A circle family and a near inclusion

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

If a unitary \(u\) satisfies \(uMu^*\subseteq N\), then every contraction \(x\in M\) lies within \(2\|u-I\|\) of the element \(uxu^*\) of \(N\), so \(\gamma(M,N)\le2\|u-I\|\). Theorem 2.4 of Close operator algebras says that the converse fails without a uniform rate: one-sided gaps can tend to \(0\) while every implementing unitary stays far from \(I\). For an injective source algebra \(M\) the converse holds with a linear bound: if \(\gamma(M,N)<\gamma<1/100\), Christensen's theorem gives a unitary \(u\) with \(uMu^*\subseteq N\) and \(\|u-I\|\le150\gamma\) (Theorem 3.1 of Averaging and the completely positive criterion). This lesson and Almost commuting matrices and small embeddings prove Theorem 2.4 [OpenAI-289c].

The idea is to encode a pair of unitary matrices \(U,V\) in a pair of von Neumann algebras. The present lesson builds, for every such pair, algebras \(M\) and \(N\) with \(W^*MW\subseteq N\) for an explicit unitary \(W\), and \[ \gamma(M,N)\le C\,\|UV-VU\| \] with an absolute constant \(C\) (Proposition 3.2). The next lesson shows that, for Voiculescu's almost commuting matrices, a unitary \(u\) with \(uMu^*\subseteq N\) and \(\|u-I\|\) small would produce a commuting pair of unitaries near \((U,V)\), which does not exist.

The encoding needs representations of the free group that depend on a point \(z\) of the unit circle through Fourier coefficients with uniformly summable moments, so that \(z\) can be replaced by a unitary matrix (Section 1). They come from the deformation of A free-group deformation and close separable algebras by a twist along the homomorphism that counts generators.

We use: Close operator algebras (Definition 1.2); A free-group deformation and close separable algebras (Lemmas 2.1 and 4.1, and the formulas for \(T\) in the proof of Proposition 2.2); and the double commutant theorem.

1. A circle family of representations

Let \(F\) be the free group on \(s_1,s_2,\dots\), let \(K=\ell^2(F)\) with basis \((\delta_x)\), and let \(\lambda_F,\rho_F\) be the left and right regular representations of \(F\). Let \(p_e\) be the projection onto \(\mathbb C\delta_e\) and \(\mathbb T=\{z\in\mathbb C:|z|=1\}\). Fix \(0<t<1\) and \(r=\sqrt{1-t^2}\), and let \(\pi_t\) and the vectors \(\xi_x\) be as in Lemma 2.1 of A free-group deformation and close separable algebras, with \(F\) in place of the group called \(G\) there.

Lemma 1.1. Let \(g\in F\) have prefixes \(g_0=e,g_1,\dots,g_l=g\), let \(M_g\) be the span of \(f_a=\delta_{g_a}\) for \(0\le a\le l\), and let \(T_g=\pi_t(g)\lambda_F(g)^*\). Then \(T_g\) is the identity on \(M_g^\perp\), maps \(M_g\) into itself, and \[ |\langle T_gf_b,f_a\rangle|\le t^{|a-b|}\qquad(0\le a,b\le l). \]

Proof. For \(g=e\), \(T_g=I\). Otherwise the proof of Proposition 2.2 of that lesson gives \(T_g\delta_y=\delta_y\) for \(y\) not a prefix of \(g\), \(T_gf_b=r\xi_{g_b}-tf_{b+1}\) for \(b<l\), and \(T_gf_l=\xi_g\), where \[ \xi_{g_b}=t^bf_0+r\sum_{a=1}^bt^{b-a}f_a . \] So the nonzero entries of column \(b\) are \(rt^b\) or \(t^l\) in row \(0\), \(r^2t^{b-a}\) or \(rt^{l-a}\) in rows \(1\le a\le b\), and \(-t\) in row \(b+1\). Since \(r\le1\), each has absolute value at most \(t^{|a-b|}\). \(\square\)

