\subsection{Retained phases and inverse branches}
\label{sec:era-branches}

The amplitude now determines how much of the phase interval
we retain. Smaller walls permit both a smaller margin at
the ends of an era and a larger number of possible era
lengths. We choose these two restrictions together, so
that the length omitted from the initial phase interval
is controlled by the same margin.

Fix $0<\eta\leq\delta$ and $J\geq20$ in the common
parameter range of Theorems~\ref{thm:era-output}
and~\ref{thm:era-interval-inverse}. Choose
\begin{equation}
 0<\sigma<\frac1{2J},\qquad
 \beta=B^\sigma,\qquad N=\lfloor1/\beta\rfloor.
 \label{eq:era-margin-choice}
\end{equation}
The retained digits will be $1,\ldots,N$, corresponding
to $m=0,\ldots,N-1$ reflections.

\begin{lemma}
\label{lem:era-parameter-choice}
For every $\varepsilon>0$ and $C\geq0$, there is
$B_*\in(0,1)$ such that $0<B\leq B_*$ implies
\begin{gather}
 0<\beta<\frac1{160},\qquad
 B\leq\varepsilon(\beta/2)^{2J},\qquad
 CB<\beta/4,\notag\\
 B\,2(m+2-\beta)\leq\varepsilon
 \quad\hbox{whenever }m+1\leq1/\beta.
 \label{eq:era-parameter-bounds}
\end{gather}
In particular, the last inequality holds for every
$0\leq m<N$.
\end{lemma}

\begin{proof}
All the powers of $B$ in
\begin{equation*}
 4B^{1-\sigma}<\varepsilon,\qquad
 2^{2J}B^{1-2J\sigma}<\varepsilon,\qquad
 4CB^{1-\sigma}<1,\qquad B^\sigma<1/160
\end{equation*}
have positive exponents. They therefore hold below one
positive threshold, which we may take smaller than one.
The identities
\[
 B=B^{1-\sigma}\beta
   =B^{1-2J\sigma}\beta^{2J}
\]
give the terminal smallness and phase margin
in~\eqref{eq:era-parameter-bounds}.
If $(m+1)\beta\leq1$ and $0<\beta<1$, then
\[
 \beta\,2(m+2-\beta)
   =2\bigl((m+1)\beta+\beta-\beta^2\bigr)\leq4.
\]
Multiplication by $B^{1-\sigma}$ proves the radius
condition. Finally $m<N$ implies
$m+1\leq N\leq1/\beta$.
\end{proof}

We shall use graphs defined on every radius $r>2$.
Call
\[
 \Gamma(r)=(r,U(r),W(r))
\]
\emph{admissible at amplitude $B$} if $U,W$ are $C^1$
on $(2,\infty)$ and, throughout that interval,
\begin{equation}
 U,W>0,\qquad U+W\leq B,\qquad
 |U'|\leq\sqrt U,\quad |W'|\leq\sqrt W.
 \label{eq:admissible-era-graph}
\end{equation}
Lemma~\ref{lem:era-parameter-choice} places each
restriction of such a graph to $I_m$ in the domain of
Theorem~\ref{thm:era-interval-inverse}, for $m<N$.

For the moment, let $m,\beta,B$ and the graph satisfy
the hypotheses of Theorem~\ref{thm:era-interval-inverse}.
Write $\zeta_m$ for the radius inverse constructed there,
and put
\begin{equation}
 K_\beta=[2\beta,1-2\beta],\qquad
 h_m(y)=\frac2{\zeta_m(y)},\qquad
 \varphi_m=P_m\circ\Gamma.
 \label{eq:physical-inverse-branch}
\end{equation}
Thus $h_m$ maps the next entrance phase $y$ to the
initial phase $x=2/r$. Recall $g_n(y)=1/(n+y)$.

\begin{proposition}
\label{prop:era-gauss-branch}
The map $h_m$ is strictly decreasing on $K_\beta$.
For every $y\in K_\beta$, it is $C^1$ near $y$ and
\begin{gather}
 \frac1{m+2}<h_m(y)<\frac1{m+1},\qquad
 \varphi_m(2/h_m(y))=y,\qquad h_m'(y)<0,\notag\\
 |h_m(y)-g_{m+1}(y)|\leq\frac{CB}{(m+1)^2},\qquad
 \left|\log\frac{|h_m'(y)|}{|g_{m+1}'(y)|}\right|
       \leq5C\sqrt B.
 \label{eq:era-gauss-branch-estimates}
\end{gather}
Here $C$ is the constant in
Theorem~\ref{thm:era-interval-inverse}.
\end{proposition}

\begin{proof}
Set $n=m+1$ and $q(y)=\zeta_m(y)/2$. The interval
inverse gives
\begin{equation}
 n<q(y)<n+1,\qquad q'(y)>0,\qquad
 |q(y)-(n+y)|\leq E,\qquad |\log q'(y)|\leq D,
 \label{eq:era-half-radius-inverse}
\end{equation}
where $E=CB$ and $D=C\sqrt B$. In particular,
$h_m=1/q$ is locally $C^1$ and
\[
 h_m'(y)=-\frac{q'(y)}{q(y)^2}<0.
\]
Its range and its inverse identity follow from
\eqref{eq:era-half-radius-inverse} and
$2/h_m(y)=\zeta_m(y)$. The mean value theorem proves
strict decrease on the whole interval.

