Kadison’s similarity theorem through uniform derivation estimates

Proof. The assertion is immediate when \Delta=0. Otherwise put d=\|\Delta\|>0, and consider the upper triangular map

It is unital and multiplicative by (6). Its diagonal map is contractive, and its off-diagonal map has norm one; hence \|\Phi\|\le2. The off-diagonal rectangular compression of \Phi_{p,q}, from K^q to F^p, is d^{-1}\Delta_{p,q}. The row and column estimates for \Phi give (7).□

Proof. We use the triangular-metric argument recalled in Ozawa [18, Lemma 1.2], allowing different domain and codomain representations and keeping track of the implementer norm. For \Delta=0, take V=0. Otherwise the map \Phi in (8) is completely bounded, with

Choose the similarity S supplied by Paulsen’s theorem, and put D=S^*S. The fact that S\Phi(\cdot)S^{-1} preserves adjoints is equivalent to

Write D=(D_{ij})_{i,j=1}^2 relative to F\oplus K. Its first row in (11) gives

Since D\ge\|D^{-1}\|^{-1}I, its compression D_{11} is positive invertible on F, with \|D_{11}^{-1}\|\le\|D^{-1}\|. Its inverse also commutes with \lambda(A). Hence

satisfies (10), by (12), and

If either Hilbert space is zero, \Delta=0, so it has already been covered.□

Proof. Put N=P', and choose a cyclic vector \Omega for P. It separates N: if x\in N and x\Omega=0, then xa\Omega=ax\Omega=0 for every a\in P, whence x=0. Let L=\overline{N\Omega}\subset K. This subspace reduces N, and restriction r:N\to\mathcal B(L) is faithful and normal. Indeed, an element in its kernel annihilates \Omega; normality follows from compression to a reducing subspace. Thus it is a normal *-isomorphism onto the von Neumann algebra r(N), and its inverse is normal. The vector \Omega is cyclic and separating for r(N).

The standard-form theorem for a representation with a cyclic separating vector realizes every normal positive functional on r(N) as a vector functional, with its vector in the natural cone [11, Lemma 2.10(1), pp. 278–279]. Apply it, through the normal isomorphism r, to

We obtain \zeta\in L\subset K such that \langle x\zeta,\zeta\rangle=\omega(x) for every x\in N. In particular \|\zeta\|^2=\omega(1)=1. The assignment

is isometric, since both squared norms equal \omega(x^*x). It extends to an isometry from \overline{N\zeta} into K^h. Extend it by zero on \overline{N\zeta}^{\perp}, obtaining a contraction v:K\to K^h. The subspace \overline{N\zeta} reduces N, and the isometry intertwines the N-actions. Therefore vx=x^{(h)}v for every x\in N. Each coordinate v_i lies in N'=P, and v\zeta=\xi, as required. All these constructions, including the standard-form assertion, are valid on nonseparable Hilbert spaces.□

Proof. Zero spaces and the zero derivation are immediate, so assume they do not occur. Put P=\sigma(A)''. Fix h\ge1, X\in M_h(A)_1, and a unit vector \xi\in K^h. Lemma 2.3 gives a unit vector \zeta\in K and a contraction v\in M_{h,1}(P) with v\zeta=\xi.

Matrix Kaplansky density gives a net of column contractions z_\alpha\in M_{h,1}(\sigma(A)) converging strongly to v. To see the rectangular assertion directly, regard v as the first column of a matrix in M_h(P), with all other columns zero. Approximate that matrix strongly by contractions in M_h(\sigma(A)), and compress to its first input coordinate. The *-image \sigma(A) is the C^*-quotient A/\ker\sigma, and the same quotient identification holds at each square matrix level. Consequently each z_\alpha lifts to w_\alpha\in M_{h,1}(A) with

For example, first lift the square matrix whose first column is z_\alpha and whose other columns vanish to a matrix of norm less than 2, then compress its first column. Only fixed slack in the quotient norm is being used.

Entrywise application of (6) gives

Since \lambda is completely contractive, Lemma 2.1 and (15) imply

At the fixed finite level h, \Delta_h(X) is already a bounded operator. We may therefore take the limit in (15), obtaining the same bound on \|\Delta_h(X)\xi\|. Taking the supremum over \xi,X,h proves (13). The quotient argument did not require \Delta to vanish on \ker\sigma; its identity was applied to the chosen lifts in A. Lemma 2.2 and (9) now give (14).□

Proof. Put E=L^2(D), E^\circ=E\ominus\mathbb C1, and B^\circ=L^2(A_0)\ominus\mathbb C1. Regrouping reduced words by their first and last A_0-letters gives the D–D bimodule decomposition

The endpoints in E are unrestricted; expanding each as \mathbb C1\oplus E^\circ recovers every reduced word containing an A_0-letter exactly once. Left and right D act on the first and last factors, respectively.

Choose a Haar unitary v in a diffuse abelian subalgebra of D and put u_n=v^n. These unitaries converge ultraweakly to zero in D. Indeed, the restriction of any normal functional to W^*(v)\cong L^\infty(\mathbb T) has an L^1(\mathbb T) density, whose Fourier coefficients tend to zero. Consequently L_{u_n} and R_{u_n^*} converge weakly to zero on E: first test against bounded vectors in D, and then use their L^2 density. On elementary tensors the matrix coefficients of

therefore tend to zero. The same holds on the whole complement by density and uniform boundedness. A D-central vector in this complement is fixed by all these operators, and hence is zero.

