A zero-entropy system without a smooth positive-volume model

Proof. Consider a block of length 2^l, and let U and V be the averages on its left and right halves. Its own average is W=(U+V)/2. The identity

and stationarity give

The last equality also follows pointwise from V-W=-(U-W). Thus v_l\geq0, and telescoping the differences of second moments proves the asserted bound on their sum. No independence of the two halves is used.

We next compare the cumulative-sum polygon on [0,2^h] with its chord. For a=0,\ldots,h, let F_a be the piecewise-linear interpolant that agrees with the cumulative sum at the multiples of 2^{h-a}. Then F_0 is the chord and F_h agrees with every integer-time cumulative sum. Passing from F_a to F_{a+1} inserts the midpoint of each of 2^a intervals of length 2^{h-a}. If U_j,W_j are respectively the left-child and parent averages on the jth such interval, the displacement at its midpoint is 2^{h-a-1}(U_j-W_j). The difference of the interpolants is linear between the old and new vertices, so

Bounding a squared maximum by the sum of squares and using stationarity of each interval gives

The triangle inequality, first for the polygonal sup norm and then in L^2, proves (2.1).

Extend the sequence (\sqrt{v_l})_{l\geq1} by zero to nonpositive indices. Its squared \ell^2 norm is at most \E Y_0^2. The right side of (2.1) is its convolution with the summable sequence (2^{-a/2-1})_{a\geq0}. The triangle inequality for the \ell^2 norms of its translates therefore proves (2.2). Finally, stationarity identifies the distribution of each translated block with that of the block just considered.□

Proof. Apply Lemma 2.1 to each g_j. Let c_{j,h} be the right side of (2.1) for that process, and put \alpha_h^2=\sum_{j=1}^d c_{j,h}^2. The lemma gives \sum_h\alpha_h^2<\infty. For example, take

This sequence has the properties in (2.5). The interval [-n,n) has length 2^{k+1}. Stationarity and the squared Markov inequality give

Define \widetilde{\mathcal G}_k by the complementary non-strict inequalities. These are finitely many continuous inequalities on a compact bundle, so this set and its projection are compact. Moreover

The pushforward identity \pi_*\widetilde\nu=\nu gives the asserted base-measure estimate. No measurable choice of a frame over each point of \mathcal G_k is needed. Finally, subtracting the chord error at u from the chord error at v gives (2.7), with the same rates \lambda_j.□

Proof. Diagonal normalization. Write

The chord hypothesis gives \|D_u\|,\|D_u^{-1}\|\leq e^{bn}. The matrices

are upper triangular with constant diagonal (e^{\lambda_1},\ldots,e^{\lambda_d}) and entries bounded by e^{Cbn}. Since the rates are averages of logarithms of diagonal entries,

Their inverse matrices also have entries bounded by e^{Cbn}.

We record how triangular products are estimated. If m is fixed and upper-triangular m-by-m matrices have constant diagonal e^{\alpha_1},\ldots,e^{\alpha_m}, bounded |\alpha_i|, and off-diagonal entries at most e^{Cbn}, a term in any product entry can change its coordinate index strictly at most m-1 times. There are at most a fixed constant times (2n+1)^{m-1} choices for those times and the intermediate indices. If the product length is l and a term has a\leq m-1 strict changes, its diagonal factors contribute at most e^{\alpha_{\max}(l-a)}. The omitted a diagonal factors cost at most a fixed factor e^{a|\alpha_{\max}|} after e^{\alpha_{\max}l} is factored out. The off-diagonal factors contribute e^{C_mbn}. The polynomial number of terms is absorbed by \ln(2n+1)\leq bn. Thus the product norm is at most e^{\alpha_{\max}l+C_mbn}.

Apply this argument to \mathord{\bigwedge}^q\widehat A_t in the coordinate-wedge basis, ordered lexicographically. This matrix is upper triangular of fixed dimension \binom dq, and its diagonal entries are \exp(\sum_{j\in I}\lambda_j) for q-element index sets I. The largest such sum is \sum_{j=1}^q\lambda_{(j)}. For the reverse inequality, the diagonal of the product indexed by a maximizing set I is exactly \exp(l\sum_{j\in I}\lambda_j); that matrix entry is no larger than the operator norm. Converting back multiplies the exterior norms only by the two endpoint gauge bounds. This proves (3.1). Applying the same argument to inverse normalized steps in reverse time proves its backward version. The cases q=0,d follow as well: the former is the identity, and the latter is the determinant line.

Removing entries between the groups. We next conjugate \widehat A_t by time-dependent upper unitriangular matrices to make the two coordinate index groups invariant. Process the superdiagonals in increasing order of their distance j-i from the diagonal. At a position i<j belonging to different groups, suppose the current entry is h_t. Use an elementary change Q_t=\Id+u_tE_{ij}, where E_{ij} has its only nonzero entry, equal to one, in position (i,j). In Q_{t+1}^{-1}\widehat A_tQ_t the new entry at this position is

To set it to zero, solve

If q<1, prescribe u_{-n}=0 and solve forward:

If q>1, prescribe u_n=0 and solve backward:

In each formula the coefficients have modulus at most one, so |u_t|\leq2n\max_a|f_a|. Different groups cannot have q=1. This bound avoids dividing by the possibly small separation of rates.

Right multiplication by Q_t changes column j by a multiple of column i, and left multiplication by Q_{t+1}^{-1} changes row i by a multiple of row j. Upper triangularity shows that every affected off-diagonal position has distance at least j-i; at that distance only (i,j) changes. The possible quadratic cross term is zero because E_{ij}\widehat A_tE_{ij}=(\widehat A_t)_{ji}E_{ij}=0. Thus a later operation cannot restore an entry on an earlier superdiagonal.

