Invariant-projection counterexamples for every irrational rotation

Proof. The commutation equation is preserved by addition, multiplication and scalar multiplication, and the identity commutes with T. If A_\lambda\to A strongly and A_\lambda T=TA_\lambda, then for every \zeta\in\mathcal H both sides converge, giving AT\zeta=TA\zeta. Thus the algebra is strongly closed. If every bounded operator commuted with T, then every rank-one orthogonal projection would commute with T, so every nonzero vector would be an eigenvector. Applying this to x,y,x+y for linearly independent vectors shows that the eigenvalues are equal; hence T would be scalar. Properness follows. The last assertion is precisely the two definitions.□

Proof. Every unitary W\in\mathcal M' and its inverse commute with T. Hyperinvariance therefore makes the subspace reducing for W. Its orthogonal projection commutes with all such unitaries and hence belongs to \mathcal M''=\mathcal M. Since T\in\{T\}', its range is invariant under T.□

Proof. The tracial representation of R_\theta is separable and infinite-dimensional. Apply Theorem 1.1, Corollary 6.2, and Lemma 1.3 there. A nonzero quasinilpotent operator is nonscalar, so Lemma 1.2 applies. Unitary equivalence transfers these properties to any separable infinite-dimensional complex Hilbert space.□

Proof. For the first orientation the function of v is a constant height at most m^{13/5} on an interval, so the variation bound is immediate. For the second, nonzero values require v<s in representatives s,v\in[0,1): if v>s, then \Delta(s-v)\le v-s<v, whereas h<1. On 0<v<s, the conditions \Delta(s-v)\ge m^{-2} and v<h\Delta(s-v) describe an interval. The reciprocal of \Delta(s-v) has at most two monotone pieces there, with the same height bound. This proves the asserted uniform bound in both orientations.

In the first orientation \bar u is the cutoff indicator, and

The change (s,v)\mapsto(s-v,v) preserves double integrals, giving the same total mass and square integral for the second orientation. To bound its averaged density, first omit the cutoff. The resulting density d is the law of z+t\Delta(z) for independent uniform z\in[0,1) and t\in(0,h). For fixed t this is an increasing piecewise-linear bijection of [0,1], with slopes 1+t and 1-t. Thus \int d=1 and |d-1|\le h/(1-h), whence

Restoring the cutoff only decreases this nonnegative density; also d\le(1-h)^{-1}\le2. These observations give the upper bounds in (2.2), while Cauchy–Schwarz gives the lower bound from its exact mass. The cancellation of the linear term here is essential when multiplying estimates over m coordinates.

Periodizing a uniform interval of length n/N_0 gives (2.3); in particular p_n\le2 for large m. For the separated powers and any permitted nonzero frequency pair, direct integration on [1/N_0,2/N_0] gives

To pass from characters to rectangles, convolve each interval indicator with the Fejér kernel

Expanding the squared geometric sum shows that F_H is nonnegative, has mass one, and has frequencies of magnitude below H. The convolved indicators lie in [0,1], preserve their means, and have Fourier coefficients of modulus at most one. Outside strips of width \eta=m^{-80} about interval endpoints, their error is O((H\eta)^{-1}), by integrating the kernel tail. The marginal bound p_n,p_{n'}\le2 shows that replacing both indicators in a product costs O(\eta+(H\eta)^{-1}), both for the actual joint law and for Haar product measure. The remaining Fourier error is O(H^2m^{-1000}). This proves (2.4); no joint density, or independence of the sampled powers, is used.

Finally, a bounded-variation function is, up to a null set, a constant plus an integral of indicators of intervals [a,1) against its complex Stieltjes measure. The constant and measure have total size at most B(f); the same holds for g. Apply (2.4) to the resulting products of interval indicators, allowing the full interval. Integration against total variation gives (2.5). Changes at the countably many possible discontinuities do not matter, because the marginals have densities.□

Proof. The admissible polynomial pairs form a nonempty compact set. For each pair, the first nonzero exponents of A and B are at most zero, and their sum belongs to [-2K,0]. The corresponding product coefficient is nonzero. The continuous displayed maximum therefore has a positive minimum on the compact set.□

Proof. Suppose K_j,G_j and all earlier blocks are fixed. Apply Lemma 2.2 with K=K_j, D=D_j and \delta=1/j to choose \varepsilon_j. Choose L_j\ge\max(1,L_{j-1}) to satisfy the second inequality in (2.10) (omit L_{j-1} when j=1), set C_j=\max(L_j,K_j^2), and then choose an integer G_{j+1} with \kappa_{j+1}\sqrt{G_{j+1}}\ge L_j. All requirements of () now hold for sufficiently large m=m_j. In what follows earlier data remain fixed while m\to\infty; in particular p_*,q_{p_*},\ell_j and the finite list of shifts t are fixed. Write N=mK_j, Q=m^{(7/10)m}, a=G_j/m, and b=m^{-19/20}.

Separated sampling powers. Equidistribution gives at least a fixed positive multiple of \ell_j Q integers n\in[Q/2,Q] with \{n\theta-t\}\in(0,\ell_j) for each required t, once m is sufficiently large. Set H=m^{300} and R=Nm^{1000}. We can select at least m^{0.69m} such powers, separately for every t, so that distinct selected powers obey

Indeed, if v'\ne0, each previously chosen n excludes at most 2R+2 integers n' for each frequency pair (v,v'). There are at most (2H+1)^2 pairs, and N\le m^2. A greedy selection thus loses only a polynomial factor in m from the available number. Since Q=m^{0.70m}, it eventually leaves at least m^{0.69m} powers. If v'=0 and v\ne0, then |vn|\ge Q/2>R for large m, so this case causes no exclusion. Fix the resulting sets \mathcal N_{j,t}.

