A free-group deformation and close separable algebras

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

Convergent sequences and a relative tensor norm reduced Theorem 2.3 of Close operator algebras to a construction inside one Hilbert space \(L\): a unital C*-algebra \(D\) with trivial center, and two unital subalgebras \(P,Q\subseteq D\), conjugate by a unitary close to \(I\), such that \(P\) has the spatial tensor-norm identity relative to \(D\) and \(Q\) does not. This lesson carries out the construction with the free group on countably many generators and so proves Theorem 2.3 [OpenAI-289b, Sections 3 to 5].

The algebra \(P\) is generated by the operators \(\lambda(g)\otimes\lambda(g)\), where \(\lambda\) is the left regular representation. For \(Q\), the second factor is replaced by a unitary representation \(\pi_t\) that is uniformly close to \(\lambda\) on the whole group (Proposition 2.2) but has diagonal coefficient \(t\) at every generator, where \(\lambda\) has coefficient \(0\). The identity notices this coefficient: summed over \(m\) generators, a product with the commutant of \(D\) has norm at least \(mt\), while the corresponding tensor has norm at most \(2\sqrt m\) (Proposition 5.2).

We use: Close operator algebras (Theorem 2.3, proved here) and Convergent sequences and a relative tensor norm (Proposition 1.1, Definition 2.1 and Corollary 2.3).

1. Reduced words and the regular representations

Let \(G\) be the free group on the generators \(s_1,s_2,\dots\). Every element \(x\in G\) is a unique reduced word \(x=a_1a_2\cdots a_l\) in the letters \(s_j^{\pm1}\), with no letter next to its inverse; \(|x|=l\) is its length, and \(e\) is the empty word. The prefixes of \(x\) are \[ x_k=a_1\cdots a_k\qquad(0\le k\le l), \] so \(x_0=e\) and \(x_l=x\). For \(x\ne e\) let \(\bar x=x_{l-1}\) be \(x\) with its last letter removed.

Lemma 1.1. Let \(x,y\in G\), let \(a\) be their longest common prefix, and write \(x=ab\), \(y=ac\). Then \(x^{-1}y=b^{-1}c\) as reduced words, so \(|x^{-1}y|=|b|+|c|\).

Proof. If \(b\) or \(c\) is empty this is clear. Otherwise the first letters of \(b\) and \(c\) differ, by maximality of \(a\). The word \(b^{-1}c\) has the inverse of the first letter of \(b\) next to the first letter of \(c\); these are not inverse to each other, and the words \(b^{-1}\) and \(c\) are reduced. \(\square\)

Let \(K=\ell^2(G)\), with orthonormal basis \((\delta_x)_{x\in G}\). It is separable because \(G\) is countable. The left and right regular representations are \[ \lambda(g)\delta_x=\delta_{gx},\qquad\rho(g)\delta_x=\delta_{xg^{-1}} . \] Both are unitary representations of \(G\), and \(\lambda(g)\rho(h)=\rho(h)\lambda(g)\) for all \(g,h\), since both sides send \(\delta_x\) to \(\delta_{gxh^{-1}}\). In particular \(\rho(G)\subseteq\lambda(G)'\).

2. A deformation of the left regular representation

Fix \(0<t<1\) and put \(r=\sqrt{1-t^2}\). For \(x\in G\) with reduced word of length \(l\) and prefixes \(x_k\), define \[ \xi_x=t^l\delta_e+r\sum_{k=1}^lt^{l-k}\delta_{x_k}. \] Then \(\xi_e=\delta_e\), and for \(x\ne e\) \[ \xi_x=t\,\xi_{\bar x}+r\,\delta_x . \] Each \(\xi_x\) is a unit vector: \(\|\xi_x\|^2\)\(=t^{2l}+r^2(1+t^2+\dots+t^{2l-2})\)\(=t^{2l}+(1-t^{2l})=1\).

Lemma 2.1.

  1. \(\langle\xi_x,\xi_y\rangle=t^{|x^{-1}y|}\) for all \(x,y\in G\).
  2. The linear span of the \(\xi_x\) equals the linear span of the \(\delta_x\), so it is dense in \(K\).
  3. For each \(g\in G\) there is a unique unitary \(\pi_t(g)\) of \(K\) with \(\pi_t(g)\xi_x=\xi_{gx}\) for all \(x\), and \(\pi_t\) is a unitary representation of \(G\).
  4. \(\pi_t(s_j)\delta_e=t\delta_e+r\delta_{s_j}\); in particular \(\langle\pi_t(s_j)\delta_e,\delta_e\rangle=t\) for every \(j\).