There are at most d(d-1)/2 operations. At each operation the bound above, the current entry bounds, and (3.6) bound the shear and its inverse by e^{Cbn}. Multiplying or adding a fixed number of such quantities only changes C; the factor 2n is absorbed by \ln(2n+1)\leq bn. Induction over this fixed number of operations therefore gives matrices C_t with

where B_t is upper triangular, has diagonal e^{\lambda_j}, and has no entries between the two groups. Its entries and those of B_t^{-1} are bounded by e^{Cbn}. All constants are independent of the length of the interval.

Let \R^F be the coordinate span with indices in F and \R^{F^c} its coordinate complement. Set

These spaces are invariant under the original matrices. Their projections are conjugates by C_t of coordinate projections, and hence have norm at most e^{Cbn} after changing C. The two groups may be interlaced in the original coordinate order. Each induced block is nevertheless upper triangular in its inherited order; a fixed permutation makes the two blocks consecutive if desired. The triangular-product estimate within these blocks proves (3.2).

Exterior volumes and mixed factors. The fast rates are exactly the p largest rates. The general exterior bound already proves the first inequality of (3.3). The normalized fast block has determinant e^{\Lambda l} on an interval of length l. To account explicitly for its possibly oblique basis, define

For p=0 this Gram determinant is one. The endpoint bounds in (3.8) imply e^{-Cbn}\leq V_F(t)\leq e^{Cbn}. The intrinsic volume expansion is exactly

This proves both volume bounds in (3.3). Applying the same argument to the inverse slow block gives (3.4), including either sign of \Omega.

The normalized block decomposition induces an invariant decomposition of each exterior power according to the number of fast and slow factors. In degree p, every sector with a slow factor has diagonal rate sum at most \Lambda-2g: at least one fast rate has been replaced by a slow rate. The triangular-product estimate therefore bounds this entire sector by e^{(\Lambda-2g)l+Cbn}. If one original vector is slow, split each of the other vectors into its fast and slow parts. There are only a fixed number of terms, each still having a slow factor. The projection norms and endpoint gauges are absorbed into e^{Cbn}, proving (3.5). Reversing time and exchanging the groups gives the backward statement. The empty-group cases require none of these mixed-factor assertions and follow from the stated conventions.□

Proof. Short products in common charts. Choose one reference point in a starting bin. Uniform Lipschitz bounds for S and S^{-1} give, for another point in that bin,

with a fixed A_0>1. Since L=o(Bn), this tends to zero uniformly. At each time on either short reference orbit choose a chart whose smaller subchart contains the reference point. All nearby orbits then remain in the same larger chart at that time. Use the prescribed bin chart at the initial time. The one-step coordinate derivatives and their inverses are bounded by a fixed constant, and their differences are at most C A_0^{|t|}h_0. The telescoping product identity consequently gives

Here P_x^+ and P_x^- denote DS^L_x and DS^{-L}_x, respectively, in these common initial and terminal charts. For example, summing the L terms in the product difference bounds it by C L A_1^{2L}h_0, which has the stated form. The inverse product bound gives the minimum stretch. On a manifold with boundary the same proof uses the nested half-box charts and uniform one-sided smooth bounds; equivalently the finitely many local expressions may be extended to their surrounding boxes.

The singular values are Lipschitz with respect to operator norm. Since \zeta_k/m_k\to0, (3.9) implies, after increasing the constants if necessary,

For completeness, the singular-value Lipschitz bound follows from the min–max characterization, because \|Pv\| and \|Qv\| differ by at most \|P-Q\| on unit vectors. The logarithmic inequality then follows from the lower bound m_k.

A threshold for every witness. Let \lambda_{(j)}(e) be the sorted rates of a good witness e. Subtracting (3.1) for exterior degrees j and j-1 gives

This argument uses the no-gap part of Proposition 3.1; a threshold has not yet been chosen. Fix one good witness e_* in U. Equations (3.10) and (3.11) show that every good witness in U, including every other witness over the same point, satisfies

Consider the d+1 candidate thresholds (1+8j)g, 0\leq j\leq d. A real number lies within distance 3g of at most one candidate. The d representative rates therefore leave a candidate whose distance from all of them is greater than 3g. For large k, all the other sorted rate lists differ by less than g, so this candidate separates every list with a margin greater than 2g. It has the weaker required margin g, and the number of rates above it is constant. Fix this candidate as \gamma_U.

Comparison of subspaces using one short map. Assume first that 0<p_U<d and put q=d-p_U. Use the single forward short map P_*=P_{\pi(e_*)}^+. Let W be its bottom q right-singular subspace. Its top p_U singular values are at least \exp((\gamma_U+g)L-Cbn) by (3.11). If v is a unit vector in the slow space of any good witness e, then (3.2) and (3.9) give

Orthogonality of the images of right-singular vectors implies

In the last bound, constants absorb |\gamma_U|=O_d(g) and the other terms bounded by a constant times L. This is at most e^{-c_1gL} for some c_1>0 and large k.

To pass from this directed estimate to a subspace estimate, observe that if \|\Pi_{W^\perp}|E\|\leq\epsilon<1 and \dim E=\dim W, then \Pi_W|E is an isomorphism. The space E is a graph over W of norm at most \epsilon/\sqrt{1-\epsilon^2}. The corresponding orthogonal projectors differ by at most a fixed multiple of \epsilon when \epsilon\leq1/2. Thus every slow space is within C e^{-c_1gL} of this same W; the triangle inequality compares any two slow spaces. No continuity assertion about singular vectors or chosen good frames has been used.

For fast spaces use the single backward short map P_{\pi(e_*)}^-. Its bottom p_U right-singular space is the comparison space. The top q singular values are at least \exp(-(\gamma_U-g)L-Cbn), whereas a unit fast vector has backward stretch at most \exp(-(\gamma_U+g)L+Cbn). The identical calculation gives the error bound

This proves (i) after reducing c to absorb fixed factors. In particular, the error is o(e^{-C_0bn}) for every fixed C_0, by bn=o(gL).

