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\title{Invariant-projection counterexamples\\for every irrational rotation}
\author{OpenAI}
\date{September 2026}
\begin{document}
\maketitle
\begin{abstract}
Fix any irrational angle $\theta$. We build a continuous weight
$f:\T\to[0,1]$ with exactly one zero and infinite negative log-integral,
such that the weighted rotation $T_f=Uf(V)$ in the hyperfinite
$\mathrm{II}_1$ factor $R_\theta$ has no nontrivial invariant projection.
So the answer to a question of Zhu, Fang, and Shi is no, and the angle is
prescribed before the weight. The operator is also nonzero and
norm-quasinilpotent. Every ingredient is proved here.
\end{abstract}
\tableofcontents
\section{Introduction and the answer}\label{sec:intro}
\subsection{The question}
Fix an irrational number $\theta\in(0,1)$. Let $\T$ be the unit circle
and $m$ its Haar probability measure. Let $U,V$ be the unitaries
generating the hyperfinite $\mathrm{II}_1$ factor $R_\theta$ with
$VU=e^{2\pi i\theta}UV$, and let $\tau$ be its trace, normalised so that
$\tau(1)=1$ and $\tau(U^kV^\ell)=0$ unless $k=\ell=0$.
For a continuous function $f$ on $\T$, write $T_f=U\,f(V)$. Here $f(V)$
means the image of $f$ under the Borel functional calculus of $V$.
\begin{definition}[Affiliated invariant subspace]
A closed subspace $\mathcal K\subseteq L^2(R_\theta,\tau)$ is
\emph{nontrivially affiliated} with $R_\theta$ if its orthogonal projection
$p$ satisfies
\[
p\in R_\theta,\qquad 0<\tau(p)<1,\qquad (1-p)\,T_f\,p=0 .
\]
We do \emph{not} assume $pT_f=T_fp$; the projection $p$ need only
annihilate the ``outgoing'' part of $T_f$.
\end{definition}
The \emph{Fuglede--Kadison determinant} of the scalar weight is
\[
\Delta(f(V))=\exp\!\Bigl(\int_\T\log|f|\,dm\Bigr),\qquad \log 0=-\infty .
\]
Zhu, Fang and Shi \cite{zfs} asked whether, when $f$ is nonzero almost
everywhere, $|f|$ is nonconstant, and $\Delta(f(V))=0$, the operator
$T_f$ must have a nontrivial affiliated invariant subspace. Their paper
answers the complementary case $\Delta(f(V))>0$ positively
\cite[Cor.~5.5]{zfs}. We answer the zero-determinant case negatively,
and we do so for each prescribed irrational angle.
\begin{theorem}[Main theorem]\label{thm:main}
For every irrational $\theta\in(0,1)$ there is a continuous
$f:\T\to[0,1]$ with
\[
f^{-1}(\{0\})=\{1\},\qquad \int_\T\log f\;dm=-\infty,
\]
such that no $p\in R_\theta$ with $0<\tau(p)<1$ satisfies
$(1-p)\,T_f\,p=0$.
\end{theorem}
The construction produces $f$ only after $\theta$ is fixed and never
uses a Diophantine hypothesis on $\theta$. Because the ambient space is
a factor representation, the theorem forbids only those invariant
subspaces whose projection lies in $R_\theta$; arbitrary closed
invariant subspaces of the underlying Hilbert space are not covered by
this statement.
\subsection{Why a naive approach fails}
For ``nice'' weights, the operator $T_f$ sits in a factor with a rich
trace, and the standard route to invariant subspaces uses the
Brown measure. The Brown measure of $T_f$ for measurable $f$ is uniform
on a circle of radius $\Delta(f(V))$ \cite[Thm.~5.4]{dykema-schultz}.
At determinant zero the Brown measure collapses to the point mass
$\delta_0$, so the projection-valued spectral theorem gives no
nontrivial projection. The Brown measure can of course collapse
without killing invariant subspaces: Dykema--Haagerup produce a whole
family of them for the quasinilpotent DT-operator
\cite{dykema-haagerup}, and Tucci produces quasinilpotent generators
with nontrivial affiliated invariant subspaces \cite{tucci2008}.
So the determinant gives no positive answer at value~$0$, and the problem
is genuinely about the operator, not about the Brown measure. The new
input is a construction that prevents invariant projections from ever
arising.
\subsection{Hyperinvariant subspaces as a corollary}
The same construction yields a quasinilpotent operator on any separable
infinite-dimensional Hilbert space with a transitive commutant, which is
a statement at the level of operator theory independent of the factor.
Recall that a closed subspace is \emph{hyperinvariant} for $T$ if every
bounded operator commuting with $T$ leaves it invariant, and that an
operator algebra is \emph{transitive} if its only common invariant
subspaces are $0$ and the whole space.
\begin{lemma}[Commutant algebra]\label{lem:commutant-algebra}
If $T$ is nonscalar then
$\{T\}'=\{A:AT=TA\}$ is a proper, strong-operator closed, unital
complex algebra. It is transitive if and only if $T$ has no nontrivial
hyperinvariant subspace.
\end{lemma}
\begin{proof}
The commutation equation is preserved by all algebraic operations, so
$\{T\}'$ is a unital algebra. If $A_n\to A$ strongly and $A_nT=TA_n$,
then $A_nT\zeta\to AT\zeta$ and $TA_n\zeta\to TA\zeta$ for every
$\zeta$, so $AT=TA$ and the algebra is strongly closed. If every
bounded operator commuted with $T$, then every rank-one projection
$e\otimes e$ would commute with $T$. So every nonzero $\zeta$ would
be an eigenvector of $T$. For linearly independent $\zeta,\eta$, the
vector $\zeta+\eta$ is again an eigenvector, and $T\zeta,T\eta$ are
proportional to $\zeta,\eta$ and to $\zeta+\eta$ simultaneously.
Hence the eigenvalues agree, and $T$ is scalar. That gives properness.
The transitivity statement is a restatement of the definition: a common
closed invariant subspace of $\{T\}'$ is precisely a hyperinvariant
subspace of $T$.
\end{proof}
\begin{lemma}\label{lem:hyperinvariant-projection}
Let $M\subseteq B(\mathcal H)$ be a von Neumann algebra and $T\in M$.
If $\mathcal K$ is a hyperinvariant subspace for $T$, its orthogonal
projection belongs to $M$ and its range is invariant under $T$.
\end{lemma}
\begin{proof}
Any unitary $W\in M'$ commutes with $T$, and the same is true of
$W^{-1}$. Hyperinvariance makes $\mathcal K$ reducing for every such
$W$. The orthogonal projection of $\mathcal K$ therefore commutes
with all $W\in M'$ and thus with all of $M'$. Von Neumann's double
commutant theorem gives $M''=M$, so the projection is in $M$. Since
$T\in\{T\}$, invariance of $\mathcal K$ under $T$ follows from
hyperinvariance.
\end{proof}
\begin{corollary}\label{cor:transitive-commutant}
Every separable infinite-dimensional complex Hilbert space carries a
nonzero norm-quasinilpotent operator whose commutant is a proper,
strongly closed, transitive unital algebra.
\end{corollary}
\begin{proof}
Apply Theorem~\ref{thm:main} and Corollary~\ref{cor:quasinilpotent} to
any irrational $\theta$, in the tracial Hilbert space $L^2(R_\theta,\tau)$.
That space is separable infinite-dimensional, and
Lemma~\ref{lem:hyperinvariant-projection} puts every hyperinvariant
projection inside $R_\theta$. A nonzero quasinilpotent operator is
nonscalar, so Lemma~\ref{lem:commutant-algebra} applies. Any
separable infinite-dimensional Hilbert space is unitarily equivalent
to $L^2(R_\theta,\tau)$, which transfers the result.
\end{proof}
\begin{remark}\label{rem:ckl}
Corollary~\ref{cor:transitive-commutant} contradicts the hyperinvariant
subspace claim of Ch\^{o}, Ko, and Lee \cite[Cor.~2.8]{cho-ko-lee} at their
parameter $\mu=1$: for the nonzero quasinilpotent $T$ above and any
conjugation $C$, their hypotheses apply to $I+T$, and $\{I+T\}'=\{T\}'$,
so $I+T$ inherits no nontrivial hyperinvariant subspace. Only that
assertion is contested; other statements of their article are not
affected.
\end{remark}
\subsection{How the construction runs}
The proof combines four ideas. We give the summary in advance; each idea
is then done carefully in its own section.
First, \emph{an obstruction in a data-rich model}. Suppose $T_f$ had an
invariant projection $p$. In the crossed-product model of $R_\theta$ a
projection becomes a field of projections $x\mapsto p(x)$ over the
circle. The trace identity and ergodicity force the diagonal entry
$r(x)=\langle p(x)e_0,e_0\rangle$ to lie strictly between $0$ and $1$
almost everywhere. Invariance becomes an orthogonality identity between
vectors attached to $x$ and $y$ for ordered pairs $(x,y)$ on overlapping
weighted orbits.
Second, \emph{ordered couplings}. We need many such pairs, even after
fixing either endpoint, and we need the endpoint marginals to be
uniformly square-integrable. A mixed-radix expansion of the circle
gives independent digits. Coupling each digit with small, controlled
``top'' and ``bottom'' probabilities gives ordered pairs whose
marginals stay in a fixed ball of $L^2(\mu)$ and whose labels stay
independent after conditioning on either endpoint.
Third, \emph{a finite-convolution obstruction}. Suppose we have identity
$\sum_{v\in\Z}A_vB_{-n-v}=0$ for $0\le n\le 2K$ on an event of
probability at least $c$. The central coefficients $A_0=r(x)$ and
$B_0=1-r(y)$ are bounded away from zero. A compactness argument makes
some finite convolution coefficient large, while the strong
attenuation of the weight keeps the complementary tail small almost
everywhere. The two estimates cannot both hold.
Fourth, \emph{a transfer to the continuous weight}. The measurable
weight used in the obstruction is smoothed and relocated to get a
continuous weight with one zero. The change is a constant plus a
measurable coboundary. The coboundary gives an (unbounded) multiplier
implementing an intertwining, and an affiliated-operator lemma shows
that trace-preserving invariant projections pass over. This gives the
continuous $f$ of Theorem~\ref{thm:main}.
The rest of the paper follows these four steps.
% body continues below
\section{Setting: the circle, the rotation, the factor}
\label{sec:setting}
Throughout we use the additive circle $X=\R/\Z$, its Haar probability
$\mu$, and the irrational rotation $\sigma x=x+\theta$. We write
$\{t\}\in[0,1)$ for the fractional part of $t\in\R$. All functions
below are Borel; all identities that hold ``almost everywhere'' are
understood on a conull set whose countably many needed translates we
discard. Because that bookkeeping is harmless, we use it without
further remark.
\subsection{The crossed-product model}
Let $K=\ell^2(\Z)$ with orthonormal basis $(e_k)_{k\in\Z}$, and let
$S e_k=e_{k+1}$. Put $\mathcal H=L^2(X,\mu;K)$. Define two families
of operators on $\mathcal H$:
\[
(U\zeta)(x)=S\,\zeta(x),\qquad
\pi(a)\zeta(x)_{k}=a(\sigma^{k}x)\zeta(x)_{k}\quad(a\in L^\infty(X)).
\]
Write $M=W^*\bigl(U,\pi(L^\infty(X))\bigr)$. Let $V=\pi(x\mapsto e^{2\pi i x})$.
\begin{lemma}[The factor]\label{lem:factor-basic}
The operators satisfy $VU=e^{2\pi i\theta}UV$, and the vector state
associated with the constant field $e_0$ is a faithful normal trace on $M$:
\[
\tau(1)=1,\qquad \tau(U^k V^\ell)=0\quad((k,\ell)\ne(0,0)).
\]
The pair $(M,\tau)$ is unitarily equivalent to $(R_\theta,\tau)$.
\end{lemma}
\begin{proof}
The relation $VU=e^{2\pi i\theta}UV$ is checked on each fiber:
\[
(VU\zeta)(x)=\pi(V)S\zeta(x)=S\zeta(x)\cdot e^{2\pi i(x+\theta)},
\]
\[
(UV\zeta)(x)=S\pi(V)\zeta(x)=S\bigl(e^{2\pi i x}\zeta(x)\bigr)
=e^{2\pi i x}\bigl(e^{2\pi i\theta}S\zeta(x)\bigr),
\]
and the two sides agree. The vectors $U^k V^\ell e_0$ span a dense
subspace of $\mathcal H$ by ordinary Fourier theory, and
$\langle U^kV^\ell e_0,e_0\rangle=\delta_{k\ell}\delta_{00}$ because
$U^kV^\ell e_0(x)=e^{2\pi i\ell x}e_k$ and $e_0$ has coordinate
vector $e_0$ throughout $X$. The commutation relation lets us move every
$U$ past every $V$ at the price of a scalar, so every product of the
generators reduces to a scalar multiple of a monomial. Therefore the
moments of $(U,V)$ in $\mathcal H$ match those in the GNS representation
of $(R_\theta,\tau)$. Riemann--Lebesgue or, equivalently, the elementary
Fourier argument identifies the map $U^kV^\ell e_0\mapsto U^kV^\ell$
with a unitary between the dense linear spans, and it extends to a
unitary between the Hilbert spaces. It conjugates $M$ to $R_\theta$ and
the vector state at $e_0$ to $\tau$.
\end{proof}
For clarity we shall use $U,V,\pi,\tau$ in the concrete model and never
need the abstract symbol $R_\theta$ again. The element $T_f$ of the
introduction becomes $T_f=U\pi(f)$.
\subsection{Fields of operators}
The operators in $M$ commute with every scalar multiplication
$M_g\zeta(x)=g(x)\zeta(x)$, $g\in L^\infty(X)$. This has a strong
consequence: every element of $M$ decomposes fiberwise. The next lemma
is the standard one; we give a proof for completeness.
\begin{lemma}[Field decomposition]\label{lem:field-decomposition}
Let $A\in B(\mathcal H)$ commute with $M_g$ for every $g\in L^\infty(X)$.
There is an essentially bounded field $x\mapsto A(x)\in B(K)$, unique
almost everywhere, such that
\[
(A\zeta)(x)=A(x)\,\zeta(x),\qquad \|A(x)\|\le\|A\|\quad\hbox{a.e.},
\]
and every column $x\mapsto A(x)v$ is measurable for every $v\in K$.
Products and adjoints are computed fiberwise.
\end{lemma}
\begin{proof}
Let $B$ be a bounded operator on $L^2(X)$ commuting with $M_g$ for all
$g$. Applying $B\mathbf 1_E=\mathbf 1_E b$ with $b=B1$ and using the
boundedness of $B$ gives, for every measurable $E$,
\[
\int_E|b|^2\,d\mu\le\|B\|^2\mu(E),
\]
so $|b|\le\|B\|$ a.e., and $Bf=bf$ first on simple functions and then
on all of $L^2(X)$. Applying this to the matrix entries of $A$ between
the coordinate copies of $L^2(X)$ gives measurable functions $a_{ij}$
with $a_{ij}\in L^\infty(X)$. Choose a countable family of finitely
supported vectors $v$ with rational coordinates and commute $A$ past
$M_{\mathbf 1_E}$:
\[
\int_E\bigl\|\textstyle\sum_j a_{ij}(y)v_j\bigr\|^2\,d\mu(y)
\le\|A\|^2\|v\|^2\mu(E).
