Universal strong Kadison–Kastler stability

Proof. Since A' is weak-star closed, its preannihilator in the trace class computes the quotient norm \dist(h,A'). Let \varphi be a trace-class functional of norm at most one vanishing on A'. A singular-value decomposition gives vectors \xi,\eta\in\ell^2\otimes H, each of norm at most one, with

This uses only countably many vectors even when H is nonseparable. Let p project onto \overline{(I\otimes A')\xi}. This subspace reduces I\otimes A', so every matrix entry of p belongs to A. Moreover p\xi=\xi and p\eta=0. Therefore

If q_n projects onto the first n coordinates of \ell^2, the compression of this commutator is [I_n\otimes h,q_npq_n], of norm at most \norm{\ad_h|_A}_{\cb} because q_npq_n\in M_n(A)_1. Testing on finitely supported vectors and then using density proves the same bound for the full commutator. Taking the supremum over \varphi proves (2.2).

For h\in A'_1, approximate each contraction in M_n(B) by one in M_n(A), with arbitrary slack above \gamma. It follows that \norm{\ad_h|_B}_{\cb}\le2\gamma. The first assertion gives distance at most 2\gamma to B', and normalization of approximants gives distance at most 4\gamma to B'_1. Reverse the algebras to finish.□

Proof. If e=0, take u=I. Otherwise let

The von Neumann image \pi(P) has the quotient matrix norms of P. The commutator of S with \pi has cb norm at most e, hence the same is true on \pi(P) by quotient lifts with arbitrarily small norm slack. By Lemma 2.2 choose D\in\pi(P)' with \norm{D-S}<2e. Its lower off-diagonal block T satisfies

Thus T is invertible and T^*T commutes with \rho(P). Its polar unitary u=T|T|^{-1} intertwines the representations. The singular values of T lie between 1-r and 1+r, where r=\norm{T-I}, so

□

Proof. Complete isometry of \rho gives \norm{F_n(X)}\ge(1-e)\norm X at every matrix size. Thus F is injective, its range V is norm closed, and \norm{F^{-1}:V\to P}_{\cb}\le(1-e)^{-1}. For y\in Q_1, match y with \rho(x) for x\in P_1, allowing arbitrary slack, to obtain \dist(y,V)\le g+e<1. If V were proper, the quotient map Q\to Q/V would have norm one, contradicting this bound on Q_1. Hence F is onto.

Transport multiplication from Q to P:

It is associative and preserves the involution. For rectangular X,Y, insert \rho(XY)=\rho(X)\rho(Y) to see that

Here the first difference is bounded by e\norm{XY} and the product difference by 2e\norm X\norm Y, since F and \rho are complete contractions. Equation (2.3) therefore gives

The cited theorem gives an involution-preserving L with \norm{L-\id}_{\cb}\le\lambda:=30e/(1-e)<1. Then \theta=FL^{-1} is a bijective *-homomorphism, and

Bijective multiplicativity makes \theta unital. A *-isomorphism between von Neumann algebras is an order isomorphism on their selfadjoint parts, so it preserves bounded increasing suprema and is normal. Apply Lemma 2.3, since 32e<1/4. When e=0, the preceding range argument gives Q=\rho(P) directly.□

Proof. Compress a contraction of M_n(A\barotimes\B(K)) by increasing finite-rank projections on K. In a corner of rank r, choose an opposite contraction in M_{nr}(B) within \gamma plus arbitrarily small slack, and extend it by zero on the complement. The compressions of the original contraction converge strongly to it. A bounded WOT cluster of the opposite contractions is in M_n(B\barotimes\B(K))_1 and is within \gamma, by WOT closedness of the norm bound, choosing the approximation slack to tend to zero along the compressions. Reverse the argument and take the supremum over n.

Now start with a contraction in M_n(A\barotimes D). The first assertion supplies a nearby contraction in M_n(B\barotimes\B(K)). Average it by conjugation over the unitary group of each finite-dimensional stage of D'. The input commutes with these groups, so the bound is preserved. A WOT cluster of the averages commutes with D' and still lies in M_n(B\barotimes\B(K)). The von Neumann tensor commutation theorem [4, Theorem III.4.5.8 and Corollary III.4.5.9, pp. 319–320] places it in M_n(B\barotimes D). Reverse and amplify as before.□

Proof. The coefficient formula defines a bounded sesquilinear form and hence an operator. Positivity at every matrix size and unitality follow by applying the positive scalar mean to matrix quadratic forms; the domain has the pointwise operations on bounded operator families. Fixed left and right factors pass through the coefficient formula. Testing the linear equations defining a weak-operator closed subspace proves range preservation. Testing vector coefficients at each rectangular size gives the norm bound. For the action average, left invariance makes the range invariant, while averaging fixes each invariant operator. An invariant algebra is preserved by the preceding range property. Predual duality gives the same argument for the general action on P, including preservation of matrix norm bounds.□

Proof. There is nothing to prove if \delta=0. Otherwise we use the triangular homomorphism associated with a derivation, as in [21, p. 216]:

This is a unital homomorphism with \norm{\Phi}\le2. For \xi\in H, the closed subspace

is invariant for \Phi(A), and the restricted homomorphism is cyclic. Haagerup’s cyclic similarity theorem gives \norm{\Phi|_{K_\xi}}_{\cb}\le\norm{\Phi|_{K_\xi}}^3\le8; see [21, Theorem 1.1 and its proof, pp. 217, 225]. Apply its column amplification to (0,\xi)\in K_\xi, and retain the upper component of each output. This gives

Taking the supremum over unit vectors proves the claim.□

Proof. The functionals a\mapsto\tau(ba), b\in P, are norm dense in the predual P_*. Indeed, their annihilator in (P_*)^*=P is zero, by faithfulness of the trace, so Hahn–Banach gives the density. For a normal vector functional \varphi, approximate \varphi in norm by one of these functionals. If \norm{x_j}\le K_0, then

This proves (i). For (ii), the defining identity of the conditional expectation gives

The same predual-density argument, using boundedness, extends the identity to every normal functional.□

Proof. We estimate the centered words first, then restore their scalar parts. The decomposition below uses the creation, annihilation, and diagonal actions in the free-product Hilbert space, as in the framework developed by Voiculescu [43] and the reduced-free-product inequalities of Ricard and Xu [39, Section 1 and Fact 2.6]. The corresponding bound for words of fixed length in free groups is Haagerup’s inequality [20, Lemma 1.4]. Here the prescribed label multiplicities require the additional tuple-matrix and square-summation estimates proved below.

Reduced words and their cuts. We first describe the Hilbert space on which the word norm will be estimated. Replace \mathcal A by W^*(A_1,\ldots,A_r) and use its faithful left representation on L^2(\mathcal A,\tau). Set

Freeness identifies this representation space with

Indeed, map a tensor to the corresponding product applied to 1. Its inner products are the tensor inner products: in the trace of the adjoint of one reduced word times another, pair the adjacent letters at the junction, split their product into its trace and centered part, and iterate. Freeness eliminates the centered terms until the copy sequences agree and all letters have been paired. Distinct summands are orthogonal. Splitting each letter into its scalar and centered parts shows that reduced words span the algebra generated by the copies, hence a dense subspace of L^2(\mathcal A,\tau).

Write h_a^{(j)}=e_a^{(j)}-d^{-1}1. Define the label map

The normalized matrix trace gives

where \mathbf1 is the vector all of whose entries are one.

Left multiplication by h_a^{(j)} creates the first letter h_a^{(j)} on a reduced tensor not beginning in copy j. On a tensor b\otimes\xi beginning in that copy, it has two parts: annihilation, equal to \tau(h_a^{(j)}b)\xi, and preservation of the first leg, with operator

Here P_j^\circ is the orthogonal projection onto H_j^\circ inside L^2(A_j,\tau), and L denotes left multiplication. These formulas also apply to the vacuum, on which only creation occurs.

Consider a fully centered word h_{a_r}^{(r)}\cdots h_{a_1}^{(1)}. The rightmost letter acts first. Since the copy indices 1,\ldots,r are distinct, once a creation or preservation occurs, every later operation must be a creation. Thus the word is a sum of cuts of two kinds: p annihilations followed by q=r-p creations, or p annihilations, one preservation, and q=r-p-1 creations.

For completeness, the domain restrictions in these cuts can be made explicit. For a set S of copy indices, let \mathcal F_{\notin S} denote the vacuum plus all reduced tensors whose first copy index is outside S. For a cut without preservation put

Its source and target spaces are canonically embedded copies of

respectively; empty tensor products mean \C. Excluding copy p makes the input reduced, and excluding copy p+1 permits the first creation. For a preservation cut put j=p+1 and E=\mathcal F_{\notin\{j\}}. Its source and target spaces are

All embeddings are isometries; a cut is zero on the orthogonal complement of its indicated source space. In particular these restrictions depend on the cut alone and never on the labels.

Distinct labels and tuple matrices. We now suppose that W has distinct labels and sum the fully centered words with multiset W. In each cut, the annihilation map is the tensor product of D_1^*,\ldots,D_p^*, with the identity on the remaining factors. The creation map uses the D_j in decreasing copy order. Their norms are at most d^{-p/2} and d^{-q/2}. Between them is a matrix on label tuples, scalar-valued without preservation and operator-valued on H_j^\circ with preservation. Tensor order reversals are unitary. Together with the isometric source and target embeddings, this gives a factorization of each cut whose norm is bounded by

The tail factor carries the identity throughout.

Without preservation, the tuple matrix has entry one exactly when the input and output tuples together list W. Group tuples by their underlying subsets. Each input subset of size p is paired only with its complementary output subset of size q, giving mutually orthogonal all-ones blocks of size q! by p!. The matrix norm is therefore \sqrt{p!q!}. Summing these cuts gives at most

For a preservation cut, p+q=r-1. Summing and replicating over the orders of the input and output tuples introduces a factor \sqrt{p!q!}. The remaining matrix is indexed by subsets B\subset W of size p on the input and subsets A\subset W of size q on the output. Its entry is K_a^{(j)} when A\cap B is empty and \{a\}=W\setminus(A\cup B); other entries vanish.

To estimate this matrix, first replace K_a^{(j)} by the projection L_{e_a^{(j)}} on L^2(A_j,\tau) and call the resulting matrix T. For a fixed row A, the admissible columns B give distinct missing labels a, so their output ranges are orthogonal. For a fixed column B, the admissible rows also give distinct a, so the projection domains are orthogonal. Thus, for an input (\xi_B)_B,

Consequently \norm{T}\leq1. Compressing every input and output leg to H_j^\circ preserves this bound. By (3.4), the remaining scalar correction is d^{-1} times the zero-one matrix of admissible pairs, tensored with I_{H_j^\circ}. This matrix has p+1 entries in each row and q+1 in each column; Cauchy–Schwarz in each row bounds its norm by \sqrt{(p+1)(q+1)}. The preservation tuple matrix therefore has norm at most

Using (3.5) and summing the preservation cuts gives

The last sum is at most five, uniformly in r. To see this, put m=r-1. For m=0,1,2,3 the sums are respectively 1, 2, 2+1/\sqrt2, and 2+2/\sqrt3. For m\geq4, unimodality of the binomial coefficients gives

Scalar parts and repeated labels. It remains to restore the scalar parts of the projections and to treat repeated labels. We record a bound for a single centered word of length l\geq1 in distinct copy positions, with arbitrary labels. The same cuts now have scalar coefficients rather than tuple sums. Each creation or annihilation has norm at most d^{-1/2}, and each preservation has norm at most one, since it is a compression of multiplication by e_a-d^{-1}1. There are l preservation cuts and l+1 cuts without preservation. Hence

This estimate does not require the labels to be distinct.

Expand a projection word of length r using e_a=d^{-1}1+h_a. A term retaining l<r centered letters contains the scalar factor d^{-(r-l)}. For 1\leq l<r, (3.8) bounds it by

The term retaining no centered letters has norm d^{-r}\leq d^{-r/2}. Thus the difference between B_W and its fully centered sum has norm at most

because at most r! ordered tuples have multiset W.