Let \(q:F\to\mathbb Z\) be the homomorphism with \(q(s_j)=1\) for all \(j\), and for \(z\in\mathbb T\) let \(D_z\) be the diagonal unitary \(D_z\delta_x=z^{-q(x)}\delta_x\). Put \[ c_0=\frac{1+t}{1-t},\qquad c_k=\frac{2t^{|k|}}{1-t}\quad(k\ne0). \]

Proposition 1.2 (the circle family). For \(z\in\mathbb T\) and \(g\in F\) put \[ \pi_z(g)=z^{q(g)}D_z\pi_t(g)D_z^* . \]

  1. Each \(\pi_z\) is a unitary representation of \(F\).
  2. There are operators \(A_k(g)\), for \(k\in\mathbb Z\) and \(g\in F\), not depending on \(z\), with \(A_k(g)=0\) for \(|k|>|g|\), \(\|A_k(g)\|\le c_k\), and \[ \pi_z(g)=\sum_{k\in\mathbb Z}z^kA_k(g). \]
  3. \(\pi_z(s_j)\delta_e=tz\delta_e+r\delta_{s_j}\), and \(\pi_z(s_j)\to tz\,p_e\) in the weak operator topology as \(j\to\infty\).

Proof. (1) \(g\mapsto z^{q(g)}\) is a homomorphism into \(\mathbb T\), and \(g\mapsto D_z\pi_t(g)D_z^*\) is a unitary representation; their product is one.

(2) Since \(D_z\lambda_F(g)D_z^*\delta_x=z^{q(x)-q(gx)}\delta_{gx}\)\(=z^{-q(g)}\lambda_F(g)\delta_x\), \[ \pi_z(g)=\bigl(D_zT_gD_z^*\bigr)\lambda_F(g). \] The diagonal unitary \(D_z\) preserves \(M_g\) and \(M_g^\perp\), so \(D_zT_gD_z^*\) is the identity on \(M_g^\perp\), and on \(M_g\) its entry in row \(a\), column \(b\) is \(z^{q(g_b)-q(g_a)}\langle T_gf_b,f_a\rangle\). Consecutive prefixes differ by one letter, so \(|q(g_b)-q(g_a)|\le|a-b|\). Let \(B_k(g)\) be the operator whose matrix on \(M_g\) keeps the entries with \(q(g_b)-q(g_a)=k\) and replaces the others by \(0\), plus the identity of \(M_g^\perp\) when \(k=0\). Then \(D_zT_gD_z^*=\sum_kz^kB_k(g)\), a finite sum, and we put \(A_k(g)=B_k(g)\lambda_F(g)\).

To bound \(\|B_k(g)\|\), recall the Schur test: if every row and every column of a matrix \((m_{ab})\) has \(\sum|m_{ab}|\le\sigma\), then its norm is at most \(\sigma\), because \[ \begin{aligned} &\Bigl|\sum_{a,b}m_{ab}x_b\bar y_a\Bigr|\\ &\quad \le\sum_{a,b}|m_{ab}|\,\frac{|x_b|^2+|y_a|^2}2\le\sigma \end{aligned} \] for unit vectors \(x,y\). In a row \(a\) of the matrix of \(B_k(g)\) on \(M_g\), the nonzero entries lie in columns \(b\) with \(|a-b|\ge|k|\); there are at most two columns at each distance \(m\ge1\), one at distance \(0\), and the entries at distance \(m\) are at most \(t^m\) by Lemma 1.1. So the row sums, and likewise the column sums, are at most \(1+2t/(1-t)=c_0\) for \(k=0\) and \(2\sum_{m\ge|k|}t^m=c_k\) for \(k\ne0\). On \(M_g^\perp\), \(B_0(g)\) is the identity and \(c_0\ge1\). Hence \(\|A_k(g)\|=\|B_k(g)\|\le c_k\), and \(B_k(g)=0\) when \(|k|>l\).

