Proof. The assertion is immediate when \(\Delta=0\). Otherwise put \(d=\|\Delta\|>0\), and consider the upper triangular map \[ \Phi(a)= \begin{pmatrix} \lambda(a)&d^{-1}\Delta(a)\\ 0&\sigma(a) \end{pmatrix} \in\mathcal B(F\oplus K). \tag{8}\] 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 (2006, 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 \[\|\Phi\|_{\mathrm{cb}} \le 1+\|\Delta\|_{\mathrm{cb}}/d=:G.\] 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 \[ D\Phi(a)=\Phi(a^*)^*D\qquad(a\in A). \tag{11}\] Write \(D=(D_{ij})_{i,j=1}^2\) relative to \(F\oplus K\). Its first row in (11) gives \[\begin{align*} D_{11}\lambda(a)&=\lambda(a)D_{11},\\ d^{-1}D_{11}\Delta(a)+D_{12}\sigma(a)&=\lambda(a)D_{12}. \tag{12}\end{align*}\] 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 \[V=-dD_{11}^{-1}D_{12}\] satisfies (10), by (12), and \[\|V\| \le d\,\|D^{-1}\|\|D\| =d\,\bigl(\|S^{-1}\|\|S\|\bigr)^2 \le dG^2.\] 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 (Haagerup 1975, Lemma 2.10(1), pp. 278–279). Apply it, through the normal isomorphism \(r\), to \[\omega(x)=\sum_{i=1}^h\langle x\xi_i,\xi_i\rangle, \qquad x\in N.\] 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 \[x\zeta\mathrel{% \BeginAccSupp{method=hex,unicode,ActualText=27FC}% \begingroup \let\mathrel{% \BeginAccSupp{method=hex,unicode,ActualText=21A6}% \mapsto \EndAccSupp{}}\mapsto \let\longrightarrow\longrightarrow \longmapsto \endgroup \EndAccSupp{}}(x\xi_1,\ldots,x\xi_h),\qquad x\in N,\] 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 6 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 \[ \|w_\alpha\|\le2,\qquad \sigma_{h,1}(w_\alpha)=z_\alpha,\qquad \sigma_{h,1}(w_\alpha)\zeta\longrightarrow\xi. \tag{15}\] 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 \[\Delta_h(X)\sigma_{h,1}(w_\alpha) =\Delta_{h,1}(Xw_\alpha) -\lambda_h(X)\Delta_{h,1}(w_\alpha).\] Since \(\lambda\) is completely contractive, Lemma 4 and (15) imply \[\bigl\|\Delta_h(X)\sigma_{h,1}(w_\alpha)\zeta\bigr\| \le c_{\rm row}\|\Delta\| \bigl(\|Xw_\alpha\|+\|\lambda_h(X)\|\|w_\alpha\|\bigr) \le4c_{\rm row}\|\Delta\|.\] 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 5 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 \[L^2(\mathcal L)\ominus L^2(D) \cong E\otimes\mathcal V\otimes E, \qquad \mathcal V= \bigoplus_{n\geq1} B^\circ\otimes(E^\circ\otimes B^\circ)^{\otimes(n-1)}.\] 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 \[L_{u_n}R_{u_n^*} =L_{u_n}\otimes I_{\mathcal V}\otimes R_{u_n^*}\] 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: \[G_j=\int_{\mathcal U(D_j)}W_uGW_u^*\,d\mu_j(u).\] This weak operator integral is well defined. The unitary representation \(u\mathrel{% \BeginAccSupp{method=hex,unicode,ActualText=21A6}% \mapsto \EndAccSupp{}}W_u\) 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, \[P_{L^2(A_0)}[W_uGW_u^*,L_s]\big|_{L^2(A_0)} =P_{L^2(A_0)}[G,L_s]\big|_{L^2(A_0)}.\] 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 \[F(a,x,z)= \left\langle Y(1\otimes x_1^\circ\otimes\cdots\otimes x_k^\circ), a^*\otimes(z_1^*)^\circ\otimes\cdots\otimes(z_l^*)^\circ \right\rangle.\] 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 \[\|b\|=\|T\| \leq\|Y\|\prod_i\|x_i^\circ\|_2 \prod_j\|(z_j^*)^\circ\|_2.\] Since \(F(a,x,z)=\tau(ab)\), this yields the endpoint estimate \[ |F(a,x,z)|\leq \|Y\|\,\|a\|_1 \prod_{i=1}^k\|x_i\|_2 \prod_{j=1}^l\|z_j\|_2. \tag{19}\] 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 \[\operatorname{Tr}\bigl(A(v_0\otimes\cdots\otimes v_{p-1})\bigr) \qquad (A\in\operatorname{End}((\mathbb C^m)^{\otimes p})),\] 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 \(m\geq p\) (Grinko and Ozols 2024, Theorem 3.1). Taking their trace pairings gives the cycle-trace forms \[\Phi_\pi(v_0,\ldots,v_{p-1}) =\prod_{(i_1\,\cdots\,i_r)\text{ a cycle of }\pi} \tau(v_{i_1}\cdots v_{i_r}), \qquad \pi\in S_p.\] 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 \[F\big|_{D_j^p}=\sum_{\pi\in S_p}\lambda_{\pi,j}\Phi_\pi.\] 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 \[\Phi_\rho(v_0,\ldots,v_{p-1}) =\begin{cases} m^{-c(\pi)},&\rho=\pi,\\ 0,&\rho\ne\pi, \end{cases}\] 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 \[|\lambda_\pi|m^{-c(\pi)} \leq\|Y\|m^{-1-(k+l)/2}.\] The coefficient is fixed while the stage size tends to infinity. It follows that \[\lambda_\pi=0\qquad\text{if}\qquad c(\pi)<1+\frac{k+l}{2}.\]