For distinct labels, combining (3.6), (3.7), and (3.9) yields

For example, choose an integer

Then the two excess terms, after division by \sqrt{(r-1)!}\,d^{-(r-1)/2}, are each at most one, proving the constant seven in (3.1).

For repeated labels, apply (3.8) to every fully centered summand and use (3.9). Since d^{-r/2}\leq d^{-(r-1)/2}, it suffices to take

These choices depend only on r and complete the proof.□

Proof. Fix r and a label a. There are \binom{d-1}{r-1} distinct-label multisets containing a, and their contribution is

For r=1 this is the entire sum. Suppose r\geq2. A repeated-label multiset has support of cardinality t\leq r-1. If it contains a, its other t-1 support labels can be chosen in \binom{d-1}{t-1} ways. Its positive multiplicities, with total r, can be chosen in \binom{r-1}{t-1} ways. Consequently the number of such multisets is at most

Their contribution to (3.10) is therefore at most C_r^2 A_r/d. Choose d_0(N) to be an integer at least all the d_r for 1\leq r\leq N and all the numbers C_r^2A_r/15 for 2\leq r\leq N. The latter condition is empty when N=1. The sum is then at most 49+15=64, uniformly in a and simultaneously for the finitely many lengths r\leq N.□

Proof. We construct the free unitary in two stages, then control its representatives.

Ultraproduct facts and the independence input. We first record the two facts about tracial ultraproducts used in the construction. A tracial ultraproduct of finite factors is a factor. Indeed, in any finite factor F the scalar \tau(x)1 belongs to the 2-norm closed convex hull of \{uxu^*:u\in\U(F)\}. To see this, take the unique element of least 2-norm in that closed convex hull. It is fixed by every unitary conjugation. It still belongs to F, because the convex hull is uniformly bounded in operator norm and the same is true of its 2-norm closure. It is therefore scalar, and its trace identifies it as \tau(x)1. In particular,

If (x_j)_\omega is central in an ultraproduct, the supremum on the right tends to zero along \omega: otherwise choosing an almost maximizing unitary in each offending coordinate would give a unitary sequence not commuting with (x_j)_\omega. Thus every central element is scalar. The ultraproduct of matrix factors whose sizes tend to infinity is diffuse, since their projections have traces approaching any prescribed number in [0,1]. An ultrapower of a \II_1 factor is likewise diffuse, since it contains the constant factor. These ultraproducts are therefore \II_1 factors.

Second, every unitary in such an ultraproduct lifts to coordinate unitaries. If (x_j)_\omega is unitary, then \lim_\omega\norm{x_j^*x_j-1}_2=0. Extend the partial isometry in the polar decomposition of x_j to a unitary u_j. This extension exists because the two complementary support projections have the same trace in the finite factor. Functional calculus gives

so (u_j)_\omega=(x_j)_\omega.

We use the following form of Popa [36, Theorem 0.1(b)]. Let \mathcal F=\prod_\omega F_j, where the F_j are finite factors with \dim F_j\longrightarrow\infty, allowing infinite dimensions. For a separable amenable von Neumann subalgebra B_1\subseteq\mathcal F, put Q=B_1'\cap\mathcal F and D=Q'\cap\mathcal F. If \mathcal X\subseteq\ker E_D is a separable linear subspace, there is a diffuse abelian von Neumann algebra A\subseteq Q such that

whenever r\geq1, x_0\in\mathcal X\cup\{1\}, x_1,\ldots,x_r\in\mathcal X, and a_1,\ldots,a_r\in A\cap\ker\tau. Here separability is for the 2-norm, as specified in [36, Section 1.1]. We use the displayed form with its final x_r and establish the additional endpoints needed below.

In our applications D is a full matrix algebra, A\subseteq D'\cap\mathcal F, and \mathcal X is selfadjoint. For a\in A, bimodularity implies E_D(a)\in Z(D), whence E_D(a)=\tau(a)1. Taking adjoints in (3.12) also treats alternating words that start with an \mathcal X-letter and end with an A-letter. For a word with both endpoints in A, write

with all displayed letters centered in their respective senses. For b\in D, traciality and [A,D]=0 give

Split a_ra_0 into its trace and a trace-zero element of A. The latter term vanishes by (3.12) with x_0=1; the scalar term vanishes by the same formula when r\geq2, and by E_D(x_1)=0 when r=1. Thus E_D(W)=0. A single centered A-letter has already been covered. Hence all alternating endpoint patterns used below are valid.

A finite-matrix model. We first construct a matrix-algebra model that is free from B=M_n(\C)\otimes1. In

apply Popa’s theorem with B_1=\C1, so that Q=\mathcal F_0 and D=\C1, and with \mathcal X=B\cap\ker\tau. The resulting diffuse abelian algebra is scalar free from B by the endpoint argument. It contains a Haar unitary \widetilde s_0: successively bisect projections in this diffuse abelian algebra to obtain a selfadjoint element uniformly distributed on [0,1], and exponentiate it by t\mapsto e^{2\pi it}. Choose unitary lifts \widetilde s_{0,j}\in M_n(\C)\otimes M_j(\C).

For each j, choose a unital matrix subalgebra R_j^0\cong M_j(\C) of M. Such a subalgebra is obtained by subdividing 1 into j projections of trace 1/j and choosing matrix units using equivalence of equal-trace projections in a finite factor. The coordinate embeddings M_n(\C)\otimes M_j(\C)\hookrightarrow M_n(M) preserve the normalized traces and fix B. They induce a trace-preserving embedding of \mathcal F_0 into

Thus the image s_0 of \widetilde s_0 is Haar and scalar free from B, and it has representatives s_{0,j}\in\U(M_n(R_j^0)).

Freeness from the constant algebra. Next we move this model into free position relative to the whole constant algebra P. The equality P^\omega=M_n(M^\omega) follows entry by entry from \norm{(x_{ab})}_2^2=n^{-1}\sum_{a,b}\norm{x_{ab}}_2^2. Since M^\omega is a factor,

Put T=W^*(P,s_0)\subseteq\mathcal P and \mathcal X=T\cap\ker E_B. This is a 2-norm separable selfadjoint linear subspace: the separably acting algebra P is countably generated as a von Neumann algebra, and adjoining one unitary preserves this property. Equivalently, bounded polynomial approximations in a countable generating set give 2-norm density. Apply Popa’s theorem with B_1=B and this \mathcal X. Choose a Haar unitary w in the resulting abelian algebra A\subseteq I_n\otimes M^\omega.

Let P_1=W^*(B,s_0) and P_2=wP_1w^*. Both contain B, since w commutes with B. Trace testing against B shows that

We claim that P_2 and P are free over B. In an alternating product of B-centered elements from these algebras, write each P_2-letter as wyw^* with y\in P_1\cap\ker E_B. The resulting word alternates letters w,w^* from A\cap\ker\tau with letters in \mathcal X. It has zero E_B by (3.12) and the endpoint extensions proved above. A product consisting of one letter has zero expectation by its assumed centering. This proves the claim.

Set s=ws_0w^* and C_s=W^*(s). It is Haar, and C_s is scalar free from B, because conjugation by w fixes B and preserves the trace. To deduce scalar freeness from P, first observe that

for m\geq1, c_i\in C_s\cap\ker\tau, and b_i\in B\cap\ker\tau. Indeed, test the trace against an arbitrary b\in B and split b into its scalar and centered parts; scalar freeness makes both resulting traces zero. By bimodularity, arbitrary B-letters may also be attached at the two endpoints in (3.13).

Single letters have trace zero by centering. Now take any mixed alternating scalar-centered word from C_s and P, and split each P-letter x as

The first summand is B-centered, and the second lies in B\cap\ker\tau. In every expanded term, group maximal blocks of C_s-letters and the selected B-letters. Each such block lies in P_2 and is B-centered by (3.13), including any B-letters at the endpoints. These blocks alternate with B-centered P-letters, so freeness over B makes their product have zero expectation. If there is no remaining B-centered P-letter, the term is one block covered by (3.13). Taking traces proves that C_s and P are scalar free.

Controlled representatives. Finally write w=I_n\otimes z with z\in\U(M^\omega), and choose unitary lifts z_j\in\U(M). Then

have all the asserted properties. For the commutator estimate, average h over the compact unitary group of R_j:

This operator commutes with R_j, and \norm{h-h^{(j)}}\leq\delta. Therefore I_n\otimes h^{(j)} commutes with s_j, and

□

Proof. Put v=sU. It suffices first to show

for r\geq1, nonzero integers k_i, and x_i\in P\cap\ker\tau. These cyclic mixed words suffice: if an alternating word starts and ends in the same algebra, cyclically bring those two letters together, split their product into its scalar and centered parts, and induct on word length. Regard an expanded trace word as a cyclic sequence of letters s and s^* separated by elements of P. The expansion of (sU)^k uses (sU)^k itself for k>0 and (U^*s^*)^{-k} for k<0. At a transition from an s-letter to an s^*-letter the separator is Ux_iU^*; at a transition from s^* to s it is x_i. Thus every separator between opposite signs is centered, including the cyclic junction.

Split every other separator into its scalar and centered parts. In each resulting term, removal of a scalar separator merges only powers of s with the same sign. Consequently every surviving power has nonzero exponent, and every surviving separator is a centered element of P. If any separator remains, freeness of s from P gives trace zero. If no separator remains, all signs were the same and the term is a scalar multiple of a nonzero power of s, again with trace zero. The argument for \tau(v^k) is identical, with all signs the same from the outset. This proves the displayed identities.

Centered Laurent polynomials in v are 2-norm dense, with bounded approximants, in the centered part of W^*(v): use the Fejér sums of the corresponding bounded functions on the circle. The displayed identities therefore imply scalar freeness of W^*(v) and P. Here bounded 2-norm approximation can be passed through each trace word by telescoping and the inequality \abs{\tau(a y b)}\leq\norm{a}\norm{y}_2\norm{b}. Finally apply the proved assertion to s^* and U^* and take adjoints to obtain the assertion for Us.□

Proof. Choose a symmetric probability measure \mu on G with \mu(g)>0 for every g. Let X=G^{\N\cup\{0\}} with product probability \mathbb P=\mu^{\N\cup\{0\}}. On X define

These maps commute. The first-coordinate laws of their pushforwards are respectively g\mu and \mu*\mu, while all later coordinates remain independent with law \mu. Full support therefore makes both pushforwards equivalent to \mathbb P.

Let \mathcal I be the completed sigma algebra of sets E satisfying S^{-1}E=E modulo null sets. It is countably generated modulo null sets. Indeed the measure algebra of X is separable for the metric \mathbb P(E\mathbin\triangle F), so its subspace \mathcal I is separable. Choose a countable dense family E_j in that subspace. For each E\in\mathcal I, a sequence from this family with summable symmetric-difference errors converges almost everywhere in indicators to 1_E. Thus the E_j generate \mathcal I after completion. The G-action preserves \mathcal I, because it commutes with S. Include all translates gE_j and set

Equip D with its Borel probability measure and, when taking measurable functions, its completion. The coordinate-permutation action (h x)_{j,g}=x_{j,h^{-1}g} satisfies b(h\omega)=h b(\omega). It is nonsingular, since the action on X is nonsingular. We have b\circ S=b almost everywhere. Pullback by b identifies L^\infty(D) with the S-fixed subalgebra of L^\infty(X): one inclusion follows from b\circ S=b, and the other from the generation of \mathcal I. This description also applies to scalar L^2 functions. Countably many representative identities can always be made simultaneous by deleting a null set.

For F\in L^\infty(G\times D) put

The map J is well defined on equivalence classes: there are only countably many group coordinates, and each section has a \nu-null exceptional set. These maps are positive unital contractions. Take a simultaneous pointwise weak-star cluster subnet of the maps T_n, using the product of the weak-star compact balls \prod_F\{a\in L^\infty(X):\norm{a}\le\norm{F}\}. Denote its limit by T; it is linear, positive, unital and contractive.

