Close Separable C*-Algebras Without Spatial Conjugacy

Proof. A norm-convergent sequence is bounded, so (3) defines a diagonal operator on H, with norm \sup_n\norm{x_n}. Algebra operations and adjoints preserve the limit condition, and the constant identity belongs to the algebra. To check norm closedness, suppose x^{(k)}\to x in the supremum norm, with coordinate limits p_k\in P. Then

Thus p_k\to p\in P. The estimate

shows that x_n\to p. Moreover,

The two summands are norm-separable: P is a subspace of the separable metric space D, and finite-support sequences from a countable dense subset of D are dense in c_0(\N,D). Their sum is norm-separable.

For the metric assertion, write \alpha(P,Q)=\sup_{p\in P_1}\dist(p,Q_1). Given a contraction x\in\E_D(P) with limit p and \eta>0, choose q\in Q_1 with \norm{p-q}<\alpha(P,Q)+\eta, and choose N so that \sup_{n>N}\norm{x_n-p}<\eta. Keeping the first N entries of x and putting q thereafter gives a contraction y\in\E_D(Q) with \norm{x-y}<\alpha(P,Q)+2\eta. This proves the upper bound for the directed distance. Conversely, for a constant contraction (p,p,\ldots) and any contraction y\in\E_D(Q) with limit q\in Q_1,

Taking infima and suprema proves equality of the directed distances. Interchanging P,Q gives (4).

Let e_n be the projection onto the nth copy of L. Every e_n is a central element of \E_D(P). Conversely, a central element (x_n) commutes with all finite-support D-valued sequences, so each x_n lies in Z(D)=\C\id_L. Thus the center consists exactly of convergent scalar sequences. Every nonzero central projection contains an e_n, and hence the minimal nonzero central projections are precisely the e_n. The same holds for \E_D(Q).

If S\E_D(P)S^*=\E_D(Q), conjugation by S therefore permutes the projections e_n. Its restriction from the nth coordinate to its image is a unitary U_n:L\to L. The corresponding coordinate corners of both algebras are exactly D, since every finitely supported D-valued sequence is allowed. Consequently U_nDU_n^*=D. For b\in Q, pull back the constant sequence (b,b,\ldots) under S. Its nth entry is U_n^*bU_n, regardless of the coordinate permutation. It belongs to \E_D(P), so its entries converge in norm to some p_b\in P. This is equivalent to (5) and proves (ii).

Finally, an operator commuting with \E_D(P) commutes with all e_n, so it is block diagonal. Commutation with the finite-support D corners means that each block lies in D'. These conditions are also sufficient, and do not depend on P. Thus both commutants equal

This proves (iii).□

Proof. If UDU^*=D, then UD'U^*=D'. For q_i=Up_iU^* and z_i\in D', apply (7) to p_i and U^*z_iU. Conjugate its product sum by U and its tensor sum by U\otimes U. This gives (7) for q_i,z_i.

Now fix finite families b_i\in Q and z_i\in D'. Put q_{i,n}=U_np_{b_i}U_n^*. For each n, the first part gives the identity for the entire list (q_{i,n})_i, since its entries all lie in U_nPU_n^*. Both errors

are at most \sum_i\norm{q_{i,n}-b_i}\norm{z_i}, which tends to zero. Taking the limit proves the identity for the prescribed families in Q and D'.□

Proof. A spatial conjugacy would give the unitaries in Proposition 2.1(ii), contradicting Lemma 2.3. For the distance bound, p\mapsto WpW^* bijects the closed unit balls and

Apply Proposition 2.1(i).□

Proof. Use the tree of reduced words with edges given by right multiplication by a generator or its inverse, rooted at e. Its distance is d(x,y)=|x^{-1}y|, so left translations preserve distance. For x\ne e, let p(x) be the neighbor of x toward e, obtained by deleting the last letter of its reduced word. Recursively define

These are unit vectors supported on their root paths. If z is the last common vertex of the root paths to x and y, the common-support components are t^{d(x,z)}\xi_z and t^{d(y,z)}\xi_z, respectively. The remaining supports are disjoint. Hence

Also \delta_e=\xi_e and \delta_x=(\xi_x-t\xi_{p(x)})/r for x\ne e, so their span is dense. The assignment \xi_x\mapsto\xi_{gx} preserves every finite Gram matrix by (12). It therefore defines an isometry on the span, and the inverse assignment for g^{-1} makes the extension unitary. Denote this unitary by \pi_t(g). The representation law holds on the dense span and hence on K. Since \pi_t(s_i)\delta_e=\xi_{s_i}=t\delta_e+r\delta_{s_i}, (9) follows.

