Matrix commutators in finite factors

Written by Claude Opus 5.5 (Anthropic), October 2026. Self-checked by the writing AI. Public domain (CC0).

The free corner estimate concerns a special algebra: a free product of an increasing union of matrix algebras with an arbitrary finite algebra. This lesson transfers it to every II\(_1\) factor \(M\) with separable predual acting on finitely many copies of its standard space [OpenAI-288, Section 6]. The result: if an operator \(Y\) satisfies \(\|[Y,a]\|\le\delta\|a\|\) for all \(a\in M\), then \(\|[Y^{(h)},X]\|\le(3c_1+2)\delta\|X\|\) for every matrix \(X\in M_h(M)\), with a constant independent of \(M\), of the number of copies and of \(h\).

Four steps lead there. A matrix test is encoded as a single commutator with a trace-zero symmetry \(s\) of \(P=M_m(M)\). Averaging over an irreducible subfactor \(R\subseteq M\) generated by increasing matrix algebras produces an operator commuting exactly with \(Q=M_m(R)\). In the tracial ultrapower \(P^\omega\), Popa's free independence theorem provides a Haar unitary \(u\) of \(Q^\omega\) free from \(P\); the conjugate \(D=uQu^*\) is then free from \(P\) and generated by increasing matrix algebras. Finally the averaged operator is extended to the Hilbert-space ultrapower and compressed to the Hilbert space of \(W^*(D,P)\), where the free corner estimate applies; on the constant vectors this compression is the original operator.

We use: Theorem 1.1 of Invariant tensors and the word structure (the free corner estimate, with \(c_1=3\pi^2(1+4c_0)\)), proved through The Hankel test; Lemma 4.1 of Words in free subalgebras; from Haar unitaries, freeness and tracial ultraproducts, tracial ultraproducts (Section 4, Lemmas 4.1 and 4.2), Lemma 2.3 and Corollary 2.4; Theorem 4.2 of Free independence in ultraproducts (Popa's free independence); Proposition 8.4 of Projections and types (splitting off a matrix algebra), its comparison theorem Theorem 5.5 and Proposition 14.2; trace-preserving conditional expectations, Theorem 9.1 of Integration for a trace; Kaplansky's density theorem, Theorem 7.1; and a left-invariant Haar probability measure on a compact group, Theorem 8.3 of Haar measure on locally compact groups. For a von Neumann algebra \(P\subseteq B(K)\) and \(Y\in B(K)\) we define \(g_P(Y)\) by (0.1): \[ g_P(Y)=\sup_{a\in P,\ \|a\|\le1}\|Ya-aY\|. \]

1. Irreducible subfactors generated by matrix algebras

All II\(_1\) factors below carry their unique faithful normal tracial state \(\tau\).

Lemma 1.1. Let \(N\) be a II\(_1\) factor, \(y=y^*\in N\) and \(\eta>0\). There is a unital full matrix subalgebra \(F\subseteq N\) of size at least \(2\) with \(\|E_{F'\cap N}(y)-\tau(y)1\|_2<\eta\). Moreover \(F'\cap N\) is a II\(_1\) factor, and \(E_{F'\cap N}(f)=\tau(f)1\) for \(f\in F\).

Proof. If \(F\cong M_b(\mathbb C)\) is a unital subalgebra with matrix units \(f_{ij}\), Proposition 8.4 of the projections lesson identifies \(N\) with \(M_b(\mathbb C)\otimes f_{11}Nf_{11}\), \(F\) with \(M_b(\mathbb C)\otimes1\), and \(F'\cap N\) with \(1\otimes f_{11}Nf_{11}\), a II\(_1\) factor. The trace of \(N\) is the product of the normalized trace of \(M_b\) and the normalized trace of the corner, since that product is a faithful normal tracial state; hence \(E_{F'\cap N}(x\otimes1)=\operatorname{tr}(x)1\), which is the last assertion.