Trial phases and the spread condition. For i\in J_j, take independent uniform \lambda_i\in[1/N,2/N]. For each integer k in the range of (2.8), the law of \{k\lambda_i\} has density at most 5: it is the periodization of a uniform interval of length |k|/N\ge1/4, whose density is at most 1+N/|k|. The probability that an index fails the strict interior version of (2.8) is therefore at most 10b. If fewer than m/20 indices succeed, more than 19m/20 fail. Independence and a union bound give failure probability at most 2^m(10b)^{(19/20)m}. A further union over both signs of k, up to m^{(4/5)m}, tends to zero: the coefficient of m\log m in its logarithm is 4/5-(19/20)^2<0. The lower range in the proposition requires no probabilistic choice. When G_jK_j\le|k|\le N/4, every trial phase has a\le|k|\lambda_i\le1/2; the same observation will hold for the eventual \alpha_i.

One-coordinate and two-power estimates. Fix an orientation, put u=w_m^{(r')}, and set \bar u(s)=\int_0^1u(s,v)\,dv and A_*=\int_0^1\bar u(s)^2\,ds. Lemma 2.1, with N_0=N, gives \|\bar u\|_\infty\le2, (1-2m^{-2})^2\le A_*\le1+4h^2 and \iint u^2\le4h^{-1}\log m, where h=m^{-3/5}. Its bounded-variation estimate also gives, uniformly in s,

Indeed, the error from (2.5) is O(m^{-60}). The one-power density estimate (2.3) likewise gives \mathbb E[u(s,\{n\lambda_i\})\bar u(s)] =\bar u(s)^2+O(m^{-50}), since n\ge Q/2.

The product sampling test. We now have one-coordinate and two-power estimates. Their role is to control the second moment of an average of products over a whole block; the diagonal terms will be absorbed by the large number of sampling powers, while the distinct-power terms approach the ideal product law. For independent Haar variables z=(z_i)_{i\in J_j}, and a fixed power set of size P, put

By independence of the trial phases over i, the distinct-power terms of \mathbb E\|Z\|_2^2, and the terms of \mathbb E\langle Z,Z^{\rm id}\rangle, are products of m factors A_*+O(m^{-50}). They equal A_*^m+O(m^{-49}) by telescoping, because (1+4h^2+O(m^{-50}))^m stays bounded. The equal-power terms are each at most (8h^{-1}\log m)^m by (2.2) and the one-phase density bound, and their total averaging weight is 1/P. Thus

since P\ge m^{0.69m} and h^{-1}=m^{0.60}. Also

by (2.2) and mh^2=m^{-1/5}\to0. It follows that \mathbb E\|Z-1\|_2^2\to0. There are only finitely many t at this stage. Markov’s inequality, the two orientations, and the spread estimate therefore let us fix trial phases with strict spread inequalities and sampling norms below 1/(2j), simultaneously for every test. We can also avoid the finitely many equalities n\lambda_i\in\Z and the endpoints 1/N,2/N, since each has probability zero.

Realizing the phases by circle characters. Now fix this m and these trial phases. Successively choose q_i as multiples of the preceding frequency so that \{q_i\theta\} is as close to \lambda_i as desired and q_i/q_{i-1} is as large as desired. This is possible because the positive multiples of q_{i-1}\theta are dense. More formally, make a sequence of such finite block choices for which all new ratios tend to infinity and all \alpha_i\to\lambda_i. The Haar law of (u_i(x))_{i\in J_j} then converges weakly to product Haar measure. For each fixed nonzero integer character vector, the last nonzero coefficient times its frequency dominates the sum of the earlier terms, once the ratios are sufficiently large. That Fourier coefficient is therefore exactly zero eventually. Trigonometric approximation proves the weak convergence. The joint law including the deterministic phases converges to product Haar in the coordinates and the fixed vector of trial phases.

For fixed m the finitely many squared sampling tests are bounded functions of these coordinates and phases. Their discontinuities have zero limiting measure: every selected \{n\lambda_i\} avoids the phase cut, and for that phase only finitely many coordinate values can lie on a cutoff or interval boundary. Weak convergence therefore gives convergence of their integrals. This familiar extension from continuous tests follows, for example, by continuous upper and lower approximations outside open neighborhoods of the null discontinuity sets. It applies to the squared differences from one as well. Choose a sufficiently late finite block choice. Its sampling norms are below 1/j, all the strict spread and angle bounds persist, and all frequency ratios determined so far satisfy their required bounds. The ratio connecting the previous block to this one can be made large enough to meet that block’s outstanding condition; the ratio at the end of this block is imposed at the next stage. This completes the induction.□

Proof. If B\in B(L^2(Y,\nu)) commutes with all multipliers, put b=B1. For every measurable set E,

Thus |b|\le\|B\| almost everywhere. The identity Bf=bf follows first for simple functions and then, by density, for all f\in L^2(Y). Apply this observation to the matrix entries of A between the coordinate copies of L^2(Y), obtaining measurable functions a_{ij}.

Let v range over the countable set of finitely supported vectors with rational real and imaginary parts. Commutation with M_{\mathbf1_E} and the operator bound give

Consequently the integrand is at most \|A\|^2\|v\|^2 almost everywhere. Outside the union of these countably many null sets, the matrix defines an operator A(y) on K with norm at most \|A\|. Its columns, and then its values on every vector, are measurable by finite-coordinate approximation. The represented field agrees with A on finite-coordinate simple functions, hence on all of L^2(Y;K).

Equality of two such fields follows by applying their represented operators to each constant basis field. Products act fiberwise by the representation formula, and the entries of the adjoint field are \overline{a_{ji}}. This proves the assertions about products and adjoints. The identities p=p^*=p^2 then give orthogonal projections on a common conull set.□

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 latter range has zero zeroth coordinate. Since (D_yv)_0=f(y)v_0 and f(y)>0, every vector in the former range also has zero zeroth coordinate, so r(y)=0. If r(y)=1, then e_0 belongs to the former range, and its image f(y)e_0 shows that r(\sigma y)=1.