Endpoint comparisons. Keep the threshold of U. For paths starting in U and ending at time n in a common bin V, choose one representative path for this pair of bins. Apply the preceding short-product comparison from their time-n points backward for length L. Their initial closeness for this comparison is supplied by V, not by propagating the starting-bin closeness for n steps. All intervals [n-L,n] lie inside the good interval, and all paths use the same group dimension and threshold. In particular, (3.1) applies on each [n-L,n] with the same full-segment rates, so the singular-value comparison and the backward argument just proved compare their fast spaces at n. At -n, the identical all-subinterval estimate for forward short maps on [-n,-n+L] gives the slow-space assertion.

Finally, if E and its chosen representative F have equal dimension and \|\Pi_E-\Pi_F\|\leq\epsilon, then for v\in E,

The restriction is onto F by equality of dimensions, and its inverse has norm at most (1-\epsilon^2)^{-1/2}\leq2 for large k. If one group is empty, the associated spaces are identically zero or the full coordinate space, so all comparisons and projection conclusions follow directly.□

Proof. The exterior power is multiplicative. Multilinearity in its p columns, together with the one-step bound K+1 for the perturbed matrices, gives

Expand the product using either the unperturbed exterior matrix or E_t at each time. A term with q\geq1 insertions has q+1 unperturbed intervals whose total length is m-q. Allowing empty intervals, the hypothesis bounds that term by

There are \binom mq choices of the inserted positions. Sum over q=1,\ldots,m and use -\sigma\leq|\sigma| to prove (3.12). This calculation explains why negative \sigma, as may occur for backward exterior volumes, is harmless: each removed time step contributes only the displayed factor e^{-\sigma}.

For the final assertion put u=C\epsilon e^{a+|\sigma|} and use (1+u)^m-1\leq mu e^{mu}. After division by e^{\sigma m-C_0bn} the bound is at most

The logarithm of its decisive first factor is at most -3Bn/2+O(bn)+\ln(2n), which tends to -\infty by (2.9). The quantity in the second exponential tends to zero for the same reason. All constants are fixed, proving the uniform little-oh assertion. For exterior degree zero the two products are both the identity, so the result holds without an expansion.□

Proof. Uniform coordinates. At each reference location choose one of finitely many slightly larger regular patches containing a fixed-radius neighborhood of that location. At time zero both strings use \chi_Q; at the terminal times they use the endpoint charts already specified. Shrinking the fixed radius once, the one-step coordinate maps \chi_{t+1}\circ S\circ\chi_t^{-1} and their first two derivatives are uniformly bounded wherever they will be used. The inverse one-step maps also have uniform first and second derivative bounds. For example, differentiating the inverse identity bounds their second derivatives by a fixed multiple of the forward second-derivative bound and the cube of the inverse first-derivative bound. This follows from a finite atlas with nested compact patches and smoothness on compact M; it does not depend on the number of steps. A patch can be chosen separately at each internal time. No iterate S^n is assigned a uniform second derivative bound.

Use displacement variables v_t^+ about S^tx, 0\le t\le n, and v_t^- about S^ty, -n\le t\le0. Write A_t^+,A_t^- for the derivative steps in these charts. Matching the two locations at zero and imposing the terminal equations gives the linear system

The matching datum for the actual problem is m=\chi_Q(y)-\chi_Q(x); its norm is O(r). The actual terminal data are zero.

The inverse in weighted coordinates. Let \gamma=\gamma_Q and put \widetilde v_t=e^{-\gamma t}v_t. Use the same weighting on equation data, with the step datum at its output time t+1. By Proposition 3.1, the weighted propagators on slow spaces forward and fast spaces backward have norm at most e^{Cbn}e^{-g|t-s|}. The associated projections have norm at most e^{Cbn}. Geometric sums cost at most C/g\le e^{Cbn}, by (4.5) and the floor on b.

On the future half, prescribe a slow vector a\in E_0^s(x). Variation of constants gives its slow part at every time by propagating a forward and adding the projected step data. If s_n is the resulting slow value at time n, the fast value there is uniquely

Propagate c_n and the fast step data backward. All forced terms have norm at most e^{Cbn} times the sup norm of the weighted data. The dependence of the resulting fast value at zero on a is the map

Here \widetilde A(t,s)=e^{-\gamma(t-s)}A(t,s). Both decaying factors are present in (4.14); there is no assumption that the endpoint projection annihilates the slow space. Thus \widetilde v_0^+=a+K_+a+q_+, with q_+ bounded by the weighted data as above.

On the past half prescribe b'\in E_0^f(y), propagate fast backward, use the slow endpoint projection to fix the slow value at -n, and then propagate slow forward. This gives \widetilde v_0^-=b'+K_-b'+q_-, where

The map (a,b')\mapsto a-b' from E_0^s(x)\oplus E_0^f(y) to the starting coordinate space has inverse norm at most e^{Cbn}. Indeed the corresponding reference spaces are complementary with this bound, and their perturbation is e^{-cgL}; multiplication by the reference inverse still gives a perturbation tending to zero. The terms K_+a-K_-b' are smaller still. A Neumann series therefore solves the matching equation with inverse cost e^{Cbn}. This proves an e^{Cbn} inverse bound for the full weighted system. Removing the weights costs at most e^{2\gamma(n+1)}, so its unweighted inverse \mathcal L^{-1} satisfies

If a group is empty, omit its variables, propagation formula, and terminal datum. The matching map on the remaining spaces and the same bounds apply, including the cases p_Q=0,d.