If z\in D'\cap\mathcal L, its component perpendicular to L^2(D) is D-central, because the displayed decomposition is a bimodule decomposition. Thus z=E_D(z)\in D, where E_D is the trace-preserving conditional expectation. It follows that z\in Z(D)=\mathbb C1. The center of \mathcal L is contained in this relative commutant. Its faithful finite trace and its diffuse subfactor D then make \mathcal L a {\rm II}_1 factor.□

Proof. Average the original operator over each compact matrix unitary group:

This weak operator integral is well defined. The unitary representation u\mapsto W_u is strongly continuous on each group, as is seen first on bounded reduced words and then by density. Every conjugate in the integral commutes with L_D. It also obeys (16), since \theta_u permutes the D-corners of trace t. Convexity and weak operator closedness of norm balls preserve these properties under integration and under a weak operator cluster limit \widetilde G as j\to\infty. Such a limit exists because \|G_j\|\leq\|G\|. For fixed u\in\mathcal U(D_i), all G_j with j\geq i commute with W_u, by Haar invariance. Thus \widetilde G has every stated symmetry.

The operators W_u fix L^2(A_0) pointwise and commute with L_s. For each conjugate, and therefore for the averages and their limit,

Finally, [G^*,L_x]=-[G,L_{x^*}]^*, so the hypothesis for a self-adjoint corner algebra also gives its adjoint version. The same applies to \widetilde G.□

Proof. For a vector or algebra element x\in E, put x^\circ=x-\tau(x)1. On unrestricted variables a,x_1,\ldots,x_k,z_1,\ldots,z_l\in D, define the complex multilinear form

It is multilinear with our convention that inner products are linear in the first argument. Fixing the tail vectors and taking their Hilbert-space slice of Y gives an operator T:E\to E commuting with L_D. The tracial commutation theorem gives T=R_b for a bounded b\in D, with

Since F(a,x,z)=\tau(ab), this yields the endpoint estimate

The same slice also identifies coefficients with arbitrary bounded first endpoints: replacing 1,a^* by d,c replaces a by c^*d, because \langle db,c\rangle=\tau(c^*db).

Finite matrix invariants and stable coefficients. Put p=1+k+l and restrict all variables to a stage D_j\cong M_m(\mathbb C). The intertwining assumption says that F is invariant under simultaneous unitary conjugation. To see precisely the finite-dimensional consequence, label its variables v_0,\ldots,v_{p-1}. Every multilinear form on M_m^p has a unique expression

by the nondegenerate trace pairing. Conjugation invariance is equivalent to A commuting with every u^{\otimes p}. Finite-dimensional Schur–Weyl duality identifies this commutant with the span of the tensor permutation operators; they are linearly independent when m\geq p [10, Theorem 3.1]. Taking their trace pairings gives the cycle-trace forms

Here cycles follow the successor convention of the displayed product; reversing the convention only reindexes the permutations. The normalization from \operatorname{Tr} to \tau multiplies each form by a nonzero scalar, so these forms are also independent for m\geq p. Thus

This identity is on unrestricted matrix variables: the centering occurs inside the definition of F. Under a unital inclusion of full matrix stages, normalized traces agree, and hence every \Phi_\pi restricts to the identically defined form on the smaller stage. Uniqueness on any smaller stage of size at least p proves that \lambda_{\pi,j} is independent of j from that point onward. Denote it by \lambda_\pi.

Isolating coefficients before restricting to centered tails. The invariant-theory step has expressed every coefficient through finitely many cycle-trace forms with coefficients fixed across matrix stages. We now use the endpoint bound to remove long cycles, then Hilbert-space boundedness to remove pairs within one side. Fix \pi\in S_p. Choose distinct matrix indices for the slots and put v_i=e_{i,\pi(i)}. In a nonzero product of these matrix units, the variable in slot i must be followed by the variable whose initial index is \pi(i). Closing each product under its trace therefore forces exactly the cycles of \pi. Consequently

where c(\pi) is the number of cycles. Every chosen matrix unit has L^1 norm m^{-1} and L^2 norm m^{-1/2}, including diagonal units. Applying (19) gives

The coefficient is fixed while the stage size tends to infinity. It follows that

Now restrict the k+l tail variables to centered matrices. Any form with a singleton tail cycle vanishes. For a permutation with no such singleton, let r be the length of the cycle containing the distinguished variable a=v_0. Its number of cycles is at most

If r>1, this is below the required threshold. If r=1, the threshold is attained only when every tail cycle has length two. Thus the only remaining terms have a as a singleton and pair all tail slots. On centered tails,

where \mathcal P ranges over pairings of the tail slots. There are no terms when the number of tail slots is odd.

Using a=c^*d gives the corresponding simple-tensor coefficients with first endpoints d,c in the union of the matrix stages. These formulas extend to arbitrary endpoints in E and arbitrary centered L^2 tail letters. Indeed, that union is L^2 dense, the left side is continuous in its individual L^2 letters, and each factor on the right is continuous by |\tau(uv)|\leq\|u\|_2\|v\|_2. At this point the assertion is continuity of coefficients on simple tensors, not boundedness of each pairing on the completed tensor product.