Proof. (1) Write \(x=ab\) and \(y=ac\) as in Lemma 1.1, with \(|a|=k\). Splitting the sum defining \(\xi_x\) at the prefix \(a\) gives \[ \xi_x=t^{|b|}\xi_a+\eta_x, \] where \(\eta_x\) is supported on the prefixes of \(x\) longer than \(a\); similarly \(\xi_y=t^{|c|}\xi_a+\eta_y\). The vector \(\xi_a\) is supported on the prefixes of \(a\). A prefix of \(x\) longer than \(a\) begins with \(a\) followed by the first letter of \(b\), and a prefix of \(y\) longer than \(a\) begins with \(a\) followed by the first letter of \(c\); these letters differ. So the supports of \(\xi_a\), \(\eta_x\) and \(\eta_y\) are pairwise disjoint, and \[ \langle\xi_x,\xi_y\rangle=t^{|b|+|c|}\|\xi_a\|^2=t^{|x^{-1}y|} \] by Lemma 1.1.

(2) Each \(\xi_x\) is a finite combination of basis vectors, and conversely \(\delta_e=\xi_e\) and \(\delta_x=(\xi_x-t\xi_{\bar x})/r\) for \(x\ne e\).

(3) For finitely many coefficients \(c_x\), part 1 and \((gx)^{-1}(gy)=x^{-1}y\) give \[ \begin{aligned} \Bigl\|\sum_xc_x\xi_{gx}\Bigr\|^2&=\sum_{x,y}c_x\bar c_yt^{|x^{-1}y|}\\ &=\Bigl\|\sum_xc_x\xi_x\Bigr\|^2 . \end{aligned} \] So \(\sum_xc_x\xi_x\mapsto\sum_xc_x\xi_{gx}\) is a well-defined isometry of the dense span of part 2 onto itself, and it extends uniquely to an isometry \(\pi_t(g)\) of \(K\). On the vectors \(\xi_x\), \(\pi_t(g)\pi_t(h)=\pi_t(gh)\) and \(\pi_t(e)=I\), hence everywhere by continuity. So \(\pi_t(g)\) has the inverse \(\pi_t(g^{-1})\) and is unitary.

(4) \(\pi_t(s_j)\delta_e=\pi_t(s_j)\xi_e=\xi_{s_j}=t\delta_e+r\delta_{s_j}\). \(\square\)

For \(t=0\) the same formulas give \(\xi_x=\delta_x\) and the left regular representation; we put \(\pi_0=\lambda\). Representations of this kind go back to Pytlik and Szwarc. Burger, Ozawa and Thom use a family of them to show that the left regular representation of the free group on countably many generators lies on a path of representations that is continuous in the uniform topology and whose other members are irreducible and pairwise inequivalent [BOT, Theorem 4.6]. The estimate needed here is the following.

Proposition 2.2. For \(0<t<1\) put \[ c(t)=t+t^2+\frac t{1-t}. \] Then \(\|\pi_t(g)-\lambda(g)\|\le c(t)\) for every \(g\in G\). In particular \(c(t)\le4t\) for \(t\le1/2\).

Proof. For \(g=e\) both sides are \(I\). Let \(g\ne e\) have length \(l\) and prefixes \(g_0=e,g_1,\dots,g_l=g\), and put \(T=\pi_t(g)\lambda(g)^*\), so that \(\|\pi_t(g)-\lambda(g)\|=\|T-I\|\). For \(y\ne g\), the element \(u=g^{-1}y\) is not \(e\); if \(a\) is its last letter, then \(\delta_u=(\xi_u-t\xi_{ua^{-1}})/r\) and \[ T\delta_y=\pi_t(g)\delta_u=\frac{\xi_y-t\,\xi_{ya^{-1}}}r . \] Vectors off the prefixes of \(g\). Let \(y\) not be a prefix of \(g\); then \(y\ne e\). Write \(y=bc\) and \(g=bd\) with \(b\) the longest common prefix, so \(c\ne e\). By Lemma 1.1, \(u=d^{-1}c\) as a reduced word, so its last letter is the last letter of \(c\), which is the last letter of \(y\). Hence \(ya^{-1}=\bar y\), and \(T\delta_y=(\xi_y-t\xi_{\bar y})/r=\delta_y\).