Approximate \(y\) in \(\|\cdot\|_2\) by a spectral step function \(y_0=\sum_{i=1}^k\lambda_ip_i\) with \(\sum_ip_i=1\). For a large integer \(b\), split off \(\lfloor b\tau(p_i)\rfloor\) mutually orthogonal subprojections of trace \(1/b\) inside each \(p_i\) (Corollary 2.4 of the Haar unitary lesson). The remaining projection has trace \(\big(b-\sum_i\lfloor b\tau(p_i)\rfloor\big)/b<k/b\), an integer multiple of \(1/b\), so it splits into projections of trace \(1/b\) as well. The resulting \(b\) projections are mutually equivalent (Lemma 2.3 of the Haar unitary lesson) and add up to \(1\), so they are the diagonal of matrix units of a unital \(F\cong M_b(\mathbb C)\). Let \(y_1\in F\) take the value \(\lambda_i\) on the pieces split off from \(p_i\) and \(0\) on the remaining pieces. Then \(\|y-y_1\|_2\le\|y-y_0\|_2+\|y_0\|\sqrt{k/b}\). Since \(E=E_{F'\cap N}\) is contractive for \(\|\cdot\|_2\), \(E(y_1)=\tau(y_1)1\) and \(|\tau(y-y_1)|\le\|y-y_1\|_2\), \[ \|E(y)-\tau(y)1\|_2\le2\|y-y_1\|_2 , \] which is small for a good spectral approximation and then large \(b\). \(\square\)

Lemma 1.2 (Popa). Every II\(_1\) factor \(M\) with separable predual contains a subfactor \(R\) with \(R'\cap M=\mathbb C1\), generated by an increasing sequence \(F_1\subseteq F_2\subseteq\cdots\) of unital full matrix subalgebras whose sizes tend to infinity.

Proof. \(L^2(M)\) is separable, so the self-adjoint part of the unit ball of \(M\) has a countable \(\|\cdot\|_2\)-dense subset \(\{x_j\}\). Enumerate the pairs \((j,k)\in\mathbb N^2\). Start with \(F_0=\mathbb C1\). Given a unital full matrix subalgebra \(F_n\), the relative commutant \(N_n=F_n'\cap M\) is a II\(_1\) factor (Lemma 1.1). For the next pair \((j,k)\), apply Lemma 1.1 in \(N_n\) to \(y=E_{N_n}(x_j)\) and \(\eta=1/k\), obtaining \(F\subseteq N_n\), and put \(F_{n+1}=F_n\vee F\cong F_n\otimes F\). Then \(N_{n+1}=F'\cap N_n\), its size at least doubles, and since \(E_{N_{n+1}}=E_{N_{n+1}}E_{N_n}\) and \(\tau(E_{N_n}(x_j))=\tau(x_j)\), \[ \|E_{N_{n+1}}(x_j)-\tau(x_j)1\|_2<1/k . \] For later stages the inequality persists, because \(E_{N_{n'}}=E_{N_{n'}}E_{N_{n+1}}\) for \(n'\ge n+1\) and conditional expectations are \(\|\cdot\|_2\)-contractive. Let \(R=(\bigcup_nF_n)''\). Then \(R'\cap M=\bigcap_nN_n=:N\), and \(E_N=E_NE_{N_n}\) for every \(n\), so \(\|E_N(x_j)-\tau(x_j)1\|_2<1/k\) for every pair \((j,k)\): \(E_N(x_j)=\tau(x_j)1\). By density \(E_N(x)=\tau(x)1\) for all \(x\in M\), so \(N=\mathbb C1\). In particular \(R\) is a factor; it is finite and contains matrix algebras of unbounded size, so it is of type II\(_1\). \(\square\)

2. Free position in the ultrapower

Fix a free ultrafilter \(\omega\) on \(\mathbb N\). For a tracial von Neumann algebra \(P\), \(P^\omega\) is its tracial ultrapower, with \(P\subseteq P^\omega\) as constant sequences; for a subalgebra \(Q\subseteq P\), \(Q^\omega\subseteq P^\omega\) is the subalgebra of classes of bounded sequences in \(Q\) (Lemma 4.1 of the Haar unitary lesson).

Lemma 2.1 (free position). Let \(Q\subseteq P\) be an irreducible inclusion of II\(_1\) factors, \(P\) with separable predual. There is a Haar unitary \(u\in Q^\omega\) such that \(W^*(u)\) and \(P\) are free in \(P^\omega\). Consequently \(D=uQu^*\subseteq Q^\omega\) is free from \(P\); and if \(Q\) is generated by an increasing sequence of unital full matrix algebras of unbounded sizes, so is \(D\).