Pullback by a nonsingular measurable map is weak-star continuous on L^\infty. To see this directly, for q\in L^1 the measure E\mapsto\int_{S^{-1}E}q\,d\mathbb P is absolutely continuous with respect to \mathbb P; its Radon–Nikodym density is the required preadjoint applied to q. Its L^1 norm is at most \norm{q}_1. Consequently the telescoping bound

implies (TF)\circ S=TF. Write TF=(mF)\circ b. Moreover J\lambda_g F=(JF)\circ g^{-1} and the group action commutes with S. Weak-star continuity of its nonsingular pullback gives the first identity in (4.1). The identity b\circ S=b gives T_n(fF)=(f\circ b)T_nF; multiplication by a fixed bounded function is weak-star continuous, proving the second identity.

For the product mean use X^2, its two commuting shifts S_1,S_2, and rectangular averages

A simultaneous weak-star cluster is fixed by both shifts. These common fixed functions are exactly the functions of (b(\omega),b(\omega')). For completeness, a fixed scalar L^2(X^2) function has almost every section in the closed subspace L^2(\mathcal I)\subset L^2(X) in each variable, by Fubini and the preceding identification. The two orthogonal projections onto L^2(\mathcal I)\otimes L^2(X) and L^2(X)\otimes L^2(\mathcal I) commute; their product has range L^2(\mathcal I)\otimes L^2(\mathcal I). This proves the assertion. The preceding equivariance and module arguments now apply in both coordinates.

Here are details of the operator-field extension; they also explain why no normality of m is needed. On a countable rational complex vector subspace dense in K, define the coefficients of m(F)(x) by applying the scalar m to \ip{F(t,x)\xi}{\eta}. Outside a single null set these coefficients are sesquilinear and bounded by \norm{F}\norm{\xi}\norm{\eta}. They therefore define a bounded operator, and extension by density supplies all coefficients. Positivity of the scalar mean implies that this extension preserves adjoints; the norm bound gives contractivity. The coefficient identities also prove equivariance.

The scalar module identity implies the same coefficient formula when the test vectors depend measurably on x. First take countably valued bounded vector fields: on each member of the countable joint partition, localize using the scalar module identity and use the fixed-vector formula. Then approximate arbitrary bounded measurable vector fields uniformly by countably valued ones, using separability of K and the norm bound. Testing BFC on fixed vectors with the variable vectors C(x)\xi and B(x)^*\eta proves (4.2). In particular, if F takes values in a von Neumann algebra A\subset\B(K), its mean commutes with each constant operator in A'. Choose a countable strongly dense subset of the unit ball of A' to make this simultaneous almost everywhere; then m(F)(x)\in A''=A. Every step applies also to m^{(2)}.□

Proof. We give a quadratic improvement and then iterate it. For the current family define

The defect inequality implies \norm{H_g-V_g}_\infty\le e, hence \norm{A_g-V_g}_\infty\le e and \norm{H_g-A_g}_\infty\le2e. All H_g are unitary and all A_g are contractions. There is the exact factorization

Using the left and right module identities to cancel the linear terms in the product gives

Both right sides have norm at most 4e^2. The same calculation gives \norm{I-A_gA_g^*}\le4e^2. Since A_g is within e<1 of a unitary, it is invertible in L^\infty(D,A). Its unitary polar part V'_g=A_g(A_g^*A_g)^{-1/2} satisfies

Functional calculus preserves measurability: here it can be obtained by uniform polynomial approximation on a fixed compact interval away from zero. The new defect is at most 16e^2, since changing the three terms in A_g\alpha_g(A_h)-A_{gh} to their polar parts costs at most 12e^2. Also \sup_g\norm{V'_g-V_g}_\infty\le e+4e^2.

Starting with e_0=e, the successive defects satisfy e_{j+1}\le16e_j^2\le e_j/4. Thus the families converge uniformly in g and in essential operator norm to unitary fields W_g. Their defects tend to zero, and the total change is bounded by

If e=0, the original family already works. Countability of G and nonsingularity permit deletion of all exceptional sets and their translates. Finally the cocycle identity with g=h=1 makes the unitary W_1 idempotent, so W_1=I.□

Proof. Set B=m_t(\Sigma_tU_t^*). Then \norm{B-I}_\infty\le\delta, so B is invertible. Equivariance, the module identities and the cocycle identity give

Thus BU_s=\Sigma_s\alpha_s(B). Taking adjoints and products shows B^*B=U_s\alpha_s(B^*B)U_s^*, so the same identity holds after replacing B by its unitary polar part L. The inequalities (1-\delta)I\le|B|\le(1+\delta)I give \norm{L-B}_\infty\le\delta, proving the stated bound.□

Proof. For almost every (x,y) let C_{x,y} be the closed convex hull in L^2(A,\tau) of the countable set

It has a unique vector c(x,y) of least Hilbert norm. To prove measurability explicitly, enumerate its rational convex combinations as z_j(x,y) and put d(x,y)=\inf_j\norm{z_j(x,y)}_2. These are measurable functions: bounded weakly measurable operator fields are strongly measurable as vectors in the separable trace Hilbert space. For each n, choose the first j with \norm{z_j(x,y)}_2^2<d(x,y)^2+2^{-n}. The choices are measurable. For two such choices, the parallelogram identity and the lower bound d on the norm of their midpoint give

They converge pointwise in L^2 to the measurable field c.

Every z_j is an operator contraction with \norm{z_j-I}\le r. These inequalities persist under bounded weak-operator limits, and weak-operator compactness in the trace representation identifies any such limit with the L^2 limit by its value on the trace vector. Consequently c(x,y)\in A, \norm{c(x,y)}\le1 and \norm{c(x,y)-I}\le r. On bounded subsets, trace-Hilbert measurability implies weak operator measurability in the trace representation: test first on the dense vectors p,q\in A\subset L^2(A,\tau), where the coefficients are \tau(q^*cp), and then use the uniform operator bound. Thus c is also a measurable bounded operator field.

The linear isometry

from the trace Hilbert space to itself sends C_{g^{-1}x,g^{-1}y} onto C_{x,y}. Indeed the cocycle identity sends its generator with index t to the generator with index gt. Uniqueness of the least-norm vector gives the asserted equivariance for c. Since \norm{c-I}\le r<1, its unitary polar part a is measurable, satisfies \norm{a-I}\le2r, and has the same transformation law: polar decomposition respects multiplication on either side by unitaries. Finally \norm{W_g(x)W_g(y)^*-I}\le2\delta follows by comparing both W_g(x) and W_g(y) with u_g.□

Proof. Use bilateral independent coordinates (\omega_j)_{j\in\Z} with law \mu, and let \theta be the left shift, so (\theta\omega)_j=\omega_{j+1}. Define

Symmetry of \mu makes b_+ and b_- independent with common law \nu. The identities b\circ S=b and equivariance give

For the first identity apply b\circ S=b to the forward sequence; for the second, apply it to (\omega_0^{-1},\omega_{-1}^{-1},\ldots) and then use equivariance. All these identities hold almost everywhere, including after every integer shift. Consequently

Let f_m be the Hilbert-valued conditional expectation of f on coordinates [-m,m]. Then \eps_m=\norm{f-f_m}_2\to0 by density of finite-coordinate functions in the product L^2 space. Fix n>2m+1 and put

Thus A, P and B depend respectively on the disjoint coordinate sets [-m,m], [m+1,n-m-1] and [n-m,n+m]; they are independent. Equation (4.6) and unitarity yield

Set a_0=\E A and b_0=\E B. Independence and orthogonality of centered Hilbert-valued random variables give

For example, condition first on P: the centered terms P(A-a_0) and B-b_0 have zero mean, are independent, and their squared norms are unchanged by P. It follows that

Undoing the first transport shows \norm{f-Q^{-1}a_0}_2\le3\eps_m. This approximation depends only on coordinates [0,m]. Undoing the other transport shows \norm{f\circ\theta^n-Rb_0}_2\le3\eps_m. After shifting back by n, the approximating function depends only on coordinates [-m,-1]. Hence f belongs both to the closed L^2 subspace of functions of the nonnegative coordinates and to that of functions of the negative coordinates. The two sigma algebras are independent, so their common subspace consists of constants: if f is measurable with respect to the second one, its conditional expectation on the first is its constant expectation, and this conditional expectation also equals f. Write f=v almost everywhere. The joint law of (b_+,b_-) is \nu\times\nu, so f_0=v almost everywhere. Its equivariance and nonsingularity imply \sigma(g)v=v for every g.□

Proof. Choose t with d(M,N)<t<1/10. If v\in N were a proper isometry, choose x\in M_1 with \norm{x-v}<t. Then x^*x\ge(1-t)^2I, so the polar part w of x is an isometry in M. Finiteness makes w unitary. Since 1-t\le |x|\le1, \norm{w-v}\le\norm{w-x}+\norm{x-v}<2t<1. This is impossible: a proper isometry within distance less than one of a unitary would be invertible. Thus N is finite.

Suppose that q\in Z(N) is a projection. Choose a\in M_1 with \norm{a-q}<t. For v\in\U(M) choose b\in N_1 with \norm{v-b}<t. Since q commutes with b,

The trace-Hilbert closed convex hull of the conjugates vav^* contains \tau(a)I. Indeed, its unique vector of least norm is conjugacy invariant, is bounded by one by weak compactness of the operator ball, and is therefore scalar; its trace is \tau(a). Consequently \norm{q-\tau(a)I}\le3t<1/2. A nontrivial projection has distance 1/2 from the scalars, so q is zero or one. The spectral projections of every central selfadjoint element are therefore trivial, proving Z(N)=\C I.□

Proof. We first make a single selfadjoint element nearly scalar under a matrix average. Let P be a \II_1 factor, y=y^*\in P, and \eps>0. Choose a spectral step approximation y_0=\sum_{i=1}^l\lambda_i p_i, where \sum_i p_i=I. For a sufficiently large integer b, split \lfloor b\tau(p_i)\rfloor orthogonal projections of trace 1/b from each p_i. The remaining projection has trace

and also splits into projections of trace 1/b. The resulting b equivalent projections extend to matrix units for a unital F\cong M_b(\C) in P. Let y_1\in F be diagonal, equal to \lambda_i on the pieces taken from p_i and zero on the remainder. Then

Haar averaging over \U(F) sends y_1 to \tau(y_1)I. Contractivity of this average in L^2 gives

by first choosing y_0 and then b\ge2.

Choose an L^2-dense sequence (x_j) in the selfadjoint unit ball of M, and enumerate the requirements (j,k)\in\N^2. Start with Q_0=\C I. Given a full matrix algebra Q_n, its relative commutant P_n=Q_n'\cap M is a \II_1 factor: matrix units identify M with a matrix algebra over one of its corners and P_n with that corner. For the next requirement (j,k) apply (5.1) in P_n to E_{P_n}(x_j), with error 1/k. If F\subseteq P_n is the resulting matrix algebra, set Q_{n+1}=Q_n\vee F. These commuting matrix algebras generate a full matrix algebra of at least twice the size of Q_n, and

Hence \norm{E_{P_{n+1}}(x_j)-\tau(x_j)I}_2<1/k, and all earlier bounds persist under subsequent expectations. For R=(\bigcup_nQ_n)'' the expectation onto R'\cap M=\bigcap_nP_n thus sends every x_j to \tau(x_j)I. Density gives R'\cap M=\C I. In particular R is a factor; it is finite and has unital matrix subalgebras of unbounded size, so it is of type \II_1. Its increasing matrix presentation makes it hyperfinite.□

Proof. The commutant of (M\cup\{e_R\})'' is J_MRJ_M. Indeed, an element of M' is right multiplication by an element of M. If it commutes with e_R, applying it to \Omega shows that its right multiplier lies in L^2(R)\cap M=R, where the equality follows from the trace expectation. Conversely right multiplication by R reduces L^2(R). The same argument applies to N. Equality of the basic constructions therefore gives J_MRJ_M=J_NRJ_N. For each r\in R the two displayed operators take \Omega to r^*\Omega. The vector \Omega is separating for the common right algebra, since it is cyclic for M, so the operators are equal. Applying these identities twice proves the last assertion.□

Proof. Denote the supremum by K. Start with bounded a,b\in M with \norm{a}_2,\norm{b}_2\le1, without an operator-norm bound. By (5.2), finite convex averages of conjugations by unitaries of R can bring both aa^* and bb^* within one of their scalar traces in L^2. One first takes a sufficiently high Haar average and then approximates its integral by a finite sum; rational weights and repetitions give equal weights. Choose such a family (d_i). Independently choose a family (e_l) so that the averages of e_l^*a^*ae_l and e_l^*b^*be_l have the same property. Enumerate the Cartesian product of the two families by k=1,\ldots,m and put a_k=d_i a e_l, b_k=d_i b e_l. Their covariance averages satisfy

and the same bounds hold for L_b,Q_b.