Excluding pairings within one side. Choose an infinite orthonormal family (h_n) of self-adjoint centered vectors in E. For example, normalized real trigonometric functions of a Haar unitary give such a family. Then \tau(h_nh_m)=\delta_{nm}. Fix a pairing \mathcal P, with r_s pairs entirely in the source slots and r_t entirely in the test slots. Assign a disjoint infinite pool of the h_n to each pair. For each source–test pair, use one fixed member of its pool in both slots. For each source–source pair, sum the source tensor over M members of its pool, placing the same member in its two slots, independently for the different pairs. Form the test tensor analogously for the test–test pairs. Take both first endpoints to be 1.

In any term of the expanded coefficient every label occurs exactly twice, in the positions assigned by \mathcal P, and different pairs have different labels. Hence all other pairings contribute zero. The resulting coefficient is exactly \lambda_{\mathcal P}M^{r_s+r_t}. The source summands are orthonormal, as are the test summands, so the product of the two tensor norms is M^{(r_s+r_t)/2}. Boundedness of Y gives

Letting M\to\infty proves \lambda_{\mathcal P}=0 whenever r_s+r_t>0. The argument also applies when one side has no tail slots.

All surviving pairs therefore join a source slot to a test slot, which requires k=l. If these numbers differ, all simple-tensor coefficients vanish and Y=0. Otherwise these pairings are indexed by bijections \sigma\in S_k, and the coefficients are

Since \tau(x_i z_j)=\langle x_i,z_j^*\rangle, these are precisely the coefficients of I_E\otimes\sum_\sigma a_\sigma U_\sigma. Equality against all elementary test tensors, and their density, proves the asserted operator identity.□

Proof. Let P_k be the level projection. It commutes with L_D and all W_u. For \alpha\in J_k and \beta\in J_l, take the scalar slice in these factors of P_lGP_k to obtain an operator

It satisfies the hypotheses of Lemma 3.4. For k\ne l every such slice vanishes, so P_lGP_k=0. Thus the levels reduce G.

Fix a unit vector h\in E^\circ and let V_k:J_k\to\mathcal H_k be the isometry V_k\alpha=1\otimes h^{\otimes k}\otimes\alpha. Define B_k=V_k^*GV_k. On an equal-level scalar slice, the lemma gives

Every U_\sigma fixes x^{\otimes k} for every x\in E^\circ. Consequently the last scalar is independent of the unit letter used to define V_k. The same slice identity, now tested against arbitrary endpoint and tail vectors and arbitrary \beta, gives the first equality in (20). These tests determine the whole output on level k, and the other levels have already been excluded. This also proves the assertion for every endpoint d_0\in E, with no boundedness requirement on its representative. No self-adjointness of x is used: the relevant cross contraction is \tau(xz)=\langle x,z^*\rangle.

The same argument applies to G^*. Its induced operator on J_k is V_k^*G^*V_k=(V_k^*GV_k)^*=B_k^*, proving the second equality.□

Proof. The case r=0 is immediate. Fix r\ne0 and a projection e\in D with \tau(e)=t. The factoriality established in Lemma 3.2 and comparison of projections give r'\le e and a partial isometry v\in\cL such that v^*v=r and vv^*=r'. Right multiplication by v^* is a unitary left e\cL e-module map

Indeed its inverse is right multiplication by v, and

These equalities also hold for L^2 vectors by continuity. The vector r' is cyclic for the left action of e\cL e on eHr', since (e\cL e)r'=e\cL r' is dense in that Hilbert space. Thus eHr is a cyclic domain representation of the unital algebra e\cL e.

Since G commutes with L_e, its restriction defines a bounded map G_e:eHr\to eH. On these domain and codomain representations the map

is a rectangular derivation of norm at most \epsilon, by (16). Lemma 2.4 therefore supplies an exact intertwiner T:eHr\to eH with

The value t=1/N allows an extension that preserves this norm. Choose v_1,\ldots,v_N\in D with

The maps (\xi_i)\mapsto\sum_i v_i\xi_i identify (eHr)^N with Hr and (eH)^N with H unitarily. Define

In these coordinates \widetilde T is T^{(N)}, and G|_{Hr} is G_e^{(N)}, because G commutes with each L_{v_i}. Consequently

For a\in\cL, its left action has matrix entries L_{v_i^*av_j} in both coordinate systems. Every v_i^*av_j lies in e\cL e, so the intertwining identity for T implies \widetilde T L_a|_{Hr}=L_a\widetilde T.

Extend \widetilde T by zero on H(1-r). Since R_r commutes with all left multiplications, this extension belongs to L(\cL)' and hence is R_b for a bounded b\in\cL, by the tracial commutation theorem. Its vanishing on H(1-r) says R_b=R_bR_r, which, with the convention R_b\xi=\xi b, gives b=rb. This proves (21). Finally G^* commutes with L_D and [G^*,L_a]=-[G,L_{a^*}]^*, so all the same hypotheses and estimates hold for G^*.□

Proof. Fix k\ge1 and first take v\in A_0; the zero vector is harmless. Choose a nonzero projection q\in D with a=\tau(q)\le t. The left support of qv is dominated by q. Its right support r is equivalent to its left support by polar decomposition, so \tau(r)\le a\le t. Choose b=rb as in Lemma 4.1 for this r, and put

The level-zero identity in (20) gives G(qv)=qB_0v, and the tensor decomposition gives \|qv\|_2=\sqrt a\,\|v\|_2. It follows that

The vector B_0v need only belong to L^2(A_0); every expression here uses bounded left multiplication on this vector.

