64c3814
==== TABLE OF CONTENTS ====
A-finite-entropy-system-without-a-smooth-positive-volume-model//01-introduction.tex
A-finite-entropy-system-without-a-smooth-positive-volume-model//02-stationary.tex
A-finite-entropy-system-without-a-smooth-positive-volume-model//03-linear.tex
A-finite-entropy-system-without-a-smooth-positive-volume-model//04-switching.tex
A-finite-entropy-system-without-a-smooth-positive-volume-model//05-words.tex
A-finite-entropy-system-without-a-smooth-positive-volume-model//06-process.tex
A-finite-entropy-system-without-a-smooth-positive-volume-model//07-obstruction.tex
A-finite-entropy-system-without-a-smooth-positive-volume-model//main.tex
A-finite-entropy-system-without-a-smooth-positive-volume-model//preamble.tex
A-finite-entropy-system-without-a-smooth-positive-volume-model//references.bib
===========================
==== BEGIN A-finite-entropy-system-without-a-smooth-positive-volume-model//01-introduction.tex ====
\section{Introduction}\label{sec:introduction}
The smooth realization problem asks which measure-preserving systems can
be represented by smooth dynamics on a compact manifold.
Here a \emph{smooth positive-volume model} of an invertible
probability-preserving transformation $(X,\mu,T)$ consists of a compact
smooth manifold $M$ of finite dimension, a smooth positive probability
volume $\nu$, and a $C^\infty$ diffeomorphism $S$ preserving $\nu$,
together with a measure-preserving isomorphism
\[
F:X_0\longrightarrow M_0,\qquad F\circ T=S\circ F,
\]
on invariant conull measurable sets. Both $F$ and $F^{-1}$ are measurable;
no continuity is required. Here smooth positive volume means a probability
measure with strictly positive smooth local densities,
$d\nu=f(x)|dx_1\cdots dx_d|$, where $f>0$ is smooth.
The manifold may have a smooth boundary, with these densities smooth up
to the boundary. No orientation of $M$ or orientation preservation by
$S$ is required. The manifold and its dimension may depend on the
original system.
A standard nonatomic probability space means a probability space
isomorphic, modulo null sets, to the unit interval with Lebesgue
probability.
Finite entropy is a necessary restriction in the smooth realization
problem. For example, Ruelle's differentiable entropy inequality
\cite[Theorem~2]{Ruelle1978} gives finite entropy from compact derivative
bounds. The question addressed here is
whether this entropy restriction suffices when the invariant measure
must be a smooth positive density.
For invertible ergodic transformations of standard nonatomic
probability spaces, two positive realization results help locate this
restriction.
Lind and Thouvenot realized finite-entropy ergodic transformations by
Lebesgue-preserving homeomorphisms of the two-torus
\cite{LindThouvenot1977}. They also proved that a hyperbolic
two-torus automorphism represents every such system with measure entropy
below the automorphism's topological entropy, using a suitable invariant
Borel probability of full support; Quas and Soo recall this earlier result in
\cite[\S1]{QuasSoo2016}. Quas and Soo's Theorem~2 extends the result to
toral automorphisms with no root-of-unity eigenvalue.
Choosing a sufficiently high iterate of a hyperbolic automorphism
therefore gives a smooth model for any prescribed finite-entropy
system, with an invariant Borel probability of full support.
Full support does not require a smooth positive density.
The requirement on the invariant measure is essential. A realization
with an arbitrary invariant Borel probability, which may be singular,
does not by itself give a smooth positive-volume model. This distinction is part of the
classical formulation discussed by Katok and Thouvenot
\cite[pp.~324--326]{KatokThouvenot1997}. They developed slow entropy
invariants and constructed ergodic $\Z^2$ actions having no Lipschitz
realization on a compact space of finite box dimension, even when the
model may carry an arbitrary invariant Borel probability
\cite[Proposition~4 and Corollary~2]{KatokThouvenot1997}.
The commuting generators must be realized together in that problem;
its obstruction does not by itself answer the question for one
transformation. Foreman and Weiss
\cite[p.~2607]{ForemanWeiss2022} also discuss finite-entropy realization,
in a formulation allowing an invariant measure merely equivalent to
volume. Our conclusion concerns the smooth positive-density version;
we do not assert an obstruction for every equivalent-to-volume
invariant measure. Their anti-classification theorems concern the complexity of the
measure-isomorphism relation among smooth transformations, including
an ergodic area-preserving torus subclass
\cite[Theorems~1--2 and Corollary~4]{ForemanWeiss2022}.
They do not assert that a particular abstract finite-entropy system
lacks such a model. Subsequent anti-classification results treat
weakly mixing smooth transformations \cite[Theorem~B]{Kunde2024} and mixing
zero-entropy smooth transformations \cite[Theorem~2]{GerberKunde2025Mixing};
these too study equivalence inside families of smooth models.
We give a negative answer to the smooth-volume realization question,
already among systems of zero entropy.
For a finite probability vector $p$, its Shannon entropy is
\[
H(p)=-\sum_a p(a)\log_2 p(a),\qquad 0\log_2 0=0.
\]
The entropy $H_\mu(\mathcal P)$ of a finite measurable partition is
the entropy of its atom probabilities; the entropy of a finite random
variable is likewise the entropy of its law.
The Kolmogorov--Sinai entropy of $T$ is
\[
h_\mu(T)=\sup_{\mathcal P}\lim_{n\to\infty}
\frac1n H_\mu\!\left(\bigvee_{j=0}^{n-1}T^{-j}\mathcal P\right),
\]
where $\mathcal P$ ranges over finite measurable partitions.
\begin{theorem}\label{thm:main}
There exists an ergodic invertible measure-preserving transformation
$(X,\mu,T)$ of a standard nonatomic probability space such that
$h_\mu(T)=0$ and $(X,\mu,T)$ is not measurably conjugate, on invariant conull sets, to
any $C^\infty$ diffeomorphism of a compact finite-dimensional manifold
preserving a strictly positive smooth probability density. The manifold
may have smooth boundary and need not be orientable.
\end{theorem}
The proof isolates a property of smooth positive volume that remains
useful after an arbitrary measurable change of coordinates.
At a dyadic time $n=2^k$, partition the manifold into small cells,
choose a cell with its volume as probability, and choose two independent
points $x,y$ from its normalized volume. On most such pairs, one can
find a point $z$ whose future shadows that of $x$ and whose past shadows
that of $y$. More than the existence of this point is needed:
the law of $z$ on the retained pairs is bounded by twice the original
volume. The number of cells is at most $\exp(CB_kn)$ for a sequence
with $\sum_k B_k^{16}<\infty$.
The precise statement is \Cref{thm:switching}.
This switch estimate is the main analytic ingredient. Its proof begins
with a square-summable dyadic estimate for stationary diagonal
derivative logs, rather than an assumed rate in an ergodic theorem.
Finite-interval triangular changes of basis give the linear
estimates needed to solve a boundary problem joining the two orbit
halves. A relative exterior-power estimate controls the change in
small endpoint minors, and a determinant identity and global
injectivity control the law of the switch.
The quantitative summability and the bound on the output measure are
the features that will obstruct realization.
We construct the opposing symbolic system from finite sets
$\mathcal W_i$ of words of lengths $m_i$. Each word at the next level is
a concatenation of words at the preceding level, and the system
records their compatible alignment phases. Word labels are nonzero
vectors over $\F_2$. Certain pairs of labels separated by a power-of-two
number of blocks must be orthogonal for a fixed alternating form.
These constraints coexist with high entropy of short label lists and
with a positive symbol-distance separation between distinct or
misaligned words.
The entropy of the label lists makes two independent lists unlikely
to satisfy most of the orthogonality tests. A smooth switch would
force their phases and most of their labels to agree with the
corresponding past and future lists of a single legal symbolic point.
Those lists do satisfy the tests. Summability of the smooth partition
cost, together with the growth of the word scales, supplies times at
which conditioning on a cell loses too little entropy to reconcile
the two conclusions.
The order of choices is important: the symbolic system is fixed first.
Only then is a hypothetical smooth model, with any one finite
dimension, considered. The scale argument absorbs all constants
depending on that model. Thus the construction does not select a
different counterexample for each dimension.
\paragraph{Organization.}
\Cref{sec:stationary,sec:linear,sec:switching} prove the switching
estimate. \Cref{sec:words} constructs the finite word families.
\Cref{sec:process} supplies an ergodic probability on their phase
extension and establishes the window entropy and graph constraints.
\Cref{sec:obstruction} compares the two laws of one graph-passing
statistic and proves \Cref{thm:main}.
\paragraph{Conventions.}
Logarithms in entropies and combinatorial estimates are to base $2$;
natural logarithms are written $\ln$, and derivative growth is written
with $\exp$ or $e$. Total variation distance between finite laws $p,q$
is $\TV(p,q)=\tfrac12\sum_a|p(a)-q(a)|$.
For finite random variables, conditional entropy means
\[
H(Y\mid Z)=\sum_{z:\Pr(Z=z)>0}\Pr(Z=z)\,
H\bigl(\mathcal L(Y\mid Z=z)\bigr).
\]
Conditional partition entropy uses the same definition for atom labels.
We write $H_{\mathrm{bin}}(s)=-s\log_2 s-(1-s)\log_2(1-s)$
for binary entropy, with the same zero-mass convention at the endpoints.
Constants denoted by $C$ may change from line to line.
In the smooth part they depend only on the fixed smooth data,
dimension, and a fixed exceptional-mass tolerance, unless otherwise
specified. All asymptotic estimates there are uniform over the
points, bins, and finite-interval frames satisfying the displayed
hypotheses. The empty determinant and empty product equal one.
==== END A-finite-entropy-system-without-a-smooth-positive-volume-model//01-introduction.tex ====
==== BEGIN A-finite-entropy-system-without-a-smooth-positive-volume-model//02-stationary.tex ====
\section{Stationary estimates for derivative growth}\label{sec:stationary}
The switching construction requires control of derivatives on a long finite
orbit segment. The accuracy may depend on the length of the segment, but
the squared errors must be summable over dyadic lengths. We first obtain
this summability for an arbitrary bounded stationary process. Applying
the result to a triangular representation of the derivative then gives
large sets of orbit segments with the required control.
\subsection{Deviation from a dyadic chord}
Let $(Y_t)_{t\in\Z}$ be a real stationary process with
$\E Y_0^2<\infty$. Stationarity means that translating all indices in
any finite list does not change its joint distribution. For $h\geq 0$
define the block average and normalized maximal chord deviation by
\[
A_h=2^{-h}\sum_{t=0}^{2^h-1}Y_t,
\qquad
Z_h=2^{-h}\max_{0\leq u\leq 2^h}
\left|\sum_{t=0}^{u-1}Y_t-uA_h\right|.
\]
The maximum is over integer $u$, and an empty sum is zero.
In particular $Z_0=0$. The chord here uses the average on the
particular block under consideration; it is not a line whose slope is
the mean of the stationary process.
\begin{lemma}[Dyadic chord estimate]\label{lem:dyadic}
For the process above, put
\[
v_l=\E A_{l-1}^2-\E A_l^2,\qquad l\geq 1.
\]
Then $v_l\geq 0$, $\sum_{l\geq 1}v_l\leq\E Y_0^2$, and
\begin{equation}\label{eq:dyadic-convolution}
\|Z_h\|_{L^2}
\leq \frac12\sum_{a=0}^{h-1}2^{-a/2}\sqrt{v_{h-a}}
\qquad(h\geq 1).
\end{equation}
Consequently
\begin{equation}\label{eq:dyadic-square-sum}
\sum_{h\geq 1}\|Z_h\|_{L^2}^2
\leq C\,\E Y_0^2
\end{equation}
for an absolute constant $C$. The same estimates hold for a block
beginning at any deterministic integer time.
\end{lemma}
\begin{proof}
Consider a block of length $2^l$, and let $U$ and $V$ be the averages on
its left and right halves. Its own average is $W=(U+V)/2$. The identity
\[
\frac{(U-W)^2+(V-W)^2}{2}
=\frac{U^2+V^2}{2}-W^2
\]
and stationarity give
\[
v_l
=\E\frac{(U-W)^2+(V-W)^2}{2}
=\E(U-W)^2.
\]
The last equality also follows pointwise from $V-W=-(U-W)$.
Thus $v_l\geq0$, and telescoping the differences of second moments proves
the asserted bound on their sum. No independence of the two halves is
used.
We next compare the cumulative-sum polygon on $[0,2^h]$ with its chord.
For $a=0,\ldots,h$, let $F_a$ be the piecewise-linear interpolant that
agrees with the cumulative sum at the multiples of $2^{h-a}$.
Then $F_0$ is the chord and $F_h$ agrees with every integer-time
cumulative sum. Passing from $F_a$ to $F_{a+1}$ inserts the midpoint
of each of $2^a$ intervals of length $2^{h-a}$.
If $U_j,W_j$ are respectively the left-child and parent averages on
the $j$th such interval, the displacement at its midpoint is
$2^{h-a-1}(U_j-W_j)$.
The difference of the interpolants is linear between the old and new
vertices, so
\[
2^{-h}\|F_{a+1}-F_a\|_\infty
\leq 2^{-a-1}\max_{1\leq j\leq 2^a}|U_j-W_j|.
\]
Bounding a squared maximum by the sum of squares and using stationarity
of each interval gives
\[
\left\|2^{-h}\|F_{a+1}-F_a\|_\infty\right\|_{L^2}
\leq 2^{-a-1}\bigl(2^a v_{h-a}\bigr)^{1/2}.
\]
The triangle inequality, first for the polygonal sup norm and then in
$L^2$, proves \eqref{eq:dyadic-convolution}.
Extend the sequence $(\sqrt{v_l})_{l\geq1}$ by zero to nonpositive
indices. Its squared $\ell^2$ norm is at most $\E Y_0^2$.
The right side of \eqref{eq:dyadic-convolution} is its convolution
with the summable sequence $(2^{-a/2-1})_{a\geq0}$.
The triangle inequality for the $\ell^2$ norms of its translates
therefore proves \eqref{eq:dyadic-square-sum}.
Finally, stationarity identifies the distribution of each translated
block with that of the block just considered.
\end{proof}
\subsection{A stationary triangular derivative representation}
Fix a compact smooth manifold $M$ of dimension $d\geq1$, a smooth
Riemannian metric, a smooth diffeomorphism $S:M\to M$, and an
$S$-invariant probability measure $\nu$. In the switching argument
$\nu$ will have a smooth positive density. That extra assumption is
not needed in this section. The constructions also apply to a compact
manifold with smooth boundary.
The compact frame extension and triangular cocycle have a classical
precedent in Oseledets's proof of the multiplicative ergodic theorem
\cite[\S4, pp.~208--209, equations~(6)--(10)]{Oseledets1968}.
We use the full orthogonal frame bundle and a positive-diagonal upper
triangular convention. The finite-time estimates below follow from the
stationary chord lemma; we do not invoke a limiting Lyapunov splitting
or a convergence rate in the multiplicative ergodic theorem.
The orthonormal frame bundle $\mathcal F M$ consists of the linear
isometries $e:\R^d\to T_xM$, with bundle projection $\pi(e)=x$.
For each frame there is a unique positive-diagonal QR factorization
\begin{equation}\label{eq:frame-qr}
D S_x\circ e=\widetilde S(e)\,R(e),
\end{equation}
where $\widetilde S(e)$ is an orthonormal frame at $Sx$ and $R(e)$
is upper triangular with positive diagonal.
This is ordinary Gram--Schmidt in the indicated order, with the length
of each new perpendicular component chosen positive.
It depends continuously, indeed smoothly, on the frame and the base
point.
The map $\widetilde S$ is invertible. If
$DS_x e=e'R$, then
$(DS_x)^{-1}e'=D(S^{-1})_{Sx}e'=eR^{-1}$, where the inverse
linear map has domain $T_{Sx}M$ and codomain $T_xM$.
The matrix $R^{-1}$ is again upper triangular with
positive diagonal. Taking positive-diagonal QR after applying the
inverse derivative therefore recovers $e$. This also proves
continuity of the inverse. The construction uses the full orthonormal
frame bundle and does not require a global frame or an orientation.
There is a $\widetilde S$-invariant probability $\widetilde\nu$ whose
projection is $\nu$. To see this, first put the normalized
orthogonal-group Haar measure on every fiber over $\nu$.
Orthogonal changes of local trivialization preserve this fiber
measure, so the resulting probability is defined without choosing a
global section. Each of its iterates under $\widetilde S$ projects
to $\nu$. Cesaro averages have a weakly convergent subsequence,
because $\mathcal F M$ is compact. The difference between the
pushforward of an average and the average itself tends to zero
against every continuous function, so its limit is invariant.
Continuity of $\pi$ preserves the marginal $\nu$ in the limit.
Compactness and invertibility provide a constant $A_*>1$ such that
\begin{equation}\label{eq:qr-uniform-bounds}
\|R(e)\|,\ \|R(e)^{-1}\|\leq A_*
\qquad(e\in\mathcal F M).
\end{equation}
In particular $A_*^{-1}\leq R(e)_{jj}\leq A_*$.
The functions
\[
g_j(e)=\ln R(e)_{jj},\qquad 1\leq j\leq d,
\]
are continuous and bounded, and
$(g_j(\widetilde S^t e))_{t\in\Z}$ is stationary under
$\widetilde\nu$.
\begin{proposition}[Good finite orbit segments]\label{prop:good-frames}
For each $\eta>0$ there is a positive sequence $(b_k)_{k\geq1}$ with
\begin{equation}\label{eq:b-summability}
b_k\geq 2^{-k/10},
\qquad
\sum_{k\geq1}b_k^2<\infty,
\end{equation}
and compact sets $\widetilde{\mathcal G}_k\subset\mathcal F M$ and
$\mathcal G_k=\pi(\widetilde{\mathcal G}_k)\subset M$ such that
\[
\widetilde\nu(\widetilde{\mathcal G}_k)\geq1-\eta/100,
\qquad \nu(\mathcal G_k)\geq1-\eta/100.
\]
For $n=2^k$, $b=b_k$, and $e\in\widetilde{\mathcal G}_k$, define
\[
A_t=R(\widetilde S^t e)\quad(-n\leq t<n),
\qquad
\lambda_j=\frac1{2n}\sum_{t=-n}^{n-1}\ln(A_t)_{jj}.
\]
Then for every $j$ and every integer $-n\leq u\leq n$,
\begin{equation}\label{eq:good-chord}
\left|\sum_{t=-n}^{u-1}\ln(A_t)_{jj}
-(u+n)\lambda_j\right|\leq bn.