Thus \sigma^{-1}\{r=0\}\subseteq\{r=0\} and \{r=1\}\subseteq\sigma^{-1}\{r=1\} modulo null sets. Probability preservation makes both inclusions equalities modulo null sets. Ergodicity makes each event null or conull. The trace formula and 0<\tau(p)<1 exclude both conull cases.□

Proof. First replace every diagonal entry by its minimum with B, so that the diagonal field is bounded everywhere. Given an arbitrarily small \varepsilon>0, simultaneous Lusin approximation supplies a compact Q\subset X on which every F_k and every vector map x\mapsto p(x)e_k is continuous, while \mu(X\setminus Q) is sufficiently small to make the \beta_l-mass outside Q^2 less than \varepsilon for all l. For completeness, the simultaneous assertion follows by applying Lusin’s property to the countably many maps with summable exceptional measures and intersecting the resulting compact sets. The usual property for maps into separable metric spaces follows by finite-valued approximation and measure regularity.

On Q^2, the capped diagonal field is strongly continuous: continuity on the basis vectors and its uniform bound imply continuity on every vector. The same argument applies to p(x) and p(y). Products of strongly continuous, uniformly bounded fields are strongly continuous. Therefore the zero set Z_Q of the capped version of (3.3) in Q^2 is compact. The assumed identity implies \beta_l(Z_Q)\ge\beta_l(X^2)-\varepsilon. The compact-set upper bound under weak convergence and convergence of total masses give

The capped global zero set thus has full \beta-measure, since \varepsilon was arbitrary. It is measurable by its countably many matrix entries. Finally the cap changes no diagonal entry \beta-almost everywhere.□

Proof. Fix A,t. In every sufficiently late stage j, put a=d_{j-1}\ge A. Since q_A divides q_a, the sampling set \mathcal N_{j,t} occurs in Proposition 2.3. Write w=w_{m_j} and w^{(2)}=w_{m_j}^{(2)} for this stage’s kernels in (2.1). Average over n\in\mathcal N_{j,t} the graph measures x\mapsto(x,\sigma^n x) with starting density

By (2.11), both marginal densities tend to 1 in L^2(\mu). For the second marginal this uses the second orientation w^{(2)}(z,v)=w(z-v,v). As every translation n\theta lies within \ell_j of t and \ell_j\to0, testing continuous functions on X^2 shows that these measures converge weakly to the Haar-first graph measure of x\mapsto x+t. At every point of positive density,

so there is no wrap and all current-block coordinates are ordered.

For every selected n, remove the same set of starting points, namely those satisfying at least one of

The removed Haar measure is at most

The L^2 bound on the averaged first marginal and Cauchy–Schwarz show that the removed weighted mass tends to zero. Thus the trimmed measures have the same weak limit. Each trimmed marginal is dominated by its untrimmed counterpart, so both retain uniform L^2 bounds and hence uniform absolute continuity.

For A\le i\le a, the number q_i t is integral. The coordinate advance is therefore q_i\{n\theta-t\}<q_a\ell_j, and trimming prevents a wrap. For future coordinates it is n\alpha_i<2Q_j/N_s, again without a wrap. Together with the current block, this orders every coordinate i\ge A. Hence, on all trimmed measures,

The bound is independent of j and v. It also holds on the limiting graph, because u_i(x+t)=u_i(x) for i\ge A. The positive-orbit identity holds on the approximants, so Lemma 3.3 gives (3.10). There are only countably many pairs (A,t) under consideration.□

Proof. The mean-one assumption gives \|h_i\|_2^2=1+\|h_i-1\|_2^2. Hence each finite product density obeys

For a bounded cylinder function \varphi depending on the first n coordinates, integration under \nu is therefore bounded in absolute value by e^{C/2}\|\varphi\|_{L^2(\mu)}. These cylinder functions are dense in L^2(\mu), so the integration functional extends continuously to that space. The Riesz representation theorem gives a real H\in L^2(\mu) with norm at most e^{C/2} representing the extension. The finite signed measure H\mu and the probability measure \nu agree on all cylinder sets. Uniqueness of finite measures on the product sigma-algebra gives H\mu=\nu; thus H\ge0 almost everywhere and \int H\,d\mu=1. Applying the first assertion to h_i=h_i^{\epsilon_i} proves the hybrid assertion.□

Proof. Either raw marginal density is at most 10. For each width its top and bottom contributions together are at most 4\rho_s(w)/w, the tag contributes at most 2, and the diagonal component contributes at most 1.

We also need a sharper integral estimate. 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. For a tagged interval the corresponding bound is 2m_s. Express a raw marginal density minus 1 as the sum of its at most five weighted mean-zero perturbations. The finite-sum Cauchy–Schwarz inequality gives

In fact the four width terms contribute at most 40\kappa_s^2/m_s, and the tag at most 10/m_s. Passing from raw variables to digits replaces d_i by its average on each radix bin, so Jensen’s inequality preserves (4.2).

Use the uniform law at each digit as the reference probability measure. The pairs of digits are independent across coordinates, and each endpoint digit density has mean one. Since |J_s|=m_s, summing (4.2) minus one over the coordinates gives the common bound 50(\sum_s\kappa_s^2+1), for either endpoint and every deterministic choice between them. The initial digit has density one. Lemma 4.1 therefore gives the bound

Outside the null sets of ambiguous expansions, the mixed-radix map is a measure-preserving bijection from the reference digit space to X. Transporting the density through this map gives a Haar density with the same L^2 norm. This proves the assertion for both endpoints and all the deterministic hybrids.□

Proof. Fix D. The rational shifts y^{(D)}-x have only finitely many possible values. Lemma 3.4 holds outside a common Haar-null set for all these shifts, so the absolute continuity of the x marginal implies the required identity at (x,y^{(D)}). This remains valid even though the selected shift depends on the pair.