Let z be a Haar unitary of qDq for its normalized trace, and set x_n=a^{-1/2}z^n for n\ge1. Then

The weak convergence follows, for example, from orthonormality of the positive and negative powers. The reduced words \xi_n=(sx_n)^kv have norm \|v\|_2 and right support at most r, because their last x_n satisfies x_nq=x_n. With A_n=(sx_n)^{k-1}s, we have

Thus (21) yields

To estimate the second term, write (18) as H=E\otimes F, where

and let P_{\mathrm{sc}} be projection onto \C1\otimes F. This projection keeps every word level and selects the scalar component of its first D factor. On this decomposition left multiplication by x_n acts on E. With inner products linear in the first variable, define \ell_n(\eta)=\langle\eta,x_n^*\rangle. Then

For each elementary tensor this expression tends to zero in norm, since x_n^*\rightharpoonup0. Approximating the fixed vector y by finite tensor sums, and using the common bound \|\ell_n\|=1, proves

On (1-P_{\mathrm{sc}})H, multiplication by A_n prepends a reduced centered word ending in s to a leading centered D letter. Each prefix letter has L^2 norm one, so this restriction is an isometry. On P_{\mathrm{sc}}H we use the bound \|L_{A_n}\|\le a^{-(k-1)/2}. Also qy=y gives

Consequently

and (25) implies

Here q,k,v,b,y are all fixed before n tends to infinity. The possibly large prefix bound therefore disappears with the vanishing coefficient; a convergence rate uniform in a or k is unnecessary.

By the full-output identity (20), the left side of (24) equals

for every n: its endpoint and repeated D factors all have norm one. Taking the limit superior proves (22) for bounded v. Density of A_0 in B and boundedness of the operators extend it to all v\in B. Applying the same argument to G^*, using the adjoint identities of Lemma 3.5, gives the final assertion.□

Proof. Lemma 4.2 and (20) give

with zero error for k=0. Each error lies in level k by Lemma 3.5. Their squared norms therefore add on finite orthogonal sums. Passing to the closure, and repeating the argument for G^*, gives the full-codomain estimates

Write R=G\iota-\iota J, R'=G^*\iota-\iota J^*, and T_h=\iota^*L_h\iota. Taking the adjoint of the second error identity gives \iota^*G=J\iota^*+(R')^*. A direct expansion now yields

Apply (16) to the first term and (28) to the other two. Their bounds are \epsilon\|h\|, 2c_0\epsilon\|h\|, and 2c_0\epsilon\|h\|, proving (27).□

Proof. Every group word has the form w^l or w^ld, l\in\mathbb Z. Since dp=p and dw^ld=w^{-l},

An even function of w commutes with d and hence with p. Thus f\mapsto f(w)p is a normal *-homomorphism on the even-function algebra, with unit p. The group trace gives

initially for trigonometric polynomials and then for bounded functions by L^2 approximation. In particular the map is faithful, and its range is a von Neumann algebra. The compressed group-word span is strongly dense in p\mathcal C p and consists of even Laurent polynomials times p. This proves (29) and the isomorphism assertion. It also gives

For freeness, let a_1,\ldots,a_j\in pDp be centered for \tau_p. Then \tau(a_i)=0 and a_id=a_i. The centered trigonometric span in p\mathcal C p is generated by the elements pw^lp, l\ge1. For l_1,\ldots,l_j\ge1, absorbing the corner projections and using traciality gives

All the displayed D letters are centered, and s is centered and free from the whole algebra D, by \cL=D*A_0. Thus the last word is a reduced alternating word of trace zero. Fejér means of an even bounded zero-mean function are even zero-mean trigonometric polynomials, are uniformly bounded, and converge in L^2(m). Approximating each corner letter in this manner extends the vanishing to all cyclic alternating products of even length. Products of odd alternating length have their endpoints in the same algebra. Move one endpoint cyclically next to the other, split their product into its centered and scalar parts, and apply the even-length case and induction on the length. The length-one case is centeredness. This proves freeness.□

Proof. Put \mathcal A=pDp, \mathcal Z=p\mathcal C p, and a=e-up\in\mathcal A, so \tau_p(a)=0. Since p\in\mathcal C, bimodularity shows that E_{\mathcal C}(ebe) belongs to \mathcal Z. On p\cL p, its restriction is the trace-preserving expectation onto \mathcal Z for \tau_p. Expand

Freeness gives zero expectation for ab and ba. For the last term, test against any c\in\mathcal Z and write c=c^\circ+\tau_p(c)p. Freeness and traciality give

Thus E_{\mathcal Z}(aba)=0 also. This proves the expectation formula.

For the norm estimate work in

This representation is faithful, so its operator norm is the algebra norm. The free-word decomposition, grouping a possible initial \mathcal A letter, takes the form

where F_0^\circ is the closed span of reduced words starting in the centered L^2 subspace \mathcal Z^\circ. Let Q_0 be the projection of F_0 onto the complement of the vacuum, and let \Pi project in H_c onto words starting in \mathcal Z^\circ. Then

where 1_{\mathcal A}=p has L^2(\tau_p) norm one. Consequently

The following three blocks are the standard creation, annihilation, and diagonal decomposition of a centered free-product letter; see Ricard and Xu [26, Section 1, Lemmas 2.1–2.3 and Corollary 2.4]. We give the bounds needed here directly. Let T=L_b on H_c. Because b is centered, on the vacuum and on words starting in \mathcal A^\circ its left action simply prepends the letter b. Therefore

