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\title{Positive Metric Entropy for the Standard Map\\
       at Large Parameters}
\author{OpenAI}
\date{September 23, 2026}

\begin{document}
\maketitle

\begin{abstract}
The standard sine map of the two-dimensional torus is a one-line formula
with an entire universe inside it. We prove that, for every sufficiently
large positive parameter, this map tears up area at a positive rate: a
set of positive area is genuinely chaotic, the top Lyapunov exponent is
strictly positive there, and the Kolmogorov--Sinai entropy with respect
to normalized area is strictly positive. This answers affirmatively the
positive-parameter version of Sinai's entropy conjecture, in the
formulation of Berger and Turaev, and gives the stronger
all-sufficiently-large-parameter conclusion.
\end{abstract}

\tableofcontents

\section{Preface: what this paper is about}
\label{sec:preface}

This section is a plain-language tour. Nothing here is used in the
proof; a reader who already knows the topic may skip directly to
\cref{sec:intro}.

\subsection*{The torus}

Start with the unit square $[0,1]\times[0,1]$. Glue the left edge to
the right edge, and the bottom edge to the top. The result is a flat
doughnut called the \emph{torus}, written $\T^2=\R^2/\Z^2$. A point on
the torus is a pair $(x,y)$ that we are free to shift by an integer in
each coordinate, so $(0.3,0.7)$ and $(1.3,-0.3)$ are the same point.
The torus inherits area from the unit square, and this \emph{normalized
area}, usually denoted $m$, is the only measure in this paper. When we
say a set has positive area we mean $m(A)>0$.

\subsection*{Maps and chaos}

A \emph{map} on the torus is a rule $f:\T^2\to\T^2$ giving each point
$z$ a next point $f(z)$. Iterating it produces the \emph{orbit}
$z,f(z),f^2(z),f^3(z),\ldots$ of $z$. Some orbits look tame: they may
return close to where they started after a while, or wander into a nice
pattern. Other orbits are chaotic: two starting points that begin very
close end up very far apart, and no matter how finely you prepare the
initial state, you cannot predict the long-time future.

A standard way to measure chaos quantitatively at a single point $z$
is via the \emph{Lyapunov exponent}. Take a small tangent vector $v$ at
$z$ and push it forward by the derivative matrices $Df^n(z)$. If the
length typically grows like
\[
 \norm{Df^n(z)v}\sim e^{\lambda n}\norm v,
\]
then $\lambda=\lambda(z,v)$ is the Lyapunov exponent in the $v$
direction. A positive exponent means that nearby points separate
exponentially fast; this is the technical meaning of chaos.

\subsection*{The standard map}

The central object of this paper is the \emph{standard sine map}, or
simply \emph{standard map}:
\begin{equation}\label{preface:map}
 f_k(x,y)=\bigl(x+y+k\sin(2\pi x),\;y+k\sin(2\pi x)\bigr)
 \pmod{\Z^2},\qquad k\in\R.
\end{equation}
The map is area preserving, because at each step the derivative is a
shear of determinant one. What does the formula do? Write
$\phi(x)=2x+k\sin(2\pi x)$ and change coordinates by
$(x,y)\mapsto(x,x-y)$. This change of coordinates is an area-preserving
linear map, and it rewrites \eqref{preface:map} as
\begin{equation}\label{preface:rec}
 q_{i+1}=\phi(q_i)-q_{i-1}\pmod1,\qquad
 \phi(x)=2x+k\sin(2\pi x).
\end{equation}
When $k=0$, $\phi(x)=2x$ and the recurrence is linear; the second
coordinate stays put and the first advances by $y$ at every step. The
torus is chopped into horizontal circles, each rotating at a fixed
speed. Very tame. When $k>0$ a \emph{twist} appears: the shear is
proportional to $k\sin(2\pi x)$, so points at different $x$ are
kicked by different amounts. For small $k$ the KAM theorem still
guarantees that most orbits are tame, trapped on smooth invariant
curves. For large $k$, numerics and partial theory suggest that the
opposite happens and chaos dominates.

\subsection*{From exponent to entropy}

A 2-by-2 determinant-one matrix has singular values $D$ and $D^{-1}$
with $D\ge1$. Since the standard map preserves area, at almost every
$z$ the two Lyapunov exponents are $\lambda_+(z)$ and $-\lambda_+(z)$,
for a number $\lambda_+(z)\ge0$. A point is chaotic if
$\lambda_+(z)>0$.

There is a global invariant that measures chaos \emph{on average},
called the \emph{Kolmogorov-Sinai entropy} $h_m(f_k)$ of $f_k$ with
respect to normalized area $m$. Entropy of $1$ bit per step, for
instance, means the map produces one bit of new information about the
initial state at every step, in the sense of information theory.
Pesin's formula, valid for any smooth area-preserving diffeomorphism of
a compact surface, connects chaos on a set to entropy:
\begin{equation}\label{preface:pesin}
 h_m(f_k)=\int_{\T^2}\lambda_+(z)\dd m(z).
\end{equation}
So positive entropy is exactly positive average chaos. Proving positive
entropy means proving that a set of starting points of positive area
has strictly positive top Lyapunov exponent.

\subsection*{The conjecture and our result}

Sinai, and later Berger and Turaev in the precise formulation we
follow, conjectured that the standard map has positive metric entropy
for parameters $k$ of positive Lebesgue measure, and in particular for
all sufficiently large $k$. Our main theorem resolves the strongest
version.

\begin{theorem}[Main theorem]\label{preface:main}
There exists a constant $k_0>0$ such that, for every $k\ge k_0$,
\[
 h_m(f_k)>0.
\]
Equivalently, for every $k\ge k_0$ the top Lyapunov exponent
$\lambda_+$ is positive on a set of positive area.
\end{theorem}

The theorem deliberately claims no more than this. It does not claim
that $\lambda_+(z)>0$ at every $z$, or that $m$ is ergodic, or a
uniform positive lower bound on entropy. It claims that a positive-area
set of starting points is chaotic, and \eqref{preface:pesin} upgrades
that to strictly positive entropy. The threshold $k_0$ is huge and not
made explicit.

There is a corollary worth stating already. Under the same hypothesis
as the main theorem, there is a positive-area invariant set $E$ on
which $f_k$ is nonuniformly hyperbolic and ergodic, and the return map
to a finite union of pieces is Bernoulli. We give this as
\cref{main:components} below.

\subsection*{How to read the rest of the paper}

The technical proof has two quantitative ingredients, a multiscale
limit construction, and one contradiction argument.
\begin{itemize}[leftmargin=*]
\item \Cref{sec:intro} states the main theorem, discusses where it sits
  in the literature, and previews the proof.
\item \Cref{sec:setup} builds the algebra: transfer matrices,
  growth logarithms, and the expected shortfall $e_n$.
\item \Cref{sec:cancelation,sec:local-scarcity} prove the first
  quantitative input, a probabilistic bound on nearly-maximal
  cancellations. It says that two long products that grow almost as
  fast as possible from a common point, but whose concatenation barely
  grows at all, are exponentially unlikely.
\item \Cref{sec:bridge} proves the second quantitative input, a
  product estimate across a fast bridge. It says that if a long
  interval grows at nearly maximal rate, then nonnegative weights at
  the two ends behave almost independently on average.
\item \Cref{sec:scales,sec:shape} build the multiscale limit
  measures and classify their support.
\item \Cref{sec:end} assembles a signed deficit that telescopes to
  a positive value, decomposes it into a terminal and an invariant
  contribution, and derives a contradiction from the two estimates.
\end{itemize}

\section{Introduction and statement of the main theorem}
\label{sec:intro}

Let $m$ be normalized Lebesgue area on $\T^2=\R^2/\Z^2$. The standard
sine map \eqref{preface:map}, with parameter $k\in\R$, is an
area-preserving $C^\infty$ diffeomorphism of the torus. We consider
its Kolmogorov-Sinai entropy $h_m(f_k)$ with respect to $m$.

It is important to distinguish two notions of complexity. A map can
have huge \emph{topological} entropy while the area measure sees only
rare orbits; and a map with a hyperbolic invariant set of full
Hausdorff dimension can still have zero area on that set. The theorem
below is about area: it is a statement about where the map actually
tears a positive fraction of the torus apart, not merely somewhere.

\begin{theorem}\label{main:entropy}
There exists $k_0>0$ such that, for every $k\ge k_0$,
\[
 h_m(f_k)>0.
\]
Equivalently, for every such $k$ the largest Lyapunov exponent
$\lambda_+$ of $f_k$ is positive on a set of positive area $m$.
\end{theorem}

In particular the parameter set with positive metric entropy contains
an interval of positive length. This resolves affirmatively the
Berger--Turaev formulation of Sinai's positive metric entropy
conjecture (see \cite{BergerTuraev2019,Obata2021}) and yields the
stronger all-sufficiently-large-parameter conclusion. Berger and
Turaev's recurrence $(x,y)\mapsto(2x-y+a\sin(2\pi x),x)$ is conjugate
to \eqref{preface:map} by $(x,y)\mapsto(x,x-y)$, with $a=k$.

\paragraph{What this theorem does and does not say.} The assertion is
about both parameter measure and, at each parameter, phase-space
area. It says nothing about ergodicity, almost-everywhere positivity
of exponents, or a uniform positive lower bound on entropy.

\paragraph{Literature context.} The standard map is Chirikov's
prototype model for conservative instability \cite{Chirikov1979}.
At large parameters, Duarte constructed hyperbolic basic sets and
exhibited elliptic-island and tangency phenomena in the standard
family \cite{Duarte1994}. Gorodetski \cite{Gorodetski2012} produced
transitive invariant sets of full Hausdorff dimension for a residual
set of large parameters. These results reveal extensive deterministic
structure, but none implies positive-area chaos: hyperbolic sets may
have zero area, full Hausdorff dimension does not imply positive
area, and residual parameter sets may have zero Lebesgue measure.

Other entropy statements concern different notions. Obata
\cite{Obata2021} proved uniqueness of the measure of maximal entropy
for sufficiently large parameters, without identifying it as area.
Oliveira \cite{Oliveira2024} proved positive topological entropy for
$k\ne2/\pi$ in the normalization of \eqref{preface:map}; topological
entropy does not by itself give positive entropy for the area measure.
Berger and Turaev \cite{BergerTuraev2019} proved Herman's
positive-entropy conjecture by conservative perturbations, including
perturbations that are not amplitude changes in \eqref{preface:map}.
Related hyperbolicity appears in Anosov-driven skew products
\cite{BergerCarrasco2014} and randomly perturbed standard maps
\cite{BlumenthalXueYoung2017}; those change the dynamical system or
its probability law. Our question concerns Lebesgue area for the
exact sine map, at each sufficiently large amplitude.

\paragraph{Pesin's formula.} We use the classical entropy formula,
not a new entropy theory. The multiplicative ergodic theorem
\cite{Oseledets1968} gives almost-everywhere Lyapunov exponents for
any $f_k$. The determinant-one condition makes them $\lambda_+(z)$
and $-\lambda_+(z)$ with $\lambda_+(z)\ge0$. Pesin's formula in the
absolutely continuous invariant-measure form
\cite{Pesin1977,Mane1981,Mane1983Errata} gives
\begin{equation}\label{main:pesin}
 h_m(f_k)=\int_{\T^2}\lambda_+(z)\dd m(z).
\end{equation}
The map is smooth, the surface is compact and boundaryless, and $m$
is an absolutely continuous invariant probability. The derivative
and its logarithm are bounded at each fixed parameter, so all
integrability hypotheses hold. Neither ergodicity nor a prior
nonzero-exponent assumption is required. The task reduces to
positivity of $\lambda_+$ on a set of positive area.

\begin{corollary}[Nonuniformly hyperbolic ergodic component]
\label{main:components}
For each $k\ge k_0$ there is a measurable $f_k$-invariant set
$E=E(k)$ with $m(E)>0$ on which
$(E,m(E)^{-1}m|_E,f_k)$ is ergodic with one positive and one negative
Lyapunov exponent almost everywhere. There is $N=N(k,E)\ge1$ and a
measurable partition $E=E_0\sqcup\cdots\sqcup E_{N-1}$ with
$f_k(E_j)=E_{j+1\bmod N}$ such that
$(E_j,m(E_j)^{-1}m|_{E_j},f_k^N)$ is Bernoulli for each $j$. All
invariance and partition statements hold modulo $m$-null sets, and
$m(E_j)=m(E)/N>0$.
\end{corollary}

\begin{proof}
Fix $k\ge k_0$ and let $H_k$ be the invariant Oseledets set where
$\lambda_+>0$. It has positive area by \cref{main:entropy}, and
$\det Df_k=1$ makes the other exponent $-\lambda_+<0$ there. The
normalized restriction $\nu_k=m(H_k)^{-1}m|_{H_k}$ is an absolutely
continuous invariant probability on the compact torus with no zero
Lyapunov exponents. Pesin's ergodic and Bernoulli decomposition
theorems in the form stated in \cite[Theorems 3.3--3.4]{KatokBurns1994}
apply to $\nu_k$ and yield a positive-$\nu_k$ ergodic component $E$
with a finite cyclic partition as asserted. Since
$\nu_k(E)=m(E)/m(H_k)>0$, the component has positive area.
Normalizing $\nu_k$ on $E$ or on a partition piece agrees with
normalizing $m$. Area preservation gives $m(E_j)=m(E)/N$.
\end{proof}

\subsection*{Proof mechanism}

Suppose toward a contradiction that arbitrarily large parameters
have $\lambda_+=0$ almost everywhere. In transfer coordinates, the
derivative $Df_k$ becomes a product of $2\times2$ determinant-one
matrices $T_{i,j}$. Put $M=2\pi k+4$ (a uniform bound for one step
and its inverse) and $g(i,j)=\log_M\norm{T_{i,j}}$. Then
$0\le g(i,j)\le|i-j|$ and the triangle inequality holds by
submultiplicativity of the norm, so $g$ is a stationary
\emph{semimetric} on orbit time. The expected \emph{shortfall}
\[
 e_n=\E\Bigl[1-\frac{g(0,n)}n\Bigr]\in[0,1]
\]
measures how far the derivative is, on average, from maximum growth
at scale $n$. At a $\lambda_+=0$ parameter, $e_n\to1$ as
$n\to\infty$; at fixed $n$, $e_n\to0$ as $k\to\infty$ because the
linear part of $\phi$ dominates the bounded sinusoid. Between those
regimes, $e_n$ first grows. There is a dyadic window where it grows
regularly; normalizing by the terminal shortfall inside the window
produces rescaled distance arrays
$g_n(s,t)=g(\lfloor ns\rfloor,\lfloor nt\rfloor)/n$ whose laws have
locally finite vague limits on a compact space of semimetric arrays.

Two quantitative estimates control the possible limits. The
combination is enough to force any non-affine limit into one of
three shapes (entirely slow, one unit-speed exterior ray, or two
unit-speed exterior rays). A capped shortfall $\Phi$ is then used:
$\Phi$ is smooth, nearly linear with small slope near zero, and
equal to $1$ on uniformly slow intervals. Summing a symmetric second
difference of $\Phi$ over the dyadic window telescopes to a positive
value, but decomposes into a terminal contribution that is strictly
smaller (a nonzero terminal measure keeps a slow interval visible at
arbitrarily fine scale) and an invariant remainder that is
nonpositive (translation and dilation-force averaging). The two
bounds contradict the positive telescoping value.

The two quantitative estimates are the following.

\emph{Cancellation.} Two long products that grow almost at maximal
rate from a common center, but whose concatenation is nearly as
short as possible, occur with probability exponentially small in the
length. The mechanism: such a cancellation supplies a single
linearized solution that is tiny at both endpoints; matched expansion
scales on its two sides give many local tests; a uniform analytic
scarcity estimate at matched scales converts those into exponential
decay. Precedent for finite-time direction and tangency geometry in
the standard family is in Bloor--Luzzatto \cite{BloorLuzzatto2009};
the matched-scale estimate used here is proved from scratch.

\emph{Bridge.} On a bridge $[a,b]$ that grows at nearly maximal
rate, nonnegative observations near $a$ and near $b$ behave almost
independently on average, up to an exponential error in the bridge
length. The mechanism is a change of variables: finite Dirichlet
problems build two transverse contracting graph families; a pair of
transverse slices bounds the desired weighted integral by a multiple
of a bound near the opposite end. This is a product estimate and does
not presuppose mixing. Blumenthal \cite{Blumenthal2021} obtained
fixed-parameter correlation bounds useful over a finite parameter-
dependent window; those do not give the weighted estimate across
arbitrary long fast bridges.

\paragraph{Organization.} \Cref{sec:setup} fixes transfer coordinates
and growth conventions. \Cref{sec:cancelation,sec:local-scarcity}
prove cancellation scarcity and its local analytic estimate.
\Cref{sec:bridge} proves the product estimate across a fast bridge.
\Cref{sec:scales} constructs the dyadic limit measures;
\Cref{sec:shape} classifies their support; \Cref{sec:end} completes
the capped-deficit balance. Everything used is proved here; no
external standard-map entropy assertion is imported.

\section{Transfer matrices and growth shortfalls}
\label{sec:setup}

Throughout the proof, the parameter $k$ is positive and sufficiently
large. Set
\begin{equation}\label{setup:parameters}
 K=2\pi k,\qquad M=K+4,\qquad
 \phi(x)=2x+k\sin(2\pi x),\qquad
 v(x)=\phi'(x)=2+K\cos(2\pi x).
\end{equation}
Growth logarithms are to base $M$. Natural logarithms are retained in
the definitions of entropy and Lyapunov exponents. Unless a dependence
is stated, a constant denoted $C$ is allowed to increase from one
occurrence to the next, and does not depend on the orbit, the integer
times, or $M$.

Write $(x_i,y_i)=f_k^i(z)$ for all $i\in\Z$, and put $q_i=x_i\in\T$.
The identity $x_i-y_i=x_{i-1}$ shows that the fixed torus automorphism
$(x,y)\mapsto(x,x-y)$ replaces the state by $(q_i,q_{i-1})$. It
preserves normalized area, and the recurrence becomes
\begin{equation}\label{setup:recurrence}
 q_{i+1}=\phi(q_i)-q_{i-1}\pmod1.
\end{equation}
When varying an orbit locally we choose initial real lifts and iterate
this same recurrence over $\R$, without reducing modulo one. Its
linearization is
\begin{equation}\label{setup:transfer}
 \binom{u_{i+1}}{u_i}
 =A_i\binom{u_i}{u_{i-1}},\qquad
 A_i=\begin{pmatrix}v(q_i)&-1\\1&0\end{pmatrix}.
\end{equation}
For $a,b\in\Z$, let $T_{a,b}$ carry the linearized pair at time $a$ to
the pair at time $b$. Thus $T_{a,c}=T_{b,c}T_{a,b}$ and
$T_{b,a}=T_{a,b}^{-1}$.

All matrix and vector norms are Euclidean unless otherwise stated.
Every transfer matrix has determinant one, so its singular values are
$D,D^{-1}$ with $D\ge1$; in particular
\begin{equation}\label{setup:norm-bounds}
 \norm{T_{a,b}}=\norm{T_{b,a}},\qquad
 1\le\norm{T_{a,b}}\le M^{|b-a|}.
\end{equation}
Indeed both $A_i$ and $A_i^{-1}$ have Euclidean norm at most $M$. The
same bound holds in the maximum norm: their largest absolute row sum
is at most $K+3<M$. Also
\begin{equation}\label{setup:derivatives}
 |\phi'|\le M,\qquad |\phi^{(p)}|\le C_pM\quad(p\ge2)
\end{equation}
for every fixed derivative order $p$. These estimates apply equally on
all real lifts. The fixed change of coordinates does not alter
Lyapunov exponents, since conjugating a derivative product by a fixed
invertible matrix changes its logarithmic norm by a bounded amount.