Proof. The centred part \(P^\circ\) of \(P\) is separable for \(\|\cdot\|_2\); let \(X_0\subseteq P^\circ\) be countable and \(\|\cdot\|_2\)-dense. Theorem 4.2 of the free independence lesson, applied to the constant irreducible inclusions \(Q\subseteq P\) and to \(X_0\subseteq P^\omega\), gives a Haar unitary \(u\in Q^\omega\) with \[ \begin{aligned} &\tau(x_0u^{n_1}x_1u^{n_2}\cdots u^{n_j}x_j)\\ &\quad =0 \end{aligned}\tag{2.1} \] for \(j\ge1\), nonzero integers \(n_i\), \(x_1,\dots,x_{j-1}\in X_0\) and \(x_0,x_j\in X_0\cup\{1\}\).

Let \(y_1\cdots y_n\) be a centred alternating word in \(W^*(u)\) and \(P\). A centred \(a\in W^*(u)\) is an \(\|\cdot\|_2\)-limit of centred Laurent polynomials in \(u\): by Kaplansky's density theorem it is an \(\|\cdot\|_2\)-limit of elements of the C*-algebra generated by \(u\), hence of Laurent polynomials \(L\), and \(L-\tau(L)1\) is a combination of powers \(u^{n}\), \(n\ne0\), at \(\|\cdot\|_2\)-distance at most \(2\|L-a\|_2\) from \(a\). A centred \(x\in P\) is an \(\|\cdot\|_2\)-limit of elements of \(X_0\). The trace of a product is continuous in each factor for \(\|\cdot\|_2\) when the other factors are fixed and bounded, since \(|\tau(XYZ)|\le\|ZX\|\,\|Y\|_2\). Replacing the letters one at a time, the trace of the word is a limit of traces of combinations of words (2.1), so it is \(0\). Hence \(W^*(u)\) and \(P\) are free.

By Lemma 4.1 of the free words lesson, applied in \(P^\omega\) with \(B=Q\), the algebras \(D=uQu^*\) and \(P\) are free. Since \(u\in Q^\omega\), \(D\subseteq Q^\omega\). If \(Q\) is generated by unital full matrix algebras \(F_n\), then \(D\) is generated by the full matrix algebras \(uF_nu^*\). \(\square\)

3. The Hilbert-space ultrapower

Let \(P\) be a tracial von Neumann algebra. The Hilbert ultrapower \(\mathcal H_\omega\) of \(L^2(P)\) is the space of bounded sequences \((\xi_i)\) in \(L^2(P)\) modulo those with \(\lim_\omega\|\xi_i\|_2=0\), with \(\langle(\xi_i),(\eta_i)\rangle=\lim_\omega\langle\xi_i,\eta_i\rangle\); it is a Hilbert space. A bounded sequence of operators \((T_i)\) on \(L^2(P)\) acts on \(\mathcal H_\omega\) coordinatewise, with norm at most \(\sup_i\|T_i\|\). The class of a bounded sequence \((a_i)\) in \(P\) in \(P^\omega\) and its class in \(\mathcal H_\omega\) have the same norm \(\lim_\omega\|a_i\|_2\), so \(P^\omega\ni[a_i]\mapsto(a_i)\) extends to an isometry of \(L^2(P^\omega)\) onto a closed subspace \(\mathcal C\subseteq\mathcal H_\omega\).

Lemma 3.1.

  1. For a bounded sequence \(a=(a_i)\) in \(P\), the coordinatewise left multiplication \(L_a\) on \(\mathcal H_\omega\) and \(L_{a^*}=L_a^*\) preserve \(\mathcal C\); on \(\mathcal C=L^2(P^\omega)\), \(L_a\) is left multiplication by \([a_i]\in P^\omega\).
  2. Let \(S\in B(L^2(P))\), let \(S^\omega\) be its coordinatewise action, and \(\widehat S=P_{\mathcal C}S^\omega|_{\mathcal C}\). If \(S\) commutes with \(L_q\) for every \(q\) in a subalgebra \(Q\subseteq P\), then \(\widehat S\) commutes with \(L_x\) for every \(x\in Q^\omega\).
  3. The map \(\xi\mapsto(\xi,\xi,\dots)\) is an isometry of \(L^2(P)\) into \(\mathcal C\), and \(S^\omega\) maps the constant sequence of \(\xi\) to the constant sequence of \(S\xi\).

