A counterexample to Naimark’s problem in ZFC

Proof. In a nontrivial convex decomposition of \widetilde h, both restrictions to B equal h by purity. Uniqueness then makes both states equal to \widetilde h.□

Proof. Realize the two states by unit vectors \xi,\eta in a common irreducible representation \pi. On their finite-dimensional span choose a unitary carrying \xi to \eta and a self-adjoint logarithm of that unitary. Kadison transitivity, including its self-adjoint form [7, Theorem 1], supplies a self-adjoint a\in B whose action on this span is the chosen logarithm. The span reduces \pi(a), so u=e^{ia} satisfies \pi(u)\xi=\eta. If h is represented by \xi and k by \eta, then k=h\circ\Ad(u^*). Conjugating their unique extensions by the same unitary proves the last assertion.□

Proof. Every state of B satisfying the projection constraints extends to a state of C with the same constraints, hence to H. It must therefore equal h. Every extension of h also satisfies the constraints and must equal H.□

Proof. Put

Here L is a closed left ideal and N is a C^*-subalgebra. If a\in L, then a^*a\in N, since h((a^*a)^2)\le\|a\|^2h(a^*a)=0.

First, h is the only state annihilating N. If another state k annihilates N, then L is contained in the kernel of a\mapsto\pi_k(a)\xi_k. The map Q_h:B\to H_h, Q_h(a)=\pi_h(a)\xi_h, is onto by one-vector Kadison transitivity. This assertion concerns the whole Hilbert space, even when it is nonseparable. Since \ker Q_h=L, the open mapping theorem for B/L\to H_h shows that

defines a bounded operator T:H_h\to H_k. It intertwines the representations. Thus T^*T is scalar by irreducibility of \pi_h; its value on \xi_h shows that T^*T=1. Taking vector coefficients gives k=h.

We next construct the decreasing net. Order N_+ by the usual operator order, a directed order, and set

Functional calculus gives e_s\in N, and operator monotonicity of inversion makes (e_s) increasing. For b\in N and s\ge m(bb^*+b^*b),

the same estimate holds for \|b(1-e_s)\|^2. Thus (e_s) is an approximate identity of N, and h(x_s)=1.

Let q be the strong limit of (e_s), the support projection of N. Every normal state supported on 1-q annihilates N, and hence is h; conversely, h(1-q)=1. The corner (1-q)B^{**}(1-q) therefore has exactly one normal state. Normal states separate its self-adjoint elements, so this corner is scalar. It follows that 1-q=r_h is minimal and x_s\downarrow r_h. The identity r_har_h=h(a)r_h follows by applying h to this scalar corner.

Finally, let \rho_s be arbitrary states of B. Every weak-star cluster point of the positive functionals a\mapsto\rho_s(x_sax_s) annihilates N, by the approximate-identity estimates. A nonzero cluster point, after normalization, is therefore h. For self-adjoint a, a failure of (2.1) would give a subnet and norming states \rho_s with a cluster point nonzero on a-h(a)1, a contradiction. Real and imaginary parts give the conclusion for arbitrary a.□

Proof. The infimum of the P_n in B^{**} supports exactly one normal state, namely h. As in the preceding proof, it is the minimal projection r_h. For arbitrary states \rho_n, every cluster point \psi of \rho_n(P_n\,\cdot\,P_n) vanishes on 1-P_m for each fixed m. If \psi\ne0, its normalization satisfies every peaking constraint and therefore equals h. The same norming-state argument proves excision. For the last assertion use nesting:

□

Proof. The normal extension of \pi_h annihilates r_k. Otherwise a unit vector in the range of \pi_h^{**}(r_k) would represent k, by r_kar_k=k(a)r_k. Every nonzero vector in an irreducible representation is cyclic, so this would imply h\sim k.