The probability $\PP$ and expectation $\E$ always refer to normalized
area of the initial state. At every fixed time $i$, the pair
$(q_i,q_{i-1})$ has uniform product distribution on $\T^2$. The
process is stationary under all integer time shifts. This does not
assert independence of coordinates observed at different times.

We record growth and its expected shortfall by
\begin{equation}\label{setup:distance-deficit}
 g(i,j)=\log_M\norm{T_{i,j}},\qquad
 e_n=\E\left[1-\frac{g(0,n)}n\right]\quad(n\ge1).
\end{equation}
The transfer identities make $g$ a stationary semimetric on orbit time:
it is symmetric, vanishes on the diagonal, and satisfies the triangle
inequality. Distinct times may have zero distance. The one-step bound
gives
\begin{equation}\label{setup:distance-bound}
 0\le g(i,j)\le|i-j|,
\end{equation}
so $e_n\in[0,1]$ measures the average loss from the upper growth rate
over an interval of length $n$. These definitions do not assume that
Lyapunov exponents vanish; that assumption enters only in the
contradiction argument of \cref{sec:scales}.

For nonzero vectors $a,b\in\R^2$, $a\wedge b$ denotes their
determinant. For unit vectors its absolute value is the absolute sine
of their angle and depends only on the unoriented lines. The absolute
sines satisfy the triangle inequality
\begin{equation}\label{setup:sine-triangle}
 |a\wedge c|\le |a\wedge b|+|b\wedge c|
\end{equation}
for unit vectors: express the two outer vectors in the orthonormal
basis given by $b$ and a perpendicular vector, and expand.

The following elementary observation converts a small total growth
deficit into good growth on many subintervals.

\begin{lemma}[Shortfall count]\label{setup:shortfall-count}
Let $z_0,\ldots,z_N$ be real numbers with $z_j-z_{j-1}\le1$, and set
\[
 \mathcal D=N-(z_N-z_0)
 =\sum_{j=1}^N\bigl(1-z_j+z_{j-1}\bigr).
\]
For $c<1$, the number of indices $j$ for which some $0\le l<j$
satisfies $z_j-z_l<c(j-l)$ is at most $C\mathcal D/(1-c)$, with an
absolute constant $C$. The same conclusion holds for the number of
indices $j$ with some $j<l\le N$ satisfying
$z_l-z_j<c(l-j)$.
\end{lemma}

\begin{proof}
For each bad index choose one witnessing interval. Its length is a
positive integer, and the sum of the nonnegative step shortfalls over
it exceeds $(1-c)$ times its length. Select disjoint intervals greedily
in decreasing order of length, discarding each interval that
intersects one already selected. Regard intervals as half-open when
deciding disjointness, so the shortfalls over selected intervals have
disjoint sets of steps.

Every discarded interval meets a selected interval of at least its
length and is contained in the concentric threefold enlargement of
that selected interval. The bad endpoints are therefore covered by
these enlargements. An interval of integer length $L\ge1$ contains at
most $3L+1$ integer points in its threefold enlargement. If the
selected lengths are $L_1,\ldots,L_s$, the number of bad indices is
at most $4\sum_rL_r$. Disjointness and nonnegativity give
$(1-c)\sum_rL_r\le\mathcal D$. This proves the first statement; the
same selection applied to forward witnessing intervals proves the
second.
\end{proof}

We will use the count both for logarithms of vector norms, whose
one-step increments are at most one by \cref{setup:norm-bounds}, and
for the logarithms of successive maximum norms of solutions of
\cref{setup:transfer}. Neither use requires a lower bound on each
individual increment.

\section{Nearly full-rate cancellations}
\label{sec:cancelation}

The first quantitative input concerns two long products that expand
almost as fast as possible from a common center, but whose
concatenation barely grows. The shape of such a configuration forces
a single linearized solution to be small at both distant endpoints.
Matched expansion scales on its two sides provide many independent
local tests, and a uniform analytic scarcity estimate turns those
tests into exponential probability decay.

\begin{theorem}[Cancellation estimate]\label{can:theorem}
There are constants $\delta>0$, $C<\infty$, $\gamma>0$, and
$M_0<\infty$ such that, for every $M\ge M_0$, every $i\in\Z$, and
every $n\in\N$,
\begin{equation}\label{can:event-bound}
 \PP\left\{
 \begin{array}{l}
 \min\bigl(\log_M\norm{T_{i,i-n}},
           \log_M\norm{T_{i,i+n}}\bigr)\ge(1-\delta)n,\\[2pt]
 \log_M\norm{T_{i-n,i+n}}\le\delta n
 \end{array}
 \right\}
 \le C\exp(-\gamma n).
\end{equation}
All four constants are independent of the center, the length, and
$M$ in the indicated range.
\end{theorem}

Throughout this section and the next, fix
\begin{equation}\label{can:c}
 c=\frac{99}{100}.
\end{equation}
Constants used before the final choice of $\delta$ are independent of
$\delta$, $n$, and $M$, unless a dependence is stated explicitly.

\subsection*{A solution small at both endpoints}

By stationarity it is enough to consider $i=0$. Suppose the event in
\cref{can:event-bound} occurs. Write
\[
 D_- = \norm{T_{0,-n}},\qquad D_+ = \norm{T_{0,n}},
\]
and let $a_-$ and $a_+$ be unit most-expanded input directions for
these two products. If $b$ is a unit vector perpendicular to $a_-$,
then
\[
 \norm{T_{0,-n}b}=D_-^{-1},\qquad
 \norm{T_{0,n}b}\ge D_+\abs{a_-\wedge a_+}.
\]
The first equality uses determinant one: the singular values of
either product are its norm and the reciprocal of its norm. Applying
the full product to the vector $T_{0,-n}b$ gives
\[
 \norm{T_{-n,n}}
 \ge D_-D_+\abs{a_-\wedge a_+},
 \qquad
 \abs{a_-\wedge a_+}\le M^{-(2-3\delta)n}.
\]
Resolving $b$ in the singular input directions of $T_{0,n}$ also gives
the upper bound
\[
 \norm{T_{0,n}b}
 \le D_+^{-1}+D_+\abs{a_-\wedge a_+}
 \le 2M^{-(1-3\delta)n}.
\]
The left endpoint has the smaller bound $D_-^{-1}$. Thus the
solution of the linearized recurrence with central pair $b$ is small
at both endpoints. One of its entries at times $0$ and $-1$ has
absolute value at least $1/\sqrt2$. Choose that time $e\in\{0,-1\}$
and divide the whole solution by its value there. The normalized
solution satisfies
\begin{equation}\label{can:small-solution}
 u_e=1,\qquad \norm{(u_0,u_{-1})}\le\sqrt2,\qquad
 \norm{(u_n,u_{n-1})},\ \norm{(u_{-n},u_{-n-1})}
 \le 3M^{-(1-3\delta)n}.
\end{equation}
This construction is only used as an existence statement; no
measurable selection of a singular direction is required.

\subsection*{Outward paths and good indices}

Fix $e$ and $q_e$. Use $\theta=q_{e-1}$ as the other circle
coordinate, so that $q_{e+1}=\phi(q_e)-\theta$ modulo one. The two
outward paths are
\[
 q_j^-=q_{e-j},\qquad q_j^+=q_{e+j}\qquad(j\ge0).
\]
On either path positions and every linearized solution obey the same
second-order recurrence. In calculations involving one path, omit the
sign and write $q_j,u_j,v_j=v(q_j)$, with $u_0=1$. Define
\begin{equation}\label{can:t-definition}
 t_0=0,\qquad t_1=1,\qquad t_{j+1}=v_jt_j-t_{j-1}.
\end{equation}
For local real lifts of the positions,
\[
 \partial_\theta q_j^-=t_j^-,\qquad
 \partial_\theta q_j^+=-t_j^+.
\]
The sign difference comes only from the initial positions at index
one; it does not change the recurrence or the Wronskian identity
\begin{equation}\label{can:wronskian}
 u_{j-1}t_j-u_jt_{j-1}=1\qquad(j\ge1).
\end{equation}
Indeed the left side is one at $j=1$ and the recurrence preserves it.
A common normalized solution on the two paths has
\begin{equation}\label{can:center-equation}
 u_1^-+u_1^+=v(q_e).
\end{equation}
Conversely, two outward solutions with value one at index zero and
satisfying \cref{can:center-equation} assemble into one solution of
the original recurrence across the center.

On each path the two small endpoint entries in
\cref{can:small-solution} are indexed by $N-1,N$. More explicitly, the
forward and backward lengths are $N_+=n-e$ and $N_-=n+e+1$,
respectively. Define pair-growth logarithms
\[
 \tau_j=\log_M\max(\abs{t_j},\abs{t_{j-1}}),\quad
 \upsilon_j=\log_M\max(\abs{u_j},\abs{u_{j-1}})
 \quad(1\le j\le N).
\]
Both pairs are nonzero, so these logarithms are defined. The bounds
for one step and its inverse in the maximum norm give
\begin{equation}\label{can:log-increments}
 \abs{\tau_{j+1}-\tau_j}\le1,\qquad
 \abs{\upsilon_{j+1}-\upsilon_j}\le1.
\end{equation}

\begin{definition}\label{can:good-definition}
An index $2\le j\le N-2$ is \emph{good} for the selected normalized
solution if
\begin{equation}\label{can:good}
 \begin{aligned}
 \tau_j-\tau_l&\ge c(j-l)&& (1\le l<j),\\
 \tau_{j+1}-\tau_j&\ge c,\\
 \upsilon_j-\upsilon_{j+1}&\ge c,
 \qquad\upsilon_{j+1}-\upsilon_{j+2}&\ge c.
 \end{aligned}
\end{equation}
For a path extending beyond a selected index, the same inequalities
define goodness without reference to a particular terminal length
$N$.
\end{definition}

\begin{lemma}\label{can:many-good}
There is a constant $A$ such that, on each outward path associated to
\cref{can:small-solution}, the number of indices in $\{2,\ldots,N-2\}$
which are not good is at most
\begin{equation}\label{can:bad-count}
 \frac{A(\delta n+1)}{1-c}.
\end{equation}
The estimate holds uniformly for $M\ge e$.
\end{lemma}

\begin{proof}
At index $N$, \cref{can:wronskian} and the endpoint bound imply
\[
 1\le 2\max(\abs{u_N},\abs{u_{N-1}})
          \max(\abs{t_N},\abs{t_{N-1}}),
\]
and hence
\[
 \tau_N\ge(1-3\delta)n-\log_M6,\qquad
 \upsilon_N\le-(1-3\delta)n+\log_M3.
\]
Since $\tau_1=0$, $\upsilon_1\ge0$, and $N-1\le n$, the total
shortfalls of the paths $\tau$ and $-\upsilon$ satisfy
\begin{align*}
 \sum_{j=1}^{N-1}\bigl[1-(\tau_{j+1}-\tau_j)\bigr]
 &\le3\delta n+\log6,\\
 \sum_{j=1}^{N-1}\bigl[1-(\upsilon_j-\upsilon_{j+1})\bigr]
 &\le3\delta n+\log3.
\end{align*}
Every summand is nonnegative by \cref{can:log-increments}. Apply
\cref{setup:shortfall-count} to $\tau$ to count indices where the
first line of \cref{can:good} fails. An individual increment smaller
than $c$ costs at least $1-c$ from the relevant shortfall sum. This
counts failures of the second line; applied to $-\upsilon$, it counts
failures of either of the two inequalities in the last line, with a
factor of at most two because a single increment can be tested at two
neighboring indices. Summing these bounds proves \cref{can:bad-count}.
\end{proof}

At a good index let $L_j=\abs{t_j}$. The first line of \cref{can:good}
with $l=j-1$ makes $t_j$ the dominant entry of its pair, and the next
increase makes $t_{j+1}$ dominant at the next pair. Consequently
\begin{equation}\label{can:good-t}
 \begin{gathered}
 L_j=M^{\tau_j},\qquad
 \abs{t_l}\le L_jM^{-c(j-l)}\quad(1\le l<j),\\
 \abs{t_{j+1}}\ge M^cL_j.
 \end{gathered}
\end{equation}
The two decreases for the $u$-pairs similarly give
\[
 \abs{u_{j+1}}\le M^{-c}\abs{u_j}
 \le M^{-2c}\abs{u_{j-1}}.
\]
Using \cref{can:wronskian} and $\abs{t_{j-1}/t_j}\le M^{-c}$, we
obtain, after increasing a fixed lower bound for $M$,
\begin{equation}\label{can:good-u}
 \abs{u_{j-1}}L_j
 \le 1+M^{-2c}\abs{u_{j-1}}L_j,
 \qquad \abs{u_{j-1}}\le\frac2{L_j}.
\end{equation}

For $t_j\ne0$, let $R_j$ be the initial value at index one of the
solution having value one at index zero and value zero at index $j$.
This solution exists and is unique: changing its initial value at
index one by $a$ changes its value at index $j$ by $at_j$. Its value
at index $j-1$ is $1/t_j$, by \cref{can:wronskian}. Its value at index
$j+1$ is therefore $-1/t_j$, which proves
\[
 R_{j+1}-R_j=\frac1{t_jt_{j+1}}.
\]
For any normalized solution, varying its initial value at index one
also gives $u_1-R_{j+1}=u_{j+1}/t_{j+1}$. At a good index,
\cref{can:good-t,can:good-u} thus yield
\begin{equation}\label{can:boundary-expansion}
 \begin{aligned}
 u_1
 &=R_{j+1}+\frac{u_{j+1}}{t_{j+1}}\\
 &=R_j+\frac1{t_j^2(v_j-t_{j-1}/t_j)}
  +O\left(\frac{M^{-3c}}{L_j^2}\right).
 \end{aligned}
\end{equation}
The implied constant is absolute and independent of the index.

\subsection*{Matching the two expansion scales}

\begin{lemma}\label{can:matching}
There are fixed constants $\delta_0>0$, $c_1>0$, $C_1<\infty$, and
$n_*<\infty$ with the following property. If $0<\delta\le\delta_0$ is
fixed, increasing $n_*$ if necessary, every configuration in
\cref{can:small-solution} with $n\ge n_*$ has at least $c_1n$ pairs
$(j_m^+,j_m^-)$ of good indices such that both index sequences
increase strictly and
\begin{equation}\label{can:matching-condition}
 \abs{\tau_{j_m^+}^+-\tau_{j_m^-}^-}\le\frac35.
\end{equation}
The collection of all possible increasing index-pair patterns for a
given $n$ has cardinality at most $\exp(C_1n)$.
\end{lemma}

\begin{proof}
Choose a small fixed $\beta>0$, for example $\beta=10^{-3}$. By
\cref{can:bad-count}, choosing $\delta_0$ sufficiently small and then
$n_*$ sufficiently large makes the bad-index count on each path at
most $\beta n$. The total-shortfall estimate also gives, at every
index,
\[
 j-1-3\delta n-\log6\le\tau_j\le j-1.
\]
Shrink $\delta_0$ once more and increase $n_*$ so that, for
$n/3\le j\le n/2$, the corresponding $\tau_j$ lies in
$[n/4,3n/4]$, while on the minus path
\[
 \tau_{N-2}^-\ge\tau_N^--2>3n/4+1.
\]
There are at least $n/6-2-\beta n$ good plus indices in this range.

For each such index put $a=\tau_j^+$. Starting from $\tau_1^-=0$, let
$k$ be the first minus index with $\tau_k^-\ge a$. We have
$\tau_{k-1}^-<a\le\tau_k^-$ and
$\tau_k^--\tau_{k-1}^-\le1$. Thus at least one of these two values is
within $1/2$ of $a$. The chosen index is in $\{2,\ldots,N-2\}$: the
large lower bound on $a$ excludes the first two indices, and the
displayed lower bound at $N-2$ provides the upper bound. If the chosen
minus index is bad, discard this plus index.

Good plus indices $j<j'$ have $\tau_{j'}^+-\tau_j^+\ge c(j'-j)$.
At most two such logarithms can belong to an interval of length one,
since $2c>1$. A fixed bad minus index can therefore be responsible for
at most two discarded plus indices. At least $n/6-2-3\beta n$
candidates remain.

Thin these candidates in increasing plus order by retaining the first
one, then the next whose plus logarithm is at least two larger than
the last retained logarithm, and continuing. Successive candidate
logarithms are separated by at least $c$, so each retention removes
at most three candidates. A fixed positive fraction of $n$ remains;
for example $c_1=10^{-2}$ works after shrinking $\delta_0$ if needed.
The matched minus logarithms increase strictly, because each is within
$1/2$ of the corresponding plus logarithm and retained plus
logarithms have gaps at least two. For two good minus indices,
increasing index implies strictly increasing logarithm by
\cref{can:good}. Strict increase of their logarithms therefore forces
strict increase of their indices as well. The matching error is at
most $1/2$, which implies \cref{can:matching-condition}.

Each path contains at most $n+1$ available indices. An increasing
paired pattern is determined by a subset of indices on each path, with
the two subsets of equal cardinality paired in order. There are at
most $2^{2n+2}$ such choices. Enlarging an absolute $C_1$ gives the
stated exponential bound.
\end{proof}

\subsection*{The local input}

The analytic fact used at each matched pair is a uniform scarcity
statement, proved in \cref{sec:local-scarcity}.

\begin{lemma}[Local scarcity]\label{can:local-scarcity}
For every $\eta>0$ there are $\eta'>0$ and $M_\eta<\infty$ such that
the following holds for $M\ge M_\eta$. Fix $q_e$, a base coordinate
$\theta_*$, and indices $j_+,j_-\ge2$ on the two outward paths.
Suppose both indices are good at $\theta_*$ for a common normalized
solution, and their logarithms satisfy \cref{can:matching-condition}.
Let
\[
 L_+=\abs{t_{j_+}^+(\theta_*)},\qquad
 L_-=\abs{t_{j_-}^-(\theta_*)},\qquad
 L=\max(L_+,L_-).
\]
Let $\mathcal S\subset[-1,1]$ consist of the $h$ for which, at
$\theta=\theta_*+h/L$, the same two indices are good for some common
solution with $u_e=1$. The solution is allowed to depend on $h$, and
neither endpoint smallness nor a new matching condition is required at
these other points. Then
\begin{equation}\label{can:scarcity-bound}
 \abs{\{x\in\R:\dist(x,\mathcal S)<\eta'\}}\le\eta.
\end{equation}
Here and below the size of a subset of the line or circle is its
Lebesgue measure. The constants are uniform over the central position,
the base coordinate, and the selected indices.
\end{lemma}

\subsection*{From local scarcity to exponential decay}

\begin{proof}[Proof of \cref{can:theorem}]
Fix once and for all a sufficiently small $\delta>0$ allowed by
\cref{can:matching}. Its constants $c_1,C_1,n_*$ are now fixed.
Choose $\eta>0$ so small that
\begin{equation}\label{can:eta-choice}
 5\eta<1,\qquad
 c_1\log\frac1{5\eta}>C_1+1.
\end{equation}
Obtain $\eta'$ and $M_\eta$ from \cref{can:local-scarcity}. Increase a
fixed lower bound $M_0$ so that all estimates above hold, that
$M_0\ge M_\eta$, and that
\begin{equation}\label{can:radius-choice}
 M_0^{-c}<\frac14,\qquad M_0^{-c}/5<\eta'.
\end{equation}
These choices are independent of $n$, the center, and $M\ge M_0$.

Fix $e\in\{0,-1\}$ and $q_e$. It is enough to consider patterns
containing exactly $r=\lceil c_1n\rceil$ pairs, since a longer pattern
can be truncated. For a fixed such increasing pattern, let $\mathcal F$
be the set of $\theta$ admitting one common normalized solution with
all of the following properties: the original central pair has norm at
most $\sqrt2$; both endpoint pairs satisfy \cref{can:small-solution};
every pair of selected indices is good and satisfies
\cref{can:matching-condition}. One solution must satisfy all conditions
of a pattern at once.