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 \[1+\frac{k+l-(r-1)}2.\] 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, \[F(a,x,z)=\tau(a) \sum_{\mathcal P}\lambda_{\mathcal P} \prod_{\{i,j\}\in\mathcal P}\tau(v_i v_j),\] 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 \[|\lambda_{\mathcal P}|M^{r_s+r_t} \leq\|Y\|M^{(r_s+r_t)/2}.\] 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 \[\tau(c^*d)\sum_{\sigma\in S_k}a_\sigma \prod_{i=1}^k\tau(x_i z_{\sigma(i)}).\] 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 \[Y_{\alpha,\beta}: E\otimes(E^\circ)^{\otimes k} \longrightarrow E\otimes(E^\circ)^{\otimes l}, \qquad \|Y_{\alpha,\beta}\|\leq\|G\|\|\alpha\|\|\beta\|.\] It satisfies the hypotheses of Lemma 12. 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 \[Y_{\alpha,\beta} =I_E\otimes\sum_{\sigma\in S_k} a_\sigma(\alpha,\beta)U_\sigma, \qquad \langle B_k\alpha,\beta\rangle =\sum_{\sigma\in S_k}a_\sigma(\alpha,\beta).\] 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. Every group word has the form \(w^l\) or \(w^ld\), \(l\in\mathbb Z\). Since \(dp=p\) and \(dw^ld=w^{-l}\), \[pw^ldp=pw^lp=\frac{w^l+w^{-l}}2p.\] An even function of \(w\) commutes with \(d\) and hence with \(p\). Thus \(f\mathrel{% \BeginAccSupp{method=hex,unicode,ActualText=21A6}% \mapsto \EndAccSupp{}}f(w)p\) is a normal \(*\)-homomorphism on the even-function algebra, with unit \(p\). The group trace gives \[\tau_p(f(w)p) =2\tau(f(w)p)=\tau(f(w))+\tau(f(w)d)=\int_{\mathbb T}f\,dm,\] 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 \[\|f(w)p\|=\|f\|_\infty,\qquad \|f(w)p\|_{2,\tau_p}=\|f\|_{L^2(m)}.\]