\]
Hence the matrix acts as a bounded operator $A(y)$ on $K$ with norm at
most $\|A\|$ on a conull set. Each column is measurable by
finite-coordinate approximation; the resulting field represents $A$ on
finite-coordinate simple functions and hence on all of $\mathcal H$.
Uniqueness and the product and adjoint formulas are consequences of
evaluating on the constant fields $e_k$; details are immediate from
the matrix formula.
\end{proof}
Now let $J$ be the antiunitary on $\mathcal H$ defined by
$(J\zeta)(x)=S\zeta(\sigma x)$. Direct computation shows that both $U$
and every $\pi(a)$ commute with $J$. Therefore $M$ commutes with $J$,
and the field of any $A\in M$ satisfies a covariance identity.
\begin{lemma}[Covariance]\label{lem:covariance}
For every $A\in M$ and almost every $x\in X$,
\[
A(\sigma x)=S^{-1}A(x)S .
\]
\end{lemma}
\begin{proof}
Evaluate $JA=AJ$ on the constant basis fields. On the one hand
$(JA)e_k((x))$ means $(J(Ae_k))(x)=S(Ae_k)(\sigma x)$. On the other
hand $(AJ)e_k=Je_k=e_{k+1}$, so its value at $x$ is $A(x)e_{k+1}$.
Therefore $S(Ae_k)(\sigma x)=A(x)e_{k+1}$, i.e.
\[
A(\sigma x)e_k=S^{-1}A(x)e_{k+1}
\]
for every $k$ and almost every $x$. Reindexing gives
$A(\sigma x)=S^{-1}A(x)S$: the two bounded operators agree on the basis
$(e_k)_k$ on a conull set.
\end{proof}
The trace formula follows.
\begin{lemma}[Trace formula]\label{lem:trace-formula}
For every $A\in M$,
\[
\tau(A)=\int_X\langle A(x)e_0,e_0\rangle\,d\mu(x)
=\int_XA(x)_{00}\,d\mu(x).
\]
\end{lemma}
\begin{proof}
Both sides are weakly continuous in $A$ on the unit ball, and both agree
with $\tau$ on the monomials $U^kV^\ell=\pi(e^{2\pi i\ell x})S^k$. For
these $A$, the fields are $A(x)_{00}=e^{2\pi i\ell x}$ when $k=0$ and $0$
otherwise, so the displayed integral reproduces $\tau$ on the
generators.
\end{proof}
\subsection{Ergodicity and the strict diagonal}
Irrational rotation is ergodic. We record the two points we use. If
$h\in L^2(X)$ and $h\circ\sigma=h$, then all Fourier coefficients
$\hat h(k)$ with $k\ne0$ satisfy $\hat h(k)=e^{2\pi ik\theta}\hat h(k)$,
so they vanish. In particular $h$ is constant.
\begin{lemma}[Strict diagonal]\label{lem:strict-diagonal}
Let $\sigma$ be an invertible ergodic probability-preserving
transformation of $(Y,\nu)$. On $L^2(Y,\nu;K)$ let
$(U\zeta)(y)=S\zeta(y)$ and $\pi(a)(y)e_k=a(\sigma^ky)e_k$. Let $M$
be a von Neumann algebra containing $U$ and $\pi(L^\infty(Y))$ and
consisting of measurable fields with covariance
$A(\sigma y)=S^{-1}A(y)S$. Suppose $\tau(A)=\int\langle A(y)e_0,e_0\rangle$
defines a faithful normal trace. Let $f\in L^\infty(Y,\nu)$ be
positive a.e., $T=U\pi(f)$, and $p\in M$ an orthogonal projection with
$(1-p)Tp=0$. If $0<\tau(p)<1$, then $0<\langle p(y)e_0,e_0\rangle<1$
a.e.
\end{lemma}
\begin{proof}
Put $r(y)=\langle p(y)e_0,e_0\rangle$ and
$D_y=\operatorname{diag}_k f(\sigma^ky)$. Invariance and covariance
give $D_y\operatorname{ran}p(y)\subseteq\operatorname{ran}p(\sigma y)$.
If $r(\sigma y)=0$, every vector in the range of $p(\sigma y)$ has
zeroth coordinate $0$. Since $(D_yv)_0=f(y)v_0$ and $f(y)>0$, every
vector in $\operatorname{ran}p(y)$ also has zeroth coordinate $0$;
hence $r(y)=0$. Symmetrically, if $r(y)=1$ then $e_0\in\ran p(y)$,
and its image $D_y e_0=f(y)e_0$ is in $\ran p(\sigma y)$, so
$r(\sigma y)=1$. Thus $\sigma^{-1}\{r=0\}\subseteq\{r=0\}$ and
$\{r=1\}\subseteq\sigma^{-1}\{r=1\}$ up to null sets. Preserving $\nu$
turns both inclusions into equalities. Ergodicity forces each event
to be null or conull, and the trace inequality
$0<\int r\,d\nu=\tau(p)<1$ excludes both alternatives.
\end{proof}
\subsection{The compact-transfer lemma}
Later, an identity that holds on a rich family of pairs will need to be
passed to a weak limit. Discontinuous projection fields can be
approximated by their restrictions to a large compact set, provided the
marginals of the approximating measures do not concentrate on small
sets.
\begin{definition}
A family of finite positive Borel measures on the compact metric space
$X$ is \emph{uniformly absolutely continuous} with respect to $\mu$ if
for each $\varepsilon>0$ there is $\delta>0$ such that every Borel set of
$\mu$-measure less than $\delta$ has measure less than $\varepsilon$
for every member of the family.
\end{definition}
\begin{lemma}[Compact transfer]\label{lem:compact-transfer}
Let $p(x)$ be a Borel field of orthogonal projections on $K$ and let
$F_k:X\to(0,\infty)$, $k\in\Z$, be finite Borel functions. Suppose
finite positive Borel measures $\beta_n$ on $X^2$ converge weakly to
$\beta$. Assume that both families of marginals of $\beta_n$ are
uniformly absolutely continuous with respect to $\mu$, and the two
marginals of $\beta$ are absolutely continuous. Suppose further that
some $0<B<\infty$ satisfies
\[
F_k(y)/F_k(x)\le B\quad(k\in\Z)
\]
almost everywhere for each $\beta_n$ and for $\beta$. If
\[
(1-p(y))\diag_k\bigl(F_k(y)/F_k(x)\bigr)p(x)=0
\qquad\beta_n\text{-a.e.\ for every }n,
\]
then the identity holds $\beta$-a.e.
\end{lemma}
\begin{proof}
First cap every diagonal entry at $B$; the cap changes no entry on a
conull set of any $\beta_n$ or of $\beta$ and preserves the identity.
Now cap-version is bounded everywhere. For $\varepsilon>0$, apply
simultaneous Lusin approximation to the countably many maps
$x\mapsto F_k(x)$, $x\mapsto p(x)e_k$. On a compact $Q\subseteq X$,
all of them are continuous, and $\mu(X\setminus Q)$ is small enough
that $\beta_n(X^2\setminus Q^2)$ and $\beta(X^2\setminus Q^2)$ are
both less than $\varepsilon$. The simultaneous statement follows by
applying the usual Lusin property to the countably many maps with
summable exceptional measures and intersecting the compact sets.
On $Q^2$, the capped diagonal field
$x,y\mapsto\diag_k(F_k(y)/F_k(x))$ is strongly continuous:
$F_k(y)/F_k(x)$ is a continuous scalar function on $Q^2$, the field
$p(x)$ is strongly continuous on $Q$, and strong continuity plus a
uniform bound implies strong continuity of the products. Therefore
the zero set $Z_Q$ of the capped version of the identity on $Q^2$ is
closed and hence compact. The assumption gives
$\beta_n(Z_Q)\ge\beta_n(X^2)-\varepsilon$. The portmanteau inequality
under weak convergence, together with $\beta_n(X^2)\to\beta(X^2)$, gives
$\beta(Z_Q)\ge\beta(X^2)-\varepsilon$. Since $\varepsilon$ was
arbitrary, $\beta$ concentrates on the global zero set of the capped
field. The cap and the original field agree $\beta$-a.e.\ by the
hypothesis on $\beta$.
\end{proof}
\section{Scalar lemmas for the sampling kernels}
\label{sec:scalars}
This section collects the finite computations that the sampling scheme
needs. None of them involves the factor or the angle; they are only
about two probability spaces. Once proved, the same lemmas are reused
by the product model of the appendix.
\subsection{The sampling kernel}
For an integer $m\ge200$ set
\begin{equation}\label{eq:h-def}
h=m^{-3/5},\qquad \Delta(z)=\min\bigl(\{z\},1-\{z\}\bigr)
\quad(z\in\R).
\end{equation}
Define on two circles a function $w_m$:
\begin{equation}\label{eq:sampling-kernel}
w_m(z,v)=\mathbf 1_{\{\Delta(z)\ge m^{-2}\}}\,
\frac{\mathbf 1_{\{0<v<h\Delta(z)\}}}{h\Delta(z)},
\end{equation}
where $v$ is represented in $[0,1)$ and the expression is zero when the
cutoff fails. The first kernel $w_m$ forces a small positive move of
size $v$ with no wrap around zero, while conditioning on
$\Delta(z)\ge m^{-2}$. Set
\[
w_m^{(1)}(z,v)=w_m(z,v),\qquad w_m^{(2)}(z,v)=w_m(z-v,v).
\]
Here $w_m^{(2)}$ is the same density measured from the endpoint $z$;
equivalently, the pair $(z-v,z)$ has the first orientation.
The kernel has several elementary bounds. Around the cut at $0$ we
pay a cutoff penalty of order $m^{-2}$, and the height of the kernel
is of order $h^{-1}$. The next lemma makes those quantitative.
\begin{lemma}[Kernel estimates]\label{lem:kernel-estimates}
Let $m\ge200$, $h=m^{-3/5}$, $u=w_m^{(r)}$ for $r\in\{1,2\}$, and
$\bar u(s)=\int_0^1u(s,v)\,dv$.
\begin{enumerate}
\item Uniformly in $s$,
\[
\|u(s,\cdot)\|_\infty+\Var\bigl(u(s,\cdot)\bigr)=O(m^{13/5}).
\]
\item
\[
\|\bar u\|_\infty\le2,\qquad \int_0^1\bar u=1-2m^{-2},
\]
\[
(1-2m^{-2})^2\le\int_0^1\bar u^2
\le1+\frac{h^2}{(1-h)^2},
\]
\[
\iint_{[0,1]^2}u^2=\frac2h\log\frac{m^2}{2}.
\]
\end{enumerate}
The constants are absolute for large $m$.
\end{lemma}
\begin{proof}
For $r=1$ the function $u(s,\cdot)$ is a constant of height
$h^{-1}\Delta(s)^{-1}\le m^{13/5}$ on the interval
$(0,h\Delta(s))$, and its total variation is twice that height.
For $r=2$, nonzero values require $v<s$ in representatives $s,v\in[0,1)$:
if $v>s$ then $\Delta(s-v)\le v-s<v$, while $h<1$. On $0<v<s$ the
conditions $\Delta(s-v)\ge m^{-2}$ and $v<h\Delta(s-v)$ describe an
interval (possibly empty). The reciprocal of $\Delta(s-v)$ has at most
two monotone pieces there, and each piece is bounded by $m^2$. So the
same height bound holds for $r=2$, with total variation at most twice
that bound.
For $r=1$, $\bar u$ is the cutoff indicator of mass $1-2m^{-2}$, and
\[
\iint u^2=\frac2h\int_{m^{-2}}^{1/2}\frac{ds}{s}
=\frac{2}{h}\log\frac{m^2}{2}.
\]
The bijection $(s,v)\mapsto(s-v,v)$ preserves the double integral, so
$r=2$ has the same mass and the same squared integral. To bound
$\int\bar u^2$ for $r=2$, drop the cutoff and consider the law of
$y=z+t\Delta(z)$ for independent uniform $z\in[0,1)$ and
$t\in(0,h)$. For fixed $t$ this is a piecewise-linear bijection of
$[0,1]$ with slopes $1+t$ and $1-t$, so its density $d_t$ obeys
$\int d_t=1$ and $|d_t-1|\le h/(1-h)$. The mixture over $t$ has
$\int d=1$ and
\[
\int d^2=1+\int(d-1)^2\le1+\frac{h^2}{(1-h)^2}.
\]
Restoring the cutoff only lowers the nonnegative density. Also
$d\le(1-h)^{-1}\le2$. Cauchy--Schwarz for the lower bound on
$\int\bar u^2$ uses the exact mass $1-2m^{-2}$.
\end{proof}
\subsection{Separated-phase estimates}
\label{sec:sep-phase}
For fixed $\theta$ later we use phases $\{n\alpha\}$ where $\alpha$ is
either random on a short interval or a fixed small number that is the
$\{q\theta\}$ of a large multiple of $\theta$. We need the two phases
to look like independent uniforms, even though they are not. The next
lemma states exactly the estimates we need.
\begin{lemma}[Separated phases]\label{lem:separated-phases}
Let $0<N_0\le N\le m^2$ and let $\lambda$ be uniform on $[1/N_0,2/N_0]$.
Let $n\in[Q/2,Q]\cap\N$ with $Q=m^{(7/10)m}$.
\begin{enumerate}
\item[(i)] The density $p_n$ of $\{n\lambda\}$ satisfies
$|p_n-1|\le N_0/n$ almost everywhere.
\item[(ii)] Let $H=m^{300}$. Suppose $n,n'$ are positive, distinct,
and obey the separation condition
\[
|vn+v'n'|>Nm^{1000}
\quad\bigl(0<|v|+|v'|,\ |v|,|v'|\le H,\ v,v'\in\Z\bigr).
\]
Then for intervals $I,J\subseteq[0,1)$,
\[
\bigl|\mathbb P(\{n\lambda\}\in I,\{n'\lambda\}\in J)
-|I|\,|J|\bigr|=O(m^{-80}).
\]
\item[(iii)] Under the same hypotheses, for complex functions $f,g$ of
bounded variation on $[0,1]$,
\[
\bigl|\mathbb E[f(\{n\lambda\})g(\{n'\lambda\})]
-f_0g_0\bigr|
\le C m^{-80}B(f)B(g),
\]
where $f_0=\int f$, $g_0=\int g$, and
$B(f)=\|f\|_\infty+\Var f$.
\end{enumerate}
The constants are absolute for large $m$.
\end{lemma}
\begin{proof}
(i) Multiplying $\lambda$ by $n$ periodises the interval
$[n/N_0,2n/N_0]$ modulo one. Its projection to $[0,1)$ has density at
most $1+n/N_0$ and at least $1-n/N_0$; since $n\ge Q/2>N_0$ this
gives (i).
(ii) Excluding the finitely many frequency pairs $(v,v')$ with
$|v|,|v'|\le H$, direct integration on $[1/N_0,2/N_0]$ gives
\[
\bigl|\mathbb E e^{2\pi i(vn+v'n')\lambda}\bigr|
\le\frac{N_0}{\pi|vn+v'n'|}
\le\frac1\pi m^{-1000}.
\]
Convolve each interval indicator with the Fej\'er kernel
$F_H(s)=H^{-1}\bigl(\sin(\pi Hs)/\sin(\pi s)\bigr)^2$. The Fej\'er
kernel is nonnegative, has mass one, and has frequency support in
$|k|<H$. The convolved indicators have means $|I|$ and $|J|$ and
Fourier coefficients of modulus at most one. Outside strips of width
$\eta=m^{-80}$ about the endpoints, their error is
$O((H\eta)^{-1})$. The marginal bound (i) shows that the strips cost
at most $O(\eta)$. The remaining Fourier error is at most
$H^2m^{-1000}$, which is negligible. This gives (ii).