Use the same independent fair signs in the two sums

Bimodularity and sign orthogonality give

Put \norm{x}_4^4=\tau((x^*x)^2). The three pairings in the fourth moment give exactly

Tracial Cauchy–Schwarz gives \abs{\tau((a_k^*a_l)^2)}\le\tau(a_ka_k^*a_la_l^*). Dropping the last nonpositive term therefore yields

If A=w|A|, set A_s=w|A|1_{[0,s]}(|A|) and define B_s similarly. These operators have norm at most s, their L^2 norms do not increase, and

Substituting them in (5.4) and applying Cauchy–Schwarz in probability gives

Take s=4\sqrt{12}, and then take the supremum over the dense bounded L^2 unit vectors a,b. We obtain \norm{V}\le192K+\norm{V}/2, proving the assertion.□

Proof. Set T=J_NJ_M-I. The common right-action identity implies that T commutes with both actions of R. For x\in M_1 choose y\in N_1 with \norm{x-y}<b. Then

Thus the supremum in Lemma 5.6 is at most 2b. That lemma gives \norm{T}\le768b, and \norm{T}=\norm{J_M-J_N}.□

Proof. Write the map as \theta:\mathcal A\to N, where N is the target factor. Preservation of the Jordan product gives preservation of xyx, since xyx=2x\circ(x\circ y)-x^2\circ y, where x\circ y=(xy+yx)/2. Polarizing in x gives preservation of xyz+zyx. Fix a,b, and set

Compute \theta(ab\,c\,ba+ba\,c\,ab) in two ways. The polarized triple identity gives DCE+ECD, where C=\theta(c); applying the unpolarized identity first to the inner and then to the outer products gives ABCBA+BACAB. Expanding their equality yields

where surjectivity allows every C in the target. Substitute C=C_1kC_2, and use the same identity with C_1 and C_2 to move h past the adjacent factors. The two resulting terms are equal, so

A factor is prime: xNy=\{0\} implies x=0 or y=0, since the central supports of the support projections of nonzero x,y are both one. If k\ne0, primeness first gives hC_1k=0 for every C_1, and then h=0. Thus, for each pair (a,b), either \theta(ab)=\theta(a)\theta(b) or \theta(ab)=\theta(b)\theta(a).

For fixed a, the sets of b satisfying these two respective identities are additive subgroups whose union is the whole additive group. A group cannot be the union of two proper subgroups: choose an element in each but not the other, and their sum belongs to neither. One identity therefore holds for all b. The same two-subgroup argument, now for those a for which each identity holds for every b, makes one orientation hold globally.□

Proof. Choose a countable strongly dense subgroup G\subseteq\U(M). For each g\in G choose a unitary v_g\in N with \norm{v_g-g}\le3b: a contraction approximant to g is invertible for b<1, and its polar part has this bound. Thus

Apply Lemma 4.2 to these constant fields on the boundary space (D,\nu) of Lemma 4.1. For sufficiently small absolute b it gives measurable unitary fields W_g:D\to\U(N) satisfying

where C_1 is absolute.

To obtain this correspondence on unitary vectors, we will compare the cocycle with the original left and right actions, and then compare its two boundary values within N. The first comparison uses the separate action on D^2. Form its product cocycle and the constant representation

The left and right N actions commute, so \Sigma is an exact cocycle for the separate G^2 action. By Lemma 5.7,

The product version of Lemma 4.3 gives a measurable unitary field L:D^2\to\U(K), with \norm{L-I}_\infty\le C_2b, such that

The constants C_1,C_2, and all subsequent C_i in this proof, are absolute.

The field L takes values in \U(K); its intertwining identity alone does not supply trace vectors of unitaries in N. For this we use the diagonal comparison. Equation (5.5) gives

For sufficiently small b, Lemma 4.4 therefore gives a measurable a:D^2\to\U(N) with \norm{a-I}_\infty\le C_3b and

We can now combine the two comparisons: the vector field f(x,y)=L(x,y)^*a(x,y)\Omega is equivariant for the diagonal representation g\mapsto U_{g,g}. Indeed right multiplication by W_g(y)^* on trace vectors is J_NW_g(y)J_N, so (5.7) and (5.6) give f(x,y)=U_{g,g}f(g^{-1}x,g^{-1}y). By Lemma 4.5, f is a constant invariant vector. The invariant vectors for g\mapsto gJ_MgJ_M are exactly \C\Omega. To see this, strong density extends invariance from G to \U(M); averaging over \U(Q_j) acts on x\Omega as E_j(x)\Omega and converges to the projection onto \C\Omega by (5.2). Hence

Discard one null set and all its countably many G^2 translates, so that the preceding identities hold at a chosen pair (x,y) and at all its translates. Define the Hilbert-space unitary

It fixes \Omega. Using (5.6) for (g,1) and (5.8) at (g^{-1}x,y) gives

where b_g\in\U(N) and \norm{b_g-g}\le C_4b uniformly in g.

Every u\in\U(M) is the strong limit of a sequence g_j\in G. The vectors b_{g_j}\Omega=S_0g_j\Omega are Cauchy in trace L^2(N). For unitaries, traciality gives \norm{b_{g_j}^*-b_{g_l}^*}_2=\norm{b_{g_j}-b_{g_l}}_2. Uniformly bounded L^2 convergence in the standard representation implies strong convergence: test first on the dense right-multiplier vectors N'\Omega and then use boundedness. Thus b_{g_j} and their adjoints converge strongly, and their limits are a unitary b_u\in N and its adjoint. We obtain

where the norm inequality follows by weak closedness of the appropriate operator ball.

Every element of M is a linear combination of unitaries, so the rule

defines a linear map into N. It is well-defined because \Omega is separating for N. Writing a contraction as its real and imaginary selfadjoint parts, each an average of two unitaries, gives

In particular \vartheta is bounded, and it sends every unitary to a unitary. For z=z^* expand \vartheta(e^{itz})^*\vartheta(e^{itz})=I in operator norm. The coefficient of t says \vartheta(z)^*=\vartheta(z). Applying this also to z^2 and comparing the coefficient of t^2 gives \vartheta(z^2)=\vartheta(z)^2. Complex linearity and polarization show that \vartheta preserves adjoints and the Jordan product.

When 2C_4b<1, (5.11) makes its range closed and makes \vartheta injective. Every contraction of N has distance at most (1+2C_4)b from this range, by the original gap and (5.11). For (1+2C_4)b<1 the range must be all of N: otherwise the quotient map by this proper closed linear subspace has norm one, contrary to that bound.

Lemma 5.8 shows that this surjective Jordan map is either multiplicative or antimultiplicative. The latter possibility is excluded by (5.11). Indeed, for unitaries z,w\in M antimultiplicativity would give

A unital copy of M_2(\C) in M contains anticommuting selfadjoint unitaries, whose commutator has norm two. Taking 6C_4b<2 excludes the anti case. Thus \vartheta is a unital star-isomorphism, and it is normal because a bijective order isomorphism of von Neumann algebras preserves all bounded increasing suprema. This proves the proposition, with C_{\mathrm{st}}=2C_4 and one absolute sufficiently small b_0.□

Proof of Theorem 5.1. Choose t> d(M,N) below the absolute thresholds specified below. By Lemma 5.2, N is a finite factor. If either factor is a matrix algebra, Theorem 2.1 for the mutual near inclusions gives the required near-identity conjugacy, with an absolute linear bound. We may therefore suppose both factors are of type \II_1.

Choose R\subseteq M from Lemma 5.3. By Theorem 2.1, there is a unitary w with \norm{w-I}\le150t and wRw^*\subseteq N. Replacing N by N_0=w^*Nw makes R common, and

Take g=302t and apply Proposition 5.4, provided g<1.74\cdot10^{-13}. Write M_s=\pi(M), N_s=\rho(N_0). Their common subfactor remains irreducible in M_s by faithfulness. By Lemma 5.5 its right actions agree, and d(M_s,N_s)\le2\beta, with \beta<50948\sqrt g. Choose a strict standard-gap bound b=3\cdot50948\sqrt g. If b<b_0, Proposition 5.9 supplies a normal star-isomorphism \vartheta_s:M_s\to N_s with \norm{\vartheta_s-\id}\le C_{\mathrm{st}}b on M_s.

Transport it to the original representation:

For x\in M_1 choose y\in(N_0)_1 with \norm{x-y}<g. The original-contraction comparison in Proposition 5.4 and faithfulness of \rho give

Let e(t)=C_{\mathrm{st}}b+50948\sqrt{302t}+604t. This is an absolute function tending to zero.

On the original H\oplus H, consider the faithful normal paired representation x\mapsto x\oplus\theta(x) of M, and the selfadjoint switch s(\xi,\eta)=(\eta,\xi). Its scalar commutator norm on that factor is at most e(t). Theorem 3.1 therefore gives

At each matrix level the switch commutator has off-diagonal block [x_{ij}-\theta(x_{ij})], so equivalently \norm{\theta-\id}_{\cb}\le C_{\mathrm{amp}}e(t). For sufficiently small absolute t, Lemma 2.3 gives a unitary v\in\B(H) with vMv^*=N_0 and \norm{v-I}\le C_{\mathrm{pair}}C_{\mathrm{amp}}e(t), for an absolute constant C_{\mathrm{pair}}. Then u=wv conjugates M onto N, and

All thresholds used above are absolute. Besides t<1/10, they are the bounds in Theorem 2.1 and Lemma 2.3 and the two bounds in Propositions 5.4 and 5.9. Choose t_{\mathrm{fin}} so that they hold for every 0<t<t_{\mathrm{fin}}, and take \eta_{\mathrm{fin}} to be the maximum of the displayed bound and the matrix-factor bound. These bounds are nondecreasing functions of t and tend to zero. This establishes the claimed uniformity on the original Hilbert space.□

Proof. For each n\geq1 choose isometries s_1,\ldots,s_n in that common copy such that s_i^*s_j=\delta_{ij}1 and \sum_i s_is_i^*=1. The operator V:H^n\to H given by V(\xi_i)=\sum_i s_i\xi_i is unitary. For either P=M or P=N,

Indeed the inverse coefficient formula is x_{ij}=s_i^*xs_j. The same unitary therefore identifies the two amplified unit balls with the two original unit balls. Taking the supremum over n proves the assertion.□

Proof. Swapping two sites of the same type is implemented by a unitary in a finite site algebra. This unitary preserves the product state, so it belongs to D_\rho. Suppose x\in D_\rho'\cap D. Given a finite site element a, choose bounded finite site approximants x_j with \norm{(x-x_j)\Omega_\rho}\to0. A product of the indicated swaps moves a to an element a_j supported away from x_j. Since x commutes with the implementing unitary and the state is invariant,

State Cauchy–Schwarz bounds the error in the first product by \rho(aa^*)^{1/2}\norm{(x-x_j)\Omega_\rho}; the other error tends to zero as well. Hence \rho(ax)=\rho(a)\rho(x) for every finite site a. Density gives (x-\rho(x)1)\Omega_\rho=0, and faithfulness gives x=\rho(x)1. In particular D is a factor.

Choose a faithful normal state \psi on A and put P=A\barotimes D with state \psi\otimes\rho. In the regular representation of its modular crossed product on L^2(\R,H_P), the generators are

Here H_P can be the standard space of P. Let Z be central in the crossed product. After Fourier transform in t, commutation with every \lambda_s makes Z a decomposable operator. The crossed product also commutes with the constant representation of P'. Consequently its decomposable field takes values in P. Commutation with the constant operators 1\otimes D_\rho puts this field in A\otimes1: normal slices and D_\rho'\cap D=\C1 prove this relative-commutant identity.