Self-adjointness gives the same bound for (1-\Pi)T\Pi, and \|\Pi T\Pi\|\le\|b\|. Decomposing T into these three blocks, using \|L_e\|\le1, and applying (31), we obtain

This proves (30).□

Completion of the proof of Theorem 3.1. Put \theta=u/4=1/(2N), so 0<\theta\le1/4, and use angular coordinates z=e^{i\alpha}, -\pi<\alpha\le\pi, to define

This function is real, even, and of mean zero. The trace formula in (29) gives

Consequently (30) gives

For every l\ge1 the Fourier coefficient is

If 0\le j,k<N, then 1\le j+k+1\le2N-1 and (j+k+1)\theta<1<\pi/2. The inequality \sin x\ge2x/\pi on [0,\pi/2] therefore gives

Testing the leading N by N block on its constant unit vector, and using \theta N=1/2, proves the uniform bound

Combine (34), (27), (35), and (36) to obtain

Cancelling u>0 yields

The level-zero case of (20) identifies G|_B with B_0, so this is exactly the required compression estimate for the symmetrized operator. Lemma 3.3 preserves the compression of [G,L_s] between vectors of B. The same estimate therefore holds for the original operator in Theorem 3.1, completing its proof.□

Proof. We first give an approximation step in an arbitrary \mathrm{II}_1 factor N. If y=y^*\in N and \eta>0, there is a unital full matrix algebra F\subset N, of size at least two, such that

Indeed, approximate y in L^2 by a spectral step y_0=\sum_{i=1}^k\lambda_i p_i, where the p_i partition the identity. For a large integer b, split off \lfloor b\tau(p_i)\rfloor subprojections of trace 1/b in each p_i. The remaining projection has trace

Its numerator is an integer, so it too splits into projections of trace 1/b. These pieces form a partition of the identity by b equivalent projections and hence extend to matrix units for a unital F\cong M_b(\C). Define y_1\in F to have value \lambda_i on the pieces taken from p_i, and value zero on the remaining pieces. Then

The expectation of y_1 onto F'\cap N is \tau(y_1)1. Contractivity of this expectation on L^2, and \abs{\tau(y-y_1)}\leq\norm{y-y_1}_2, give the required estimate by choosing the spectral approximation and then b.

Choose a countable L^2-dense subset \{x_j:j\geq1\} of the self-adjoint unit ball of M, and enumerate all requirements (j,k)\in\N^2. Begin with F_0=\C1. If the current stage is a unital full matrix algebra F_n, then N_n=F_n'\cap M is a \mathrm{II}_1 factor: matrix units identify M with a matrix algebra over a corner and N_n with that corner. For the next requirement (j,k), apply the approximation step in N_n to y=E_{N_n}(x_j), with error less than 1/k, obtaining F\subset N_n. Set F_{n+1}=F_n\vee F. This is again a full matrix algebra, its size is at least twice that of F_n, and N_{n+1}=F'\cap N_n. The identity E_{N_{n+1}}=E_{N_{n+1}}E_{N_n} gives

All previously obtained bounds persist because the relative commutants decrease and their trace expectations compose.

Let R=(\bigcup_nF_n)'' and N=R'\cap M=\bigcap_nN_n. Since E_N=E_NE_{N_n}, every requirement remains valid with N in place of N_{n+1}. Thus E_N(x_j)=\tau(x_j)1 for every j. Density gives E_N(x)=\tau(x)1 for all x\in M, so N=\C1. In particular R is a factor. It is finite and contains full matrix algebras of unbounded size, hence is of type \mathrm{II}_1, with the asserted increasing-matrix presentation.□

Proof. We use the ultraproduct-of-subalgebras case of Popa’s independence theorem [25, Theorem 0.1(a); see also Theorem 4.3]. For von Neumann subalgebras A,B\subset M, the notation A\prec_M B means Popa’s intertwining of a nonzero corner of A into a finite matrix amplification of B. Its hypotheses are finite factors M_i with \dim M_i\longrightarrow\infty, and unital von Neumann subalgebras Q_i\subset M_i satisfying Q_i\nprec_{M_i}(Q_i'\cap M_i). Write \mathbf M=\prod_\omega M_i, \mathbf Q=\prod_\omega Q_i, and C=\mathbf Q'\cap\mathbf M. Given an L^2-separable subspace X\subset\mathbf M\ominus C, the theorem supplies a diffuse abelian von Neumann algebra A\subset\mathbf Q such that

whenever n\geq1, x_0\in X\cup\{1\}, x_i\in X for 1\leq i\leq n, and a_i\in A\ominus\C1. In particular, if C=\C1 and T\subset\mathbf M is a separable-predual von Neumann subalgebra, take X=T\ominus\C1. The moment identities, traciality, and decomposition into scalar and centered parts show that A is free from T. Here separability is for L^2, as stipulated in Popa [25, Section 1.1].

We check these hypotheses for the constant sequences M_i=P and Q_i=Q. The dimension condition holds because these factors are infinite-dimensional. If Q\prec_P\C1, the intertwining data would include a nonzero normal *-homomorphism from a nonzero corner of Q into a finite-dimensional matrix algebra. Such a homomorphism from a factor is faithful, whereas a \mathrm{II}_1 factor has arbitrarily large families of nonzero orthogonal projections. This is impossible. Since Q'\cap P=\C1, the non-intertwining hypothesis follows.