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 \[\begin{align*} &\tau_p(a_1pw^{l_1}p\cdots a_jpw^{l_j}p)\\ &\qquad=2\tau(a_1w^{l_1}\cdots a_jw^{l_j})\\ &\qquad=2\tau\big(a_1s(ds)^{l_1-1}\cdots a_js(ds)^{l_j-1}\big)=0. \end{align*}\] All the displayed \(D\) letters are centered, and \(s\) is centered and free from the whole algebra \(D\), by \(\mathcal L=D*A_0\). 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 \(p\mathcal Lp\), its restriction is the trace-preserving expectation onto \(\mathcal Z\) for \(\tau_p\). Expand \[ebe=u^2b+u(ab+ba)+aba.\] 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 \[\tau_p(c^\circ aba)=0, \qquad \tau_p(aba)=\tau_p(ba^2) =\tau_p(b)\tau_p(a^2)=0.\] Thus \(E_{\mathcal Z}(aba)=0\) also. This proves the expectation formula.

For the norm estimate work in \[H_c=L^2(W^*(\mathcal A,\mathcal Z),\tau_p).\] 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 \[H_c=L^2(\mathcal A,\tau_p)\otimes F_0, \qquad F_0=\mathbb C\Omega\oplus F_0^\circ,\] 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 \[\Pi=|1_{\mathcal A}\rangle\langle1_{\mathcal A}|\otimes Q_0, \qquad L_e=\lambda_{\mathcal A}(e)\otimes I_{F_0},\] where \(1_{\mathcal A}=p\) has \(L^2(\tau_p)\) norm one. Consequently \[ \|L_e\Pi\|=\|\Pi L_e\|\le\|e\|_{2,\tau_p}=\sqrt u. \tag{31}\]

The following three blocks are the standard creation, annihilation, and diagonal decomposition of a centered free-product letter; see Ricard and Xu (2006, sec. 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 \[(1-\Pi)T(1-\Pi)=0,\qquad \|\Pi T(1-\Pi)\|\le\|b\|_{2,\tau_p}.\] 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 \[\begin{align*} \|ebe\|=\|L_eTL_e\| &\le\|L_e\Pi T(1-\Pi)L_e\| +\|L_e(1-\Pi)T\Pi L_e\| +\|L_e\Pi T\Pi L_e\|\\ &\le\sqrt u\,\|b\|_{2,\tau_p} +\sqrt u\,\|b\|_{2,\tau_p}+u\|b\|. \end{align*}\] This proves (30). ◻

Completion of the proof of Theorem 9. 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 \[f(e^{i\alpha})=\mathbf1_{[-\theta,\theta]}(\alpha)-\frac\theta\pi.\] This function is real, even, and of mean zero. The trace formula in (29) gives \[\|b_f\|\le1,\qquad \|b_f\|_{2,\tau_p}^2 =\frac\theta\pi\left(1-\frac\theta\pi\right)\le u.\] Consequently (30) gives \[ \|h\|=\|eb_fe\|\le3u. \tag{35}\] For every \(l\ge1\) the Fourier coefficient is \[\gamma_l=\frac{\sin(l\theta)}{\pi l}.\] 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 \[(H_f)_{jk}\ge\frac{2\theta}{\pi^2}.\] Testing the leading \(N\) by \(N\) block on its constant unit vector, and using \(\theta N=1/2\), proves the uniform bound \[ \|H_f\|\ge\frac{2\theta N}{\pi^2}\ge\frac1{2\pi^2}. \tag{36}\] Combine (34), (27), (35), and (36) to obtain \[\frac{u}{2\pi^2}\|[B_0,L_s|_B]\| \le3u(1+4c_0)\epsilon.\] Cancelling \(u>0\) yields \[\|[B_0,L_s|_B]\| \le6\pi^2(1+4c_0)\epsilon=c_1\epsilon.\] 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 11 preserves the compression of \([G,L_s]\) between vectors of \(B\). The same estimate therefore holds for the original operator in Theorem 9, completing its proof. ◻