It remains to prove that the perturbation is uniformly small over all group elements, rather than just over the generators. Fix g\ne e, write h_0=e,h_1,\ldots,h_l=g for its root path, and put T=\pi_t(g)\lambda(g)^*. For y\ne g, (11) gives

The vertex g p(g^{-1}y) is the neighbor of y toward g. If y is off the path [e,g], this neighbor is also p(y), the neighbor toward e; see Figure 1. Thus T\delta_y=\delta_y off that path. On the path, the neighbor toward g of h_k, for k<l, is h_{k+1}, so

Let M_g=\spanop\{\delta_{h_k}:0\le k\le l\}. On this space define the truncated shift S\delta_{h_k}=\delta_{h_{k+1}} for k<l and S\delta_{h_l}=0, and put X\delta_{h_k}=\xi_{h_k}. For R=\operatorname{diag}(1,r,\ldots,r), the recursion gives

Indeed, its kth column is t^k\delta_e+r\sum_{j=1}^k t^{k-j}\delta_{h_j}. Since \norm{S}\le1 and \norm{R}=1,

Writing R'=\operatorname{diag}(r,\ldots,r,1), (14) becomes T|_{M_g}=XR'-tS. Therefore

The same bound holds on K, since T is the identity on M_g^\perp. Multiplication by \lambda(g) proves (10). For g=e the difference is zero, so the supremum includes this case as well.□

Proof. The operator is the orthogonal direct sum of the unitaries \sigma(x). On \delta_x\otimes v, the left side of (17) equals

□

Proof. Commutation with all \id_K\otimes E_{xy} forces a bounded operator to be Y\otimes\id_K: commutation with the diagonal matrix units removes off-diagonal blocks, and commutation with the remaining units makes the diagonal blocks identical. The first-factor generators then impose exactly Y\in\lambda(G)'. Such operators commute with all the second-factor generators, proving (20). In particular, \rho(g)\otimes\id_K\in D'.

If Y\otimes\id_K\in Z(D), then Y commutes with \lambda(G) by (20). It also commutes with \rho(G), since Y\otimes\id_K belongs to D and the right translations belong to D'. Thus Y\delta_e is invariant under \lambda(g)\rho(g), which acts on the basis by conjugation x\mapsto gxg^{-1}. Every nonidentity conjugacy class in G is infinite: conjugate a fixed nonidentity reduced word by distinct generators absent from that word. A square-summable vector invariant under conjugation must therefore be supported at e. Write Y\delta_e=\alpha\delta_e. Commuting with left translations gives Y\delta_x=\alpha\delta_x for every x, proving Z(D)=\C\id_L.

For the final assertion, Lemma 4.1 shows that W=W_{\pi_t}W_{\pi_0}^* sends every Q_0(g) to Q_t(g) by conjugation. Moreover, the direct-sum description of these unitaries gives

□

Proof. Define a unitary on L by

It commutes with every element of D' by (20). On elementary vectors one computes

Thus VP_0V^*=\id_K\otimes C^*(\lambda(G)). Given finite families p_i\in P_0 and z_i\in D', write

Conjugation by V gives

On L\otimes L=K^{\otimes4}, conjugation by V\otimes\id_L gives

The last equality permutes the four tensor factors and uses the fact that tensoring with a nonzero identity operator does not change norm. This proves (7).□

Proof. Let F_i be the orthogonal projection onto the closed span of words beginning with s_i^{-1}. The F_i are pairwise orthogonal. Write \lambda(s_i)=A_i+B_i, where

There is no cancellation when s_i is prepended to a word not beginning with s_i^{-1}. Thus the range of A_i consists of vectors supported on words beginning with s_i, and these ranges are mutually orthogonal. For every v\in K,

while

Adding the two bounds proves (22).□

Proof. Choose an integer m>4/t^2, so mt>2\sqrt m, and take

These belong to P_t and D', respectively. Since \lambda(s_i)\rho(s_i)\delta_e=\delta_e, (9) implies

On the other hand, the tensor sum on L\otimes L is

Apply Lemma 4.1 to the representation \sigma=\pi_t\otimes\rho\otimes1 on K^{\otimes3}, where 1 is the trivial representation on the last copy of K. Together with Lemma 4.4, it gives

Equations (23) and (24) contradict the identity for this finite list.□

Proof of Theorem 1.1. Given \varepsilon>0, choose 0<t<1 so small that 2c(t)<\varepsilon. Use the algebras P_0,P_t\subseteq D\subseteq\B(L) from Section 4, and put

The space H is separable. Lemma 4.2 supplies the scalar center of D, so Proposition 2.1 shows that A,B are unital norm-separable C^*-algebras with common identity and that A''=B''. The same lemma supplies a unitary W with WP_0W^*=P_t and \norm{W-\id_L}\le c(t). By Corollary 2.4,

Finally, Lemmas 4.3 and 4.5 show that the relative tensor-norm identity holds for P_0 but not for P_t. Corollary 2.4 therefore excludes every unitary conjugating A onto B.□