(iii) A bounded-variation function $f$ on $[0,1]$ differs, on a conull
set, by a constant of size $\|f\|_\infty$ plus an integral of interval
indicators against the total-variation measure of $f$. Apply (ii) to
the resulting products of interval indicators and integrate against
total variation. This gives the stated bound with the constant $C$
absorbing the integral of the two-variable function against
$\nu_f\times\nu_g$.
\end{proof}
\subsection{A nonzero finite convolution}
Later, our obstruction produces a finite part of a convolution of two
sequences. The compactness lemma below is the reason this finite part
cannot vanish when the central coefficients are bounded away from zero.
\begin{lemma}[Nonzero finite convolution]\label{lem:polynomial-coefficient}
For every integer $K\ge0$, $D\ge\delta>0$ there is
$\varepsilon(K,D,\delta)>0$ with the following property. Let
$A,B$ be complex Laurent polynomials supported in $[-K,K]$, with
coefficients of modulus at most $D$, and with $|A_0|,|B_0|\ge\delta$.
Then
\[
\max_{0\le n\le2K}|(AB)_{-n}|\ge\varepsilon(K,D,\delta).
\]
\end{lemma}
\begin{proof}
The set of admissible polynomial pairs is a nonempty compact subset of
a finite-dimensional complex vector space. For each pair the first
nonzero exponents of $A$ and $B$ are at most zero, and their sum lies
in $[-2K,0]$. The coefficient of the product at that sum is the
product of those two nonzero coefficients, hence nonzero. The
continuous function $A,B\mapsto\max_{0\le n\le2K}|(AB)_{-n}|$ thus has
a strictly positive value at every admissible pair; on the compact set
it attains a positive minimum which we call $\varepsilon$.
\end{proof}
\subsection{Two elementary compactness facts}
Finally, two short facts we use without comment. The first is that
under irrational rotation, uniform convergence to the integral holds
in the continuous category. The second is the Fourier version of the
same statement, uniform over all starting points. We record them in
one lemma.
\begin{lemma}[Averaging under irrational rotation]\label{lem:weyl}
Let $F$ be a continuous complex function on $X$, and let
$\overline F=\int_XF\,d\mu$. Then
\[
\sup_{x\in X}\Bigl|\frac1N\sum_{k=0}^{N-1}F(\sigma^kx)-\overline F\Bigr|
\to0\quad(N\to\infty).
\]
The convergence is uniform in $F$ for a fixed compact family, and it is
also uniform in the starting point $x$. The same statement holds with
$\theta$ replaced by any nonzero integer multiple of $\theta$.
\end{lemma}
\begin{proof}
For $F(x)=e^{2\pi inx}$, the average along the orbit is
$N^{-1}\sum_{k=0}^{N-1}e^{2\pi ink\theta}e^{2\pi inx}$, which is the
product of a geometric sum and a character. Since $n\theta$ is not an
integer for $n\ne0$, the geometric sum tends to zero. Uniform
trigonometric approximation reduces an arbitrary continuous $F$ to the
character case, and the final uniform-in-$x$ statement uses the fact
that the character terms are uniformly bounded in $x$.
\end{proof}
\section{The construction parameters}
\label{sec:parameters}
The construction selects, for the fixed $\theta$, a list of frequencies
$q_i$ and a block structure $J_j=\{d_{j-1}+1,\dots,d_j\}$ of sizes
$m_j$. The frequencies are integer multiples with rapidly growing
ratios, so the phases $\{q_i\theta\}$ are as regular as possible. The
blocks and their phases give the finite sampling tests of the next
section.
The next proposition fixes all the parameters; every later section
will reference its items. The proof of the proposition is chosen
recursively and only needs bare irrationality.
\begin{proposition}[Parameter list]\label{prop:parameters}
For the fixed $\theta$ there exist positive integers $m_j,K_j,G_j$,
$q_i$, positive reals $L_j,C_j,\varepsilon_j$, and consecutive blocks
$J_j=\{d_{j-1}+1,\dots,d_j\}$ of size $m_j$, with $d_0=0$, having all
of the following properties. Set
\[
q_0=1,\quad u_i(x)=\{q_ix\},\quad \alpha_i=\{q_i\theta\},
\]
\[
\kappa_j=(100j)^{-1},\quad N_j=m_jK_j,\quad
a_j=G_j/m_j,\quad b_j=m_j^{-19/20},\quad Q_j=m_j^{(7/10)m_j}.
\]
\begin{itemize}
\item
Frequencies are strictly increasing and divisible:
\[
q_{i+1}/q_i\ge N_j,\qquad q_i\mid q_{i+1}\qquad(i\in J_j),
\]
\[
1/N_j<\alpha_i<2/N_j\qquad(i\in J_j).
\]
\item
Spread of phases: for every integer $k$ with
$N_j/4\le|k|\le m_j^{(4/5)m_j}$, at least $m_j/20$ indices $i\in J_j$
satisfy
\[
1-\{k\alpha_i\}\in[b_j,1-b_j].
\]
For every $|k|$ with $G_jK_j\le|k|\le N_j/4$, every $i\in J_j$ has
$1-\{k\alpha_i\}\in[a_j,1-a_j]$.
\item
Block sizes and cutoffs:
\[
K_1=100,\quad G_1=1,\quad
m_j\ge\max(K_j,2^j,4G_j,200),
\]
\[
K_{j+1}=\lceil m_j^{(3/4)m_j}\rceil>2K_j,\qquad
G_{j+1}K_{j+1}<m_j^{(4/5)m_j},
\]
\[
\kappa_jm_j^{1/40}\ge L_j,\qquad
\kappa_{j+1}\sqrt{G_{j+1}}\ge L_j,\qquad
h^*_{j+1}/K_{j+1}\le2^{-j},
\]
where $h^*_j$ are integers determined in the sampling section.
\item
Costs and error constants:
\[
L_j\ge1 \text{ nondecreasing},\qquad C_j=\max(L_j,K_j^2),\qquad
D_j=\exp\Bigl((K_j+1)\sum_{s<j}C_sm_s\Bigr),
\]
and, putting $\varepsilon(K,D,\delta)$ as in
Lemma~\ref{lem:polynomial-coefficient},
\[
\varepsilon_j=\varepsilon(K_j,D_j,1/j),\qquad
(2K_j+1)3e^{-L_j/10000}\le\varepsilon_j/j.
\]
\item
Sampling sets. Put $p_*=d_{j-1}$ and
$\ell_j=\bigl((j+1)^2(1+p_*)q_{p_*}\bigr)^{-1}$. For every
$t=r/q_{p_*}$ with $r\in\Z$ and $0\le r<q_{p_*}$ there is a finite
nonempty set $\mathcal N_{j,t}\subseteq[Q_j/2,Q_j]\cap\N$ with
$\{n\theta-t\}\in(0,\ell_j)$, and for $r'=1,2$,
\[
\Bigl\|\frac1{|\mathcal N_{j,t}|}\sum_{n\in\mathcal N_{j,t}}
\prod_{i\in J_j}w_{m_j}^{(r')}\bigl(u_i(x),\{n\alpha_i\}\bigr)-1
\Bigr\|_{L^2(\mu)}<\frac1j.
\]
\end{itemize}
\end{proposition}
\begin{proof}
We construct the data stage by stage. Fix $\theta$. We also need
auxiliary continuous bump functions for the continuity argument;
describe them now so they are available later. Let $O_j\subseteq X$ be
the open arc of radius $1/(10j)$ about zero. Choose a continuous
nonnegative bump $\psi$ supported in $O_1$ and positive near the
center of $O_1$. Because the positive orbit of $\theta$ is dense,
finitely many translates $x\mapsto\psi(\sigma^nx)$, $0\le n\le h^*_j$,
have a strictly positive sum on $X$; divide that sum to obtain
continuous functions $\phi_{j,n}\ge0$ with
\begin{equation}\label{eq:partition-unity}
\sum_{n=0}^{h^*_j}\phi_{j,n}=1,\qquad
\sigma^n(\operatorname{supp}\phi_{j,n})\subseteq O_j.
\end{equation}
All the $h^*_j$ and $\phi_{j,n}$ are fixed before the parameter choices
below, so the constants $h^*_j$ are available as upper bounds.
Assume all data of stages $1,\dots,j-1$ are fixed. Choose
$\varepsilon_j>0$ via Lemma~\ref{lem:polynomial-coefficient} with
$K=K_j$, $D=D_j$, $\delta=1/j$. Choose $L_j\ge\max(1,L_{j-1})$ large
enough that
\[
(2K_j+1)3e^{-L_j/10000}\le\varepsilon_j/j,
\]
set $C_j=\max(L_j,K_j^2)$, and choose $G_{j+1}\ge1$ with
$\kappa_{j+1}\sqrt{G_{j+1}}\ge L_j$. All of this is possible without
choosing $m_j$.
Now choose $m=m_j$ large enough that
\[
m\ge\max(K_j,2^j,4G_j,200),\qquad
\kappa_jm^{1/40}\ge L_j,\qquad
G_{j+1}\lceil m^{(3/4)m}\rceil<m^{(4/5)m}.
\]
The last inequality holds for large $m$ by the already chosen $G_{j+1}$
and $K_{j+1}=\lceil m^{(3/4)m}\rceil$. Declare this $m_j$ and put
$K_{j+1}=\lceil m_j^{(3/4)m_j}\rceil$.
At each stage the remaining data are chosen conditional on the sizes.
Set $m=m_j$, $N=N_j=mK_j$, $Q=Q_j=m^{(7/10)m}$,
$a=a_j=G_j/m$, $b=b_j=m^{-19/20}$.
\emph{Trial phases.} Let $(\lambda_i)_{i\in J_j}$ be independent and
uniform on $[1/N,2/N]$. The lower phase range
$G_jK_j\le|k|\le N/4$ is automatic: since $G_jK_j\ge G_j\cdot K_j$ and
$K_j\ge1$, for such $|k|$ and $i\in J_j$ we have
\[
a\le|k|\lambda_i\le\frac{|k|}{N}\le\frac14,
\]
and similarly for negative $|k|$, so $1-\{k\lambda_i\}\in[a,1-a]$
automatically. This will survive the passage from $\lambda_i$ to
$\alpha_i$ below.
\emph{The upper phase range.} For $|k|$ with
$N/4\le|k|\le m^{(4/5)m}$, the law of $\{k\lambda_i\}$ is the periodised
uniform on an interval of length $|k|/N\ge1/4$, so it has density at
most $1+N/|k|\le5$. The probability of missing the range $[b,1-b]$ is
at most $10b$. Independence across
$J_j$ and a union over subsets of $J_j$ of size more than $19m/20$ show
that for a fixed $k$ the failure probability is at most
$2^m(10b)^{(19/20)m}$. A further union over both signs and over at
most $2m^{(4/5)m}$ values of $k$ gives a failure probability of at most
\[
2m^{(4/5)m}\cdot2^m(10b)^{(19/20)m}
=\exp\bigl(-\tfrac{41}{400}m\log m+O(m)\bigr).
\]
Markov's inequality then produces a deterministic choice of trial
phases for which the two phase assertions hold at stage $j$, with all
$2^m$ possible misses managed simultaneously. Fix once and for all
such a choice $(\lambda_i)_{i\in J_j}$; the density bound at most $2$
used in (iii) of Lemma~\ref{lem:separated-phases} then transfers to
the actual $\alpha_i$ below.
\subsubsection*{Separated sampling powers}
Fix $t=r/q_{p_*}$ as in the proposition. Write $\ell=\ell_j$. By the
equidistribution statement of Lemma~\ref{lem:weyl} applied to the
interval $(t,t+\ell)$ in $X$, for every sufficiently large $m$ at
least $\ell Q/3$ of the integers $n\in[Q/2,Q]$ satisfy
\[
\{n\theta-t\}\in(0,\ell).
\]
Put $H=m^{300}$ and $R=Nm^{1000}$. Select a subset
$\mathcal N_{j,t}\subseteq[Q/2,Q]$ by greedy scan: go through the
integers $n\in[Q/2,Q]$ in increasing order, and accept $n$ if
\[
|v n+v'n'|>R
\quad\bigl(|v|,|v'|\le H,\ (v,v')\neq(0,0),\ v,v'\in\Z\bigr)
\]
for every previously accepted $n'$. Each accepted integer forbids at
most $(2H+1)^2(2R+2)\le m^{1605}$ further integers. Therefore if we
accept $P$ integers the scan uses at most $Pm^{1605}$ lost integers.
Choose $P=m^{0.69m}$. Since $Q=m^{0.7m}$ and $\ell$ is fixed at this
stage, the total intake $\ell Q/3$ is at least $P+Pm^{1605}$ for
large $m$, so the greedy scan succeeds. This gives $\mathcal N_{j,t}$
with the required displacement and separation.
\subsubsection*{The product sampling test}
Fix an orientation $r'\in\{1,2\}$, the set $\mathcal N=\mathcal N_{j,t}$
of size $P\ge m^{0.69m}$, and the trial phases $(\lambda_i)_{i\in J_j}$
from above. Put $d=m$, $u=w_m^{(r')}$, $\bar u(s)=\int_0^1u(s,v)\,dv$,
$A_*=\int_0^1\bar u^2$. Lemma~\ref{lem:kernel-estimates} gives
\[
(1-2m^{-2})^2\le A_*\le1+ch^2,\qquad
\iint u^2\le ch^{-1}\log m.
\]
For $z=(z_i)_{i\in J_j}\in[0,1]^d$ set
\[
Z(z)=\frac1P\sum_{n\in\mathcal N}\prod_{i\in J_j}
u\bigl(z_i,\{n\lambda_i\}\bigr),\qquad
Z_{\rm id}(z)=\prod_{i\in J_j}\bar u(z_i).
\]
The $z_i$ are independent uniform on $[0,1]$ and the $\lambda_i$ are
the fixed trial phases; we compute expectations jointly.
For $n\neq n'$ and fixed $s$, the function
$F_{n,n'}(s)=\mathbb E_{\lambda_i}[u(s,\{n\lambda_i\})u(s,\{n'\lambda_i\})]$
satisfies $|F_{n,n'}(s)-\bar u(s)^2|\le cm^{-50}$ uniformly in $s$,
by item (iii) of Lemma~\ref{lem:separated-phases} applied to the
bounded-variation function $u(s,\cdot)$. For $n=n'$,
$\mathbb E_{\lambda_i}u(s,\{n\lambda_i\})^2\le2\iint u^2\le ch^{-1}\log m$
by item (i). Also, item (i) gives
$\int_0^1\bar u(s)\mathbb E_{\lambda_i}u(s,\{n\lambda_i\})\,ds
=A_*+O(m^{-50})$ since $\|\bar u\|_\infty\le2$ and $n\ge Q/2$.
Finally $\int_0^1\bar u=1-2m^{-2}$.
\emph{Square of the sum.} Expand
\[
\mathbb E_z\|Z\|_2^2=\frac1{P^2}\sum_{n,n'}\prod_{i\in J_j}
\int_0^1F_{n,n'}(s)\,ds .