\end{equation}
Consequently, on every subinterval $-n\leq u\leq v\leq n$,
\begin{equation}\label{eq:good-subinterval}
\left|\sum_{t=u}^{v-1}\ln(A_t)_{jj}
-(v-u)\lambda_j\right|\leq2bn.
\end{equation}
\end{proposition}
\begin{proof}
Apply \Cref{lem:dyadic} to each $g_j$.
Let $c_{j,h}$ be the right side of
\eqref{eq:dyadic-convolution} for that process, and put
$\alpha_h^2=\sum_{j=1}^d c_{j,h}^2$.
The lemma gives $\sum_h\alpha_h^2<\infty$.
For example, take
\[
b_k=40\eta^{-1/2}\alpha_{k+1}+2^{-k/10}.
\]
This sequence has the properties in \eqref{eq:b-summability}.
The interval $[-n,n)$ has length $2^{k+1}$.
Stationarity and the squared Markov inequality give
\[
\widetilde\nu\!\left(
\max_{\substack{1\leq j\leq d\\-n\leq u\leq n}}
\left|\sum_{t=-n}^{u-1}g_j(\widetilde S^t e)
-(u+n)\lambda_j\right|>b_kn
\right)
\leq \frac{4\alpha_{k+1}^2}{b_k^2}
\leq\frac{\eta}{400}.
\]
Define $\widetilde{\mathcal G}_k$ by the complementary non-strict
inequalities. These are finitely many continuous inequalities on a
compact bundle, so this set and its projection are compact.
Moreover
\[
\pi^{-1}(M\setminus\mathcal G_k)
\subset\mathcal F M\setminus\widetilde{\mathcal G}_k.
\]
The pushforward identity $\pi_*\widetilde\nu=\nu$ gives the asserted
base-measure estimate. No measurable choice of a frame over each
point of $\mathcal G_k$ is needed.
Finally, subtracting the chord error at $u$ from the chord error at $v$
gives \eqref{eq:good-subinterval}, with the same rates $\lambda_j$.
\end{proof}
We call a frame in $\widetilde{\mathcal G}_k$ good at scale $k$,
and a base point in $\mathcal G_k$ good at that scale.
Every good base point has a good witness frame. The estimates below
hold for every such witness; witnesses at different points or scales
need not be chosen coherently. We do not require an individual point
to be good at all sufficiently large scales.
The stationary argument has now supplied the summable quantity that
will control the cost of switching. Fix $\eta$ and its sequence
$(b_k)$, and use the following parameters at scale $n=2^k$:
\begin{equation}\label{eq:smooth-scales}
\begin{aligned}
b&=b_k,& B&=b^{1/8},& g&=b^{1/2},\\
L&=\lceil b^{1/4}n\rceil,&
h_0&=e^{-Bn},&r&=e^{-2Bn},&\delta&=e^{-3Bn/2}.
\end{aligned}
\end{equation}
The parameter $b$ controls the diagonal chord error $bn$;
$g$ will be the separation margin for the finite-interval rates,
$L$ is the short comparison time, and $B$ sets the spatial scales
and the upper bound $CBn$ for the logarithm of the number of
partition cells. Here $h_0$ and $r$ will be spatial lengths; $\delta$ will be a
neighborhood radius for an orbit construction.
Since $b_k\to0$ and $bn\geq n^{9/10}$,
\begin{equation}\label{eq:smooth-hierarchy}
\ln n=o(bn),\quad bn=o(gL),\quad gL=o(gn),\quad
gn=o(L),\quad L=o(Bn).
\end{equation}
The ceiling in $L$ has no effect on these comparisons.
We also have
\[
\sum_k B_k^{16}=\sum_k b_k^2<\infty,
\qquad Bn\geq n^{79/80}.
\]
All subsequent asymptotic estimates are for $k\to\infty$.
Constants may depend on the fixed smooth data, dimension, and $\eta$,
but not on the point, its good witness, or the scale.
==== END A-finite-entropy-system-without-a-smooth-positive-volume-model//02-stationary.tex ====
==== BEGIN A-finite-entropy-system-without-a-smooth-positive-volume-model//03-linear.tex ====
\section{Linear estimates on finite orbit segments}\label{sec:linear}
We now turn the diagonal control of
\Cref{prop:good-frames} into estimates for the full derivative.
There are three steps. A finite-interval change of basis separates
groups of diagonal rates, short derivatives identify nearly the same
spaces at nearby base points, and a product-expansion estimate controls
small perturbations relative to the possibly small exterior volumes.
These statements will supply the linear data for switching orbit
segments.
\subsection{Triangular products and separation of rates}
For invertible matrices $A_t$ indexed by $-n\leq t<n$, write
\[
A(t,s)=A_{t-1}\cdots A_s\quad(t>s),\qquad
A(s,s)=\Id,\qquad A(s,t)=A(t,s)^{-1}.
\]
All vector norms and exterior-power norms in the next proposition are
Euclidean operator norms. If $E$ is a $q$-dimensional subspace and
$A$ is injective, let
\[
J_q(A|E)=\|(\mathord{\bigwedge}^q A)\xi\|,
\]
where $\xi$ is either unit simple $q$-vector spanning $E$.
This is the intrinsic $q$-dimensional volume expansion. We set
$J_0=1$, and the zeroth exterior power is the identity on $\R$.
\begin{proposition}[Finite-interval splitting]\label{prop:finite-splitting}
Fix $d\geq1$ and $A_*>1$. There is a constant $C$, depending only on
these two quantities, with the following properties.
Let $n\geq2$, $b>0$, and $\ln(2n+1)\leq bn$.
Suppose that $A_t$, $-n\leq t<n$, are upper triangular with positive
diagonal and
\[
\|A_t\|,\ \|A_t^{-1}\|\leq A_*,
\qquad
\left|\sum_{t=-n}^{u-1}\ln(A_t)_{jj}
-(u+n)\lambda_j\right|\leq bn
\]
for $-n\leq u\leq n$ and $1\leq j\leq d$, where
$\lambda_j=(2n)^{-1}\sum_{t=-n}^{n-1}\ln(A_t)_{jj}$.
First, order the rates as
$\lambda_{(1)}\geq\cdots\geq\lambda_{(d)}$.
For every $0\leq q\leq d$ and every $-n\leq s\leq t\leq n$,
\begin{equation}\label{eq:exterior-rate-comparison}
\left|\ln\|\mathord{\bigwedge}^q A(t,s)\|
-(t-s)\sum_{j=1}^q\lambda_{(j)}\right|\leq Cbn.
\end{equation}
The analogous estimate for $A(s,t)$ uses the $q$ largest members of
$\{-\lambda_1,\ldots,-\lambda_d\}$.
Next, suppose that real $\gamma$ and $g>0$ separate all the rates:
each $\lambda_j$ is at most $\gamma-g$ or at least $\gamma+g$.
Let $F$ be the set of indices in the second group, let
$p=|F|$, and put
\[
\Lambda=\sum_{j\in F}\lambda_j,\qquad
\Omega=-\sum_{j\notin F}\lambda_j.
\]
There are decompositions
$\R^d=E_t^s\oplus E_t^f$, $-n\leq t\leq n$, invariant under each
$A_t$, with $\dim E_t^f=p$, whose two projections have norm at most
$e^{Cbn}$. For $s\leq t$ and $l=t-s$ they satisfy
\begin{align}
\|A(t,s)|E_s^s\|&\leq e^{(\gamma-g)l+Cbn},
&
\|A(s,t)|E_t^f\|&\leq e^{-(\gamma+g)l+Cbn},
\label{eq:split-propagation}\\
\|\mathord{\bigwedge}^p A(t,s)\|&\leq e^{\Lambda l+Cbn},
&
e^{\Lambda l-Cbn}\leq
J_p(A(t,s)|E_s^f)&\leq e^{\Lambda l+Cbn},
\label{eq:fast-volume}\\
\|\mathord{\bigwedge}^{d-p} A(s,t)\|&\leq e^{\Omega l+Cbn},
&
e^{\Omega l-Cbn}\leq
J_{d-p}(A(s,t)|E_t^s)&\leq e^{\Omega l+Cbn}.
\label{eq:slow-backward-volume}
\end{align}
If $0<p<d$ and at least one of $v_1,\ldots,v_p$ belongs to $E_s^s$,
then
\begin{equation}\label{eq:mixed-exterior}
\|A(t,s)v_1\wedge\cdots\wedge A(t,s)v_p\|
\leq e^{(\Lambda-2g)l+Cbn}\prod_{j=1}^p\|v_j\|.
\end{equation}
The backward analogue has exterior degree $d-p$, exponent
$(\Omega-2g)l+Cbn$, and at least one factor in $E_t^f$.
When a group is empty its space is $\{0\}$, all statements on that
space are vacuous, and the exterior-degree-zero conventions apply.
\end{proposition}
\begin{proof}
\emph{Diagonal normalization.}
Write
\[
e_j(u)=\sum_{t=-n}^{u-1}\bigl(\ln(A_t)_{jj}-\lambda_j\bigr),
\qquad
D_u=\diag(e^{e_1(u)},\ldots,e^{e_d(u)}).
\]
The chord hypothesis gives
$\|D_u\|,\|D_u^{-1}\|\leq e^{bn}$.
The matrices
\[
\widehat A_t=D_{t+1}^{-1}A_tD_t
\]
are upper triangular with constant diagonal
$(e^{\lambda_1},\ldots,e^{\lambda_d})$ and entries bounded by
$e^{Cbn}$. Since the rates are averages of logarithms of diagonal
entries,
\begin{equation}\label{eq:bounded-rates}
|\lambda_j|\leq\ln A_*.
\end{equation}
Their inverse matrices also have entries bounded by $e^{Cbn}$.
We record how triangular products are estimated. If $m$ is fixed and
upper-triangular $m$-by-$m$ matrices have constant diagonal
$e^{\alpha_1},\ldots,e^{\alpha_m}$, bounded $|\alpha_i|$, and
off-diagonal entries at most $e^{Cbn}$, a term in any product entry
can change its coordinate index strictly at most $m-1$ times.
There are at most a fixed constant times $(2n+1)^{m-1}$ choices
for those times and the intermediate indices.
If the product length is $l$ and a term has $a\leq m-1$ strict
changes, its diagonal factors contribute at most
$e^{\alpha_{\max}(l-a)}$.
The omitted $a$ diagonal factors cost at most a fixed factor
$e^{a|\alpha_{\max}|}$ after $e^{\alpha_{\max}l}$ is factored out.
The off-diagonal factors contribute $e^{C_mbn}$.
The polynomial number of terms is absorbed by
$\ln(2n+1)\leq bn$. Thus the product norm is at most
$e^{\alpha_{\max}l+C_mbn}$.
Apply this argument to $\mathord{\bigwedge}^q\widehat A_t$ in the
coordinate-wedge basis, ordered lexicographically.
This matrix is upper triangular of fixed dimension $\binom dq$,
and its diagonal entries are
$\exp(\sum_{j\in I}\lambda_j)$ for $q$-element index sets $I$.
The largest such sum is $\sum_{j=1}^q\lambda_{(j)}$.
For the reverse inequality, the diagonal of the product indexed by
a maximizing set $I$ is exactly
$\exp(l\sum_{j\in I}\lambda_j)$; that matrix entry is no larger
than the operator norm. Converting back multiplies the exterior
norms only by the two endpoint gauge bounds.
This proves \eqref{eq:exterior-rate-comparison}.
Applying the same argument to inverse normalized steps in reverse
time proves its backward version. The cases $q=0,d$ follow as well:
the former is the identity, and the latter is the determinant line.
\emph{Removing entries between the groups.}
We next conjugate $\widehat A_t$ by time-dependent upper
unitriangular matrices to make the two coordinate index groups
invariant. Process the superdiagonals in increasing order of their
distance $j-i$ from the diagonal.
At a position $i<j$ belonging to different groups, suppose the
current entry is $h_t$. Use an elementary change
$Q_t=\Id+u_tE_{ij}$, where $E_{ij}$ has its only nonzero entry,
equal to one, in position $(i,j)$.
In $Q_{t+1}^{-1}\widehat A_tQ_t$ the new entry at this position is
\[
h_t+e^{\lambda_i}u_t-e^{\lambda_j}u_{t+1}.
\]
To set it to zero, solve
\begin{equation}\label{eq:shear-recurrence}
u_{t+1}=q u_t+f_t,\qquad
q=e^{\lambda_i-\lambda_j},\qquad f_t=e^{-\lambda_j}h_t.
\end{equation}
If $q<1$, prescribe $u_{-n}=0$ and solve forward:
\[
u_t=\sum_{a=-n}^{t-1}q^{\,t-1-a}f_a.
\]
If $q>1$, prescribe $u_n=0$ and solve backward:
\[
u_t=-\sum_{a=t}^{n-1}q^{\,t-1-a}f_a.
\]
In each formula the coefficients have modulus at most one, so
$|u_t|\leq2n\max_a|f_a|$.
Different groups cannot have $q=1$.
This bound avoids dividing by the possibly small separation of rates.
Right multiplication by $Q_t$ changes column $j$ by a multiple of
column $i$, and left multiplication by $Q_{t+1}^{-1}$ changes row
$i$ by a multiple of row $j$.
Upper triangularity shows that every affected off-diagonal position
has distance at least $j-i$; at that distance only $(i,j)$ changes.
The possible quadratic cross term is zero because
$E_{ij}\widehat A_tE_{ij}=(\widehat A_t)_{ji}E_{ij}=0$.
Thus a later operation cannot restore an entry on an earlier
superdiagonal.
There are at most $d(d-1)/2$ operations.
At each operation the bound above, the current entry bounds, and
\eqref{eq:bounded-rates} bound the shear and its inverse by
$e^{Cbn}$. Multiplying or adding a fixed number of such quantities
only changes $C$; the factor $2n$ is absorbed by $\ln(2n+1)\leq bn$.
Induction over this fixed number of operations therefore gives
matrices $C_t$ with
\begin{equation}\label{eq:finite-gauge}
\|C_t\|,\ \|C_t^{-1}\|\leq e^{Cbn},
\qquad
B_t=C_{t+1}^{-1}A_tC_t,
\end{equation}
where $B_t$ is upper triangular, has diagonal $e^{\lambda_j}$,
and has no entries between the two groups. Its entries and those of
$B_t^{-1}$ are bounded by $e^{Cbn}$.
All constants are independent of the length of the interval.
Let $\R^F$ be the coordinate span with indices in $F$ and
$\R^{F^c}$ its coordinate complement. Set
\[
E_t^f=C_t\R^F,\qquad E_t^s=C_t\R^{F^c}.
\]
These spaces are invariant under the original matrices.
Their projections are conjugates by $C_t$ of coordinate projections,
and hence have norm at most $e^{Cbn}$ after changing $C$.
The two groups may be interlaced in the original coordinate order.
Each induced block is nevertheless upper triangular in its inherited
order; a fixed permutation makes the two blocks consecutive if desired.
The triangular-product estimate within these blocks proves
\eqref{eq:split-propagation}.
\emph{Exterior volumes and mixed factors.}
The fast rates are exactly the $p$ largest rates. The general exterior
bound already proves the first inequality of \eqref{eq:fast-volume}.
The normalized fast block has determinant $e^{\Lambda l}$ on an
interval of length $l$.
To account explicitly for its possibly oblique basis, define
\[
V_F(t)=
\sqrt{\det\bigl((C_t|_{\R^F})^*(C_t|_{\R^F})\bigr)}.
\]
For $p=0$ this Gram determinant is one. The endpoint bounds in
\eqref{eq:finite-gauge} imply
$e^{-Cbn}\leq V_F(t)\leq e^{Cbn}$.
The intrinsic volume expansion is exactly
\[
J_p(A(t,s)|E_s^f)
=e^{\Lambda l}\frac{V_F(t)}{V_F(s)}.
\]
This proves both volume bounds in \eqref{eq:fast-volume}.
Applying the same argument to the inverse slow block gives
\eqref{eq:slow-backward-volume}, including either sign of $\Omega$.
The normalized block decomposition induces an invariant decomposition
of each exterior power according to the number of fast and slow
factors. In degree $p$, every sector with a slow factor has diagonal
rate sum at most $\Lambda-2g$: at least one fast rate has been
replaced by a slow rate.
The triangular-product estimate therefore bounds this entire sector
by $e^{(\Lambda-2g)l+Cbn}$.
If one original vector is slow, split each of the other vectors into
its fast and slow parts. There are only a fixed number of terms, each
still having a slow factor. The projection norms and endpoint gauges
are absorbed into $e^{Cbn}$, proving \eqref{eq:mixed-exterior}.
Reversing time and exchanging the groups gives the backward statement.
The empty-group cases require none of these mixed-factor assertions
and follow from the stated conventions.
\end{proof}
The estimates concern every subinterval of one good finite trajectory.
The change of basis is made once on that trajectory; only its values
at the two ends of a subinterval enter a product estimate.
In particular, a gauge loss is not multiplied once for each time step.
For a good frame in \Cref{prop:good-frames}, the matrices of the
derivative in the moving orthonormal frames satisfy the hypotheses
above for all sufficiently large $k$.
The proposition therefore gives tangent spaces
$E_t^s,E_t^f\subset T_{S^t x}M$ and the corresponding derivative
estimates. We use this notation also after expressing the spaces in
chart coordinates. Coordinate maps from a fixed uniformly regular
finite atlas conjugate every product only at its endpoints, so their
bounded norms and inverse norms change the constants in the estimates,
but not their form.
\subsection{Spaces that can be chosen uniformly in a spatial bin}
We need spaces at nearby starting points to agree much more accurately
than their possible small angle. The comparison uses short derivative
products and the parameters in \eqref{eq:smooth-scales}.
Fix a finite family of coordinate charts with smaller relatively
compact subcharts covering $M$.
Their coordinate maps, inverse maps, and the local expressions of
$S$ and $S^{-1}$ have uniform bounds through order two on the relevant
compact sets. Each spatial bin below is contained in one of the
smaller subcharts, has coordinate diameter at most $C h_0$, and
uses that specified chart to identify its tangent spaces with $\R^d$.
At a boundary of $M$ use nested half-box charts.
For equal-dimensional subspaces $E,F\subset\R^d$, measure their
distance by $\|\Pi_E-\Pi_F\|$, where $\Pi_E,\Pi_F$ are Euclidean
orthogonal projections. This distance concerns the subspaces and
does not require a choice of bases in them.
\begin{lemma}[Common spaces in a bin]\label{lem:bin-spaces}
For all sufficiently large $k$, each spatial bin $U$ meeting
$\mathcal G_k$ has a number
\[
\gamma_U\in[g,(8d+1)g]
\]
such that every good witness frame over $U$ has all its rates at most
$\gamma_U-g$ or at least $\gamma_U+g$, with the same number $p_U$
in the second group.