Vectors at the prefixes. For \(k<l\), \(u=g^{-1}g_k\) is the inverse of the last \(l-k\) letters of \(g\), whose last letter is the inverse of the \((k+1)\)-st letter of \(g\). So \(g_ka^{-1}=g_{k+1}\), and with \(\xi_{g_{k+1}}=t\xi_{g_k}+r\delta_{g_{k+1}}\), \[ T\delta_{g_k}=\frac{\xi_{g_k}-t\xi_{g_{k+1}}}r=r\,\xi_{g_k}-t\,\delta_{g_{k+1}} . \] Finally \(T\delta_g=\pi_t(g)\delta_e=\xi_g\).

The estimate. Let \(M\) be the span of \(f_k=\delta_{g_k}\), \(0\le k\le l\). Then \(T-I\) vanishes on \(M^\perp\) and maps \(M\) into \(M\), so \(\|T-I\|\) is the norm of \(T-I\) on \(M\). On \(M\) let \(S\) be the shift with \(Sf_k=f_{k+1}\) for \(k<l\) and \(Sf_l=0\), let \(R\) be diagonal with entries \(1,r,\dots,r\), and \(R'\) diagonal with entries \(r,\dots,r,1\). Since \[ \xi_{g_k}=\sum_{j=0}^kt^{k-j}\,R f_j, \] the operator \(X:f_k\mapsto\xi_{g_k}\) is \(X=R(I+N)\) with \(N=\sum_{j=1}^lt^jS^{*j}\). The formulas for \(T\) say \(T=XR'-tS\) on \(M\). Hence \[ T-I=(RR'-I)+RNR'-tS . \] The diagonal entries of \(RR'\) are \(r\) and \(r^2\), so \(\|RR'-I\|\le1-r^2=t^2\). Also \(\|N\|\le\sum_{j\ge1}t^j=t/(1-t)\), and \(\|R\|,\|R'\|,\|S\|\le1\). This gives \(\|T-I\|\le c(t)\). For \(t\le1/2\), \(t/(1-t)\le2t\) and \(t^2\le t/2\). \(\square\)

3. Absorption and free generators

Lemma 3.1 (Fell's absorption). Let \(\sigma\) be a unitary representation of \(G\) on a Hilbert space \(J\), and define \(W_\sigma\) on \(K\otimes J\) by \(W_\sigma(\delta_x\otimes v)=\delta_x\otimes\sigma(x)v\). Then \(W_\sigma\) is unitary and \[ \begin{gathered} W_\sigma(\lambda(g)\otimes I_J)W_\sigma^*=\lambda(g)\otimes\sigma(g)\\ (g\in G). \end{gathered} \]

Proof. \(K\otimes J\) is the orthogonal sum of the subspaces \(\delta_x\otimes J\), and \(W_\sigma\) acts on each as the unitary \(\sigma(x)\); so \(W_\sigma\) is unitary, with \(W_\sigma^*(\delta_x\otimes v)=\delta_x\otimes\sigma(x)^*v\). On \(\delta_x\otimes v\), the left side gives \(\delta_{gx}\otimes\sigma(gx)\sigma(x)^*v=\delta_{gx}\otimes\sigma(g)v\). \(\square\)

Lemma 3.2. For every \(m\ge1\), \[ \Bigl\|\sum_{j=1}^m\lambda(s_j)\Bigr\|\le2\sqrt m . \]

Proof. Let \(F_j\) be the projection onto the closed span of the \(\delta_x\) for which the reduced word \(x\) begins with \(s_j^{-1}\). The \(F_j\) are pairwise orthogonal. Write \(\lambda(s_j)=A_j+B_j\) with \(A_j=\lambda(s_j)(I-F_j)\) and \(B_j=\lambda(s_j)F_j\). If \(x\) does not begin with \(s_j^{-1}\), then \(s_jx\) is reduced as written and begins with \(s_j\). So \(A_j\) maps into the span of the words beginning with \(s_j\), and the ranges of different \(A_j\) are orthogonal. For \(v\in K\), \[ \begin{aligned} \Bigl\|\sum_jA_jv\Bigr\|^2&=\sum_j\|(I-F_j)v\|^2\\ &\le m\|v\|^2 . \end{aligned} \] By the Cauchy–Schwarz inequality and orthogonality of the \(F_j\), \[ \Bigl\|\sum_jB_jv\Bigr\|\le\sum_j\|F_jv\|\le\sqrt m\,\|v\| . \] Adding the two bounds gives the claim. \(\square\)