\]
The distinct-pair terms have total weight $1-1/P$; each factor is
$A_*+O(m^{-50})$, so their product is $A_*^d+O(dm^{-50})$ by
telescoping, using $A_*^d\le(1+ch^2)^d\le2$. The self-pair terms have
total weight $1/P$; each is at most $(ch^{-1}\log m)^d$. Thus
\[
\mathbb E_z\|Z\|_2^2
\le(A_*^d+O(dm^{-50}))\bigl(1-1/P\bigr)
+\frac{(ch^{-1}\log m)^d}{P}.
\]
Similarly $\mathbb E_z\langle Z,Z_{\rm id}\rangle=A_*^d+O(dm^{-50})$
and $\mathbb E_z\|Z_{\rm id}\|_2^2=A_*^d$. Therefore
\[
\mathbb E_z\|Z-Z_{\rm id}\|_2^2
\le O(dm^{-50})+\frac{(ch^{-1}\log m)^d}{P}
+\frac{2A_*^d}{P},
\]
and each of the last two terms is exponentially small because
$P\ge m^{0.69m}$, $h^{-1}=m^{0.6}$, and $A_*^d\le2$.
Also
\[
\mathbb E_z\|Z_{\rm id}-1\|_2^2
=A_*^d-2(1-2m^{-2})^d+1\longrightarrow0,
\]
since $A_*^d\to1$ by $dh^2=m^{-1/5}\to0$ and $(1-2m^{-2})^d\to1$.
Hence $\mathbb E_z\|Z-1\|_2^2\to0$. By Markov's inequality, at
stage $j$ we can choose the trial phases making
$\mathbb E_z\|Z-1\|_2^2<1/(4j^2)$ for each of the two orientations.
Fix such trial phases.
\subsubsection*{Realizing the trial phases by circle characters}
Fix $m=m_j$ and the trial phases $\lambda_i$. Choose $q_i$
successively over $i\in J_j$: given $q_{i-1}$ and target $\lambda_i$,
choose $q_i$ a multiple of $q_{i-1}$ with $q_i/q_{i-1}\ge N_j$ and
$\{q_i\theta\}$ so close to $\lambda_i$ that
\[
|\{n\alpha_i\}-\{n\lambda_i\}|\le m^{-10}
\qquad(n\in\mathcal N_{j,t},\ i\in J_j),
\]
where $\alpha_i=\{q_i\theta\}$. This is possible because positive
multiples of $q_{i-1}\theta$ are dense in $\T$; the finitely many
constraints above define an open set of choices.
For each $n\in\mathcal N_{j,t}$, the product
$\prod_i u(z_i,\{n\alpha_i\})$ differs from the product
$\prod_i u(z_i,\{n\lambda_i\})$ by $o(1)$ uniformly in $z$. The
perturbation is bounded by $d\cdot m^{-10}$ times the total variation
of each factor (Lemma~\ref{lem:kernel-estimates}). So the $L^2(\mu)$
norm of the sampling average with $\alpha_i$ is within $1/(2j)$ of
the norm with $\lambda_i$, and the sampling inequality holds with
$\alpha_i$.
The strict phase statements hold: the trial-phase bounds
$1-\{k\lambda_i\}\in[b_j,1-b_j]$ and $\in[a_j,1-a_j]$ together with
$|\alpha_i-\lambda_i|\le m^{-10}\ll b_j$ give the corresponding
bounds for $\alpha_i$. The ratio conditions $q_{i+1}/q_i\ge N_j$ hold
by construction. The ratio from this block into the next one is
imposed at the next stage; no present condition depends on it.
The proof of Proposition~\ref{prop:parameters} is complete.
\end{proof}
\subsubsection*{Consequences of the parameter choice}
\label{sec:parameter-consequences}
From $K_{j+1}>2K_j$ and $m_j\ge K_j$ one gets
\[
\sum_{s>j}1/K_s\le2/K_{j+1},\qquad
\sum_j(1+h^*_j)/K_j<\infty,\qquad
Q_j/K_{j+1}\longrightarrow0 .
\]
The last ratio is $m_j^{(7/10)m_j}/m_j^{(3/4)m_j}
=m_j^{-m_j/20}\to0$. The other bounds $m_j\ge d_{j-1}+200$,
$m_j\ge K_j$, $m_j\ge 2^j$ hold by the choice of $m_j$.
\section{The measurable weight and its orbits}
\label{sec:measurable}
We now convert the parameters into an operator and an identity between
projection fields at different points of the circle. The weight below
is only measurable, with finitely many pieces; the continuous version
comes later. All of the analysis in this section and the next two is
carried out for this measurable weight.
\subsection{The weight}
Keep the parameters of Proposition~\ref{prop:parameters}. Define
\begin{equation}\label{eq:measurable-weight}
g_*(x)=\sum_{j\ge1}C_j\sum_{i\in J_j}
\lfloor u_i(x)+\alpha_i\rfloor,
\qquad f_*(x)=e^{-g_*(x)},\qquad T_*=U\pi(f_*).
\end{equation}
\begin{lemma}[The weight is well defined]\label{lem:weight-finite}
The sum in \eqref{eq:measurable-weight} is finite almost everywhere and
$0<f_*\le1$. Also $f_*>0$ almost everywhere.
\end{lemma}
\begin{proof}
Let $E_i=\{x:\lfloor u_i(x)+\alpha_i\rfloor\ne0\}$, so that
$\mu(E_i)=\alpha_i$. Then
\[
\sum_i\mu(E_i)=\sum_j|J_j|\cdot\frac{2}{N_j}
=2\sum_j\frac1{K_j}<\infty,
\]
using $\alpha_i<2/N_j$ and $N_j=m_jK_j$. The elementary first
Borel--Cantelli lemma, which here just says that the measure of a union
in the tail is at most the sum of tail measures, gives
$\mu(\limsup_iE_i)=0$. Off that null set only finitely many floors are
nonzero, so $g_*(x)<\infty$. Since $g_*\ge0$, we have $0<f_*\le1$;
because $g_*(x)<\infty$ almost everywhere, $f_*>0$ almost everywhere.
\end{proof}
\subsection{Cocycles and ratios}
For $k\in\Z$ put
\begin{equation}\label{eq:cocycle-definition}
F_k(x)=\exp\Bigl(-\sum_j C_j\sum_{i\in J_j}
\lfloor u_i(x)+k\alpha_i\rfloor\Bigr).
\end{equation}
\begin{lemma}[Cocycle identity]\label{lem:cocycle-identity}
There is an invariant conull set on which $F_0=1$, all $F_k$ are
finite and positive, and
\[
F_{v+n}(x)=F_v(x)\,F_n(\sigma^v x)\qquad(v,n\in\Z).
\]
Also $F_1=f_*$.
\end{lemma}
\begin{proof}
Fix $k$. For large $i$ one has $|k|\alpha_i<1$, so the probability
that $\lfloor u_i+k\alpha_i\rfloor\ne0$ is exactly $|k|\alpha_i$. The
sum $\sum_i|k|\alpha_i\le2|k|\sum_jK_j^{-1}<\infty$, so
Borel--Cantelli gives $F_k<\infty$ almost everywhere. Take the
countable intersection over $k\in\Z$ and then the countable intersection
of all its $\sigma$-translates; call the resulting invariant conull set
$X_0$. On $X_0$ all $F_k$ are finite. The elementary identity
$\lfloor t+(k+n)\alpha\rfloor=\lfloor t+k\alpha\rfloor+\lfloor\{t+k\alpha\}+n\alpha\rfloor$
gives
$F_{k+n}(x)=F_k(x)F_n(\sigma^kx)$ termwise. Since $F_k\ge0$ and $F_k$
is positive on $X_0$, the identity makes sense there. Values off
$X_0$ are irrelevant for the trace and may be extended arbitrarily.
\end{proof}
We shall change $F_k$ on the exceptional set if needed, without further
remark. On $X_0$ define the ratio
\begin{equation}\label{eq:ratio}
R_{x,y}(k)=\frac{F_k(y)}{F_k(x)},\qquad x,y\in X_0,\ k\in\Z.
\end{equation}
If $u_i(x)\le u_i(y)$ for all $i$, then each floor in the numerator is
at least the corresponding floor in the denominator, so
$0<R_{x,y}(k)\le1$. More precisely, for $i\in J_s$ the $i$th factor
contributes $e^{-C_s}$ to $R_{x,y}(k)$ exactly when
\begin{equation}\label{eq:crossing-condition}
u_i(x)<1-\{k\alpha_i\}\le u_i(y).
\end{equation}
Otherwise it contributes $1$. When $k\alpha_i$ is integral, the event
is empty.
\subsection{From invariance to orbit identities}
Let $p\in M$ be an orthogonal projection with $(1-p)T_*p=0$. Then
$\ran p(x)$ is invariant under $T_*(x)$ fiberwise, so
\[
T_*(x)^n\ran p(x)\subseteq\ran p(\sigma^nx)\qquad(n\ge0).
\]
By the standard cocycle formula,
$T_*(x)^n=S^n\operatorname{diag}_vF_n(\sigma^vx)$. Applying this
and dividing by $F_n(x)>0$, and using the cocycle identity to write
$F_n(\sigma^vx)/F_n(x)=F_v(\sigma^nx)/F_v(x)=R_{x,y}(v)$ for
$y=\sigma^nx$, we obtain
\begin{equation}\label{eq:positive-orbit-identity}
(1-p(y))\operatorname{diag}_vR_{x,y}(v)\,p(x)=0
\qquad(y=\sigma^nx,\ n\ge1).
\end{equation}
This is the elementary identity that everything else will push beyond
orbits.
\begin{remark}
The division by $F_n(x)$ in the previous display is legitimate because
$F_n$ is strictly positive on $X_0$; the resulting diagonal need not
be uniformly bounded in $x$ (it is bounded above by $1/F_n(x)$).
\end{remark}
We shall use the following consequence of Lemma~\ref{lem:strict-diagonal}.
\begin{lemma}\label{lem:diagonal-interior}
If $p\in M$ is an orthogonal projection with $(1-p)T_*p=0$ and
$0<\tau(p)<1$, then for almost every $x\in X$,
\[
0<r(x):=\langle p(x)e_0,e_0\rangle<1.
\]
\end{lemma}
\begin{proof}
Apply Lemma~\ref{lem:strict-diagonal} with $Y=X$, $f=f_*$, the field
decomposition of Lemma~\ref{lem:field-decomposition}, the covariance of
Lemma~\ref{lem:covariance}, and the trace of
Lemma~\ref{lem:trace-formula}. The hypothesis $f_*>0$ of that lemma
holds by Lemma~\ref{lem:weight-finite}.
\end{proof}
\subsection{The rational-graph lemma}
The next lemma is where the sampling identities enter. It takes the
orbit identity \eqref{eq:positive-orbit-identity} -- a statement about
$y=\sigma^nx$ with $n\ge1$ -- and extends it to shifts
$y=x+t$ where $t$ is an integer multiple of $1/q_A$ for some $A$. The
family of such $t$ is large enough to feed the digit couplings of
Section~\ref{sec:couplings}.
\begin{lemma}[Rational-graph identity]\label{lem:rational-graphs}
Let $p\in M$ be a projection with $(1-p)T_*p=0$. For every integer
$A\ge1$ and every $t\in X$ of the form $t=r/q_A$ with $r\in\Z$
and $0\le r<q_A$, the identity
\begin{equation}\label{eq:rational-graph}
(1-p(x+t))\operatorname{diag}_vR_{x,x+t}(v)\,p(x)=0
\end{equation}
holds $\mu$-almost everywhere in $x$. The conull set can be chosen
simultaneously over all such $(A,t)$.
\end{lemma}
\begin{proof}
Fix $A$, $t=r/q_A$. For a fixed large stage $j$ put
$a=d_{j-1}\ge A$, so $q_A\mid q_a$ and $\mathcal N_{j,t}$ is defined
by Proposition~\ref{prop:parameters}. Write $m=m_j$, $w=w_m$,
$w^{(2)}=w_m^{(2)}$ for this stage's kernel and its reflected version.
Consider the family of finite graph measures on $X^2$,
\[
\beta_{j,n}(dx,dy)=\prod_{i\in J_j}w(u_i(x),\{n\alpha_i\})\,d\mu(x)
\qquad(y=\sigma^nx,\ n\in\mathcal N_{j,t}),
\]
and their complete average
$\beta_j=|\mathcal N_{j,t}|^{-1}\sum_{n\in\mathcal N_{j,t}}\beta_{j,n}$.
The first marginal of $\beta_{j,n}$ has density
$\prod_i\bar w(u_i(x))$ with respect to $\mu$, and the second marginal
has density $\prod_i\bar w^{(2)}(u_i(x))$ up to the coordinate change
$y=\sigma^nx$. By item (5) of
Proposition~\ref{prop:parameters},
\[
\|\rho_{j,n}^{(1)}-1\|_{L^2(\mu)}<\frac1j,\qquad
\|\rho_{j,n}^{(2)}-1\|_{L^2(\mu)}<\frac1j,
\]
uniformly in $n\in\mathcal N_{j,t}$. The triangle inequality then
shows the same bounds for the averaged marginals
$\rho_j^{(1)},\rho_j^{(2)}$ of $\beta_j$.
Because every $n\theta$ is within $\ell_j$ of $t$ and $\ell_j\to0$,
testing continuous functions on $X^2$ shows that
$\beta_j\Rightarrow\beta$, the graph measure of $x\mapsto x+t$ with
first marginal $\mu$; the second marginal of $\beta$ is
$\mu(\cdot-t)$. On the support of each $\beta_{j,n}$ (positive
density $\mu$-a.e.), the current-block coordinates satisfy
\[
0<\{n\alpha_i\}<m^{-3/5}\Delta(u_i(x))\qquad(i\in J_j),
\]
by the definition of $w$. In the first block the coordinates are
therefore ordered without wrap; and for future blocks we impose that
ordering by trimming below.
For each selected $n$ and for fixed $A,t$, remove from the measure
$\beta_{j,n}$ the set of starting points satisfying any of
\[
u_i(x)\ge1-q_a\ell_j\quad(A\le i\le a),
\qquad
u_i(x)\ge1-2Q_j/N_s\quad(i\in J_s,\ s>j).
\]
The removed Haar measure is at most
\[
a q_a\ell_j+2Q_j\sum_{s>j}1/K_s
\le (j+1)^{-2}+4Q_j/K_{j+1}\to0.
\]
The $L^2$ bound $\|\rho_j^{(1)}-1\|_{L^2(\mu)}<1/j$ and
Cauchy--Schwarz show the same for the $\beta_j$-mass removed. The
trimmed measures therefore have the same weak limit $\beta$, and their
marginals are dominated pointwise by the untrimmed ones; so the
marginals remain uniformly $L^2$-bounded and uniformly absolutely
continuous with respect to $\mu$.
\emph{Order at every coordinate $i\ge A$.} For $A\le i\le a$ we have
$q_it\in\Z$, so the $i$th coordinate advances by
$q_i\{n\theta-t\}<q_a\ell_j$; trimming prevents a wrap at $1$. For
future coordinates $i\in J_s$ with $s>j$, the advance is
$\{n\alpha_i\}<2Q_j/N_s$; again trimming prevents a wrap. For the
current block the advance is already handled by the definition of $w$
and by the strict positivity bound. Hence every coordinate
$i\ge A$ is ordered:
\[
u_i(x)\le u_i(y)\qquad(i\ge A,\ y=\sigma^nx+\text{trim}).