Apply \Cref{prop:finite-splitting} using this threshold.
There is $c>0$, independent of the bin and the witnesses, such that:
\begin{enumerate}[label=\textup{(\roman*)}]
\item The spaces $E_0^s$ from any two good witnesses over $U$ are
within $e^{-c gL}$ of one another in its coordinate chart.
The same holds for the spaces $E_0^f$.
\item For good paths starting in $U$ whose time-$n$ endpoints lie
in a common endpoint bin $V$, their spaces $E_n^f$ are within
$e^{-c gL}$ of one another in the chart of $V$.
For paths whose time-$(-n)$ endpoints lie in a common bin,
the same assertion holds for $E_{-n}^s$.
\end{enumerate}
In particular, fix one representative space for each nonempty
combination of bins in \textup{(ii)}. Orthogonal projection onto it,
restricted to any corresponding actual endpoint space, is an
isomorphism with inverse norm at most $2$.
If $p_U=0$ or $p_U=d$, the relevant zero or full spaces agree exactly.
\end{lemma}
\begin{proof}
\emph{Short products in common charts.}
Choose one reference point in a starting bin.
Uniform Lipschitz bounds for $S$ and $S^{-1}$ give, for another point
in that bin,
\[
d_M(S^t x,S^t y)\leq C A_0^{|t|}h_0
\qquad(|t|\leq L)
\]
with a fixed $A_0>1$.
Since $L=o(Bn)$, this tends to zero uniformly.
At each time on either short reference orbit choose a chart whose
smaller subchart contains the reference point.
All nearby orbits then remain in the same larger chart at that time.
Use the prescribed bin chart at the initial time.
The one-step coordinate derivatives and their inverses are bounded by
a fixed constant, and their differences are at most
$C A_0^{|t|}h_0$.
The telescoping product identity consequently gives
\begin{equation}\label{eq:short-product-comparison}
\|P_x^\pm-P_y^\pm\|\leq e^{-Bn+CL}=:\zeta_k,\qquad
\sigma_{\min}(P_x^\pm)\geq e^{-CL}=:m_k.
\end{equation}
Here $P_x^+$ and $P_x^-$ denote $DS^L_x$ and $DS^{-L}_x$,
respectively, in these common initial and terminal charts.
For example, summing the $L$ terms in the product difference bounds
it by $C L A_1^{2L}h_0$, which has the stated form.
The inverse product bound gives the minimum stretch.
On a manifold with boundary the same proof uses the nested half-box
charts and uniform one-sided smooth bounds; equivalently the finitely
many local expressions may be extended to their surrounding boxes.
The singular values are Lipschitz with respect to operator norm.
Since $\zeta_k/m_k\to0$, \eqref{eq:short-product-comparison} implies,
after increasing the constants if necessary,
\begin{equation}\label{eq:short-log-singular}
|\ln\sigma_j(P_x^+)-\ln\sigma_j(P_y^+)|
\leq 2\zeta_k/m_k .
\end{equation}
For completeness, the singular-value Lipschitz bound follows from
the min--max characterization, because $\|Pv\|$ and $\|Qv\|$
differ by at most $\|P-Q\|$ on unit vectors.
The logarithmic inequality then follows from the lower bound $m_k$.
\emph{A threshold for every witness.}
Let $\lambda_{(j)}(e)$ be the sorted rates of a good witness $e$.
Subtracting \eqref{eq:exterior-rate-comparison} for exterior degrees
$j$ and $j-1$ gives
\begin{equation}\label{eq:singular-rate}
\left|L^{-1}\ln\sigma_j(P_{\pi(e)}^+)
-\lambda_{(j)}(e)\right|
\leq Cbn/L.
\end{equation}
This argument uses the no-gap part of
\Cref{prop:finite-splitting}; a threshold has not yet been chosen.
Fix one good witness $e_*$ in $U$.
Equations \eqref{eq:short-log-singular} and \eqref{eq:singular-rate}
show that every good witness in $U$, including every other witness
over the same point, satisfies
\[
\max_j|\lambda_{(j)}(e)-\lambda_{(j)}(e_*)|
\leq Cbn/L+2\zeta_k/(m_kL)=o(g).
\]
Consider the $d+1$ candidate thresholds $(1+8j)g$, $0\leq j\leq d$.
A real number lies within distance $3g$ of at most one candidate.
The $d$ representative rates therefore leave a candidate whose
distance from all of them is greater than $3g$.
For large $k$, all the other sorted rate lists differ by less than
$g$, so this candidate separates every list with a margin greater
than $2g$. It has the weaker required margin $g$, and the number of
rates above it is constant. Fix this candidate as $\gamma_U$.
\emph{Comparison of subspaces using one short map.}
Assume first that $0<p_U<d$ and put $q=d-p_U$.
Use the single forward short map $P_*=P_{\pi(e_*)}^+$.
Let $W$ be its bottom $q$ right-singular subspace.
Its top $p_U$ singular values are at least
$\exp((\gamma_U+g)L-Cbn)$ by \eqref{eq:singular-rate}.
If $v$ is a unit vector in the slow space of any good witness $e$,
then \eqref{eq:split-propagation} and
\eqref{eq:short-product-comparison} give
\[
\|P_*v\|\leq e^{(\gamma_U-g)L+Cbn}+\zeta_k.
\]
Orthogonality of the images of right-singular vectors implies
\[
\|\Pi_{W^\perp}v\|
\leq e^{-2gL+Cbn}
+\zeta_k e^{-(\gamma_U+g)L+Cbn}
\leq e^{-2gL+Cbn}+e^{-Bn+CL}.
\]
In the last bound, constants absorb
$|\gamma_U|=O_d(g)$ and the other terms bounded by a constant times
$L$. This is at most $e^{-c_1gL}$ for some $c_1>0$ and large $k$.
To pass from this directed estimate to a subspace estimate, observe
that if $\|\Pi_{W^\perp}|E\|\leq\epsilon<1$ and
$\dim E=\dim W$, then $\Pi_W|E$ is an isomorphism.
The space $E$ is a graph over $W$ of norm at most
$\epsilon/\sqrt{1-\epsilon^2}$.
The corresponding orthogonal projectors differ by at most a fixed
multiple of $\epsilon$ when $\epsilon\leq1/2$.
Thus every slow space is within $C e^{-c_1gL}$ of this same $W$;
the triangle inequality compares any two slow spaces.
No continuity assertion about singular vectors or chosen good frames
has been used.
For fast spaces use the single backward short map
$P_{\pi(e_*)}^-$.
Its bottom $p_U$ right-singular space is the comparison space.
The top $q$ singular values are at least
$\exp(-(\gamma_U-g)L-Cbn)$, whereas a unit fast vector has backward
stretch at most $\exp(-(\gamma_U+g)L+Cbn)$.
The identical calculation gives the error bound
\[
e^{-2gL+Cbn}
+\zeta_k e^{(\gamma_U-g)L+Cbn}
\leq e^{-2gL+Cbn}+e^{-Bn+CL}.
\]
This proves \textup{(i)} after reducing $c$ to absorb fixed factors.
In particular, the error is $o(e^{-C_0bn})$ for every fixed $C_0$,
by $bn=o(gL)$.
\emph{Endpoint comparisons.}
Keep the threshold of $U$.
For paths starting in $U$ and ending at time $n$ in a common bin $V$,
choose one representative path for this pair of bins.
Apply the preceding short-product comparison from their time-$n$
points backward for length $L$.
Their initial closeness for this comparison is supplied by $V$,
not by propagating the starting-bin closeness for $n$ steps.
All intervals $[n-L,n]$ lie inside the good interval, and all paths
use the same group dimension and threshold.
In particular, \eqref{eq:exterior-rate-comparison} applies on each
$[n-L,n]$ with the same full-segment rates, so the singular-value
comparison and the backward argument just proved compare their fast
spaces at $n$. At $-n$, the identical all-subinterval estimate for
forward short maps on $[-n,-n+L]$ gives the slow-space assertion.
Finally, if $E$ and its chosen representative $F$ have equal dimension
and $\|\Pi_E-\Pi_F\|\leq\epsilon$, then for $v\in E$,
\[
\|\Pi_F v\|\geq\sqrt{1-\epsilon^2}\,\|v\|.
\]
The restriction is onto $F$ by equality of dimensions, and its inverse
has norm at most $(1-\epsilon^2)^{-1/2}\leq2$ for large $k$.
If one group is empty, the associated spaces are identically zero or
the full coordinate space, so all comparisons and projection
conclusions follow directly.
\end{proof}
For later use, choose one good witness in each nonempty starting bin
and orthonormal bases within each of its two spaces at time zero.
The matrix having these two bases as columns has norm bounded by a
dimension-only constant and inverse norm at most $e^{Cbn}$, by the
projection bounds of \Cref{prop:finite-splitting}.
It gives fixed product coordinates for that bin.
Likewise \Cref{lem:bin-spaces} supplies fixed endpoint projections
for each relevant pair of bins.
All these choices are finite at a fixed scale. Their constancy within
the indicated bins, rather than any continuity of choices between
bins, is what the switching construction will use.
\subsection{Relative stability of exterior products}
The bounds on every unperturbed subinterval also control a perturbed
long product. This must be a relative estimate: a leading volume can
be exponentially small, so continuity with an absolute error alone is
insufficient.
\begin{lemma}[Exterior-product perturbation]\label{lem:relative-exterior}
Fix $d\geq1$ and $K\geq1$. Let $0\leq p\leq d$, let $m\geq1$,
and let $A_0,\ldots,A_{m-1}$ and
$\widehat A_0,\ldots,\widehat A_{m-1}$ be real $d$-by-$d$ matrices
such that
\[
\|A_t\|\leq K,\qquad
\|\widehat A_t-A_t\|\leq\epsilon\leq1.
\]
Use the same product notation as above for nonnegative indices.
Suppose $a\geq0$, $\sigma\in\R$, and
\[
\|\mathord{\bigwedge}^p A(t,s)\|
\leq e^{\sigma(t-s)+a}
\qquad(0\leq s\leq t\leq m).
\]
There is $C=C(d,K)$ such that
\begin{equation}\label{eq:exterior-perturbation}
\left\|\mathord{\bigwedge}^p\widehat A(m,0)
-\mathord{\bigwedge}^p A(m,0)\right\|
\leq e^{\sigma m+a}
\left[(1+C\epsilon e^{a+|\sigma|})^m-1\right].
\end{equation}
For $p=0$ the left side is zero.
In particular, use the scales \eqref{eq:smooth-scales}, take
$m\leq2n$, $a\leq C_1bn$, $|\sigma|\leq C_2$, and
$\epsilon\leq C_3e^{-3Bn/2}$, with fixed constants.
Then for every fixed $C_0>0$ the left side of
\eqref{eq:exterior-perturbation} is
\begin{equation}\label{eq:relative-exterior-small}
o\!\left(e^{\sigma m-C_0bn}\right),
\end{equation}
uniformly for all matrices satisfying these hypotheses.
\end{lemma}
\begin{proof}
The exterior power is multiplicative. Multilinearity in its $p$
columns, together with the one-step bound $K+1$ for the perturbed
matrices, gives
\[
E_t:=\mathord{\bigwedge}^p\widehat A_t
-\mathord{\bigwedge}^p A_t,
\qquad \|E_t\|\leq C\epsilon.
\]
Expand the product using either the unperturbed exterior matrix or
$E_t$ at each time. A term with $q\geq1$ insertions has $q+1$
unperturbed intervals whose total length is $m-q$.
Allowing empty intervals, the hypothesis bounds that term by
\[
e^{\sigma(m-q)+(q+1)a}(C\epsilon)^q
=e^{\sigma m+a}(C\epsilon e^{a-\sigma})^q.
\]
There are $\binom mq$ choices of the inserted positions.
Sum over $q=1,\ldots,m$ and use $-\sigma\leq|\sigma|$ to prove
\eqref{eq:exterior-perturbation}. This calculation explains why
negative $\sigma$, as may occur for backward exterior volumes, is
harmless: each removed time step contributes only the displayed
factor $e^{-\sigma}$.
For the final assertion put
$u=C\epsilon e^{a+|\sigma|}$ and use
$(1+u)^m-1\leq mu e^{mu}$.
After division by $e^{\sigma m-C_0bn}$ the bound is at most
\[
C m\epsilon e^{(2C_1+C_0)bn+C_2}
\exp\!\bigl(Cm\epsilon e^{C_1bn+C_2}\bigr).
\]
The logarithm of its decisive first factor is at most
$-3Bn/2+O(bn)+\ln(2n)$, which tends to $-\infty$ by
\eqref{eq:smooth-hierarchy}. The quantity in the second exponential
tends to zero for the same reason.
All constants are fixed, proving the uniform little-oh assertion.
For exterior degree zero the two products are both the identity, so
the result holds without an expansion.
\end{proof}
The lemma applies to forward degree-$p$ products with
$\sigma=\Lambda$ and to backward degree-$(d-p)$ products with
$\sigma=\Omega$ from \Cref{prop:finite-splitting}.
Together with \Cref{lem:bin-spaces}, it supplies uniform linear
and relative-volume estimates for orbit segments whose starting
points occupy one small spatial bin.
==== END A-finite-entropy-system-without-a-smooth-positive-volume-model//03-linear.tex ====
==== BEGIN A-finite-entropy-system-without-a-smooth-positive-volume-model//04-switching.tex ====
\section{Switching a past and a future with controlled volume}
\label{sec:switching}
The finite-interval estimates now give a geometric operation on pairs of
nearby points. It joins the future of one point to the past of the other.
We must control the distribution of the resulting point as well as its
orbit: this requires constructing both complementary joins and estimating
their joint Jacobian.
The joining geometry is related to the canonical local-product
coordinates of hyperbolic dynamics \cite[\S I.7]{Smale1967}.
Solving a linear difference equation and then a nonlinear contraction
also has a standard shadowing precedent
\cite[Lemma~3.2 and Proposition~4.1]{ChowLinPalmer1989}.
Here the slow group may include zero or positive rates; we prove the
finite-interval boundary inverse directly. The measure inequality
requires the further determinant and injectivity arguments below.
\begin{theorem}[Smooth switching]\label{thm:switching}
Let $M$ be a compact smooth manifold of positive finite dimension, let
$\nu$ be a smooth positive probability volume in the density sense of
\Cref{sec:introduction}, and let $S:M\to M$
be a smooth diffeomorphism preserving $\nu$. Fix a Riemannian distance
$d_M$ and $\eta>0$. There exist positive numbers $B_k$ with
$\sum_k B_k^{16}<\infty$ and a constant $C$ such that the following
holds for all sufficiently large $k$, with $n=2^k$.
There is a finite measurable partition $\mathcal D_k$ of $M$ with
$\#\mathcal D_k\le \exp(CB_kn)$. Define the probability measure
\begin{equation}\label{eq:pair-law}
\rho_k=\sum_{\substack{Q\in\mathcal D_k\\\nu(Q)>0}}
\frac{(\nu|_Q)\otimes(\nu|_Q)}{\nu(Q)}.
\end{equation}
Thus one first samples a cell with its $\nu$-weight and then samples
two independent points in that cell. There are a measurable set
$G_k\subset M\times M$, with $\rho_k(G_k)\ge1-\eta$, and a measurable
map $z_k:G_k\to M$ such that
\begin{equation}\label{eq:switch-law}
(z_k)_*(\rho_k|_{G_k})\le 2\nu
\end{equation}
and
\begin{equation}\label{eq:switch-shadow}
\sup_{(x,y)\in G_k}\left[
\max_{0\le t\le n}d_M(S^t z_k(x,y),S^t x)
+\max_{-n\le t\le0}d_M(S^t z_k(x,y),S^t y)
\right]\longrightarrow0.
\end{equation}
The measure on the left of \eqref{eq:switch-law} is not normalized
after restriction to $G_k$. The same conclusions hold if $M$ has a
smooth boundary.
\end{theorem}
The proof occupies the rest of this section. Its first two steps produce
two complementary switches $(z,w)$. The last two steps show that their
pair map is injective and has volume Jacobian $1+o(1)$. This is what
will give \eqref{eq:switch-law}.
\subsection{Grids, retained pairs, and fixed endpoint projections}
We may assume $0<\eta<1$. Apply \Cref{prop:good-frames} with parameter
$\eta/2$. It supplies $b=b_k\ge n^{-1/10}$,
$\sum_k b_k^2<\infty$, and a compact set of good base points,
denoted here by $\mathcal A_k$ to distinguish it from the retained pair set,
with suitable frame paths on $[-n,n]$, such that
$\nu(M\setminus\mathcal A_k)<\eta/100$. Set
\begin{equation}\label{eq:switch-scales}
B=b^{1/8},\quad h_0=e^{-Bn},\quad r=e^{-2Bn},\quad
L=\lceil b^{1/4}n\rceil,\quad g=b^{1/2},\quad
\delta=e^{-3Bn/2}.
\end{equation}
Here and below constants depend only on the fixed smooth data, dimension,
and $\eta$. They never depend on a point, bin, frame witness, or $k$.
In particular,
\begin{equation}\label{eq:switch-scale-order}
\log n=o(bn),\quad bn=o(gL),\quad gL=o(gn),\quad
gn=o(L),\quad L=o(Bn).
\end{equation}
These relations absorb every fixed power of a quantity bounded by
$e^{Cbn}$ or $e^{Cgn}$ at the places where it is used below.
Choose a finite system of coordinate patches with smaller coordinate
boxes covering $M$, each smaller box having closure in its patch.
Assign points to the first smaller box containing them. This gives
disjoint chart regions whose boundaries lie in finitely many smooth
coordinate faces. The coordinate maps, their inverses, and the chart
density of $\nu$ have uniform bounds on slightly larger compact patches;
the density has a uniform positive lower bound. Within each region
use the grid of coordinate cubes of side $h_0$, keeping only cubes whose
closures lie in that region. We call these cubes \emph{coarse bins}.
The discarded part has volume $O(h_0)$: it is contained in a fixed
multiple of the $h_0$-neighborhood of the finitely many region boundaries.
All comparisons at a common starting or endpoint bin use its one fixed
coordinate map.
For a starting bin $Q$ meeting $\mathcal A_k$, \Cref{lem:bin-spaces}
gives a common threshold $\gamma_Q\in[g,Cg]$, a fast dimension $p_Q$,
and reference spaces $E_Q^s,E_Q^f$ at time zero. The spaces of every
good frame path beginning in $Q$ are within $e^{-cgL}$ of these spaces.
Their complementary projections and the inverse of a basis orthonormal
within each space have norm at most $e^{Cbn}$. The threshold and the
reference spaces are chosen once for this bin. This involves finitely
many choices at each scale, not a choice of a frame as a measurable
function on $\mathcal A_k$.