Proof. (1) The coordinatewise product of two bounded sequences in \(P\) is a bounded sequence in \(P\), so \(L_a\) and \(L_{a^*}\) preserve the dense subspace of \(\mathcal C\) formed by classes of bounded algebra sequences, hence \(\mathcal C\). If \(\lim_\omega\|a_i\|_2=0\), then \(\|a_ib_i\|_2\le\|a_i\|_2\sup_i\|b_i\|\to0\) for bounded \((b_i)\), so \(L_a\) vanishes on \(\mathcal C\); thus on \(\mathcal C\) it depends only on \([a_i]\) and is left multiplication by it. (2) Every \(x\in Q^\omega\) has a bounded representing sequence \((q_i)\) in \(Q\), and \(S^\omega\) commutes with \(L_{(q_i)}\) coordinatewise. Since \(\mathcal C\) reduces \(L_{(q_i)}\), compression to \(\mathcal C\) preserves this commutation. (3) For bounded \(a\in P\) the constant sequence lies in \(\mathcal C\); constant sequences of \(L^2\)-vectors are limits of these. \(\square\)

4. Finite standard multiplicities

Proposition 4.1. Let \(M\) be a II\(_1\) factor with separable predual, \(r\ge1\), and let \(M\) act on \(H_r=L^2(M)^r\) by left multiplication in every copy. For every \(Y\in B(H_r)\), every \(h\ge1\) and every \(X\in M_h(M)\), acting entrywise on \(H_r^h\), \[ \begin{aligned} &\|[Y^{(h)},X]\|\\ &\quad \le(3c_1+2)\,g_M(Y)\,\|X\|. \end{aligned}\tag{4.1} \]

Proof. Put \(\delta=g_M(Y)\) and assume \(\|X\|\le1\).

Step 1: one symmetry. Choose \(h_0\ge h\) with \(m=2h_0\ge r\). Extend \(Y\) by \(0\) to \(H_m\supseteq H_r\); the copy projection commutes with the action of \(M\), so the extension still has \(g_M(Y)=\delta\). Let \(X'=X\oplus0\in M_{h_0}(M)\), \(d_1=(1-X'X'^*)^{1/2}\) and \(d_2=(1-X'^*X')^{1/2}\), and, in \(P=M_m(M)=M_2(M_{h_0}(M))\), \[ s=\begin{pmatrix}d_1&X'\\X'^*&-d_2\end{pmatrix}. \] Functional calculus gives \(d_1X'=X'd_2\), so \(s=s^*=s^{-1}\). The normalized trace \(\tau_P=\frac1m\sum_i\tau(x_{ii})\) gives \(X'X'^*\) and \(X'^*X'\) the same moments, hence, by polynomial approximation on \([0,1]\), \((1-X'X'^*)^{1/2}\) and \((1-X'^*X')^{1/2}\) the same trace; so \(\tau_P(s)=0\). Write vectors of \(H_m^m\) as arrays \((\xi_{ij})\), \(i\) the amplification index and \(j\) the copy index. The map \((\xi_{ij})\mapsto\sqrt m\,(\xi_{ij})\) is a unitary of \(H_m^m\) onto \(L^2(P)\), because \(\|(\xi_{ij})\|_{L^2(P)}^2=\frac1m\sum_{i,j}\|\xi_{ij}\|_2^2\); under it, the matrix action of \(P\) is left multiplication. The upper right block of \([Y^{(m)},L_s]\), with respect to the splitting of the index \(i\) into two halves, is \([Y^{(h_0)},X']\); compressing to the first \(h\) rows and the first \(r\) copies gives \([Y^{(h)},X]\). So \[ \begin{aligned} &\|[Y^{(h)},X]\|\\ &\quad \le\|[Y^{(m)},L_s]\| . \end{aligned}\tag{4.2} \]

Step 2: averaging over \(R\). Let \(R\subseteq M\) be as in Lemma 1.2, with stages \(F_n\). Average \(Y\) over the compact unitary group of \(F_n\) acting on \(H_m\) by common left multiplication, \(Y_n=\int L_vYL_v^*\,dv\), and let \(Y_0\) be a weak operator cluster point of \((Y_n)\). Since \(\|L_vYL_v^*-Y\|=\|[L_v,Y]\|\le\delta\), also \(\|Y_n-Y\|\le\delta\) and \(\|Y_0-Y\|\le\delta\). An average over a later stage commutes with the earlier stages, so \(Y_0\) commutes with \(\bigcup_nF_n\) and, by strong density, with \(R\). Finally \(g_M(Y_0)\le g_M(Y)+2\|Y_0-Y\|\le3\delta\).