\]
The bound extends to the limit $\beta$ since
$u_i(x+t)=u_i(x)$ for $i\ge A$ (recall $t=r/q_A$ and $q_A\mid q_i$
for $i\ge A$).
\emph{Uniform bound on the diagonal.} Combining the ordering with
\eqref{eq:crossing-condition}, every ratio satisfies, for
$(x,y)$-pairs in the trimmed graph or in the limit graph,
\[
R_{x,y}(v)\le\exp\Bigl(\sum_{i<A}C_{s(i)}\Bigr)
\qquad(v\in\Z),
\]
where $s(i)$ is the block of $i$. This bound is independent of $j$
and of $v$, and the number $\sum_{i<A}C_{s(i)}$ is finite because only
finitely many $i<A$ occur.
\emph{Transfer to the limit.} On each trimmed approximant, the
positive-orbit identity \eqref{eq:positive-orbit-identity} holds, so
\[
(1-p(y))\operatorname{diag}_vR_{x,y}(v)\,p(x)=0
\]
$\beta_j$-almost everywhere. Both marginal families of $\beta_j$ are
uniformly absolutely continuous (indeed, uniformly $L^2$-bounded on
the same order), both limit marginals are absolutely continuous with
respect to $\mu$, and the diagonal is bounded by an absolute factor
$B=\exp(\sum_{i<A}C_{s(i)})$. So the compact-transfer
Lemma~\ref{lem:compact-transfer} applies and gives the identity
$\beta$-almost everywhere, i.e.\ \eqref{eq:rational-graph}.
Only countably many pairs $(A,t)$ occur, and each is handled by the
argument above. Intersect the resulting conull sets; this gives a
single exceptional set, which we again call the conull set.
\end{proof}
\section{Ordered couplings via mixed-radix digits}
\label{sec:couplings}
Lemma~\ref{lem:rational-graphs} gives the identity \eqref{eq:rational-graph}
at rational shifts $x\mapsto x+t$ with $t$ a multiple of $1/q_A$. We
now want many pairs $(x,y)$ with $y-x$ not rational, and with both
endpoints having bounded marginal densities. The reason is that the
invariance identity alone does not furnish enough threshold crossings:
along orbits, most ratios $R_{x,y}(k)$ are close to $1$, and we need a
mechanism that forces many drops of size $e^{-C_s}$. The digit
construction supplies that mechanism. It is also where the sharp
behaviour of $\sqrt{w/m}$ enters.
\subsection{The mixed-radix expansion}
Write $r_i^*=q_{i+1}/q_i$ (an integer). A uniform point $x\in X$ has
the expansion
\begin{equation}\label{eq:digit-expansion}
x=\frac{A_0}{q_1}+\sum_{i\ge1}\frac{A_i}{q_{i+1}},
\qquad
A_0\in\{0,\dots,q_1-1\},\quad
A_i\in\{0,\dots,r_i^*-1\},
\end{equation}
with independent uniform digits. On a finite prefix this is the
standard mixed-radix expansion of an interval of length
$1/(q_1q_2\cdots q_{n+1})$; taking the limit in the natural decreasing
cylinder filtration gives the asserted representation. The measure
preservation is immediate from the cylinder lengths; the ambiguity at
cylinder boundaries is a countable union of null sets, which we
discard. In particular $A_i=\lfloor r_i^*u_i(x)\rfloor$.
We shall couple the digits of the two endpoints. Because
$u_i(x)$ depends on the digit tail starting at $A_i$ (after multiplying
by $q_i$), equal digits propagate as equal coordinate values, and small
perturbations of the first digit give small perturbations of the
coordinate.
\subsection{Square-integrable product densities}
We need to bound the endpoint marginal densities in $L^2(\mu)$. The
relevant estimate is therefore one about a certain product of
coordinate densities.
\begin{lemma}[Product density bound]\label{lem:product-density-bound}
Let $(\Omega_i,\mathcal F_i,\mu_i)_{i\ge1}$ be probability spaces and
$h_i\ge0$ with $\int h_i\,d\mu_i=1$. If
\[
\sum_{i\ge1}\|h_i-1\|_{L^2(\mu_i)}^2\le C<\infty,
\]
then the product measure $\nu=\bigotimes_i(h_i\mu_i)$ has density with
respect to $\mu=\bigotimes_i\mu_i$ of $L^2(\mu)$-norm at most $e^{C/2}$.
In particular, if two densities $h_i^0,h_i^1$ satisfy
$\|h_i^a-1\|_{L^2(\mu_i)}^2\le b_i$ for both $a=0,1$ and
$\sum_ib_i\le C$, then every deterministic hybrid obtained by choosing
$h_i^{a_i}$ with $a_i\in\{0,1\}$ also has $L^2$ density at most
$e^{C/2}$.
\end{lemma}
\begin{proof}
By $\int h_i=1$, $\|h_i\|_2^2=1+\|h_i-1\|_2^2$. The finite product
density of order $n$ therefore satisfies
\[
\Bigl\|\prod_{i=1}^nh_i\Bigr\|_{L^2}^2
=\prod_{i=1}^n\bigl(1+\|h_i-1\|_2^2\bigr)
\le e^{C}.
\]
For every bounded cylinder $\varphi$ depending on $n$ coordinates,
\[
\Bigl|\int\varphi\,d\nu\Bigr|
\le e^{C/2}\|\varphi\|_{L^2(\mu)} .
\]
Cylinder functions are dense in $L^2(\mu)$, so the integration functional
extends continuously. By Riesz representation there is a real
$H\in L^2(\mu)$ of norm at most $e^{C/2}$ that represents the
extension. On cylinder sets $H\mu$ and $\nu$ agree; uniqueness of
finite measures on the product sigma-algebra gives $\nu=H\mu$.
Hence $H\ge0$ a.e., $\int H=1$, and $h_i\leftrightarrow h_i^{\epsilon_i}$
shows the hybrid statement.
\end{proof}
\subsection{The coupling law}
Now fix the block $j$. We build a probability law $\mathbb P_j$ on
pairs $(x,y)$ together with some auxiliary labels. The mixture is
indexed by a block $s$, but differs at $s=j$ (tagged) from $s\ne j$. At
block $s$ write $m=m_s$ and
\[
\rho_s(w)=\kappa_s\sqrt{w/m}\qquad(w=a_s,\ b_s).
\]
Note $\rho_s(w)/w=\kappa_s/\sqrt{mw}\le\kappa_s$ by $a_s,b_s\ge1/m$.
At every index $i\in J_s$ choose a raw pair $(X_i,Y_i)\in[0,1)^2$ by the
following mixture, independently over $i$:
\begin{itemize}
\item with probability $\rho_s(a_s)$, take $X_i$ uniform on $[0,1-a_s/2)$
and $Y_i$ uniform on $[1-a_s/2,1)$, independently;
\item with probability $\rho_s(a_s)$, take $X_i$ uniform on $[0,a_s/2)$
and $Y_i$ uniform on $[a_s/2,1)$, independently;
\item with probability $\rho_s(b_s)$, do the same with $a_s$
replaced by $b_s$;
\item with probability $\rho_s(b_s)$, do the same with $a_s$
replaced by $b_s$, bottom orientation;
\item if $s=j$, with probability $1/m_s$ take \emph{tagged}:
$X_i$ uniform on $[0,1/(2m_s))$ and $Y_i$ uniform on
$[1-1/(2m_s),1)$;
\item with the remaining probability
$q_{j,s}=1-2\rho_s(a_s)-2\rho_s(b_s)-\mathbf 1_{\{s=j\}}/m_s$,
set $X_i=Y_i$ uniform on $[0,1)$.
\end{itemize}
The four width labels are kept distinct even if $a_s=b_s$ numerically.
The parameter bounds give $\rho_s(w)/w\le\kappa_s\le1/100$ and
$m_s\ge200$, so $q_{j,s}\ge0.98$.
Define digits
\[
A_i=\lfloor r_i^*X_i\rfloor,\qquad B_i=\lfloor r_i^*Y_i\rfloor,
\]
and form $x$ from $(A_i)$ and $y$ from $(B_i)$ by
\eqref{eq:digit-expansion}. Each endpoint law is therefore absolutely
continuous with respect to $\mu$, and the two share the same initial
digit $A_0$.
Because every component satisfies $X_i\le Y_i$ coordinatewise,
\[
X_i\le Y_i\ \Rightarrow\ A_i\le B_i\ \Rightarrow\ u_i(x)\le u_i(y)
\qquad(i\ge1).
\]
The losses in passing from the raw pair $(X_i,Y_i)$ to $(u_i(x),u_i(y))$
are bounded by $1/r_i^*$:
\begin{equation}\label{eq:raw-to-digit}
|u_i(x)-X_i|\le\frac1{r_i^*},\qquad
|u_i(y)-Y_i|\le\frac1{r_i^*}.
\end{equation}
\subsection{Uniform marginal bounds}
We need to bound the two endpoint marginals in $L^2(\mu)$, uniformly in
$j$, and also every deterministic hybrid (choosing the $x$-digit or the
$y$-digit at each coordinate) in $L^2(\mu)$.
\begin{lemma}[Marginal bound]\label{lem:digit-marginals}
There is a finite constant $H_0$, independent of $j$, such that the
marginal laws of $x$ and of $y$ under $\mathbb P_j$, and every
deterministic hybrid of their coordinate densities, have densities with
respect to $\mu$ of $L^2$-norm at most $H_0$.
\end{lemma}
\begin{proof}
At every coordinate, a raw marginal density is at most $10$: each of
the four width components has height at most $2/w$, and their total
contribution is $4\rho_s(w)/w\cdot\max(1,1/\cdot)$; the tag contributes
$2$; the diagonal term contributes $1$. A more careful accounting
gives
\[
0<q_{j,s}\le h_i\le1+\sum_w4\frac{\rho_s(w)}{w}
+\mathbf 1_{\{s=j\}}\le2+8\kappa_s<10.
\]
We need the sharper $L^2$ bound. A uniform density on an interval of
length at least $w/2$ differs from $1$ by a mean-zero function of
squared $L^2$-norm at most $2/w$. A tagged interval contributes a
mean-zero perturbation of $L^2$-norm at most $2m_s$. Expressing
$h_i-1$ as the sum of its mean-zero pieces and applying Cauchy--Schwarz
to these at most five summands gives
\[
\|h_i-1\|_2^2
\le20\sum_w\frac{\rho_s(w)^2}{w}
+5\mathbf 1_{\{s=j\}}/m_s
\le20\Bigl(\frac{2\kappa_s^2}{m_s}
+\frac{\mathbf 1_{\{s=j\}}}{m_s}\Bigr).
\]
This bound holds for either endpoint and for the hybrid choice between
them, because the four width contributions have the same form for
either endpoint.
Passing from raw coordinates to digits replaces $h_i$ by its average on
each radix bin, so Jensen's inequality gives the same bound. Summing
over the $m_s$ coordinates of every block gives a total
\[
\sum_{i\ge1}\|h_i-1\|_2^2\le20\sum_s\bigl(2\kappa_s^2+2^{-s}\bigr)
\le C_0<\infty,
\]
since $\sum_s\kappa_s^2<\infty$ and $\sum_s2^{-s}<\infty$.
Lemma~\ref{lem:product-density-bound} gives a product density of
$L^2$-norm at most
\[
H_0=\exp(C_0/2)<\infty,
\]
uniformly in $j$ and in the choice of deterministic hybrid. Transporting
through the mixed-radix bijection preserves the $L^2$-norm because that
bijection is measure-preserving.
\end{proof}
The initial digit $A_0$ has density $1$ in the reference measure; the
hybrid statement therefore covers every possible combination of
"$x$-marginal" and "$y$-marginal" choices on the digits.
\subsection{The identity on the ordered pairs}
We can now state and prove the transfer of
Lemma~\ref{lem:rational-graphs} to $\mathbb P_j$-almost every pair.
\begin{proposition}[Identity on the coupled pairs]
\label{prop:ordered-couplings}
Let $p\in M$ be a projection with $(1-p)T_*p=0$. For every stage $j$,
\[
(1-p(y))\operatorname{diag}_{k\in\Z}\bigl(R_{x,y}(k)\bigr)p(x)=0
\quad\mathbb P_j\text{-a.e.},
\]
and also
\[
0<R_{x,y}(k)\le1\qquad(k\in\Z)\qquad\mathbb P_j\text{-a.e.}
\]
\end{proposition}
\begin{proof}
\emph{Reduce to rational shifts.} For $D\ge1$, let $y^{(D)}$ be the
point whose digits in positions $1,\dots,D$ are those of $y$, and whose
later digits are those of $x$. Then
$y^{(D)}-x$ is a multiple of $1/q_{D+1}$ by
\eqref{eq:digit-expansion}, and the coordinate order $X_i\le Y_i$
gives $u_i(x)\le u_i(y^{(D)})$ for every $i\ge1$. The map
$y^{(D)}\to y$ is measurable, in fact continuous for
$x$ fixed and the mixed-radix topology (equivalently, the product
topology on digit tails).
Lemma~\ref{lem:rational-graphs} gives the identity \eqref{eq:rational-graph}
for each rational shift $y^{(D)}-x$ outside a common null set $N_D
\subseteq X$ of starting points, since only countably many $(D,t)$ are
involved. The $x$-marginal of $\mathbb P_j$ is absolutely continuous
with respect to $\mu$ (Lemma~\ref{lem:digit-marginals}), so
$\mathbb P_j(x\in N_D)=0$. Thus the identity holds at $(x,y^{(D)})$
$\mathbb P_j$-almost everywhere for every fixed $D$.
\emph{Ordering and uniform diagonal bound.} For each fixed pair, the
coordinate order $u_i(x)\le u_i(y)$ (or $u_i(x)\le u_i(y^{(D)})$)
gives $R_{x,y}(k)\le1$ and $R_{x,y^{(D)}}(k)\le1$ by
\eqref{eq:crossing-condition}. Therefore the diagonal is uniformly
bounded by $1$ on the approximants and on the limit.
\emph{Weak convergence and transfer.} Let $\beta_D$ be the law of
$(x,y^{(D)})$ under $\mathbb P_j$. As $D\to\infty$, we have
$y^{(D)}\to y$ in the product topology on $X$; because $X^2$ is compact
metric and $y^{(D)}$ converges to $y$, $\beta_D\Rightarrow\nu_j$, the
law of $(x,y)$ under $\mathbb P_j$. The marginals of $\beta_D$ are
uniformly absolutely continuous: their first marginal is the
$x$-marginal of $\nu_j$, and their second marginal is a hybrid
$L^2$-bounded family as in Lemma~\ref{lem:digit-marginals}. Both
marginals of $\nu_j$ are absolutely continuous. Hence the hypotheses
of Lemma~\ref{lem:compact-transfer} are satisfied with $B=1$, and the
identity passes to the limit:
\[
(1-p(y))\operatorname{diag}_kR_{x,y}(k)p(x)=0
\qquad\nu_j\text{-a.e.}
\]
Finally, the diagonal bound $R_{x,y}(k)\le1$ extends to the limit as
well, since the pairs are ordered.
\end{proof}
\section{Conditional labels and forced crossings}
\label{sec:conditional}
The couplings give $\P_j$-almost every pair, but we also need fine
conditional control: even after one endpoint is fixed, the coordinate
labels of the other endpoint stay usefully random. The present section
quantifies that.