The set $\mathcal F$ is measurable, in fact compact. To see this
without selecting solutions, parameterize the central pair by its entry
other than the fixed entry $u_e=1$. The norm bound puts that parameter
in $[-1,1]$. All later solution entries and recurrence coefficients
depend continuously on this parameter and on the circle coordinate
$\theta$. Each pair entering a logarithm is nonzero, by invertibility
of the recurrence. The endpoint, goodness, and matching conditions are
closed inequalities on the compact product of the circle with
$[-1,1]$. Their projection is therefore compact.

For $\theta\in\mathcal F$ and $1\le m\le r$, define
\[
 L_m(\theta)=\max\left(
 \abs{t_{j_m^+}^+(\theta)},\abs{t_{j_m^-}^-(\theta)}\right),
 \quad s_m(\theta)=L_m(\theta)^{-1},
\]
and let
\[
 V_m=\bigcup_{\theta\in\mathcal F}
 B\bigl(\theta,s_m(\theta)/10\bigr)
\]
be an open subset of the circle. Goodness at the later of two selected
indices gives, separately on both paths,
\[
 \abs{t_{j_{m+1}^\pm}^\pm(\theta)}
 \ge M^{c(j_{m+1}^\pm-j_m^\pm)}
      \abs{t_{j_m^\pm}^\pm(\theta)}.
\]
Since both index sequences increase strictly,
\begin{equation}\label{can:radius-decay}
 s_{m+1}(\theta)\le M^{-c}s_m(\theta),
 \qquad s_m(\theta)\le M^{-c}.
\end{equation}
For fixed $M,n$, these positive radii also have a positive lower bound:
the one-step bound gives $\abs{t_j}\le M^{j-1}\le M^n$ at all indices
in question, so $s_m\ge M^{-n}$.

We claim
\begin{equation}\label{can:cover-contraction}
 \abs{V_{m+1}}\le5\eta\abs{V_m}\qquad(1\le m<r).
\end{equation}
Select disjoint balls from the family defining $V_m$ by the usual
greedy procedure: at each stage select a remaining ball whose radius
is at least half the supremum of the remaining radii, then remove all
remaining balls meeting it, assigning each to the selected ball. The
selected balls are disjoint and have a fixed positive lower radius, so
only finitely many are selected on the circle and the procedure must
exhaust the family. Every assigned ball meets its selected ball and
has radius at most twice that selected ball.

Consider one selected center $\theta_*$ and write $s=s_m(\theta_*)$.
The centers of its assigned balls are at circle distance at most
$3s/10$ from $\theta_*$. The window of radius $s$ is an injective lift
interval by \cref{can:radius-choice,can:radius-decay}. All assigned
centers belong to $\mathcal F$, so are successful for the same
selected pair of indices in the local test based at $\theta_*$. A
next-level ball centered at an assigned $\theta$ has radius at most
\[
 \frac{s_{m+1}(\theta)}{10}
 \le \frac{M^{-c}s_m(\theta)}{10}
 \le \frac{M^{-c}s}{5}.
\]
After rescaling the lifted coordinate by $1/s$, the union of all these
next-level balls is contained in the $\eta'$-neighborhood of the
successful set in \cref{can:local-scarcity}, by
\cref{can:radius-choice}. It has unscaled length at most $\eta s$.
Summing over selected balls,
\[
 \abs{V_{m+1}}\le\eta\sum_{\text{selected}}s
 =5\eta\sum_{\text{selected}}
  \abs{B(\theta_*,s/10)}
 \le5\eta\abs{V_m},
\]
as claimed. The ball selected at a center is included in its own
assigned group, so no next-level balls are omitted.

Since $\mathcal F\subset V_m$ for every $m$ and $\abs{V_1}\le1$,
iteration gives
\[
 \abs{\mathcal F}\le(5\eta)^{r-1}
 \le(5\eta)^{-1}
  \exp\left(-c_1n\log\frac1{5\eta}\right).
\]
There are at most $\exp(C_1n)$ patterns. By \cref{can:eta-choice}, their
union has circle measure at most $C\exp(-\gamma n)$ for a fixed
$\gamma>0$, uniformly in the fixed coordinate $q_e$. The pair
$(q_e,q_{e-1})$ has normalized product Lebesgue law. Integrating the
conditional bound over $q_e$ and summing the two choices of $e$
therefore gives the same form of bound for the original event, for
every $n\ge n_*$. Every point of that event is covered, by
\cref{can:small-solution,can:matching}. Finally enlarge $C$ to cover
the finitely many positive integers $n<n_*$, whose event probabilities
are at most one. Stationarity restores every center $i$.
\end{proof}

\section{Local scarcity at matched scales}
\label{sec:local-scarcity}

This section proves \cref{can:local-scarcity}. We use the two outward
paths and the notation established in the cancellation argument: in
particular $c=99/100$, the central coordinate is fixed, and the
variable is $\theta=q_{e-1}$. The selected indices are denoted by
$j_+$ and $j_-$. Every assertion about an interval of $\theta$ uses
real lifts of the recurrence, and the constants below are independent
of the selected indices and the base configuration.

After rescaling, the central relation \cref{can:center-equation}
becomes an approximate identity between a cubic polynomial and two
shifted reciprocal-cosine terms. A positive-measure Hausdorff limit of
the closures of successful sets would force an analytic identity. We
rule out that identity even when the two reciprocal terms nearly
cancel or one phase slope tends to zero. The derivative estimates in
the first two subsections supply the uniform remainders needed for
these two degeneracies.

\subsection*{Control on a single rescaled path}

Fix either outward path and write $j$ for its selected index. Put
\[
 L_j=\abs{t_j(\theta_*)},\qquad \xi=L_j(\theta-\theta_*),\qquad
 D=\frac{\dd}{\dd\xi}.
\]
The initial position $q_0$ is constant and $q_1$ is affine in $\xi$,
with derivative $\pm L_j^{-1}$. By \cref{can:good-t}, the base
linearized solution satisfies
\begin{equation}\label{loc:base-growth}
 \frac{\abs{t_l(\theta_*)}}{L_j}\le M^{-c(j-l)}\quad(1\le l\le j).
\end{equation}
We will prove, throughout $\abs{\xi}\le1$, that
\begin{align}
 \abs{q_l(\xi)-q_l(0)}+\abs{Dq_l(\xi)}
 &\le C M^{-c(j-l)}, \label{loc:first-derivatives}\\
 \abs{D^p q_l(\xi)}
 &\le C_p M^{-pc(j-l)}M^{1-pc}
 \quad(2\le p\le5), \label{loc:higher-derivatives}
\end{align}
for $0\le l\le j$. The terms with $l=0$ vanish.

Write $B_l=M^{-c(j-l)}$ and $\Delta_l=q_l(\xi)-q_l(0)$. Taylor's
formula for the recurrence, followed by propagation with the base
linearized matrices, gives
\begin{equation}\label{loc:nonlinear-comparison}
 \abs{\Delta_l}
 \le \abs{\xi}\frac{\abs{t_l(\theta_*)}}{L_j}
       +C\sum_{d=1}^{l-1}M^{l-d}\abs{\Delta_d}^2.
\end{equation}
The quadratic error at site $d$ is at most $CM\abs{\Delta_d}^2$, and
the remaining $l-d-1$ steps have norm at most $M^{l-d-1}$. For $l=1$
the estimate $\abs{\Delta_l}\le2B_l$ follows directly from
\cref{loc:base-growth}. If it holds for all earlier indices, the sum
in \cref{loc:nonlinear-comparison} is bounded by
\[
 4C B_l^2\sum_{a\ge1}M^{(1-2c)a}.
\]
Since $B_l\le1$ and $1-2c<0$, this is at most $B_l$ for all
sufficiently large $M$. Induction proves the required displacement
bound, uniformly in $j$.

Compare $Dq_l(\xi)$ with the base variation
$\pm t_l(\theta_*)/L_j$. Their initial data agree; their difference is
propagated by the base matrices with forcing
\[
 \bigl(v(q_d(\xi))-v(q_d(0))\bigr)Dq_d(\xi),
\]
whose absolute value is at most $CM\abs{\Delta_d}\abs{Dq_d}$.
Induction with $\abs{Dq_d}\le2B_d$ uses the same geometric sum and
proves \cref{loc:first-derivatives}.

The higher derivative induction uses the recurrence at the varied
orbit. For $p\ge2$,
\[
 D^p q_{l+1}=v(q_l)D^p q_l-D^p q_{l-1}+F_{p,l},
\]
where $F_{p,l}$ is a finite sum, with coefficients depending only on
$p$, of terms
\[
 \phi^{(s)}(q_l)\prod_{i=1}^{s}D^{m_i}q_l,
 \qquad 2\le s\le p,\quad m_i\ge1,\quad
 \sum_{i=1}^{s}m_i=p.
\]
Each $m_i$ is strictly smaller than $p$. The first derivative bound
and the induction hypothesis give $\abs{D^{m_i}q_l}\le
C_{m_i}B_l^{m_i}$, since $M^{1-m_ic}\le1$ whenever $m_i\ge2$.
Consequently $\abs{F_{p,l}}\le C_p M B_l^p$. The initial higher
derivatives at sites $0,1$ vanish. Propagation by the varied matrices
gives
\[
 \abs{D^p q_l}\le C_p\sum_{d=1}^{l-1}M^{l-d}B_d^p
 \le C_p B_l^p\sum_{a\ge1}M^{(1-pc)a}
 \le C'_p B_l^p M^{1-pc}.
\]
This proves \cref{loc:higher-derivatives} successively for
$p=2,3,4,5$.

At the selected site $Dq_j(0)=\pm1$. The case $p=2$ shows that
$Dq_j(\xi)=Dq_j(0)+O(M^{1-2c})$. For all sufficiently large $M$,
\begin{equation}\label{loc:varied-t}
 \abs{t_j(\xi)}\ge L_j/2,\qquad
 \abs{t_l(\xi)}\le C L_j M^{-c(j-l)}\quad(l<j).
\end{equation}
The quotient
\begin{equation}\label{loc:quotient}
 E=\frac{t_{j-1}}{t_j}=\frac{Dq_{j-1}}{Dq_j}
 \quad\hbox{satisfies}\quad
 \abs{D^p E}\le C_p M^{-c}\quad(0\le p\le4).
\end{equation}
The denominator is bounded away from zero. Every derivative of its
reciprocal of order at most four is bounded by the already proved
bounds through $D^5q_j$. Every derivative of the numerator through
order four contains a derivative of $q_{j-1}$ with a factor at most
$CM^{-c}$. The product rule proves \cref{loc:quotient}.

\subsection*{A fourth-derivative bound for the boundary value}

Fourth-order control makes the boundary value replaceable by a cubic
polynomial with a small remainder that stays negligible even after
normalizing the first nonzero correction when the two reciprocal
cosine terms nearly cancel.

Let $R_j$ be the initial value at site $1$ of the solution $U$ with
$U_0=1$ and $U_j=0$. This solution exists throughout our interval
because $t_j$ does not vanish. We claim that, with $\rho=4/5$,
\begin{equation}\label{loc:boundary-fourth}
 L_j^2\abs{D^4R_j}\le C M^{-2-\rho}.
\end{equation}

Consider the $(j-1)$-by-$(j-1)$ tridiagonal matrix $B$ with diagonal
$v(q_a)$, $1\le a\le j-1$, and both adjacent diagonals equal to $-1$.
Its determinant is $t_j$: expansion along the last row gives the same
recursion and initial conditions as those for $t$. The interior
boundary problem is $BU=e_1$, so $R_j=(B^{-1})_{11}$. More
explicitly, its inverse is
\begin{equation}\label{loc:green}
 G_{ab}=t_{\min(a,b)}U_{\max(a,b)}\qquad(G=B^{-1}).
\end{equation}
Fix column $b$. Its entries solve the homogeneous recurrence except
at row $b$ and are zero at boundary sites $0,j$. At row $b$ the
output of $B$ is $t_{b+1}U_b-t_bU_{b+1}=1$, by the constant Wronskian
and the data $t_0=0,t_1=1,U_0=1$. So the column is $B^{-1}e_b$.

At the right boundary the Wronskian gives $U_{j-1}=1/t_j$. Backward
propagation and \cref{loc:varied-t} give
\begin{equation}\label{loc:boundary-backwards}
 L_j\abs{U_a}\le 2M^{j-a-1}.
\end{equation}
For $d_a=j-a\ge1$, \cref{loc:green,loc:varied-t,loc:boundary-backwards}
give
\begin{equation}\label{loc:green-size}
 \abs{G_{ab}}\le C
 M^{\min(d_a,d_b)-1-c\max(d_a,d_b)}.
\end{equation}
The chain rule and the bounds for derivatives of $q_a$ also give
\begin{equation}\label{loc:potential-derivatives}
 \abs{D^p v(q_a)}\le C_p M^{1-cp d_a}
 \quad(1\le p\le4).
\end{equation}
Every term contains a derivative of $v$ bounded by $C_pM$ and
derivatives of $q_a$ of total order $p$.

Starting with $DG=-G(DB)G$, differentiate three more times and take
the $(1,1)$ entry. The result is a finite linear combination, with
fixed integer coefficients, of sums of terms
\begin{equation}\label{loc:inverse-chains}
 U_{a_1}(D^{p_1}v_{a_1})G_{a_1a_2}
 (D^{p_2}v_{a_2})\cdots
 G_{a_{s-1}a_s}(D^{p_s}v_{a_s})U_{a_s},
 \qquad p_i\ge1,\quad \sum_{i=1}^{s}p_i=4.
\end{equation}
Here $v_a=v(q_a)$, $1\le s\le4$, and each site index is summed
independently from $1$ to $j-1$. Differentiating a potential factor
increases its derivative order; differentiating an inverse factor
inserts another $-G(DB)G$. Also $G_{1a}=U_a$ because $t_1=1$.

Put $d_i=j-a_i$. After multiplication by $L_j^2$, the exponent of $M$
bounding a term of \cref{loc:inverse-chains} is
\begin{align*}
 H&=-1+d_1+d_s+
 \sum_{i=1}^{s-1}\bigl(\min(d_i,d_{i+1})-c\max(d_i,d_{i+1})\bigr)
       -c\sum_{i=1}^{s}p_id_i\\
 &\le -1+2\min_i d_i-(3c-2)\sum_{i=1}^{s}p_id_i\\
 &\le -1-\frac{47}{100}\sum_{i=1}^{s}p_id_i.
\end{align*}
The first inequality writes each adjacency term as
$-\abs{d_i-d_{i+1}}+(1-c)\max(d_i,d_{i+1})$; total variation is at
least $d_1+d_s-2\min_i d_i$, and the sum of maxima is at most
$2\sum_i d_i\le2\sum_i p_id_i$. The last uses $4\min_i d_i\le\sum_i
p_id_i$ and $c=99/100$. Summing over sites is harmless: for every
composition of $4$,
\[
 \sum_{d_1,\ldots,d_s\ge1}
 M^{-1-(47/100)\sum_i p_id_i}
 =M^{-1}\prod_{i=1}^{s}
 \frac{M^{-(47/100)p_i}}{1-M^{-(47/100)p_i}}
 \le C M^{-72/25}.
\]
Only finitely many such compositions and chain types occur. Since
$72/25>2+4/5$, this proves \cref{loc:boundary-fourth}.

\subsection*{The cubic relation at successful points}

Return to the two paths and put $L=\max(L_{j_+},L_{j_-})$ and
$h=L(\theta-\theta_*)$. Number the paths so that the first has the
larger base scale. Write $Q_i(h)$ for their selected positions and
$E_i(h)$ for the quotients in \cref{loc:quotient}. With
$b_i=Q_i'(0)$ and $d_i=Q_i''(0)$, the preceding estimates give
\begin{align}
 Q_i(h)&=z_i+b_ih+\tfrac12d_ih^2+\mathcal E_i(h),
 &\norm{\mathcal E_i}_{C^1([-1,1])}
   &\le C\abs{b_i}^3M^{1-3c}, \label{loc:phase-taylor}\\
 \abs{d_i}&\le C\abs{b_i}^2M^{1-2c},
 &\abs{b_1}&=1,\quad
 \abs{b_2}=r\in[M^{-3/5},1], \label{loc:phase-slopes}\\
 \abs{\partial_h^pE_i}&\le C_pM^{-c}\abs{b_i}^p
 &&(0\le p\le4). \label{loc:scaled-quotient}
\end{align}
The scale change is $\xi=(L_{j_i}/L)h=\abs{b_i}h$. It also gives, for
$2\le p\le5$,
\begin{equation}\label{loc:scaled-phase-derivatives}
 \abs{Q_i^{(p)}}\le C_p\abs{b_i}^pM^{1-pc},\qquad
 \abs{Q_i'}\asymp\abs{b_i}.
\end{equation}

At a successful point there is a common solution with central value
$1$, and its two outward values at site $1$ add to $v(q_e)$. Apply
\cref{can:boundary-expansion} to both paths at that point. Since
$\abs{t_{j_i}(\theta)}=L\abs{Q_i'(h)}$ and
$v(Q_i)-E_i=K(\cos(2\pi Q_i)+(2-E_i)/K)$, multiplication by $KL^2$
yields
\[
 KL^2\Bigl(v(q_e)-\sum_{i=1}^{2}R_{j_i}(\theta)\Bigr)
 =\sum_{i=1}^{2}
 \frac{1}{(Q_i')^2(\cos(2\pi Q_i)+(2-E_i)/K)}
 +O(r^{-2}M^{1-3c}).
\]
The error bound uses \cref{loc:scaled-phase-derivatives} to compare
the varied $t_{j_i}$ with their base sizes. Replace the left side by
its cubic Taylor polynomial $P(h)$ at $0$. The contribution of each
$R_{j_i}$ to the Taylor remainder is at most
\[
 C K L^2\abs{b_i}^4 L_{j_i}^{-2}M^{-2-\rho}
 =C K\abs{b_i}^2M^{-2-\rho}.
\]
Consequently every successful $h\in[-1,1]$ satisfies
\begin{equation}\label{loc:cubic-relation}
 P(h)=\sum_{i=1}^{2}
 \frac{1}{(Q_i'(h))^2
       (\cos(2\pi Q_i(h))+(2-E_i(h))/K)}
       +O(M^{-1-\rho}+r^{-2}M^{1-3c}).
\end{equation}
The polynomial may depend on the configuration. Its degree is at most
three. The displayed error is uniform on the successful set, and
tends to zero uniformly over all configurations, since
$r^{-2}M^{1-3c}\le M^{-77/100}$.

\subsection*{Polynomial extraction from a Hausdorff limit}

Consider any sequence of configurations with $M\to\infty$ and
nonempty successful sets $\mathcal S_M$. Pass to a subsequence on
which their closures converge in Hausdorff distance to a nonempty
compact set $F\subset[-1,1]$. We prove that $F$ has measure zero.
Suppose to the contrary that it has positive measure.

We record the compactness argument for the polynomials that will be
used below. Suppose, after an indicated subtraction or multiplication
of \cref{loc:cubic-relation}, a polynomial $P_M^*$ of degree at most
three agrees at successful points, up to an error tending to zero,
with functions converging uniformly on every compact subinterval of
$[-1,1]$ away from a fixed finite set of limiting poles in that
interval. Choose four distinct points of $F$ outside those poles.
Hausdorff convergence and approximation of points in the closure
supply successful points converging to these four points. The values
of $P_M^*$ there are bounded. The associated Vandermonde matrices
converge to an invertible matrix, so all four coefficients of $P_M^*$
are bounded. Pass to a further subsequence on which the polynomials
converge coefficientwise. For every point of $F$ away from the poles,
the same approximation argument shows equality with the limiting
function. In particular, equality holds on a set of positive measure.
Clearing the finitely many denominator factors used in that case
gives an identity between real analytic functions on $\R$: the
difference has an accumulation point of zeros and hence vanishes
identically. Passing to further subsequences throughout does not
change the fixed Hausdorff limit $F$.

