p^{-p/(p-1)}$.
By Theorem~\ref{T:Frobenius descendant2}, each irreducible subquotient $W$
of $\varphi_* V_1$ satisfies $IR(W) = p^{-p/(p-1)}$;
hence $H^1(W \otimes V'_2) = 0$ by the previous case, so
$H^1(\varphi_* V_1 \otimes V'_2) = 0$ by the snake lemma.

By Lemma~\ref{L:push pull},
\begin{align*}
\varphi_* V_1 \otimes \varphi_* V_2 
&\cong \oplus_{m=0}^{p-1} (\varphi_* V_1 \otimes W_m \otimes V'_2) \\
&\cong (\varphi_* V_1 \otimes V'_2)^{\oplus p}.
\end{align*}
(The last isomorphism uses the fact that
$\varphi_* V_1 \cong \varphi_* V_1 \otimes W_m$.)
This yields $H^1(\varphi_* V_1 \otimes \varphi_* V_2) = 0$;
since 
$\varphi_* (V_1 \otimes V_2)$ is a direct summand of
$\varphi_* V_1 \otimes \varphi_* V_2$ (again by Lemma~\ref{L:push pull}),
$H^1(\varphi_* (V_1 \otimes  V_2)) = 0$.
By Lemma~\ref{L:push pull} once more,
$H^1(V_1 \otimes V_2) = H^1(\varphi_* (V_1 \otimes V_2)) = 0$.

In the general case, $1 \geq IR(V_2) > IR(V_1)$.
If $IR(V_1) > p^{-1/(p-1)}$, then Theorem~\ref{T:Frobenius antecedent} 
implies that $V_1, V_2$ have Frobenius antecedents $V'_1, V'_2$,
and that any extension $0 \to V_1 \to V \to V_2^\dual \to 0$
itself is the pullback of an extension
$0 \to V'_1 \to V' \to (V'_2)^\dual \to 0$.
To show that any extension of the first type splits, it suffices to
do so for the second type; that is, we may
reduce from $V_1,V_2$ to $V'_1, V'_2$. By repeating this
enough times, we get to a
situation where $IR(V_1) \leq p^{-1/(p-1)}$. We may then apply
the previous cases.
\end{proof}

From here, the proof of the following theorem is purely formal.
\begin{theorem}[Strong decomposition theorem] \label{T:decomposition}
Let $V$ be a finite differential module over $F_\rho$. Then there exists
a decomposition 
\[
V = \bigoplus_{s \in (0,1]} V_s
\]
where every subquotient $W_s$ of $V_s$ satisfies $IR(W_s) = s$.
\end{theorem}
\begin{proof}
We induct on $\dim V$; we need only consider $V$ not irreducible.
Choose a short exact sequence $0 \to U_1 \to V \to U_2 \to 0$ with $U_2$
irreducible. Split $U_1 = \oplus_{s \in (0,1]} U_{1,s}$ where every
subquotient $W_s$ of $U_{1,s}$ satisfies $IR(W_s) = s$.
For each $s \neq IR(U_2)$, we have $H^1(U_2^\dual \otimes U_{1,s}) = 0$
by repeated application of Proposition~\ref{P:split irreducible seq}
plus the snake lemma. Consequently, we have
\[
V = V' \oplus \bigoplus_{s \neq IR(U_2)} U_{1,s},
\]
where $0 \to U_{1,IR(U_2)} \to V' \to U_2 \to 0$ is exact.
\end{proof}
As with the original decomposition theorem, we obtain
the following corollaries. 
\begin{cor}
Let $V$ be a finite differential module over $F_\rho$ whose intrinsic
subsidiary radii are all less than $1$. Then $H^0(V) = H^1(V) = 0$.
\end{cor}
\begin{cor}
With $V = \oplus_{s \in (0,1]} V_s$ as in Theorem~\ref{T:decomposition},
we have $H^i(V) = H^i(V_1)$ for $i=0,1$.
\end{cor}
This suggests that the difficulties in computing $H^0$ and $H^1$ 
arise in the case of intrinsic generic radius 1. We will 
pursue a closer study of this case in a later
unit.