Throughout $j\ge1$ is fixed. For $i\in J_s$ we label each raw pair
$(X_i,Y_i)$ by the component that produced it: \emph{top} at width
$w$ ($w\in\{a_s,b_s\}$), \emph{bottom} at width $w$, the \emph{tag}
(only when $s=j$), and \emph{diagonal}. Each label is a $\sigma$-finite
event in the probability space of the mixture. We shall condition on the
entire $x$-endpoint, and separately on the entire $y$-endpoint.
\begin{lemma}[Conditional labels]\label{lem:conditional-labels}
Conditional on $x$, the coordinate labels are independent, and the
posterior of each label at index $i$ depends only on the digit $A_i$.
The same holds conditional on $y$, with $B_i$. Moreover, for
$i\in J_s$, $w\in\{a_s,b_s\}$ and $u\in[w,1-w]$,
\begin{align}
\P_j\bigl(\text{top label of width }w\text{ at }i\mid x\bigr)
&\ge\frac{\rho_s(w)}{10}
&&\text{on }\{u_i(x)<u\},
\label{eq:conditional-crossing-top}\\
\P_j\bigl(\text{bottom label of width }w\text{ at }i\mid y\bigr)
&\ge\frac{\rho_s(w)}{10}
&&\text{on }\{u_i(y)\ge u\}.
\label{eq:conditional-crossing-bot}
\end{align}
The labels above are chosen within the indicated width, so the event
counts the two orientations separately. In the first case, the top
label forces $u_i(y)\ge u$; in the second, the bottom label forces
$u_i(x)<u$.
\end{lemma}
\begin{proof}
The endpoint $x$ determines every digit $A_i$ almost surely: the mixed
radix map is injective outside the countable ambiguity set. Once all
$A_i$ are fixed, the digit columns are independent across $i$, so the
labels are too.
For a finite collection of labels $\ell_1,\dots,\ell_n$ at distinct
indices $i_1,\dots,i_n$, the conditional probability is, by the
independence across coordinates and the product formula for
conditioning,
\[
\prod_{k=1}^n\frac{\P_j(\text{label }\ell_k\text{ at }i_k)\cdot
\P_j(A_{i_k}=a_k\mid \ell_k\text{ at }i_k)}{\P_j(A_{i_k}=a_k)} .
\]
The product verifies first on cylinder events (via conditional
independence of coordinates) and extends by the monotone class theorem.
Since each factor depends only on the corresponding digit, the
asserted conditional independence holds.
\emph{Top label conditional on $x$.} Suppose $u_i(x)<u\le1-w$. Because
$1/r_i^*\le1/N_s<w/2$, the entire radix bin of $X_i$ that contains
$u_i(x)$ lies in $[0,1-w/2)$. On that bin, the top component of width
$w$ has $X_i$-density $(1-w/2)^{-1}\ge1$, while the total raw marginal
density $h_i^x$ is at most $10$. Therefore, by
\eqref{eq:conditional-crossing-top} written in the mixture form,
\[
\P_j(\text{top label of width }w\text{ at }i\mid x_i=t)
\ge\frac{\rho_s(w)\cdot 1/2}{h_i^x(t)}
\ge\frac{\rho_s(w)}{10},
\]
on the bin. Under this label $X_i\in[0,1-w/2)$, so
by \eqref{eq:raw-to-digit}
\[
u_i(y)\ge1-w/2-1/r_i^*>1-w\ge u.
\]
That is, the top label of width $w$ forces $u_i(y)\ge u$.
\emph{Bottom label conditional on $y$.} Suppose $u_i(y)\ge u\ge w$.
Then the radix bin of $Y_i$ that contains $u_i(y)$ lies in $[w/2,1)$.
On that bin the bottom component of width $w$ has $Y_i$-density
$(1-w/2)^{-1}\ge1$, and $h_i^y\le10$, so
\[
\P_j(\text{bottom label of width }w\text{ at }i\mid y_i=t)
\ge\frac{\rho_s(w)}{10}.
\]
Under this label, $Y_i\in[w/2,1)$, and by \eqref{eq:raw-to-digit}
\[
u_i(x)\le w/2+1/r_i^*<w\le u .
\]
That is, the bottom label of width $w$ forces $u_i(x)<u$.
The strict inequalities at the bin boundaries are included because
$1/r_i^*<w/2$ strictly.
\end{proof}
\section{The convolution obstruction}
\label{sec:obstruction}
The invariance identity gives an exact convolution relation between two
sequences $A$ and $B$ attached to the endpoints. We will compare two
estimates: a lower bound on a finite part of that convolution coming from
the parameters, and an upper bound on the complementary part coming from
the tail attenuation of the weight. The two cannot coexist; that is the
contradiction.
We separate the abstract statement from its application so that the
product model of the appendix can use the same lemma. The probability
space in the abstract statement may include auxiliary labels besides the
endpoints, and the tail estimate is stated with the conditional
probability bounds of Lemma~\ref{lem:conditional-labels}.
\subsection{An abstract obstruction lemma}
\begin{lemma}[Conditional convolution obstruction]
\label{lem:conditions-convolution}
Let $K\ge1$ be an integer, $D\ge a>0$, $c>0$, and $\eta,\delta\ge0$.
Let $\xi(x),\gamma(y)$ be measurable vectors in the unit ball of
$\ell^2(\Z)$ on a probability space with endpoints $x,y$, and let
$F_k>0$ be finite measurable functions of an endpoint with $F_0=1$.
Put
\[
A_v=\frac{\xi_v(x)}{F_v(x)},\qquad
B_w=F_{-w}(y)\overline{\gamma_{-w}(y)},\qquad
R(k)=\frac{F_k(y)}{F_k(x)}.
\]
Assume the following three conditions hold almost surely:
\begin{enumerate}
\item[(i)] For $0\le n\le2K$ and every $v\in\Z$,
\[
\sum_{v\in\Z}A_vB_{-n-v}=0,
\qquad
0<\frac{F_{n+v}(y)}{F_v(x)}
\le\min\{R(v),R(n+v)\}\le1.
\]
\item[(ii)] On an event $E$ of probability at least $c$,
$|A_v|,|B_v|\le D$ for every $|v|\le K$, and
$|A_0|,|B_0|\ge a$.
\item[(iii)] For each $|k|>K$ there are events
$H_k^{(1)},H_k^{(2)}$ with
\[
R(k)\le\eta+\mathbf 1_{H_k^{(1)}}+\mathbf 1_{H_k^{(2)}}\quad\text{on }E,
\qquad
\P(H_k^{(1)}\mid x)\le\delta,\quad
\P(H_k^{(2)}\mid y)\le\delta.
\]
The conditional bounds are under the full probability law, before
restriction to $E$.
\end{enumerate}
Let $\varepsilon(K,D,a)>0$ be the constant of
Lemma~\ref{lem:polynomial-coefficient}. Then
\[
c\,\varepsilon(K,D,a)
\le (2K+1)(\eta+2\sqrt\delta).
\]
\end{lemma}
\begin{proof}
Condition (i) and Cauchy--Schwarz give absolute convergence of every
convolution $\sum_vA_vB_{-n-v}$.
For $0\le n\le2K$ set
\[
c_n=\sum_{\substack{|v|\le K\\|n+v|\le K}}A_vB_{-n-v}.
\]
On $E$, Lemma~\ref{lem:polynomial-coefficient} applied to the truncated
Laurent polynomials
\[
A^{[K]}(z)=\sum_{|v|\le K}A_vz^v,\qquad
B^{[K]}(z)=\sum_{|w|\le K}B_wz^w
\]
gives $\sum_{n=0}^{2K}|c_n|\ge\varepsilon(K,D,a)$. Its expected value
on $E$ is at least $c\,\varepsilon(K,D,a)$.
Fix $n$. For each omitted index $v$ (meaning $|v|>K$ or $|n+v|>K$),
choose the larger of the two indices: $k(v)=v$ when $|v|>K$, and
$k(v)=n+v$ when $|v|\le K$. By construction $|k(v)|>K$. Using the
zero convolution in (i) and the positivity bound in (i), we get
\[
\mathbf 1_E|c_n|
\le\eta+\sum_{b=1}^2\sum_{v\text{ omitted}}
|\xi_v(x)\gamma_{n+v}(y)|\mathbf 1_{H_{k(v)}^{(b)}} .
\]
The unmasked tail $\sum|\xi_v(x)\gamma_{n+v}(y)|$ is bounded by one via
Cauchy--Schwarz and $\|\xi(x)\|_{2}\le1$, $\|\gamma(y)\|_{2}\le1$.
That bound is absorbed into $\eta$.
For the first exceptional sum, apply Cauchy--Schwarz on probability
times counting measure, and condition first on $x$:
\[
\E\Bigl[\sum_{v\text{ omitted}}|\xi_v(x)\gamma_{n+v}(y)|
\mathbf 1_{H_{k(v)}^{(1)}}\Bigr]
\le\sqrt{\delta}\cdot1 .
\]
Concretely, the first factor of the Cauchy--Schwarz bound is
\[
\E\sum_{v\text{ omitted}}|\xi_v(x)|^2\mathbf 1_{H_{k(v)}^{(1)}}
=\E\Bigl[\mathbf 1_{H_{k(v)}^{(1)}}\sum_{v}|\xi_v(x)|^2\Bigr]
\le\delta,
\]
using the conditional bound on each deterministic $k(v)$, and the second
factor is
$\E\sum_v|\gamma_{n+v}(y)|^2\le1$. For the other exceptional sum put
its indicator with $|\gamma_{n+v}(y)|^2$ and condition first on $y$.
Everything is nonnegative, so Tonelli's theorem justifies all the
interchanges of sum and integral. No simultaneous union bound over the
individually exceptional events is required: each $H_k^{(b)}$ is handled
separately, and the conditional bound transfers to expectations via
Tonelli.
Taking expectations in the previous display and summing over
$0\le n\le2K$ gives
\[
c\,\varepsilon(K,D,a)
\le (2K+1)(\eta+2\sqrt\delta).
\]
\end{proof}
\subsection{The exact convolution identities}
Now return to the circle. Suppose, for contradiction, that $p\in M$ is
a projection with $0<\tau(p)<1$ and $(1-p)T_*p=0$. Set
\[
\xi(x)=p(x)e_0,\qquad \gamma(y)=(1-p(y))e_0,\qquad
r(x)=p(x)_{0,0}=\langle p(x)e_0,e_0\rangle .
\]
Both $\xi$ and $\gamma$ have norm at most $1$ at every point. By
Lemma~\ref{lem:diagonal-interior}, $0<r(x)<1$ a.e.\
(because $\tau(p)= \int r\,d\mu$ and $0<\tau(p)<1$).
Fix a stage $j$. Use the coupling law $\mathbb P_j$ of Section
\ref{sec:couplings}. By Proposition~\ref{prop:ordered-couplings}, for
$\mathbb P_j$-almost every pair the vector
$\operatorname{diag}_vR_{x,y}(v)\xi(x)$ lies in $\ran p(y)$. The fiber
$T_*(y)$ preserves $\ran p(y)$, hence so do all its nonnegative powers.
Coordinatewise, $T_*(y)$ multiplies the $w$-th coordinate by
$F_1(\sigma^w y)=f_*(\sigma^w y)$, so the $n$-th power multiplies the
$(n+v)$-th coordinate of $\operatorname{diag}_vR_{x,y}(v)\xi(x)$ by
$F_n(\sigma^{n+v}y)\cdot R_{x,y}(v)$, giving
\[
\frac{F_n(\sigma^{n+v}y)}{F_v(x)}\xi_v(x)
=\frac{F_{n+v}(y)}{F_v(x)}\xi_v(x).
\]
Pairing with $\gamma(y)$ gives the exact convolution identity
\begin{equation}\label{eq:convolution-identity}
0=\sum_{v\in\Z}\frac{F_{n+v}(y)}{F_v(x)}
\xi_v(x)\overline{\gamma_{n+v}(y)}
=\sum_{v\in\Z}A_vB_{-n-v}\qquad(n\ge0),
\end{equation}
where
\[
A_v=\frac{\xi_v(x)}{F_v(x)},\qquad
B_w=F_{-w}(y)\overline{\gamma_{-w}(y)}.
\]
For $n\ge0$, the estimate
\[
0<\frac{F_{n+v}(y)}{F_v(x)}\le\min\{R_{x,y}(v),R_{x,y}(n+v)\}\le1
\]
follows from $F_{n+v}(y)\le F_v(y)$ and $F_v(x)\ge F_{n+v}(x)$. The
absolute convergence of the sum uses Cauchy--Schwarz and
$\sum|\xi_v(x)\gamma_{n+v}(y)|\le1$.
\subsection{An event of uniformly positive probability}
The event $E_j$ below is where the central coefficients $A_0,B_0$ are
bounded away from zero and the finite coefficients $A_v,B_v$ stay
bounded. We need it to have probability bounded below by a constant
independent of $j$.
Fix $j$; abbreviate $K=K_j$, $m=m_j$. On the probability space of
$\mathbb P_j$ define $E_j$ by the following three conditions:
\begin{itemize}
\item For every $i\in J_j$, $X_i<1-3/m$ and $Y_i>3/m$, and at least one
$i\in J_j$ is \emph{tagged}.
\item For every $s>j$ and $i\in J_s$, $X_i<1-3K/N_s$ and
$Y_i>3K/N_s$.
\item $r(x)\ge1/j$ and $1-r(y)\ge1/j$.
\end{itemize}
We will show that $\P_j(E_j)\ge c_*$ for a positive constant $c_*$
independent of $j$ for large $j$; the eventual threshold may depend on
$p$.
At a current-block coordinate, a tagged choice has probability $1/m$ and
satisfies the first bullet. If not tagged, an admissible nontagged pair
satisfies the same bullet with probability at least $1-61/m$; the
probability of an inadmissible pair (outside both strips) is at most
$60/m$, because each raw marginal density is at most $10$ and each
exclusion costs at most $6/m$ of mass. Independence across the block
gives that exactly one tag, with admissible nontagged values elsewhere,
has probability at least
\[
m\cdot\frac 1m\cdot(1-61/m)^{m-1}\ge e^{-90}
\]
for $m\ge200$, using $\log(1-t)\ge-t/(1-t)$.
The future-block exclusions cost at most
\[
60K_j\sum_{s>j}1/K_s\le 120K_j/K_{j+1}\longrightarrow0 .
\]
This is a countable union over future coordinates, controlled by the
convergence of $\sum_{s>j}1/K_s$.
Finally, the trace conditions. By Lemma~\ref{lem:digit-marginals} and
the Cauchy--Schwarz bound on absolutely continuous marginals,
\[
\P_j(x:r(x)<1/j)\le H_0\mu(x:r(x)<1/j)^{1/2},
\]
\[
\P_j(y:1-r(y)<1/j)\le H_0\mu(y:1-r(y)<1/j)^{1/2},
\]
both of which tend to zero as $j\to\infty$, since $0<r<1$ a.e.\ and
$H_0<\infty$.
Therefore $\P_j(E_j)\ge c_*:=e^{-90}/2$ for all large $j$. Define, for
$0\le n\le 2K$,
\begin{equation}\label{eq:truncated-convolution}
c_{j,n}=\sum_{\substack{|v|\le K\\|n+v|\le K}}A_vB_{-n-v}.
