Expected amenable subalgebras preserving core commutants

Proof. Apply [10, Theorem D, (4)\Rightarrow(6)] to Theorem 1.1.□

Proof. Suppose that the theorem fails, and choose (M,\varphi) as in (6). Put H=L^2(M,\varphi) and X=\log\Delta_\varphi. Let e^{isY} implement \gamma_s on H, so that e^{isY}(a\Omega)=\gamma_s(a)\Omega. Since the flow preserves the state and commutes with its modular action, X and Y strongly commute.

Two products after averaging. The scalar fixed algebra M_\varphi implies that the weak* averages of \sigma_t(x) over [-T,T] converge to \varphi(x)1. Indeed, any cluster point is modularly invariant and has state value \varphi(x). We use this factorization to turn the two orders of multiplication into isometries.

Let m be the symmetric translation-invariant mean obtained from these interval averages as T\to\infty along an ultrafilter. Quotient bounded measurable H-valued functions by the null space of m(\|f(t)\|^2), and complete to a Hilbert space \mathcal H_m. Define on elementary tensors

The inner products of the images of x\Omega\otimes a\Omega and y\Omega\otimes b\Omega are, respectively,

Both equal \varphi(y^*x)\varphi(b^*a). By linearity and completion, R,L:H\otimes H\to\mathcal H_m are isometries. Translation (\tau_u f)(t)=f(t+u) is unitary on \mathcal H_m and satisfies

From eigenstates to the product identity. We claim that

The binormal identity first compares the two sides on vectors whose X-spectrum lies in a short interval. Translation covariance then lets us combine these comparisons without accumulating an error over the intervals.

For a unit vector h\in H, define the state

The modular averages just computed give \Phi_h\circ\lambda=\Phi_h\circ\rho=\varphi; here e_M=1. Moreover, \Phi_h(f(X))=\langle f(X)h,h\rangle. For modular-analytic a\in M, put a_+=\sigma_{i/2}(a), so that \Omega a_+=a\Omega. The definitions of R and L give

These identities extend from h=x\Omega to every h\in H by continuity.

Suppose the spectral support of a sequence of unit vectors h is contained in intervals shrinking to s. Every weak* accumulation state of \Phi_h then has spectral evaluation at s. By (5),

Both vectors on the left have squared norm \|a\Omega\|^2, so their difference tends to zero.

This convergence is uniform when the centers s range over a fixed compact set and the interval lengths tend to zero. Otherwise choose violating vectors and a convergent subsequence of centers, and take a weak* accumulation state. The preceding argument gives a contradiction. In this step the varying test \gamma_s(a^*) may be replaced by its limit: the common left marginal \varphi and Cauchy–Schwarz bound the change by

which tends to zero. Thus no operator-norm continuity of \gamma is needed.

Analytic vectors a\Omega are dense in H, and all maps being compared are isometric. Hence the same local conclusion holds for a fixed \eta\in H, with \gamma_s(a)\Omega replaced by e^{isY}\eta. Spectral calculus gives

whenever \supp_X(h)\subset[s-\delta,s+\delta] and \|h\|=1. Consequently, for T=R-L\exp(iX\otimes Y), a fixed compact interval I, and fixed \eta, there is a function \alpha_{I,\eta}(\delta)\to0 such that

when h has spectral support in an interval of length at most \delta contained in I.

Translation covariance shows that T^*T commutes with the spectral projections of X\otimes1. For \supp_X(h)\subset I, partition I into finitely many disjoint Borel intervals I_j of length at most \delta and write h_j=\one_{I_j}(X)h. The vectors T(h_j\otimes\eta) are orthogonal, so

Letting \delta\downarrow0 and then using density proves (7).