Since $q(y)\geq n\geq1$ and $y\geq0$,
\[
 |h_m(y)-g_n(y)|
 =\frac{|q(y)-(n+y)|}{q(y)(n+y)}
 \leq\frac E{n^2}.
\]
For the derivatives, the exact ratio is
\begin{equation}
 \frac{|h_m'(y)|}{|g_n'(y)|}
       =q'(y)\left(\frac{n+y}{q(y)}\right)^2.
 \label{eq:era-reciprocal-derivative}
\end{equation}
The positive factor $t=(n+y)/q(y)$ satisfies
$|t-1|\leq E/q(y)\leq E$. The phase margin gives
$E<\beta/4<1/4$; hence
$1-1/t\leq\log t\leq t-1$ implies $|\log t|\leq2E$.
Taking logarithms in~\eqref{eq:era-reciprocal-derivative}
therefore yields an error at most $D+4E$.
The input conditions imply $B\leq\varepsilon\leq10^{-4}$,
so $B\leq\sqrt B$ and $D+4E\leq5C\sqrt B$.
\end{proof}

For an admissible graph and the choices
\eqref{eq:era-margin-choice}, define its retained phases by
\begin{equation}
 \mathcal V_\Gamma(B)
   =\bigcup_{m=0}^{N-1}
       \bigl(h_m(1-2\beta),h_m(2\beta)\bigr).
 \label{eq:era-retained-phases}
\end{equation}
These are exactly the images of the open interval
$(2\beta,1-2\beta)$ under the retained inverse branches.

\begin{theorem}
\label{thm:era-retained-phases}
There are $C\geq1$ and $B_*\in(0,1)$, depending only
on $\eta,J,\sigma$, such that every graph admissible at
$0<B\leq B_*$ has the following properties.
One has $0<\beta<1/160$, and each $m<N$ admits the
inverse branch and all the estimates of
Proposition~\ref{prop:era-gauss-branch}.
Its closed image is
\begin{equation}
 h_m(K_\beta)
   =[h_m(1-2\beta),h_m(2\beta)].
 \label{eq:era-closed-branch-image}
\end{equation}
The open branch images are pairwise disjoint, and
\begin{equation}
 \lambda\bigl((0,1)\setminus\mathcal V_\Gamma(B)\bigr)
       \leq10B^\sigma.
 \label{eq:era-one-step-loss}
\end{equation}
The omitted set is Borel.
\end{theorem}

\begin{proof}
Choose $\varepsilon,C$ for the interval inverse and use
Lemma~\ref{lem:era-parameter-choice} to fix $B_*$.
Every $I_m$, $m<N$, lies in $(2,\infty)$, and the
admissible graph obeys all the required bounds there.
Proposition~\ref{prop:era-gauss-branch} therefore applies
with the same constants. Continuity and strict decrease
give both the closed image~\eqref{eq:era-closed-branch-image}
and its open-interval version.

Put $L_m=h_m(2\beta)$, $R_m=h_m(1-2\beta)$.
Each interval $(R_m,L_m)$ lies inside
$(1/(m+2),1/(m+1))$, so these intervals are disjoint
subsets of $(0,1)$. Their union is open.
It remains to estimate its total length.
For $n=m+1$, the elementary difference formula gives
\[
 |g_n(s)-g_n(t)|\leq\frac{|s-t|}{n^2},
 \qquad s,t\geq0.
\]
Apply this at $(s,t)=(2\beta,0)$ and
$(1-2\beta,1)$, and add the two endpoint errors from
\eqref{eq:era-gauss-branch-estimates}. With $E=CB$,
\begin{equation}
 L_m-R_m\geq
 \frac1{m+1}-\frac1{m+2}
       -\frac{4\beta+2E}{(m+1)^2}.
 \label{eq:era-branch-length}
\end{equation}
The sums in this inequality satisfy
\[
 \sum_{m=0}^{N-1}
    \left(\frac1{m+1}-\frac1{m+2}\right)
       =1-\frac1{N+1},\qquad
 \sum_{m=0}^{N-1}\frac1{(m+1)^2}\leq2.
\]
For the second bound, sum
$n^{-2}\leq2(n^{-1}-(n+1)^{-1})$ over $1\leq n\leq N$.
Disjointness and~\eqref{eq:era-branch-length} now give
\begin{equation*}
 \lambda\bigl((0,1)\setminus\mathcal V_\Gamma(B)\bigr)
       \leq\frac1{N+1}+8\beta+4E.
\end{equation*}
Since $N=\lfloor1/\beta\rfloor$, one has
$1/(N+1)<\beta$. Also $E<\beta/4$.
The right-hand side is therefore less than $10\beta$,
which proves~\eqref{eq:era-one-step-loss}.
\end{proof}

The estimate includes the tail of digits larger than $N$
and the gaps at both ends of every retained digit interval.
It uses only the branch endpoints; logarithmic derivative
control is reserved for the distortion of successive
compositions.