Contraction with a larger uniqueness radius. Taylor expansion of each one-step map gives a residual \mathcal R(v) with

on the sup norm ball of radius \delta. Every residual coordinate depends on just its one step, so the constant has no factor n. The matching and terminal equations are affine and have zero nonlinear residual. The nonlinear problem is v=\mathcal L^{-1}(a_0+\mathcal R(v)), where \|a_0\|_\infty\le Cr. The scale relations give

For large k the displayed map sends the \delta-ball into its \delta/2-ball and has Lipschitz constant at most 1/4. It has a unique fixed point in the larger ball, with \|v\|_\infty\le 2CQ_kr. The exact one-step equations and the matching equation make this array one genuine orbit; call its point at zero z. Chart bounds give (4.11). In the boundary case its distance from each reference orbit is o(h_0), so it stays inside M.

Time-zero coordinates and wall margins. For the linear solution with zero step and terminal data, the two initial displacement spaces are the graphs in (4.14) and (4.15). In the reference product coordinates they differ from the slow and fast coordinate spaces by at most e^{Cbn}(e^{-cgL}+e^{-2gn+Cbn}). Solving their matching equation therefore differs from ideal exchange of the fast and slow coordinates by at most

All basis conversion and matching inverse factors are included in the constant C. The nonlinear solution differs from the linear one by at most CQ_k(2CQ_kr)^2=e^{Cgn}r^2 in ordinary coordinates, and still by e^{Cgn}r^2=o(r) in product coordinates. This proves (4.12). The proof for w is identical with the two inputs exchanged.

Increase k so that the error in each product coordinate is less than \tau r/4. The ideal exchanged point has distance at least \tau r from every product face because both inputs do. Thus z,w have coordinate margins at least 3\tau r/4 in their original cell. Similarly, (4.11) in the fixed endpoint charts is less than \tau h_0/4, so each corresponding endpoint retains margin at least 3\tau h_0/4 in its original coarse bin. These strict margins will recover all labels from the outputs. The point obtained does not depend on the auxiliary internal charts or on the frame witnesses used to justify the estimates. Two such constructions have the same endpoint equations and sharp radius e^{Cgn}r=o(\delta). In either one’s charts both are therefore in its uniqueness ball, and hence coincide.□

Proof. Put p=p_Q. For the future map at x, use the actual fast and slow spaces first, with orthonormal bases within each. The fast p-minor has absolute value at least e^{\Lambda_x n-Cbn}, where \Lambda_x is the sum of its fast rates. This follows from the fast volume lower bound in Proposition 3.1 and from the uniformly invertible restriction of the endpoint projection. Every other maximal minor contains a slow starting vector and is at most e^{(\Lambda_x-2g)n+Cbn} when both groups occur. Replacing the actual bases with the common reference basis changes their spaces by e^{-cgL}. After a change of orthonormal basis within each space, multilinearity and the full exterior norm bound show that the leading minor stays nonzero and that all other maximal minors are o(1) times its absolute value. Factors from the inverse of the possibly oblique starting basis are at most e^{Cbn}. Call the resulting leading absolute minor A_x. More explicitly, when 0<p<d, every nonleading maximal future minor \Delta in the common reference basis satisfies

The first term comes from the mixed-exterior estimate. The second comes from the change of spaces, using the full exterior upper bound and the leading-minor lower bound. Increasing C includes the oblique-basis factors. If p=0 or p=d, there are no nonleading maximal future minors, so this comparison is unnecessary.

Define A_y in the same way for the future of y. For the past maps the leading columns are slow, and the backward exterior rate is

Denote their leading absolute minors by B_x,B_y. They have lower bounds e^{\Omega_x n-Cbn} and e^{\Omega_y n-Cbn}, respectively, and all other maximal minors are relatively negligible. No sign is prescribed for \Lambda_x or \Omega_x in a perturbation estimate.

We must transfer the future minors from x to z relative to A_x, even if this leading quantity is small. Use the reference charts along the future of x for both nearby orbits. By (4.11), their one-step derivatives differ by at most C\delta for large k. On every unperturbed subinterval the p-th exterior product has norm at most e^{\Lambda_x\ell+Cbn}. Applying Lemma 3.3 gives, for every fixed C_0,

where the derivatives are represented in these common starting and endpoint coordinates. To see the scale in this application, an expansion term with q\ge1 altered factors is bounded by

The binomial sum, divided by e^{\Lambda_xn-C_0bn}, tends to zero because n\delta e^{Cbn}\to0. The bounded factor e^{q|\Lambda_x|} covers either sign of the rate. The same reasoning backward uses \Omega_y and e^{q|\Omega_y|}. Fixed endpoint projections and the common starting basis preserve (4.21), after increasing C_0 if an inverse basis factor is needed. Hence at z the leading future and past minors are A_x(1+o(1)) and B_y(1+o(1)), with the same relative smallness of all the other minors. At w they are A_y and B_x.

Expand each full determinant along its p future rows. There is one leading product, from fast future columns and slow past columns. Every other term has a relatively negligible maximal minor. In the common starting basis the four absolute determinants are therefore

All leading factors are positive. In particular these maps are locally invertible. If p=0, future targets and minors are empty and A_x=A_y=1; if p=d, the past minors are B_x=B_y=1. The same determinants and conclusions apply, without mixed minors of the missing kind.