In the starting chart let $E_Q:\R^{d-p_Q}\times\R^{p_Q}\to\R^d$
be the matrix with these orthonormal slow and fast columns, respectively.
Tile the chart by the parallelepipeds $E_Q([0,r)^d+rj)$, $j\in\Z^d$,
and keep the ones whose closures lie in $Q$. Translations of the grid
are immaterial. These are the \emph{product cells}. Although
$\|E_Q^{-1}\|\le e^{Cbn}$, the forward norm satisfies
$\|E_Q\|\le\sqrt2$: for $(a,b)$,
$|E_Q(a,b)|\le |a|+|b|$. Thus every cell has physical diameter at
most $C_dr$ and volume at least $e^{-Cbn}r^d$ up to a fixed factor.
The total number of full cells is consequently at most
\begin{equation}\label{eq:cell-count}
C e^{Cbn}r^{-d}\le e^{CBn}.
\end{equation}
Every point of a coarse bin at distance greater than $C_dr$ from its
walls belongs to a full product cell. Hence the partial-cell loss,
summed over all coarse bins, is $O(r/h_0)=o(1)$.
Use one additional cell for all omitted points and for bins not meeting
$\mathcal A_k$. Assign grid boundaries arbitrarily; they are null.
This completes the partition $\mathcal D_k$ without changing
\eqref{eq:cell-count}.
For two points $x=(x_s,x_f)$ and $y=(y_s,y_f)$ in the product
coordinates of one cell, the ideal exchange
\[
z_0=(y_s,x_f),\qquad w_0=(x_s,y_f)
\]
stays in that cell. These points combine the slow coordinate of one
input with the fast coordinate of the other. For the pairs retained
below, we will construct actual orbit joins $z,w$ whose product
coordinates differ from $z_0,w_0$ by $o(r)$.
Removing thin coordinate strips will therefore keep those
joins in their original cell. This is why the cells follow the two
reference spaces, which need not be orthogonal; see
\Cref{fig:ideal-cell-exchange}.
\begin{figure}[htbp]
\centering
\begin{tikzpicture}[x=.9cm,y=.9cm,>=Stealth,
point/.style={circle,fill,inner sep=1.6pt}]
\draw[thick] (0,0)--(4.8,0)--(6.3,2.8)--(1.5,2.8)--cycle;
\draw[dashed,gray] (1.104,0)--(2.604,2.8);
\draw[dashed,gray] (3.504,0)--(5.004,2.8);
\draw[dashed,gray] (.39,.728)--(5.19,.728);
\draw[dashed,gray] (1.05,1.96)--(5.85,1.96);
\node[point,label=above left:{$x$}] at (2.154,1.96) {};
\node[point,label=below right:{$y$}] at (3.894,.728) {};
\node[point,label=above right:{$z_0$}] at (4.554,1.96) {};
\node[point,label=below left:{$w_0$}] at (1.494,.728) {};
\draw[->] (0,-.4)--(1.5,-.4) node[midway,below] {slow};
\draw[->] (-.45,0)--(.05,.933) node[midway,above left] {fast};
\end{tikzpicture}
\caption{A two-dimensional schematic of the ideal exchange in a
product cell. Dashed lines hold one product coordinate fixed.
The two coordinate directions may be oblique. For retained pairs, the
actual joins are proved to differ from $z_0,w_0$ by $o(r)$ in
\Cref{lem:switch-orbits}.}
\label{fig:ideal-cell-exchange}
\end{figure}
Fix a small number $\tau>0$, independent of $k$. The interior of a
product cell retained at this stage consists of points whose every
product coordinate is at least $\tau r$ from its two cell faces.
The discarded coordinate strips have at most $2d\tau$ of its Euclidean
volume. The constant determinant of $E_Q$ cancels in this fraction.
Moreover, the ratio between the largest and smallest smooth density on
the physical cell is $1+O(r)$, uniformly in $Q$. Thus the total smooth
volume of these strips is at most $2d\tau+o(1)$. The corresponding
strips of width $\tau h_0$ in coarse bins have the same bound. These
are estimates for the original measure $\nu$, without conditioning on
membership in $\mathcal A_k$.
If $M$ has boundary, also exclude points having any iterate at an integer
time in $[-n,n]$ at distance at most $c_0h_0$ from $\partial M$, where
$c_0>0$ is fixed. A smooth boundary collar has volume $O(h_0)$, so
invariance bounds this loss by $O(nh_0)=o(1)$. The latter limit follows
from $Bn\ge n^{79/80}$. Local coordinate extensions across the boundary
may be used for uniform derivative bounds; all orbit tubes constructed
below lie in the original manifold.
Let $R_k$ be the Borel set of points in $\mathcal A_k$ that lie in the
retained interiors of full product cells, whose two endpoints $S^{\pm n}u$
lie in coarse bins at coordinate distance at least $\tau h_0$ from
their faces, and that satisfy the boundary provision when needed.
Bins not meeting $\mathcal A_k$ contain no such point. The preceding
estimates and invariance give
\begin{equation}\label{eq:retained-point-loss}
\nu(M\setminus R_k)\le \eta/100+C_d\tau+o(1).
\end{equation}
Choose $\tau$ so that $C_d\tau<\eta/100$, and then increase $k$ so
that the right-hand side is less than $\eta/4$.
Both marginals of \eqref{eq:pair-law} are exactly $\nu$. Consequently,
for
\begin{equation}\label{eq:retained-pairs}
G_k=(R_k\times R_k)\cap
\bigcup_{Q'\text{ a full product cell}}(Q'\times Q'),
\qquad \rho_k(G_k)\ge1-\eta/2\ge1-\eta.
\end{equation}
The measure $\rho_k$ is supported on pairs in the same partition cell,
and $R_k$ meets only full product cells. Hence
$\rho_k(G_k)=\rho_k(R_k\times R_k)$, and the two marginal bounds give
the displayed estimate. No uniform distribution of good points inside
individual cells is required.
We next fix the maps that will specify the switch. For each starting
bin $Q$ and endpoint bin $U$ reached at time $n$ by a point of
$\mathcal A_k\cap Q$, choose a representative endpoint fast space.
Let $P^+_{Q,U}:\R^d\to\R^{p_Q}$ be its orthogonal projection, followed
by a fixed orthonormal identification with $\R^{p_Q}$. Similarly, for
an endpoint bin $V$ reached at time $-n$, define
$P^-_{Q,V}:\R^d\to\R^{d-p_Q}$ from a representative slow space.
By the endpoint conclusion of \Cref{lem:bin-spaces}, these projections
restrict to isomorphisms on the actual fast and slow endpoint spaces
of every corresponding good path, with inverse norm at most $2$ for
large $k$. The maps depend only on the displayed bins and the starting
threshold. They do not depend on a later choice of input in those bins.
Zero-dimensional projections and determinants have their usual empty
space meanings.
For a retained pair $(x,y)$, write $Q$ for its coarse starting bin and
$U_x,V_x,U_y,V_y$ for the coarse bins of
$S^nx,S^{-n}x,S^ny,S^{-n}y$, respectively. If $\chi_U$ denotes the
coordinate map attached to a bin, define the following smooth maps on
the open sets where the coordinates make sense:
\begin{equation}\label{eq:endpoint-maps}
\begin{aligned}
F_x(u)&=P^+_{Q,U_x}\chi_{U_x}(S^nu),&
P_x(u)&=P^-_{Q,V_x}\chi_{V_x}(S^{-n}u),\\
F_y(u)&=P^+_{Q,U_y}\chi_{U_y}(S^nu),&
P_y(u)&=P^-_{Q,V_y}\chi_{V_y}(S^{-n}u).
\end{aligned}
\end{equation}
The subscripts denote fixed bin labels. We seek $z,w$ satisfying
\begin{equation}\label{eq:four-switch-equations}
F_x(z)=F_x(x),\quad P_y(z)=P_y(y),\qquad
F_y(w)=F_y(y),\quad P_x(w)=P_x(x).
\end{equation}
The next step both solves these equations near the prescribed halves
and gives a larger neighborhood in which each solution is unique.
\Cref{fig:switch-data} displays the required exchange of endpoint data.
\begin{figure}[htbp]
\centering
\begin{tikzpicture}[>=Stealth,
point/.style={circle,draw,minimum size=7mm,inner sep=1pt},
future/.style={->,thick,blue!65!black},
past/.style={->,thick,dashed,orange!75!black}]
\node[point] (x) at (-3,0.85) {$x$};
\node[point] (y) at (-3,-0.85) {$y$};
\node[point] (z) at (3,0.85) {$z$};
\node[point] (w) at (3,-0.85) {$w$};
\draw[future] (x) -- node[above] {$F_x$} (z);
\draw[future] (y) -- node[below] {$F_y$} (w);
\draw[past] (x) -- node[pos=0.27,above,sloped] {$P_x$} (w);
\draw[past] (y) -- node[pos=0.27,below,sloped] {$P_y$} (z);
\end{tikzpicture}
\caption{The endpoint equations for the complementary switches.
Solid arrows copy projected future coordinates; dashed arrows copy
projected past coordinates. The arrows represent endpoint data,
not trajectories in the manifold.}
\label{fig:switch-data}
\end{figure}
\subsection{A boundary inverse and a nonlinear orbit}
\begin{lemma}\label{lem:switch-orbits}
For every sufficiently large $k$ and every $(x,y)\in G_k$, the first
two equations of \eqref{eq:four-switch-equations} have a solution $z$
with
\begin{equation}\label{eq:sharp-shadow}
\max_{0\le t\le n}d_M(S^tz,S^tx)
+\max_{-n\le t\le0}d_M(S^tz,S^ty)
\le e^{Cgn}r.
\end{equation}
In fixed regular coordinates along these halves the solution is unique
among displacement arrays of sup norm at most $\delta$.
The complementary solution $w$ has the analogous properties with $x,y$
exchanged. In the product coordinates of the common cell, if
$x=(x_s,x_f)$ and $y=(y_s,y_f)$, then
\begin{equation}\label{eq:ideal-switch}
z=(y_s,x_f)+o(r),\qquad w=(x_s,y_f)+o(r),
\end{equation}
uniformly over all retained pairs. Both outputs remain in the original
product cell, and their relevant endpoints remain in the respective
four original endpoint bins.
\end{lemma}
\begin{proof}
\emph{Uniform coordinates.}
At each reference location choose one of finitely many slightly larger
regular patches containing a fixed-radius neighborhood of that location.
At time zero both strings use $\chi_Q$; at the terminal times they use
the endpoint charts already specified. Shrinking the fixed radius once,
the one-step coordinate maps
$\chi_{t+1}\circ S\circ\chi_t^{-1}$ and their first two derivatives
are uniformly bounded wherever they will be used. The inverse
one-step maps also have uniform first and second derivative bounds.
For example, differentiating the inverse identity bounds their second
derivatives by a fixed multiple of the forward second-derivative bound
and the cube of the inverse first-derivative bound. This
follows from a finite atlas with nested compact patches and smoothness
on compact $M$; it does not depend on the number of steps. A patch
can be chosen separately at each internal time. No iterate $S^n$
is assigned a uniform second derivative bound.
Use displacement variables $v_t^+$ about $S^tx$, $0\le t\le n$,
and $v_t^-$ about $S^ty$, $-n\le t\le0$. Write $A_t^+,A_t^-$
for the derivative steps in these charts. Matching the two locations
at zero and imposing the terminal equations gives the linear system
\begin{equation}\label{eq:linear-boundary-system}
\begin{aligned}
v_{t+1}^{\pm}-A_t^{\pm}v_t^{\pm}&=f_t^{\pm},\\
v_0^+-v_0^-&=m,\\
P^+_{Q,U_x}v_n^+&=h_+,\qquad
P^-_{Q,V_y}v_{-n}^-=h_-.
\end{aligned}
\end{equation}
The matching datum for the actual problem is
$m=\chi_Q(y)-\chi_Q(x)$; its norm is $O(r)$.
The actual terminal data are zero.
\emph{The inverse in weighted coordinates.}
Let $\gamma=\gamma_Q$ and put $\widetilde v_t=e^{-\gamma t}v_t$.
Use the same weighting on equation data, with the step datum at its
output time $t+1$. By \Cref{prop:finite-splitting}, the weighted
propagators on slow spaces forward and fast spaces backward have norm
at most $e^{Cbn}e^{-g|t-s|}$. The associated projections have norm
at most $e^{Cbn}$. Geometric sums cost at most
$C/g\le e^{Cbn}$, by \eqref{eq:switch-scale-order} and the floor on $b$.
On the future half, prescribe a slow vector $a\in E_0^s(x)$.
Variation of constants gives its slow part at every time by propagating
$a$ forward and adding the projected step data. If $s_n$ is the
resulting slow value at time $n$, the fast value there is uniquely
\[
c_n=(P^+_{Q,U_x}|_{E_n^f(x)})^{-1}
(\widetilde h_+-P^+_{Q,U_x}s_n).
\]
Propagate $c_n$ and the fast step data backward. All forced terms have
norm at most $e^{Cbn}$ times the sup norm of the weighted data.
The dependence of the resulting fast value at zero on $a$ is the map
\begin{equation}\label{eq:future-kernel-graph}
K_+=-\widetilde A_x(0,n)|_{E_n^f(x)}
(P^+_{Q,U_x}|_{E_n^f(x)})^{-1}
P^+_{Q,U_x}\widetilde A_x(n,0)|_{E_0^s(x)},
\qquad \|K_+\|\le e^{-2gn+Cbn}.
\end{equation}
Here $\widetilde A(t,s)=e^{-\gamma(t-s)}A(t,s)$.
Both decaying factors are present in \eqref{eq:future-kernel-graph};
there is no assumption that the endpoint projection annihilates the
slow space. Thus $\widetilde v_0^+=a+K_+a+q_+$, with $q_+$ bounded
by the weighted data as above.
On the past half prescribe $b'\in E_0^f(y)$, propagate fast backward,
use the slow endpoint projection to fix the slow value at $-n$, and
then propagate slow forward. This gives
$\widetilde v_0^-=b'+K_-b'+q_-$, where
\begin{equation}\label{eq:past-kernel-graph}
\|K_-\|\le e^{-2gn+Cbn},\qquad
\|q_-\|\le e^{Cbn}\|\text{weighted data}\|_{\infty}.
\end{equation}
The map $(a,b')\mapsto a-b'$ from $E_0^s(x)\oplus E_0^f(y)$ to
the starting coordinate space has inverse norm at most $e^{Cbn}$.
Indeed the corresponding reference spaces are complementary with this
bound, and their perturbation is $e^{-cgL}$; multiplication by the
reference inverse still gives a perturbation tending to zero.
The terms $K_+a-K_-b'$ are smaller still. A Neumann series therefore
solves the matching equation with inverse cost $e^{Cbn}$.
This proves an $e^{Cbn}$ inverse bound for the full weighted system.
Removing the weights costs at most $e^{2\gamma(n+1)}$, so its
unweighted inverse $\mathcal L^{-1}$ satisfies
\begin{equation}\label{eq:boundary-inverse}
\|\mathcal L^{-1}\|_{\infty\to\infty}\le Q_k,
\qquad Q_k=e^{Cgn}.
\end{equation}
If a group is empty, omit its variables, propagation formula, and
terminal datum. The matching map on the remaining spaces and the
same bounds apply, including the cases $p_Q=0,d$.
\emph{Contraction with a larger uniqueness radius.}
Taylor expansion of each one-step map gives a residual
$\mathcal R(v)$ with
\begin{equation}\label{eq:one-step-remainder}
\|\mathcal R(v)\|_\infty\le C\|v\|_\infty^2,
\qquad
\|\mathcal R(v)-\mathcal R(u)\|_\infty
\le C\delta\|v-u\|_\infty
\end{equation}
on the sup norm ball of radius $\delta$. Every residual coordinate
depends on just its one step, so the constant has no factor $n$.
The matching and terminal equations are affine and have zero nonlinear
residual. The nonlinear problem is
$v=\mathcal L^{-1}(a_0+\mathcal R(v))$, where
$\|a_0\|_\infty\le Cr$.
The scale relations give
\[
CQ_k\delta\longrightarrow0,\qquad Q_kr/\delta\longrightarrow0.
\]
For large $k$ the displayed map sends the $\delta$-ball into its
$\delta/2$-ball and has Lipschitz constant at most $1/4$.
It has a unique fixed point in the larger ball, with
$\|v\|_\infty\le 2CQ_kr$. The exact one-step equations and the
matching equation make this array one genuine orbit; call its point
at zero $z$. Chart bounds give \eqref{eq:sharp-shadow}.
In the boundary case its distance from each reference orbit is
$o(h_0)$, so it stays inside $M$.
\emph{Time-zero coordinates and wall margins.}
For the linear solution with zero step and terminal data, the two
initial displacement spaces are the graphs in
\eqref{eq:future-kernel-graph} and \eqref{eq:past-kernel-graph}.
In the reference product coordinates they differ from the slow and
fast coordinate spaces by at most
$e^{Cbn}(e^{-cgL}+e^{-2gn+Cbn})$. Solving their matching equation
therefore differs from ideal exchange of the fast and slow coordinates
by at most
\begin{equation}\label{eq:linear-switch-error}
e^{Cbn}\bigl(e^{-cgL}+e^{-2gn+Cbn}\bigr)r=o(r).
\end{equation}
All basis conversion and matching inverse factors are included in the
constant $C$. The nonlinear solution differs from the linear one by
at most $CQ_k(2CQ_kr)^2=e^{Cgn}r^2$ in ordinary coordinates, and still
by $e^{Cgn}r^2=o(r)$ in product coordinates. This proves
\eqref{eq:ideal-switch}. The proof for $w$ is identical with the
two inputs exchanged.
Increase $k$ so that the error in each product coordinate is less than
$\tau r/4$. The ideal exchanged point has distance at least $\tau r$
from every product face because both inputs do. Thus $z,w$ have
coordinate margins at least $3\tau r/4$ in their original cell.
Similarly, \eqref{eq:sharp-shadow} in the fixed endpoint charts is
less than $\tau h_0/4$, so each corresponding endpoint retains margin
at least $3\tau h_0/4$ in its original coarse bin. These strict
margins will recover all labels from the outputs.
The point obtained does not depend on the auxiliary internal charts or
on the frame witnesses used to justify the estimates. Two such
constructions have the same endpoint equations and sharp radius
$e^{Cgn}r=o(\delta)$. In either one's charts both are therefore in
its uniqueness ball, and hence coincide.