Reflection forces a lattice. Define (J_0f)(t)=\Delta^{it}f(-t). Symmetry of m makes J_0 a unitary involution on \mathcal H_m. On elementary tensors,

Similarly J_0L=R\Sigma, where \Sigma flips the input factors. Applying J_0 to (7) and composing with \Sigma gives L=R\exp(iY\otimes X). Substitute this back into (7) and cancel the isometry R to obtain

Let S\subset\R^2 be the support of the joint spectral measure of (X,Y). The tensor-product spectral measure has support S\times S, so the continuous phase in (8) satisfies

If S contains two linearly independent points, pairing with them confines S to the inverse image of (2\pi\mathbb Z)^2 under an invertible linear map. Thus S is countable. If S spans a non-axis line, choose a nonzero (r_0,s_0)\in S; then r_0s_0\ne0, and any point t(r_0,s_0)\in S satisfies 2tr_0s_0\in2\pi\mathbb Z. Again S is countable. In both cases the joint spectral measure is supported on countably many points, so X has pure point spectrum. The remaining possibilities put S on a coordinate axis and give X=0 or Y=0.

Excluding the spectral alternatives. A modular action with scalar fixed algebra has no Hilbert-space eigenvectors other than multiples of \Omega; we recall the argument of [12, Lemma 2.1]. Let P_r be the spectral projection of X at r. For x\in M, the bounded character averages

satisfy \|x_T\|\le\|x\| and x_T\Omega\to P_r(x\Omega). A weak* cluster point y\in M represents this vector, since \varphi(z^*x_T) converges to \langle P_r(x\Omega),z\Omega\rangle for every z\in M. Thus y\Omega=P_r(x\Omega), and the separating property of \Omega gives \sigma_t(y)=e^{irt}y. Density of M\Omega shows that any nonzero eigenspace is detected by a nonzero such eigenoperator.

For this y, both y^*y and yy^* lie in M_\varphi=\C1. Hence y is a nonzero scalar multiple of a unitary u. Since \sigma_{-i}(u)=e^r u, the KMS identity gives

Therefore r=0, and then y itself is scalar. It follows that the only eigenvectors of X are multiples of \Omega. They cannot span H because M is nontrivial. This excludes pure point spectrum, including X=0. Finally, Y=0 would make the bicentralizer flow trivial on M, contradicting its ergodicity. All alternatives are excluded, proving the theorem.□

Proof. Applying E_N to the asymptotic commutation condition gives E_N(B)\subset\B(N,\varphi)=\C1, by Theorem 3.1. State preservation determines the scalar. For x\in N and b\in B,

which characterizes E_B(x)=\varphi(x)1. Using E_B for the first product and E_N for the second gives

Polarization and completion prove the assertion.□

Proof.

Construction and marginals. Take finitely many elementary tensors. For T\ge0, apply the recurrence simultaneously to their two indices. At each stage the operator matrix is multiplied on the left and right by the adjoint and the original of the diagonal matrix of that position’s labels. Its positivity is preserved, and complete positivity of \theta_{-d_j} preserves it at the next stage. Taking the final \Omega coefficients therefore gives a positive scalar kernel. For T=1, the identity E_B|_N=\varphi(\cdot)1 reduces this kernel to

the Gram kernel of the elementary tensors. Applying positivity to 0\le T\le\|T\|1 now bounds the quadratic form by \|T\| times the tensor norm squared. Thus the form depends only on the tensor vector, extends to the completion, and defines a positive unital map. The same argument with an additional matrix index proves complete positivity.

For T=\lambda(a), the first map produces \gamma_{-d_n}(E_B(a))\in B with state value \varphi(a). At a subsequent stage, an element b\in B is replaced by \gamma_{-d_{j-1}}(E_B(y_j^*bx_j))\in B. Its state value is

using E_N|_B=\varphi(\cdot)1. Iterating this scalar identity gives the left marginal in (16). For the right marginal, the full right algebra is fixed by each \theta, hence belongs to its multiplicative domain. Right multipliers commute with the labels and pass through every stage, while E_B|_N=\varphi(\cdot)1 factors the labels as above. For e_N, use [e_N,N]=0 and the fact that the fixed projection e_N also belongs to every map’s multiplicative domain. Since e_N\Omega=\Omega, the recurrence gives the same Gram kernel as for T=1. These computations establish all of (16).