\subsection*{Nonzero limiting slopes and opposite cosines}

First suppose $r$ has a positive limit $r_0$. Take convergent
subsequences of the phases modulo integers and of the slopes. The
first slope has a fixed sign after a further subsequence, so write
its value as $b\in\{-1,1\}$. By
\cref{loc:phase-taylor,loc:scaled-quotient}, the two terms on the
right of \cref{loc:cubic-relation} converge, away from their limiting
poles, to
\[
 \sec(2\pi(z+bh))+
 r_0^{-2}\sec(2\pi(z'+b'h)),\qquad \abs{b'}=r_0.
\]
Polynomial extraction and multiplication by both cosines show that
the zero sets of these cosines on $\R$ coincide: at a zero of either
cosine, the other must vanish because its coefficient is positive.
Their zero sets are arithmetic progressions with respective spacings
$1/(2\abs{b})$ and $1/(2\abs{b'})$, so $r_0=1$. The two cosines are
therefore equal or opposite. Equality would leave twice a secant,
which cannot equal a polynomial: multiply by the cosine and evaluate
at a zero of it. Only the opposite case remains.

Reflect the second phase if necessary and translate it by an odd
half-integer. The resulting phase $\widetilde Q_2$ has the same
limiting affine phase as $Q_1$, while
\[
 \cos(2\pi Q_2)=-\cos(2\pi\widetilde Q_2),\qquad
 (Q_2')^2=(\widetilde Q_2')^2.
\]
By \cref{loc:phase-taylor},
\begin{equation}\label{loc:phase-difference}
 \widetilde Q_2-Q_1=p(h)+O_{C^1}(M^{1-3c}),\qquad
 \deg p\le2,\qquad \norm{p}_{\rm coeff}\longrightarrow0.
\end{equation}
Here $\norm{p}_{\rm coeff}$ is the maximum absolute value of its
three coefficients. Define
\[
 \ell=\max(\norm{p}_{\rm coeff},K^{-1}).
\]
Then $\ell\to0$, but every error of order $M^{1-3c}=M^{-197/100}$
is $o(\ell)$. The error in \cref{loc:cubic-relation} is also
$o(\ell)$ in the present case, since $r\to1$ and $\rho>0$.

Expand on a compact subinterval away from the zeros of the common
limiting cosine. Put
\[
 f(x,y)=y^{-2}\sec(2\pi x),\qquad A=2\pi(z_1+b_1h).
\]
The denominators in \cref{loc:cubic-relation} stay uniformly away
from zero there for large $M$. Their shifts are
$2/K+O(M^{-c}/K)$. Expanding the reciprocals, with their respective
cosine signs, gives the sum of the two terms as
\begin{align*}
 &f(Q_1,Q_1')-f(\widetilde Q_2,\widetilde Q_2')\\
 &\quad-\frac2K\left((Q_1')^{-2}\sec^2(2\pi Q_1)
            +(\widetilde Q_2')^{-2}\sec^2(2\pi\widetilde Q_2)\right)
       +o(\ell).
\end{align*}
The reciprocal expansion has error $O(K^{-2}+M^{-c}/K)=o(\ell)$.
It is essential to subtract the two $f$ terms at the actual common
phase before replacing that phase by an affine one. Taylor expansion
of $f$, using \cref{loc:phase-difference}, gives
\[
 f(Q_1,Q_1')-f(\widetilde Q_2,\widetilde Q_2')
 =-f_x(Q_1,Q_1')p-f_y(Q_1,Q_1')p'+o(\ell).
\]
The quadratic error is $O(\ell^2)$ and the phase remainder is
$O(M^{1-3c})$. Moreover $(Q_1,Q_1')=(z_1+b_1h,b_1)+O(M^{1-2c})$;
replacing the derivatives of $f$ by their values at this affine pair
produces only $O(\ell M^{1-2c})=o(\ell)$. Similarly, the preceding
$K^{-1}$ term can be evaluated at the common affine pair with error
$o(K^{-1})$. We conclude that \cref{loc:cubic-relation} becomes
\begin{equation}\label{loc:opposite-expansion}
 P(h)=\frac{2p'}{b_1^3}\sec A
      -\frac{2\pi p}{b_1^2}\sin A\,\sec^2 A
      -\frac4{K b_1^2}\sec^2 A+o(\ell).
\end{equation}

Divide by $\ell$ and take a subsequence such that $p/\ell\to\bar p$
coefficientwise and $K^{-1}/\ell\to\beta$. At least one is nonzero,
because $\max(\norm{p/\ell}_{\rm coeff},K^{-1}/\ell)=1$. Polynomial
extraction applies to $P/\ell$. Multiply its limiting identity by
$b^2\cos^2(2\pi(z+bh))$ and evaluate at every zero of that cosine.
One obtains
\begin{equation}\label{loc:alternating-obstruction}
 2\pi\bar p(h)\sin(2\pi(z+bh))+4\beta=0
 \quad\hbox{at every cosine zero.}
\end{equation}
The zeros form an unbounded arithmetic progression and their sine
values alternate between $1$ and $-1$. Thus $\bar p$ is bounded on
that progression and must be constant. The alternating signs then
force both that constant and $\beta$ to be zero, contradicting their
normalization. This excludes every positive limiting slope.

\subsection*{A vanishing second slope}

It remains to treat $r\to0$. After a subsequence, the first term of
\cref{loc:cubic-relation} converges away from its poles to
$\sec(2\pi(z+bh))$, with $\abs{b}=1$. Write the second term as
\begin{equation}\label{loc:slow-denominator}
 g(h)=\frac1{r^2S(h)},\qquad
 S(h)=\left(\frac{Q_2'(h)}r\right)^2
       \left(\cos(2\pi Q_2(h))+\frac{2-E_2(h)}K\right),
 \qquad w=\max(\abs{S(0)},r).
\end{equation}
Uniformly on $[-1,1]$,
\begin{equation}\label{loc:slow-derivatives}
 \abs{S^{(m)}}\le C_m r^m\quad(1\le m\le4).
\end{equation}
To see every derivative involved, set $a=Q_2'/r$. Then $a$ is
bounded and bounded away from zero, and
\[
 \abs{a^{(m)}}\le C_m r^m M^{1-(m+1)c}\le C_m r^m
 \quad(1\le m\le4)
\]
by \cref{loc:scaled-phase-derivatives}. The derivatives through
order four of the cosine factor are $O(r^m)$ by the chain rule; those
of the shift are bounded by $C_mr^mM^{-c}/K$ by
\cref{loc:scaled-quotient}. The product rule proves
\cref{loc:slow-derivatives}. This is the point requiring the fifth
derivative of $Q_2$.

If $w\to0$, there is also the sharper fact
\begin{equation}\label{loc:slow-nonzero-slope}
 \frac{\abs{S'(0)}}r\longrightarrow2\pi.
\end{equation}
Indeed $a(0)^2=1$ and $\abs{Q_2'(0)}=r$. The cosine with its shift
equals $S(0)\to0$, so $\abs{\sin(2\pi Q_2(0))}\to1$. Differentiating
the factor $a^2$ contributes $O(rwM^{1-2c})=o(r)$; differentiating the
shift contributes $O(rM^{-c}/K)=o(r)$. The remaining contribution is
$-2\pi\sin(2\pi Q_2(0))Q_2'(0)$, which proves the assertion.

Take a subsequence on which $r/w$ has a limit. If this limit is
positive, then $w\to0$, and Taylor's formula gives
\[
 \frac{S(h)}w=\frac{S(0)}w+\frac{S'(0)}w h+O(r^2/w).
\]
Here $r^2/w\le r\to0$. The coefficients are bounded; by
\cref{loc:slow-nonzero-slope}, a further subsequence gives a limiting
affine polynomial $A_0(h)$ with nonzero slope. Multiply
\cref{loc:cubic-relation} by $r^2w$. Its first term and its error
tend to zero locally away from the fast poles, while its second term
converges to $1/A_0(h)$ away from the one possible zero of $A_0$.
Polynomial extraction would therefore give a polynomial of degree at
most three equal to $1/A_0$. Multiplication by $A_0$ and evaluation
at its real zero is a contradiction.

We may assume therefore that $r/w\to0$. Then $w=\abs{S(0)}$
eventually, and
\begin{equation}\label{loc:almost-constant-denominator}
 \frac{S(h)}{S(0)}=1+O(r/w)\longrightarrow1
 \quad\hbox{uniformly on }[-1,1].
\end{equation}
In particular $S$ has no zeros there for large $M$. Let $T_3g$ be
the cubic Taylor polynomial of $g$ at $0$. If $w$ stays bounded away
from zero, the reciprocal derivative formula and
\cref{loc:slow-derivatives} give $g^{(4)}=O(r^2)$ uniformly. Thus
$g-T_3g=O(r^2)$. Subtracting $T_3g$ from \cref{loc:cubic-relation}
leaves only the fast secant in the limit. It cannot agree with a cubic
polynomial: multiply by its cosine and evaluate at a zero.

The last case is $w\to0$ and $r/w\to0$. To track the possibly large
remainder of $g$, differentiate explicitly:
\begin{equation}\label{loc:reciprocal-fourth}
 g^{(4)}=\frac1{r^2}\left(
 \frac{24(S')^4}{S^5}
 -\frac{36(S')^2S''}{S^4}
 +\frac{6(S'')^2+8S'S'''}{S^3}
 -\frac{S''''}{S^2}\right).
\end{equation}
By \cref{loc:slow-derivatives,loc:slow-nonzero-slope},
$S'(h)/S'(0)=1+O(r)$ uniformly. Together with
\cref{loc:almost-constant-denominator}, this makes the first term
in \cref{loc:reciprocal-fourth} equal to
\[
 (24+o(1))\frac{S'(0)^4}{r^2S(0)^5}.
\]
Relative to that term, the remaining three displayed contributions
are respectively $O(w)$, $O(w^2)$, and $O(w^3)$: for example the
first ratio is bounded by $C\abs{S}\abs{S''}/\abs{S'}^2\le Cw$.
All these comparisons are uniform on $[-1,1]$. Put
\[
 a_M=\frac{S'(0)^4}{r^2S(0)^5}.
\]
Taylor's formula with integral remainder, valid on both sides of
zero, now yields
\begin{equation}\label{loc:quartic-remainder}
 g(h)-T_3g(h)=a_Mh^4+o(\abs{a_M})
 \quad\hbox{uniformly on }[-1,1].
\end{equation}
If $\abs{a_M}\to\infty$, subtract $T_3g$ from
\cref{loc:cubic-relation} and divide by $a_M$. The fast term tends to
zero away from its poles and the original error also tends to zero
after this division. Polynomial extraction would identify a
polynomial of degree at most three with $h^4$, impossible.
Otherwise pass to a subsequence with $a_M\to a\in\R$. Without
division, the same subtraction gives a limiting identity between a
cubic polynomial and
\[
 \sec(2\pi(z+bh))+ah^4.
\]
Move the quartic term to the polynomial side and multiply by the
cosine. Evaluation at any of its zeros is again a contradiction.
These alternatives exhaust the vanishing-slope case.

\subsection*{Uniform neighborhoods of successful sets}

We have proved that every Hausdorff limit of nonempty closures of
successful sets along $M\to\infty$ has measure zero. This implies the
uniform assertion of \cref{can:local-scarcity}, including its
neighborhood width. Indeed, if the assertion failed for some $\eta>0$,
then for every integer $n$ one could choose $M_n\ge n$ and a
successful set $\mathcal S_n\subset[-1,1]$ such that
\[
 \operatorname{Leb}_{\R}\{x:\dist(x,\mathcal S_n)<1/n\}>\eta.
\]
These sets are nonempty. Compactness of the nonempty compact subsets
of $[-1,1]$ for Hausdorff distance supplies a subsequence of their
closures with limit $F$. For every $\varepsilon>0$, eventually
\[
 \{x:\dist(x,\mathcal S_n)<1/n\}
 \subset\{x:\dist(x,F)<\varepsilon\}.
\]
Thus every $\varepsilon$-neighborhood of $F$ has measure at least
$\eta$. Continuity of Lebesgue measure from above for bounded
neighborhoods implies $\operatorname{Leb}_{\R}(F)\ge\eta$, contrary
to the result just proved. The empty successful set satisfies the
claimed estimate automatically. This completes the proof of
\cref{can:local-scarcity}.

\section{A product estimate across a fast bridge}
\label{sec:bridge}

The second quantitative input compares observations near opposite ends
of a long interval on which the linearized product grows almost at its
maximum rate. The comparison comes from two transverse families of
contracting graphs in the middle of the interval. All measures in this
section are the normalized area measure, expressed in the coordinates
$(q_i,q_{i-1})$ when convenient, whenever they define probabilities or
expectations. Label lengths and slice integrals use ordinary
one-dimensional Lebesgue measure in their displayed coordinates.

For distinct integers $d,e$, define the shortfall
\begin{equation}\label{bridge:shortfalls}
 \mathfrak h_{d,e}
   =1-\frac{\log_M\norm{T_{d,e}}}{\abs{d-e}}.
\end{equation}
These quantities belong to $[0,1]$. An \emph{observation} with bound
$(R,L,B)$ is a function of at most $R$ such quantities whose defining
function on $[0,1]^R$ has absolute value at most $B$ and Lipschitz
constant at most $L$ for the maximum norm. Allowing fewer than $R$
arguments is harmless. The defining functions and the integer indices
may vary from one application to another, provided these bounds are
preserved. Equivalently, the bounds may be imposed only on the realized
range $\mathcal R\subseteq[0,1]^R$ of the arguments. Indeed, a function
$f$ on that range extends by
$u\mapsto\inf_{v\in\mathcal R}(f(v)+L\norm{u-v}_\infty)$; clipping this
extension to $[-B,B]$ preserves its values on $\mathcal R$ and its
Lipschitz bound.

\begin{theorem}[Product estimate across a fast bridge]
\label{bridge:theorem}
There are absolute constants $c_b\in(0,1)$, $\kappa>0$, and $\sigma>0$
with the following property. Fix observation bounds $(R,L,B)$ and a
threshold $h>0$. For sufficiently large $M$ and sufficiently large
$n=b-a$, let $W,H$ be observations with these bounds, with $W\ge0$.
All indices occurring in one observation are within $\kappa n$ of $a$,
and all indices occurring in the other are within $\kappa n$ of $b$.
Either assignment of $W,H$ to the two ends is allowed. Then
\begin{equation}\label{bridge:product}
 \E\!\left[
   \one_{\{\log_M\norm{T_{a,b}}\geq c_b n\}}
   W\one_{\{H>h\}}
 \right]
 \leq
 C\bigl(\E W+M^{-\sigma n}\bigr)\PP(H>h/2).
\end{equation}
The constants $c_b,\kappa,\sigma$ are independent of the observation
bounds and threshold. The constant $C$ and the lower size requirements
for $M,n$ may depend on $(R,L,B)$ and $h$, but not on the indices,
the defining functions, or the bridge location.
\end{theorem}

We first construct the graphs needed in the proof. Finite-time
contracting directions and curves have been developed in stable-manifold
constructions under weak hyperbolicity hypotheses \cite{HollandLuzzatto2006}.
Here the graphs are level sets of a distant orbit coordinate, obtained
from finite Dirichlet problems; we prove their extent and transverse
distortion directly. We use $c=0.98$ throughout the construction. A
direction is understood modulo sign.

\begin{lemma}[Contracting graphs and their distortion]
\label{bridge:graphs}
For sufficiently large $M$, suppose $B_+\geq2$ and, at a fixed state
with central coordinates $(x,y)=(q_0,q_{-1})$,
\begin{equation}\label{bridge:prefix-growth}
 D_l:=\norm{T_{0,l}}\geq M^{cl}
 \qquad(1\leq l\leq B_+).
\end{equation}
For at least one choice $j\in\{B_+-1,B_+\}$ there is a smooth graph
through this state, parametrized by $q_{-1}=y+r$, $\abs r\leq1/4$, on
which the lifted coordinate $q_j$ is fixed. On this graph,
\begin{equation}\label{bridge:graph-contraction}
 \abs{q_p(r)-q_p(0)}+
 \abs{\frac{\dd q_p(r)}{\dd r}}
 \leq C M^{-0.8(p+1)}
 \qquad(0\leq p<j).
\end{equation}
The corresponding endpoint pair at time $B_+$ changes by at most
$C M^{-0.8(B_+-1)}$. The partial derivative
$\partial q_j/\partial q_0$, with $q_{-1}$ held independent, is nonzero
throughout the graph, and the absolute values of this partial derivative
at any two points of the graph have a ratio between two absolute
positive constants.

For a fixed choice of $j$, one may require the closed eligibility
condition
\begin{equation}\label{bridge:row-choice}
 \abs{\partial q_j/\partial q_0}\geq D_{B_+}/3
 \quad\text{at the initial point}.
\end{equation}
The graphs depend continuously on eligible initial points. There is an
identical backward statement, with almost horizontal graphs in
$(x,y)$, obtained by using $(q_{-1},q_0)$ as the ordered initial pair.
\end{lemma}

\begin{proof}
Let $s_l$ be a most contracted unit input direction for $T_{0,l}$. Since
a single inverse step has norm at most $M$,
\[
 \norm{T_{0,l}s_{l+1}}\leq\frac{M}{D_{l+1}}.
\]
The projection onto the expanded input direction for $T_{0,l}$ therefore
gives
\begin{equation}\label{bridge:direction-step}
 \abs{s_{l+1}\wedge s_l}\leq\frac{M}{D_lD_{l+1}}.
\end{equation}
The absolute sine between two unoriented lines satisfies the triangle
inequality. Summing \cref{bridge:direction-step}, and using
\cref{bridge:prefix-growth}, yields
\[
 \abs{s_{B_+}\wedge s_l}
 \leq C M^{1-c-2cl}
 \qquad(1\leq l\leq B_+).
\]
Resolve $s_{B_+}$ into the singular input directions for $T_{0,l}$.
Using $D_l\leq M^l$ as well as its lower bound gives
\begin{equation}\label{bridge:stable-prefix}
 \norm{T_{0,l}s_{B_+}}
 \leq M^{-cl}+C M^{l+1-c-2cl}
 \leq C M^{-0.94l}.
\end{equation}
The last inequality follows from
$1-c+(1-2c)l=0.02-0.96l\leq-0.94l$ for $l\geq1$.

The first component of an input vector becomes the second component
after one step. Thus \cref{bridge:stable-prefix} at $l=1$ shows that
$s_{B_+}$ is nearly vertical. Let $t$ solve the variational recurrence
with $t_{-1}=0$, $t_0=1$. Horizontal input has a projection bounded
away from zero onto the expanded input direction for $T_{0,B_+}$. For
large $M$, at least one of the two coordinates $t_{B_+}$ and $t_{B_+-1}$
consequently has magnitude at least $D_{B_+}/3$. Choose the
corresponding $j$. This proves that the two closed choices in
\cref{bridge:row-choice} cover the states under consideration.

The row of $T_{0,B_+}$ selecting $q_j$ has norm at least
$\abs{t_j}\geq D_{B_+}/3$, and its value on $s_{B_+}$ has magnitude at
most $D_{B_+}^{-1}$. Its kernel direction therefore differs from
$s_{B_+}$ by $O(D_{B_+}^{-2})$. Normalize that direction so that its
second component is 1, and let $U$ denote its variational solution.
Then $U_{-1}=1$, $U_j=0$, and, for $0\leq p<j$,
\begin{equation}\label{bridge:linear-dirichlet}
 \abs{U_p}\leq C M^{-0.9(p+1)}.
\end{equation}
Indeed, apply \cref{bridge:stable-prefix} at $l=p+1$ to its main
component. The error component is bounded by
\[
 C M^{p+1}D_{B_+}^{-2}
 \leq C M^{-0.96(p+1)},
\]
because $p+1\leq B_+$. Normalization changes these bounds only by an
absolute factor.