Consequently h(ay_{\nu_0}a^*)\to0 as \nu_0 increases. For \nu\ge\nu_0, the inequalities y_\nu^2\le y_\nu\le y_{\nu_0} give

Choose \nu_0 first, then use excision to choose \lambda_0. This proves smallness on a rectangular tail and hence the product-net limit. Lemma 2.5 gives the stated projection case.□

Proof. The space H_h is infinite-dimensional. Choose orthonormal unit vectors \eta_j and unitaries u_j\in B with \pi_h(u_j)\xi_h=\eta_j, as in Lemma 2.2. The projections u_jr_hu_j^* are pairwise orthogonal, since

They all have trace \tau(r_h), so m\tau(r_h)\le1 for every positive integer m.□

Proof. For such an expectation, bimodularity gives \tau(b^*E(a))=\tau(b^*a) for every b\in B. Two expectations therefore have difference d\in B satisfying \tau(d^*d)=0, so they agree. Faithfulness follows from \tau(E(a^*a))=\tau(a^*a).□

Proof. The d_n are nonzero orthogonal projections. Their closed linear span is isomorphic to c_0(\mathbb N_0), and adjoining the unit gives the claimed description of D. Indeed p_n=1-\sum_{j<n}d_j and \|p_n\|=1, since \tau(p_n)>0. Multiplying by p_n shows that 1 has distance at least one from every linear combination of d_0,\ldots,d_{n-1}.

The values in (2.3) define a continuous function: by d_n\le p_n and Lemma 2.5,

Every coordinate of E is a state, so E is unital and completely positive: positivity of each matrix amplification is tested pointwise in the commutative range. The map fixes D. To check bimodularity, an element of D acts scalarly on each d_n; at infinity, h|_D is the evaluation character. Cauchy–Schwarz for h then gives h(ba)=h(b)h(a) and h(ab)=h(a)h(b) for b\in D. These coordinate identities prove bimodularity.

Finally, \sum_n\tau(d_n)=1 and the trace on D has no mass at infinity. Summing the finite coordinates gives \tau(E(a))=\tau(a), since the remaining trace satisfies |\tau(p_Na)|\le\|a\|\tau(p_N)\to0.□

Proof. Let J be a proper closed two-sided ideal of C. The quotient map q:C\to C/J is injective on B, since B is simple and q(1)\ne0. Every pure state h of B therefore extends through this quotient to a state of C annihilating J. By (U), this extension is h\circ E. Thus h(E(a))=0 for every positive a\in J and every pure h on B. Pure states separate positive elements, so E(a)=0; faithfulness gives a=0. Hence J=0.□

Proof. For i\in\{F,G\}, complete i as a right Hilbert D-module L_i with inner product

Left multiplication by a\in i is bounded, since

and its adjoint is left multiplication by a^*. The expectation gives an orthogonal decomposition

Both summands are invariant under left and right multiplication by D. Form the Hilbert D-module

Here \otimes_D is the interior tensor product of Hilbert modules; the left D-action on a word acts on its first factor. The first summand D is called the vacuum summand, with distinguished vector 1.

Let K^{(i)} be the vacuum summand together with all words not starting with i. The decomposition of L_i identifies

Represent i by left multiplication on the first tensor factor. The two representations agree on D. They are faithful: an element a annihilating the vacuum satisfies E_D^i(a^*a)=0, so its faithful factor trace gives a=0. Let Z be the C^*-algebra they generate in the adjointable operators on K.

A centered element of i acting on K^{(i)} adds its own tensor factor at the front. Thus a nonempty alternating centered product has zero vacuum compression. Centering arbitrary factor elements shows that D and these products span a dense subspace of Z. Vacuum compression consequently takes values in the left copy of D and defines the conditional expectation E_D.

Write \tau_D for the common trace on D. We next prove that \tau_Z=\tau_D\circ E_D is tracial and faithful. For traciality, compare \tau_Z(aw) and \tau_Z(wa) when a belongs to one factor and w is an alternating centered word. If w has length at least two, both values vanish. Indeed, multiplication at a same-factor end, followed by centering that end product, leaves either an alternating centered word or a D-coefficient times a nonempty such word. At a different-factor end, first center a and use D-bimodularity. For a one-letter word, use the factor trace when the factors agree; when they differ, both values are zero. For w\in D, use the factor trace. Density and repeated cyclic movement of a factor now prove traciality on all of Z.