4. The ambient algebra and two close subalgebras

Let \(L=K\otimes K\), a separable Hilbert space, and let \(E_{xy}\) be the matrix units of \(K\): \(E_{xy}\delta_z=\delta_x\) if \(z=y\), and \(0\) otherwise. For \(a\in\{0,t\}\) and \(g\in G\) put \[ \begin{gathered} Q_a(g)=\lambda(g)\otimes\pi_a(g),\\ P_a=C^*\bigl(Q_a(g):g\in G\bigr). \end{gathered} \] Let \(D\subseteq B(L)\) be the C*-algebra generated by the operators \[ \begin{gathered} \lambda(g)\otimes I,\\ I\otimes\lambda(g),\\ I\otimes\pi_t(g),\\ I\otimes E_{xy}\\ (g,x,y\in G). \end{gathered} \] Since \(Q_a(g)=(\lambda(g)\otimes I)(I\otimes\pi_a(g))\), both \(P_0\) and \(P_t\) are contained in \(D\), and all three algebras contain \(I=Q_a(e)\). Each is generated by a countable set, so each is norm separable.

Lemma 4.1.

  1. \(D'=\{Y\otimes I:Y\in\lambda(G)'\}\).
  2. \(Z(D)=\mathbb CI\).

Proof. (1) For \(x\in G\) let \(J_x:K\to L\) be the isometry \(J_xv=v\otimes\delta_x\). Then \(I\otimes E_{xy}=J_xJ_y^*\), and every \(T\in B(L)\) is determined by the blocks \(T_{xy}=J_x^*TJ_y\). Suppose \(T\) commutes with all \(I\otimes E_{xy}\). Multiplying \(TJ_xJ_y^*=J_xJ_y^*T\) by \(J_u^*\) on the left and by \(J_y\) on the right gives \(T_{ux}=0\) for \(u\ne x\) and \(T_{xx}=T_{yy}\). So all diagonal blocks equal one operator \(Y\), the others vanish, and \(T=Y\otimes I\). Commuting with \(\lambda(g)\otimes I\) then means \(Y\in\lambda(G)'\). Conversely, \(Y\otimes I\) with \(Y\in\lambda(G)'\) commutes with every generator of \(D\).

(2) Let \(Y\otimes I\) be central in \(D\). By part 1, \(Y\) commutes with \(\lambda(G)\). Since \(\rho(G)\subseteq\lambda(G)'\), the operators \(\rho(g)\otimes I\) lie in \(D'\), and the central element \(Y\otimes I\in D\) commutes with them; so \(Y\) commutes with \(\rho(G)\) as well. The vector \(f=Y\delta_e\) therefore satisfies \(\lambda(g)\rho(g)f=Y\lambda(g)\rho(g)\delta_e=f\), that is, \(f(g^{-1}xg)=f(x)\) for all \(g,x\). Every \(x\ne e\) has infinitely many conjugates: if neither \(s_j\) nor \(s_j^{-1}\) occurs in the reduced word \(x\), then \(s_jxs_j^{-1}\) is reduced as written, and different such \(j\), of which there are infinitely many, give different words. A square-summable function constant on these classes vanishes off \(e\), so \(f=\alpha\delta_e\) for some \(\alpha\in\mathbb C\). Then \(Y\delta_x=Y\lambda(x)\delta_e=\lambda(x)f=\alpha\delta_x\) for all \(x\), so \(Y=\alpha I\). \(\square\)

Lemma 4.2. The unitary \(W=W_{\pi_t}W_{\pi_0}^*\) of \(L\), formed with Lemma 3.1 for \(J=K\), satisfies \[ WP_0W^*=P_t,\qquad\|W-I\|\le c(t). \]

Proof. By Lemma 3.1, conjugation by \(W_{\pi_0}^*\) sends \(Q_0(g)=\lambda(g)\otimes\lambda(g)\) to \(\lambda(g)\otimes I\), and conjugation by \(W_{\pi_t}\) sends this to \(Q_t(g)\). So \(WQ_0(g)W^*=Q_t(g)\) for all \(g\), and \(WP_0W^*=P_t\). Both \(W_{\pi_t}\) and \(W_{\pi_0}\) act on \(\delta_x\otimes K\) as \(\pi_t(x)\), respectively \(\lambda(x)\), so by Proposition 2.2 \[ \begin{gathered} \begin{aligned} \|W-I\|&=\|W_{\pi_t}-W_{\pi_0}\|\\ &=\sup_x\|\pi_t(x)-\lambda(x)\|\\ &\le c(t). \end{aligned}\\ \square \end{gathered} \]

5. The identity holds for one subalgebra and fails for the other

Proposition 5.1. \(P_0\) has the spatial tensor-norm identity relative to \(D\).