We now solve the nonlinear boundary problem. Let $\xi_p$ be the change
of the lifted position $q_p$, with boundary data $\xi_{-1}=r$ and
$\xi_j=0$. The tridiagonal matrix on $0,\ldots,j-1$ with diagonal
$v(q_p)$ and adjacent entries $-1$ has determinant $t_j\ne0$. Its
inverse is
\begin{equation}\label{bridge:green}
 G_{lp}=t_{\min(l,p)}U_{\max(l,p)}.
\end{equation}
For completeness, the proposed column solves the homogeneous
recurrence except at its source. At that source the output is 1,
because the Wronskian $t_{p+1}U_p-t_pU_{p+1}$ equals its boundary
value 1. Both boundary values of the proposed column are zero. This
verifies \cref{bridge:green} directly.

Subtract the original orbit recurrence from the new one, and move the
nonlinear Taylor remainder to the right side. With its sign absorbed
into $F_p$, the boundary problem becomes
\begin{equation}\label{bridge:fixed-point}
 \xi_l=U_l r+\sum_{p=0}^{j-1}G_{lp}F_p(\xi_p),
 \qquad
 \abs{F_p(u)}\leq C M u^2,
 \quad
 \abs{F'_p(u)}\leq C M\abs u.
\end{equation}
Use the finite-dimensional weighted norm
\[
 \norm{\xi}_*=\max_{0\leq l<j}
          M^{0.8(l+1)}\abs{\xi_l}.
\]
On a ball of fixed radius $R_0$, the Lipschitz bound for the nonlinear
part of \cref{bridge:fixed-point}, before weighting its output at
index $l$, is bounded by a constant depending on $R_0$ times
\[
 \sum_{p\leq l}
 M^{p-0.9(l+1)+1-1.6(p+1)}
 +\sum_{p>l}
 M^{l-0.9(p+1)+1-1.6(p+1)}.
\]
Here $\abs{t_p}\leq M^p$ and \cref{bridge:linear-dirichlet} were
used. After multiplying by $M^{0.8(l+1)}$, the exponents are,
respectively,
\[
 -0.1(l+1)-0.6(p+1),
 \qquad
 1.8(l+1)-2.5(p+1).
\]
Their sums tend uniformly to zero as $M\to\infty$, independently of
$j$. The linear term $Ur$ has bounded weighted norm for $\abs r\leq1/4$.
Choose $R_0$ larger than twice this bound and then choose $M$ large.
The right side of \cref{bridge:fixed-point} maps this ball to itself
and is a contraction there. Its unique fixed point gives the desired
graph on the whole parameter interval.

The same estimates show that
$I-G\diag(F'_p(\xi_p))$ has a uniformly bounded inverse in this
weighted norm. Multiplication by the original Dirichlet matrix
identifies it with the varied linearized Dirichlet problem. That
problem is therefore invertible, and the finite-dimensional implicit
function theorem gives smooth dependence on $r$. Differentiating
\cref{bridge:fixed-point} solves with right side $U$, which has
bounded weighted norm. This proves \cref{bridge:graph-contraction}.
The contraction and its fixed point also depend continuously on the
eligible initial data: all the coefficients are continuous there, and
the contraction constant and ball are uniform. This proves the
asserted continuous dependence.

If $j=B_+$, the endpoint pair estimate follows directly from the
fixed boundary value and \cref{bridge:graph-contraction}. If
$j=B_+-1$, the recurrence and the fixed value of $q_{B_+-1}$ give
$\xi_{B_+}=-\xi_{B_+-2}$, proving the same estimate without an extra
factor of $M$.

It remains to control transverse distortion. At the varied orbit let
$t_p(r)$ again have initial values $t_{-1}(r)=0$, $t_0(r)=1$.
Invertibility of the varied Dirichlet problem gives $t_j(r)\ne0$.
The graph derivative $U_p(r)=\dd q_p(r)/\dd r$ has boundary values 1
and 0 at indices $-1,j$ and solves that varied problem. Its Green
diagonal is therefore $t_p(r)U_p(r)$. Differentiating the determinant
of the varied Dirichlet matrix gives the exact identity
\begin{equation}\label{bridge:determinant-distortion}
 \frac{\dd}{\dd r}\ln\abs{t_j(r)}
 =\sum_{p=0}^{j-1}t_p(r)U_p(r)^2\phi''(q_p(r)).
\end{equation}
The derivative of a diagonal entry is $\phi''(q_p(r))U_p(r)$,
explaining the second factor of $U_p$. Since $\abs{t_p(r)}\leq M^p$
for every varied orbit, \cref{bridge:graph-contraction} bounds the
absolute value of \cref{bridge:determinant-distortion} by
\[
 C\sum_{p\geq0}M^p M^{-1.6(p+1)}M
 =C\sum_{p\geq0}M^{-0.6(p+1)}\leq C.
\]
Integration over $\abs r\leq1/4$ proves the bounded-ratio claim.

For backward evolution set $\widetilde q_p=q_{-1-p}$. The recurrence
has the same form in these variables. At step $l$ its ordered pair is
the swapped original pair at time $-l$, and the initial pair is
$(q_{-1},q_0)$. Swapping coordinates is an isometry, so the backward
growth hypotheses and all norm estimates are identical. The resulting
graphs vary $q_0$ and have small slope in the $q_{-1}$ coordinate.
\end{proof}

\begin{proof}[Proof of \cref{bridge:theorem}]
We first find many usable middle times. On the bridge event choose a
unit vector realizing $\norm{T_{a,b}}$ and follow its norms along the
bridge. The increments of their logarithms to base $M$ are at most 1,
and their total shortfall from 1 is at most $(1-c_b)n$. By
\cref{setup:shortfall-count}, if $c_b$ is sufficiently close to 1,
at least $\eta n$ indices $i$ in the middle third, for an absolute
$\eta>0$, satisfy
\begin{equation}
 \begin{aligned}
 \norm{T_{i,i+l}}&\geq M^{cl}
       &&(1\leq l\leq b-i),\\
 \norm{T_{i,i-l}}&\geq M^{cl}
       &&(1\leq l\leq i-a).
 \end{aligned}
 \label{bridge:middle-eligibility}
\end{equation}
In detail, remove indices admitting a forward interval starting there
or a forward interval ending there whose average log increment is
less than $c$. The counting bound removes at most
$C(1-c_b)n/(1-c)$ indices. At a retained index, the chosen vector
bounds the forward operator norm on either kind of interval from
below. On an interval ending at $i$, use also equality of the forward
and inverse operator norms to obtain the second inequality in
\cref{bridge:middle-eligibility}. For large $n$, the integer rounding
of the middle third does not affect a fixed positive proportion.

It suffices to establish \cref{bridge:product}, with a uniform
constant, when the bridge indicator is replaced by the indicator of
\cref{bridge:middle-eligibility} for one fixed middle index $i$.
Indeed, the bridge indicator is at most $(\eta n)^{-1}$ times the sum
of these middle-index indicators; multiplying by the nonnegative
weight and summing then costs only an absolute constant.

Shift $i$ to time 0. Put $B_+=b-i$ and $B_-=i-a$; both are at least
$n/3-O(1)$. Apply \cref{bridge:graphs} forward and backward. There
are two possible endpoint-coordinate choices in each direction.
We work with one fixed pair of choices and eventually sum over the
four possibilities. Write $Q(x,y)$ for the lifted forward coordinate
held fixed, and $F(x,y)$ for the lifted backward coordinate held
fixed. On forward graphs $Q_x\ne0$; on backward graphs $F_y\ne0$.
For large $M$, all forward slopes $\abs{\dd x/\dd y}$ and backward
slopes $\abs{\dd y/\dd x}$ are less than a common absolute number
$a_0<1/2$. The relevant transverse partial derivatives have uniformly
bounded absolute ratios along each graph.

\paragraph{Transport of endpoint observations.}
Along a forward graph the endpoint pair at time $b$ changes by at
most $CM^{-0.8(B_+-1)}$. A single forward or inverse map is
$M$-Lipschitz on the lifted coordinates. Thus pair differences
throughout the window of radius $\kappa n$ about $b$ are at most
\[
 C M^{\kappa n+O(1)}M^{-0.8(B_+-1)}.
\]
The one-step matrix difference is at most $CM$ times the position
difference. Since forward and inverse product norms agree, the
endpoints of each observation interval may first be put in increasing
order. For a product of length at most $2\kappa n$, the
telescoping product identity therefore bounds its matrix difference
by a length factor times
\[
 C M^{3\kappa n+O(1)}M^{-0.8(B_+-1)}.
\]
Every product norm compared here is at least 1. The elementary
inequalities $\abs{\norm A-\norm B}\leq\norm{A-B}$ and
$\abs{\log u-\log v}\leq\abs{u-v}$ for $u,v\geq1$ consequently
give the same exponential control for the normalized logarithms in
\cref{bridge:shortfalls}. The positive integer interval lengths in
their denominators cause no additional loss.

Choose fixed $\kappa,\sigma>0$ with
\begin{equation}
 3\kappa+\sigma<0.8/3.
 \label{bridge:error-exponents}
\end{equation}
The strict margin absorbs the length factor and the bounded powers
of $M$ when $n$ is large. After increasing the size requirements to
account for the observation bounds, each observation at the forward
end changes by at most
\begin{equation}
 \delta_n=C M^{-\sigma n}
 \label{bridge:observation-error}
\end{equation}
along a forward graph. The same statement holds for an observation
at the backward end along a backward graph. All these estimates are
uniform in the chosen middle index, endpoint-coordinate choices,
and observation indices. Increase $M,n$ further so that
$\delta_n<h/2$.

The two transverse coordinate families used next are depicted in the
original source; here we describe them in words. One family consists
of the almost vertical forward graphs, the other of the almost
horizontal backward graphs. They cross at the middle-time torus.

\paragraph{Forward labels and their area.}
Assume first that $H$ is at the forward end and $W$ at the backward
end. Cover the middle-time torus by the finitely many closed squares
of a grid of side $s_0=1/32$ in a fundamental domain. Overlaps on
boundaries have area zero and cause no problem for an upper bound.
Fix one square $I_x\times I_y$ and choose $y_c\in I_y$. All position
functions in this square and its small neighborhood are computed
from the same real lifted recurrence starting at $(x,y)$.

Every forward graph through an eligible point in the square extends
over all of $I_y$, with a margin, since its parameter range has length
$1/2$ and $s_0<1/4$. Label it by its intercept $s$ at $y=y_c$ and
write it as $(X(s,y),y)$. Two such graphs which meet have the same
value of $Q$ and coincide locally because $Q_x\ne0$. They then
coincide on the whole common interval: the coincidence set is both
open, by implicit uniqueness, and closed, by continuity. In
particular distinct labels have disjoint images at every fixed $y$.

Let $E$ be the labels of forward graphs through forward-eligible
points of this square for which $H\geq h$. We use closed eligibility
conditions, including \cref{bridge:row-choice}. They and $H\geq h$
define a compact subset of the closed square. Continuous dependence
of the graph in \cref{bridge:graphs} makes its intercept a continuous
function of the initial point. Thus $E$ is compact. The graph is
uniquely determined by its intercept, so these constructions define
$X$ unambiguously on $E\times I_y$.

For these labels, every point of the graph over $I_y$ has $H>h/2$ by
\cref{bridge:observation-error}. The identity
\[
 Q(X(s,y),y)=Q(s,y_c)
\]
and implicit differentiation give
\begin{equation}
 X_s(s,y)=\frac{Q_x(s,y_c)}{Q_x(X(s,y),y)}.
 \label{bridge:label-jacobian}
\end{equation}
The derivative in \cref{bridge:label-jacobian} is interpreted through
a local smooth extension when $E$ has empty interior; the existence
of that extension is justified below. The distortion assertion of
\cref{bridge:graphs} bounds its absolute value above and below by
positive constants.

The map $(s,y)\mapsto(X(s,y),y)$ on $E\times I_y$ is injective and
has Jacobian comparable to 1. Its image stays in a fixed bounded
lift: the small slope bounds the displacement from the square by an
absolute constant. Projection of this bounded lift to the torus has
bounded multiplicity. Change of variables therefore yields
\begin{equation}
 \abs E\leq C\PP(H>h/2).
 \label{bridge:label-area}
\end{equation}
The factor $\abs{I_y}=s_0$ has been absorbed into the constant.

\paragraph{A transverse sweep for the weight.}
Fix $s\in E$. Let $D_s\subseteq I_y$ consist of the parameters $y$
for which $z(y)=(X(s,y),y)$ lies in the square and is backward-eligible
in the fixed backward endpoint choice. This is a compact set. Carry
each $z(y)$ along its backward graph to first coordinate $u\in I_x$,
and denote the resulting second coordinate by $\Theta_s(u,y)$.
The carry is defined throughout $I_x$, with a margin.

For each fixed $u$, the map $y\mapsto\Theta_s(u,y)$ is injective on
$D_s$. If two carried points agreed, their backward graphs would
coincide throughout $I_x$ by regular-level uniqueness. A single such
graph cannot meet the chosen forward graph in two distinct points:
two intersections would satisfy both
$\abs{\Delta x}\leq a_0\abs{\Delta y}$ and
$\abs{\Delta y}\leq a_0\abs{\Delta x}$, forcing equality of the
points. This proves the asserted injectivity.

The equation defining the carry is
\[
 F(u,\Theta_s(u,y))=F(X(s,y),y).
\]
Differentiation with respect to the starting parameter gives
\begin{equation}
 \partial_y\Theta_s(u,y)
 =\frac{F_y(z(y))+F_x(z(y))X_y(s,y)}
        {F_y(u,\Theta_s(u,y))}.
 \label{bridge:sweep-jacobian}
\end{equation}
The ratio of the two $F_y$ terms is uniformly bounded above and
below along the backward graph. Moreover $\abs{F_x/F_y}\leq a_0$
at its starting point, since $-F_x/F_y$ is its slope, and
$\abs{X_y}\leq a_0$. Hence the absolute value of
\cref{bridge:sweep-jacobian} is uniformly bounded above and below
by positive constants.

The endpoint observation estimate gives, for all $u\in I_x$ and
$y\in D_s$,
\[
 W(X(s,y),y)\leq W(u,\Theta_s(u,y))+\delta_n.
\]
Apply change of variables to the injective map
$(u,y)\mapsto(u,\Theta_s(u,y))$ on $I_x\times D_s$.
Its image lies in a bounded lift. Since $W\geq0$, its integral over
that image is at most a fixed multiple of $\E W$. Using
\cref{bridge:sweep-jacobian} and then dividing by $\abs{I_x}$ gives
\begin{equation}
 \int_{D_s}W(X(s,y),y)\,\dd y
 \leq C\bigl(\E W+M^{-\sigma n}\bigr).
 \label{bridge:conditional-weight}
\end{equation}

\paragraph{Changes of variables on the eligible subsets.}
Here are details ensuring the preceding changes of variables do not
require the eligible sets to have interior. Consider first one forward
graph with label $s$. It is a compact regular level arc over $I_y$ and
extends with a margin beyond this interval. The partial derivative
$Q_x$ is nonzero on the arc. The implicit function theorem in finitely
many overlapping charts along the arc therefore continues it smoothly
for intercepts in an open neighborhood of $s$. The continued level is
prescribed by $Q(s',y_c)$ for each nearby intercept $s'$. Uniqueness
in the overlapping charts makes these local continuations consistent.
Every nearby eligible label follows the same continuation, because
its regular graph passes through that intercept. Thus $X$ on
$E\times I_y$ is locally the restriction of a smooth function of
$(s,y)$, and \cref{bridge:label-jacobian} holds at all required
points.

For the transverse sweep, fix $s$ and a point of $D_s$. Its backward
graph is likewise a compact regular arc, with $F_y\ne0$, extending
with margin over $I_x$. The same continuation argument applies to
the smoothly varying starting points $(X(s,y),y)$ on the fixed
forward graph. It gives a local smooth extension of
$\Theta_s(u,y)$ in both variables, agreeing with all nearby eligible
carries by uniqueness. The maps used in both area comparisons are
consequently restrictions of local smooth diffeomorphisms, with the
displayed nonvanishing Jacobians.

A countable cover by such charts can be split into disjoint Borel
pieces. On each piece ordinary change of variables applies to its
measurable subset; the images of the pieces are disjoint because
the full restricted maps are injective. Summing proves the formulas
on the compact eligible domains. This also establishes the
measurability of all integrands and swept regions used above.
No extension of eligibility to a neighborhood is assumed.

\paragraph{Completion of the area comparison.}
Every point of the fixed square satisfying both eligibility
conditions and $H>h$ belongs to a forward graph with label in $E$.
Change variables from $x$ to this label at each $y$, using
\cref{bridge:label-jacobian}. Drop conditions other than square
membership and backward eligibility, which is allowed because the
weight is nonnegative. The desired weighted integral over this
square and this pair of choices is bounded by
\[
 C\int_E\int_{D_s}W(X(s,y),y)\,\dd y\,\dd s.
\]
Combining \cref{bridge:conditional-weight,bridge:label-area} bounds
this by the right side of \cref{bridge:product}.

If $H$ is at the backward end, begin instead with the labels of the
almost horizontal backward graphs and sweep them in the $x$
direction to bound their label measure by $\PP(H>h/2)$. For each
such graph, transport its eligible points along the almost vertical
forward graphs to estimate $W$. The proof just given applies with
$x,y$ and $Q,F$ interchanged: the graph extents, distortion bounds,
small-slope injectivity argument, and local chart construction are
the same. This proves the estimate for the other assignment of the
two observations as well.

Finally sum over the finitely many squares and the at most four
endpoint-coordinate choices. The constants remain uniform. Average
over the usable middle indices as explained after
\cref{bridge:middle-eligibility}. This establishes
\cref{bridge:product} and completes the proof.
\end{proof}

\section{Product distances and measures across scales}
\label{sec:scales}

Suppose, toward a contradiction, that there is a sequence of parameters
$k\to+\infty$ for which the top Lyapunov exponent vanishes almost
everywhere. All limits in this section are taken along this sequence,
passing to subsequences when indicated, except where a fixed-parameter
limit in orbit length is explicitly specified. The structural constants
in \cref{can:theorem,bridge:theorem} are fixed throughout; the bridge
estimate's auxiliary constants may depend on its observation bounds.
We use the growth semimetric $g$ and expected shortfall $e_n$ from
\cref{setup:distance-deficit}; their parameter dependence remains
suppressed. At each parameter in the sequence, the zero-exponent
assumption and bounded convergence give
\begin{equation}\label{scale:long-deficit}
 e_n\longrightarrow1\qquad(n\to\infty).
\end{equation}
By contrast, the shortfall at every fixed length tends to zero as the
parameter grows. We select a dyadic range in which the shortfall first
increases, and sum the loss between each whole interval and its two
halves over that range. This sum telescopes, but its limiting mass
must still be controlled near arrays with one constant growth rate.
The local tests and measures below are designed to retain that mass.

\subsection*{Four-point comparison and truncation}

\begin{lemma}[Four-point comparison]\label{scale:fourpoint}
Fix a base index $i$ and write
\[
 D_j=\norm{T_{i,j}},\qquad
 R_{jp}=M^{g(j,p)-g(i,j)-g(i,p)}.
\]
There is an absolute constant $C_0\ge1$ such that, for every orbit,
every sufficiently large parameter, and every choice of integer indices,
\begin{equation}\label{scale:R-fourpoint}
 R_{jp}\le C_0\max\{R_{jl},R_{lp}\}.
\end{equation}
\end{lemma}

\begin{proof}
Choose a most expanded unit input direction $a_j$ for $T_{i,j}$; if
$D_j=1$, any singular direction may be used. Set
\[
 Q_{jp}=\max\{|a_j\wedge a_p|,D_j^{-2},D_p^{-2}\}.
\]
We claim
\begin{equation}\label{scale:R-comparison}
 Q_{jp}\le R_{jp}\le3Q_{jp}.
\end{equation}
To obtain the sine lower bound, use a unit input at the base
perpendicular to $a_p$. Its norm at $p$ is $D_p^{-1}$, whereas its norm
at $j$ is at least $D_j|a_j\wedge a_p|$. The transfer from $p$ to $j$
therefore has norm at least $D_jD_p|a_j\wedge a_p|$. The inequalities
\[
 \norm{T_{p,j}}\ge D_j/D_p,\qquad
 \norm{T_{p,j}}\ge D_p/D_j
\]
give the other two lower bounds; the second uses equality of the norms
of a determinant-one matrix and its inverse.