\end{proof}
\subsection{Relative minors and the joint determinant}
We have constructed complementary orbit joins. A single join need not
have a convenient Jacobian, so we now keep both of them. Their endpoint
coordinates are permuted, which will cancel the large expansion and
contraction factors.
For the fixed labels of a retained pair define
\begin{equation}\label{eq:joint-endpoint-map}
H_{ab}(u)=(F_a(u),P_b(u)),\qquad a,b\in\{x,y\}.
\end{equation}
Each target has dimension $p_Q+(d-p_Q)=d$ and uses the fixed
orthonormal coordinates in the projection ranges.
\begin{lemma}\label{lem:switch-jacobian}
All four derivatives $DH_{xx}(x)$, $DH_{yy}(y)$,
$DH_{xy}(z)$, and $DH_{yx}(w)$ are nonsingular. Locally the equations
\eqref{eq:four-switch-equations} define a smooth pair map
$\Phi(x,y)=(z,w)$ with
\begin{equation}\label{eq:pair-jacobian}
J_{\nu\otimes\nu}\Phi(x,y)=1+o(1),
\end{equation}
uniformly on the retained pairs. The conclusion includes empty slow
or fast groups.
\end{lemma}
\begin{proof}
Put $p=p_Q$. For the future map at $x$, use the actual fast and slow
spaces first, with orthonormal bases within each. The fast $p$-minor
has absolute value at least $e^{\Lambda_x n-Cbn}$, where
$\Lambda_x$ is the sum of its fast rates. This follows from the
fast volume lower bound in \Cref{prop:finite-splitting} and from the
uniformly invertible restriction of the endpoint projection.
Every other maximal minor contains a slow starting vector and is at
most $e^{(\Lambda_x-2g)n+Cbn}$ when both groups occur.
Replacing the actual bases with the common reference basis changes
their spaces by $e^{-cgL}$. After a change of orthonormal basis within
each space, multilinearity and the full exterior norm bound show that
the leading minor stays nonzero and that all other maximal minors
are $o(1)$ times its absolute value. Factors from the inverse of the
possibly oblique starting basis are at most $e^{Cbn}$. Call the
resulting leading absolute minor $A_x$. More explicitly, when
$0<p<d$, every nonleading maximal future minor $\Delta$ in the common
reference basis satisfies
\[
\frac{|\Delta|}{A_x}
\le C\left(e^{-2gn+Cbn}+e^{-cgL+Cbn}\right)=o(1).
\]
The first term comes from the mixed-exterior estimate. The second
comes from the change of spaces, using the full exterior upper bound
and the leading-minor lower bound. Increasing $C$ includes the
oblique-basis factors. If $p=0$ or $p=d$, there are no nonleading
maximal future minors, so this comparison is unnecessary.
Define $A_y$ in the same way for the future of $y$.
For the past maps the leading
columns are slow, and the backward exterior rate is
\[
\Omega_x=-\sum_{\text{slow at }x}\lambda_j(x).
\]
Denote their leading absolute minors by $B_x,B_y$.
They have lower bounds $e^{\Omega_x n-Cbn}$ and
$e^{\Omega_y n-Cbn}$, respectively, and all other maximal minors are
relatively negligible. No sign is prescribed for $\Lambda_x$ or
$\Omega_x$ in a perturbation estimate.
We must transfer the future minors from $x$ to $z$ relative to $A_x$,
even if this leading quantity is small. Use the reference charts
along the future of $x$ for both nearby orbits. By
\eqref{eq:sharp-shadow}, their one-step derivatives differ by at most
$C\delta$ for large $k$. On every unperturbed subinterval the
$p$-th exterior product has norm at most
$e^{\Lambda_x\ell+Cbn}$.
Applying \Cref{lem:relative-exterior} gives, for every fixed $C_0$,
\begin{equation}\label{eq:relative-minor-transfer}
\left\|\bigwedge^pD(S^n)_z-
\bigwedge^pD(S^n)_x\right\|
=o\bigl(e^{\Lambda_xn-C_0bn}\bigr),
\end{equation}
where the derivatives are represented in these common starting and
endpoint coordinates. To see the scale in this application, an
expansion term with $q\ge1$ altered factors is bounded by
\[
e^{\Lambda_x(n-q)+(q+1)C_1bn}(C\delta)^q.
\]
The binomial sum, divided by $e^{\Lambda_xn-C_0bn}$, tends to zero
because $n\delta e^{Cbn}\to0$. The bounded factor
$e^{q|\Lambda_x|}$ covers either sign of the rate. The same reasoning
backward uses $\Omega_y$ and $e^{q|\Omega_y|}$.
Fixed endpoint projections and the common starting basis preserve
\eqref{eq:relative-minor-transfer}, after increasing $C_0$ if an
inverse basis factor is needed. Hence at $z$ the leading future and
past minors are $A_x(1+o(1))$ and $B_y(1+o(1))$, with the same relative
smallness of all the other minors. At $w$ they are $A_y$ and $B_x$.
Expand each full determinant along its $p$ future rows. There is one
leading product, from fast future columns and slow past columns.
Every other term has a relatively negligible maximal minor. In the
common starting basis the four absolute determinants are therefore
\begin{equation}\label{eq:four-leading-products}
\begin{array}{c|cccc}
\text{map and point}&H_{xx}(x)&H_{yy}(y)&H_{xy}(z)&H_{yx}(w)\\ \hline
\text{absolute determinant}&
A_xB_x(1+o(1))&A_yB_y(1+o(1))&
A_xB_y(1+o(1))&A_yB_x(1+o(1)).
\end{array}
\end{equation}
All leading factors are positive. In particular these maps are
locally invertible. If $p=0$, future targets and minors are empty and
$A_x=A_y=1$; if $p=d$, the past minors are $B_x=B_y=1$. The same
determinants and conclusions apply, without mixed minors of the
missing kind.
The defining equations say that
$(H_{xy}(z),H_{yx}(w))$ is a fixed permutation of
$(H_{xx}(x),H_{yy}(y))$. Differentiation gives the exact identity
\begin{equation}\label{eq:exact-switch-determinant}
|\det D\Phi(x,y)|=
\frac{|\det DH_{xx}(x)|\,|\det DH_{yy}(y)|}
{|\det DH_{xy}(z)|\,|\det DH_{yx}(w)|}.
\end{equation}
Each full derivative represented in the common product basis differs
from its coordinate derivative by the factor $\det E_Q$. There are
two such factors in the numerator and two in the denominator, so
they cancel exactly. Substituting \eqref{eq:four-leading-products}
cancels $A_x,B_x,A_y,B_y$ individually, with no comparison between the
rates of $x$ and $y$. Thus the coordinate Jacobian is $1+o(1)$.
All four starting points lie in the same cell of diameter $O(r)$.
The ratio of the two output smooth densities to the two input densities
is therefore $1+o(1)$ as well. This proves
\eqref{eq:pair-jacobian}.
\end{proof}
\subsection{Measurability, recovered labels, and global injectivity}
The Jacobian calculation is local. To convert it into the desired
measure inequality we must ensure that no multiplicity arises from
different choices of bins or local branches.
\begin{lemma}\label{lem:switch-injective}
The map $\Phi:G_k\to M\times M$, $\Phi(x,y)=(z,w)$, is measurable
and injective. Its restriction to countably many measurable pieces
agrees with smooth local diffeomorphisms having the Jacobian in
\eqref{eq:pair-jacobian}.
\end{lemma}
\begin{proof}
First fix a retained pair. Its cell and four endpoint labels are
locally constant among retained inputs near it, because of the positive
wall margins. By \Cref{lem:switch-jacobian}, the implicit equations
define a smooth local pair branch near its outputs. This branch
agrees with the constructed switch on nearby retained inputs.
Indeed, for this fixed $k$, continuity of the finitely many maps
$S^t$, $|t|\le n$, makes their reference halves arbitrarily close
when their inputs are close. The constructed solutions have sharp
radius $e^{Cgn}r=o(\delta)$ about those halves. Both the old local
implicit branch and the construction for the nearby pair therefore
lie in the old $\delta$-tube, and satisfy the same fixed endpoint
equations with the nearby right-hand sides. The contraction uniqueness
argument identifies them even for these changed data: two solutions
$u,v$ in the old ball satisfy
$\mathcal L(v-u)=\mathcal R(v)-\mathcal R(u)$, so
$\|v-u\|_\infty\le CQ_k\delta\|v-u\|_\infty$ forces equality.
Both candidate arrays here use the old reference halves. Their
matching datum is therefore the same old coordinate difference;
only the prescribed endpoint values change with the nearby inputs.
The charts along the old pair may be kept fixed in this neighborhood. In particular no regularity of a choice of good
frame witnesses is needed.
The retained set is Borel, and $M\times M$ is second countable.
A countable subcollection of these relative neighborhoods covers it.
Subtracting earlier members partitions it into countably many Borel
pieces. On each piece $\Phi$ is the restriction of the indicated
smooth local diffeomorphism; this proves measurability and supplies
the pieces for change of variables.
For injectivity suppose
$\Phi(x,y)=\Phi(x',y')=(z,w)$. The interiors of product cells are
disjoint, and both outputs stayed in the original cell. Thus the
two input pairs have the same product cell and hence the same starting
bin $Q$, threshold, and dimensions. The four endpoint-bin labels
are recovered from the ordered output pair by
\begin{equation}\label{eq:recovered-bin-labels}
(U_x,V_x,U_y,V_y)=
\bigl(\operatorname{bin}(S^nz),\operatorname{bin}(S^{-n}w),
\operatorname{bin}(S^nw),\operatorname{bin}(S^{-n}z)\bigr).
\end{equation}
All four endpoints have positive margins in their bins. Therefore
these labels, and exactly the four projections and coordinate maps in
\eqref{eq:endpoint-maps}, are the same for the competing inputs.
On the future half, both $x$ and $x'$ are within $e^{Cgn}r$ of $z$;
on the past half both are within that distance of $w$. Consequently
their full orbits on $[-n,n]$ are within $2e^{Cgn}r=o(\delta)$ of
each other. Moreover the recovered equations give
\[
F_x(x')=F_x(z)=F_x(x),\qquad
P_x(x')=P_x(w)=P_x(x).
\]
Apply the boundary uniqueness argument of \Cref{lem:switch-orbits}
with both reference halves equal to the true good orbit of $x$.
It uses $x$'s endpoint projections, zero matching datum, and zero
terminal data. The orbit of $x$ is one solution. In the same charts
the orbit of $x'$ is another solution in its $\delta$-ball, since its
distance is $o(\delta)$ even after fixed chart conversions. Uniqueness
gives $x'=x$. The same argument gives $y'=y$.
Only the true input orbits were required to be good; neither $z$ nor
$w$ was required to belong to $\mathcal A_k$ or to $R_k$.
\end{proof}
\begin{proof}[Proof of \Cref{thm:switching}]
The construction above applies for all sufficiently large $k$.
Equation \eqref{eq:switch-scales} gives
$\sum_kB_k^{16}=\sum_kb_k^2<\infty$; any finitely many unused terms
can be chosen positively. The partition bound follows from
\eqref{eq:cell-count}, retention from \eqref{eq:retained-pairs}, and
uniform shadowing from \eqref{eq:sharp-shadow}.
It remains to prove the output measure bound. On the square of a
positive-volume cell $Q'$, the original law $\rho_k$ has density
$1/\nu(Q')$ with respect to $\nu\otimes\nu$. The map $\Phi$ sends
retained pairs in that square into the same square. By
\Cref{lem:switch-jacobian}, its inverse volume Jacobian is at most $2$
for all sufficiently large $k$. Apply change of variables on the
countably many pieces of \Cref{lem:switch-injective}. Their images
are disjoint by global injectivity. Summing therefore gives
\[
\Phi_*(\rho_k|_{G_k})\le 2\rho_k.
\]
The factor $1/\nu(Q')$ is unchanged. Restriction to endpoint-label
pieces or to $G_k$ is never followed by normalization, and zero-volume
cells contribute nothing. Taking first marginals proves
\eqref{eq:switch-law}, because the first marginal of $\rho_k$ is
$\nu$. The construction already handled the optional smooth boundary.
\end{proof}
==== END A-finite-entropy-system-without-a-smooth-positive-volume-model//04-switching.tex ====
==== BEGIN A-finite-entropy-system-without-a-smooth-positive-volume-model//05-words.tex ====
\section{Words with prescribed relations and recognizable boundaries}
\label{sec:words}
We now construct finite families of words that will define the symbolic
system. At each level, short lists of constituent labels have large
entropy, although specified pairs of labels satisfy a bilinear relation.
We also require that a word can be recognized, together with its alignment,
from a sufficiently accurate copy. The construction in this section is
entirely finite at each step; no invariant probability is chosen yet.
Nested-word constructions and probabilistic control of frequencies and
shifted comparisons have precedents in
\cite[\S4.1 and Proposition~43]{ForemanRudolphWeiss2011}.
Related word and phase frameworks appear in
\cite[\S\S4.2.3--4.2.4]{ForemanWeiss2022}. Our graph constraints and the
per-word condition below are proved separately; in particular, we do
not require every lower-level word to occur in every parent or import
unique ergodicity from those constructions.
All positions in a finite word are numbered from zero. For a word $u$,
write $u[t,t+q)$ for its consecutive subword of length $q$ starting at $t$.
For two words of the same length, $d_{\mathrm H}(u,v)$ denotes the number
of coordinates at which they differ. Thus this distance counts symbols
or constituent labels according to the alphabet being used. For
probabilities $p,q$ on a finite set, we use
\[
\TV(p,q)=\frac12\sum_a |p(a)-q(a)|.
\]
Write $\mathcal L(Y)$ for the distribution of a random variable $Y$.
\subsection{Parameters and the four word properties}
Each new word will concatenate words from the preceding family.
We control two sampling experiments: choosing a word uniformly from
its family, and fixing a parent at the next level and choosing one of
its constituents uniformly. We compare the laws of windows at the
same internal offset in these two chosen words. Requiring this
approximation for every parent will allow arbitrary parent weights
in the eventual invariant measure.
The construction first samples constituent lists satisfying the
relations inside independent groups. It then keeps lists satisfying
the per-parent approximation, and finally selects a family with the
required window distributions and distances. The parameters below
provide enough groups and candidate words for these selections.
Fix
\begin{equation}\label{eq:word-parameters}
\varepsilon_i=2^{-i-2},\qquad
L_i=2^{2s_i}-1,\qquad \ell_i=\log L_i,\qquad
K_i=\lfloor\sqrt{\ell_i}\rfloor,
\end{equation}
and put
\begin{equation}\label{eq:word-lengths}
N_i=2^{K_i},\qquad H_i=2^{\lfloor K_i/2\rfloor},\qquad
D_i^*=2^{\lfloor K_i/8\rfloor},\qquad
m_1=1,\qquad m_{i+1}=m_iN_i.
\end{equation}
Starting from the integer $s_1$, define $2s_{i+1}$ to be the greatest even
integer not exceeding
\begin{equation}\label{eq:word-recursion}
X_i:=\frac{\varepsilon_i}{100}N_i\ell_i.
\end{equation}
The proof below shows that this recursion is well defined at every level
if $K_1\ge256$. There is no need to optimize this initial size.
Let $\mathcal A_i=\F_2^{2s_i}\setminus\{0\}$. On the full vector space
use the nondegenerate alternating form
\[
[u,v]_i=\sum_{a=1}^{s_i}
\bigl(u_a v_{s_i+a}+u_{s_i+a}v_a\bigr)\in\F_2.
\]
The fixed symbol alphabet is $\mathcal A=\mathcal A_1$.
Each family $\mathcal W_i\subset\mathcal A^{m_i}$ will have $L_i$ members.
Once it has been constructed, choose a bijection with $\mathcal A_i$;
this equips its word labels with the form $[\cdot,\cdot]_i$.
In particular, a word label and its vector label are identified only
through this fixed bijection.
Whenever $w$ is a concatenation of $N_i$ members of $\mathcal W_i$, let
\[
c_i(w)=(u_0,\ldots,u_{N_i-1})\in\mathcal W_i^{N_i}
\]
be its list of constituents. The list is unambiguous because all members
of $\mathcal W_i$ have the same length and are distinct.
Here $m_i$ is the symbol length, $L_i$ counts distinct level-$i$
words, and $N_i$ counts constituent occurrences in a level-$(i+1)$
word, with repetitions allowed. In the properties below, $H_i$ is
the group length and $D_i^*$ bounds the window lengths and prescribed
power-of-two lags; these lengths and lags are measured in level-$i$
constituent positions.
We fix the approximation tolerance
\begin{equation}\label{eq:word-tv-tolerance}
\vartheta=10^{-4}.
\end{equation}
\begin{proposition}[Finite word families]\label{prop:words}
Choose $s_1$ so that $K_1\ge256$, and define the parameters by
\eqref{eq:word-parameters}--\eqref{eq:word-recursion}.
There are families $\mathcal W_i\subset\mathcal A^{m_i}$ of cardinality
$L_i$, with $\mathcal W_1=\mathcal A$, such that every
$w\in\mathcal W_{i+1}$ is a concatenation of $N_i$ members of
$\mathcal W_i$ and the following properties hold.
\begin{enumerate}[label=\textup{(W\arabic*)},ref=W\arabic*,leftmargin=*]
\item\label{word:graph}\textbf{Relations within groups.}
Write $c_i(w)=(u_0,\ldots,u_{N_i-1})$.
For every $0\le b<N_i/H_i$, every
$0\le k\le\lfloor K_i/8\rfloor$, and every
$0\le r<r+2^k<H_i$,
\[
[u_{bH_i+r},u_{bH_i+r+2^k}]_i=0.
\]
\item\label{word:distance}\textbf{Distance between words and shifted words.}
If $v,w\in\mathcal W_{i+1}$ are distinct, then
\[
d_{\mathrm H}\bigl(c_i(v),c_i(w)\bigr)
\ge(1-\varepsilon_i)N_i.
\]
For all $v,w,z\in\mathcal W_{i+1}$, with repetitions allowed, and all
$1\le t<N_i$,
\[
d_{\mathrm H}\bigl((c_i(v)c_i(w))[t,t+N_i),c_i(z)\bigr)
\ge(1-\varepsilon_i)N_i.
\]
\item\label{word:window}\textbf{Distributions of short windows.}
Let $U$ be uniform on $\mathcal W_{i+1}$.
For each $0\le t\le N_i-D_i^*$ there is a probability $Q_{i,t}$ on
$\mathcal W_i^{D_i^*}$ such that
\[
\TV\bigl(\mathcal L(c_i(U)[t,t+D_i^*)),Q_{i,t}\bigr)
\le\vartheta.