The coordinate order and (3.7) bound every ratio by 1, both for these pairs and for their limit (x,y). Their measures converge weakly by the almost-sure convergence y^{(D)}\to y. Lemma 4.2 supplies uniform L^2 bounds for both marginals. In particular their masses on a set of Haar measure \delta are at most H_0\sqrt\delta. Lemma 3.3 now passes the identity to the limit.□

Proof. The entire endpoint determines all its digits almost surely. For any finite collection of labels, conditioning on all x-digits gives the product of the individual posteriors

One verifies this formula first against cylinder tests in the digit sequence, using independence of coordinate pairs, and then against all bounded digit-measurable tests by the monotone class theorem. The common initial digit is independent of those pairs. This proves conditional independence given x; the proof for y is the same. In particular, the dependence of u_i(x) on later digits does not change the label posteriors after the entire endpoint has been conditioned on.

The parameters satisfy

If u_i(x)<u\le1-w, the whole radix bin containing X_i lies in [0,1-w/2). On that bin the top component’s X_i-density is at least 1, while the total raw marginal density is at most 10. Its posterior probability is therefore at least \rho_s(w)/10. Under this label, (4.3) gives

For the second assertion, u_i(y)\ge u\ge w places the entire Y_i bin in [w/2,1). The bottom component again has density at least 1 there, so its posterior is at least \rho_s(w)/10. Its X_i value satisfies

These estimates include thresholds at radix-bin endpoints.□

Proof. The multiplier bound in (5.1) and Cauchy–Schwarz make every convolution there absolutely convergent. Put

On E, Lemma 2.2 applied to the two truncated Laurent polynomials gives \sum_{n=0}^{2K}|c_n|\ge\varepsilon(K,D,a). Its expected value on E is therefore at least c\varepsilon(K,D,a).

Fix n. For an omitted index v, choose k(v)=v if |v|>K, and k(v)=n+v otherwise. This is a deterministic choice with |k(v)|>K. The zero convolution and the two multiplier bounds give

The unmasked sum of products of vector coordinates is at most one, which accounts for \eta. We dropped \mathbf1_E from the nonnegative exceptional terms so that their full-law conditional bounds remain applicable.

For the first exceptional sum, Cauchy–Schwarz on probability times counting measure bounds its expectation by

Indeed, condition the first factor’s square on x, apply (5.2) to each deterministic k(v), and use \sum_v|\xi_v(x)|^2\le1. The second factor is at most one. For the other exceptional sum put its indicator with |\gamma_{n+v}(y)|^2 and condition on y, giving the same bound. Tonelli’s theorem justifies every interchange of the nonnegative sums and integrals. No sum of the individual exceptional probabilities is required. Taking expectations in (5.4) and summing in n proves (5.3).□

Proof. At a current-block coordinate, a tag has probability 1/m and satisfies both endpoint exclusions. Since each raw marginal density is at most 10, the probability of an inadmissible outcome is at most 60/m. Thus a nontagged admissible outcome has probability at least 1-61/m. Independence across the block shows that exactly one tag and admissible nontagged outcomes at the other coordinates have probability at least

For example, the last estimate follows from \log(1-t)\ge-t/(1-t) with t=61/m.

The total probability excluded by the future-block inequalities is at most

By (3.9), 0<r<1 Haar-almost everywhere. Lemma 4.2 therefore bounds the probability lost from the last line of the definition of E_j by

Consequently c_*=e^{-90}/2 works for all sufficiently large j.

On E_j, rounding changes the raw endpoint inequalities by at most 1/N_s. Since K\ge1, they imply

For 0\le v\le K, the bound \alpha_i<2/N_s makes the floors of u_i(x)+v\alpha_i and u_i(y)-v\alpha_i zero in every such block. Earlier-block floors have absolute value at most K+1. Hence

For v<0 these two quantities are at most 1, by monotonicity and F_0=1. Since the vector coordinates have modulus at most 1, all coefficients of the Laurent polynomials

have modulus at most D_j. Their constant coefficients are respectively r(x) and 1-r(y), both at least 1/j on E_j. The coefficient of z^{-n} in their product is c_{j,n}. The polynomial bound (2.10) now gives \sum_{n=0}^{2K}|c_{j,n}|\ge\varepsilon_j on E_j. Integration proves the second assertion.□

Proof. First suppose K_j<|k|\le N_j/4. A tagged coordinate in J_j satisfies, after rounding,

Here \beta=|k|\alpha_i lies in (1/m_j,1/2). The threshold 1-\{k\alpha_i\} equals 1-\beta for k>0 and \beta for k<0. It is crossed in either case, so (3.7) gives R_{x,y}(k)\le e^{-C_j}\le e^{-L_j} on E_j. Set both exceptional events to be empty in this range.

We next cover all larger |k|. Put M_s=m_s^{(4/5)m_s}. The intervals

cover [N_j/4,\infty). Indeed the next left endpoint is below the preceding right endpoint by (), 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 belong to [a_s,1-a_s] at every coordinate of J_s. In its upper part, (2.8) places them in [b_s,1-b_s] at at least m_s/20 coordinates. The initial interval uses the upper-part assertion for s=j.

Thus for each fixed k in this tail we can choose, deterministically, a block J_s, a width w\in\{a_s,b_s\}, and a set I\subset J_s of at least m_s/20 indices such that

The lower-part choice always has s>j. In either type of choice,

For w=b_s the first quantity is \kappa_sm_s^{1/40}\ge L_s\ge L_j; for w=a_s it is \kappa_s\sqrt{G_s}\ge L_{s-1}\ge L_j.

Let H_k be the event of no crossing at any index in I, and define

Conditional on x, the first event forbids the top label at every index in the indicated x-measurable set. Conditional independence and Lemma 4.4 therefore give

If the x-measurable set has fewer than |I|/2 elements, the event is empty, so the same bound holds.

For the other conditional estimate, note explicitly that coordinate order and the absence of crossings imply

Thus H_k^{(2)} requires at least |I|/2 indices with u_i(y)\ge u_i^*. Conditional on y, the bottom labels at all those indices are forbidden. The same product estimate proves the second bound in (5.9). These arguments use the original law, without conditioning on E_j. Figure 1 displays the complementary split.