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 \(a\in L^1(M)_+\) (Hiai 2020, Corollary 5.14 and Remark 5.15). The map on cyclic orbits \[x\xi\mathrel{% \BeginAccSupp{method=hex,unicode,ActualText=27FC}% \begingroup \let\mathrel{% \BeginAccSupp{method=hex,unicode,ActualText=21A6}% \mapsto \EndAccSupp{}}\mapsto \let\longrightarrow\longrightarrow \longmapsto \endgroup \EndAccSupp{}}xa^{1/2}\in L^2(M)\] 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 \(\widetilde K=L^2(M)^{\oplus\mathbb N}\). These are standard consequences of the normal representation theory of finite von Neumann algebras; see Takesaki (1979).

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, \[[p_nYp_n,L_x]=p_n[Y,L_x]p_n\qquad(x\in M).\] Proposition 21 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 \(x\mathrel{% \BeginAccSupp{method=hex,unicode,ActualText=21A6}% \mapsto \EndAccSupp{}}[Y,x]\). With \(\delta=g_P(Y)\), they give \[\lVert YV-VY^{(h)}\rVert\le c_{\mathrm{row}}\delta, \qquad \lVert Y^{(h)}V^*-V^*Y\rVert\le c_{\mathrm{row}}\delta.\] For \(X\in M_h(P)\), the identity \[\begin{align*} [Y,VXV^*] &=(YV-VY^{(h)})XV^* +V[Y^{(h)},X]V^*\\ &\hspace{2em}+VX(Y^{(h)}V^*-V^*Y) \end{align*}\] and \(\lVert VXV^*\rVert=\lVert X\rVert\) imply \[\lVert[Y^{(h)},X]\rVert \le(1+2c_{\mathrm{row}})\delta\lVert X\rVert.\] 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 \(\lVert Y-Y_0\rVert\le\delta\). Then \[[Y^{(h)},X]=[(Y-Y_0)^{(h)},X],\] 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 \[ \lVert Y_1-Y\rVert\le\delta, \qquad g_P(Y_1)\le3\delta. \tag{39}\] 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 \(g_P(Y_1)\le g_P(Y)+2\lVert Y_1-Y\rVert\).

Use the central direct-integral decomposition on a separable Hilbert space (Takesaki 1979); precise section locators for the factor fibers, decomposable operators, and norm formula are Blackadar (2006, secs. III.1.6.1–4 and III.5.1.10–15):

\[K=\int^{\oplus}K_\alpha\,d\mu(\alpha), \qquad P=\int^{\oplus}P_\alpha\,d\mu(\alpha),\] 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 \[ g_{P_\alpha}\bigl(Y_1(\alpha)\bigr)\le3\delta \quad\text{for almost every }\alpha. \tag{40}\]

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 \[c_b=b(1+b^*b)^{-1/2}\in P.\] 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 \[c\mathrel{% \BeginAccSupp{method=hex,unicode,ActualText=27FC}% \begingroup \let\mathrel{% \BeginAccSupp{method=hex,unicode,ActualText=21A6}% \mapsto \EndAccSupp{}}\mapsto \let\longrightarrow\longrightarrow \longmapsto \endgroup \EndAccSupp{}}c(1-c^*c)^{-1/2}.\] 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 \(\lVert[Y_1,c_b]\rVert\le3\delta\) 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 22 on that conull set. For any fixed \(X\in M_h(P)\), decomposability of its entries and the same norm formula give \[\lVert[Y_1^{(h)},X]\rVert \le3C_{\mathrm{fac}}\delta\lVert X\rVert.\] Restoring \(Y\) costs at most \(2\delta\lVert X\rVert\) by (39), giving \(C=3C_{\mathrm{fac}}+2\). ◻