For the upper bound, transport a unit vector from $p$ back to the
base. Its components along $a_p$ and $a_p^\perp$ have absolute values
at most $D_p^{-1}$ and $D_p$, respectively. Also
\[
 \norm{T_{i,j}a_p^\perp}
 \le D_j|a_j\wedge a_p|+D_j^{-1}.
\]
It follows that
\[
 R_{jp}\le D_p^{-2}+|a_j\wedge a_p|+D_j^{-2},
\]
proving \cref{scale:R-comparison}. By \cref{setup:sine-triangle},
$Q_{jp}\le2\max\{Q_{jl},Q_{lp}\}$, so $C_0=6$ is permissible.
\end{proof}

\begin{lemma}[Truncation of a cancellation]\label{scale:truncation}
Fix integers $i$, $n\ge1$, and $1\le s\le n$, and put
\[
 L=i-n,\quad R=i+n,\quad L'=i-s,\quad R'=i+s,
\]
\[
 D=n-\min\{g(i,L),g(i,R)\},\qquad
 \ell=g(i,L)+g(i,R)-g(L,R).
\]
Use the quantities $R_{uv}$ of \cref{scale:fourpoint} with base $i$.
Then
\begin{align}
 g(i,L'),g(i,R')&\ge s-D,
 \label{scale:cut-wings}\\
 R_{L'R'}&\le C_0^2\max\{R_{LR},M^{-2s+2D}\}.
 \label{scale:cut-R}
\end{align}
There is a constant $A$ independent of the parameter and indices such
that the cut cancellation loss satisfies
\begin{equation}\label{scale:cut-loss}
 g(i,L')+g(i,R')-g(L',R')
 \ge\min\{\ell,2s-2D\}-A,
\end{equation}
and consequently
\begin{equation}\label{scale:cut-full}
 g(L',R')\le2s-\min\{\ell,2s-2D\}+A.
\end{equation}
\end{lemma}

\begin{proof}
The triangle inequality and \cref{setup:distance-bound} give
\[
 g(i,L')\ge g(i,L)-g(L,L')\ge(n-D)-(n-s)=s-D,
\]
and similarly on the right. In particular,
\[
 R_{LL'}\le M^{(n-s)-(n-D)-(s-D)}=M^{-2s+2D},
 \qquad R_{RR'}\le M^{-2s+2D}.
\]
Apply \cref{scale:R-fourpoint} twice along the chain $L',L,R,R'$ to
obtain \cref{scale:cut-R}. Taking negative base-$M$ logarithms gives
\cref{scale:cut-loss}, with any fixed upper bound for $2\log_M C_0$ as
$A$. Such a bound is uniform after imposing, for example, $M\ge2$.
Finally, each cut wing has length at most $s$, which yields
\cref{scale:cut-full}.
\end{proof}

\begin{lemma}[Doubling bound]\label{scale:doubling}
For all sufficiently large parameters and every integer $n\ge1$,
\begin{equation}\label{scale:doubling-equation}
 e_n\le e_{2n}\le C e_n+\frac{C}{n},
\end{equation}
with a constant independent of $n$ and the parameter.
\end{lemma}

\begin{proof}
Use the notation of \cref{scale:truncation}. Both $D$ and $\ell$ are
nonnegative, and $\ell\le2n$. Let $\delta>0$ be the fixed constant in
\cref{can:theorem}. Choose a fixed $H$ sufficiently large that
\[
 \ell>H(1+D),\qquad s=\lfloor\ell/2\rfloor
 \quad\Longrightarrow\quad
 s\ge1,\quad 2D+A\le\delta s.
\]
This is possible since $s\ge\ell/2-1$. Also $s\le n$ and $2s\le\ell$.
Thus \cref{scale:cut-wings,scale:cut-full} imply
\[
 \min\{g(i,i-s),g(i,i+s)\}\ge(1-\delta)s,
 \qquad g(i-s,i+s)\le2D+A\le\delta s.
\]
This is the event bounded by \cref{can:theorem}, with spacing $s$. On
the part where this integer $s$ is selected, $\ell<2s+2$. Writing the
constants of that theorem as $C_{\rm can},\gamma>0$,
\[
 \E\bigl[\ell\,\one_{\{\ell>H(1+D)\}}\bigr]
 \le C_{\rm can}\sum_{s=1}^{\infty}(2s+2)e^{-\gamma s}
 \le C.
\]
The complementary contribution is at most $H(1+\E D)$, hence
\begin{equation}\label{scale:mean-cancellation}
 \E\ell\le C(1+\E D).
\end{equation}
Stationarity gives
\[
 \E D\le\E[n-g(i,L)]+\E[n-g(i,R)]=2ne_n,
 \qquad e_{2n}=e_n+\frac{\E\ell}{2n}.
\]
These identities and \cref{scale:mean-cancellation} prove the result.
In particular the lower inequality uses only $\ell\ge0$.
\end{proof}

\subsection*{Selecting the first growing deficit}

\begin{lemma}[Critical scales]\label{scale:selection}
Along the assumed zero-exponent parameter sequence, there are powers of
two $n_0=n_0(K)$ and $N=N(K)>n_0$, and numbers $\epsilon=e_N$, such
that
\begin{equation}\label{scale:selected-limits}
 n_0\to\infty,\qquad \epsilon\to0,\qquad
 \epsilon n_0\to\infty,
\end{equation}
and, for every fixed integer $m\in\Z$ (once the lengths below are
positive integers),
\begin{equation}\label{scale:selected-boundaries}
 e_{2^m n_0}=o(\epsilon),\qquad
 e_{2^m N}=O_m(\epsilon).
\end{equation}
All scale sums below run over powers of two.
\end{lemma}

\begin{proof}
First, for every fixed integer $a\ge1$, we have $e_a\to0$ as
$K\to\infty$. To see this, fix $0<\zeta<1$. The uniform marginal law
of each $q_i$ implies
\[
 \PP\{|v(q_i)|<M^{1-\zeta}\}\longrightarrow0.
\]
Indeed $v(x)=2+K\cos(2\pi x)$, and the length of the set on which
$|\cos(2\pi x)|\le(M^{1-\zeta}+2)/K$ tends to zero. A union bound
therefore shows that, with probability tending to one, all $a$
coefficients in the product satisfy $|v(q_i)|\ge M^{1-\zeta}$. Start
the linear recurrence with $(u_0,u_{-1})=(1,0)$. On this event, for
$M$ large, induction gives
\[
 |u_{j+1}|\ge(M^{1-\zeta}-1)|u_j|\ge|u_j|
 \quad(0\le j<a).
\]
Consequently $g(0,a)/a\ge\log_M(M^{1-\zeta}-1)$ there. The deficit
always lies in $[0,1]$, so $\limsup_{K\to\infty}e_a\le\zeta$. Letting
$\zeta\downarrow0$ proves the claim.

Choose powers of two $n_0\to\infty$ sufficiently slowly that
$e_{n_0}\to0$, by diagonal selection over the fixed lengths
$1,2,4,\ldots$. Set
\[
 \theta_K=(n_0^{-1}+e_{n_0})^{1/2}.
\]
For all sufficiently large $K$, $e_{n_0}<\theta_K<1$. By
\cref{scale:long-deficit}, there is a first power of two $N\ge n_0$ for
which $e_N>\theta_K$. It satisfies $N>n_0$ and $e_{N/2}\le\theta_K$.
Applying \cref{scale:doubling} once gives
\[
 \theta_K<\epsilon=e_N\le C\theta_K+C/n_0\longrightarrow0.
\]
Also $\epsilon n_0\ge\theta_K n_0\ge\sqrt{n_0}\to\infty$. For fixed
$m\ge0$, iteration of \cref{scale:doubling} yields
\[
 e_{2^m n_0}\le C_m(e_{n_0}+n_0^{-1})
 =C_m\theta_K^2=o(\epsilon),
\]
and $e_{2^m N}\le C_m(\epsilon+N^{-1})=O_m(\epsilon)$. For a fixed
negative $m$, the lengths $2^m n_0$ and $2^mN$ are eventually positive
integers. Dyadic monotonicity bounds their deficits by
$e_{n_0}=o(\epsilon)$ and $e_N=\epsilon$, respectively. This proves
every assertion.
\end{proof}

\subsection*{A compact space of distance arrays}

The next construction records shortfall increments as measures on
rescaled distance arrays. Its output will be locally finite limits
away from affine arrays, together with a dilation balance; the
discarded affine part requires separate control in the final argument.

Let $\mathcal D=\Z[1/2]$ denote the dyadic rational numbers. Define
$X$ to be the set of symmetric semimetric arrays
$d:\mathcal D\times\mathcal D\to[0,\infty)$ satisfying
\begin{equation}\label{scale:ambient-bound}
 d(s,t)\le|s-t|+2.
\end{equation}
Give $X$ the topology of pointwise convergence. It is a closed subset
of a countable product of compact intervals, and hence is compact and
metrizable. Its compact subset of affine arrays is
\[
 \mathcal A=\{d_w:0\le w\le1\},\qquad d_w(s,t)=w|t-s|.
\]
These are precisely the arrays of this form that belong to $X$: the
bound at arbitrarily large separations forces $w\le1$. Write
$\mathcal U=X\setminus\mathcal A$.

For a positive integer $n$, define the random array
\begin{equation}\label{scale:rescaled-array}
 g_n(s,t)=\frac{g(\lfloor ns\rfloor,\lfloor nt\rfloor)}{n}
 \qquad(s,t\in\mathcal D).
\end{equation}
It belongs to $X$, since its distances are at most $|s-t|+1/n$.
Introduce the finite measures
\begin{equation}\label{scale:prelimit-measures}
 \mu_K=\epsilon^{-1}\sum_{\substack{n_0\le n<N\\n\text{ a power of }2}}
                  \Law(g_n),\qquad
 \nu_K=\epsilon^{-1}\Law(g_N),\qquad
 \xi_K=\epsilon^{-1}\Law(g_{n_0}).
\end{equation}
Their total masses need not remain bounded. We obtain limits on
$\mathcal U$ by controlling local departures from affinity.

For an interval $I=[s,t]$ with distinct dyadic endpoints, let $I_l,I_r$
be its left and right halves and define continuous functions on $X$ by
\begin{equation}\label{scale:tests}
 h(I)=1-\frac{d(s,t)}{t-s},\qquad
 J(I)=h(I)-\frac{h(I_l)+h(I_r)}{2},\qquad
 V(I)=\bigl(h(I_l)-h(I_r)\bigr)^2.
\end{equation}
The functions $J(I)$ and $V(I)$ are nonnegative on all of $X$: the
first is the triangle inequality, the second a square. The shortfall
$h(I)$ itself need not be nonnegative on all of $X$. It does belong to
$[0,1]$ on a sample $g_n$ whenever the endpoints multiplied by $n$
are integers. For any fixed dyadic interval, this alignment holds at
every scale $n\ge n_0$ occurring here once $K$ is sufficiently large.

\begin{lemma}[Bounds for local tests]\label{scale:local-mass}
Let $I$ have dyadic endpoints and length $p=2^m$, where $m\in\Z$ is
fixed. Then
\begin{equation}\label{scale:sum-test-bound}
 \int\bigl(J(I)+V(I)\bigr)\,\dd\mu_K=O_I(1).
\end{equation}
Uniformly over the individual dyadic scales $n_0\le n\le N$,
\begin{equation}\label{scale:individual-test-bound}
 \epsilon^{-1}\E\bigl[h(I)(g_n)+J(I)(g_n)+V(I)(g_n)\bigr]
 =O_I(1).
\end{equation}
At the initial scale the stronger bound holds:
\begin{equation}\label{scale:initial-test-bound}
 \epsilon^{-1}\E\bigl[h(I)(g_{n_0})+J(I)(g_{n_0})
                         +V(I)(g_{n_0})\bigr]=o(1).
\end{equation}
All three statements concern sufficiently large $K$, so the indicated
endpoints and half-endpoints align.
\end{lemma}

\begin{proof}
By stationarity,
\[
 \E h(I)(g_n)=e_{pn},\qquad
 \E J(I)(g_n)=e_{pn}-e_{pn/2}.
\]
Summation over consecutive dyadic scales gives the exact identity
\begin{equation}\label{scale:J-telescope}
 \sum_{\substack{n_0\le n<N\\n\text{ a power of }2}}
       \E J(I)(g_n)=e_{pN/2}-e_{pn_0/2}.
\end{equation}
By \cref{scale:selection} and dyadic monotonicity, its nonnegative
right side is $O_I(\epsilon)$.

Put $a=h(I_l)$, $b=h(I_r)$, and $\bar h=(a+b)/2$. At the aligned
sampled arrays, $0\le\bar h\le h(I)\le1$, and therefore
\begin{align*}
 \frac{V(I)}4
 &=\frac{a^2+b^2}{2}-\bar h^2\\
 &\le\frac{a^2+b^2}{2}-h(I)^2+2J(I).
\end{align*}
Write $b_q=\E\bigl[(1-g(0,q)/q)^2\bigr]$ for integer $q\ge1$. By
stationarity and another dyadic telescoping,
\[
 \frac14\sum_n\E V(I)(g_n)
 \le b_{pn_0/2}-b_{pN/2}
      +2\bigl(e_{pN/2}-e_{pn_0/2}\bigr)
 \le b_{pn_0/2}+2e_{pN/2}.
\]
Here and in this display only, the unqualified sum is over the scales
in \cref{scale:J-telescope}. Since $0\le b_q\le e_q$, the last
expression is $O_I(\epsilon)$. This proves \cref{scale:sum-test-bound}.

At any single aligned scale, $J(I)\le h(I)$ and
$V(I)=(a-b)^2\le a+b$. Hence the expectation in
\cref{scale:individual-test-bound} is at most $2e_{pn}+2e_{pn/2}$.
Dyadic monotonicity bounds these terms by the corresponding
terminal-scale deficits, which are $O_I(\epsilon)$ by
\cref{scale:selection}. At $n=n_0$ they are $o(\epsilon)$, again by
that lemma, including its consequence for negative shifts.
\end{proof}

\begin{lemma}[Detection of affinity]\label{scale:affinity-tests}
As $I$ ranges over intervals with dyadic endpoints and power-of-two
length, the common zero set of $J(I)+V(I)$ is exactly $\mathcal A$.
Consequently, every compact subset $F\subset\mathcal U$ admits finitely
many such intervals $I_1,\ldots,I_b$ and a number $c_F>0$ such that
\begin{equation}\label{scale:finite-test-cover}
 \sum_{a=1}^{b}\bigl(J(I_a)+V(I_a)\bigr)\ge c_F
 \quad\hbox{on }F.
\end{equation}
\end{lemma}

\begin{proof}
Every affine array annihilates these tests. Conversely, suppose they
all vanish. For one of the stated intervals $I$, the equalities
$J(I)=V(I)=0$ say that its two halves have the same speed as $I$ and
that their distances add to the full distance. Repeating this argument
shows that every edge of every dyadic subdivision of $I$ has distance
$w_I$ times its length, with $w_I=d(s,t)/(t-s)$. The triangle
inequality gives the same upper bound between any two subdivision
nodes. Applying the triangle inequality across the full interval,
using those nodes as intermediate points, gives the reverse bound, and
therefore equality. Every dyadic point of $I$ is a subdivision node at
some level. Apply this argument to $[-2^a,2^a]$ for arbitrarily large
integers $a$. The common pair $0,1$ fixes one speed on all these
intervals, so the array is globally affine. The ambient bound forces
that speed to lie in $[0,1]$.

For the last assertion, each point of $F$ has a positive test. The
corresponding open sets cover $F$, and a finite subcover suffices.
Their sum has a positive minimum on the compact set $F$.
\end{proof}

\subsection*{Locally finite limits and their symmetries}

The limiting inequality \cref{scale:tree-inequality} below is the
zero-hyperbolic Gromov-product condition: after identifying points at
zero semimetric distance, $B_v(s,t)/2$ is the Gromov product based at
$v$; see \cite[Section 2.2, Definition 2.10]{Bestvina1997}. The
determinant-one estimate and the passage to the limiting arrays are
proved here; no real-tree representation theorem is used.

\begin{proposition}[Limits away from affinity]\label{scale:limits}
After passing to a subsequence, the restrictions of $\mu_K$ and $\nu_K$
to $\mathcal U$ converge vaguely to locally finite nonnegative Borel
measures $\mu$ and $\nu$, respectively. Moreover,
\begin{equation}\label{scale:initial-negligible}
 \xi_K\longrightarrow0\quad\hbox{vaguely on }\mathcal U.
\end{equation}
The measures $\mu$ and $\nu$ are supported on the closed class
$X_0\subset X$ consisting of arrays satisfying $d(s,t)\le|s-t|$ and the
inequalities
\begin{equation}\label{scale:tree-inequality}
 B_v(s,t)\ge\min\{B_v(s,p),B_v(p,t)\},\qquad
 B_v(s,t)=d(v,s)+d(v,t)-d(s,t)
\end{equation}
for every $v,s,t,p\in\mathcal D$. Every member of $X_0$ extends uniquely
to a continuous semimetric on $\R$, obeying the same inequalities at all
real arguments. Both limiting measures are invariant under
\[
 (\tau_r d)(s,t)=d(s+r,t+r)\qquad(r\in\mathcal D).
\]
For every interval $I$ as in \cref{scale:local-mass},
\begin{equation}\label{scale:limiting-test-bound}
 \int(J(I)+V(I))\,\dd\mu<\infty,
 \qquad
 \int(h(I)+J(I)+V(I))\,\dd\nu<\infty.
\end{equation}
In particular,
\begin{equation}\label{scale:terminal-shortfall}
 \int h([0,1])\,\dd\nu\le1.
\end{equation}
\end{proposition}

\begin{proof}
For each compact $F\subset\mathcal U$, combine
\cref{scale:finite-test-cover} with
\cref{scale:sum-test-bound,scale:individual-test-bound} to obtain
uniform bounds on $\mu_K(F)$ and $\nu_K(F)$. Using
\cref{scale:initial-test-bound} instead gives $\xi_K(F)=o(1)$. The
latter proves \cref{scale:initial-negligible}, since a continuous
compactly supported test is bounded and has compact support.

Here is an explicit way to extract the local limits. Fix a compatible
metric on $X$ and let $a(d)=\dist(d,\mathcal A)$. Choose a continuous
nondecreasing function $\chi:[0,\infty)\to[0,1]$ equal to zero on
$[0,1/2]$ and equal to one on $[1,\infty)$. Set
$\chi_j(d)=\chi(2^j a(d))$. These functions increase to one on
$\mathcal U$, have compact support there, and satisfy $\chi_{j+1}=1$ on
$\supp\chi_j$. The finite measures $\chi_j\mu_K$ and $\chi_j\nu_K$ have
bounded total masses for each $j$. Weak compactness for finite measures
on compact metric spaces, followed by diagonal extraction, gives weak
limits for all of them. Their restrictions agree on open regions where
the corresponding cutoffs are one: this follows by testing against
continuous functions supported in such a region, for which the
prelimit integrals agree. These compatible restrictions define locally
finite measures on $\mathcal U$. Every compactly supported continuous
function is supported where some $\chi_j$ is one, so these are the
asserted vague limits.