Outside H_k, at least one chosen coordinate crosses. All coordinates are ordered, so (3.7) and (5.11) give R_{x,y}(k)\le e^{-L_j}. On H_k the ratio is at most 1. This proves (5.10) and the lemma.□

Proof. Every commutant unitary preserves \operatorname{ran}p and \operatorname{Dom}H, and commutes with H on that domain. It therefore maps the displayed closed space onto itself. Its projection commutes with the commutant, proving P\in\mathcal M. If \zeta\in\operatorname{ran}p\cap\operatorname{Dom}H, both invariance assumptions give A\zeta\in\operatorname{ran}p\cap\operatorname{Dom}H, and BH\zeta=HA\zeta. Thus B preserves the unclosed image and, by boundedness, its closure.

It remains to prove the trace assertion; bounded similarity cannot be assumed. Put E_N=\mathbf1_{[1/N,N]}(H). Since H is injective, E_N\uparrow1. For q_N=p\wedge E_N, the bounded element Hq_N=(HE_N)q_N of \mathcal M is bounded below by 1/N on its initial space. Its image is closed, and its polar decomposition identifies the initial projection q_N with its final projection, which is dominated by P. Therefore

The last inequality follows directly from the polar decomposition of (1-E_N)p: its initial support is p-(p\wedge E_N) and its final support is at most 1-E_N. Letting N increase gives \tau(P)\ge\tau(p).

For the reverse inequality, set q'_N=(1-p)\wedge E_N. The bounded operator H^{-1}q'_N=(H^{-1}E_N)q'_N is likewise bounded below on its initial space and has closed image. This image is orthogonal to \operatorname{ran}P. Indeed, for \eta\in\operatorname{ran}q'_N and \zeta\in\operatorname{ran}p\cap\operatorname{Dom}H, self-adjointness gives

here H^{-1}\eta lies in \operatorname{Dom}H. Polar decomposition and the same meet estimate yield

Passing to the limit proves the opposite inequality for \tau(P).□

Proof. For M>0, the function \ell_M(x)=\log\max(f(e^{2\pi i x}),e^{-M}) is continuous. Its integrals decrease to -\infty by (6.6). Irrational rotation averages of a continuous function converge uniformly to its integral: this follows for Fourier monomials from their geometric sums and then for continuous functions by uniform trigonometric approximation. Hence, for any a>0, first choosing M with \int_X\ell_M\,d\mu<-2a and then taking n sufficiently large gives

The expression on the left is \|T_f^n\|, because the power is a unitary U^n times multiplication by that continuous product. The spectral-radius formula yields radius zero, since a was arbitrary. Finally T_f\ne0 because f>0 almost everywhere.□

Proof. In the first range, put t=|k|\theta_i. The angle bounds give

For k>0 the threshold is 1-t, and for k<0 it is t. Both belong to [a_j,1-a_j] because a_j\le1/4.

For the second range, the uniform law of k\theta_i on an interval of length |k|/N_j has, after reduction modulo one, a density bounded by

This follows by counting the integer translates covering a point: the covering multiplicity of an interval of length r is at most r+1. The statement also applies to negative k by reflection. Thus a given coordinate misses [b_j,1-b_j] with probability at most 10b_j. If fewer than m/20 coordinates satisfy the assertion, more than 19m/20 coordinates miss the interval. Independence over i\in J_j and a union bound over the possible sets of misses show that, for a fixed k, this failure probability is at most 2^m(10b_j)^{(19/20)m} for large m. A further union bound over the at most 2m^{(4/5)m} nonzero integers in the range gives

These probabilities are summable because m_j\ge2^j. The first Borel–Cantelli Lemma now gives the second assertion for all sufficiently large j, simultaneously for every integer in each stated range.□

Proof. We give the construction and its estimates before choosing the probability-one event. The notation O(\cdot) in this proof has absolute constants, uniform in the coordinate, the selected powers, and the interval endpoints involved in a rectangle test.

We average many separated orbit powers so that both marginal densities approach those of an ideal model with independent uniform phases. The coordinate weights impose triangular laws on I and concentrate near the diagonal elsewhere, while their marginal bounds remain uniform in the growing product. A final restriction to ordered tail coordinates removes asymptotically negligible mass.

Separated powers and phase discrepancy

Take l sufficiently large that m_l\ge200, as required by the scalar lemma. At stage l write m=m_l, d=d_l, and set

For all sufficiently large l we can choose, deterministically, a set \mathcal N_l of integers in [Q/2,Q] such that

Indeed, given a previously selected n, a fixed frequency pair with v'\ne0 forbids an interval of possible n' of length 2R/|v'|\le2R, hence at most 2R+2 integers. There are O(H^2) frequency pairs. Since N_l\le m^2, each selection therefore excludes at most O(m^{1602}) integers. When v'=0 and v\ne0, the inequality is automatic because n\ge Q/2>R eventually; likewise, a zero first frequency creates no obstruction. The constraints are symmetric in the selected powers, since the full set of frequency pairs is used. Greedy selection from the \asymp Q available integers yields at least cQ/m^{1602}\ge m^{0.69m} choices for large m.

For a coordinate i\in J_s with s\le l, the angle \theta_i is uniform on [1/N_s,2/N_s]. Apply the separated-phase part of Lemma 2.1 with

The powers just selected satisfy its separation hypothesis. Thus t_n=\{n\theta_i\} has a density \varphi_{i,n} satisfying

In particular, that density is at most two for large l. For distinct selected powers and functions g,g' of bounded variation, the same lemma gives

No independence between t_n and t_{n'} is asserted. The imported estimate uses their Fourier coefficients and one-dimensional marginal densities, and so applies even when their joint law is singular. We shall also use the direct consequence of (A.9)

The weighted measures and both marginal orientations

Fix a finite I and restrict to sufficiently large l that I\subset\{1,\ldots,d\}. Set

For 0\le s,t<1 define

The second line is defined to be zero if \Delta(s)<\alpha. Let \widetilde\eta_{I,l} be the average, over n\in\mathcal N_l, of the push-forward under x\mapsto(x,\sigma_\theta^n x) of

Each weight is bounded for fixed l, so these are finite positive measures. They are carried by pairs satisfying x_i<y_i for i\le d, since h\Delta(x_i)\le h(1-x_i)<1-x_i and h<1.