(3) The operator \(T_{s_j}\) is the identity off the span of \(\delta_e\) and \(\delta_{s_j}\), and \(T_{s_j}\delta_{s_j}=\xi_{s_j}=t\delta_e+r\delta_{s_j}\). So \[ \begin{aligned} \pi_z(s_j)\delta_e&=D_zT_{s_j}D_z^*\delta_{s_j}\\ &=z\,D_z(t\delta_e+r\delta_{s_j})\\ &=tz\delta_e+r\delta_{s_j}. \end{aligned} \] For \(x\ne e\) and all \(j\) with \(s_j\ne x^{-1}\), \(s_jx\notin\{e,s_j\}\), so \(\pi_z(s_j)\delta_x=\delta_{s_jx}\). For fixed \(x,y\in F\), \(\langle\pi_z(s_j)\delta_x,\delta_y\rangle\) therefore tends to \(tz\) if \(x=y=e\) and to \(0\) otherwise, which are the matrix entries of \(tz\,p_e\). The operators \(\pi_z(s_j)\) are unitaries, so convergence of the matrix entries implies weak operator convergence. \(\square\)

The constants enter through the two sums \[ \begin{gathered} S_0=\sum_kc_k=\frac{1+t}{1-t}+\frac{4t}{(1-t)^2},\\ S_1=\sum_k|k|\,c_k=\frac{4t}{(1-t)^3}. \end{gathered} \] For \(t=1/2\): \(c_0=3\), \(c_k=2^{2-|k|}\) for \(k\ne0\), \(S_0=11\) and \(S_1=16\).

2. Substituting unitary matrices for the circle parameter

Let \(d\ge1\) and let \(U,V\) be unitaries on \(\mathbb C^d\). On \(L=K\otimes K\otimes\mathbb C^d\) define, for \(g\in F\), \[ \begin{gathered} a(g)=\sum_kA_k(g)\otimes I\otimes U^k,\\ b(g)=\sum_kI\otimes A_k(g)\otimes V^k . \end{gathered} \] Both sums are finite.

Lemma 2.1. \(a\) and \(b\) are unitary representations of \(F\) on \(L\), and as \(j\to\infty\), in the weak operator topology, \[ \begin{gathered} a(s_j)\to t\,p_e\otimes I\otimes U,\\ b(s_j)\to t\,I\otimes p_e\otimes V . \end{gathered} \]

Proof. Write \(U=\sum_zzP_z\), a finite sum over the eigenvalues \(z\in\mathbb T\) of \(U\), with spectral projections \(P_z\). Then \(U^k=\sum_zz^kP_z\), and Proposition 1.2(2) gives \[ a(g)=\sum_z\pi_z(g)\otimes I\otimes P_z . \] The \(P_z\) are pairwise orthogonal with sum \(I\), so \(a\) is a unitary representation because each \(\pi_z\) is one. By Proposition 1.2(3), \(\pi_z(s_j)\otimes I\otimes P_z\to tz\,p_e\otimes I\otimes P_z\) weakly: the matrix entries between elementary tensors converge, and the operators are bounded. Summing over \(z\) gives the limit \(t\,p_e\otimes I\otimes U\). The same argument with the eigenvalues of \(V\) applies to \(b\). \(\square\)

The ranges of \(a\) and \(b\) need not commute (Exercise 4.4); this is the point of the construction. Their failure to commute is controlled by the following elementary estimate.

Lemma 2.2. For unitaries \(U,V\) and integers \(m,n\), \[ \begin{aligned} &\|U^mV^n-V^nU^m\|\\ &\quad \le|m|\,|n|\,\|UV-VU\| . \end{aligned} \]

Proof. Write \([X,Y]=XY-YX\) and \(\nu=\|[U,V]\|\). Since \([U^{-1},V]=-U^{-1}[U,V]U^{-1}\), \(\|[U^{\pm1},V]\|=\nu\). For \(m\ge1\), \([U^m,V]=\sum_{i=0}^{m-1}U^i[U,V]U^{m-1-i}\), so \(\|[U^m,V]\|\le m\nu\), and the same holds with \(U^{-1}\) in place of \(U\). Hence \(\|[U^m,V]\|\le|m|\nu\) for every integer \(m\). Applying this to the unitaries \(V\) and \(U^m\) gives \(\|[U^m,V^n]\|\le|n|\,\|[U^m,V]\|\le|m||n|\nu\). \(\square\)