An off-diagonal matrix unit at a \nu-site is a modular eigenoperator with frequency \log\nu or -\log\nu. Its regular representation, after Fourier transform, is that matrix unit followed by a frequency translation. If a(\zeta)\otimes1 denotes the field of Z, commutation therefore gives

The generated subgroup is countable and dense in \R. Test the field on countably many vector coefficients. Translation continuity in local L^1 extends its invariance to every real translation, and convolution with compactly supported approximate identities shows that each coefficient is constant. Thus Z=a_0\otimes1 is a constant field. Commutation with \pi(A\otimes1), using strong continuity of the modular action and a countable strongly dense set, gives a_0\in A'\cap A=\C1. The core is a factor, so the continuous-core criterion makes P a type \III_1 factor.

Finally, adjoining another countable family of \lambda-sites and reindexing that family gives the asserted absorption. All these tensor products preserve the displayed product states. The usual identification of the two-ratio product with the hyperfinite \III_1 factor is also recorded in [22, Example 1.6, p. 103]; the argument above establishes the additional assertion for every A.□

Proof. A countably decomposable infinite factor is properly infinite and contains a unital copy of \B(\ell^2): split its identity into countably many equivalent projections and choose corresponding matrix units [4, III.1.3.5 and III.1.5.1, pp. 243, 247]. Theorem 2.1 embeds this amenable subfactor of M_0 into N_0 by a unitary of norm distance at most 150t from 1. Conjugating N_0 in the inverse direction gives u_0 for which the subfactor is common. This also proves that N_0 is infinite. The adjusted ordinary gap is at most 301t. Lemma 6.1 makes this also a cb gap. Lemma 2.5 transfers it to the tensor pair because D' is hyperfinite. By Lemma 6.2, both tensor factors are of type \III_1 and absorb R_\lambda.

For the moment write N_1=(u_0N_0u_0^*)\barotimes D. The bicentralizer and expected-subfactor results just stated give an irreducible hyperfinite \II_1 factor R\subseteq M and a faithful normal state \varphi on M centralizing R. The given representation of M has a cyclic separating vector realizing \varphi. Here is the representation point needed for this assertion. A normal representation on a separable space splits as a countable sum of cyclic normal representations. Each cyclic summand embeds in the standard representation by the standard-form vector implementing its vector functional [19, Lemma 2.10(1), p. 278]. A countable amplification of the standard representation is unitarily equivalent to it, since its type \III commutant contains a countable family of isometries with ranges summing to 1. Finally, every nonzero projection in that countably decomposable type \III commutant is equivalent to 1 [4, Theorem III.1.7.9]. Thus every nonzero normal representation here is equivalent to the standard one. Transport the standard vector for \varphi to obtain the desired \Omega on the actual space H\otimes K. No conjugation of either algebra is made in this choice.

Put S=J_MRJ_M\subseteq M'. Centralization gives

First apply Theorem 2.1 to R\subseteq M and its near inclusion in N_1. It gives a unitary v_1 with v_1Rv_1^*\subseteq N_1 and \norm{v_1-1}=O(t). Put N_2=v_1^*N_1v_1, so R\subseteq M\cap N_2. Cb gaps change by at most twice the norm distance of the adjusting unitary from 1. Thus Lemma 2.2 gives d(M',N_2')=O(t). Apply Theorem 2.1 to S\subseteq M' and its near inclusion in N_2'. It gives a unitary v_2 with

Set N=v_2^*N_2v_2. Since v_2 commutes with R, this algebra still contains R, and now S\subseteq N'. The algebra M, its state \varphi, and the vector \Omega have remained fixed. The unitary in the statement is v=v_2^*v_1^*, with \norm{v-1}=O(t), and the resulting cb gap is O(t).

We next check irreducibility and the vector, rather than assume they survive an arbitrary small perturbation. Choose increasing matrix subfactors R_j strongly generating R. Haar averaging over \U(R_j) converges weakly on M to x\mapsto\varphi(x)1. Indeed every bounded cluster point commutes with all R_j, hence is scalar, and the averages preserve \varphi. If a nontrivial projection p\in R'\cap N existed, a close approximant x\in M would therefore give

The left side is 1/2. Taking t sufficiently small proves R'\cap N=\C1. The same argument in the commutants, with S and the vector state on M', proves S'\cap N'=\C1.

Let p be the projection onto \overline{N'\Omega}. It belongs to N. By (6.1), this subspace reduces R: for r\in R and x'\in N', the vector rx'\Omega=x'r\Omega lies in \overline{N'\Omega}, and the same holds for r^*. Thus p\in R'\cap N, so p=1. Similarly the projection onto \overline{N\Omega} belongs to S'\cap N' and is 1. Hence \Omega is cyclic and separating for N. The common commuting right action in (6.1) shows that its state on N is invariant under conjugation by \U(R), and hence centralizes R. The two operators J_MrJ_M,J_NrJ_N\in N' now agree on \Omega; since \Omega is separating for N', they agree everywhere. All conditions in Definition 6.3 are satisfied.□

Proof. For s\in R put R_s=J_Ms^*J_M=J_Ns^*J_N. On either graph, the operation g_c(x)\mapsto g_c(rxs) for r,s\in\U(R) is the restriction of the common unitary

Both graph subspaces reduce these unitaries. Thus T=(1-E_c^N)E_c^M commutes with them.

A bounded element retaining the graph signal. First take c=\sqrt{k}, k\in\N. If T\ne0, choose x\in M with

Such vectors are dense in the graph. Write

Choose independent Haar unitaries u_i,v_i\in\U(R_j) and independent signs \epsilon_i\in\{-1,1\}, and put

For fixed unitaries, the vectors Tg_c(x_i) all have the same norm. The sign second moment and the elementary Hilbert fourth-moment bound give

Consequently, with sign probability at least 1/12,

For completeness, this probability bound follows by applying Cauchy–Schwarz to the part of a nonnegative random variable above half its mean: \mathbb P(Z\geq\E Z/2)\geq(\E Z)^2/(4\E Z^2).

We need a simultaneous bound on the fourth moments of y. The Haar averages

converge weakly to \varphi(a)1 as in the preceding proof. For distinct i,l, centralizer cycling and Haar invariance give the exact identities

For the first identity, cycle v_i^* to the right and average over the Haar unitary v_iv_l^*. For the second, cycle v_i^* and v_i away and average over u_i^*u_l. Each formula contains just one weakly convergent average. The three sign-pairing terms in \varphi((y^*y)^2) have expected limits

For the third, state Cauchy–Schwarz gives \abs{\varphi(a^2)}\leq\varphi(aa^*)^{1/2}\varphi(a^*a)^{1/2}, and the two expected factors agree by interchanging i,l. The all-equal indices contribute m^{-1}\varphi((x^*x)^2). The same calculation for yy^* interchanges \alpha,\beta. For the fixed x,k, first choose m so that the weighted diagonal contribution satisfies

The limiting weighted off-diagonal contribution is at most \alpha^2+k\beta^2+2(k+1)\alpha\beta\leq6, because \alpha+k\beta=1 and k\geq1. Now choose j large enough that the weighted off-diagonal contribution is at most 7. Since \E\norm{g_c(y)}^2=1, these choices give

Markov’s inequality bounds the probability that the bracket in (6.4) exceeds 240 by 1/24. Together with (6.3), this gives a realization with both properties. Fix such a y, and clip its singular values at a fixed constant K_1:

Functional calculus on both sides of the polar decomposition gives

Choose K_1 once so that \sqrt{240}/K_1\leq1/(4\sqrt2). Writing a_0=1/(4\sqrt2), the signal loss satisfies

Thus the choice of K_1 is independent of the size of \norm{T}, and

A bounded row retaining the normalized signal. We have produced a bounded element retaining the signal, but a single bounded-element approximation would still cost \sqrt{k}\gamma on the second graph leg. To remove that factor, replicate z into the row

where the r_i are independent Haar unitaries in a sufficiently large R_j. For rows of graph vectors use the normalized Hilbert norm

Equivariance makes the normalized signal of this row at least a_0\norm{T} for every choice of the r_i. Put \alpha_z=\varphi(z^*z) and \beta_z=\varphi(zz^*). Haar averaging and centralizer cycling give

For example, the second off-diagonal term averages zz^* under r_i^*r_l between z^* and z. By (6.5), the limits are bounded by an absolute constant: use \varphi((zz^*)^2)\leq K_1^2\beta_z and \varphi((z^*z)^2)\leq K_1^2\alpha_z for the diagonals. Choose the stage and then a realization with

where C_1 is absolute.

Clip the whole row on the left:

so \norm{X}\leq K_2. If Q=1-f_{K_2}(A^{1/2}), then Q^2\leq A/K_2^2 and Y-X=QY. As c^2=k, its normalized squared graph error is

Choose the absolute K_2 so that \sqrt{C_1}/K_2\leq a_0/2. The same proportional error estimate for the row gives

Comparison at one matrix level and all rescalings. Cb closeness gives a row Z\in M_{1,k}(N) with \norm{X-Z}\leq K_2\gamma, with arbitrarily small slack if needed. For any row [e_i] of norm at most \eta,

Here the graph-vector formula itself makes sense for all operators on H, whether or not they lie in M or N. Because g_c(X_i)\in\operatorname{ran}E_c^M and g_c(Z_i)\in\operatorname{ran}E_c^N,

Together with (6.7), this bounds \norm{(1-E_c^N)E_c^M} by an absolute multiple of \gamma. If T=0 the same assertion is immediate. Interchanging M,N gives the reverse bound, hence the claimed projection bound for c=\sqrt{k}.

The antiunitary flip (\xi,\overline\eta)\mapsto(\eta,\overline\xi) takes g_c(x) to c g_{1/c}(x^*), so the bound also holds for c=1/\sqrt{k}. For c\geq1 choose k=\lfloor c^2\rfloor; then c/\sqrt{k}\in[1,\sqrt2]. Scaling the second leg by this ratio has condition number at most \sqrt2 and increases either one-sided subspace gap by at most that factor. Reciprocal scaling handles 0<c<1. Finally, for two orthogonal projections E,F,

This proves the uniform assertion.□

Proof. The upper-left block of the projection onto a closed operator graph is (1+A^*A)^{-1}. Applied to (6.2), this gives (1+c^2\Delta_P)^{-1}. With g(s)=(1+e^s)^{-1}, Proposition 6.5 consequently yields

Let p=1_{(-\infty,a-1/2]}(L_N) and q=1_{[a+1/2,\infty)}(L_M), and regard pq as an operator qH\to pH. Let A and B be the restrictions of g(L_N-a) and g(L_M-a) to those respective subspaces. Then

For D=A(pq)-(pq)B, integration of the derivative of e^{-tA}(pq)e^{tB} gives

The denominator is a fixed positive number. Reversing the two algebras proves the other estimate.□

Proof. Put d=\norm{\ad_h|_P}. Choose a unital copy S of \B(\ell^2) in P, with matrix units (e_{ij}) whose diagonal sum is I. The finite-dimensional unital algebras

increase and generate S. Average h over \U(S_n) and take a weak-operator cluster point h_0. Each average is self-adjoint, within d of h, and has scalar commutator bound at most d on P. These properties pass to the cluster point. Haar invariance at every earlier stage gives h_0\in S'.

For any k, choose isometries s_1,\ldots,s_k\in S with s_i^*s_j=\delta_{ij}I and \sum_i s_is_i^*=I. The unitary V:H^k\to H, V(\xi_i)=\sum_i s_i\xi_i, identifies M_k(P) with P and h_0^{(k)} with h_0. Consequently

By Lemma 2.2, \dist(h_0,P')\le d, so in fact \dist(h,P')\le2d. The stated larger constant will be convenient.□

Proof. Suppose there is no common K. By scaling a counterexample and subtracting a self-adjoint sum from P+P', we can choose factors P_j\subset\B(H_j) and self-adjoint h_j with

Taking real parts justifies using a self-adjoint approximating sum. Its subtraction does not change either distance in the assertion, because its commutator with P_j belongs to P_j.