We will treat the marginals both without a test and after restriction to a cylinder rectangle

where the E_i^x,E_i^y are intervals with rational endpoints and all but finitely many are the whole circle. Take l large enough that all tested coordinates are at most d. A missing test is represented by using the whole circle for every interval. For the first and second marginal, respectively, define

Changing variables y=\sigma_\theta^n x preserves Haar measure. It follows that the two tested marginal densities are exactly the averages of the products of the corresponding u_i^{(r)}. There is no factor from coordinates past d, even in the second orientation, since their unweighted Haar law is translation invariant.

We check the extra factors introduced by triangular laws and interval tests. Write u_i for either orientation and put

Uniformly in s, the coordinate, and the test intervals, we have

For an untested triangular factor, the bound is O(1): in the first orientation its support in t is 0<t<1-s, and in the second it is 0<t\le s. The possible value at t=s affects no integral. For the untested near-diagonal factors, Lemma 2.1 gives the bound O(m^{13/5}). To include the tests, in either orientation the extra factor is a constant in \{0,1\} times an interval indicator translated or reflected modulo one. Its supremum is at most one and its variation is bounded by an absolute constant. The inequality

therefore proves (A.15), including all jumps at the test endpoints.

The moment bounds needed below are

For an untested triangular factor the two averages are exactly 1-s and s, respectively. In either orientation its total integral and squared integral are 1/2, while A_i=1/3. For an untested near-diagonal factor, Lemma 2.1 gives

Since tests only decrease the nonnegative factors and their averages, these statements imply (A.16) for every tested factor. Their actual masses need not agree; in particular, the near-diagonal integral identities are not being imposed on the triangular factors or on the tested kernels. The quadratic error O(h^2) is what keeps the product of the moment bounds controlled as the number of coordinates grows.

The estimate in the growing product dimension

For either fixed orientation and a fixed rectangle test or no test, write

We prove the estimate

The independence used here is independence across coordinates \theta_i. In a single coordinate two phases may still be dependent, as already accounted for in (A.10).

For distinct n,n', Tonelli’s Theorem, product Haar measure, and coordinate independence yield the exact factorization

By (A.15) and (A.10), its ith factor is A_i+e_i with |e_i|\le c m^{-50}. Similarly,

Here (A.11) and \bar u_i\le2 bound the error in each factor from A_i by 4N_l/n=O(m^{-50}). Thus both sorts of factors have the same stated error bound.

The product errors do not require any lower bound on A_i. Telescoping gives

because (A.6) gives d\le1.01m, and hence dh^2\le1.01m^{-1/5}. In particular, A\le(1+ch^2)^d=O(1).

For a self-pairing, use only the one-phase density bound at most 2 and (A.16):

To see the cancellation of all other main terms explicitly, expand the squared norm. The distinct-pair terms have total averaging weight 1-1/P_l and main term (1-1/P_l)A. The cross terms with Z_l^0 have main term -2A, and the final squared norm of Z_l^0 is A. Their sum is -A/P_l\le0. By (A.18), their remaining contribution has absolute value O(dm^{-50}). The self-pairings have total weight 1/P_l. Dropping -A/P_l proves the slightly weaker bound (A.17).

This estimate is summable in l. Indeed,

for large m, whereas P_l\ge m^{0.69m}. Thus

whose sum is finite by m_l\ge2^l.

One probability-one event and the weak limit

For each fixed finite I, orientation, and rational cylinder rectangle or absent test, Tonelli’s Theorem applied to (A.19) shows that

where only the sufficiently large admissible stages are included. In particular, \|Z_l-Z_l^0\|_2\to0. Finite subsets of \N form a countable family, as do finite cylinder rectangles with rational endpoints, and there are only two orientations. Intersect these countably many probability-one events. For the remainder of the proof take any \theta in this intersection.

Without a test, Z_l is a marginal density of \widetilde\eta_{I,l}. Since \|Z_l^0\|_2^2=A=O(1), the L^2 convergence just proved bounds both marginal norms uniformly in l. We may omit finitely many initial stages; alternatively, each of their densities is bounded for fixed l, so they can be included by enlarging the bound. The total masses of \widetilde\eta_{I,l} are consequently uniformly bounded.

For a fixed cylinder rectangle \mathcal R, using the first orientation with its test shows that

The ideal integral has a direct product interpretation on the first d coordinates. For i\in I, integration of w_i(x_i,t) in x_i,t gives exactly the triangular measure under y_i=x_i+t. For i\notin I, first take uniform x_i, discard \Delta(x_i)<\alpha, and, otherwise, take t uniform in (0,h\Delta(x_i)), setting y_i=x_i+t. Before the discard this is a probability law (the value at the two endpoint values of x_i may be defined arbitrarily), and |y_i-x_i|\le h/2. The total loss caused by all cutoffs is at most

by the union bound, with the finite triangular factors having mass at most one. Each fixed tested coordinate outside I therefore converges to its uniform diagonal law. More explicitly, outside an h-neighborhood of a tested interval’s endpoints, membership of y_i in that interval agrees with membership of x_i; the neighborhood has Lebesgue measure O(h). Since a fixed rectangle tests only finitely many coordinates, all such errors vanish. Untested coordinates contribute only their masses, and the total cutoff estimate controls their product even as d increases. It follows that

These limits imply weak convergence on X\times X. Here are the topological details. The boundaries of the interval rectangles have zero \lambda_I measure, since every single-coordinate marginal of \lambda_I is absolutely continuous. Every continuous function on the compact product can be uniformly approximated by a continuous function of finitely many coordinates: in a summable product metric, replace the tail coordinates by fixed points and use uniform continuity. On the resulting finite torus, a sufficiently fine rational grid gives a step function, a finite linear combination of rational rectangle indicators, with arbitrarily small uniform error. Thus (A.20) and the uniform bound on total masses give convergence of integrals for every continuous function. In particular,