3. The two algebras and the near inclusion

Let \(G=F\times F\), \(S=\ell^2(G)\) with basis \((\delta_x)_{x\in G}\), and let \(\lambda,\rho\) be the left and right regular representations of \(G\) on \(S\). On \(H=S\otimes L\), every operator \(T\) has blocks \(T_{x,y}\in B(L)\), defined by \(\langle T_{x,y}v,w\rangle=\langle T(\delta_y\otimes v),\delta_x\otimes w\rangle\). For \(x=(x_1,x_2)\in G\) put \[ f(x)=a(x_1)\,b(x_2), \] a unitary of \(L\), and let \(W\) be the block diagonal unitary of \(H\) with \(W(\delta_x\otimes v)=\delta_x\otimes f(x)v\). Define \[ \begin{gathered} M=W\bigl(\lambda(G)''\otimes I_L\bigr)W^*,\\ N=\bigl(\rho(G)\otimes I_L\bigr)', \end{gathered} \] where \(\lambda(G)''\otimes I_L\) denotes the set of operators \(X\otimes I_L\) with \(X\in\lambda(G)''\).

Lemma 3.1. \(M\) and \(N\) are von Neumann algebras on \(H\) containing \(I_H\), and \(W^*MW\subseteq N\). An operator \(T\) on \(H\) lies in \(N\) exactly when \(T_{xq,yq}=T_{x,y}\) for all \(x,y,q\in G\).

Proof. \(N\) is the commutant of a self-adjoint set, so it is a von Neumann algebra. An operator commuting with all \(I_S\otimes Y\), \(Y\in B(L)\), has the form \(X\otimes I_L\): with matrix units of \(L\) relative to an orthonormal basis, this is the block argument in the proof of Lemma 4.1(1) of A free-group deformation and close separable algebras. It then commutes with \(\lambda(G)'\otimes I_L\) exactly when \(X\in\lambda(G)''\). So \(\lambda(G)''\otimes I_L\) is the commutant of \((\lambda(G)'\otimes I_L)\cup(I_S\otimes B(L))\), a von Neumann algebra, and so is its unitary conjugate \(M\). Since \(\rho(G)\subseteq\lambda(G)'\), every \(X\in\lambda(G)''\) commutes with \(\rho(G)\); so \(W^*MW=\lambda(G)''\otimes I_L\subseteq N\). Finally, \((\rho(q)\otimes I_L)(\delta_y\otimes v)=\delta_{yq^{-1}}\otimes v\), so \(T\) commutes with \(\rho(q)\otimes I_L\) exactly when \(T_{xq^{-1},yq^{-1}}=T_{x,y}\) for all \(x,y\). \(\square\)

So \(M\) is spatially embedded in \(N\) by \(W^*\). The next proposition shows that \(M\) is also nearly included in \(N\) when \(U\) and \(V\) nearly commute.

Proposition 3.2. Let \(\nu=\|UV-VU\|\) and \(C=2S_0^2S_1^2\). For every \(X\in\lambda(G)''\) there is \(Y\in N\) with \[ \bigl\|W(X\otimes I_L)W^*-Y\bigr\|\le C\nu\|X\| . \] For \(X=\lambda(g)\), \(g\in G\), one can take \(Y=\lambda(g)\otimes f(g)\). In particular \(\gamma(M,N)\le C\nu\), and \(C=61952\) for \(t=1/2\).