The modular unitary group. First extend the spectral formula for each individual \theta_{-q} from C_0(\R) to the functions s\mapsto e^{its}. Its restriction to C_0(X_M) is a homomorphism, so that algebra belongs to its multiplicative domain. For f\in C_0(\R), it follows that

The ranges of these cutoffs span a dense subspace of \HH. Consequently \theta_{-q}(e^{itX_M})=e^{it(X_M+q)}; these unitaries also belong to the multiplicative domain. No limiting operation has been passed through \theta_{-q}.

In the kernel recurrence, commute e^{itX_M} to the right across each x_j, using e^{itX_M}x_j=\sigma_t(x_j)e^{itX_M}. The next map replaces the left factor y_j^*\sigma_t(x_j)\in N by its state value. The resulting coefficient is

which proves

A nonnormal map need not preserve weak operator integrals, so this identity alone does not yet justify (17). We next show that the coefficient states needed to recover continuous functions are supported inside a compact spectral interval.

Spectral cutoffs and continuous functions. Call x\in N bandlimited if it admits a finite bound L\ge0 such that, for every interval J,

Such elements are strongly dense. Indeed, convolve the modular action with an approximate identity having smooth, compactly supported Fourier transforms. If the Fourier support of a smoothing kernel lies in [-L,L], the resulting element has the displayed transfer property: testing e^{itX_M}xe^{-itX_M}=\sigma_t(x) against scalar Fourier kernels shows that spectral pieces separated by more than L have zero matrix coefficient.

Two operations in the kernel preserve a suitable interval on which a positive operator vanishes. Suppose that A\ge0 annihilates the spectral subspace associated with an open interval I. For f\in C_c(I), f(X_M)A=0, so the multiplicative-domain identity gives

The cutoff ranges span the spectral subspace for I-q; thus \theta_{-q}(A) annihilates that translated interval. If x has transfer in [-L,L], the preceding transfer rule also shows that x^*Ax annihilates every interval J with J+[-L,L]\subset I.

Apply these rules to a diagonal kernel with bandlimited labels x_j, of transfer bounds L_j, starting from A=\one_{\{|X_M|>R\}}. Translating by each -d_j and shrinking by each L_j leaves an open interval about zero whenever

The final operator then annihilates \Omega, whose modular energy is zero. Hence its diagonal kernel coefficient vanishes. For a finite sum of bandlimited elementary tensors, choose one R that works for every term; positivity and Cauchy–Schwarz make the off-diagonal coefficients vanish as well. These tensor sums form a dense subspace of H^{\otimes n}.

For a unit vector \xi in this subspace, set \Phi_\xi(T)=\langle Z(T)\xi,\xi\rangle and P_R=\one_{[-R,R]}(X_M). We have proved \Phi_\xi(1-P_R)=0. Cauchy–Schwarz therefore yields

The transfer property applied to \Omega also puts x_j\Omega in the X-spectral subspace for [-L_j,L_j]. Our choice of R therefore places the spectral distribution of S_n+d on \xi in [-R,R] as well. Both sides of (19), evaluated on \xi, are thus determined by functions restricted to this same compact interval. Finite linear combinations of the exponentials form a uniformly dense self-adjoint algebra in C([-R,R]). Uniform approximation now proves

Continuity in \xi and polarization prove (17) on the whole tensor product. The radius R may depend on the labels, shifts, and tensor length: only the resulting exact identity will enter the uniform estimate.

The word coefficient. First, (11) and state preservation of E_B give

The identity follows first for \zeta=a\Omega and then by density. It shows that moving an output test backwards through U keeps its last factor in B\Omega. To compute a word coefficient, take elementary tensors and record the adjoints of the successive test labels as

Each c_j belongs to B. Because the right multiplier is fixed and lies in the multiplicative domain of every \theta, the recurrence for T=b^*\rho(a_+) gives T_j=c_j\rho(a_+). Its final coefficient is \varphi(c_0a).