Proof. Let \(V\) be the unitary of \(L\) with \(V(v\otimes\delta_x)=\lambda(x)^*v\otimes\delta_x\). For \(Y\in\lambda(G)'\), \(Y\lambda(x)^*=\lambda(x)^*Y\), so \(V\) commutes with \(D'\) by Lemma 4.1(1). For \(g\in G\), \[ \begin{aligned} &VQ_0(g)V^*(v\otimes\delta_x)\\ &\quad =V\bigl(\lambda(gx)v\otimes\delta_{gx}\bigr)=v\otimes\delta_{gx}, \end{aligned} \] so \(VQ_0(g)V^*=I\otimes\lambda(g)\) and \(VP_0V^*=I\otimes C^*(\lambda(G))\). Let \(p_1,\dots,p_m\in P_0\) and \(z_1,\dots,z_m\in D'\), and write \(Vp_iV^*=I\otimes X_i\) and \(z_i=Y_i\otimes I\). Since \(V\) commutes with each \(z_i\), \[ V\Bigl(\sum_ip_iz_i\Bigr)V^*=\sum_iY_i\otimes X_i . \] On \(L\otimes L=K\otimes K\otimes K\otimes K\), conjugation by \(V\otimes I_L\) turns \(\sum_ip_i\otimes z_i\) into \(\sum_iI\otimes X_i\otimes Y_i\otimes I\). Exchanging the two middle factors by a unitary and dropping the outer identity factors does not change norms. Both sides of Definition 2.1 therefore equal \(\|\sum_iY_i\otimes X_i\|\). \(\square\)

Proposition 5.2. For \(0<t<1\), \(P_t\) does not have the spatial tensor-norm identity relative to \(D\).

Proof. Choose an integer \(m>4/t^2\); then \(2\sqrt m<mt\). For \(1\le i\le m\) put \(p_i=Q_t(s_i)\in P_t\) and \(z_i=\rho(s_i)\otimes I\), which lies in \(D'\) by Lemma 4.1(1). Since \(\lambda(s_i)\rho(s_i)\delta_e=\delta_e\), Lemma 2.1(4) gives \[ p_iz_i(\delta_e\otimes\delta_e)=\delta_e\otimes(t\delta_e+r\delta_{s_i}), \] so the vector state at \(\delta_e\otimes\delta_e\) takes the value \(mt\) on \(\sum_ip_iz_i\), and \(\|\sum_ip_iz_i\|\ge mt\). On the other side, \[ \begin{gathered} \sum_ip_i\otimes z_i=\sum_i\lambda(s_i)\otimes\sigma(s_i),\\ \sigma(g)=\pi_t(g)\otimes\rho(g)\otimes I, \end{gathered} \] where \(\sigma\) is a unitary representation of \(G\) on \(K\otimes K\otimes K\). By Lemma 3.1 this operator is unitarily equivalent to \((\sum_i\lambda(s_i))\otimes I\), so Lemma 3.2 gives \(\|\sum_ip_i\otimes z_i\|\le2\sqrt m<mt\). \(\square\)

6. Proof of Theorem 2.3

Let \(\varepsilon>0\), and choose \(t\) with \(0<t\le1/2\) and \(8t<\varepsilon\). Construct \(D\), \(P_0\) and \(P_t\) on \(L\) as in Section 4, and on the separable space \(H=\bigoplus_{n\ge1}L\) put \[ A=\mathcal E_D(P_0),\qquad B=\mathcal E_D(P_t). \] By Lemma 4.1(2), \(D\) has trivial center, so Proposition 1.1 of Convergent sequences and a relative tensor norm applies: \(A\) and \(B\) are norm-separable C*-algebras containing \(I_H\), and \(A''=B''\). Lemma 4.2 and Corollary 2.3 of that lesson give \[ d(A,B)\le2\|W-I\|\le2c(t)\le8t<\varepsilon . \] By Propositions 5.1 and 5.2, \(P_0\) has the spatial tensor-norm identity relative to \(D\) and \(P_t\) does not, so by the same corollary no unitary of \(H\) conjugates \(A\) onto \(B\). \(\square\)

The subalgebras \(P_0\) and \(P_t\) themselves are conjugate by \(W\). By Lemma 2.2(1) of Convergent sequences and a relative tensor norm and Propositions 5.1 and 5.2, no unitary normalizing \(D\) conjugates \(P_0\) onto \(P_t\); in particular \(W\) does not normalize \(D\).