The defining equations say that (H_{xy}(z),H_{yx}(w)) is a fixed permutation of (H_{xx}(x),H_{yy}(y)). Differentiation gives the exact identity

Each full derivative represented in the common product basis differs from its coordinate derivative by the factor \det E_Q. There are two such factors in the numerator and two in the denominator, so they cancel exactly. Substituting (4.22) cancels A_x,B_x,A_y,B_y individually, with no comparison between the rates of x and y. Thus the coordinate Jacobian is 1+o(1). All four starting points lie in the same cell of diameter O(r). The ratio of the two output smooth densities to the two input densities is therefore 1+o(1) as well. This proves (4.20).□

Proof. First fix a retained pair. Its cell and four endpoint labels are locally constant among retained inputs near it, because of the positive wall margins. By Lemma 4.3, the implicit equations define a smooth local pair branch near its outputs. This branch agrees with the constructed switch on nearby retained inputs. Indeed, for this fixed k, continuity of the finitely many maps S^t, |t|\le n, makes their reference halves arbitrarily close when their inputs are close. The constructed solutions have sharp radius e^{Cgn}r=o(\delta) about those halves. Both the old local implicit branch and the construction for the nearby pair therefore lie in the old \delta-tube, and satisfy the same fixed endpoint equations with the nearby right-hand sides. The contraction uniqueness argument identifies them even for these changed data: two solutions u,v in the old ball satisfy \mathcal L(v-u)=\mathcal R(v)-\mathcal R(u), so \|v-u\|_\infty\le CQ_k\delta\|v-u\|_\infty forces equality. Both candidate arrays here use the old reference halves. Their matching datum is therefore the same old coordinate difference; only the prescribed endpoint values change with the nearby inputs. The charts along the old pair may be kept fixed in this neighborhood. In particular no regularity of a choice of good frame witnesses is needed.

The retained set is Borel, and M\times M is second countable. A countable subcollection of these relative neighborhoods covers it. Subtracting earlier members partitions it into countably many Borel pieces. On each piece \Phi is the restriction of the indicated smooth local diffeomorphism; this proves measurability and supplies the pieces for change of variables.

For injectivity suppose \Phi(x,y)=\Phi(x',y')=(z,w). The interiors of product cells are disjoint, and both outputs stayed in the original cell. Thus the two input pairs have the same product cell and hence the same starting bin Q, threshold, and dimensions. The four endpoint-bin labels are recovered from the ordered output pair by

All four endpoints have positive margins in their bins. Therefore these labels, and exactly the four projections and coordinate maps in (4.9), are the same for the competing inputs.

On the future half, both x and x' are within e^{Cgn}r of z; on the past half both are within that distance of w. Consequently their full orbits on [-n,n] are within 2e^{Cgn}r=o(\delta) of each other. Moreover the recovered equations give

Apply the boundary uniqueness argument of Lemma 4.2 with both reference halves equal to the true good orbit of x. It uses x’s endpoint projections, zero matching datum, and zero terminal data. The orbit of x is one solution. In the same charts the orbit of x' is another solution in its \delta-ball, since its distance is o(\delta) even after fixed chart conversions. Uniqueness gives x'=x. The same argument gives y'=y. Only the true input orbits were required to be good; neither z nor w was required to belong to \mathcal A_k or to R_k.□

Proof of Theorem 4.1. The construction above applies for all sufficiently large k. Equation (4.4) gives \sum_kB_k^{16}=\sum_kb_k^2<\infty; any finitely many unused terms can be chosen positively. The partition bound follows from (4.6), retention from (4.8), and uniform shadowing from (4.11).

It remains to prove the output measure bound. On the square of a positive-volume cell Q', the original law \rho_k has density 1/\nu(Q') with respect to \nu\otimes\nu. The map \Phi sends retained pairs in that square into the same square. By Lemma 4.3, its inverse volume Jacobian is at most 2 for all sufficiently large k. Apply change of variables on the countably many pieces of Lemma 4.4. Their images are disjoint by global injectivity. Summing therefore gives

The factor 1/\nu(Q') is unchanged. Restriction to endpoint-label pieces or to G_k is never followed by normalization, and zero-volume cells contribute nothing. Taking first marginals proves (4.2), because the first marginal of \rho_k is \nu. The construction already handled the optional smooth boundary.□

Proof. For \widehat p=q^{-1}\sum_bY_b, independence between vectors gives

Apply the inequality \E|Z|\le(\E|Z|^2)^{1/2} coordinatewise and then Cauchy–Schwarz to the sum over the A coordinates.□

Proof of Proposition 5.1. Growth of the fixed parameters. Whenever X_i\ge6, rounding in (5.3) gives

Indeed, X_i-2<2s_{i+1}\le X_i, and 2s_{i+1}-1<\log(2^{2s_{i+1}}-1)<2s_{i+1} when 2s_{i+1}\ge2. We claim inductively that

The first two inequalities hold at i=1. If they hold at level i, then \ell_i\ge K_i^2 implies

Consequently K_{i+1}\ge2^{K_i/2}-1\ge2^{K_i/3}\ge K_i^2\ge4K_i. These elementary inequalities hold for every K_i\ge256; for example, K/3-2\log K is increasing there and is positive at 256. Since \varepsilon_{i+1}=\varepsilon_i/2, the second bound in (5.7) propagates as well. In particular, all parameters are positive and defined, H_i divides N_i, and D_i^*\le H_i.

Sampling the relations. Suppose that \mathcal W_i and its vector labeling have been fixed. Generate a list (U_0,\ldots,U_{N_i-1}) in independent consecutive groups of H_i positions. Within each group, proceed from left to right. Choose each new label uniformly among the nonzero vectors orthogonal to all earlier labels at the lags 2^k, 0\le k\le\lfloor K_i/8\rfloor, that stay inside the group. There are at most \lfloor K_i/8\rfloor+1\le K_i linear equations. Their common kernel has codimension at most K_i, whether or not the equations are independent. As 2s_i\ge\ell_i\ge K_i^2, the number of nonzero choices is at least

Denote this unconditioned law on \mathcal W_i^{N_i} by P_i^0. Every sampled list satisfies the relations in Property (W1).

For later use, set \zeta_i=2^{2K_i}/L_i within this proof. The conditional probability of any specified label at any position, given all preceding positions, is at most \zeta_i. If r_1<\cdots<r_{d'} are specified positions, conditioning first at r_{d'} and then successively at the earlier specified positions gives

All unspecified labels are integrated out. This argument does not require independence within a group.