Proof of Theorem 2. Let \(P\subset\mathcal B(K)\) 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 \(\mathcal B(K)\) 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 \(a\in A_X\), \[\lVert[Y_0,\sigma(a)]\rVert\le g_P(Y)\lVert a\rVert.\] The \(C^*\)-quotient norm of the image gives, for every \(b\in\sigma(A_X)\), \[\lVert[Y_0,b]\rVert \le g_P(Y)\inf_{\sigma(a)=b}\lVert a\rVert =g_P(Y)\lVert b\rVert.\] By Kaplansky density the same scalar commutator bound holds on \(P_0\). Proposition 23, applied to \(P_0\) and \(\sigma_h(X)\), now yields \[\lVert[Y_0^{(h)},\sigma_h(X)]\rVert \le C g_P(Y)\lVert\sigma_h(X)\rVert \le C g_P(Y)\lVert X\rVert.\] 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 \(\sigma\): \[K=\bigoplus_{i\in I}K_i.\] 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 \(a\mathrel{% \BeginAccSupp{method=hex,unicode,ActualText=21A6}% \mapsto \EndAccSupp{}}\Delta(a)|_{K_i}\) is a rectangular derivation with cyclic domain representation \(\sigma|_{K_i}\) and full output representation \(\sigma\). Lemma 7 gives a bounded map \(V_i:K_i\to K\) such that \[\Delta(a)|_{K_i}=V_i\sigma(a)|_{K_i}-\sigma(a)V_i.\] For a finite subset \(F\subset I\), put \[q_F=\sum_{i\in F}q_i, \qquad V_F=\sum_{i\in F}V_iq_i\in\mathcal B(K).\] Every \(V_F\) is bounded, and the reducing property gives the exact identity \[ [V_F,\sigma(a)]=\Delta(a)q_F\qquad(a\in A). \tag{41}\] 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 \[\lVert[V_F,b]\rVert\le\lVert\Delta\rVert\lVert b\rVert.\] This uses the \(C^*\)-quotient norm of the \(*\)-image and does not require \(\sigma\) to be faithful. Kaplansky density then implies \[g_{\sigma(A)''}(V_F)\le\lVert\Delta\rVert.\]

Apply Theorem 2 to \(V_F\) and \(\sigma(A)''\). For every \(X\in M_h(A)\), \[\begin{align*} \lVert\Delta_h(X)q_F^{(h)}\rVert &=\lVert[V_F^{(h)},\sigma_h(X)]\rVert\\ &\le C\lVert\Delta\rVert\lVert\sigma_h(X)\rVert \le C\lVert\Delta\rVert\lVert X\rVert. \end{align*}\] 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 5 then supplies a bounded implementing operator. ◻

Proof of Theorem 1. For every faithful unital \(*\)-representation of \(A\), each bounded derivation into the ambient bounded operators is inner by Proposition 24. Thus condition (2) of Theorem 8 holds for \(A\). Its equivalent similarity property gives, for every bounded unital homomorphism \(\pi:A\to\mathcal B(H)\), 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 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 \[\|[T,e]\|=\max\{\|(1-e)Te\|,\|eT(1-e)\|\}\le\alpha_M(T),\] 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. (2010, Equation (2.1)), gives \[2\mathop{\mathrm{dist}}(T,M)=\|\operatorname{ad}_T|_P\|_{\mathrm{cb}}, \qquad \operatorname{ad}_T(a)=[T,a],\] since \(P'=M\). Taking the supremum over finite matrix sizes in Theorem 2 therefore yields \[2\mathop{\mathrm{dist}}(T,M)\le Cg_P(T)\le4C\alpha_M(T),\] which proves Equation (2). ◻