We also verify the full ultrapower relative commutant:

For a uniformly bounded representative (x_i) of an element of this commutant, commuting with all sequences of unitaries from Q implies

Otherwise one could choose a witnessing unitary at each coordinate on an \omega-large set. The L^2-closed convex hull of the Q-unitary orbit of x_i has a unique vector of minimum norm. It is Q-central, has trace \tau(x_i), and belongs to P: the hull stays in the operator-norm ball of radius \norm{x_i}, which is L^2-closed by weak-* compactness and trace pairings. Irreducibility makes this vector \tau(x_i)1. Every point in the hull is within d_i of x_i, so \norm{x_i-\tau(x_i)1}_2\leq d_i. Bounded scalar sequences equal their scalar ultralimit in the tracial ultrapower. This proves the claim, which concerns all of Q^\omega, rather than merely the constant copy of Q.

Choose a Haar unitary u_1 in the diffuse abelian algebra obtained by applying Popa’s theorem to the constant P. Then u_1\in Q^\omega is free from P. The algebra W^*(P,u_1) has separable L^2 space: countably many words in generators of P, in u_1, and in their adjoints have linear span dense in L^2. A second application of the same theorem therefore gives a Haar unitary u_2\in Q^\omega free from W^*(P,u_1). Expanding centered reduced words shows that P,W^*(u_1),W^*(u_2) are jointly free. Thus W^*(u_1,u_2) is free from P and, by its free Haar moments, is trace-isomorphic to L(\mathbb F_2). This is a separable-predual \mathrm{II}_1 factor: the nontrivial conjugacy classes of \mathbb F_2 are infinite, so a central square-summable Fourier coefficient function is supported at the identity. Apply Lemma 6.1 inside W^*(u_1,u_2) to obtain D. Freeness passes to its subalgebra D, proving the lemma. The ambient tracial ultrapower throughout is P^\omega.□

Proof. Put \delta=g_M(Y). By homogeneity it suffices to consider \norm{X}\leq1. We first arrange a symmetry test on a square array of standard copies. Choose h_0\geq h so that m=2h_0\geq r, extend Y by zero to H_m, and retain the notation Y. The copy projection reduces the common action of M, so this extension still has scalar commutator bound \delta. For X'=X\oplus0\in M_{h_0}(M), define in P=M_m(M)

Functional calculus gives (1-X'X'^*)^{1/2}X'=X'(1-X'^*X')^{1/2}, and hence s=s^*=s^{-1}. Traciality gives equal moments for X'X'^* and X'^*X', including the zeroth moment; polynomial approximation on [0,1] gives equal traces for their displayed square roots. Consequently \tau_P(s)=0.

Write vectors of H_m^m as arrays (\xi_{ij}), with i the outer amplification coordinate and j the copy coordinate. The map (\xi_{ij})\mapsto\sqrt m\,(\xi_{ij}) is a unitary from H_m^m to L^2(P), since \norm{(\xi_{ij})}_2^2=m^{-1}\sum_{i,j}\norm{\xi_{ij}}_2^2 on the latter space. Under this identification, the matrix action of P is left multiplication. The upper-right block of [Y^{(m)},L_s] is the commutator with X'. Restricting its row coordinates to the original h and its copy coordinates to the original r recovers [Y^{(h)},X] for the original operator. It is therefore enough to bound [Y^{(m)},L_s].

Choose R\subset M from Lemma 6.1, with matrix stages F_n. On H_m, average Y over the compact unitary group of each F_n, using common left multiplication, and take a weak operator cluster point Y_0 as n tends to infinity. For every such unitary v, \norm{L_vYL_v^*-Y}\leq\delta. The averages, and hence their cluster point, are within \delta of Y. An average over a later stage commutes with every earlier stage, so

The first identity extends from the matrix stages to R by strong density; the norm bounds pass to the cluster point by weak operator closedness of norm balls.

On L^2(P)=H_m^m, set S=Y_0^{(m)} and Q=M_m(R). Matrix units and R'\cap M=\C1 imply Q'\cap P=\C1. Moreover S commutes with L_Q, since it has the same operator Y_0 on every row and Y_0 commutes with every entry from R. For e_0=e_{11}\otimes1, the space e_0L^2(P) is one row of m standard copies, with e_0Pe_0\cong M acting by common left multiplication. Since S commutes with L_{e_0}, it follows on the full Hilbert space that

If f\in Q is a projection with \tau_P(f)\leq1/m, projection comparison in the factor Q gives a unitary v\in Q with f\leq ve_0v^*. Conjugating by L_v, which commutes with S, yields

We have obtained the same small-corner estimate for every projection of Q of trace at most 1/m. The next step transports that estimate to the ultrapower, where Lemma 6.2 supplies the free subfactor. Fix a nonprincipal ultrafilter \omega on \N. Recall the tracial quotient and distinguish it from the Hilbert ultrapower:

The Hilbert ultrapower uses bounded Hilbert-space sequences modulo those whose Hilbert norms tend to zero along \omega. Sending the class of a uniformly operator-bounded sequence (a_i) to its Hilbert-ultrapower class is a trace-preserving isometry on the dense subspace P^\omega\subset L^2(P^\omega). Its extension identifies \mathcal C=L^2(P^\omega) with a closed subspace of \mathcal H.