\end{equation}
\begin{lemma}[Finite lower bound]
\label{lem:truncation-bound}
There is a constant $c_*>0$ such that, for all sufficiently large $j$,
\[
\P_j(E_j)\ge c_*,\qquad
\E_j\Bigl[\mathbf 1_{E_j}\sum_{n=0}^{2K_j}|c_{j,n}|\Bigr]
\ge c_*\varepsilon_j .
\]
The point past which the second inequality holds may depend on $p$.
\end{lemma}
\begin{proof}
The first inequality is the argument above. For the second, work on
$E_j$. Rounding from $(X_i,Y_i)$ to $(u_i(x),u_i(y))$ moves each raw
coordinate by at most $1/r_i^*\le1/N_s$, while the endpoint strips in
$E_j$ have width $3K/N_s$. Hence, for $i\in J_s$ with $s\ge j$,
\[
u_i(x)<1-2K/N_s,\qquad u_i(y)>2K/N_s .
\]
Consequently, for $0\le v\le K$ and $\alpha_i<2/N_s$,
\[
\lfloor u_i(x)+v\alpha_i\rfloor=\lfloor u_i(y)-v\alpha_i\rfloor=0,
\qquad i\in J_s,\ s\ge j.
\]
For the finitely many earlier blocks $s<j$, each floor has absolute
value at most $K+1$, because $0<\alpha_i<1$. Hence
\[
1/F_v(x)\le D_j:=\exp\Bigl((K+1)\sum_{s<j}C_sm_s\Bigr),\qquad
F_{-v}(y)\le D_j,\qquad 0\le v\le K .
\]
For $v<0$, monotonicity gives $F_{-v}(x)\ge1$ and $F_v(y)\le1$, so
$|A_{-v}|\le1$ and $|B_{-v}|\le1$.
Therefore every coefficient of the Laurent polynomials
\[
A^{[K]}(z)=\sum_{|v|\le K}A_vz^v,\qquad B^{[K]}(z)=\sum_{|v|\le K}B_vz^v
\]
has modulus at most $D_j$. The constant coefficients are
$A_0=r(x)$ and $B_0=1-r(y)$, both at least $1/j$ on $E_j$. The
coefficient of $z^{-n}$ in $A^{[K]}(z)B^{[K]}(z)$ is exactly
$c_{j,n}$. Lemma~\ref{lem:polynomial-coefficient} with
$K=K_j$, $D=D_j$, $\delta=1/j$ now gives
$\sum_{n=0}^{2K_j}|c_{j,n}|\ge\varepsilon_j$ on $E_j$. Integrating over
$E_j$ gives the second inequality.
\end{proof}
\subsection{Tail attenuation}
The complementary part of the convolution consists of the terms with
$|v|>K$ or $|n+v|>K$; the largest such index has magnitude $>K$. We
bound $R_{x,y}(k)$ for $|k|>K_j$ using the conditional-labels lemma.
\begin{lemma}[Tail attenuation]\label{lem:tail-attenuation}
For every $j$ and every integer $k$ with $|k|>K_j$ there are measurable
events $H_k^{(1)},H_k^{(2)}$ with
\[
\P_j(H_k^{(1)}\mid x)\le e^{-L_j/400},\qquad
\P_j(H_k^{(2)}\mid y)\le e^{-L_j/400}\quad\text{a.s.},
\]
and
\begin{equation}\label{eq:tail-ratio-bound}
R_{x,y}(k)\le e^{-L_j}+\mathbf 1_{H_k^{(1)}}+\mathbf 1_{H_k^{(2)}}
\quad\text{on }E_j .
\end{equation}
The conditional bounds hold under the full probability law, before any
restriction to $E_j$.
\end{lemma}
\begin{proof}
First suppose $K_j<|k|\le N_j/4$. Then on $E_j$, the tagged coordinate
$i\in J_j$ has $X_i<1/m_j$ and $Y_i>1-1/m_j$. Therefore, after rounding,
\[
u_i(x)<1/m_j,\qquad u_i(y)>1-1/m_j .
\]
The phase $\beta=|k|\alpha_i$ lies in $[1/m_j,1/2]$: from the lower bound
$\alpha_i>1/N_j$ and $|k|>K_j$ we have $|k|\alpha_i>K_j/N_j=1/m_j$, and
from $\alpha_i<2/N_j$ and $|k|\le N_j/4$ we have
$|k|\alpha_i<1/2$. The threshold $1-\{k\alpha_i\}$ equals
$1-\beta$ for $k>0$ and $\beta$ for $k<0$, and either is in
$[1-1/m_j,1-1/m_j]$ or $[1/m_j,1/m_j]$, hence within the allowed
interval $[u_i(x),u_i(y)]$. So \eqref{eq:crossing-condition} applies
and gives
$R_{x,y}(k)\le e^{-C_j}\le e^{-L_j}$ on $E_j$, since $C_j\ge L_j$.
Take $H_k^{(1)}=H_k^{(2)}=\emptyset$ in this range.
Now let $|k|>N_j/4$. Set $M_s=m_s^{(4/5)m_s}$. The intervals
\[
[N_j/4,M_j],\qquad [G_sK_s,M_s]\quad(s>j)
\]
cover $[N_j/4,\infty)$: the next left endpoint is below the previous
right endpoint by the scale conditions, the right endpoints tend to
infinity, and $G_sK_s\le N_s/4$ because $m_s\ge4G_s$. Split each later
interval at $N_s/4$. In its lower part the thresholds
$1-\{k\alpha_i\}$ belong to $[a_s,1-a_s]$ at every coordinate: this is
the lower phase assertion of Proposition~\ref{prop:parameters}. In its
upper part, the spread assertion places them in $[b_s,1-b_s]$ at least
$m_s/20$ times. The initial interval uses the upper-part assertion for
$s=j$.
Therefore, for each fixed $k$ in the tail, there is a deterministic
block $J_s$, a width $w\in\{a_s,b_s\}$, and a set $I\subseteq J_s$ of
at least $m_s/20$ indices such that
\[
u_i^*:=1-\{k\alpha_i\}\in[w,1-w]\qquad(i\in I).
\]
The lower-part choice has $s>j$; the upper-part choice could have $s\ge
j$. In either case, the intensity satisfies
\[
m_s\rho_s(w)\ge L_j,\qquad C_s\ge L_j,
\]
because $m_s\rho_s(b_s)=\kappa_s m_s^{1/40}\ge L_s$, and
$m_s\rho_s(a_s)=\kappa_s\sqrt{G_s}\ge L_{s-1}$.
Let $H_k\subseteq E_j$ be the event of no crossing at any index in $I$.
Define
\[
H_k^{(1)}=H_k\cap
\bigl\{\#\{i\in I:u_i(x)<u_i^*\}\ge|I|/2\bigr\},
\qquad
H_k^{(2)}=H_k\setminus H_k^{(1)}.
\]
The set $I$ is fixed, so the first event forbids the \emph{top} label
$w$ at every index in the $x$-measurable set
$\{i\in I:u_i(x)<u_i^*\}$. Conditional independence and the
conditional-crossing bound \eqref{eq:conditional-crossing-top} give,
whenever the $x$-measurable set has $\ge|I|/2$ elements,
\[
\P_j(H_k^{(1)}\mid x)
\le(1-\rho_s(w)/10)^{|I|/2}
\le e^{-m_s\rho_s(w)/400}
\le e^{-L_j/400}.
\]
If the set has fewer than $|I|/2$ elements, $H_k^{(1)}=\emptyset$, so
the same bound holds.
For the second estimate, coordinate order and the absence of crossings
on $H_k$ imply, for every $i\in I$,
\[
u_i(x)<u^*_i\quad\Longleftrightarrow\quad u_i(y)<u^*_i .
\]
Thus $H_k^{(2)}$ requires at least $|I|/2$ indices with $u_i(y)\ge
u^*_i$. Conditional on $y$, the \emph{bottom} labels at all those
indices are forbidden. Conditional independence and
\eqref{eq:conditional-crossing-bot} give
$\P_j(H_k^{(2)}\mid y)\le e^{-L_j/400}$. These estimates use the
original probability law, before restricting to $E_j$.
Finally, $H_k^{(2)}$ was defined inside $E_j$ to make the $y$-measurable
set well-defined; the conditional bound above uses only $\{y_i\ge
u^*_i\}$, which is $y$-measurable.
Outside $H_k$, at least one chosen coordinate crosses. Because all
coordinates are ordered on $E_j$, \eqref{eq:crossing-condition} and
$C_s\ge L_j$ give $R_{x,y}(k)\le e^{-L_j}$. On $H_k$, $R_{x,y}(k)\le1$.
This gives \eqref{eq:tail-ratio-bound} after replacing $H_k$ with
$H_k^{(1)}\cup H_k^{(2)}$: on $H_k\setminus(H_k^{(1)}\cup H_k^{(2)})$
there is no $H$ event, so $R_{x,y}(k)\le e^{-L_j}$; on
$H_k^{(1)}\cup H_k^{(2)}$, $R_{x,y}(k)\le1\le e^{-L_j}+\mathbf 1_{H}$.
The claimed bound follows.
\end{proof}
\subsection{Completion of the contradiction}
Apply Lemma~\ref{lem:conditions-convolution} to the law $\mathbb P_j$
with
\[
K=K_j,\quad D=D_j,\quad a=1/j,\quad c=c_*=e^{-90}/2,
\quad \eta=e^{-L_j},\quad \delta=e^{-L_j/400}.
\]
Hypotheses (i) and the first part of (ii) come from the exact
convolution identity \eqref{eq:convolution-identity} and
Lemma~\ref{lem:truncation-bound}. Hypothesis (iii) is
Lemma~\ref{lem:tail-attenuation}. The lemma therefore gives
\[
c_*\varepsilon_j
\le(2K_j+1)\bigl(e^{-L_j}+2e^{-L_j/800}\bigr).
\]
Since $e^{-L_j}\le e^{-L_j/800}$ and $2e^{-L_j/800}\le3e^{-L_j/10000}$
for $L_j\ge1$, and since $(2K_j+1)\cdot3e^{-L_j/10000}\le\varepsilon_j/j$
by the parameter choice,
\[
c_*\varepsilon_j\le\varepsilon_j/j .
\]
The constant $c_*>0$ is independent of $j$, and $\varepsilon_j>0$, so
$c_*\le1/j$ for all large $j$. Choosing $j>1/c_*$ gives a
contradiction. Therefore there is no projection $p\in M$ with
$0<\tau(p)<1$ and $(1-p)T_*p=0$. This proves the measurable case.
\section{A continuous weight with one zero}
\label{sec:continuous}
The measurable result gives no projection for the weight $f_*$. We now
replace $f_*$ by a continuous weight with exactly one zero, preserving
the obstruction. The key is that the change is a constant plus a
measurable coboundary, and the coboundary gives an intertwining between
$T_f$ and a positive scalar multiple of $T_*$, implemented by an
(unbounded) positive affiliated operator. An affiliated-operator lemma
shows that trace-preserving invariant projections pass over.
\subsection{The affiliated-transport lemma}
A positive self-adjoint operator is \emph{affiliated} with a von Neumann
algebra $M$ if all its spectral projections belong to $M$. Such an
operator commutes with every commutant unitary, including on its domain.
The next lemma is our mechanism for transferring projections.
\begin{lemma}[Affiliated transport]\label{lem:affiliated-transport}
Let $(M,\tau)$ be a finite von Neumann algebra on a Hilbert space with
faithful normal tracial state $\tau$. Let $H$ be an injective
positive self-adjoint operator affiliated with $M$, and let $A,B\in M$.
Suppose
\[
A(\operatorname{Dom}H)\subseteq\operatorname{Dom}H,
\qquad
HA\zeta=BH\zeta\quad(\zeta\in\operatorname{Dom}H).
\]
For every projection $p\in M$ with $A\operatorname{ran}p\subseteq
\operatorname{ran}p$, the projection $P$ onto
$\overline{H(\operatorname{ran}p\cap\operatorname{Dom}H)}$
belongs to $M$, is invariant under $B$, and satisfies $\tau(P)=\tau(p)$.
\end{lemma}
\begin{proof}
Every commutant unitary $W\in M'$ commutes with $H$ on its domain,
preserves $\operatorname{ran}p\cap\operatorname{Dom}H$, and maps the
closed subspace $\overline{H(\operatorname{ran}p\cap\operatorname{Dom}H)}$
onto itself. Its orthogonal projection therefore commutes with $W$.
Varying $W\in M'$ over all commutant unitaries gives
$P\in M''=M$.
Invariance under $B$: for $\zeta\in\operatorname{ran}p\cap\operatorname{Dom}H$,
the first assumption gives $A\zeta\in\operatorname{ran}p\cap
\operatorname{Dom}H$, and the intertwining identity gives
$BH\zeta=HA\zeta\in H(\operatorname{ran}p\cap\operatorname{Dom}H)$.
By boundedness of $B$, the closed range is invariant.
Trace: put $E_N=\one_{[1/N,N]}(H)$. Since $H$ is injective,
$E_N\uparrow\one$. Define $q_N=p\wedge E_N$ in the projection lattice
of $M$. On $\operatorname{ran}q_N$ the bounded element
$Hq_N=(HE_N)q_N\in M$ is bounded below by $1/N$ on its initial space, so
its image is closed, and the polar decomposition of $Hq_N$ identifies
the initial projection $q_N$ with its final projection, which is
dominated by $P$. Therefore
\[
\tau(P)\ge\tau(q_N).
\]
The polar decomposition of $(1-E_N)p$ has initial projection
$p-q_N=p-(p\wedge E_N)$ and final projection dominated by $1-E_N$;
since the initial and final projections of a polar decomposition have
the same trace,
\[
\tau(p)-\tau(q_N)\le\tau(1-E_N),
\quad\text{i.e.}\quad
\tau(q_N)\ge\tau(p)+\tau(E_N)-1.
\]
Hence $\tau(P)\ge\tau(p)+\tau(E_N)-1$; letting $N\uparrow\infty$ gives
$\tau(P)\ge\tau(p)$.
For the opposite inequality let $q'_N=(1-p)\wedge E_N$. The bounded
positive element $H^{-1}q_N'=(H^{-1}E_N)q'_N$ is again bounded below on
its initial space, has closed image, and has the same trace as $q_N'$.
Its image is orthogonal to $\operatorname{ran}P$: for
$\eta\in\operatorname{ran}q_N'$ and
$\zeta\in\operatorname{ran}p\cap\operatorname{Dom}H$, self-adjointness
of $H$ and the identity $H^{-1}\eta\in\operatorname{Dom}H$ give
\[
\langle H^{-1}\eta,H\zeta\rangle=\langle\eta,\zeta\rangle=0.
\]
Thus the final projection of $H^{-1}q_N'$ is at most $1-P$, giving
$\tau(q_N')\le\tau(1-P)$. Polar decomposition likewise gives
$\tau(q_N')\ge\tau(1-p)+\tau(E_N)-1$, so $\tau(1-P)\ge\tau(1-p)$ for
every $N$, i.e. $\tau(P)\le\tau(p)$. Combining with
$\tau(P)\ge\tau(p)$ gives equality.
\end{proof}
\subsection{A continuous cost via smoothing}
The first step replaces each one-coordinate cost by a smoothed version
that is continuous. Write $\delta_i=\alpha_i/4$ and, for $u\in X$,
\[
s_i(u)=\frac1{2\delta_i}\int_{-\delta_i}^{\delta_i}\{u+t\}\,dt.