Fix a free ultrafilter \omega on \N and form the norm ultraproducts

Here a norm ultraproduct is the algebra of bounded sequences modulo the sequences whose operator norms tend to zero along \omega. The algebra A is simple. Indeed, a nonzero positive element a has uniformly bounded positive representatives a_j and a number t>0 such that e_j=1_{[t,\infty)}(a_j) is nonzero on a set in \omega. In a separably acting type \III factor each nonzero projection is equivalent to I. Choose isometries v_j with v_jv_j^*=e_j on that set. Then v_j^*a_jv_j\ge tI there. The class v^*av is invertible in A and belongs to the ideal generated by a. Thus every nonzero closed ideal of A contains I.

Commutation with [h_j]\in E defines a bounded derivation D:A\to A. To see that its range is in A, choose, for every bounded sequence x_j\in P_j, elements y_j\in P_j with

The y_j are uniformly bounded, and their class is the commutator in E. Its definition is therefore independent of the choices; linearity and the derivation identity follow in E.

Sakai’s theorem for simple unital C^*-algebras gives a\in A with D(x)=[a,x] [40, Section 1, p. 259; Section 2, p. 260]. No separability of the norm ultraproduct is required. Since the h_j are self-adjoint, replacing a by (a+a^*)/2 leaves this identity unchanged. Choose uniformly bounded self-adjoint representatives a_j\in P_j. The equality of the two derivations implies

Otherwise contractions witnessing a fixed positive commutator norm could be selected at each coordinate of an ultrafilter-large set, contradicting equality on their class in A. Lemma 7.2 now gives

a contradiction.□

Proof. Write b(t)=\dist(v(t),\mathcal G_P). This is a continuous function, since distance to any nonempty subset is 1-Lipschitz. Suppose b=b(t) is smaller than a fixed absolute radius, and choose p\in\mathcal G_P with \norm{p^*v(t)-I}<b+\eta. For sufficiently small b+\eta, principal functional calculus writes p^*v(t)=e^{ih} with h=h^* and \norm{h}\le2(b+\eta). As p normalizes P, this conjugation has the same one-sided approximation bound e. The twice-integrated conjugation formula gives

It follows that

Apply Lemma 7.3 and approximate h by a self-adjoint sum a+b' with a\in P, b'\in P'. Since a and b' commute, e^{i(a+b')}=e^{ia}e^{ib'} belongs to \mathcal G_P. The inequality \norm{e^{ih}-e^{ik}}\le\norm{h-k} for self-adjoint h,k therefore shows, on letting both approximation slacks vanish, that

Choose r>0 within the radius and with 8Kr\le1/4. If Ke<r/2, the continuous function b, starting at zero, cannot first reach r: (7.2) would give r<3r/4. Thus b(t)<r throughout the interval between 0 and any chosen t\in I. Absorbing the quadratic term in (7.2) gives b(t)\le2Ke uniformly. This proves the assertion with suitably reduced e_0.□

Proof. Write w=(t-i)/(t+i) and give the unit circle normalized measure d\theta/(2\pi). The transformation

is unitary from L^2(\mathbb T,H) onto \mathcal K: the change of variables has d\theta/(2\pi)=dt/(\pi(1+t^2)). For n\geq0, J(w^n) is a linear combination of (t+i)^{-k}, 1\leq k\leq n+1, whose Fourier transforms are supported in [0,\infty). For n<0 its expression is a linear combination of (t-i)^{-k}, whose Fourier transforms are supported in (-\infty,0]. These support statements follow directly from

and its complex conjugate, interpreted under the L^2 Fourier transform. Since the images of all circle modes form an orthonormal basis, J identifies the nonnegative-mode subspace with P_0\mathcal K.

Conjugating the multiplier by J gives the circle multiplier \widetilde B(w)=B(t). Apply its invariant-subspace property to every constant vector. All negative Fourier coefficients of \widetilde B vanish. If b_n denotes its nth coefficient, defined by matrix coefficients, then

converges in operator norm locally uniformly and is operator-norm holomorphic. Its value at w=\rho e^{i\theta} is the circle Poisson integral of \widetilde B, since all its negative coefficients vanish. The Poisson kernel is positive and has integral one, so \norm{F(w)}\leq\norm{\widetilde B}_\infty. Changing variables in the circle Poisson integral gives exactly (7.3); explicitly, harmonic measure at t+ir pushes forward to r\,ds/(\pi((t-s)^2+r^2)).

For completeness, multiplication of these extensions can be checked without presupposing pointwise boundary limits. The radial Poisson integrals converge in L^2(\mathbb T,H) on each fixed vector, for both \widetilde B and \widetilde B^*, and likewise for \widetilde S and \widetilde S^*. Choose a sequence of radii tending to one along which these convergences hold pointwise for a countable dense family of vectors. Uniform boundedness extends the pointwise convergence to every vector. Along this sequence the products converge strongly-* almost everywhere to \widetilde B(w)x\widetilde S(w). Each Fourier coefficient of the bounded analytic product F(w)xG(w) therefore converges, by dominated convergence on matrix coefficients, to the corresponding coefficient of that boundary product. In particular its negative coefficients vanish, and its nonnegative coefficients are precisely the Taylor coefficients of F xG. This proves the asserted product identity. The same argument applies to products with a scalar bounded analytic function.□

Proof. The letters C below denote constants depending only on the universal constants in the stated inputs, and can increase between occurrences. All required smallness conditions are met by decreasing one universal \gamma_0 at the end.

The initial approximate normalizers. Both U_M(t) and U_N(t) normalize their respective algebras. For every matrix size k and x\in M_k(M)_1, first move U_M(t)^*xU_M(t) into M_k(N)_1, conjugate by U_N(t), and move back into M_k(M)_1. The two approximation errors are at most \gamma each, with arbitrarily small slack. Thus

Here T^{(k)}=I_{\C^k}\otimes T. Applying the same two approximations before conjugating by U_M(t) gives the corresponding assertion for W(t)^*xW(t). Consequently

The frequency nest and its signs. Use the notation \mathcal K,D,P_a introduced above. Let C_s denote multiplication by e^{ist}, and let \mathcal U_P be multiplication by U_P(t). Spectral calculus gives

The sum on the right is the selfadjoint operator obtained from the commuting spectral measures on the two tensor factors; these identities therefore include their domains. They can also be checked in the spectral representation of L_P, where \mathcal U_P is a scalar frequency translation on each fiber. Put \mathcal W=\mathcal U_N\mathcal U_M^*. Directly moving the spectral projections past these unitaries gives

After Fourier transformation the product in the middle has fiber

Its two thresholds have separation one, with midpoint a-\tfrac12-\xi. The cut estimate of Corollary 6.6 bounds its norm by C\gamma, uniformly in a and \xi. The same argument with M,N interchanged applies to C_1\mathcal W^*. In particular

Adjoin 0,I to the complete nest \{P_a:a\in\R\}. Arveson’s distance formula for its nest algebra [2, Theorem 1.1, p. 212] says that the distance of an operator T to the operators preserving every P_a\mathcal K equals \sup_a\norm{(1-P_a)TP_a}. Choose preserving operators T_B,T_S with

enlarging C to allow slack in taking the infimum.

Character averaging produces multipliers. The averaging below parallels Arveson’s construction of an expectation onto multiplication operators that preserves the frequency nest algebra on L^2(\mathbb T) [2, Proposition 5.1, pp. 228–229]. Use the invariant mean on the discrete group \R from Lemma 2.6. For T\in\B(\mathcal K) put

This UCP map has range in the commutant of the scalar characters. Since C_s^*P_aC_s=P_{a-s}, conjugation by C_s preserves the entire nest algebra. The coefficient average preserves it as well: each identity (1-P_a)C_sTC_s^*P_a=0 remains zero after taking the mean. Both target multipliers C_1\mathcal W and C_1\mathcal W^* commute with every C_s, so averaging retains both approximation bounds.

The scalar characters generate the scalar multiplication algebra L^\infty(\R). Indeed, an L^1 function annihilating all characters has zero Fourier transform and hence is zero. Their commutant on L^2(\R,H) is therefore the algebra of decomposable operators. One may see the latter assertion by taking matrix coefficients relative to a countable orthonormal basis of H: each coefficient commutes with scalar multiplication and is itself a scalar multiplier; the operator norm bound on a countable dense set of finite vectors gives a common almost-everywhere bounded operator-valued field. Thus \mathcal E(T_B) and \mathcal E(T_S) are multipliers by fields B(s),S(s) satisfying

They preserve P_0\mathcal K, so Lemma 7.5 applies. Write their extensions as B(z),S(z).

Analytic products retain the approximate algebra relations. The boundary products satisfy, in both orders,

By the product identity and the positive Poisson kernel,

Similarly, for every k and x\in M_k(M)_1,

Here is a justification that avoids choosing measurable nearest points in M_k(M). On the boundary, the inequalities follow from (7.4) and (7.8). Test against any normal functional annihilating M_k(M). The product identity identifies its value in the upper half-plane with the Poisson average of its boundary values, so the same bound holds there. Since M_k(M) is weak-* closed, its norm distance is the supremum over these annihilating normal functionals of norm at most one. This last distance formula follows from Hahn–Banach applied to the preannihilator in the trace class.

For z=t+ir with 0<r\leq\gamma define

Because |e^{-iz}|=e^r, equations (7.9)–(7.10) give, uniformly in r,t,k,

The last line is for x\in M_k(M)_1.

Polar smoothing gives a norm-continuous path. For small enough \gamma, both products in (7.12) are invertible. They provide a right and a left inverse for A_r(t), which agree, and

Together with (7.11), this places the spectrum of |A_r(t)| in [1-C\gamma,1+C\gamma]. Its polar unitary

therefore satisfies

For the second bound, first use \norm{C_r(t)-A_r(t)^{-1}} \leq\norm{A_r(t)^{-1}}\norm{A_r(t)C_r(t)-I}, and then compare A_r(t)^{-1} with V_r(t)^*. It follows from (7.13) that V_r(t) approximately normalizes M in both directions, with error C\gamma at every matrix level.

For fixed r>0, the fields A_r(t) are norm continuous by holomorphy, and their inverses are uniformly bounded. Continuous functional calculus makes V_r(t) norm continuous. Put

This is a norm-continuous unitary path with E_r(0)=I and

To check composition here, approximate V_r(0)^*xV_r(0) by an m\in M with \norm{m}\leq1+C\gamma, and apply the corresponding estimate for V_r(t) to m; the sum of the two errors is still C\gamma. Apply Lemma 7.4 to the portion of this path joining 0 to t, in either time direction. For each r,t it gives a_r(t)\in\U(M) and b_r'(t)\in\U(M') such that

where a further slack of \gamma allows for an unattained infimum. No continuous choice of these factors is needed.

The boundary comparison holds at every fixed time. Set F_0(s)=e^{is}W(s), a bounded strongly-* continuous field, and

The Poisson formula and (7.8) give \norm{A_r(t)-D_r(t)}\leq e^r\delta. Thus

At every fixed t, the Poisson approximate identity gives D_r(t)\to W(t) strongly-* as r\downarrow0. For clarity, on a fixed vector split the integral into |s-t|<h and its complement. Strong continuity controls the first piece, while the Poisson mass of the complement tends to zero. Apply the same argument to the adjoint field; the scalar prefactor converges to e^{-it}. This proves both strong limits everywhere, not just almost everywhere.

Consequently

since W(0)=I and all factors are uniformly bounded. By (7.17),

These estimates compare E_r(t) with a convergent family; they do not assert convergence of E_r(t) itself.

A UCP limit recovers an actual normalizer. Fix t. For a sequence r\downarrow0 define the UCP maps

Since b_r'(t) commutes with M, this is the restriction of \Ad(a_r(t)b_r'(t)). Equations (7.16) and (7.18) therefore imply, at every matrix level,

Indeed, for any bounded operators T,R the cb norm of the difference of the maps x\mapsto TxT^* and x\mapsto RxR^* is at most (\norm{T}+\norm{R})\norm{T-R}.