Trimming the tail to impose order everywhere

The untrimmed measures now have the required weak limit and both marginal bounds. It remains to impose order past the first d coordinates. Define \eta_{I,l} by restricting \widetilde\eta_{I,l} to the Borel set where x_i\le y_i also for every i>d. For a fixed selected n, the weight W_n depends only on the first d coordinates, so the tail x-coordinates remain independent uniform coordinates under that summand, apart from multiplication by its total mass. For a uniform tail coordinate, the probability that x_i>\{x_i+n\theta_i\} is exactly \{n\theta_i\}\le n\theta_i. Consequently, uniformly over n\in\mathcal N_l, the relative mass discarded from this summand is at most

The union bound is valid for the countable tail by increasing finite unions. More precisely, the discarded mass of a summand equals its total mass times the Haar probability of its tail failure event, because the weight and that event depend on disjoint coordinates. This statement also holds for a summand of mass zero. Averaging over n shows that the total discarded mass is bounded by (A.21) times the uniformly bounded untrimmed mass. Hence the difference of the two measures tends to zero in total variation and \eta_{I,l}\Longrightarrow\lambda_I.

Each trimmed marginal is a submeasure of the corresponding untrimmed marginal. Its Radon–Nikodym density is therefore nonnegative and bounded almost everywhere by the untrimmed density, so its L^2 norm cannot increase. Trimming preserves the finite set of positive orbit powers and now imposes order in every coordinate. All conclusions hold on the single probability-one event already constructed, for every finite I, proving the Proposition.□

Proof. Each generator commutes with scalar multiplication by any function of x. It also commutes with the unitary (J\zeta)(x)=S\zeta(\sigma x). These commutation relations therefore hold for every H\in M. Lemma 3.1 therefore represents H by an essentially bounded measurable field H(x), with products and adjoints computed fiberwise. Commutation with J gives (A.22).

Conversely, let H be an essentially bounded measurable field satisfying (A.22). All its integer-iterate covariance identities hold on one conull set. Put h_r(x)=H(x)_{r,0} for r\in\Z. Then

so the rth matrix diagonal of H is U^r\pi(h_r)\in M. To recover H from these diagonals, use the constant-field unitaries D_t e_k=e^{2\pi i kt}e_k, t\in\T. The strong integral

belongs to M, where \mathcal F_N is the Fejér kernel. The map t\mapsto D_tHD_t^* is strongly continuous and uniformly bounded. The positivity, unit integral and concentration at zero of \mathcal F_N imply H_N\to H strongly. The strong closure of M therefore proves the converse field characterization.

The displayed state is the vector state associated with the constant field e_0, hence is normal. If H\geq0 and \tau(H)=0, its 00 entry vanishes almost everywhere. Covariance then forces every diagonal entry to vanish almost everywhere. Positivity, or the inequality |H_{k,l}|^2\leq H_{k,k}H_{l,l}, implies H=0. Thus the state is faithful. For H,H'\in M, covariance and the change of variable z=\sigma^r x give

The sums over r are absolutely integrable by the row and column \ell^2 bounds and Cauchy–Schwarz. Summing proves \tau(HH')=\tau(H'H).

The rotation \sigma is ergodic by Section A.1. If H is central, it commutes with \pi(x\mapsto e^{2\pi i x_1}). In each fiber this last operator has distinct diagonal entries, because \theta_1 is irrational; hence H(x) is diagonal. Commutation with U makes all its diagonal entries equal. Equation (A.22) and ergodicity make their common value constant almost everywhere. The center of M is therefore scalar. The algebra contains the infinite-dimensional abelian algebra \pi(L^\infty(X)), so this finite factor is not a matrix algebra and is of type \mathrm{II}_1. Finally, the predual of a von Neumann algebra on a separable Hilbert space is a quotient of the separable trace class, and is separable.□

Proof. For n\geq1, direct multiplication and (A.24) give T^n=U^n\pi(F_n). All the floors defining F_n are nonnegative. For each fixed q, the estimate \lfloor x_i+n\theta_i\rfloor\geq n\theta_i-1 on the first q blocks therefore gives

Since C_s\geq K_s^2, the last expression tends to zero as q\to\infty. Also, Tonelli’s Theorem and the same angle bounds give

Since |T|=\pi(f) and the trace on \pi(L^\infty(X)) is integration against \mu, the determinant formula gives \Delta(T)=0. Positivity of f almost everywhere ensures T\ne0.□

Proof. Fix this p and a Borel version of its projection field. We may shrink X_0 to a common invariant Borel set of full measure on which covariance, the cocycle identities, and all the following countably many power identities hold. Invariance of \ran p(x) under T(x) gives invariance under T(x)^n, for each n\geq1. Applying S^{-n} and using covariance yields

Division by F_n(x)>0 changes the diagonal entries to F_k(\sigma^n x)/F_k(x) by (A.24). Proposition A.3 gives ordered orbit-pair measures converging to \lambda_I with uniformly L^2-bounded marginal densities. Their first marginals are absolutely continuous, so the countable collection of orbit identities holds almost everywhere on each approximant. The marginals of \lambda_I are absolutely continuous. The uniform L^2 bounds give uniform absolute continuity by Cauchy–Schwarz. Orderedness and (A.25) give the common cap B=1 on every ratio, under both the approximants and the limit. Lemma 3.3 now proves (A.27).□

Proof. For a top or bottom component of width w, each endpoint interval has length at least w/2, so its density is at most 2/w. A tagged endpoint density is at most m. Consequently either one-coordinate marginal density h_i satisfies

Throughout these sums the two widths are counted with their labels.