\]
This is a smoothed sawtooth: $s_i(u)=\{u\}$ outside the
$\delta_i$-neighborhood of the cut at $0$. Set
\[
h_i(x)=C_j\bigl(\alpha_i+s_i(u_i(x))-s_i(u_i(x)+\alpha_i)\bigr)
\quad(i\in J_j).
\]
The elementary identity
$\lfloor\{u\}+\alpha\bigr\rfloor=\alpha+\{u\}-\{u+\alpha\}$ (for
$0<\alpha<1$) and the smoothing identity show that $h_i\ge0$ and $h_i$
is the smoothing average of the one-coordinate contribution
$C_j\lfloor u_i(x)+\alpha_i\rfloor$. The elementary computations give
\[
\int_Xh_i\,d\mu=C_j\alpha_i,\qquad
\mu(\operatorname{supp}h_i)\le\alpha_i+2\delta_i.
\]
Let
\[
d_i^{\rm sm}(x)=C_j\bigl(s_i(u_i(x))-u_i(x)\bigr).
\]
Then the exact measurable identity
\begin{equation}\label{eq:smoothing-coboundary}
h_i(x)=C_j\lfloor u_i(x)+\alpha_i\rfloor+d_i^{\rm sm}(x)
-d_i^{\rm sm}(\sigma x)
\end{equation}
holds almost everywhere. The support of $d_i^{\rm sm}$ has measure at
most $2\delta_i$: it is a subset of the $\delta_i$-neighborhood of the
cut, and the exceptional set for both $(u_i(x))$ and $(u_i(x)+\alpha_i)$
is of measure at most $2\delta_i$.
\subsection{Moving each piece into a shrinking arc}
Recall the arcs $O_j$ and the partition of unity $\phi_{j,n}$,
$0\le n\le h^*_j$, prepared in the parameters proof, with
$\sum_n\phi_{j,n}=1$ and $\sigma^n(\operatorname{supp}\phi_{j,n})\subseteq O_j$.
Define
\[
h'_i(x)=\sum_{n=0}^{h^*_j}(h_i\phi_{j,n})(\sigma^{-n}x).
\]
This is continuous and nonnegative, supported in $O_j$, and has the same
integral as $h_i$. Telescoping gives
\begin{equation}\label{eq:relocation-coboundary}
h'_i=h_i+d'_i-d'_i\circ\sigma,
\qquad
d'_i(x)=\sum_{n=1}^{h^*_j}\sum_{r=1}^n(h_i\phi_{j,n})(\sigma^{-r}x).
\end{equation}
The transfer $d'_i$ has support in just $h^*_j$ translates of
$\operatorname{supp}h_i$, one for each possible $r\in\{1,\dots,h^*_j\}$;
consequently
\[
\sum_i\Bigl(\mu\{d_i^{\rm sm}\ne0\}+\mu\{d'_i\ne0\}\Bigr)
\le C\sum_j\frac{1+h^*_j}{K_j}<\infty,
\]
using $\mu\{\operatorname{supp}h_i\}\le\alpha_i+2\delta_i\le2\alpha_i$
and $\alpha_i<2/N_j$. Each of the transfer sums
$\sum_i(d_i^{\rm sm}+d'_i)$ has only finitely many nonzero terms almost
everywhere, including at the translated point $\sigma x$.
The sum $\sum_ih'_i$ is locally finite on $X\setminus\{0\}$ because
successive blocks are supported in $O_j\downarrow\{0\}$; it is
therefore continuous there. Its integral is infinite:
\[
\int_X\sum_ih'_i\,d\mu=\sum_jC_j\sum_{i\in J_j}\alpha_i
\ge\sum_j\frac{C_j}{K_j}=\infty
\]
by Tonelli's theorem and $\alpha_i>1/N_j$ and $C_j\ge K_j^2$.
\subsection{A background cost that diverges at the zero}
Local finiteness on $X\setminus\{0\}$ does not force divergence at $0$.
We add a background cost.
Let $W_j=2^{j+1}$. Choose continuous $0\le v_j\le1$ with $v_j(0)=1$,
supported in closed arcs $I_j$ shrinking to $\{0\}$, small enough that
$\sigma I_j,\sigma^2I_j,\dots,\sigma^{W_j}I_j$ are pairwise disjoint and
\[
\sum_jW_j\mu(I_j)<\infty.
\]
The arcs can be chosen so small because the finitely many angles
$\theta,\dots,W_j\theta$ have positive minimum separation.
Define, for $x\ne0$,
\[
g_0(x)=1+\sum_j\Bigl(v_j(x)-\frac1{W_j}\sum_{n=1}^{W_j}
v_j(\sigma^{-n}x)\Bigr).
\]
The subtracted series has norm bounded by $\sum_jW_j^{-1}\cdot W_j\cdot1
=\sum_jW_j^{-1}\cdot$ ... more precisely, using disjointness at each
$j$,
\[
\Bigl\|\frac1{W_j}\sum_{n=1}^{W_j}v_j(\sigma^{-n}\cdot)\Bigr\|_\infty
\le\frac1{W_j},
\]
so the subtracted series converges uniformly and its norm is at most
$1/2$. The positive series is locally finite away from $0$. Hence
$g_0$ is continuous on $X\setminus\{0\}$ and $g_0\ge1/2$.
As $x\to0$, the positive part $\sum_{j\le J}v_j(x)\to J$ for every
fixed $J$. The subtracted series is bounded in absolute value by
$1/2$ uniformly in $x$, so $g_0(x)\to\infty$ as $x\to0$.
Write
\[
b_0(x)=\sum_j\frac1{W_j}\sum_{n=1}^{W_j}\sum_{r=1}^n
v_j(\sigma^{-r}x).
\]
The $j$th summand is supported in $\bigcup_{r\le W_j}\sigma^rI_j$, and
$\sum_jW_j\mu(I_j)<\infty$, so $b_0$ is finite almost everywhere. On
the common conull set where $b_0\circ\sigma$ and $b_0$ are both finite
(which is conull), telescoping gives
\[
g_0=1+b_0\circ\sigma-b_0.
\]
Now define
\[
g(x)=g_0(x)+\sum_ih'_i(x)\quad(x\ne0),\qquad g(0)=\infty,
\]
\[
b(x)=b_0(x)-\sum_i(d_i^{\rm sm}(x)+d'_i(x)).
\]
The prior arguments show $g$ is nonnegative, finite and continuous on
$X\setminus\{0\}$, tends to infinity at $0$, and has infinite integral.
The sum defining $b$ is finite almost everywhere. Summing
\eqref{eq:smoothing-coboundary} and \eqref{eq:relocation-coboundary}
gives, on a conull set,
\begin{equation}\label{eq:continuous-coboundary}
g=g_*+1+b\circ\sigma-b .
\end{equation}
\subsection{The continuous weight and its determinant}
Define on the unit circle
\[
f(e^{2\pi ix})=e^{-g(x)},\qquad e^{-\infty}=0.
\]
This is a nonnegative continuous function; it vanishes exactly at the
point $1$, because $g(x)=\infty$ only at $0$, and $g(x)<\infty$
everywhere else.
\begin{lemma}\label{lem:continuous-weight-basic}
The function $f$ is continuous, nonnegative, zero exactly at $1$,
nonzero elsewhere, and
\[
\int_\T\log f\,dm=-\infty,\qquad \Delta(f(V))=0.
\]
Its modulus $|f|=f$ is nonconstant and its zero set is a single point.
\end{lemma}
\begin{proof}
Continuity and the zero set were checked. For the determinant, the
Fuglede--Kadison determinant of $f(V)$ is $\exp(\int\log f\,dm)$:
indeed, the spectral distribution of $f(V)$ is the Haar distribution of
$f$, so $\log\Delta(f(V))=\int\log f\,dm$. (This is the definition of
the determinant in a finite factor.) The integral is $-\infty$ because
$\int g\,d\mu=\infty$ and $f=e^{-g}$. Also $|U\pi(f)|=\pi(f)=f(V)$,
so the same computation applies to $T_f=U\pi(f)$.
\end{proof}
\subsection{The intertwining by an affiliated operator}
Let $H_b$ be the operator that multiplies the $k$-th fiber coordinate
by $e^{b(\sigma^kx)}$:
\[
(H_b\zeta)(x)_k=e^{b(\sigma^kx)}\zeta_k(x),
\qquad
\operatorname{Dom}H_b=\Bigl\{\zeta\in\mathcal H:
\sum_k\int e^{2b(\sigma^kx)}|\zeta_k(x)|^2\,d\mu(x)<\infty\Bigr\}.
\]
$H_b$ is positive, self-adjoint, and injective on its domain. It is
affiliated with $M$: its spectral projections are the $\pi$-images of
spectral indicators of $e^b\in L^\infty(X)$.
\begin{lemma}[Intertwining]\label{lem:intertwining}
For $\zeta\in\operatorname{Dom}H_b$,
\[
T_f\zeta\in\operatorname{Dom}H_b,
\qquad
H_bT_f\zeta=e^{-1}T_*H_b\zeta,
\]
where $e$ is Euler's number.
\end{lemma}
\begin{proof}
Recall \eqref{eq:continuous-coboundary}:
$g=g_*+1+b\circ\sigma-b$ almost everywhere. By the coordinate
calculation
\[
(T_f\zeta)(x)_k=e^{-g(\sigma^{k-1}x)}\zeta_{k-1}(x),
\]
\[
(H_bT_f\zeta)(x)_k=e^{b(\sigma^kx)}e^{-g(\sigma^{k-1}x)}\zeta_{k-1}(x).
\]
Also
\[
(T_*H_b\zeta)(x)_k=e^{-g_*(\sigma^{k-1}x)}e^{b(\sigma^{k-1}x)}
\zeta_{k-1}(x).
\]
By \eqref{eq:continuous-coboundary} evaluated at $\sigma^{k-1}x$,
\[
e^{-g(\sigma^{k-1}x)}
=e^{-g_*(\sigma^{k-1}x)-1-b(\sigma^kx)+b(\sigma^{k-1}x)}.
\]
Multiplying by $e^{b(\sigma^kx)}$ gives
\[
e^{b(\sigma^kx)}e^{-g(\sigma^{k-1}x)}
=e^{-1}e^{-g_*(\sigma^{k-1}x)}e^{b(\sigma^{k-1}x)}.
\]
Therefore $(H_bT_f\zeta)(x)_k=e^{-1}(T_*H_b\zeta)(x)_k$ for every $k$
and a.e.\ $x$. Because $T_*$ is bounded and $H_b\zeta\in L^2$
(by $\zeta\in\operatorname{Dom}H_b$), the right-hand side is in
$L^2$. Hence $H_bT_f\zeta\in L^2$, which means precisely
$\sum_k\int e^{2b(\sigma^{kx})}|(T_f\zeta)_k(x)|^2d\mu(x)<\infty$,
i.e.\ $T_f\zeta\in\operatorname{Dom}H_b$.
\end{proof}
Then Lemma~\ref{lem:affiliated-transport} with $A=T_f$, $B=e^{-1}T_*$,
$H=H_b$ sends any invariant projection of $T_f$ to an invariant
projection of $e^{-1}T_*$ with the same trace. Since $e^{-1}\ne0$,
invariance under $e^{-1}T_*$ is the same as invariance under $T_*$.
The measurable-weight result gives that no projection $P\in M$ with
$0<\tau(P)<1$ is invariant under $T_*$. Therefore no projection
$p\in M$ with $0<\tau(p)<1$ is invariant under $T_f$.
This proves Theorem~\ref{thm:main}.
\section{Quasinilpotence}
\label{sec:quasinilpotent}
The operator $T_f$ is not just projection-free; it is quasinilpotent.
This is the zero-determinant case of the spectral radius formula of
Zhu, Fang, and Shi \cite[Thm.~2.3]{zfs}. Their formula uses the
truncated logarithm and only needs uniform averaging under irrational
rotation.
\begin{corollary}\label{cor:quasinilpotent}
The operator $T_f$ is nonzero and norm-quasinilpotent, i.e.\
$\lim_{n\to\infty}\|T_f^n\|^{1/n}=0$. Also $\Delta(T_f)=0$.
\end{corollary}
\begin{proof}
$\Delta(T_f)=0$ was proved in
Lemma~\ref{lem:continuous-weight-basic}. $T_f\ne0$ because $f>0$
almost everywhere.
For $M>0$ put $\ell_M(t)=\log\max(f(e^{2\pi it}),e^{-M})$. Each
$\ell_M$ is continuous, and $\int_X\ell_M\,d\mu\downarrow-\infty$ by
$\int\log f=-\infty$. Uniform averaging under irrational rotation
(Lemma~\ref{lem:weyl}) gives
\[
\sup_{x\in X}\Bigl|\frac1n\sum_{k=0}^{n-1}\ell_M(\sigma^kx)
-\int_X\ell_M\,d\mu\Bigr|\to0\qquad(n\to\infty)
\]
for each fixed $M$. Fix $a>0$, choose $M$ large enough that
$\int_X\ell_M\,d\mu<-2a$, and then $n$ large enough that the average
stays below $-a$ for every $x$. Then, using
$f\le\max(f,e^{-M})=e^{\ell_M}$ pointwise,
\[
\sup_{x\in X}\prod_{k=0}^{n-1}f(\sigma^kx)
\le\exp\Bigl(\sup_{x\in X}\sum_{k=0}^{n-1}\ell_M(\sigma^kx)\Bigr)
\le e^{-an}.
\]
The product $\prod_{k=0}^{n-1}f(\sigma^kx)$ is exactly $\|T_f^n\|$: by
direct computation and the cocycle $F_n=f_*^n$, the operator $T_f^n$ is
$U^n$ times multiplication by the continuous product, and the norm of
$U^n\pi(h)$ on $\ell^2(\Z)\otimes L^2(X)$ is $\|h\|_\infty$. See
Lemma~\ref{lem:weyl-family} below for the interchange needed here.
That is, $\|T_f^n\|^{1/n}\le e^{-a}$ eventually. Since $a>0$ was
arbitrary, $\lim\|T_f^n\|^{1/n}=0$.
\end{proof}
\begin{lemma}[Norm formula for weighted shifts]\label{lem:weyl-family}
For any bounded Borel $h:X\to\C$,
\[
\|U\pi(h)\|_{B(\mathcal H)}=\|h\|_\infty.
\]
More generally, $\|(U\pi(f))^n\|=\|\prod_{k=0}^{n-1}f\circ\sigma^k\|_\infty$.
\end{lemma}
\begin{proof}
$(U\pi(h)\zeta)(x)_k=h(\sigma^{k-1}x)\zeta_{k-1}(x)$, so
\[
\|U\pi(h)\zeta\|^2=\sum_k\int|h(\sigma^{k-1}x)|^2|\zeta_{k-1}(x)|^2\,d\mu(x)
\le\|h\|_\infty^2\|\zeta\|^2.
\]
Equality is attained by $\zeta(x)$ supported near a point where
$|h|=\|h\|_\infty$, so the operator norm is exactly $\|h\|_\infty$. The
iterated statement follows by direct computation using
$f_*=e^{-g_*}$ and the cocycle $F_n$:
\[
(T_f^n\zeta)(x)_k=f(\sigma^{k-1}x)\cdots f(\sigma^{k-n}x)\zeta_{k-n}(x),
\]
which is $U^n$ times multiplication by $\prod_{j=0}^{n-1}f(\sigma^j\cdot)$
up to a coordinate shift. The same argument as for $U\pi(h)$ shows the
norm equals the sup of the product.
\end{proof}
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\end{document}