Compactness of the product of the weak-operator compact balls \{y\in M:\norm{y}\leq\norm{x}\}, indexed by x\in M, supplies a subnet for which \Phi_r(x) converges weakly for every x. Denote its limit by \Phi(x). Linearity, unitality and positivity at every matrix level pass to this limit, so \Phi:M\to M is UCP. No normality of \Phi is needed. Bounded strong-* continuity of multiplication gives Q_r(t)xQ_r(t)^*\to W(t)xW(t)^* strongly-* for each fixed x, and likewise at every fixed matrix level. Weak-operator lower semicontinuity of norm in (7.19) now yields

The representation \rho_t is faithful and normal, and its range has two-sided cb gap at most 2\gamma from M by (7.5). In particular the reverse ordinary gap required by Proposition 2.4 is small. That proposition supplies a unitary q_t with

Then n_t=q_tW(t) is the required unitary normalizer and \norm{n_t-W(t)}=\norm{q_t-I}\leq C\gamma. All estimates and smallness thresholds used only universal constants, which proves the uniform assertion.□

Proof. Write v=e^{ih} by the principal logarithm, with h=h^* and \norm{h}\le2\norm{v-I}. For r_0 small, the norm-convergent logarithm series on \B(\B(H)) gives

Indeed \Ad_v=\exp(i\ad_h) and \norm{\ad_h} is within the radius where logarithm and exponential are inverse analytic functions. Each term of the logarithm series preserves P, since vPv^*=P. Hence i\ad_h restricts to a bounded *-derivation of P.

The factor P is a simple unital C^*-algebra, by the same projection argument as in Lemma 7.3. Sakai’s theorem therefore gives a self-adjoint a_0\in P such that h-a_0\in P'. A scalar may be subtracted from a_0 and added to this commutant element. Choose the scalar at the midpoint of the spectrum of a_0. We claim that the resulting a_0 has \norm{a_0}\le\norm{h}.

For any self-adjoint a\in P, its spectral diameter is at most \norm{\ad_a|_P}. In fact, choose nonzero spectral projections near the two spectral endpoints and a norm-one partial isometry between nonzero subprojections of them. Compressing [a,w] by w^* shows that its norm is at least the endpoint separation minus the two spectral tolerances. Let those tolerances tend to zero. Apply this to a_0, using \norm{\ad_{a_0}|_P}=\norm{\ad_h|_P}\le2\norm{h}. After the midpoint shift, \norm{a_0}\le\norm{h} and b_0'=h-a_0 has norm at most 2\norm{h}. They commute, so v=e^{ia_0}e^{ib_0'}; these are the required factors, with total displacement at most 3\norm{h}.□

Proof. Let m_s denote the invariant operator mean just described. Put

Every F_t(s) is a unitary normalizer and \norm{F_t(s)-v_t}\le e. Hence \norm{A_t-v_t}\le e. Translation invariance and the exact factorization

give

The centered terms have norm at most 2e, and their means vanish; expanding verifies the first equality with the written operator order. Likewise

so its norm is at most 4e^2. Taking polar parts in \B(H) alone would not ensure membership in \mathcal N_P. We instead use the local product structure of exact normalizers.

By Lemma 7.7, choose, for every s,t,

All these choices can be arbitrary. Their averages a_t=m_sa_{t,s}\in P and b_t'=m_sb_{t,s}'\in P' satisfy

This follows by expanding the centered product; its two factors have norm at most 2C_0e. Unitarity of the individual factors also gives

For small e the averages are invertible, since they are within C_0e of I. Their polar unitaries \widehat a_t\in P and \widehat b_t'\in P' differ from them by at most Ce^2, with an absolute C: use \norm{|a_t|-I}\le \norm{a_t^*a_t-I}/(1+\sqrt{1-4C_0^2e^2}), and similarly on the other side. Therefore

It follows, uniformly in t,r, that

For the second inequality use \norm{A_t}\le1 and \norm{w_t}=1 to compare products, then the preceding bound on the defect of A.

Iterate this operation, always with the same absolute constants. Choose e_1 so small that every new defect is at most half the preceding one and every displacement is at most twice the preceding defect. The total displacement is then at most 4e. The families converge uniformly in t in operator norm to unitaries z_t, satisfying z_tz_r=z_{t+r}. The group \mathcal N_P is norm closed: if unitaries u_j\to u normalize P, norm closure of P gives uPu^*\subset P and, by applying the same argument to u_j^*, the reverse inclusion. Thus z_t\in\mathcal N_P for all t. The exact multiplication law implies z_0=I and z_{-t}=z_t^*, as required.□

Proof of Proposition 7.1. Put U_t=\Delta_N^{it} and W(t)=\Delta_N^{it}\Delta_M^{-it}. Lemma 7.6 supplies n_t\in\mathcal N_M with \sup_t\norm{n_t-W(t)}\le C\gamma. Set v_t=n_t\Delta_M^{it}, so v_t\in\mathcal N_M and \sup_t\norm{v_t-U_t}\le C\gamma. Since U is an exact representation, \sup_{t,s}\norm{v_tv_s-v_{t+s}}\le3C\gamma. For sufficiently small universal \gamma, apply Lemma 7.8 to obtain an exact representation z_t\in\mathcal N_M with

Take the invariant mean T=m_tz_tU_t^*. It is within C'\gamma of I and hence invertible for small \gamma. For every s, translation invariance and the two exact representation laws imply

Its polar unitary w=T|T|^{-1} therefore also satisfies z_sw=wU_s. Moreover \norm{w-I}\le2\norm{T-I}\le2C'\gamma. In particular z_s=wU_sw^*; this proves strong continuity of z without any regularity assumption on the selected normalizers.

Set q=w^*. Since z_s normalizes M, the intertwining identity gives U_s(qMq^*)U_s^*=qMq^* for every s. At each matrix level conjugation by q^{(k)} moves contractions by at most 2\norm{q-I}. Consequently

Enlarging C_{\rm mod} and reducing c_{\rm mod} to meet the finitely many universal smallness conditions proves all assertions.□

Proof. The generators \pi(x) and \lambda_r lie in P\barotimes\B(L^2(\R)) and are fixed by every \alpha_s. For the reverse inclusion, take a fixed operator X in that tensor product. If f,g\in C_c(\R), put

These integrals are understood on vectors. The operator Y remains fixed because translations commute with U_s\otimes\rho_s. Define bounded operators from \mathcal K to H by

They have norm at most \norm{f}_2 and depend continuously on t in operator norm. On compactly supported vector functions, Fubini’s theorem gives the integral kernel

The kernel is bounded and jointly norm continuous. Each value belongs to P: the evaluation operators intertwine the constant action of P', and X commutes with that action. Fixedness of Y gives

For fixed s the identity first holds almost everywhere by testing the two kernels on compactly supported vectors. Norm continuity then makes it hold at every (t,r). Consequently

To reconstruct a bounded operator from this kernel inside the generated algebra, use the masks

The function \chi_j is the Fourier transform of the probability measure with density

with the continuous value at a=0. Thus multiplication of an operator kernel by \chi_j(t-r) is a contraction: it is the weak integral of \Ad(Q_a) against this measure. Let Y_j be the masked operator. Using (8.4) and \rho_v\xi(t)=\xi(t+v),

Indeed the integrand has compact support and is strongly continuous, since v\mapsto k(0,v) is norm continuous. Its vector integral is a strong limit of finite sums of elements of C(P,U). Formula (8.3) verifies the equality on compactly supported vector functions. On these tests the kernels are bounded on a set of finite measure and \chi_j(t-r)\to1, so dominated convergence gives Y_j\to Y weakly. The uniform norm bound extends this convergence to all vector tests. Hence Y\in C(P,U).

Choose nonnegative f_n\in C_c(\R) of integral one with supports shrinking to zero. Strong continuity of translations gives \lambda(f_n)\to I and \lambda(f_n)^*\to I strongly. Applying the preceding conclusion with f=g=f_n gives \lambda(f_n)X\lambda(f_n)^*\to X strongly. This proves (8.1).

For the second assertion,

Thus \beta preserves C(P,U). An operator fixed by all \beta_a commutes with the scalar multiplication algebra on L^2(\R). If it also belongs to P\barotimes\B(L^2(\R)), it is multiplication by a bounded measurable P-valued field F(t). Here and below field measurability is weak operator measurability; on bounded sets, separability of H permits countably many coefficient tests. The remaining \alpha-fixedness says, for every s,

It follows that G(t)=U_tF(t)U_t^* is translation invariant in L^\infty(\R;\B(H)). Such a field is almost everywhere constant. To see this without choosing uncountably many exceptional sets, test against a countable dense family of vector pairs. Each scalar coefficient is translation invariant as an L^\infty class; convolving with a smooth compactly supported approximate identity gives continuous translation-invariant functions, hence constants. Their local L^1 convergence proves the assertion for the coefficients and then for the field. Since U_t normalizes P, the constant value x belongs to P. Therefore F(t)=U_t^*xU_t almost everywhere, which proves (8.2).□

Proof. By Lemma 2.5, tensoring the pair with \B(L^2(\R)) preserves the upper bound. Average on the common ambient operator algebra using V_s=U_s\otimes\rho_s in Lemma 2.6. For a contraction X in any matrix level of C(P,U), choose a contraction Y in the same matrix level of Q\barotimes\B(L^2(\R)) within the chosen gap bound. The matrix amplification of E_V fixes X, takes Y into C(Q,U) by Lemma 8.1, and preserves both the norm bound and the approximation error. Reverse P,Q. This proves the assertion at every matrix level with the same bound.□

Proof. Write \mathcal A=C(P,U) and \mathcal B=C(Q,U). Use a slightly smaller gap bound if necessary so that Lemma 8.2 gives \dcb(\mathcal A,\mathcal B)<\gamma. The semifinite factor \mathcal B contains matrix units (e_{ij})_{i,j\ge1} with \sum_i e_{ii}=I strongly and e=e_{11} finite and nonzero. Here is the usual projection construction. A faithful normal semifinite trace, together with separability of the representation, gives an increasing sequence of finite-trace projections with supremum I. Split their successive differences by trace continuity, and group the resulting pieces into countably many projections of one fixed positive finite trace. The infinite total trace ensures that the grouping exhausts I. Equal-trace comparison gives the required matrix units. Their weak closed span S is a unital copy of \B(\ell^2), hence amenable.

Apply Theorem 2.1 to S\subset_\gamma\mathcal A. For \gamma<1/100 it gives a unitary w with

Put \mathcal A_1=w^*\mathcal A w, so S\subseteq\mathcal A_1\cap\mathcal B and

Compression by the common projection e gives

The algebra B_0 is a type \II_1 factor. When \kappa<1/10, Lemma 5.2 first makes A_0 finite and then makes it a factor. Thus Theorem 5.1 applies to this pair on the separable Hilbert space e\mathcal K. Choosing its orientation appropriately gives a unitary v_0 on e\mathcal K with

Inflate this unitary using the common matrix units:

The sum converges strongly and is a unitary, because it is the direct sum of copies of v_0 on the orthogonal spaces e_{ii}\mathcal K. Moreover

For x\in\mathcal A_1, its transported matrix entries e_{1i}xe_{j1} belong to A_0. The entries of VxV^* therefore belong to B_0. Its finite matrix compressions lie in \mathcal B and converge strongly to VxV^*; hence V\mathcal A_1V^*\subseteq \mathcal B. The reverse argument gives equality. Thus

It remains to descend to H. Average the character action \beta on the entire \B(\mathcal K), obtaining a unital completely positive map E_\beta from Lemma 2.6, with V_a=Q_a. By Lemma 8.1, it maps \mathcal B into \pi(Q) and fixes \pi(x) for every x\in\B(H). Therefore

defines a unital completely positive map P\to Q. Since the same map \pi is completely isometric on all of \B(H),

Here \id_P denotes the original inclusion into \B(H). The ordinary gap d(P,Q) is less than \gamma. Apply Proposition 2.4. Its absolute thresholds hold for all sufficiently small \gamma, because \eta_{\rm fin}(t)\to0. Its displacement bound supplies \zeta(\gamma). A constant multiple of \gamma+\eta_{\rm fin}(301\gamma), enlarged to a nondecreasing function if needed, is sufficient. This proves the proposition.□

Proof. If either factor is finite, Theorem 5.1 applies, with the roles reversed if necessary. Its finite-neighbor conclusion also shows that for a sufficiently small gap both factors are finite in that case. We may therefore suppose that both are infinite.