\]
Moreover, for $1\le d'\le D_i^*$, any distinct positions
$0\le r_1<\cdots<r_{d'}<D_i^*$, and any prescribed labels
$v_1,\ldots,v_{d'}\in\mathcal W_i$,
\begin{equation}\label{eq:word-sublist}
Q_{i,t}\{u:u_{r_j}=v_j\text{ for }1\le j\le d'\}
\le 2\left(\frac{2^{2K_i}}{L_i}\right)^{d'}.
\end{equation}
\item\label{word:frequency}\textbf{Approximation in each individual word.}
Suppose $i\ge2$. For every fixed $w\in\mathcal W_{i+1}$, write
$c_i(w)=(u_0,\ldots,u_{N_i-1})$ and let $J$ be uniform on
$\{0,\ldots,N_i-1\}$. If $U$ is uniform on $\mathcal W_i$, then for
each $0\le t\le N_{i-1}-D_{i-1}^*$,
\[
\TV\left(
\mathcal L\bigl(c_{i-1}(u_J)[t,t+D_{i-1}^*)\bigr),
\mathcal L\bigl(c_{i-1}(U)[t,t+D_{i-1}^*)\bigr)
\right)\le\vartheta.
\]
The first law averages the constituents of the single fixed word $w$;
it does not average over $\mathcal W_{i+1}$.
\end{enumerate}
\end{proposition}
Property~(\ref{word:frequency}) compares the two sampling experiments
described above. The following elementary estimate controls empirical
laws both within each parent and across the selected family.
\begin{lemma}[Averages of probability vectors]\label{lem:word-tv}
Let $Y_1,\ldots,Y_q$ be independent random probability vectors on a set
of cardinality $A$, all with mean $p$. Then
\[
\E\,\TV\left(\frac1q\sum_{b=1}^qY_b,p\right)
\le\frac12\sqrt{\frac Aq}.
\]
Coordinates within an individual vector need not be independent.
\end{lemma}
\begin{proof}
For $\widehat p=q^{-1}\sum_bY_b$, independence between vectors gives
\[
\sum_a\operatorname{Var}(\widehat p(a))
=\frac1{q^2}\sum_{b,a}\operatorname{Var}(Y_b(a))
\le\frac1{q^2}\sum_b\E\sum_aY_b(a)^2
\le\frac1q.
\]
Apply the inequality $\E|Z|\le(\E|Z|^2)^{1/2}$ coordinatewise and then
Cauchy--Schwarz to the sum over the $A$ coordinates.
\end{proof}
\begin{proof}[Proof of \Cref{prop:words}]
\emph{Growth of the fixed parameters.}
Whenever $X_i\ge6$, rounding in \eqref{eq:word-recursion} gives
\begin{equation}\label{eq:word-rounding}
X_i-3<\ell_{i+1}\le X_i,\qquad
\ell_{i+1}\ge\frac{X_i}{2}
=\frac{\varepsilon_i}{200}2^{K_i}\ell_i.
\end{equation}
Indeed, $X_i-2<2s_{i+1}\le X_i$, and
$2s_{i+1}-1<\log(2^{2s_{i+1}}-1)<2s_{i+1}$ when $2s_{i+1}\ge2$.
We claim inductively that
\begin{equation}\label{eq:word-growth}
K_i\ge256,\qquad
\varepsilon_iK_i\ge32\,2^{i-1},\qquad
K_{i+1}\ge2^{K_i/3}\ge K_i^2\ge4K_i.
\end{equation}
The first two inequalities hold at $i=1$. If they hold at level $i$,
then $\ell_i\ge K_i^2$ implies
\[
X_i\ge\frac8{25}K_i2^{K_i}\ge6,
\qquad
\ell_{i+1}\ge\frac4{25}K_i2^{K_i}\ge2^{K_i}.
\]
Consequently
$K_{i+1}\ge2^{K_i/2}-1\ge2^{K_i/3}\ge K_i^2\ge4K_i$.
These elementary inequalities hold for every $K_i\ge256$;
for example, $K/3-2\log K$ is increasing there and is positive at $256$.
Since $\varepsilon_{i+1}=\varepsilon_i/2$, the second bound in
\eqref{eq:word-growth} propagates as well. In particular, all parameters
are positive and defined, $H_i$ divides $N_i$, and $D_i^*\le H_i$.
\smallskip\noindent
\emph{Sampling the relations.}
Suppose that $\mathcal W_i$ and its vector labeling have been fixed.
Generate a list $(U_0,\ldots,U_{N_i-1})$ in independent consecutive
groups of $H_i$ positions. Within each group, proceed from left to
right. Choose each new label uniformly among the nonzero vectors
orthogonal to all earlier labels at the lags
$2^k$, $0\le k\le\lfloor K_i/8\rfloor$, that stay inside the group. There are at most $\lfloor K_i/8\rfloor+1\le K_i$
linear equations. Their common kernel has codimension at most $K_i$,
whether or not the equations are independent. As $2s_i\ge\ell_i\ge K_i^2$,
the number of nonzero choices is at least
\begin{equation}\label{eq:word-choices}
2^{2s_i-K_i}-1
\ge2^{2s_i-K_i-1}
\ge\frac{L_i}{2^{2K_i}}.
\end{equation}
Denote this unconditioned law on $\mathcal W_i^{N_i}$ by $P_i^0$.
Every sampled list satisfies the relations in Property~(\ref{word:graph}).
For later use, set $\zeta_i=2^{2K_i}/L_i$ within this proof.
The conditional probability of any specified label at any position,
given all preceding positions, is at most $\zeta_i$.
If $r_1<\cdots<r_{d'}$ are specified positions, conditioning first at
$r_{d'}$ and then successively at the earlier specified positions gives
\begin{equation}\label{eq:word-raw-sublist}
P_i^0\{U_{r_j}=v_j\text{ for all }j\}\le\zeta_i^{d'}.
\end{equation}
All unspecified labels are integrated out. This argument does not
require independence within a group.
Each individual $U_r$ nevertheless has a uniform marginal on
$\mathcal W_i$. To see this, the group of linear maps preserving the
alternating form acts transitively on its nonzero vectors. Given such
a vector $v$, nondegeneracy provides $w$ with $[v,w]_i=1$; the plane
spanned by $v,w$ is nondegenerate, and its orthogonal complement carries
a nondegenerate alternating form of dimension two less. Induction
extends $v,w$ to a symplectic basis, proving transitivity by mapping one
such basis to another. The sequential rule is equivariant under these
maps, including its uniform choices, so each marginal is invariant
under a transitive action and hence uniform.
\smallskip\noindent
\emph{Imposing the approximation in each word.}
For $i\ge2$ and an offset $0\le t\le N_{i-1}-D_{i-1}^*$, let
\[
f_t(u)=c_{i-1}(u)[t,t+D_{i-1}^*),\qquad u\in\mathcal W_i,
\]
and let $p_t$ be its law under uniform choice of $u$.
For each of the $q_i=N_i/H_i$ independently sampled groups, form the
empirical probability vector of the $H_i$ values $f_t(U_r)$ in that
group. Its mean is $p_t$ by the uniform one-label marginals just proved.
The average of these independent vectors is exactly the law of
$f_t(U_J)$ when $J$ is uniform over the entire sampled list.
The alphabet of $f_t$ has cardinality at most
$A_i=L_{i-1}^{D_{i-1}^*}$.
Thus \Cref{lem:word-tv}, Markov's inequality, and a union over at most
$N_{i-1}$ offsets bound the probability of violating any requirement in
Property~(\ref{word:frequency}) by
\begin{equation}\label{eq:word-filter-budget}
F_i:=\vartheta^{-1}N_{i-1}
\sqrt{\frac{L_{i-1}^{D_{i-1}^*}}{N_i/H_i}}.
\end{equation}
Below we verify $F_i<1/2$ at every level $i\ge2$.
Let $\mathcal E_i$ be the event that all these requirements hold, and
let $P_i=P_i^0(\,\cdot\mid\mathcal E_i)$.
For $i=1$, take $\mathcal E_1$ to be the whole sample space and
$P_1=P_1^0$. In all cases $P_i^0(\mathcal E_i)\ge1/2$, so
\eqref{eq:word-raw-sublist} yields
\begin{equation}\label{eq:word-filtered-sublist}
P_i\{U_{r_j}=v_j\text{ for all }j\}\le2\zeta_i^{d'}.
\end{equation}
The factor two is paid once for the whole conditioning event.
No uniform marginal or independence between groups is asserted for
$P_i$. The relations hold on its support, as does
Property~(\ref{word:frequency}) for each individual list.
\smallskip\noindent
\emph{Selecting the finite family.}
Put $M_i=L_{i+1}$ and sample $M_i$ independent complete lists with law
$P_i$. For each offset $0\le t\le N_i-D_i^*$, let $Q_{i,t}$ be the
law under $P_i$ of the length-$D_i^*$ window at $t$.
It satisfies \eqref{eq:word-sublist} by
\eqref{eq:word-filtered-sublist}.
The empirical window law of the $M_i$ samples is an average of
independent point-mass probability vectors on an alphabet of size
$L_i^{D_i^*}$. A second application of \Cref{lem:word-tv} and Markov's
inequality, now with a union over at most $N_i$ offsets, bounds the
probability of violating any required window approximation by
\begin{equation}\label{eq:word-window-budget}
E_i:=\vartheta^{-1}N_i
\sqrt{\frac{L_i^{D_i^*}}{L_{i+1}}}.
\end{equation}
We will have $E_i<1/4$ at every level.
It remains to impose the distances, including comparisons in which
some of the sampled-word indices coincide. First use the reference
law in which the $N_i$ labels in each distinct indexed list are
independent and uniform. Consider one pairwise comparison between
distinct indices, or one comparison of a target list with a shifted
window from two source lists. In the latter case write $t$ for its
offset, with $1\le t<N_i$; any of the three indices may coincide.
For a specified set of matching positions, form a multigraph whose
vertices are variables (sample index, position) and whose edges are
the asserted equalities. Every target position occurs once. The
source positions also occur once: they use $t,\ldots,N_i-1$ in the
first source and $0,\ldots,t-1$ in the second, disjoint ranges even if
the source indices agree. The degree is therefore at most two,
counting repeated edges. There is no loop. In a pairwise comparison
the indices are distinct, and in a shifted comparison the source
position differs from the target position by $t$ modulo $N_i$.
A connected component with $v$ vertices requires $v-1$ independent
equalities and has probability $L_i^{-(v-1)}$ under the reference law.
A path has one fewer edge than vertices; a cycle has equally many,
with at least two edges. Thus every component imposes at least half
as many independent equalities as edges, including a pair of parallel
edges. If $e$ positions are specified to match, their probability is
at most $L_i^{-e/2}$.
Failure of a distance requirement entails at least
$\lceil\varepsilon_iN_i\rceil$ matching positions. Summing over their
possible sets, of which there are at most $2^{N_i}$, gives the bound
\begin{equation}\label{eq:word-reference-distance}
2^{N_i}L_i^{-\varepsilon_iN_i/2}.
\end{equation}
By \eqref{eq:word-choices}, a complete list under $P_i^0$ has likelihood
ratio at most $2^{2K_iN_i}$ relative to the reference law, and under
$P_i$ at most $2\,2^{2K_iN_i}$.
Each comparison uses at most three distinct sampled lists, so its
probability under the actual sampling law is at most
$(2\,2^{2K_iN_i})^3$ times \eqref{eq:word-reference-distance}.
Repeated indices do not introduce an additional independent list.
There are fewer than $4N_iM_i^3$ pairwise and shifted comparisons in
total. Their union failure probability is consequently at most
\begin{equation}\label{eq:word-distance-budget}
T_i:=32N_iL_{i+1}^3\,
2^{(6K_i+1)N_i}L_i^{-\varepsilon_iN_i/2}.
\end{equation}
We will verify $T_i<1/4$.
These estimates address separate requirements: the first sampling law
enforces the relations, conditioning enforces the approximation in
each individual word, and the final independent samples provide the
window laws and distances. We now check that the fixed initial size
makes all three probability bounds hold at every step.
\smallskip\noindent
\emph{Uniform bounds from one initial choice.}
For $K\ge256$, set
\[
B_0(K)=14+K+\tfrac12\,2^{K/8}(K+1)^2.
\]
The elementary inequalities
$14+K\le2^{K/8}$ and $(K+1)^2\le2^{K/8}$ hold at $256$ and persist
thereafter, as the ratios of their left sides to their right sides
decrease. Hence
\begin{equation}\label{eq:word-common-bound}
B_0(K)\le2^{K/4}\le\tfrac18\,2^{K/3}.
\end{equation}
We also have $\log(\vartheta^{-1})<14$ and
$K_i^2\le\ell_i<(K_i+1)^2$.
For the first probability bound, put $K=K_{i-1}$ and use
$\log(N_i/H_i)\ge K_i/2$ to obtain
\[
\log F_i
\le B_0(K)-\frac{K_i}{4}
\le-\frac18\,2^{K/3}.
\]
For the second, put $K=K_i$. Since $\ell_{i+1}\ge2^K$,
\[
\log E_i
\le B_0(K)-\frac{\ell_{i+1}}2
\le2^{K/4}-\frac{2^K}{2}
\le-\frac{2^K}{4}.
\]
Finally \eqref{eq:word-rounding} and
$\varepsilon_i\ell_i\ge\varepsilon_iK_i^2\ge32K_i$ give
\begin{align*}
\log T_i
&\le5+K_i+N_i
\left(6K_i+1-\frac{47}{100}\varepsilon_i\ell_i\right)\\
&\le5+K_i-9K_i2^{K_i}
\le-K_i2^{K_i}.
\end{align*}
It follows that
\begin{equation}\label{eq:word-uniform-budgets}
F_i\le2^{-2^{K_{i-1}/3}/8}<\tfrac12\quad(i\ge2),\qquad
E_i\le2^{-2^{K_i}/4}<\tfrac14,\qquad
T_i\le2^{-K_i2^{K_i}}<\tfrac14.
\end{equation}
All right sides are decreasing functions of their $K$ argument.
They apply simultaneously at every level by \eqref{eq:word-growth};
no later adjustment of $s_1$ is made.
The conditioning event thus has positive probability. Under the
conditioned law the two subsequent failures have total probability
less than $1/2$, so a tuple satisfying every requirement exists.
Its pairwise distance requirement is imposed on distinct sample
indices and has a positive lower bound. Equal sampled words would
have distance zero, so the tuple has exactly $L_{i+1}$ distinct
members. Flatten their constituent lists to form $\mathcal W_{i+1}$.
The empirical distribution is now exactly the uniform distribution
on this set, without deduplication or a change of weights.
It therefore has all four asserted properties. Choose its bijection
with $\mathcal A_{i+1}$ and continue the induction.
\end{proof}
\subsection{Recognition at every symbol offset}
The distances in \Cref{prop:words} concern constituent labels.
The next consequence converts them into distances in the fixed
alphabet $\mathcal A$, including shifts that do not respect any
lower-level boundary. In coding terminology the two conclusions below
are a positive minimum distance and a positive \emph{comma-free index}:
the latter measures distance from every nontrivial splice of two legal
words. This is the synchronization notion described in
\cite[\S1]{Tonchev2003}; the recursive lower bound here is proved directly.
\begin{lemma}[Recognition of words and alignment]\label{lem:recognition}
For the families in \Cref{prop:words}, put
\[
d_i=\prod_{j=1}^{i-1}(1-\varepsilon_j),\qquad
d_*=\prod_{j=1}^{\infty}(1-\varepsilon_j)>0.
\]
Distinct $v,w\in\mathcal W_i$ satisfy
\[
d_{\mathrm H}(v,w)\ge d_i m_i\ge d_*m_i.
\]
For all $v,w,z\in\mathcal W_i$ and every $1\le t<m_i$,
\[
d_{\mathrm H}\bigl((vw)[t,t+m_i),z\bigr)
\ge d_i m_i\ge d_*m_i.
\]
These distances count symbols in $\mathcal A$, and repeated source
words are allowed.
\end{lemma}
\begin{proof}
The finite products are bounded below by
$1-\sum_{j\ge1}\varepsilon_j=3/4$, proving that $d_*>0$.
At level one, distinct one-symbol words have distance one and the
offset assertion is vacuous.
Suppose both assertions hold at level $i$.
Distinct level-$(i+1)$ words differ in at least
$(1-\varepsilon_i)N_i$ constituent labels by
\Cref{prop:words}, Property~(\ref{word:distance}).
Each unequal constituent pair contributes at least $d_i m_i$
symbol mismatches, giving the bound $d_{i+1}m_{i+1}$.
Now consider a nonzero offset $t<m_{i+1}$ in two concatenated
level-$(i+1)$ words, compared with a third such word.
If $m_i$ divides $t$, the offset $t/m_i$ lies between $1$ and $N_i-1$.
The constituent-label offset requirement gives at least
$(1-\varepsilon_i)N_i$ unequal pairs, and the preceding distinct-word
bound again gives $d_{i+1}m_{i+1}$ symbol mismatches.
If $m_i$ does not divide $t$, each of the $N_i$ aligned target
constituents faces a length-$m_i$ source window with the same nonzero
offset modulo $m_i$. That source window crosses exactly two
consecutive legal level-$i$ words. The inductive offset assertion
therefore gives at least $d_i m_i$ mismatches on each target
constituent. It applies also when the two source constituents lie
on opposite sides of the join of the two larger source words.
Adding over the disjoint target constituents gives
$d_i m_{i+1}\ge d_{i+1}m_{i+1}$, completing the induction.
\end{proof}
\subsection{The number and size of the available scales}
The word construction must retain enough label information across
levels to compete with the summable smooth-switching costs.
The following estimates are the quantitative input for that comparison.
\begin{lemma}[Word rates]\label{lem:rates}
Let $R_i=\ell_i/m_i$. Every $m_i$ is a power of two, with
\[
\log m_i=\sum_{j<i}K_j.
\]
The rates satisfy
\begin{equation}\label{eq:word-rate-decay}
R_{i+1}\le\frac{\varepsilon_i}{100}R_i,
\qquad R_i\longrightarrow0.
\end{equation}
They also have the lower bound
\begin{equation}\label{eq:word-rate-lower}
R_i\ge R_1\,200^{-(i-1)}
2^{-(i-1)(i+4)/2}
\ge R_1\,2^{-(i-1)(i+20)/2},
\end{equation}
and
\begin{equation}\label{eq:word-rate-divergence}
\sum_{i=1}^{\infty}K_iR_i^{16}=\infty.
\end{equation}
\end{lemma}
\begin{proof}
The formula for $m_i$ follows from $m_1=1$ and $N_i=2^{K_i}$.