For a bounded sequence a=(a_i) in P, its coordinatewise left action on \mathcal H, and the adjoint action corresponding to (a_i^*), preserve the bounded-algebra-sequence subspace of \mathcal C. Thus \mathcal C reduces both actions. If a\in\mathcal I_\omega, then for every operator-bounded sequence (b_i),

It follows by density that the action on \mathcal C factors through the tracial quotient and is precisely left multiplication by [a_i]. This factoring assertion is needed only on \mathcal C, not on the whole Hilbert ultrapower.

Let S^{\omega}_{\mathrm{Hilb}} be the coordinatewise action of the fixed bounded operator S on \mathcal H, and set

Since S commutes with every left multiplier from Q, its Hilbert ultrapower commutes with every bounded left Q-sequence. Compression to the reducing subspace \mathcal C proves [\widehat S,L_{Q^\omega}]=0.

Set t=1/(2m). We claim that

To lift e, take a bounded self-adjoint representative (a_i) in Q and set f_i=1_{[1/2,\infty)}(a_i). The scalar inequality \abs{x-1_{[1/2,\infty)}(x)}\leq2\abs{x^2-x} gives \norm{f_i-a_i}_2\longrightarrow_\omega0, so [f_i]=e and \tau(f_i)\longrightarrow_\omega t. The strict gap t<1/m allows us to replace f_i by zero off an \omega-large set and assume \tau(f_i)\leq1/m for every i.

Given z\in eP^\omega e and \eta>0, the C^*-quotient norm gives a representative (b_i) with \sup_i\norm{b_i}\leq\norm{z}+\eta. The sequence z_i=f_ib_if_i represents z, lies in the required corners, and has the same norm bound. Consequently

This estimate passes to the Hilbert ultrapower. Its compression to \mathcal C is [\widehat S,L_z], since \mathcal C reduces the bounded left sequence action. Letting \eta decrease to zero proves (37).

By Lemma 6.2, choose D\subset Q^\omega free from the constant P, and put \mathcal L=W^*(D,P)\subset P^\omega. Faithfulness of the trace and the reduced-word inner products identify this algebra with the tracial free product D*P. Let G=P_{L^2(\mathcal L)}\widehat S|_{L^2(\mathcal L)}. The subspace L^2(\mathcal L) reduces left multiplication by \mathcal L. Thus G commutes with L_D, and (37) gives the corner hypotheses of Theorem 3.1 with A_0=P, \epsilon=3\delta, and t=1/(2m).

Finally, every constant L^2(P)-vector belongs to \mathcal C, by approximation by bounded vectors from P, and S^{\omega}_{\mathrm{Hilb}} sends the constant vector \xi to the constant vector S\xi. Both of these vectors also belong to L^2(\mathcal L). Therefore both compressions preserve the action of S on the constant subspace. As L_s preserves that subspace, for \xi,\eta\in L^2(P) we have exactly

Theorem 3.1, applied to the trace-zero symmetry s\in P, consequently yields on the original L^2(P)

Replacing Y_0 by Y costs at most 2\norm{Y_0-Y}\norm{s}\leq2\delta, since Hilbert amplification preserves the difference norm. The initial block compression now gives the claimed bound (3c_1+2)\delta for the original finite r,h.□

Proof. We first treat a type \mathrm{II}_1 factor M in an arbitrary normal representation on a separable Hilbert space. Its predual is separable. Every cyclic normal representation embeds into the tracial standard representation. Indeed, if \xi is a cyclic vector, its normal positive functional has a trace density a\in L^1(M)_+ [14, Corollary 5.14 and Remark 5.15]. The map on cyclic orbits

is isometric, because both squared norms equal the functional evaluated at x^*x. Its extension intertwines the representations and has a reducing range. Decomposing into orthogonal cyclic reducing subspaces, of which there are at most countably many, embeds the given representation into \widetilde K=L^2(M)^{\oplus\N}. These are standard consequences of the normal representation theory of finite von Neumann algebras; see Takesaki [28].

Extend Y by zero off the embedded reducing subspace. The scalar commutator bound is unchanged. On the countable amplification, let p_n be the projection onto the first n copies. Since p_n commutes with the common left action,

Proposition 6.3 applies to the compression of Y to p_n\widetilde K, with constant 3c_1+2. For any fixed finite matrix test, the corresponding compressed commutators converge strongly to the full commutator as n\to\infty. A uniform operator-norm bound passes to this strong limit. Restricting back to the original reducing subspace proves the assertion for type \mathrm{II}_1 factors.

If P is properly infinite, choose isometries v_1,\ldots,v_h\in P with orthogonal ranges. The row V=[v_1\ \cdots\ v_h] satisfies V^*V=I_{K^h}. Apply the row and column derivation estimates of Section 2 to x\mapsto[Y,x]. With \delta=g_P(Y), they give

For X\in M_h(P), the identity

and \norm{VXV^*}=\norm{X} imply

Here conjugation by the isometry V:K^h\to K preserves the norm of the middle term, regardless of whether its range is all of K.

Finally, if P is a finite type I factor, average Y over its compact unitary group to obtain Y_0\in P' with \norm{Y-Y_0}\le\delta. Then

which gives the constant 2. These three cases exhaust the factors and prove the proposition.□

Proof. Write \delta=g_P(Y). Average Y over the compact unitary groups of the finite-dimensional algebras generated by finite families of central projections. Direct these families by inclusion and take a cofinal weak operator cluster limit. The resulting operator Y_1 commutes with every central projection, hence with Z(P), and satisfies

For the first inequality, each unitary being averaged belongs to P, so its conjugation changes Y by at most \delta. The inequality is preserved by averages and weak operator limits. The second follows from g_P(Y_1)\le g_P(Y)+2\norm{Y_1-Y}.