Use Proposition 6.4, retaining its changes of representation explicitly. On H\otimes K it gives the fixed hyperfinite tensor factor D\subseteq\B(K) and unitaries u_0\in\B(H) and v\in\B(H\otimes K) such that

are a prepared type \III_1 pair, and, for absolute constants,

An enlargement of C_2 absorbs strict upper-bound slack. Let U_s=\Delta_N^{is} be the modular group belonging to the common vector in that preparation. By Proposition 7.1, a unitary q satisfies

The algebra N and its modular group have not changed. Consequently C(N,U) is its type \II_\infty modular core. No modular claim about the action on P is necessary. Apply Proposition 8.3 with \gamma=C_4t to obtain

All these operations have taken place on the actual tensor product of the original Hilbert space. Their inverse adjustments give

and

The right side tends to zero universally.

Choose a unit vector \xi\in K, and let \omega_\xi be its normal vector state on all of \B(K). Slicing the conjugation in (8.10) defines

This is a unital completely positive map into N_0. The slice is defined on the entire ambient tensor product, so comparison with x\otimes I_K gives at every matrix level

The reverse ordinary gap is less than t. A second application of Proposition 2.4 therefore gives the desired conjugating unitary on the original H, with displacement at most an absolute constant times r(t), for all sufficiently small t.

Choose t_{\rm fac} below the finite, preparation, modular-alignment, core-descent, and two UCP thresholds just used. Every threshold is absolute, and the only non-linear modulus used here is \eta_{\rm fin}, which is uniform and tends to zero. The maximum of \eta_{\rm fin}(t) and a sufficiently large constant multiple of r(t), enlarged if needed to be positive and nondecreasing, defines \eta_{\rm fac}. This establishes the stated uniform quantifiers.□

Proof. Fix z\in Z(M)_1 and choose y\in N_1 with \norm{z-y}<g. For v\in\U(N), choose a\in M_1 with \norm{v-a}<g. Since z commutes with a,

Dixmier’s averaging theorem for von Neumann algebras [4, Definition III.2.5.16 and Theorem III.2.5.18] says that the norm-closed convex hull of \{vyv^*:v\in\U(N)\} meets Z(N). An element of this intersection is a contraction and is within 3g of z. Interchanging M,N and then letting g\downarrow d(M,N) proves the gap estimate.

The centers are amenable. Apply the mutual conclusion of Theorem 2.1, with near-inclusion bound 3g, to obtain w. Finally, conjugation by w changes an ordinary gap by at most 2\norm{w-I}, giving the asserted bound for N_0.□

Proof. Choose separable unital C^*-subalgebras A\subseteq M and B\subseteq N whose strong closures are M,N. Such subalgebras exist because the strong topology on each bounded operator ball is metrizable and separable when H is separable. Choose countable norm-dense subsets \{a_j\}\subseteq A_1 and \{b_j\}\subseteq B_1. Off one null set, their fiber representations satisfy

We first check the unit-ball density needed below. The map A\to A(t) is a quotient *-homomorphism, possibly with a nonzero kernel. A contraction in A(t) has lifts of norm at most 1+s for every s>0. Rescaling these lifts to contractions shows that the image of A_1 is norm dense in A(t)_1. It follows that \{a_j(t)\} is norm dense in A(t)_1 and, by Kaplansky density, strongly dense in (M_t)_1. The same argument applies to B.

Put d_0=d(M,N). For every j,n\geq1, choose y_{j,n}\in N_1 such that

The essential-supremum norm formula and the countability of these choices give a common conull set on which

Fix such a t, an x\in(M_t)_1, and an n. Choose a net of a_j(t) converging strongly to x. Along a subnet, the corresponding y_{j,n}(t) converge weakly to a contraction y\in N_t. Operator-norm balls are weak-operator closed, so \norm{x-y}\leq d_0+1/n. Letting n increase gives \dist(x,(N_t)_1)\leq d_0. Make the same countable choices in the opposite direction and discard their null exceptions. This proves both near inclusions on one conull set.□

Proof. Work on the underlying standard Borel model of X, retaining the countable contraction fields a_j(t),b_j(t) from the preceding proof. Choose Borel versions of the dimension function, the trivializations on its countably many level sets, and the countably many contraction fields. Discard a Borel null set containing all exceptions to these identifications, the asserted density and algebraic properties, and the implementer hypothesis. On each remaining dimension stratum, the fibers are identified with one fixed separable Hilbert space K. Zero-dimensional fibers contribute nothing to the direct integral and may be omitted. It suffices to select implementers on each of these standard Borel strata.

The unitary group \U(\B(K)), with the strong-* topology, is Polish. Fix a dense sequence (\xi_l) in the unit sphere of K. Consider the relation on pairs (t,u) defined by \norm{u-I}\leq a and the following two families of conditions:

All these predicates are Borel. Indeed, multiplication of bounded operator fields and strong-* unitary fields is Borel, and the norm condition on u-I is a countable collection of vector-norm tests. The strong density of the contraction fields shows that the first family is equivalent to uM_tu^*\subseteq N_t, and the second to the reverse inclusion. Thus the relation describes exactly the implementers with the imposed bound. Its sections are nonempty by hypothesis.

The Jankov–von Neumann uniformization theorem [31, Theorem 18.1] gives a selection measurable for the sigma algebra generated by analytic sets, for an analytic relation between standard Borel spaces; our relation is Borel and hence analytic. Analytic sets are universally measurable [31, Theorem 21.10], so the selector is measurable for the completed measure \mu. Its direct integral is a unitary v with \norm{v-I}\leq a. For x\in M, the field v_t x_t v_t^* is a bounded measurable field in N_t, so the field-reconstruction part of central decomposition puts vxv^* in N. The reverse inclusion follows in the same way.□

Proof. The zero Hilbert space is immediate. By Theorem 8.4, choose \tau>0 such that any pair of factors on a separable Hilbert space at distance less than \tau has an implementer within a/2 of the identity. Choose \delta_{\mathrm{sep}}(a)>0 so small that Lemma 9.1 applies with g=\delta_{\mathrm{sep}}(a) and

For a pair satisfying the hypothesis, align the centers by the unitary w in that lemma, and set N_0=w^*Nw. The common-center pair M,N_0 has distance less than \tau. By Lemma 9.2, almost every fiber pair has distance less than \tau as well. Apply the factor theorem on each fiber and then Lemma 9.3, with bound a/2. The resulting unitary v satisfies

Now u=wv implements the original conjugacy and \norm{u-I}\leq\norm{w-I}+\norm{v-I}<a. Every tolerance used above is independent of the fibers and of H.□

Proof. Choose r with d(M,N)<r<g, and start with A_0=C^*(E,I), B_0=C^*(F,I). Suppose that separable unital A_j\subseteq M, B_j\subseteq N have been constructed. For each element of a countable norm-dense subset of (A_j)_1, choose an N_1-approximant within r, and adjoin all these approximants to B_j. In the same way, adjoin to A_j approximants in M_1 for a countable dense subset of (B_j)_1. Let A_{j+1},B_{j+1} be the unital C^*-algebras generated by these additions. Put

They are separable and contain the required tests. The stage unit balls are norm dense in the unit balls of these closures: approximate in norm and rescale if necessary. The choices of approximants therefore give d(A,B)\leq r<g.

Let K=\overline{\operatorname{span}(C^*(A,B)V)}. Because the algebra is separable and V finite, K is separable. It contains V and is reducing for both algebras. To transfer the gap through restriction, take x\in(A|_K)_1. For every s>0 there is a quotient lift a\in A with a|_K=x and \norm{a}\leq1+s. The contraction a/(1+s) has a B_1-approximant within r plus arbitrary slack. Restricting this approximant gives

Let s\downarrow0 and reverse the roles. Thus d(A|_K,B|_K)\leq r.

Finally, for a contraction in (A|_K)'', Kaplansky density gives a strongly convergent net of contractions from A|_K. Approximate these in (B|_K)_1 with errors at most r plus arbitrary slack, and take a bounded weak-operator cluster subnet. Its limit lies in ((B|_K)'')_1 and has the same error bound. Interchanging the algebras proves the final assertion.□

Proof of Theorem 1.1. Let \eps>0. The conclusion of Proposition 2.4 is uniform in the algebras and the Hilbert space, and its implementing-unitary bound tends to zero with the cb error. Choose a>0 and g_0>0 so small that this proposition gives an implementer within \eps of the identity whenever the reverse ordinary gap is at most g_0 and the cb error of the UCP map is at most 2a. All its smallness requirements are included in this choice. Choose

We prove the theorem for any M,N\subseteq\B(H) with d(M,N)<\delta.

If H=\{0\}, take its identity unitary. Otherwise fix a unit vector \xi_0\in H and a number g satisfying

Consider the directed set of quadruples i=(E_i,F_i,V_i,k_i), where E_i\subseteq M, F_i\subseteq N, and V_i\subseteq H are finite, \xi_0\in V_i, and k_i\in\N is positive. The order is inclusion in the three finite sets and the usual order in k_i.

For each i, apply Lemma 9.5 to obtain A_i,B_i,K_i. Set

These are unital von Neumann algebras on the nonzero separable space K_i, and d(P_i,Q_i)<g. By Theorem 9.4, there is a unitary u_i\in\B(K_i) with

Extend it by the identity on K_i^\perp to a unitary U_i\in\B(H). For all matrix sizes, conjugation satisfies

We next record what the local conjugacy says about the finite tests on the original space. If x\in E_i, then K_i reduces x, and the restriction of U_ixU_i^* belongs to Q_i with norm at most \norm{x}. Kaplansky density approximates this restriction on the finitely many coefficients from V_i by an element of B_i|_{K_i} of norm at most \norm{x} (with arbitrarily small slack if needed). Lift that element through the restriction quotient. Choosing the slack small enough gives b_{i,x}\in B_i\subseteq N such that

The displayed coefficients are coefficients on H, since all the vectors belong to K_i. At this stage the whole conjugated algebra need not lie in N; (9.2) is the property that will survive as the tests increase.

For each x\in M, the operators U_ixU_i^* lie in the weak-operator compact ball of radius \norm{x} in \B(H). Compactness of the product of these balls gives a subnet for which

Linearity and unitality pass to this limit. At each matrix size, positive operators form a weak-operator closed cone, so F is completely positive. Also, the norm ball of radius 2a\norm{X} is weak-operator closed for every fixed matrix X over M. Taking limits in (9.1) therefore gives

It remains to verify the range of F. Fix x\in M. Cofinality of the subnet implies that eventually x\in E_i, so choose b_{i,x} as in (9.2) on this tail. Every fixed pair \xi,\eta\in H eventually belongs to V_i, while k_i\to\infty. Hence (9.2) and the definition of F imply b_{i,x}\to F(x) in the weak-operator topology. As N is weak-operator closed, F(x)\in N. This holds for every x\in M, and proves that F:M\to N is UCP. Normality of this cluster map is neither asserted nor needed.

The original inclusion M\hookrightarrow\B(H) is faithful, normal, and unital. Its range has reverse ordinary gap less than \delta<g_0 from N, and (9.3) supplies the cb bound. Apply Proposition 2.4. It yields a unitary u\in\B(H) with uMu^*=N and \norm{u-I}<\eps.

The choices of a,g_0, and \delta depend only on \eps. The intervening choices of finite tests, separable subspaces, and quotient lifts change none of their bounds. Thus the same tolerance works for all algebras, all representations, and all Hilbert spaces, as claimed.□

Proof. Choose \delta from Theorem 1.1 with unitary displacement \eps/2. The resulting unitary satisfies uMu^*=N and \norm{u-I}<\eps/2. Conjugation by its diagonal amplification identifies the unit balls of M_n(M) and M_n(N) and moves each contraction by at most 2\norm{u-I}, independently of n. The same estimate applies to the commutants because uM'u^*=N'.□