The same recurrence computes the coefficient of C backwards from its output vector. Moving the output test across W_{d_j} applies \gamma_{-d_j}, and (11) moves it across U_j by replacing c_j with E_B(y_j^*c_jx_j). The last coefficient is again \varphi(c_0a). This proves (18) for elementary tensors; linearity and continuity give the general case.□

Proof of Proposition 5.1. First restrict to data for which the spectral support of S_n+d on \xi lies in [-\delta,\delta]. The state

has both marginals equal to \varphi and satisfies \Phi(e_N)=1. By (17), along any sequence of such data with \delta\to0, every weak* accumulation state satisfies \Phi(f(X_M))=f(0) for f\in C_0(\R). Proposition 2.1(i), applied to the fixed operator a^*\rho(a_+), and (18) give

for every modular-analytic a\in B. Since C is a contraction,

so this difference tends to zero. Density of analytic vectors a\Omega and contractivity give the same conclusion for each fixed \eta\in K. The conclusion is uniform in n, the shifts, and \xi: otherwise a sequence of violating choices would have an accumulation state in the state space of the same algebra \mathbf B(\HH) and contradict the argument just given.

For the ideal word, the same spectral localization gives

Combining these bounds proves the centered estimate. Finally suppose the S_n-support of \xi lies in any interval of length \delta. Choose q so that the containing interval for S_n+d+q is centered at zero, and replace the final shift d_n by d_n+q. This replaces C and D_{\mathrm{all}} by W_qC and W_qD_{\mathrm{all}}. Their difference has the same norm, and the input vector \eta is unchanged. The centered estimate therefore proves (14) with all the asserted uniformity.□

Proof. If no U_j remains, the word equals D_{\mathrm{all}}. Otherwise disintegrate the replaced positions over their X-spectral variables. On a fiber with values (x_j), each replaced D_j acts as the shift W_{x_j} on K, and the sum of the remaining energies lies in the original interval translated by -\sum x_j. The interval’s length is unchanged. The fiber vector in the remaining positions need not be a simple tensor: Proposition 5.1 applies to every such vector, with the same last factor \eta. If a fiber has spectral multiplicity, expand in an orthonormal basis of that multiplicity space and sum the homogeneous squared bounds. Integrating proves the hybrid assertion. Convex expansion proves the assertion for G_j; the coefficients of the expansion sum to one, independently of n.

Choose an even real L^2-normalized smooth function g supported in an interval of length \delta, and put p_\delta(t)=|\widehat g(t)|^2, with unitarily normalized Fourier transform. For the bounded operator T=C-D_{\mathrm{all}}, vector-valued Plancherel gives

Here Plancherel is applied to z\mapsto g(S_n-z)\xi\otimes\eta and then to its image under the bounded operator T; no commutation of T with S_n is required. The Fourier sign does not matter with this choice of g. Each vector on the right is spectrally localized as required by (14), and

Apply the homogeneous form of that estimate and integrate.□

Proof. Suppose otherwise. Density of compact-spectrum vectors gives unit vectors h\in H and \eta\in K with \supp_X(h)\subset[-l,l] and \|(U-D)(h\otimes\eta)\|>0. By strong continuity there are c>0 and 0<b\le1 such that

Use this fixed \eta in Lemma 6.1. Choose \delta so that \epsilon:=\epsilon_\eta(\delta)<c/8, and choose A\ge1 with \int_{-A}^A p_\delta(t)\,dt\ge3/4.

The repeated experiment. Fix a finite terminal length m. On one product probability space choose t with density p_\delta and mutually independent parameters

all independent of t. At position j put the fresh input e^{i(t-r_j)X}h. Starting with \eta on K, propagate using W_{s_j}G_jW_{-s_j}. Let \zeta_k denote the result after k positions, and let \chi_k be the ideal result with D_j in place of G_j. Since D_j commutes with the shifts on K,

Condition on every parameter other than t. Then \xi_k=\bigotimes_{j=1}^k e^{-ir_jX}h is a fixed unit vector, the input is e^{itS_k}\xi_k\otimes\eta, and the successive conjugations W_{s_j}G_jW_{-s_j} form an averaged word with fixed shifts summing to zero. Lemma 6.1 therefore gives

For k=0, both vectors are \eta.