\begin{cor} \label{C:tensor sub2}
If $V_1, V_2$ are irreducible
and $IR(V_1) < IR(V_2)$, 
then every irreducible subquotient $W$ of
$V_1 \otimes V_2$ satisfies 
$IR(W) = IR(V_1)$.
\end{cor}
\begin{proof}
Decompose $V_1 \otimes V_2 = \oplus_{s \in (0,1]} V_s$
according to Theorem~\ref{T:decomposition};
we have $V_s = 0$ whenever $s < IR(V_1)$.
If some $V_s$ with $s > IR(V_1)$ 
were nonzero, then $V_1 \otimes V_2$ would have an irreducible submodule
of intrinsic radius greater than $IR(V_1)$, in violation of
a result from a previous unit.
\end{proof}

\section{Integrality, or lack thereof}

It may be useful to keep in mind the following
limited integrality result for the intrinsic radius.
\begin{theorem}
Let $V$ be a finite differential module over $F_\rho$ with
intrinsic subsidiary radii $s_1, \dots, s_n$. 
Let $m$ be the largest integer such that $s_m = IR(V)$.
Then for any nonnegative integer $h$,
\[
s_1 > p^{p^{-h}/(p-1)}
 \quad \implies \quad
s_1^m \in |F^\times|^{p^{-h}} \rho^\ZZ.
\]
\end{theorem}
\begin{proof}
For $m=0$, we read this off from a Newton polygon. We reduce from $m$ to $m-1$
by applying $\varphi_*$ and invoking Theorem~\ref{T:Frobenius descendant2}.
\end{proof}

The exponent $p^{-h}$ cannot be removed; we will give an example to illustrate
this in the next unit.

\section{Off-centered Frobenius descendants}

Since pushing forward along Frobenius does not work well on a disc,
we must also consider ``off-centered''
Frobenius descendants, as follows.

For $\rho \in (p^{-1/(p-1)}, 1]$, let $F''_\rho$ be the completion of
$K((t-1)^p - 1)$ under the $\rho^p$-Gauss norm,
or equivalently, under the restriction
of the $\rho$-Gauss norm on $K(t)$.
(One could allow $K((t-\mu)^p - \mu^p)$
for any $\mu \in K$ of norm 1, but there is no loss of generality in
rescaling $t$ to reduce to the case $\mu=1$.)
For brevity, write $u = (t-1)^p - 1$.
Equip $F''_\rho$ with the derivation
\[
d'' = \frac{d}{du} = \frac{1}{du/dt} d.
\]
Given a differential module $V''$ over $F''_\rho$, we may
view $\psi^* V'' = V'' \otimes F_\rho$ as a differential module over
$F_\rho$. Given a differential module $V$ over $F_\rho$, we may
view the restriction $\psi_* V$ of $V$ along $F''_\rho \to F_\rho$
as a differential module over $F''_\rho$.

We may apply Lemma~\ref{L:p-th root} with $\eta$ replaced by $\eta+1$,
keeping in mind that $|\eta+1| = 1$ for $|\eta| \leq 1$. 
This has the net effect that
everything that holds for $\varphi$ also holds for $\psi$,
except that intrinsic radius must be replaced by generic radius.