For completeness, the support restrictions survive despite the
possibly unbounded total masses. The arrays $g_n$ satisfy
$g_n(s,t)\le|s-t|+1/n$. Taking logarithms in \cref{scale:R-fourpoint},
then dividing by $n$, shows that their version of
\cref{scale:tree-inequality} has error at most $(\log_M C_0)/n$.
These errors are bounded uniformly by $C/n_0$ at every sampled scale.
Any fixed strict violation, by a fixed positive margin, is thus absent
from every sampled array once $K$ is large. Testing on compact subsets
of its open violation set shows that it has zero limiting measure.
There are countably many dyadic arguments and positive rational
margins, which proves the stated support assertion. For an array with
$d(s,t)\le|s-t|$, the semimetric inequality gives
\[
 |d(s,t)-d(s',t')|\le|s-s'|+|t-t'|.
\]
It therefore has a unique jointly continuous extension to real
arguments; all its inequalities extend by continuity.

For fixed $r\in\mathcal D$, each prelimit law is invariant under
$\tau_r$ once $n_0r$ is an integer. Indeed
$\lfloor n(s+r)\rfloor=\lfloor ns\rfloor+nr$, and stationarity applies
simultaneously to the array. The map $\tau_r$ is a homeomorphism of
$X$ preserving $\mathcal A$, so it preserves compact support in
$\mathcal U$. Invariance therefore passes to the vague limits.

Finally, multiply any nonnegative test $J(I)+V(I)$ by $\chi_j$, pass
to the vague limit, and use \cref{scale:sum-test-bound}. Monotone
convergence as $j\to\infty$ gives the first bound in
\cref{scale:limiting-test-bound}. For $\nu$ use
\cref{scale:individual-test-bound} in the same way. Although $h(I)$ is
not nonnegative everywhere on $X$, it is nonnegative on the aligned
samples and on the supported limiting class $X_0$; these facts justify
the cutoff bounds and monotone convergence for that term as well. In
the case $I=[0,1]$ the uncut terminal integral is exactly
$e_N/\epsilon=1$, giving \cref{scale:terminal-shortfall}.
\end{proof}

Define the dilation
\begin{equation}\label{scale:dilation}
 (Sd)(s,t)=\frac{d(2s,2t)}2.
\end{equation}
It is a continuous map from $X$ into $X$ and fixes every affine array.
Its inverse need not map all of $X$ into $X$, but on $X_0$ it is the
homeomorphism
\[
 (S^{-1}d)(s,t)=2d(s/2,t/2).
\]
Both maps preserve $X_0\cap\mathcal U$.

\begin{proposition}[Positive dilation decomposition]\label{scale:decomposition}
On $\mathcal U$, with the measures supported on $X_0$ as above,
\begin{equation}\label{scale:measure-balance}
 S_*\mu=\mu+\nu.
\end{equation}
There is a locally finite nonnegative measure $\mu_\infty$, invariant
under $S$ and every dyadic translation, such that
\begin{equation}\label{scale:measure-decomposition}
 \mu=\mu_\infty+\sum_{j=1}^{\infty}(S^{-j})_*\nu,
 \qquad S_*\mu_\infty=\mu_\infty.
\end{equation}
The series is a series of positive measures dominated by $\mu$.
\end{proposition}

\begin{proof}
The floor convention in \cref{scale:rescaled-array} gives the exact
identity $Sg_n=g_{2n}$. Hence, as finite measures on $X$,
\begin{equation}\label{scale:finite-measure-balance}
 S_*\mu_K=\mu_K+\nu_K-\xi_K.
\end{equation}
To pass to the local limits, observe that
$S^{-1}(\mathcal A)=\mathcal A$: multiplication by two permutes the
dyadic rationals, so $Sd=d_w$ implies $d=d_w$. If $F\subset\mathcal U$
is compact, its preimage under $S$ is closed in compact $X$ and
disjoint from $\mathcal A$, hence compact in $\mathcal U$. Thus $S$ is
proper on $\mathcal U$, and composition with $S$ preserves continuous
compactly supported tests there. Vague convergence and
\cref{scale:initial-negligible} now give \cref{scale:measure-balance}.

We may henceforth work on $X_0\cap\mathcal U$, where $S$ is invertible.
Applying its inverse to \cref{scale:measure-balance} and iterating
gives, for every $b\ge1$,
\[
 \mu=(S^{-b})_*\mu+\sum_{j=1}^{b}(S^{-j})_*\nu.
\]
The increasing positive partial sums are dominated by $\mu$. Their
limit $\lambda=\sum_{j\ge1}(S^{-j})_*\nu$ is therefore a locally finite
measure. On relatively compact Borel sets define
$\mu_\infty=\mu-\lambda$. The definitions agree on overlaps and extend
by exhaustion to a locally finite nonnegative measure; this
construction avoids subtracting infinite total masses. Monotone
convergence for the positive series yields
\[
 S_*\lambda=\nu+\lambda.
\]
Subtract this identity from \cref{scale:measure-balance} on relatively
compact sets to obtain $S_*\mu_\infty=\mu_\infty$.

For $r\in\mathcal D$ and $j\ge1$, the identity
\[
 \tau_r S^{-j}=S^{-j}\tau_{r/2^j}
\]
and translation invariance of $\nu$ show that every summand of
$\lambda$ is invariant under $\tau_r$. The same holds for their sum
and, by local subtraction from the translation-invariant measure
$\mu$, for $\mu_\infty$. This proves all assertions.
\end{proof}

\section{Support restrictions and the geometry of limiting distances}
\label{sec:shape}

Throughout this section we retain the sequence of parameters, scales,
and measures constructed in \cref{sec:scales}. All limiting arrays
belong to $X_0$, and all measures are considered on
$\mathcal U=X\setminus\mathcal A$. The constants $\delta$ and $c_b$
are those fixed in \cref{can:theorem,bridge:theorem}; neither depends
on the scales or the parameter sequence.

For $d\in X_0$, let $P_d$ be the set of times having no open
neighborhood on which $d(x,y)=w|x-y|$ for some constant $w$, and let
$Q_d$ be the set of times having no open neighborhood on which
$d(x,y)=|x-y|$. These sets are closed, and $P_d\subset Q_d$. A
property of the distance on an interval concerns every pair of points
in that interval.

\begin{proposition}[Support exclusions]\label{shape:exclusions}
There is a measurable set of full measure for $\mu$, $\nu$, and
$\mu_\infty$ on which the following statements hold for every real
choice of their arguments.
\begin{enumerate}
\item There do not exist $r\in\R$ and $t>0$ such that
\begin{equation}\label{shape:forbidden-triple}
 d(r-t,r)>(1-\delta)t,\qquad
 d(r,r+t)>(1-\delta)t,\qquad
 d(r-t,r+t)<\delta t.
\end{equation}
\item If $s\in P_d$ and $t\in Q_d$, then
\begin{equation}\label{shape:slow-pair}
 d(s,t)\le c_b|t-s|.
\end{equation}
\end{enumerate}
The same statements hold almost everywhere for every positive measure
appearing in the decomposition of \cref{scale:decomposition}.
\end{proposition}

\begin{proof}
First fix $r,t\in\mathcal D$, $t>0$. For all sufficiently large $K$,
$nr$ and $nt$ are integers at every scale $n\ge n_0$ in our sums. The
event \cref{shape:forbidden-triple} for $g_n$ is contained in the
cancellation event of \cref{can:theorem}, at center $nr$ and spacing
$nt$. Consequently its $\mu_K$-mass is at most
\[
 \frac{C}{\epsilon}\sum_{n_0\le n<N}^{\rm dyadic}
 e^{-\gamma tn}
 \le \frac{C_t}{\epsilon}e^{-\gamma t n_0}\longrightarrow0.
\]
The last conclusion follows from $\epsilon n_0\to\infty$ in
\cref{scale:selection}. The event is open in $X$. To pass to a
locally finite limit, test its indicator from below by nonnegative
continuous functions with compact support in $\mathcal U$. This shows
that its $\mu$-mass is zero. Taking the countable union over
$r,t\in\mathcal D$ and then using continuity of $d$ excludes every
real triple: strict inequalities at a real center and spacing persist
at nearby dyadic ones.

We record explicitly the weighted consequence of the bridge estimate
needed for the second exclusion. Fix a bridge $[u,v]$ with dyadic
endpoints and length $r=v-u>0$. Fix a power-of-two interval $I$ in
the allowed window about one endpoint and another such interval $I'$
in the allowed window about the other endpoint. The two endpoints can
be interchanged. Put
\[
 W=J(I)+V(I),\qquad H=h(I'),
\]
and fix a threshold $\eta>0$. These are fixed bounded Lipschitz
functions of finitely many shortfalls, and $W\ge0$. All their
endpoints align with the integer times at every scale in the sum once
$K$ is large. By \cref{scale:individual-test-bound} and Markov's
inequality,
\[
 \sup_{n_0\le n<N}^{\rm dyadic}\PP(H(g_n)>\eta/2)
 \le C_{I',\eta}\epsilon.
\]
Applying \cref{bridge:theorem} at the integer bridge length $rn$,
summing, and using \cref{scale:local-mass}, we obtain
\begin{align}
 &\int \one_{\{d(u,v)>c_b r,\ H>\eta\}}W\,\dd\mu_K
 \nonumber\\
 &\qquad\le C\epsilon\left[
       \int W\,\dd\mu_K+
       \frac1\epsilon\sum_{n_0\le n<N}^{\rm dyadic}
                              M^{-\sigma rn}\right]
 \longrightarrow0.\label{shape:weighted-bridge}
\end{align}
The first bracketed term is bounded, and the second tends to zero: the
exponential sum is at most $2M^{-\sigma r n_0}$ for all sufficiently
large $K$, whereas $1/\epsilon=o(n_0)$. All constants and thresholds
in this application are fixed before $K$ tends to infinity. The lower
bounds on $M$ and on bridge length in \cref{bridge:theorem} therefore
hold eventually; no uniformity over varying test intervals is assumed.

Suppose now that \cref{shape:slow-pair} fails for some $s\in P_d$,
$t\in Q_d$. Necessarily $s\ne t$. Choose nearby distinct dyadic
bridge endpoints so that their distance still exceeds $c_b$ times
their temporal separation. The endpoints can be chosen close enough
to $s,t$ that sufficiently small neighborhoods of these two times fit
in the respective bridge windows. Some test $J(I)+V(I)$ with
power-of-two length and dyadic endpoints contained in the first
neighborhood is positive. To see this, choose such an interval
containing $s$ in its interior. If the tests for it and all its dyadic
subdivisions vanished, successive subdivision, equal half speeds, and
equality in the triangle inequality would make $d$ affine on that
interval, contrary to $s\in P_d$. In the second neighborhood there is
similarly a power-of-two interval $I'$ with $h(I')>0$. In fact exact
unit distance between the endpoints of any interval forces exact unit
distance on every subinterval, by the triangle inequality and
$d(x,y)\le|x-y|$; hence an interval containing $t$ in its interior
cannot have zero shortfall.

Choose a positive rational $\eta<h(I')$. The resulting strict bridge
and shortfall conditions define an open set. By
\cref{shape:weighted-bridge}, its $W\mu$-mass is zero, using
compactly supported continuous tests in $\mathcal U$ as above. Its
part with $W>0$ therefore has zero $\mu$-mass. There are only
countably many choices of the bridge, the two tests, their
orientation, and the rational threshold. Every violation is covered
by one of these choices, proving \cref{shape:slow-pair} simultaneously
for all real $s,t$.

Both properties are invariant under every translation and positive
dilation of time, including the corresponding division of distances by
the dilation factor. The measure inequality $(S^{-1})_*\nu\le\mu$
from \cref{scale:decomposition} consequently transfers them to $\nu$.
All the other component measures are dominated by $\mu$, so they
inherit them as well. This also provides a measurable full-measure
set: the exceptional sets were covered by countably many measurable
strict tests before extension to real arguments.
\end{proof}

The local-to-global propagation below uses this zero-hyperbolic
Gromov-product framework (\cite[Section 2.2]{Bestvina1997}), with an
explicit argument on the semimetric arrays rather than an invocation
of a theorem about geodesic real trees.

\begin{proposition}[Shape classification]\label{shape:classification}
Every nonaffine array in $X_0$ satisfying \cref{shape:exclusions} has
one of the following forms:
\begin{enumerate}
\item every secant has speed at most $c_b$;
\item there is one exterior ray of exact unit speed, and the
complementary closed half-line has all secant speeds at most $c_b$;
\item there are two exterior rays of exact unit speed, separated by a
closed interval $[a,b]$ of positive length, and every secant contained
in $[a,b]$ has speed at most $c_b$.
\end{enumerate}
The second possibility includes either orientation. These descriptions
do not assert additivity of distances across the slow part.
\end{proposition}

\begin{proof}
We first show that a connected component $U$ of $\R\setminus P_d$ is
affine with a single speed throughout. Its local speeds agree on
overlapping neighborhoods, and hence are a common $w\in[0,1]$ by
connectedness. A compact subinterval of $U$ has a sufficiently fine
partition $x_0<\cdots<x_m$ such that each consecutive triple lies in a
local affine neighborhood. When $w=0$, the triangle inequality
immediately gives $d(x_0,x_m)=0$. When $w>0$, propagate additivity
inductively. Suppose it holds from $x_0$ through $x_j$, and use the
four-point inequality of \cref{scale:tree-inequality} at base
$v=x_j$ in its limiting form
\begin{equation}\label{shape:tree-inequality}
 B_v(x,z)\ge\min\{B_v(x,y),B_v(y,z)\},\qquad
 B_v(x,z)=d(v,x)+d(v,z)-d(x,z).
\end{equation}
The inductive additivity gives
$B_v(x_0,x_{j-1})=2d(x_{j-1},x_j)>0$, whereas local additivity gives
$B_v(x_{j-1},x_{j+1})=0$. Apply \cref{shape:tree-inequality} to the
latter pair with intermediate point $x_0$. Nonnegativity of $B_v$
forces $B_v(x_0,x_{j+1})=0$, exactly the next additivity statement.
The first two steps are already locally affine. Thus
$d(x_0,x_m)=w(x_m-x_0)$, and varying the compact subinterval proves
the claim. In particular $P_d\ne\varnothing$ for a nonaffine array.

Suppose a component $U$ has speed $w>c_b$. It has at least one finite
endpoint $s\in P_d$, since $U=\R$ would imply affinity. If $w<1$,
every interior point $t\in U$ belongs to $Q_d$. Continuity at the
endpoint gives $d(s,t)=w|t-s|$, contradicting \cref{shape:slow-pair}.
Therefore $w=1$. Such a component cannot have two finite endpoints:
they would both belong to $P_d\subset Q_d$, and their unit mutual
speed would again contradict \cref{shape:slow-pair}. Hence every
component of speed greater than $c_b$ is an exterior unit ray; there
are at most two.

Let $I$ be the closed complement of these rays. It is the whole line,
a closed half-line, or a closed bounded interval. For $x<y$ in $I$,
if $[x,y]\cap P_d$ is empty, the two points lie in a component of
speed at most $c_b$. Otherwise let $p$ and $q$ be the first and last
points of $[x,y]\cap P_d$. The portions from $x$ to $p$ and from $q$
to $y$ have speed at most $c_b$, by the component claim and
continuity. The middle portion satisfies the same bound by
\cref{shape:slow-pair}, because $p,q\in P_d\subset Q_d$. The
triangle inequality gives $d(x,y)\le c_b(y-x)$. The only case not yet
settled is a bounded slow interval of length zero; \cref{shape:crossloss}
below excludes it.
\end{proof}

\begin{lemma}[Loss across two rays]\label{shape:crossloss}
Suppose there are two exterior unit rays with slow interval $[a,b]$
and put $L=b-a$. Then $L>0$, and, with the fixed constant
\begin{equation}\label{shape:cross-constant}
 C_*=2+\frac6\delta,
\end{equation}
every $z\in\R$ and $t>0$ satisfy
\begin{equation}\label{shape:cross-bound}
 0\le 1-\frac{d(z,z+t)}t\le\frac{C_*L}{t}.
\end{equation}
The shortfall is zero whenever the interval is wholly in either closed
unit ray.
\end{lemma}

\begin{proof}
For the moment permit $L=0$. Write $x_u=a-u$, $y_v=b+v$, where
$u,v\ge0$, and put $D=d(a,b)\le L$. Set
\[
 s=\frac{u+v-D-d(x_u,y_v)}2.
\]
If $s\le0$, the absolute shortfall of $[x_u,y_v]$ is at most
$L+D\le2L$. Suppose $s>0$. The reverse triangle inequalities imply
$d(x_u,y_v)\ge u-D-v$ and $d(x_u,y_v)\ge v-D-u$, so that
$s\le\min\{u,v\}$. At base $a$ the unit rays and triangle inequalities
give
\begin{align*}
 B_a(x_u,x_s)&=2s,\\
 B_a(x_u,y_v)&\ge u+(v-D)-d(x_u,y_v)=2s,\\
 B_a(y_v,y_s)&\ge(v-D)+(s-D)-(v-s)=2s-2D.
\end{align*}
Twice applying \cref{shape:tree-inequality} yields
$B_a(x_s,y_s)\ge2s-2D$. As $d(a,x_s)=s$ and $d(a,y_s)\le D+s$, it
follows that
\begin{equation}\label{shape:matched-ray-distance}
 d(x_s,y_s)\le3D\le3L.
\end{equation}

Use the temporal midpoint $m=(a+b)/2$ and spacing $t_*=s+L/2$. Its two
wing distances obey $d(x_s,m),d(m,y_s)\ge s-L/2$. If $L>0$ and
$s>3L/\delta$, these bounds are strictly greater than
$(1-\delta)t_*$ and \cref{shape:matched-ray-distance} is strictly
less than $\delta t_*$. This is the forbidden triple of
\cref{shape:forbidden-triple}. If $L=0$, every $s>0$ produces that
same strict contradiction. Thus always $s\le3L/\delta$. A full
crossing interval has absolute shortfall
\[
 (u+L+v)-d(x_u,y_v)=L+D+2s\le C_*L.
\]
An interval meeting only part of the slow interval has absolute
shortfall at most twice the length of that part. For example, if
$x<a\le y\le b$, then $d(x,y)\ge(a-x)-d(a,y)\ge(a-x)-(y-a)$, giving
loss at most $2(y-a)$. The other orientation is identical. An
interval inside $[a,b]$ has loss at most its length, hence at most
$L$, and intervals in the unit rays have zero loss. This proves
\cref{shape:cross-bound} in every case. If $L=0$, these conclusions
would give exact unit distance between every pair of times, an affine
array. Nonaffinity therefore forces $L>0$.
\end{proof}

\section{A capped deficit and the final balance}
\label{sec:end}

The cap will compare the slow interval of a limiting shape at fine and
coarse scales. It must equal one on uniformly slow secants. For a
two-ray shape, \cref{shape:cross-bound} makes the shortfall of a long
crossing interval small; a sufficiently small slope of the cap near
zero will make its averaged coarse contribution smaller than the
fine-scale occupancy of the slow interval. This is an integrated
comparison, not a pointwise sign assertion for the observable below.

Choose a fixed $\alpha>0$ so small that $10C_*\alpha<1$. Choose a smooth
nondecreasing function $\Phi:[0,1]\to[0,1]$ such that
\begin{equation}\label{end:cap}
 \Phi(u)=\alpha u\ \text{near }0,\qquad
 \Phi(u)=1\quad(1-c_b\le u\le1).
\end{equation}
For completeness, one can choose a nonnegative smooth derivative equal
to $\alpha$ near zero, zero near $1-c_b$, and with integral one on
$[0,1-c_b]$, by adding a nonnegative bump supported strictly between
the two endpoint neighborhoods. Integrating and continuing constantly
gives \cref{end:cap}. Positivity near zero and compactness give
constants $0<c_\Phi\le C_\Phi<\infty$ with
\begin{equation}\label{end:linear-bounds}
 c_\Phi u\le\Phi(u)\le C_\Phi u\qquad(0\le u\le1).
\end{equation}
Extend $\Phi$ smoothly to the real line, keeping it linear immediately
to the left of zero. The extension only defines continuous tests on
$X$; the arrays used below have the relevant shortfalls in $[0,1]$.