Use the central direct-integral decomposition on a separable Hilbert space [28]; precise section locators for the factor fibers, decomposable operators, and norm formula are Blackadar [1, Sections III.1.6.1–4 and III.5.1.10–15]:

where the P_\alpha are factors almost everywhere and the fibers are separable. The center acts as the scalar diagonal algebra, so its commutant consists of the decomposable operators. In particular, Y_1 has fibers Y_1(\alpha). We verify carefully that

The separable direct-integral decomposition admits countably many bounded measurable sections whose values generate P_\alpha almost everywhere; fix such a family. Take all rational complex *-polynomials in these sections, including the rational scalar constants. For each such section b\in P, form the contraction

On every fiber where the generators have the stated property, these contractions are norm dense in the contraction ball of the generated C^*-algebra. To see this, the polynomials are norm dense, the displayed transformation is norm continuous, and its range contains the open unit ball: its inverse there is

The closed unit ball is the norm closure of the open one. Kaplansky density then makes these countably many fiber contractions strongly dense in the unit ball of P_\alpha.

For each c_b, the bound \norm{[Y_1,c_b]}\le3\delta implies the same fiber bound outside a null set, by the essential-supremum norm formula. Removing the union of these countably many null sets leaves a single conull set on which every test holds. Strong approximation by the contraction tests passes the bound to every element of the fiber unit ball, proving (40).

Apply Proposition 7.1 on that conull set. For any fixed X\in M_h(P), decomposability of its entries and the same norm formula give

Restoring Y costs at most 2\delta\norm{X} by (39), giving C=3C_{\mathrm{fac}}+2.□

Proof of Theorem 1.2. Let P\subset\B(K) be arbitrary, and fix h\ge1, X\in M_h(P), and a unit vector \xi=(\xi_1,\ldots,\xi_h)\in K^h. Let A_X\subset P be the unital C^*-algebra generated by the entries of X. The C^*-algebra in \B(K) generated by these entries and Y is norm separable. The closed span of its orbits of the finitely many \xi_i is therefore a separable common reducing subspace K_0 for A_X and Y. Denote restriction of A_X to K_0 by \sigma, put Y_0=Y|_{K_0}, and set P_0=\sigma(A_X)''.

For a\in A_X,

The C^*-quotient norm of the image gives, for every b\in\sigma(A_X),

By Kaplansky density the same scalar commutator bound holds on P_0. Proposition 7.2, applied to P_0 and \sigma_h(X), now yields

Since K_0 reduces all the operators in this test and contains every component of \xi, the restricted commutator on \xi equals the original one. Taking the supremum over unit test vectors proves (1). If K=\{0\}, the estimate is vacuous.□

Proof. Decompose K into cyclic reducing subspaces for \sigma:

Such a decomposition is obtained by taking a maximal orthogonal family of cyclic reducing subspaces; a nonzero vector in its orthogonal complement would generate another one. The index set I can have any cardinality. Let q_i denote the projection onto K_i.

The map a\mapsto\Delta(a)|_{K_i} is a rectangular derivation with cyclic domain representation \sigma|_{K_i} and full output representation \sigma. Lemma 2.4 gives a bounded map V_i:K_i\to K such that

For a finite subset F\subset I, put

Every V_F is bounded, and the reducing property gives the exact identity

Only the scalar commutator norm, rather than the norm of V_F, needs to be uniform in F. Indeed, taking the infimum over lifts of an element b\in\sigma(A) in (41) gives

This uses the C^*-quotient norm of the *-image and does not require \sigma to be faithful. Kaplansky density then implies

Apply Theorem 1.2 to V_F and \sigma(A)''. For every X\in M_h(A),

The projections q_F converge strongly to the identity as a net indexed by finite subsets of I. Hence the displayed bound passes to \Delta_h(X) for every fixed h and X. This proves the claimed complete boundedness. Lemma 2.2 then supplies a bounded implementing operator.□

Proof of Theorem 1.1. For every faithful unital *-representation of A, each bounded derivation into the ambient bounded operators is inner by Proposition 7.3. Thus condition (2) of Theorem 2.5 holds for A. Its equivalent similarity property gives, for every bounded unital homomorphism \pi:A\to\B(H), a bounded invertible S with bounded inverse such that S\pi(\cdot)S^{-1} is a *-homomorphism. The identity isomorphism treats the zero Hilbert space as well.□

Proof of Corollary 1.3. Put P=M'. Since M is self-adjoint, its invariant subspaces are reducing and their orthogonal projections are exactly the projections in P. Hence \alpha_M is the usual off-diagonal distance seminorm. For every projection e\in P, the two off-diagonal blocks give

where the last inequality also uses the projection 1-e. Spectral step approximations of a self-adjoint contraction a\in P express it, modulo a scalar, as a norm limit of linear combinations of projections in P whose coefficients have total absolute value at most two. Thus \|[T,a]\|\le2\alpha_M(T). The real and imaginary parts of any contraction in P are self-adjoint contractions, so g_P(T)\le4\alpha_M(T).

Arveson’s commutant-distance formula, in the form recalled in Christensen et al. [7, Equation (2.1)], gives

since P'=M. Taking the supremum over finite matrix sizes in Theorem 1.2 therefore yields

which proves Equation (2).□