Dividing \eqref{eq:word-rounding} by $m_{i+1}=N_i m_i$ gives
\[
\frac{\varepsilon_i}{200}R_i
\le R_{i+1}\le\frac{\varepsilon_i}{100}R_i.
\]
Since $\varepsilon_i\le1/8$, the upper bound implies
$R_i\le800^{-(i-1)}R_1\to0$, proving
\eqref{eq:word-rate-decay}. Iterating the lower bound and using
$\sum_{j<i}(j+2)=(i-1)(i+4)/2$ prove the first inequality in
\eqref{eq:word-rate-lower}; the second uses $\log200<8$.
Also \eqref{eq:word-growth} implies
$K_i\ge K_1^{2^{i-1}}$. Hence
\[
\log(K_iR_i^{16})
\ge 2^{i-1}\log K_1+16\log R_1-8(i-1)(i+20)
\longrightarrow\infty.
\]
In particular the terms in \eqref{eq:word-rate-divergence} do not
tend to zero, and the series diverges.
\end{proof}
The families are now fixed. Their per-word approximation transfers
short-window entropy to any invariant measure on the resulting
sequence space, while \Cref{lem:recognition} recovers block labels
and phases from small symbol errors. We next construct that space
and prove the required measure-theoretic statements.
==== END A-finite-entropy-system-without-a-smooth-positive-volume-model//05-words.tex ====
==== BEGIN A-finite-entropy-system-without-a-smooth-positive-volume-model//06-process.tex ====
\section{An invariant process and its window laws}\label{sec:process}
We now turn the finite word families of \Cref{prop:words} into one
ergodic system. The probability on that system need not assign
uniform frequencies to the different words. The per-word frequency
condition in the construction is what will make the window estimates
valid for every invariant probability.
\subsection{The compact phase extension}
Let $\mathcal A=\mathcal W_1$, regarded as an alphabet, and let
\[
\Omega=\mathcal A^{\Z}\times\prod_{i\ge1}\Z/m_i\Z
\]
with its compact product topology. Write a point as $(\omega,a)$,
where $\omega=(\omega_t)_{t\in\Z}$ and
$a=(a_i)_{i\ge1}$. Define $X\subset\Omega$ by the following conditions:
\begin{enumerate}[label=(\roman*)]
\item $a_{i+1}\equiv a_i\pmod{m_i}$ for every $i$;
\item whenever $t\equiv-a_i\pmod{m_i}$, the word
$(\omega_t,\ldots,\omega_{t+m_i-1})$ belongs to $\mathcal W_i$.
\end{enumerate}
The coordinate $a_i\in\{0,\ldots,m_i-1\}$ records the index of the
origin in its level-$i$ word. Let
\[
T(\omega,a)=(\omega',a'),\qquad
\omega'_t=\omega_{t+1},\quad a'_i=a_i+1\pmod{m_i}.
\]
The defining conditions are closed, and $T$ is a homeomorphism of $X$.
For each $j$, repeat a word of $\mathcal W_j$ bi-infinitely with its
level-$j$ boundary at zero. Since higher-level words are concatenations
of lower-level ones, this sequence with zero phases satisfies all
conditions through level $j$. The finite-level closed sets are nested
and nonempty. Compactness proves that $X$ is nonempty.
The invariant-measure facts used here are the standard compact-system
ones; see \cite[Theorems~6.1.4--6.1.6]{DajaniDirksin2008}.
We give the construction together with the checks specific to the
phase extension.
\begin{proposition}\label{prop:system}
There is a $T$-invariant ergodic Borel probability $\mu$ on $X$.
Every $T$-invariant Borel probability on $X$ is nonatomic, is standard
modulo null sets, and has zero Kolmogorov--Sinai entropy.
\end{proposition}
\begin{proof}
For any $x\in X$, weak limits of
$n^{-1}\sum_{j=0}^{n-1}\delta_{T^jx}$ are invariant probabilities.
Indeed, the difference between the integrals of $f\circ T$ and $f$
against these averages tends to zero for each continuous $f$.
The set $\mathcal M_T(X)$ of invariant probabilities is nonempty,
compact, and convex.
Here is an explicit way to select an extreme point. Choose a
countable uniformly dense family $(f_j)$ in the continuous real
functions on $X$. Starting from $\mathcal M_T(X)$, successively
restrict to the compact face on which $\int f_j\,d\mu$ is maximal.
The nested intersection is nonempty and contains only one measure,
because the integrals of every $f_j$ are then fixed and determine a
probability. An intersection of faces is a face, so this measure is
extreme. If it had a $T$-invariant measurable set of mass strictly
between zero and one, normalized restriction to that set and its
complement would express it as a nontrivial convex combination of
invariant probabilities. Thus it is ergodic. The same argument applies
to invariant sets modulo null sets after taking Borel representatives.
For any invariant probability, $a_i$ is uniform on $\Z/m_i\Z$,
because $T$ cyclically permutes its level sets. If a measurable atom
had mass $c>0$, then for each $i$ it would be contained, modulo null
sets, in one of these level sets. Hence $c\le1/m_i$ for every $i$,
which is impossible since $m_i\to\infty$.
The space $X$ is a compact metrizable subspace of a countable product
of finite discrete sets. For completeness, its nonatomic
probability is standard in the precise sense needed here. Encode the
finite coordinates into successive disjoint blocks of ternary
digits $0,2$; this gives a continuous injective map of $X$ into a
compact subset of $[0,1]$. Let $\lambda$ be the image law and
$F(t)=\lambda([0,t])$ its continuous distribution function. Define
the measurable quantile
\[
Q(u)=\inf\{t:F(t)\ge u\},\qquad 0<u<1.
\]
Continuity gives $F(Q(u))=u$, and
$\{u:Q(u)\le t\}$ has Lebesgue measure $F(t)$, so
$Q_*\operatorname{Leb}=\lambda$ and $F_*\lambda=\operatorname{Leb}$.
Moreover $Q(F(x))\le x$ for $\lambda$-almost every $x$;
both sides have law $\lambda$. Their nonnegative difference has zero
integral, and therefore $Q(F(x))=x$ almost surely.
After restricting to measurable conull sets, $F$ and $Q$ are
inverse isomorphisms. Composing with the coordinate embedding
establishes the assertion. Passing to completed sigma fields
does not change it.
It remains to check the entropy of the entire phase extension.
A partition depending on finitely many coordinates is refined by
one that records $\omega_{-r},\ldots,\omega_r$ and $a_I$ for
some fixed $r,I$. Compatibility determines all the lower phases
from $a_I$, and the phase evolves deterministically. Fix any $j\ge I$.
The initial phase $a_j$ determines $a_I$ and the boundaries of all
level-$j$ words. The symbol interval $[-r,n-1+r]$ used in an $n$-step
name meets at most $\lceil(n+2r)/m_j\rceil+1$ such words, each in
$\mathcal W_j$. Consequently the number of names is at most
\[
m_j L_j^{\lceil(n+2r)/m_j\rceil+1}.
\]
Keeping $j$ fixed and sending $n\to\infty$ bounds the entropy rate
by $R_j=(\log L_j)/m_j$. By \Cref{lem:rates}, $R_j\to0$.
Only now let $j\to\infty$: every finite-coordinate partition has
entropy rate zero. This count uses no assumption about word frequencies.
Let $\mathcal P$ be any finite measurable partition with $A$ atoms.
Cylinder sets generate the Borel sigma field, and their algebra is
dense in measure. Thus the atom label of $\mathcal P$ can be
approximated by an $A$-valued finite-coordinate label
$\mathcal Q$ with mismatch probability $s$ arbitrarily small.
Completion of the measure only adds null-set changes.
The chain rule and conditioning give
\[
H(\mathcal P^n)
\le H(\mathcal Q^n)
+\sum_{t=0}^{n-1}H(T^{-t}\mathcal P\mid T^{-t}\mathcal Q)
=H(\mathcal Q^n)+nH(\mathcal P\mid\mathcal Q),
\]
where $\mathcal P^n=\bigvee_{t=0}^{n-1}T^{-t}\mathcal P$.
Recording first whether the two labels differ shows that
\[
H(\mathcal P\mid\mathcal Q)\le H_{\rm bin}(s)+s\log A.
\]
Entropy rates exist by subadditivity. Divide by $n$, take the limit,
and then let $s\downarrow0$. This proves $h_\mu(T)=0$.
\end{proof}
Fix henceforth an ergodic probability supplied by
\Cref{prop:system}. The next estimates use only invariance, so their
validity is unaffected by this choice.
\subsection{What alignment does to the sampling law}
For $x=(\omega,a)\in X$, let $V_i(x)\in\mathcal W_i$ be the word
containing the origin, namely the symbol string beginning at
$-a_i(x)$. Set
\[
A_i=\{x:a_i(x)=0\},\qquad
\mu_i=m_i\,\mu|_{A_i}.
\]
The law $\mu_i$ is a probability. It describes a point observed at a
level-$i$ boundary; it is not asserted to give uniform word
frequencies.
For an integer $t$, let
\[
r_i(x,t)=(-a_i(T^t x))\bmod m_i\in\{0,\ldots,m_i-1\}.
\]
The first level-$i$ boundary at or after $t$ is $t+r_i(x,t)$.
For an integer $D\ge2$, define the vector
\begin{equation}\label{eq:aligned-list}
\mathcal L_{i,D}(x,t)=
\bigl(V_i(T^{\,t+r_i(x,t)+jm_i}x)\bigr)_{j=0}^{D-2}.
\end{equation}
All its words lie in $[t,t+Dm_i)$, since
$r_i+(D-1)m_i\le Dm_i$.
\begin{lemma}\label{lem:aligned-law}
The law of \eqref{eq:aligned-list} under $\mu$ is the law under
$\mu_i$ of the first $D-1$ level-$i$ words beginning at zero.
Under $\mu_i$, the index
$J=a_{i+2}/m_i$ is uniform on
$\{0,\ldots,N_iN_{i+1}-1\}$ and is independent of $V_{i+2}$.
More precisely, for each such $j$ and $v\in\mathcal W_{i+2}$,
\begin{equation}\label{eq:parent-slot-law}
\mu_i(J=j,V_{i+2}=v)
=\frac{\mu_{i+2}(V_{i+2}=v)}{N_iN_{i+1}}.
\end{equation}
\end{lemma}
\begin{proof}
Invariance first allows us to take $t=0$. On the phase class
$a_i=r$, the map $T^{(-r)\bmod m_i}$ sends the normalized
restriction of $\mu$ bijectively and measure-preservingly onto
$\mu_i$. The first assertion follows by applying this separately
on all $m_i$ phase classes.
For the second assertion, $T^{jm_i}$ maps
$\{a_{i+2}=0,V_{i+2}=v\}$ onto
$\{a_{i+2}=jm_i,V_{i+2}=v\}$.
It does not change the containing parent word because
$0\le jm_i<m_{i+2}$. These sets have the same $\mu$-mass.
Multiply by $m_i$, and use
$m_{i+2}/m_i=N_iN_{i+1}$, to obtain
\eqref{eq:parent-slot-law}.
\end{proof}
The identity is the reason that a deterministic condition on every
parent word suffices. At a fixed internal offset in a parent,
the slot being sampled is uniform even when the parent itself is
not uniformly distributed.
\subsection{Entropy of short label lists}
We use two elementary entropy bounds. The second is the standard
maximal-coupling and mismatch-flag entropy argument; compare
\cite[Theorems~1--3]{Sason2013}. We include the weaker form needed here. If a law on a finite set
has every atom at most $M$, its entropy is at least
$\log(1/M)$. If two laws $P,Q$ on an $A$-point set have
$\TV(P,Q)\le s$, then
\begin{equation}\label{eq:tv-entropy}
H(P)\ge H(Q)-1-s\log A.
\end{equation}
Indeed, couple their labels with mismatch probability at most $s$.
The mismatch flag and the chain rule bound either conditional
entropy by $1+s\log A$. We use this simple bound rather than an
optimized continuity estimate.
\begin{proposition}\label{prop:window-entropy}
For all sufficiently large $i$, for every power of two
$2\le D\le D_i^*$, and every integer $t$,
\begin{equation}\label{eq:window-entropy}
H_\mu(\mathcal L_{i,D}(\,\cdot\,,t))
\ge .996(D-1)\ell_i,\qquad \ell_i=\log L_i.
\end{equation}
The large-$i$ threshold is independent of $D,t$ and of the
invariant probability on $X$.
\end{proposition}
\begin{proof}
Write $q=D-1$. By \Cref{lem:aligned-law} we work under $\mu_i$.
Decompose its uniform parent-slot index as
\[
J=hN_i+u,\qquad 0\le h<N_{i+1},\quad 0\le u<N_i.
\]
Conditional on the parent $v=V_{i+2}$ and the internal offset $u$,
the slot $h$ is uniform among the $N_{i+1}$ constituent
$\mathcal W_{i+1}$ words of $v$.
Call $u$ admissible if $u+D_i^*\le N_i$.
Its failure probability is
\[
\beta_i=\frac{D_i^*-1}{N_i}\le\frac{D_i^*}{N_i}.
\]
For every fixed $v$ and admissible $u$,
\Cref{prop:words}, Property~(\ref{word:frequency}), applied at level $i+1$,
puts the law of the $D_i^*$ labels at this offset within
$10^{-4}$ of their law in a uniform word of $\mathcal W_{i+1}$.
\Cref{prop:words}, Property~(\ref{word:window}), applied at level $i$,
puts that law within a further $10^{-4}$ of a law whose
every specified $q$-coordinate atom is at most
\[
M_i=2\left(\frac{2^{2K_i}}{L_i}\right)^q.
\]
Restriction to the first $q$ entries cannot increase total
variation. The comparison law may depend on $u$, but this
atom bound does not.
It follows from \eqref{eq:tv-entropy}, with $s=2\cdot10^{-4}$
and alphabet size $L_i^q$, that the actual $q$-list conditional
on $v,u$ has entropy at least
\[
q\ell_i-2K_iq-1-(1+sq\ell_i)
=q\bigl((1-s)\ell_i-2K_i\bigr)-2.
\]
Entropy is concave. Mixing over all $v$ and admissible $u$
preserves this lower bound; adding the remaining offsets loses
at most their fraction, since entropy is nonnegative. Thus
\begin{equation}\label{eq:window-entropy-explicit}
H(\mathcal L_{i,D})
\ge(1-\beta_i)
\left\{q\bigl((1-s)\ell_i-2K_i\bigr)-2\right\}.
\end{equation}
Here $\beta_i\to0$, $2K_i/\ell_i\to0$, and $2/\ell_i\to0$.
Once each of these three quantities is at most $.0005$,
division by $q\ell_i$, with $q\ge1$, gives a coefficient
at least
\[
(1-.0005)(1-.0002-.0005-.0005)>.996.
\]
This proves the uniform assertion.
\end{proof}
Only the two indicated levels of the construction enter this
argument. There is no accumulation of total variation errors
over all ancestors, and no independence assumption on the
sequence of adjacent words.
\subsection{The graph relation across two windows}
Two level-$i$ labels $u,v\in\mathcal W_i$ pass the graph test
when $[u,v]_i=0$, using their fixed vector identification.
The same alignment law gives a quantitative bound on the one
exception to the graph test: crossing a group boundary.
\begin{lemma}\label{lem:graph-windows}
Let $2\le D\le D_i^*$ be a power of two and set $n=m_iD$.
For all sufficiently large $i$, with $\mu$-probability at least
\[
1-\alpha_{i,D},\qquad
\alpha_{i,D}=\frac{2D-2}{H_i},
\]
all corresponding entries of
$\mathcal L_{i,D}(x,-n)$ and
$\mathcal L_{i,D}(x,0)$ pass the level-$i$ graph test simultaneously.
In particular $\sup_{D\le D_i^*}\alpha_{i,D}\to0$.
\end{lemma}
\begin{proof}
Because $n$ is a multiple of $m_i$, the advance to alignment
is the same in the two windows. Denote it by $q_0$.
The entry starts are
\begin{equation}\label{eq:two-list-starts}
-n+q_0+jm_i
\quad\hbox{and}\quad
q_0+jm_i,\qquad 0\le j\le D-2.
\end{equation}
Corresponding entries are separated by exactly $D$ level-$i$
positions. The first start through the last start span
$2D-1$ such positions, including the unused intervening position.
\Cref{fig:lists} illustrates these positions and their common lag
when $D=4$.
By \Cref{lem:aligned-law}, the first boundary has the
boundary-start law. Its position modulo $H_i$ is uniform:
its position in a $\mathcal W_{i+1}$ parent is uniform modulo
$N_i$, and $H_i$ divides $N_i$. Once $2D-1\le H_i$, exactly
$2D-2$ residues have a span that crosses an $H_i$-group boundary.
All parent boundaries are already group boundaries.
On every other residue all the labels in \eqref{eq:two-list-starts}
belong to one of the prescribed groups.
\Cref{prop:words}, Property~(\ref{word:graph}), applies because $D$ is an
allowed power-of-two lag.
Finally $2D_i^*/H_i\to0$ by the definitions.
\end{proof}
\begin{figure}[tb]
\centering
\begin{tikzpicture}[x=.92cm,y=1cm,>=Stealth]
\foreach \j in {0,1,2,3,4,5,6}{
\draw (\j,0) rectangle ++(1,.55);
}
\foreach \j in {0,1,2}{
\node at (\j+.5,.275) {$U_{\j}$};
\pgfmathtruncatemacro{\v}{\j+4}
\node at (\v+.5,.275) {$V_{\j}$};
\draw[->] (\j+.5,.65) to[bend left=22] (\v+.5,.65);
}
\node at (3.5,.275) {$-$};
\draw[decorate,decoration={brace,mirror,amplitude=4pt}]
(0,-.12)--(3,-.12) node[midway,below=6pt] {past list};
\draw[decorate,decoration={brace,mirror,amplitude=4pt}]
(4,-.12)--(7,-.12) node[midway,below=6pt] {future list};
\node at (3.5,1.55) {each arrow has lag $D=4$};
\end{tikzpicture}
\caption{The two aligned lists of one legal symbolic point when $D=4$,
writing $U_j,V_j$ for
their corresponding past and future entries. Each has $D-1$ entries.
The single unused block between them does not alter the lag between
corresponding entries. Their span has $2D-1$ block positions, which
explains the group-boundary error in \Cref{lem:graph-windows}.}
\label{fig:lists}
\end{figure}
We now have a fixed ergodic system with zero entropy, uniformly
large list entropy, and a graph relation across adjacent windows.
These properties are compatible: the information per symbol in a
level-$i$ list is of order $R_i$, which tends to zero. The many
available test scales will compensate for this decreasing rate.
The remaining task is to compare that relation with the
conditional independence in a smooth switch.