Suppose \tau_Z(z^*z)=0. For every t\in Z, traciality gives

The trace on D is faithful, so E_D((zt)^*(zt))=0. This is the D-valued inner product of zt1 with itself, and hence zt1=0. Products of centered letters create all elementary tensors in (3.4); thus Z1 is dense in K. It follows that z=0, proving faithfulness of \tau_Z.

Finally, let P_F be the orthogonal projection onto D\oplus L_F^\circ=L_F. This submodule reduces the F-action. A core w=b_1a_1\cdots a_{m-1}b_m as in (3.3) sends L_F into words starting with G, so P_FwP_F=0. Since P_F commutes with F, arbitrary endpoint coefficients satisfy

Successive centering shows that F and the products (3.3) span a dense subspace. Their compressions belong to the faithful left copy of F on L_F, which is norm closed. Compression therefore defines a conditional expectation E_F:Z\to F. The vacuum lies in L_F, giving E_D=E_D^F\circ E_F and trace preservation. Faithfulness follows from faithfulness of \tau_Z.□

Proof of Proposition 3.1. Take

Embed D in F by (d_1,d_2)\mapsto\operatorname{diag}(d_1,d_2) and in G by (d_1,d_2)\mapsto\operatorname{diag}(d_1,\theta^{-1}(d_2)). The normalized matrix traces agree on D, and the maps

are trace-preserving conditional expectations. Apply Lemma 3.2 to obtain Z, and retain the module notation used in its proof. The diagonal matrix projections e_i=e_{ii} are common to the two factors. Set

and identify A with e_1Fe_1. Then v^*v=vv^*=e_1. For d\in D_1, the element sds^* is the second-corner copy of \theta(d) in F, so vdv^*=\theta(d) in the first corner.

The matrix units of F identify Z with M_2(e_1Ze_1). Under this identification, the matrix algebra over C^*(A,v) contains F and s=e_{21}^Fv. It contains G as well, because G is generated by D and s,s^*. Hence H(A)=C^*(A,v). The trace on the corner is 2\tau_Z, and its restriction to A is \tau. The restriction of E_F gives the required faithful trace-preserving expectation onto A.

To check reduced words, write q=e_{12}^F, so v=qs and v^*=s^*q^*. Between consecutive G-letters the internal F-coefficients are

The first and last are off diagonal. The middle two have zero D-expectation precisely under the respective conditions in (3.2). Since s,s^* are centered, the expanded word is a product of the form (3.3), and its expectation onto F, hence onto A, is zero.

For uniqueness, split a coefficient at an opposite-sign turn as (a-E_i(a))+E_i(a). The first term is centered at that turn, while the second cancels two stable letters using the unitary relation. Repeating this operation expresses every polynomial as an A-term plus nonempty reduced words. Thus the stipulated data determine the trace of every polynomial, and in particular all moments of x^*x for each polynomial x.

For a faithful tracial state t on any C^*-algebra,

To see this, restrict t to the commutative algebra generated by x^*x and the unit. Its spectral measure has full support, by faithfulness, and its rth-root moments tend to the spectral maximum. Consequently identical polynomial trace moments give identical polynomial norms. The generator correspondence is therefore isometric and extends to the asserted isomorphism. This argument requires no expectation in the comparison extension beyond the stated trace-vanishing rule.□

Proof. Because D_i\subset A', the restrictions E_i|_{A'} remain conditional expectations onto D_i. The subalgebra C^*(A',v) inherits a faithful trace, the unitary relation and the reduced-word trace vanishing. Proposition 3.1 identifies it with H(A'). The reduction of polynomials in the preceding proof shows that E_A maps this subalgebra into A' and has the stated restriction.□

Proof. Consider C=H_k(H_j(O)). Corollary 3.3 places both H_j(O) and H_k(O) inside C. Compose the expectations C\to H_j(O)\to O, and call the result E_O. Its restriction to each smaller adjunction is that adjunction’s expectation onto O, as can be checked on its reduced words.