Each individual U_r nevertheless has a uniform marginal on \mathcal W_i. To see this, the group of linear maps preserving the alternating form acts transitively on its nonzero vectors. Given such a vector v, nondegeneracy provides w with [v,w]_i=1; the plane spanned by v,w is nondegenerate, and its orthogonal complement carries a nondegenerate alternating form of dimension two less. Induction extends v,w to a symplectic basis, proving transitivity by mapping one such basis to another. The sequential rule is equivariant under these maps, including its uniform choices, so each marginal is invariant under a transitive action and hence uniform.

Imposing the approximation in each word. For i\ge2 and an offset 0\le t\le N_{i-1}-D_{i-1}^*, let

and let p_t be its law under uniform choice of u. For each of the q_i=N_i/H_i independently sampled groups, form the empirical probability vector of the H_i values f_t(U_r) in that group. Its mean is p_t by the uniform one-label marginals just proved. The average of these independent vectors is exactly the law of f_t(U_J) when J is uniform over the entire sampled list. The alphabet of f_t has cardinality at most A_i=L_{i-1}^{D_{i-1}^*}. Thus Lemma 5.2, Markov’s inequality, and a union over at most N_{i-1} offsets bound the probability of violating any requirement in Property (W4) by

Below we verify F_i<1/2 at every level i\ge2.

Let \mathcal E_i be the event that all these requirements hold, and let P_i=P_i^0(\,\cdot\mid\mathcal E_i). For i=1, take \mathcal E_1 to be the whole sample space and P_1=P_1^0. In all cases P_i^0(\mathcal E_i)\ge1/2, so (5.9) yields

The factor two is paid once for the whole conditioning event. No uniform marginal or independence between groups is asserted for P_i. The relations hold on its support, as does Property (W4) for each individual list.

Selecting the finite family. Put M_i=L_{i+1} and sample M_i independent complete lists with law P_i. For each offset 0\le t\le N_i-D_i^*, let Q_{i,t} be the law under P_i of the length-D_i^* window at t. It satisfies (5.5) by (5.11). The empirical window law of the M_i samples is an average of independent point-mass probability vectors on an alphabet of size L_i^{D_i^*}. A second application of Lemma 5.2 and Markov’s inequality, now with a union over at most N_i offsets, bounds the probability of violating any required window approximation by

We will have E_i<1/4 at every level.

It remains to impose the distances, including comparisons in which some of the sampled-word indices coincide. First use the reference law in which the N_i labels in each distinct indexed list are independent and uniform. Consider one pairwise comparison between distinct indices, or one comparison of a target list with a shifted window from two source lists. In the latter case write t for its offset, with 1\le t<N_i; any of the three indices may coincide.

For a specified set of matching positions, form a multigraph whose vertices are variables (sample index, position) and whose edges are the asserted equalities. Every target position occurs once. The source positions also occur once: they use t,\ldots,N_i-1 in the first source and 0,\ldots,t-1 in the second, disjoint ranges even if the source indices agree. The degree is therefore at most two, counting repeated edges. There is no loop. In a pairwise comparison the indices are distinct, and in a shifted comparison the source position differs from the target position by t modulo N_i.

A connected component with v vertices requires v-1 independent equalities and has probability L_i^{-(v-1)} under the reference law. A path has one fewer edge than vertices; a cycle has equally many, with at least two edges. Thus every component imposes at least half as many independent equalities as edges, including a pair of parallel edges. If e positions are specified to match, their probability is at most L_i^{-e/2}. Failure of a distance requirement entails at least \lceil\varepsilon_iN_i\rceil matching positions. Summing over their possible sets, of which there are at most 2^{N_i}, gives the bound

By (5.8), a complete list under P_i^0 has likelihood ratio at most 2^{2K_iN_i} relative to the reference law, and under P_i at most 2\,2^{2K_iN_i}. Each comparison uses at most three distinct sampled lists, so its probability under the actual sampling law is at most (2\,2^{2K_iN_i})^3 times (5.13). Repeated indices do not introduce an additional independent list. There are fewer than 4N_iM_i^3 pairwise and shifted comparisons in total. Their union failure probability is consequently at most

We will verify T_i<1/4.

These estimates address separate requirements: the first sampling law enforces the relations, conditioning enforces the approximation in each individual word, and the final independent samples provide the window laws and distances. We now check that the fixed initial size makes all three probability bounds hold at every step.

Uniform bounds from one initial choice. For K\ge256, set

The elementary inequalities 14+K\le2^{K/8} and (K+1)^2\le2^{K/8} hold at 256 and persist thereafter, as the ratios of their left sides to their right sides decrease. Hence

We also have \log(\vartheta^{-1})<14 and K_i^2\le\ell_i<(K_i+1)^2. For the first probability bound, put K=K_{i-1} and use \log(N_i/H_i)\ge K_i/2 to obtain

For the second, put K=K_i. Since \ell_{i+1}\ge2^K,

Finally (5.6) and \varepsilon_i\ell_i\ge\varepsilon_iK_i^2\ge32K_i give

It follows that

All right sides are decreasing functions of their K argument. They apply simultaneously at every level by (5.7); no later adjustment of s_1 is made.

The conditioning event thus has positive probability. Under the conditioned law the two subsequent failures have total probability less than 1/2, so a tuple satisfying every requirement exists. Its pairwise distance requirement is imposed on distinct sample indices and has a positive lower bound. Equal sampled words would have distance zero, so the tuple has exactly L_{i+1} distinct members. Flatten their constituent lists to form \mathcal W_{i+1}. The empirical distribution is now exactly the uniform distribution on this set, without deduplication or a change of weights. It therefore has all four asserted properties. Choose its bijection with \mathcal A_{i+1} and continue the induction.□

Proof. The finite products are bounded below by 1-\sum_{j\ge1}\varepsilon_j=3/4, proving that d_*>0. At level one, distinct one-symbol words have distance one and the offset assertion is vacuous.