==== END A-finite-entropy-system-without-a-smooth-positive-volume-model//06-process.tex ====
==== BEGIN A-finite-entropy-system-without-a-smooth-positive-volume-model//07-obstruction.tex ====
\section{The obstruction to a smooth volume model}\label{sec:obstruction}
Fix the ergodic system constructed in \Cref{sec:process}.
The final argument concerns a single statistic of a pair of points:
the fraction of corresponding past and future labels that pass the
graph test. We define that statistic before restricting the pair
experiment. Its upper and lower bounds will therefore refer to the
same probability law.
\subsection*{The quantitative interface}
The smooth and symbolic constructions meet through the quantities in
\Cref{tab:interfaces}. At $n=2^k$, write $\mathcal D_k$ for the smooth
partition, $\rho$ for its cell-weighted pair law, $G$ for the retained
pairs, and $z$ for the switch. The exceptional-mass tolerance $\eta>0$
is fixed before these objects are chosen. On the symbolic side, $L_i=\#\mathcal W_i$,
$R_i=(\log L_i)/m_i$, and $K_i=\lfloor\sqrt{\log L_i}\rfloor$.
Every $m_i$ is a power of two. The symbolic tests have length $n=m_iD$,
where $D$ is a power of two
and $2^{\lceil K_i/16\rceil}\le D\le2^{\lfloor K_i/8\rfloor}$.
In a window of $Dm_i$ symbols we use the $D-1$ full words beginning
at the first level-$i$ boundary. Each input initially uses its own
alignment. The compact system recording the symbol sequence and all
compatible alignment phases has zero entropy for every invariant
probability, by \Cref{prop:system}.
\begin{table}[htbp]
\centering\small
\renewcommand{\arraystretch}{1.18}
\begin{tabular}{@{}>{\raggedright\arraybackslash}p{.19\textwidth}>{\raggedright\arraybackslash}p{.37\textwidth}>{\raggedright\arraybackslash}p{.37\textwidth}@{}}
\toprule
Interface & Smooth conclusion & Symbolic use \\
\midrule
Dyadic cost &
$\sum_k B_k^{16}<\infty$ and
$\#\mathcal D_k\le\exp(CB_k2^k)$ at every sufficiently large scale &
$\sum_i K_iR_i^{16}=\infty$ gives tests at which cell conditioning
costs $o((D-1)\log L_i)$ bits. \\
Original pair law &
Choose a cell with its volume weight, then two independent normalized
volume points; retain mass at least $1-\eta$ &
Window entropy holds for the chosen invariant measure, with arbitrary
parent-word weights. Independence is used only for the two inputs
conditional on the cell and a fixed entry. \\
Shadowing &
The switch follows one input for $0\le t\le n$ and the other for
$-n\le t\le0$, with error tending uniformly to zero &
Positive symbol-distance recognition recovers the common phase and
most labels, preserving the original entry pairing. \\
Output measure &
$z_*(\rho|_G)\le2\nu$, without normalizing the retained law &
Compact symbol cores and the simultaneous graph-window law transfer
to the switched point with controlled exceptional mass. \\
\bottomrule
\end{tabular}
\caption{The exact internal interface between
\Cref{thm:switching} and the fixed symbolic system.
All smooth constants are chosen only after a model is proposed.
The final comparison retains the original pair law $\rho$.}
\label{tab:interfaces}
\end{table}
\subsection{Choosing times with a negligible partition cost}
\begin{lemma}\label{lem:cheap-scales}
Let $(B_k)_{k\ge1}$ be any positive sequence with
$\sum_k B_k^{16}<\infty$. There exist levels $i\to\infty$ and
powers of two $D$ satisfying
\[
2^{\lceil K_i/16\rceil}\le D\le2^{\lfloor K_i/8\rfloor}
\]
such that, for $n=m_iD$,
\begin{equation}\label{eq:relative-cost}
\frac{B_{\log n}}{R_i}\longrightarrow0,
\qquad R_i=\frac{\ell_i}{m_i}.
\end{equation}
In particular any partitions with
$\log\#\mathcal D_k\le C B_k2^k$ obey
\begin{equation}\label{eq:cheap-partition}
\log\#\mathcal D_{\log n}=o((D-1)\ell_i)
\end{equation}
on this sequence, for every fixed $C$.
\end{lemma}
\begin{proof}
The lengths $m_i$ are powers of two. Put $b_i'=\log m_i$,
so $b_{i+1}'=b_i'+K_i$. The integer sets
\[
J_i=\{b_i'+j:\lceil K_i/16\rceil\le j\le
\lfloor K_i/8\rfloor\}
\]
are disjoint and, for large $i$, contain at least $K_i/32$
integers. If the ratios in \eqref{eq:relative-cost} were bounded
below by some $c>0$ on all sufficiently late tests, then
\[
\sum_kB_k^{16}
\ge c^{16}\sum_{i\ {\rm large}}|J_i|R_i^{16}
\ge \frac{c^{16}}{32}
\sum_{i\ {\rm large}}K_iR_i^{16}=\infty,
\]
contrary to the hypothesis and \Cref{lem:rates}.
Thus one can select successively later tests with ratio tending
to zero. Since
$B_{\log n}n=(B_{\log n}/R_i)D\ell_i$
and $D/(D-1)\le2$, \eqref{eq:cheap-partition} follows.
\end{proof}
Suppose for a contradiction that a smooth volume model
$(M,\nu,S)$ exists. Its dimension is positive because a
zero-dimensional compact manifold is finite and cannot carry a
nonatomic probability. Apply \Cref{thm:switching} with
$\eta=.01$. The model is now fixed, so all its constants and
its sequence $(B_k)$ are fixed.
Use \Cref{lem:cheap-scales} for this sequence. In what follows,
$i,D,n$ run through the selected tests and are taken sufficiently
large for every stated eventual estimate.
We use the measurable isomorphism to regard every finite symbolic
observation as a measurable observation on $M$. The original domain
can be replaced by the intersection of all its integer translates,
so all orbit identities hold on an invariant conull set.
Extend the observations arbitrarily on its null complement.
For each test let $C$ be the cell sampled in the switching
experiment, and let $\rho$ be the original law of $(x,y)$.
Its two marginals are exactly $\nu$, and its conditional law
given $C$ is the product of normalized volume on that cell.
\subsection{The independent upper bound}
Let
\[
U=\mathcal L_{i,D}(y,-n),\qquad
V=\mathcal L_{i,D}(x,0),\qquad q=D-1.
\]
Each list uses its own input's alignment advance. Both are defined
on every pair before a switch is selected.
Under the identification of $\mathcal W_i$ with the nonzero
vectors of $\F_2^{2s_i}$, write $[\,\cdot\,,\,\cdot\,]_i$ for
the alternating form from the word construction, and set
\begin{equation}\label{eq:passing-statistic}
Z(x,y)=\frac1q\sum_{j=0}^{q-1}
\one_{\{[U_j,V_j]_i=0\}}.
\end{equation}
For each fixed cell and each fixed $j$, the laws of $U_j$ and
$V_j$ are independent.
Apart from the diagonal pairs, the orthogonality relation gives the
classical binary symplectic graph; see \cite[Definition~3.3 and Theorem~3.4]{Seidel1973} for its
finite-geometric description and spectral scale. We include the
character calculation, keeping the diagonal pairs, and pass from
average Shannon entropy to the required graph bound by a heavy--light
split of each law.
\begin{lemma}\label{lem:graph-independent}
Let $L=2^{2s}-1$ and equip the nonzero vectors of
$\F_2^{2s}$ with a nondegenerate alternating form.
Suppose a family of pairs of laws $(p_\xi,q_\xi)$ on these
vectors is indexed by a probability space, and
\[
\E_\xi H(p_\xi)\ge .99\log L,\qquad
\E_\xi H(q_\xi)\ge .99\log L.
\]
For all sufficiently large $L$, independent samples from the
two laws, followed by averaging over $\xi$, pass the test
$[u,v]=0$ with probability at most $.62$.
\end{lemma}
\begin{proof}
For a law $p$, put
$A_p=\{u:p(u)>L^{-.8}\}$ and $\theta_p=p(A_p)$.
The set $A_p$ has at most $L^{.8}$ elements. Splitting entropy
according to this set gives
\begin{equation}\label{eq:heavy-entropy}
H(p)\le1+.8\theta_p\log L+(1-\theta_p)\log L
=1+(1-.2\theta_p)\log L.
\end{equation}
The same formula, with the usual zero-mass conventions, covers
empty sets. The entropy hypotheses imply
\[
\E_\xi\theta_{p_\xi},\ \E_\xi\theta_{q_\xi}
\le .05+\frac5{\log L}\le .055
\]
once $\log L\ge1000$.
Let $A$ be the matrix $A_{uv}=(-1)^{[u,v]}$ on nonzero
vectors. On the full vector space, the rows of the same matrix
are orthogonal: for $u\ne w$, nondegeneracy makes
$v\mapsto[u+w,v]$ a nonzero linear functional, so
$\sum_v(-1)^{[u+w,v]}=0$. Every row has squared norm $L+1$.
The full operator norm is therefore $\sqrt{L+1}$, and restriction
to the nonzero rows and columns has no larger norm.
The light subprobability $p^{\rm l}=p\,\one_{A_p^c}$
satisfies
\[
\|p^{\rm l}\|_2^2
\le L^{-.8}\sum_u p^{\rm l}(u)\le L^{-.8}.
\]
Hence
\[
\left|\sum_{u,v}p^{\rm l}(u)A_{uv}q^{\rm l}(v)\right|
\le\sqrt{L+1}\,L^{-.8}=:\beta_L.
\]
The passing mass of the light product is at most
$(1+\beta_L)/2$. All pairs with at least one heavy coordinate
have total mass at most $\theta_p+\theta_q$.
After averaging, and taking $L$ large enough that
$\beta_L\le .02$, the passing probability is at most
\[
.055+.055+\frac{1+.02}{2}=.62.
\]
The matrix includes its diagonal entries, so the automatic
relations $[u,u]=0$ require no separate exclusion.
\end{proof}
By the two original marginals and \Cref{prop:window-entropy},
\[
H_\rho(U),H_\rho(V)\ge .996q\ell_i.
\]
Conditioning each vector on the sampled cell loses at most
$H(C)\le\log\#\mathcal D_{\log n}=o(q\ell_i)$. Conditional
subadditivity consequently gives
\begin{equation}\label{eq:entry-entropies}
\frac1q\sum_{j=0}^{q-1}H(U_j\mid C)\ge .99\ell_i,
\qquad
\frac1q\sum_{j=0}^{q-1}H(V_j\mid C)\ge .99\ell_i
\end{equation}
for late tests. Apply \Cref{lem:graph-independent} with index
$\xi=(C,j)$, assigning $j$ uniform weight independently of the
cell. Independence is used conditional on this full index:
equivalently, first fix $j$, use the product law conditional
on $C$, and then average. No product assertion conditional only
on $C$ after mixing over a shared random $j$ is needed.
We obtain
\begin{equation}\label{eq:passing-upper}
\E_\rho Z\le .62.
\end{equation}
\subsection{From geometric shadowing to symbol agreement}
Let $d_*>0$ be the symbol-distance constant in
\Cref{lem:recognition}. It is fixed by the word construction.
Let $f:M\to\mathcal A$ be the transported symbol at time zero.
We first choose compact cores for this one fixed finite
partition, rather than for the growing level-$i$ label sets.
\begin{lemma}\label{lem:symbol-matching}
For all sufficiently large dyadic times in the smooth switching
estimate, the retained pairs for which
\begin{align}
\#\{0\le t<n:f(S^tz)\ne f(S^tx)\}
&<d_*n/100,\label{eq:future-symbol-match}\\
\#\{-n\le t<0:f(S^tz)\ne f(S^ty)\}
&<d_*n/100\label{eq:past-symbol-match}
\end{align}
have original $\rho$-mass at least $.96$.
\end{lemma}
\begin{proof}
Choose $\kappa<d_*/20000$. Inner regularity of the smooth
probability gives disjoint compact sets
$K_a\subset f^{-1}(a)$, for $a\in\mathcal A$, with
\[
\nu\left(M\setminus\bigcup_{a\in\mathcal A}K_a\right)<\kappa.
\]
Measurable atoms may first be replaced by Borel representatives
modulo null sets. There are finitely many cores, and the distance
between any two nonempty distinct cores is positive. If there is
only one nonempty core, the following equal-symbol implication is
automatic. Otherwise take the positive minimum of these distances.
Uniform shadowing is smaller than this fixed separation at all
large scales. Therefore two compared iterates both in the cores
have the same symbol.
Write $G$ for the retained set and
$\sigma=z_*(\rho|_G)$. The switching theorem gives
$\sigma\le2\nu$. For every integer $t$,
\[
(S^t)_*\sigma\le2(S^t)_*\nu=2\nu.
\]
The $x$ and $y$ time marginals under the original law are $\nu$.
At each future comparison time, the $\rho$-mass within $G$
where either $x$ or $z$ misses the cores is at most
$\kappa+2\kappa=3\kappa$. The same holds for the past comparison
with $y$.
The integrals over $G$ of the fractions of mismatches on either
half are therefore at most $3\kappa$. Markov's inequality at
threshold $d_*/100$, on each half, bounds their combined failure
mass by $600\kappa/d_*<.03$. Since $\rho(G)\ge .99$,
the simultaneous success mass is at least $.96$.
This is an estimate in the original law, without dividing
by $\rho(G)$.
The input marginals and $\sigma\le2\nu$ give zero mass to
the complement of the invariant conjugacy domain. Thus the
symbol strings and all orbit identities used in the argument
are defined outside a $\rho$-null set.
\end{proof}
\subsection{Recognition preserves the original entry pairing}
Consider a pair in the success event of \Cref{lem:symbol-matching}.
If the phases of $z$ and $x$ modulo $m_i$ differed, take the
$D-1$ full level-$i$ blocks of $x$ beginning at its first
aligned boundary in $[0,n)$. At each such interval, the symbol
string of $z$ is a window of length $m_i$ cut at a nonzero
offset across two consecutive legal level-$i$ words. This
remains true if those words are equal or if one extends beyond
the observed window. By \Cref{lem:recognition}, each of the
$D-1$ disjoint intervals contributes at least $d_*m_i$
mismatches. Their total is at least
\[
d_*(D-1)m_i\ge d_*n/2,
\]
contradicting \eqref{eq:future-symbol-match}.
The same reasoning in the past gives
\begin{equation}\label{eq:phase-recovery}
a_i(z)=a_i(x)=a_i(y).
\end{equation}
Here $a_i(S^{-n}u)=a_i(u)$ for every $u$ in the conjugacy
domain, since $n$ is a multiple of $m_i$.
Let $q_0$ be the common advance to alignment.
The already defined entries $U_j,V_j$ occupy, in the symbolic
orbit of $z$, the positions
\[
-n+q_0+jm_i,\qquad q_0+jm_i,\qquad 0\le j\le D-2.
\]
Their difference is exactly $n=Dm_i$. The index $j$ has not
been changed after selecting the switch.
When one of these entries differs in label from the corresponding
word of $z$, distinct-word separation in
\Cref{lem:recognition} costs at least $d_*m_i$ mismatches.
Consequently fewer than $D/100$ entries on each side can differ.
Let $E_{i,D}\subset X$ be the group-boundary exception from
\Cref{lem:graph-windows}, transported to $M$ and extended
arbitrarily on its null complement. Its original measure is at most
$\alpha_{i,D}\le2D_i^*/H_i\to0$, uniformly over the chosen
tests. The retained switch has
\[
\rho\{(x,y)\in G:z(x,y)\in E_{i,D}\}
\le2\alpha_{i,D}.
\]
Take a late test for which $2\alpha_{i,D}\le.01$.
On a subset of the original pair space of mass at least
$.96-.01$, the symbol-matching event holds and the two
$z$-lists satisfy every graph relation. Removing the fewer
than $D/100$ wrong entries on each side gives
\[
Z(x,y)\ge 1-\frac{2D}{100(D-1)}\ge.96.
\]
The statistic is nonnegative elsewhere, so
\begin{equation}\label{eq:passing-lower}
\E_\rho Z\ge(.96-.01)\cdot.96=.912.
\end{equation}
This contradicts \eqref{eq:passing-upper}.
\begin{proof}[Completion of the proof of \Cref{thm:main}]
\Cref{prop:system} supplies a standard nonatomic ergodic
invertible probability-preserving system of zero entropy.
If it had a smooth volume model, \Cref{thm:switching} and
\Cref{lem:cheap-scales} would supply the tests above.
The original pair law at each late test would satisfy both
\eqref{eq:passing-upper} and \eqref{eq:passing-lower}, which is
impossible. The dimension and all smooth constants were arbitrary
but fixed after construction of the system. Thus no model in
any finite dimension exists.
\end{proof}
\begin{remark}
The role of smooth positive volume is quantitative.
It gives a switch whose unnormalized output law is controlled
by the original invariant measure. That control allows a
measurable symbol observation to be compared along shadowing
orbits, despite the absence of regularity in the conjugacy.
The proof makes no analogous assertion for models carrying
only a singular measure or a nonsmooth density equivalent
to volume.
\end{remark}
==== END A-finite-entropy-system-without-a-smooth-positive-volume-model//07-obstruction.tex ====
==== BEGIN A-finite-entropy-system-without-a-smooth-positive-volume-model//main.tex ====
\documentclass[11pt]{article}
\input{preamble}
\title{A zero-entropy system without a smooth positive-volume model}
\author{OpenAI}
\date{September 25, 2026}
\begin{document}
\maketitle
\begin{abstract}
We construct an ergodic invertible transformation of a standard nonatomic
probability space with zero Kolmogorov--Sinai entropy that has no smooth
positive-volume model. More precisely, it is not measurably conjugate to any
$C^\infty$ diffeomorphism preserving a strictly positive smooth probability
density on a compact finite-dimensional manifold. A single example excludes
every finite dimension, including models on nonorientable manifolds and
manifolds with smooth boundary.
\end{abstract}
\tableofcontents
\clearpage
\input{sections/01-introduction}
\input{sections/02-stationary}
\input{sections/03-linear}
\input{sections/04-switching}
\input{sections/05-words}
\input{sections/06-process}
\input{sections/07-obstruction}
\bibliographystyle{plain}
\bibliography{references}
\end{document}
==== END A-finite-entropy-system-without-a-smooth-positive-volume-model//main.tex ====
==== BEGIN A-finite-entropy-system-without-a-smooth-positive-volume-model//preamble.tex ====
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==== END A-finite-entropy-system-without-a-smooth-positive-volume-model//preamble.tex ====
==== BEGIN A-finite-entropy-system-without-a-smooth-positive-volume-model//references.bib ====
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