A definite loss at each captured test. At step j=k+1, insert the fresh position into \zeta_k and apply W_{-s_j} on K; call this input v. Let v_0 be the analogous input formed from \chi_k. The parallelogram identity and contractivity give

Define the random projection

where the first factor is a scalar indicator and the second acts on the previous k positions. For k=0, interpret S_0 as zero. The projection Q commutes with U_j,D_j and the last-position shifts. Put

We claim

Here \E integrates all parameters involved, including t.

For (24), disintegrate the previous ideal vector over S_k=e. Its last factor is W_e\eta, and |e|\le kl. The interval [e-b,e+b] lies in [-B_j,B_j], so the probability over s_j of capturing it is exactly b/B_j, as illustrated in Figure 1. For |t|\le A, the probability over r_j of |t-r_j|\le b is b/(2A). On these fibers the current last factor is W_{e-s_j}\eta, so (21) applies. Integrating, and using the mass at least 3/4 of [-A,A], proves (24).

For (25), fix t and all parameters from the previous k steps, so that \zeta_k-\chi_k is fixed. No spectral support bound on this error is available. For every real e,

The spectral theorem therefore gives the quadratic-form inequality

Also \Pr_{r_j}(|t-r_j|\le b)\le b/(2A) for every fixed t. The two current sampling parameters are independent of each other and of the past. The fresh vector has norm one, while W_{-s_j} commutes with Q and does not affect its norm. Consequently

Now integrate over t and the past parameters to obtain

by (22). Since \|U_j-D_j\|\le2, this proves the claim.

The contradiction. Since Q commutes with the step operators, \|(U_j-D_j)v\|^2\ge\|(U_j-D_j)Qv\|^2. The triangle inequality in the mean-square Hilbert space, followed by (23), shows that the expected squared-norm loss at step j is at least

The constant in parentheses is positive. On the other hand, inserting a unit fresh vector and applying the outer shift do not change norm. The losses therefore telescope from one and have total at most one. Since \sum_j a_j=\infty, choosing sufficiently many steps contradicts (26). Hence U=D.□

Proof.

The algebra generated by N and B. By Theorem 6.2, L^*R=D is unitary. Since L,R are isometries, this implies R=LD, so their ranges agree. Put A=B\vee N. The common range is L^2(A,\varphi): it contains \Omega, is invariant under left N by its R description and under left B by its L description, and is contained in L^2(A,\varphi). Applying adjoints gives invariance under the generated von Neumann algebra. Both N and B are modularly invariant, so the expected algebra A has a canonically embedded core inside c(M).

Coordinates adapted to N. Use L to identify L^2(A,\varphi) with H\otimes K. In these coordinates left B acts as 1\otimes B, left N acts as D(N\otimes1)D^*, and the modular group is

Represent c(A) faithfully on H\otimes K\otimes L^2(\R) with constant base algebra and group unitaries

where l_\R is the translation representation. This is the usual regular crossed-product representation after conjugating its variable base algebra by the modular implementation.

Conjugate by D^* on the first two positions. This removes the twist from the action of N, which now acts on H alone, and transfers it to the core of B. Since W_s commutes with \Delta_{\varphi|_B}^{it}, the algebra B\rtimes\R is represented, at spectral value X=h, by

The two transformations in (30) are the same flow: on B it is \gamma_{-h}, and on l(t) it multiplies by e^{ith}. Thus these fibers are a fixed normal representation of B\rtimes\R composed with the core bicentralizer flow at -h, by (27). More explicitly, if H=\int^\oplus H_h\,d\nu(h) and \pi_0 is the fixed regular representation on K\otimes L^2(\R), the representation of an element z\in B\rtimes\R is

This normal direct-integral representation agrees with (30) on generators, so the formula applies to the entire crossed product.