Write $h_t(z)=1-d(z,z+t)/t$, and abbreviate $J=J([0,2])$ and
$V=V([0,2])$. Define
\begin{equation}\label{end:G-definition}
 G(d)=\Phi(h_2(0))-\tfrac12\Phi(h_1(0))-\tfrac12\Phi(h_1(1)).
\end{equation}
If $a=h_1(0)$, $b=h_1(1)$, and $\bar a=(a+b)/2$, then
$h_2(0)=\bar a+J$ and $V=(a-b)^2$. The mean value theorem followed by
the second-order symmetric Taylor bound yields
\begin{equation}\label{end:G-bound}
 |G|\le\|\Phi'\|J+\frac18\|\Phi''\|V\le C(J+V),
\end{equation}
where the derivative norms can be taken on the compact argument range
of these three tests on $X$. In particular $G$ is $\mu$-integrable by
\cref{scale:limiting-test-bound}.

At the terminal scale, $\int h_1(0)\,\dd\nu_K=1$. After a subsequence,
\cref{end:linear-bounds} therefore gives a limit
\begin{equation}\label{end:H-definition}
 H_\Phi=\lim_K\int\Phi(h_1(0))\,\dd\nu_K,
 \qquad c_\Phi\le H_\Phi\le C_\Phi.
\end{equation}

There are three steps to the contradiction. First, dyadic telescoping
gives $\int G\,\dd\mu=H_\Phi$, with no loss of weighted mass to
$\mathcal A$. Then the positive decomposition of
\cref{scale:decomposition} separates this integral into a terminal
contribution and the contribution of the invariant remainder. We prove
that the former is strictly less than $H_\Phi$ and the latter is
nonpositive. The cap's two prescribed behaviors enter precisely in
these comparisons.

\begin{proposition}[Exact limiting balance]\label{end:balance}
With the preceding choices,
\begin{equation}\label{end:balance-equation}
 \int G\,\dd\mu=H_\Phi>0.
\end{equation}
\end{proposition}

\begin{proof}
Stationarity makes the two half-interval terms in
\cref{end:G-definition} have equal expectation. Since
$h_2(0;g_n)=h_1(0;g_{2n})$, summing over the dyadic scales gives exactly
\[
 \int G\,\dd\mu_K
 =\int\Phi(h_1(0))\,\dd\nu_K
  -\frac1\epsilon\E\Phi(h_1(0;g_{n_0})).
\]
The final term tends to zero by
\cref{scale:initial-test-bound,end:linear-bounds}. Thus these integrals
converge to $H_\Phi$. We must show that this mass does not escape to
$\mathcal A$ under vague convergence.

The measures $J\mu_K$ and $V\mu_K$ have uniformly bounded total mass on
the compact space $X$, and the signed measures $G\mu_K$ have uniformly
bounded total variation by \cref{end:G-bound}. Take simultaneous weak
limits
\[
 J\mu_K\longrightarrow\lambda_J,\qquad
 V\mu_K\longrightarrow\lambda_V,\qquad
 G\mu_K\longrightarrow\eta.
\]
On $\mathcal U$ they agree with $J\mu$, $V\mu$, and $G\mu$
respectively: multiply any compactly supported continuous test there
by the indicated continuous weight. We show that
$\eta|_{\mathcal A}=0$ by treating all three ranges of affine speeds.
Write $d_w(s,t)=w|s-t|$.

\paragraph{Speeds $0\le w\le c_b$.}
Both $\Phi'$ and $\Phi''$ vanish at every shortfall $1-w$ in this
compact speed range, including its boundary $w=c_b$. For each $\rho>0$
there is an open neighborhood $U_\rho$ of
$\{d_w:0\le w\le c_b\}$ on which the Taylor argument for
\cref{end:G-bound} gives
\[
 |G|\le\rho(J+V).
\]
All arguments between the three relevant shortfalls are then uniformly
close to $[1-c_b,1]$, where both derivatives vanish. For a continuous
test supported in $U_\rho$, pass this domination to the weak limits. It
implies $|\eta||_{U_\rho}\le\rho(\lambda_J+\lambda_V)|_{U_\rho}$. The
latter positive measure is finite on $X$. Letting $\rho$ decrease to
zero shows that $\eta$ has zero restriction to this compact set of
affine arrays.

\paragraph{Speeds $c_b<w<1$.}
Choose a fixed positive dyadic $R$ so large that the endpoint windows of
the bridge $[0,R]$ contain $[0,2]$ and $[R,R+1]$ respectively; for
example require $\kappa R>2$. Put $W=J+V$ and $H=h([R,R+1])$. For each
positive rational $\eta_0$, the open set
\[
 U_{\eta_0}=\{d:d(0,R)>c_bR,\ h([R,R+1])>\eta_0\}
\]
has $W\mu_K$-mass tending to zero by \cref{shape:weighted-bridge}. The
open-set inequality for weak convergence of positive measures gives
$(\lambda_J+\lambda_V)(U_{\eta_0})=0$. Each $d_w$ in the current speed
range belongs to such a set by taking $0<\eta_0<1-w$. This countable
cover and \cref{end:G-bound} show that $\eta$ vanishes on all
intermediate affine speeds. The threshold is fixed separately for each
test; no threshold uniform as $w\uparrow1$ is needed.

\paragraph{The speed $w=1$.}
Near $d_1$, all three arguments in \cref{end:G-definition} are in the
linear part of $\Phi$, so $G=\alpha J$. It remains to prove
$\lambda_J(\{d_1\})=0$. Fix an inverse power of two $q\in(0,1)$.
For the product over $[0,2n]$, with center $n$, put
\[
 D=n-\min\{g(0,n),g(n,2n)\},\qquad
 \ell=g(0,n)+g(n,2n)-g(0,2n)=2nJ(g_n).
\]
The quantities $D/n$ and $\ell/n$ are continuous tests of $g_n$, both
zero at $d_1$. Thus the condition $2q-2D/n>\ell/n$ defines an open
neighborhood $U_q$ of $d_1$. Eventually $s=qn$ is an integer at every
scale in the sum. On $U_q$, the truncation estimate of
\cref{scale:truncation} gives, with its fixed additive constant $A$,
\[
 \ell\le\ell_{\rm cut}+A,
 \qquad
 J\le qJ([1-q,1+q])+\frac{A}{2n}.
\]
The shorter-interval test telescopes exactly, by stationarity:
\[
 \int J([1-q,1+q])\,\dd\mu_K
 =\frac{e_{qN}-e_{qn_0}}\epsilon\le1,
\]
where dyadic monotonicity gives $e_{qN}\le e_N=\epsilon$. Also
$\epsilon^{-1}\sum_n n^{-1}=O((\epsilon n_0)^{-1})=o(1)$. Since $J\ge0$,
\[
 \int_{U_q}J\,\dd\mu_K\le q+o(1).
\]
The open-set inequality gives $\lambda_J(U_q)\le q$, so
$\lambda_J(\{d_1\})\le q$. First take the fixed weak limit and then let
$q$ decrease through inverse powers of two to obtain zero. The local
identity $G=\alpha J$ transfers this conclusion to $\eta$.

These cases cover $\mathcal A$. Weak convergence on compact $X$
preserves the total mass of the signed measures, so $\eta(X)=H_\Phi$.
As $\eta|_{\mathcal A}=0$ and $\eta|_{\mathcal U}=G\mu$, this is
\cref{end:balance-equation}.
\end{proof}

\subsection*{The terminal measure and its backward dilates}

The decomposition in \cref{scale:decomposition}, together with
\cref{end:G-bound}, gives an absolutely convergent identity
\begin{equation}\label{end:integrated-decomposition}
 \int G\,\dd\mu
 =\int G\,\dd\mu_\infty
  +\sum_{j=1}^{\infty}\int G(S^{-j}d)\,\dd\nu(d).
\end{equation}
The sum of the integrals of $|G(S^{-j}d)|$ is at most $\int|G|\,\dd\mu$,
so all subsequent rearrangements of these component integrals are
justified.

\begin{lemma}[Strict bound for the terminal contribution]
\label{end:terminal}
The series in \cref{end:integrated-decomposition} satisfies
\begin{equation}\label{end:terminal-strict}
 \sum_{j=1}^{\infty}\int G(S^{-j}d)\,\dd\nu(d)<H_\Phi.
\end{equation}
\end{lemma}

\begin{proof}
Nonnegative compact cutoffs exhausting $\mathcal U$, the terminal
prelimit identities, and monotone convergence imply
\begin{equation}\label{end:terminal-bounds}
 \int h_1(0)\,\dd\nu\le1,
 \qquad \int\Phi(h_1(0))\,\dd\nu\le H_\Phi.
\end{equation}
For each integer $m\ge0$, subdivision and the triangle inequality give
\[
 h_1(0)\ge2^{-m}\sum_{r=0}^{2^m-1}h_{2^{-m}}(r2^{-m}).
\]
Dyadic translation invariance of $\nu$ therefore implies
$\int h_{2^{-m}}(0)\,\dd\nu\le1$. The same shortfalls inserted into
$\Phi$ have finite integrals, bounded by $C_\Phi$.

Set $A_m=\int\Phi(h_{2^{-m}}(0))\,\dd\nu$. Translation invariance at
the shift $2^{-j}$ gives
\[
 \int G(S^{-j}d)\,\dd\nu=A_{j-1}-A_j.
\]
Absolute convergence of the series implies convergence of $A_m$.
Fatou's lemma yields
\begin{align}
 \sum_{j=1}^{\infty}\int G(S^{-j}d)\,\dd\nu
 &\le A_0-
    \int\liminf_{m\to\infty}\Phi(h_{2^{-m}}(0))\,\dd\nu
 \nonumber\\
 &\le H_\Phi-
    \int\liminf_{m\to\infty}\Phi(h_{2^{-m}}(0))\,\dd\nu.
 \label{end:terminal-fatou}
\end{align}
If $\nu=0$, the series is zero and \cref{end:H-definition} proves the
strict conclusion. Otherwise every shape in its full-measure support has
a slow interval with nonempty interior by
\cref{shape:classification,shape:crossloss}. There is a dyadic $z$ in
that interior. For all sufficiently large $m$, the interval
$[z,z+2^{-m}]$ is slow, so $\Phi(h_{2^{-m}}(z))=1$ by \cref{end:cap}.

The measurable sets
\[
 E_z=\{d:\Phi(h_{2^{-m}}(z))=1\text{ for every sufficiently large }m\},
 \qquad z\in\mathcal D,
\]
cover this full-measure support. At least one has positive $\nu$-measure,
and translation invariance gives $\nu(E_0)=\nu(E_z)>0$. Consequently
the subtracted integral in \cref{end:terminal-fatou} is positive. It is
finite by Fatou's lemma and the bound $A_m\le C_\Phi$. This proves
\cref{end:terminal-strict} in all cases.
\end{proof}

\subsection*{The translation- and dilation-invariant remainder}

We next give the measurable and local-finiteness details needed to
integrate over shapes. A dyadic interval whose endpoint distance equals
its length has exact unit speed throughout, by the triangle inequality
and the Lipschitz bound. Thus the local unit-speed set is the union of
the interiors of those countably many intervals. More directly, define
extended-real endpoint functions on $X_0$ by
\begin{align}
 a(d)&=\sup\{r\in\mathcal D:
       d(p,q)=q-p\text{ for all }p,q\in\mathcal D, p<q<r\},
 \label{end:left-endpoint}\\
 b(d)&=\inf\{r\in\mathcal D:
       d(p,q)=q-p\text{ for all }p,q\in\mathcal D, r<p<q\}.
 \label{end:right-endpoint}
\end{align}
Use $\sup\varnothing=-\infty$ and $\inf\varnothing=+\infty$. Each ray
condition is a countable intersection of closed secant conditions, so
$a,b$ are measurable. On our nonaffine shapes a finite $a$ is exactly
the right endpoint of the exterior left unit ray, and a finite $b$ is
exactly the left endpoint of the exterior right unit ray. They obey,
for $r\in\mathcal D$,
\begin{equation}\label{end:endpoint-covariance}
 a(\tau_r d)=a(d)-r,\quad b(\tau_r d)=b(d)-r,
 \qquad a(Sd)=a(d)/2,\quad b(Sd)=b(d)/2.
\end{equation}
Thus the three shape classes and the length $L=b-a$ on the two-ray
class are measurable and transform as asserted.

\begin{lemma}[Nonpositive invariant contribution]\label{end:remainder}
The invariant remainder satisfies
\begin{equation}\label{end:remainder-bound}
 \int G\,\dd\mu_\infty\le0.
\end{equation}
\end{lemma}

\begin{proof}
On an entirely slow shape, each shortfall in \cref{end:G-definition} is
at least $1-c_b$, so all three capped values equal one and $G=0$.

Consider next shapes with exactly one unit ray. First take the
orientation with a left unit ray and finite interface $a$. For a
bounded interval $[A,B]$ of possible interface locations, choose one
fixed dyadic secant strictly to the left of $A$ and a second strictly
to the right of $B$. Their speed ratios on every such shape are
respectively $1$ and at most $c_b$. The arrays with these two
prescribed closed conditions form a compact subset of $X_0$ disjoint
from $\mathcal A$: an affine array would have the same speed on both
secants. Local finiteness of $\mu_\infty$ on $\mathcal U$ therefore
implies finite mass for interfaces in $[A,B]$.

Push the restriction of $\mu_\infty$ to this shape class forward by
$a$, obtaining a measure $\lambda$ on $\R$ finite on bounded intervals.
Its invariances, from \cref{end:endpoint-covariance}, give
\[
 \lambda([0,1))=\lambda([0,2))
 =\lambda([0,1))+\lambda([1,2))=2\lambda([0,1)).
\]
The first equality uses $S$-invariance; the last uses translation by
one. Finiteness forces $\lambda([0,1))=0$, and integer translates
covering $\R$ show that this entire shape class has zero mass. The same
argument with left and right reversed treats the other one-ray class.
Half-open intervals make the argument valid without any advance
assumption about endpoint atoms.

It remains to consider the two-ray class. Restrict to the length band
\[
 \mu_*:=\mu_\infty\big|_{\{\text{two rays},\ 1\le L<2\}}.
\]
This measure is invariant under dyadic translations. Its left endpoint
has finite intensity on bounded intervals, as follows explicitly.
Partition any bounded range of possible $a$ into finitely many bins
$[k/4,(k+1)/4]$. In each bin the fixed secant
$[(k+1)/4,(k+2)/4]$ lies in $[a,b]$, since $L\ge1$, and hence has ratio
at most $c_b$. A fixed dyadic secant to the left of the whole bounded
range has ratio one. The two conditions again put each bin's arrays
inside a compact subset of $\mathcal U$. A finite union gives the
claimed finite intensity. The right endpoint also has finite intensity
on bounded intervals because $1\le b-a<2$ bounds $a$ whenever it
bounds $b$.

Neither endpoint intensity has an atom. If the mass at some point were
positive, dyadic translation invariance would give the same positive
mass to infinitely many distinct points in a bounded interval,
contradicting local finiteness. Define the finite number
\[
 I_0=\mu_*\{0\le a<1\}.
\]
Every integer interval of length one has this same endpoint mass.

For $j\in\Z$, let $B_j$ be the two-ray band $2^j\le L<2^{j+1}$. Since
$S$ divides lengths by two and preserves $\mu_\infty$,
\[
 \mu_\infty|_{B_j}=(S^{-j})_*\mu_*.
\]
The bands are disjoint and cover the two-ray class. Integrability of
$|G|$ permits reindexing $j$ to $-j$, yielding
\begin{equation}\label{end:band-series}
 \int_{\rm two\ rays}G\,\dd\mu_\infty
 =\sum_{j\in\Z}\int G(S^j d)\,\dd\mu_*(d),
 \qquad
 \sum_{j\in\Z}\int|G(S^j d)|\,\dd\mu_*<\infty.
\end{equation}

Put $F_j=\int\Phi(h_{2^j}(0))\,\dd\mu_*$. These quantities are finite.
\Cref{shape:crossloss} makes the integrand zero unless $[0,2^j]$ meets
$[a,b]$, which requires $-2\le a\le2^j$; the endpoint intensity on
this bounded interval is finite. Translation invariance at the dyadic
shift $2^j$ gives
\begin{equation}\label{end:F-difference}
 \int G(S^j d)\,\dd\mu_*=F_{j+1}-F_j.
\end{equation}
The absolute convergence in \cref{end:band-series} implies that $F_j$
has finite limits both as $j\to-\infty$ and as $j\to+\infty$.

For the fine-scale limit, use the dominating integrable function
$\one_{\{-2\le a\le1\}}$ for $j\le0$. Off the null sets $\{a=0\}$ and
$\{b=0\}$, if $a<0<b$, then $[0,2^j]$ is contained in the slow
interval for all sufficiently negative $j$, and the capped shortfall
equals one. If zero lies outside the slow interval, the small interval
is eventually in a unit ray and its shortfall is zero. Dominated
convergence therefore gives
\begin{equation}\label{end:fine-limit}
 F_{\rm fine}:=\lim_{j\to-\infty}F_j
 =\mu_*\{a<0<b\}\ge\mu_*\{-1\le a<0\}=I_0.
\end{equation}
The inequality uses $L\ge1$; possible equality endpoints have zero mass.

For the coarse-scale limit, \cref{shape:cross-bound} gives
$0\le h_t(0)\le2C_*/t$ throughout this band. Once $t$ is sufficiently
large, all these shortfalls lie in the linear region of \cref{end:cap},
uniformly in the shape. Hence
\[
 \int\Phi(h_t(0))\,\dd\mu_*
 \le \frac{2\alpha C_*}{t}\,
                  \mu_*\{-2\le a\le t\}.
\]
For the large dyadic integers $t=2^j$, atomlessness and integer
translation invariance make the last mass $(t+2)I_0$ (the weaker bound
$(t+3)I_0$ would also suffice). Thus
\begin{equation}\label{end:coarse-limit}
 F_{\rm coarse}:=\lim_{j\to+\infty}F_j
 \le2\alpha C_*I_0.
\end{equation}
Summing \cref{end:F-difference} and applying
\cref{end:fine-limit,end:coarse-limit} now proves
\[
 \int_{\rm two\ rays}G\,\dd\mu_\infty
 =F_{\rm coarse}-F_{\rm fine}
 \le-(1-2\alpha C_*)I_0\le0.
\]
This argument also covers $I_0=0$; in that case integer translates of
$\{0\le a<1\}$ show that $\mu_*=0$. Together with the entirely slow and
one-ray cases, this establishes \cref{end:remainder-bound}.
\end{proof}

\begin{theorem}[Contradiction and eventual positive entropy]
\label{end:contradiction}
There is no sequence $k\to+\infty$ along which the largest Lyapunov
exponent of $f_k$ is zero almost everywhere. Consequently
\cref{main:entropy} holds.
\end{theorem}

\begin{proof}
Under the supposed zero-exponent sequence, all the preceding measures
and identities were constructed. Combining
\cref{end:integrated-decomposition,end:terminal-strict,end:remainder-bound}
gives
\[
 \int G\,\dd\mu<H_\Phi,
\]
contradicting the exact balance \cref{end:balance-equation}.

For every parameter, the top Lyapunov exponent exists almost everywhere
and is nonnegative, since the derivative cocycle has determinant one.
If positivity on a set of positive area failed at arbitrarily large
positive parameters, the exponent would therefore be zero almost
everywhere at each member of a sequence tending to $+\infty$, which has
just been excluded. There is consequently a finite $k_0$ such that, for
every $k>k_0$, the top exponent is positive on a set of positive
normalized area. Increasing the threshold gives the same statement for
every $k\ge k_0$. The entropy formula \cref{main:pesin} gives
$h_m(f_k)>0$ for every such $k$. In particular the measurable parameter
interval $[k_0+1,k_0+2]$ has positive one-dimensional Lebesgue measure
and satisfies the required assertion.
\end{proof}

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\end{document}