Suppose both assertions hold at level i. Distinct level-(i+1) words differ in at least (1-\varepsilon_i)N_i constituent labels by Proposition 5.1, Property (W2). Each unequal constituent pair contributes at least d_i m_i symbol mismatches, giving the bound d_{i+1}m_{i+1}.

Now consider a nonzero offset t<m_{i+1} in two concatenated level-(i+1) words, compared with a third such word. If m_i divides t, the offset t/m_i lies between 1 and N_i-1. The constituent-label offset requirement gives at least (1-\varepsilon_i)N_i unequal pairs, and the preceding distinct-word bound again gives d_{i+1}m_{i+1} symbol mismatches. If m_i does not divide t, each of the N_i aligned target constituents faces a length-m_i source window with the same nonzero offset modulo m_i. That source window crosses exactly two consecutive legal level-i words. The inductive offset assertion therefore gives at least d_i m_i mismatches on each target constituent. It applies also when the two source constituents lie on opposite sides of the join of the two larger source words. Adding over the disjoint target constituents gives d_i m_{i+1}\ge d_{i+1}m_{i+1}, completing the induction.□

Proof. The formula for m_i follows from m_1=1 and N_i=2^{K_i}. Dividing (5.6) by m_{i+1}=N_i m_i gives

Since \varepsilon_i\le1/8, the upper bound implies R_i\le800^{-(i-1)}R_1\to0, proving (5.17). Iterating the lower bound and using \sum_{j<i}(j+2)=(i-1)(i+4)/2 prove the first inequality in (5.18); the second uses \log200<8. Also (5.7) implies K_i\ge K_1^{2^{i-1}}. Hence

In particular the terms in (5.19) do not tend to zero, and the series diverges.□

Proof. For any x\in X, weak limits of n^{-1}\sum_{j=0}^{n-1}\delta_{T^jx} are invariant probabilities. Indeed, the difference between the integrals of f\circ T and f against these averages tends to zero for each continuous f. The set \mathcal M_T(X) of invariant probabilities is nonempty, compact, and convex.

Here is an explicit way to select an extreme point. Choose a countable uniformly dense family (f_j) in the continuous real functions on X. Starting from \mathcal M_T(X), successively restrict to the compact face on which \int f_j\,d\mu is maximal. The nested intersection is nonempty and contains only one measure, because the integrals of every f_j are then fixed and determine a probability. An intersection of faces is a face, so this measure is extreme. If it had a T-invariant measurable set of mass strictly between zero and one, normalized restriction to that set and its complement would express it as a nontrivial convex combination of invariant probabilities. Thus it is ergodic. The same argument applies to invariant sets modulo null sets after taking Borel representatives.

For any invariant probability, a_i is uniform on \Z/m_i\Z, because T cyclically permutes its level sets. If a measurable atom had mass c>0, then for each i it would be contained, modulo null sets, in one of these level sets. Hence c\le1/m_i for every i, which is impossible since m_i\to\infty.

The space X is a compact metrizable subspace of a countable product of finite discrete sets. For completeness, its nonatomic probability is standard in the precise sense needed here. Encode the finite coordinates into successive disjoint blocks of ternary digits 0,2; this gives a continuous injective map of X into a compact subset of [0,1]. Let \lambda be the image law and F(t)=\lambda([0,t]) its continuous distribution function. Define the measurable quantile

Continuity gives F(Q(u))=u, and \{u:Q(u)\le t\} has Lebesgue measure F(t), so Q_*\operatorname{Leb}=\lambda and F_*\lambda=\operatorname{Leb}. Moreover Q(F(x))\le x for \lambda-almost every x; both sides have law \lambda. Their nonnegative difference has zero integral, and therefore Q(F(x))=x almost surely. After restricting to measurable conull sets, F and Q are inverse isomorphisms. Composing with the coordinate embedding establishes the assertion. Passing to completed sigma fields does not change it.

It remains to check the entropy of the entire phase extension. A partition depending on finitely many coordinates is refined by one that records \omega_{-r},\ldots,\omega_r and a_I for some fixed r,I. Compatibility determines all the lower phases from a_I, and the phase evolves deterministically. Fix any j\ge I. The initial phase a_j determines a_I and the boundaries of all level-j words. The symbol interval [-r,n-1+r] used in an n-step name meets at most \lceil(n+2r)/m_j\rceil+1 such words, each in \mathcal W_j. Consequently the number of names is at most

Keeping j fixed and sending n\to\infty bounds the entropy rate by R_j=(\log L_j)/m_j. By Lemma 5.4, R_j\to0. Only now let j\to\infty: every finite-coordinate partition has entropy rate zero. This count uses no assumption about word frequencies.

Let \mathcal P be any finite measurable partition with A atoms. Cylinder sets generate the Borel sigma field, and their algebra is dense in measure. Thus the atom label of \mathcal P can be approximated by an A-valued finite-coordinate label \mathcal Q with mismatch probability s arbitrarily small. Completion of the measure only adds null-set changes. The chain rule and conditioning give

where \mathcal P^n=\bigvee_{t=0}^{n-1}T^{-t}\mathcal P. Recording first whether the two labels differ shows that

Entropy rates exist by subadditivity. Divide by n, take the limit, and then let s\downarrow0. This proves h_\mu(T)=0.□

Proof. Invariance first allows us to take t=0. On the phase class a_i=r, the map T^{(-r)\bmod m_i} sends the normalized restriction of \mu bijectively and measure-preservingly onto \mu_i. The first assertion follows by applying this separately on all m_i phase classes.