Proof. Write \(X_{x,y}=\langle X\delta_y,\delta_x\rangle\). Since \(X\) commutes with \(\rho(G)\), \(X_{xq,yq}=X_{x,y}\) for all \(x,y,q\). The operator \(Z=W(X\otimes I_L)W^*\) has blocks \[ Z_{x,y}=X_{x,y}\,f(x)f(y)^* . \] Expansion. Identify \(H\) with \((S\otimes K\otimes K)\otimes\mathbb C^d\). For \(k,m\in\mathbb Z\) let \(\Delta_{k,m}\) be the block diagonal operator on \(S\otimes K\otimes K\) whose block at \(x\) is \(A_k(x_1)\otimes A_m(x_2)\); then \(\|\Delta_{k,m}\|\le c_kc_m\). Put \[ B_{kmph}=\Delta_{k,m}(X\otimes I_K\otimes I_K)\Delta_{h,p}^*, \] so \(\|B_{kmph}\|\le c_kc_mc_pc_h\|X\|\), and the block of \(B_{kmph}\) at \((x,y)\) is \[ \begin{gathered} (A_k(x_1)\otimes A_m(x_2))\,X_{x,y}\\ \cdot(A_h(y_1)\otimes A_p(y_2))^* . \end{gathered} \] Expanding \(f(x)=a(x_1)b(x_2)\) and \(f(y)^*=b(y_2)^*a(y_1)^*\), \[ Z=\sum_{k,m,p,h}B_{kmph}\otimes U^kV^{m-p}U^{-h}. \] The series converges absolutely in norm, since \(\sum c_kc_mc_pc_h=S_0^4\). Both sides have the same blocks, because each block of the right side is a finite sum equal to \(Z_{x,y}\); so they are equal.

The reordered operator. Define \[ Y=\sum_{k,m,p,h}B_{kmph}\otimes U^{k-h}V^{m-p}, \] again absolutely convergent. Collecting terms, its blocks are \[ \begin{aligned} Y_{x,y}&=X_{x,y}\,a(x_1)a(y_1)^*\,b(x_2)b(y_2)^*\\ &=X_{x,y}\,f(xy^{-1}), \end{aligned} \] using that \(a\) and \(b\) are representations. Since \((xq)(yq)^{-1}=xy^{-1}\) and \(X_{xq,yq}=X_{x,y}\), Lemma 3.1 gives \(Y\in N\).

The estimate. By Lemma 2.2, \[ \begin{aligned} &\|U^kV^{m-p}U^{-h}-U^{k-h}V^{m-p}\|\\ &\quad =\|[V^{m-p},U^{-h}]\|\\ &\quad \le|m-p|\,|h|\,\nu . \end{aligned} \] Hence \(\|Z-Y\|\le\nu\|X\|\sum c_kc_mc_pc_h|m-p||h|\). Since \(|m-p|\le|m|+|p|\), the sum is at most \(S_0S_1\cdot2S_0S_1=C\).

For \(X=\lambda(g)\), \(X_{x,y}=1\) if \(x=gy\) and \(0\) otherwise, so \(Y_{x,y}=f(g)\) if \(x=gy\) and \(0\) otherwise; that is, \(Y=\lambda(g)\otimes f(g)\). Finally, every contraction of \(M\) is \(W(X\otimes I_L)W^*\) with \(X\in\lambda(G)''\) and \(\|X\|\le1\), so its distance to \(N\) is at most \(C\nu\). For \(t=1/2\), \(C=2\cdot11^2\cdot16^2=61952\). \(\square\)

If \(U\) and \(V\) commute, then \(f\) is a representation of \(G\), each \(W(\lambda(g)\otimes I_L)W^*=\lambda(g)\otimes f(g)\) lies in \(N\), and \(M\subseteq N\) (Exercise 4.2). Proposition 3.2 is a quantitative form of this, valid on the whole unit ball of \(M\) and not only on the group unitaries.

4. Exercises

Exercise 4.1. Show that \(\langle\pi_z(g)\delta_e,\delta_e\rangle=z^{q(g)}t^{|g|}\) for \(g\in F\) and \(z\in\mathbb T\).

Exercise 4.2. Suppose \(UV=VU\). Show that \(f\) is a unitary representation of \(G\), that \(W(\lambda(g)\otimes I_L)W^*=\lambda(g)\otimes f(g)\), and that \(M\subseteq N\).

Exercise 4.3. Show that \(\|\pi_z(s_j)-tz\,p_e\|\ge1\) for every \(j\), so the limit in Proposition 1.2(3) is not a norm limit.