Step 3: the corner estimate in \(P\). On \(L^2(P)=H_m^m\) put \(S=Y_0^{(m)}\) and \(Q=M_m(R)\). An element of \(P\) commuting with the matrix units has constant diagonal and zero off-diagonal entries, so \(Q'\cap P=(R'\cap M)\otimes1=\mathbb C1\); thus \(Q\subseteq P\) is an irreducible inclusion of II\(_1\) factors, and \(P\) has separable predual. The operator \(S\) commutes with \(L_Q\), because it is \(Y_0\) on every row and \(Y_0\) commutes with \(R\). For \(e_0=e_{11}\otimes1\), the space \(e_0L^2(P)\) is one row \(H_m\), \(e_0Pe_0\cong M\) acts there by common left multiplication, and \(S\) commutes with \(L_{e_0}\); so \(\|[S,L_z]\|\le3\delta\|z\|\) for \(z\in e_0Pe_0\). If \(f\in Q\) is a projection with \(\tau_P(f)\le1/m=\tau_P(e_0)\), the comparison theorem and Proposition 14.2 of the projections lesson give a unitary \(v\in Q\) with \(f\le ve_0v^*\); conjugating by \(L_v\), which commutes with \(S\), \[ \|[S,L_z]\|\le3\delta\|z\|\tag{4.3} \] for every projection \(f\in Q\) with \(\tau_P(f)\le1/m\) and every \(z\in fPf\).

Step 4: the ultrapower. Let \(\widehat S=P_{\mathcal C}S^\omega|_{\mathcal C}\) on \(\mathcal C=L^2(P^\omega)\). By Lemma 3.1(2), \(\widehat S\) commutes with \(L_{Q^\omega}\). Put \(t=1/(2m)\). We claim that \[ \|[\widehat S,L_z]\|\le3\delta\|z\|\tag{4.4} \] for every projection \(e\in Q^\omega\) with \(\tau(e)=t\) and every \(z\in eP^\omega e\). Take a bounded self-adjoint representing sequence \((a_i)\) of \(e\) in \(Q\) and put \(f_i=\mathbf 1_{[1/2,\infty)}(a_i)\in Q\). The scalar inequality \(|x-\mathbf 1_{[1/2,\infty)}(x)|\le2|x^2-x|\) gives \(\|f_i-a_i\|_2\le2\|a_i^2-a_i\|_2\to0\) along \(\omega\), so \((f_i)\) represents \(e\) and \(\tau(f_i)\to t<1/m\); replacing \(f_i\) by \(0\) off a set in \(\omega\), we may assume \(\tau_P(f_i)\le1/m\) for all \(i\). Given \(z\in eP^\omega e\) and \(\eta>0\), the quotient norm of the C*-algebra \(P^\omega\) provides a representing sequence \((b_i)\) with \(\sup_i\|b_i\|\le\|z\|+\eta\); then \(z_i=f_ib_if_i\) also represents \(z\), and by (4.3) \(\|[S,L_{z_i}]\|\le3\delta(\|z\|+\eta)\) for every \(i\). This bound holds for \([S^\omega,L_{(z_i)}]\), and its compression to the reducing subspace \(\mathcal C\) is \([\widehat S,L_z]\) by Lemma 3.1(1). Let \(\eta\to0\).