7. Exercises

Exercise 7.1. Show that \(\langle\pi_t(g)\delta_e,\delta_e\rangle=t^{|g|}\) for every \(g\in G\). Deduce that the function \(g\mapsto t^{|g|}\) is positive definite on \(G\), a fact due to Haagerup.

Exercise 7.2. Show that the linear map \(\delta_x\mapsto\xi_x\) on finitely supported vectors is not bounded. (Compare \(\sum_{j=1}^n\delta_{s_j}\) with its image.)

Exercise 7.3. Show that \(\|\sum_{j=1}^m\lambda(s_j)\|\ge\sqrt m\). Deduce that the argument of Proposition 5.2, which compares the value \(mt\) of a vector state with the tensor norm, can only succeed when \(m>1/t^2\).

Exercise 7.4. Compute \(\|\pi_t(s_1)-\lambda(s_1)\|\) exactly, and show that it is at least \(t\). So the bound of Proposition 2.2 has the right order as \(t\to0\).

Exercise 7.5. For \(t=0\), with \(p_i=Q_0(s_i)\) and \(z_i=\rho(s_i)\otimes I\), compute \(\langle\sum_ip_iz_i(\delta_e\otimes\delta_e),\delta_e\otimes\delta_e\rangle\), and compute \(\|\sum_ip_i\otimes z_i\|\) with Lemma 3.1. How do these values fit with Proposition 5.1?

8. Solutions

Solution 7.1. \(\pi_t(g)\delta_e=\pi_t(g)\xi_e=\xi_g\), and \(\langle\xi_g,\xi_e\rangle=t^{|g|}\) by Lemma 2.1(1). A diagonal coefficient \(g\mapsto\langle\pi(g)\eta,\eta\rangle\) of a unitary representation is positive definite: \(\sum_{g,h}c_g\bar c_h\langle\pi(h^{-1}g)\eta,\eta\rangle\)\(=\|\sum_gc_g\pi(g)\eta\|^2\ge0\).

Solution 7.2. The vector \(\sum_{j=1}^n\delta_{s_j}\) has norm \(\sqrt n\), and its image \(\sum_{j=1}^n\xi_{s_j}=nt\,\delta_e+r\sum_{j=1}^n\delta_{s_j}\) has norm at least \(nt\). The ratio \(t\sqrt n\) is unbounded.

Solution 7.3. \(\sum_j\lambda(s_j)\delta_e=\sum_j\delta_{s_j}\) has norm \(\sqrt m\). In Proposition 5.2 the tensor side has norm \(\|\sum_j\lambda(s_j)\|\ge\sqrt m\), so the lower bound \(mt\) for the product side exceeds it only if \(mt>\sqrt m\), that is \(m>1/t^2\).

Solution 7.4. Here \(l=1\), \(M\) is spanned by \(f_0=\delta_e\) and \(f_1=\delta_{s_1}\), and the proof of Proposition 2.2 gives \(Tf_0=rf_0-tf_1\) and \(Tf_1=tf_0+rf_1\). So \(T\) acts on \(M\) as a rotation, with eigenvalues \(r\pm it\), and \(T-I\) is normal with eigenvalues \(r-1\pm it\). Hence \[ \begin{aligned} &\|\pi_t(s_1)-\lambda(s_1)\|\\ &\quad =\sqrt{(1-r)^2+t^2}=\sqrt{2-2r}. \end{aligned} \] Since \(1-r=t^2/(1+r)\ge t^2/2\), this is at least \(t\).

Solution 7.5. \(p_iz_i(\delta_e\otimes\delta_e)=\delta_e\otimes\delta_{s_i}\), so the value is \(0\), and the vector state gives no lower bound. Lemma 3.1 with \(\sigma(g)=\lambda(g)\otimes\rho(g)\otimes I\) shows that \(\sum_ip_i\otimes z_i\) is unitarily equivalent to \((\sum_i\lambda(s_i))\otimes I\), so its norm is \(\|\sum_i\lambda(s_i)\|\), between \(\sqrt m\) and \(2\sqrt m\). Proposition 5.1 says that \(\|\sum_ip_iz_i\|\) has the same value; for \(t=0\) no vector state detects a difference, and indeed there is none.

References