Exercise 4.4. For \(v,w\in\mathbb C^d\), show that \[ \begin{aligned} &\bigl\langle[a(s_1),b(s_1)]\,\delta_e\otimes\delta_e\otimes v,\ \delta_e\otimes\delta_e\otimes w\bigr\rangle\\ &\quad =t^2\langle(UV-VU)v,w\rangle . \end{aligned} \] Deduce \(\|a(s_1)b(s_1)-b(s_1)a(s_1)\|\)\(\ge t^2\|UV-VU\|\).

Exercise 4.5. Show that \(\lambda(G)''\) commutes with \(\rho(G)\) directly from \(\rho(G)\subseteq\lambda(G)'\), and that \(\lambda(g)\otimes f(g)\in N\) for every \(g\in G\) even when \(U\) and \(V\) do not commute.

5. Solutions

Solution 4.1. \(D_z^*\delta_e=\delta_e\) because \(q(e)=0\), so \(\pi_z(g)\delta_e=z^{q(g)}D_z\xi_g\). Since \(D_z^*\delta_e=\delta_e\), \(\langle D_z\xi_g,\delta_e\rangle=\langle\xi_g,\delta_e\rangle=t^{|g|}\) by Lemma 2.1(1) of A free-group deformation and close separable algebras.

Solution 4.2. If \(UV=VU\), then every \(a(g)\) commutes with every \(b(h)\), since \(A_k(g)\otimes I\) and \(I\otimes A_m(h)\) commute and \(U^k\) commutes with \(V^m\). So \(f(xy)=a(x_1y_1)b(x_2y_2)\)\(=a(x_1)b(x_2)a(y_1)b(y_2)\)\(=f(x)f(y)\). For \(g\in G\), \(W(\lambda(g)\otimes I)W^*(\delta_x\otimes v)\)\(=\delta_{gx}\otimes f(gx)f(x)^*v\)\(=\delta_{gx}\otimes f(g)v\). The operator \(\lambda(g)\otimes f(g)\) commutes with \(\rho(G)\otimes I_L\), so it lies in \(N\). By the double commutant theorem, \(\lambda(G)''\) is the weak operator closure of the span of \(\lambda(G)\); the map \(X\mapsto W(X\otimes I_L)W^*\) is weakly continuous and \(N\) is weakly closed, so \(M\subseteq N\).

Solution 4.3. For \(x\ne e\), \(p_e\delta_x=0\), so \((\pi_z(s_j)-tz\,p_e)\delta_x=\pi_z(s_j)\delta_x\) is a unit vector.

Solution 4.4. From the proof of Lemma 2.1 and Proposition 1.2(3), \(a(s_1)(\delta_e\otimes\eta\otimes v)\)\(=t\,\delta_e\otimes\eta\otimes Uv+r\,\delta_{s_1}\otimes\eta\otimes v\) for every \(\eta\in K\), and similarly for \(b(s_1)\) in the second factor. Applying \(b(s_1)\) and then \(a(s_1)\) to \(\delta_e\otimes\delta_e\otimes v\), the component along \(\delta_e\otimes\delta_e\otimes\mathbb C^d\) is \(t^2\,\delta_e\otimes\delta_e\otimes UVv\); in the other order it is \(t^2\,\delta_e\otimes\delta_e\otimes VUv\). Taking the inner product with \(\delta_e\otimes\delta_e\otimes w\) gives the identity, and the norm bound follows by taking the supremum over unit vectors \(v,w\).

Solution 4.5. Every \(\rho(q)\) commutes with \(\lambda(G)\), so \(\rho(q)\in\lambda(G)'\), and every element of \(\lambda(G)''=(\lambda(G)')'\) commutes with \(\rho(q)\). The operator \(\lambda(g)\otimes f(g)\) has blocks \(f(g)\) at the pairs \((gy,y)\) and \(0\) elsewhere; these are invariant under \((x,y)\mapsto(xq,yq)\), so it lies in \(N\) by Lemma 3.1. What fails without commutation is the identity \(W(\lambda(g)\otimes I_L)W^*=\lambda(g)\otimes f(g)\).

References