For the second assertion, T^{jm_i} maps \{a_{i+2}=0,V_{i+2}=v\} onto \{a_{i+2}=jm_i,V_{i+2}=v\}. It does not change the containing parent word because 0\le jm_i<m_{i+2}. These sets have the same \mu-mass. Multiply by m_i, and use m_{i+2}/m_i=N_iN_{i+1}, to obtain (6.2).□

Proof. Write q=D-1. By Lemma 6.2 we work under \mu_i. Decompose its uniform parent-slot index as

Conditional on the parent v=V_{i+2} and the internal offset u, the slot h is uniform among the N_{i+1} constituent \mathcal W_{i+1} words of v. Call u admissible if u+D_i^*\le N_i. Its failure probability is

For every fixed v and admissible u, Proposition 5.1, Property (W4), applied at level i+1, puts the law of the D_i^* labels at this offset within 10^{-4} of their law in a uniform word of \mathcal W_{i+1}. Proposition 5.1, Property (W3), applied at level i, puts that law within a further 10^{-4} of a law whose every specified q-coordinate atom is at most

Restriction to the first q entries cannot increase total variation. The comparison law may depend on u, but this atom bound does not.

It follows from (6.3), with s=2\cdot10^{-4} and alphabet size L_i^q, that the actual q-list conditional on v,u has entropy at least

Entropy is concave. Mixing over all v and admissible u preserves this lower bound; adding the remaining offsets loses at most their fraction, since entropy is nonnegative. Thus

Here \beta_i\to0, 2K_i/\ell_i\to0, and 2/\ell_i\to0. Once each of these three quantities is at most .0005, division by q\ell_i, with q\ge1, gives a coefficient at least

This proves the uniform assertion.□

Proof. Because n is a multiple of m_i, the advance to alignment is the same in the two windows. Denote it by q_0. The entry starts are

Corresponding entries are separated by exactly D level-i positions. The first start through the last start span 2D-1 such positions, including the unused intervening position. Figure 3 illustrates these positions and their common lag when D=4.

By Lemma 6.2, the first boundary has the boundary-start law. Its position modulo H_i is uniform: its position in a \mathcal W_{i+1} parent is uniform modulo N_i, and H_i divides N_i. Once 2D-1\le H_i, exactly 2D-2 residues have a span that crosses an H_i-group boundary. All parent boundaries are already group boundaries. On every other residue all the labels in (6.6) belong to one of the prescribed groups. Proposition 5.1, Property (W1), applies because D is an allowed power-of-two lag. Finally 2D_i^*/H_i\to0 by the definitions.□

Proof. The lengths m_i are powers of two. Put b_i'=\log m_i, so b_{i+1}'=b_i'+K_i. The integer sets

are disjoint and, for large i, contain at least K_i/32 integers. If the ratios in (7.1) were bounded below by some c>0 on all sufficiently late tests, then

contrary to the hypothesis and Lemma 5.4. Thus one can select successively later tests with ratio tending to zero. Since B_{\log n}n=(B_{\log n}/R_i)D\ell_i and D/(D-1)\le2, (7.2) follows.□

Proof. For a law p, put A_p=\{u:p(u)>L^{-.8}\} and \theta_p=p(A_p). The set A_p has at most L^{.8} elements. Splitting entropy according to this set gives

The same formula, with the usual zero-mass conventions, covers empty sets. The entropy hypotheses imply

once \log L\ge1000.

Let A be the matrix A_{uv}=(-1)^{[u,v]} on nonzero vectors. On the full vector space, the rows of the same matrix are orthogonal: for u\ne w, nondegeneracy makes v\mapsto[u+w,v] a nonzero linear functional, so \sum_v(-1)^{[u+w,v]}=0. Every row has squared norm L+1. The full operator norm is therefore \sqrt{L+1}, and restriction to the nonzero rows and columns has no larger norm.

The light subprobability p^{\rm l}=p\,\one_{A_p^c} satisfies

Hence

The passing mass of the light product is at most (1+\beta_L)/2. All pairs with at least one heavy coordinate have total mass at most \theta_p+\theta_q. After averaging, and taking L large enough that \beta_L\le .02, the passing probability is at most

The matrix includes its diagonal entries, so the automatic relations [u,u]=0 require no separate exclusion.□

Proof. Choose \kappa<d_*/20000. Inner regularity of the smooth probability gives disjoint compact sets K_a\subset f^{-1}(a), for a\in\mathcal A, with

Measurable atoms may first be replaced by Borel representatives modulo null sets. There are finitely many cores, and the distance between any two nonempty distinct cores is positive. If there is only one nonempty core, the following equal-symbol implication is automatic. Otherwise take the positive minimum of these distances. Uniform shadowing is smaller than this fixed separation at all large scales. Therefore two compared iterates both in the cores have the same symbol.

Write G for the retained set and \sigma=z_*(\rho|_G). The switching theorem gives \sigma\le2\nu. For every integer t,

The x and y time marginals under the original law are \nu. At each future comparison time, the \rho-mass within G where either x or z misses the cores is at most \kappa+2\kappa=3\kappa. The same holds for the past comparison with y.

The integrals over G of the fractions of mismatches on either half are therefore at most 3\kappa. Markov’s inequality at threshold d_*/100, on each half, bounds their combined failure mass by 600\kappa/d_*<.03. Since \rho(G)\ge .99, the simultaneous success mass is at least .96. This is an estimate in the original law, without dividing by \rho(G).

The input marginals and \sigma\le2\nu give zero mass to the complement of the invariant conjugacy domain. Thus the symbol strings and all orbit identities used in the argument are defined outside a \rho-null set.□

Completion of the proof of Theorem 1.1. Proposition 6.1 supplies a standard nonatomic ergodic invertible probability-preserving system of zero entropy. If it had a smooth volume model, Theorem 4.1 and Lemma 7.1 would supply the tests above. The original pair law at each late test would satisfy both (7.6) and (7.10), which is impossible. The dimension and all smooth constants were arbitrary but fixed after construction of the system. Thus no model in any finite dimension exists.□