The fixed algebra. If z is fixed by the core bicentralizer flow, then \pi_0(\gamma_{-h}(z))=\pi_0(z) in every fiber. It therefore acts on K\otimes L^2(\R) alone, while N acts on H alone. Hence it commutes with N, proving

The opposite inclusion is (28), so the fixed algebras agree.□

Proof of Theorem 1.1.

Factor bottom algebras. For a type \mathrm{III}_1 bottom factor, Proposition 7.1 and [10, Proposition 8.6 and Theorem D] give the weak Dixmier property (29). The state on N is a strictly semifinite weight, as required there. For a bottom factor of any other type, the same weak Dixmier property is [10, Proposition 4.11]. These factor cases allow arbitrary upper algebras.

Disintegration over the center. For a general N, put J=\mathcal Z(N) and M_0=J'\cap M. The passage to M_0 makes J central in both algebras, so that we can apply the factor cases fiber by fiber. The modular action of \varphi=\varphi\circ E_N fixes J pointwise. Consequently M_0 is modularly invariant, and E_N restricts to a faithful normal expectation M_0\to N. Moreover,

inside the chosen core, by [10, Proposition 4.7].

Disintegrate N\subset M_0 over the common central algebra J, using the probability measure \mu determined by \varphi|_J. The bottom fibers N_z are factors because J=\mathcal Z(N); the upper fibers (M_0)_z need not be factors. The simultaneous disintegration theorem for operator-valued weights gives faithful normal fiber expectations E_z:(M_0)_z\to N_z and compatible faithful state fibers \varphi_z [8, Propositions 2.1–2.2 and Theorem 2.3]. Indeed, E_N is J-bimodular, and its fiber weights satisfy E_z(1)=1, so they are conditional expectations. The state identity \varphi=\varphi\circ E_N and the modular actions also disintegrate, as in the proof of that theorem.

To see the resulting core compatibility, write H_0=L^2(M_0,\varphi)=\int^\oplus H_z\,d\mu(z), where H_z=L^2((M_0)_z,\varphi_z). The regular representation of the core on L^2(\R,H_0)=\int^\oplus L^2(\R,H_z)\,d\mu(z) is given by

Both formulas act fiberwise. A countable family of algebra generators and the rational translation parameters generate the fiber cores; the common central algebra J is already present in the representation. Thus c(M_0)=\int^\oplus c((M_0)_z)\,d\mu(z) with the corresponding copies of N_z. Separability supplies these standard measurable fields. The factor cases and the direct-integral criterion [10, Proposition 4.12] now give the weak Dixmier property for N\subset c(M_0).

Averaging into the central commutant. We have proved the weak Dixmier property inside c(M_0). To apply it to an arbitrary x\in c(M), we first average x into this smaller core. The weak* compact convex hull of its orbit under \U(J) is preserved by the commuting affine maps \Ad(u), u\in\U(J). The compact-convex fixed-point theorem for commuting affine maps gives a common fixed point y\in J'\cap c(M)=c(M_0). One can obtain it by taking rectangular Cesàro averages for each finite family of commuting maps, then using compactness and the finite-intersection property. Since J\subset N, this point belongs to the weak* closed convex N-orbit hull of x. That hull is N-invariant. Applying weak Dixmier averaging inside c(M_0) therefore produces a point of N'\cap c(M) still in the original hull. This proves (29) for the original inclusion.

The amenable subalgebra. Finally [10, Theorem D, (3)\Rightarrow(4)] supplies an amenable subalgebra P\subset N with a faithful normal conditional expectation and P'\cap c(M)=N'\cap c(M). The expectation is faithful and normal by the meaning of “with expectation” in that theorem. Unitality is automatic: if p=1_P<1_N, bimodularity would imply F(1-p)=0, contradicting faithfulness. This proves the result.□