Step 5: the free corner estimate. Lemma 2.1, with \(Q\) generated by the full matrix algebras \(M_m(F_n)\), gives \(D=uQu^*\subseteq Q^\omega\), a II\(_1\) factor generated by increasing full matrix algebras of unbounded sizes and free from \(P\). Let \(\mathcal L=W^*(D,P)\subseteq P^\omega\), with the restriction of the trace of \(P^\omega\), and \(G=P_{L^2(\mathcal L)}\widehat S|_{L^2(\mathcal L)}\). The subspace \(L^2(\mathcal L)\subseteq\mathcal C\) reduces left multiplication by \(\mathcal L\). Hence \(G\) commutes with \(L_D\), and (4.4) gives \(\|[G,L_x]\|\le3\delta\|x\|\) for \(x\in e\mathcal Le\) and every projection \(e\in D\) of trace \(t=1/N\), \(N=2m\ge2\). The standing hypotheses of the free corner estimate hold with \(A_0=P\), and Theorem 1.1 of the word structure lesson gives \[ \|P_{L^2(P)}[G,L_s]|_{L^2(P)}\|\le3c_1\delta . \] By Lemma 3.1(3), \(L^2(P)\) sits in \(L^2(\mathcal L)\) as the constant vectors, and for \(\xi,\eta\in L^2(P)\), \[ \begin{aligned} &\langle[G,L_s]\xi,\eta\rangle=\langle S^\omega(s\xi),\eta\rangle-\langle S^\omega\xi,s\eta\rangle\\ &\quad =\langle Ss\xi,\eta\rangle-\langle S\xi,s\eta\rangle=\langle[S,L_s]\xi,\eta\rangle . \end{aligned} \] So \(\|[S,L_s]\|\le3c_1\delta\) on \(L^2(P)\), and \(\|[Y^{(m)},L_s]\|\le\|[S,L_s]\|+2\|Y^{(m)}-S\|\), which is at most \(3c_1\delta+2\delta\). With (4.2) this proves (4.1). \(\square\)

5. Exercises

Exercise 5.1. Verify that the matrix \(s\) of Step 1 satisfies \(s^2=1\), using \((1-X'X'^*)^{1/2}X'=X'(1-X'^*X')^{1/2}\), and prove this identity.

Exercise 5.2. Show that \(M_m(R)'\cap M_m(M)=(R'\cap M)\otimes1\).

Exercise 5.3. Prove \(|x-\mathbf 1_{[1/2,\infty)}(x)|\le2|x^2-x|\) for real \(x\).

Exercise 5.4. Let \(u\) be a Haar unitary with \(W^*(u)\) free from \(P\). Show that \(uPu^*\) is free from \(P\), and that \(uPu^*\) does not commute with \(P\) when \(P\ne\mathbb C1\).

6. Solutions

Solution 5.1. For polynomials \(\phi\), \(\phi(X'X'^*)X'=X'\phi(X'^*X')\), by induction on the degree; uniform approximation of \(\sqrt{1-x}\) on \([0,1]\) by polynomials gives the identity. Then the \((1,1)\) entry of \(s^2\) is \((1-X'X'^*)+X'X'^*=1\), the \((2,2)\) entry is \(X'^*X'+(1-X'^*X')=1\), and the \((1,2)\) entry is \((1-X'X'^*)^{1/2}X'\) minus \(X'(1-X'^*X')^{1/2}\), which is \(0\); similarly for \((2,1)\).

Solution 5.2. An element \(x=(x_{ij})\) commuting with the matrix units \(e_{kl}\otimes1\) satisfies \(x_{ij}=\delta_{ij}x_{11}\). It commutes with \(M_m(R)\) exactly when moreover \(x_{11}\) commutes with \(R\).

Solution 5.3. For \(x\ge\frac12\): \(|x-1|\le2|x||x-1|\) because \(2|x|\ge1\). For \(x<\frac12\): \(|x|\le2|x||x-1|\) because \(2|x-1|>1\).

Solution 5.4. Freeness is Lemma 4.1 of the free words lesson with \(B=P\). Let \(a\in P\) be centred and nonzero, and \(a'=uau^*\). The word \(a\,a'\,a^*\,a'^*\) is a centred alternating word, so its trace is \(0\). If \(a'\) commuted with \(a^*\), the trace would equal \(\tau(aa^*\,a'a'^*)\), and freeness gives \(\tau(xy)=\tau(x)\tau(y)\) for \(x\in P\), \(y\in uPu^*\) (expand both into scalar and centred parts); so it would be \(\tau(aa^*)^2>0\).

References