\begin{theorem} \label{T:Frobenius antecedent3}
Let $(V,D)$ be a finite differential module over $F_\rho$ such that
$R(V) > p^{-1/(p-1)}$. 
Then there exists a unique differential
module $(V'',D'')$ over $F''_\rho$ such that 
$V \cong \psi^* V''$ and $R(V'') > p^{-p/(p-1)}$.
For this $V''$, one has in fact $R(V'') = R(V)^p$.
\end{theorem}
\begin{theorem} \label{T:Frobenius descendant3}
Let $V$ be a finite differential module over $F_\rho$ with
extrinsic subsidiary radii $s_1, \dots, s_n$. Then
the subsidiary radii of $\psi_* V$
comprise the multiset
\[
\bigcup_{i=1}^n \begin{cases} \{ s_i^p, \,p^{-p/(p-1)} \mbox{ ($p-1$ times)}\}
& s_i > p^{-1/(p-1)} \\ \{p^{-1} s_i \mbox{ ($p$ times)}\} & 
s_i \leq p^{-1/(p-1)}. 
\end{cases}
\]
\end{theorem}
Note that one cannot expect Theorem~\ref{T:Frobenius descendant3}
to hold for $\rho < p^{-1/(p-1)}$, as in that case $p^{-p/(p-1)}$ is
too big to appear as a subsidiary radius of $\psi_* V$.

\section{Notes}

Lemma~\ref{L:p-th root} is taken from
[Ked05, \S 5.3] with some typos corrected.

The Frobenius antecedent theorem of Christol-Dwork [CD94, Th\'eor\`eme~5.4]
is slightly weaker than the one given here: it only applies
for $IR(V) > p^{-1/p}$. The trouble is that they use cyclic vectors,
which create some regular singularities which they only eliminate under
the stronger hypothesis. Theorem~\ref{T:Frobenius antecedent} as stated
there first appears in [Ked05, Theorem~6.13],
except that there uniqueness is only given if $IR(V') \geq IR(V)^p$.

To the best of my knowledge, the study of Frobenius descendants
is original to this presentation; in particular,
Theorems~\ref{T:Frobenius descendant2}
and~\ref{T:Frobenius descendant3} are original. The strong decomposition
theorem (Theorem~\ref{T:decomposition}) is also original; we do not know of
a proof without Frobenius descendants.

\section{Exercises}

\begin{enumerate}
\item
Prove Lemma~\ref{L:push pull}.
\item
Prove that for any finite differential module $V'$ over $F'_\rho$
with $IR(V') > p^{-p/(p-1)}$, $H^0(V') = H^0(\varphi^* V')$.
\end{enumerate}

\end{document}

\begin{lemma} \label{L:still irred}
Let $V'$ be a finite irreducible differential module over $F'_\rho$
with $IR(V') > p^{-p/(p-1)}$. Then $\varphi^* V'$ is also irreducible.
\end{lemma}
\begin{proof}
By Corollary~\ref{C:antecedent}, $V'$ is the Frobenius antecedent of
$\varphi^* V'$; if $\varphi^* V'$ had a nontrivial proper submodule,
its Frobenius antecedent would then be contained in $V'$.
\end{proof}

\begin{lemma} \label{L:submodule}
Let $V$ be a finite differential module over $F_\rho$ with
$IR(V) = p^{-1/(p-1)}$, and let $W'$ be an irreducible
differential submodule of $\varphi_* V$ 
with $IR(W') > p^{-p/(p-1)}$. Then $W'$ is the Frobenius antecedent
of a proper nonzero
differential submodule of $V$. In particular, $V$ cannot be irreducible.
\end{lemma}
\begin{proof}
By adjunction,
the nonzero map $W' \to \varphi_* V$ induces a nonzero map
$\varphi^* W' \to V$. 
(More explicitly, pull back $W' \to \varphi_* V$ to $\varphi^* W' \to \varphi^* 
\varphi_* V$, then compose with the canonical projection $\varphi^* 
\varphi_* V \to V$.) Since $\varphi^* W'$ is also irreducible
by Lemma~\ref{L:still irred}, the map $\varphi^* W' \to V$
must be injective. 
By Corollary~\ref{C:antecedent}, $W'$ is the Frobenius
antecedent of $\varphi^* W'$ and $IR(\varphi^* W') = IR(W')^{1/p} > IR(V)$,
so $\varphi^* W' \neq V$.
\end{proof}