If v is the uniform density on an interval of length at least w/2, then \int(v-1)=0 and \|v-1\|_2^2\le2/w. The corresponding bound for a tagged interval is m. Subtracting 1 from h_i expresses it as the sum of four weighted mean-zero perturbations and, when s=j, one tagged perturbation. Cauchy–Schwarz for these at most five summands gives

This estimate holds for either endpoint. Summing over the m_s coordinates of every block gives the bound 20(2\sum_s\kappa_s^2+1) for every deterministic selection of endpoint marginals. Each selected coordinate density is nonnegative and has integral one, and the coordinate laws are independent. Thus Lemma 4.1 supplies the asserted L^2 density with norm at most H. In particular, for each such product law \eta and every Borel set A\subseteq X,

To verify the assertions after conditioning on an entire endpoint, we describe the conditional kernels. At coordinate i\in J_s, let \ell run through the replacement labels, including the tag when present. Write \pi_\ell for its mixture probability and u_\ell,v_\ell for the normalized interval densities of its two endpoints. The marginal densities in this coordinate are

They are strictly positive. A conditional law of y_i given x_i=t is

The reverse kernel K_i^y interchanges u_\ell and v_\ell. Independence of the coordinate pairs shows, first for cylinder tests, that \bigotimes_i K_i^x(x_i,\cdot) is a conditional law of y given x. The monotone class theorem extends the defining identity to all Borel tests on X^2. The corresponding product of reverse kernels is a conditional law of x given y. These formulas prove both assertions of conditional independence.

If x_i=t<u\le1-w, then t\in[0,1-w/2), where the top component of width w has x_i-density (1-w/2)^{-1}\ge1. Its whole y_i interval lies above u. Since h_i^x\le10, its contribution to (A.38) is at least \rho_s(w)/10. This proves (A.34). If y_i=t\ge u\ge w, the bottom component has y_i-density (1-w/2)^{-1}\ge1 at t, and its whole x_i interval lies below u. The reverse kernel therefore proves (A.35). Intersecting the full-measure sets for any specified countable family of thresholds gives the final assertion.□

Proof. Fix j. For D\ge1, define

and let \beta_D be the law of (x,y^{(D)}) under \mathbb P_j. We claim that

Condition on a pattern c of mixture labels in the first D coordinates. There are finitely many patterns. Let I(c) consist of the coordinates carrying replacement labels. At each i\in I(c) the conditioned pair has density

relative to the unnormalized triangular measure \ind_{\{0\le t<z<1\}}\,dt\,dz. Here U_{c,i},V_{c,i} are the two intervals of the selected component. Their lengths are positive and their rectangle is supported on t<z. The remaining prefix coordinates are uniformly diagonal.

For the diagonal tail, let g be the full x-marginal density from Lemma A.7. Write \mu=\mu_{\le D}\otimes\mu_{>D} and define

Fubini identifies g_D as the density of x_{>D}, and Jensen’s inequality proves the norm bound. Independence across coordinates ensures that conditioning on c leaves this tail law unchanged. Therefore the conditional law of (x,y^{(D)}) has density

with respect to \lambda_{I(c)}. This follows on cylinder sets from the coordinate construction, and hence on all Borel sets. The rectangle factors form a finite product and are bounded. The tail factor is integrable and need not equal one. Summing the conditional laws over the finitely many patterns proves (A.40).

Proposition A.6 now gives the pair identity \beta_D-almost everywhere. To pass to the limit, the first marginal of \beta_D is the x-marginal of \nu_j, while its second marginal uses the y-marginals in the first D coordinates and the x-marginals thereafter. Lemma A.7 and (A.37) therefore give uniform absolute continuity of both marginal families, and absolute continuity of the two marginals of \nu_j.

On the common probability space, y^{(D)}\to y in the product topology. Since X^2 is compact metrizable, bounded convergence against continuous functions proves \beta_D\Rightarrow\nu_j. Every one of these laws is carried by ordered pairs. Their absolutely continuous marginals put both endpoints in the common conull set on which the cocycles are finite and positive. The ordered-coordinate formula for the cocycle ratios consequently gives 0<R_{x,y^{(D)}}(k)\le1 and 0<R_{x,y}(k)\le1, simultaneously for every signed k. All hypotheses of Lemma 3.3 are now satisfied, with the fixed cap B=1. That lemma transfers the identity from \beta_D to \nu_j, proving (A.39).□

Proof. Since f>0 almost everywhere, T^*T=\pi(f)^2 has zero kernel, and |T|=\pi(f). The unitary U is therefore its polar factor. On the conull set where the sum in (A.23) has finite support, the value of f is determined by a finite subset of the countable coordinate set. There are only countably many such subsets. Consequently f takes values in a countable subset of (0,1], up to a null set. Its nonnull level sets give countably many spectral projections whose sum is 1, which proves the assertion about spectral type.

The representation convention gives

For a finite F, the tuple (f\circ\sigma^{-n})_{n\in F} has countable range on a common conull set. The nonnull joint level sets therefore give an atomic partition of the generated algebra \mathcal A_F. This includes F=\varnothing, when the generated unital algebra is scalar.

For the infinite family, \mathcal A is an invariant abelian subalgebra of \pi(L^\infty(X)). The rotation is totally ergodic. Indeed, for any integer q\ne0 and any nonconstant finite-support character indexed by v=(v_i), rational independence implies q\sum_i v_i\theta_i\notin\Z. The Fourier argument used in Section A.1 thus proves ergodicity of \sigma^q.

Suppose e were an atom of \mathcal A. Its integer translates under \operatorname{Ad}U are atoms of equal positive trace. Distinct atoms in an abelian algebra are orthogonal, so only finitely many can occur. The orbit of e is therefore finite, and for some positive integer q, U^q eU^{-q}=e. Because e\in\pi(L^\infty(X)), total ergodicity forces e=1. An abelian algebra whose identity is an atom is scalar. But \mathcal A contains the nonconstant multiplication operator \pi(f): a positive almost-everywhere constant would have finite logarithmic integral, contrary to Lemma A.5. Hence \mathcal A has no atoms, as claimed.□