We prove that the two smaller adjunctions are free over O: an alternating product of elements from their respective kernels of E_O has zero E_O-expectation. Approximate each centered H_k(O)-element by finite sums of nonempty reduced k-words over O. Such approximations follow by reducing a polynomial approximation and subtracting its O-expectation. Keep the centered H_j(O)-elements as coefficients.

When two k-words are joined across a centered element b\in H_j(O), the new internal coefficient has the form aba', where a,a'\in O. Its expectation onto O is aE_O(b)a'=0, so its expectation onto either D_i^k is also zero. The resulting word is therefore a nonempty reduced k-word over H_j(O), including at every new opposite-sign turn. Initial and final H_j(O)-coefficients are allowed. Its expectation onto H_j(O) vanishes. Passing to norm limits proves the freeness assertion. An alternating product with no k-block is a single centered j-element and is already killed by E_O.

The individual expectations and this freeness determine all joint O-valued moments. Indeed, center the factors of a joint word; the fully alternating centered term vanishes, and every other term has fewer blocks after its O-coefficients are absorbed into neighbors. The resulting recursion is symmetric in j and k. Applying the same argument in the reverse construction gives equal scalar trace moments for every polynomial in O,v_j,v_k. Faithfulness and (3.6) give the desired isomorphism.□

Proof. The compatible norm-one expectations extend continuously to O and retain complete positivity, bimodularity and trace preservation. Denote these extensions by E_i. The compatible traces extend faithfully to both completions. For example, the quotient of R by its trace-null ideal is injective, hence isometric, on every earlier faithfully tracial algebra. It is therefore isometric on their dense union and on R, so that ideal is zero. The same argument applies to O.

We have R=C^*(O,v), and the unitary relation persists. Consider a nonempty reduced word with coefficients in O. Approximate its finitely many coefficients in one common earlier O_\beta. At a turn requiring E_i(a)=0, replace an approximation b by b-E_i^\beta(b). This remains in O_\beta, is exactly centered, and differs from a by at most twice the original error. Thus the given word is a norm limit of reduced words at earlier stages, all with trace zero. Proposition 3.1 identifies R with H(O). This finite-coefficient argument requires no assumption on the cofinality of \lambda.□

Proof. Lemma 2.9 supplies the expectations onto D_i. The differences p_n^i-p_{n+1}^i are nonzero, mutually orthogonal projections, so each D_i is the algebra of continuous functions on \mathbb N_0\cup\{\infty\}. Matching these differences defines \theta, and the equality of trace values makes it trace preserving. Thus the reduced HNN construction applies. Its trace and expectation already give the corresponding assertions in the theorem. The work is to prove the assertions about pure states.

Pass to the two-by-two construction. Use the notation of Lemma 3.2 and Proposition 3.1:

Write E_D^F:F\to D for the expectation, e_i=e_{ii} for the common diagonal projections, and s=e_{21}^G. Recall that H(A)=e_1Ze_1, that Z=M_2(H(A)) using the matrix units of F, and that v=e_{12}^F s satisfies vp_n^1v^*=p_n^2 in the corner. Put

Here p_n^i e_i denotes the matrix with entry p_n^i in position (i,i) and zero elsewhere. The projections P_n^i peak to \phi_i: their zeroth term is e_i, so a state taking value one on them is supported on that corner and restricts there to f_i. The states \phi_1,\phi_2 are pure and inequivalent, by matrix amplification of the GNS representations of f_1,f_2. Also F has no finite-dimensional irreducible representations. Indeed, any such representation would restrict along a\mapsto\operatorname{diag}(a,a) to a nonzero finite-dimensional representation of A, whose image would supply a finite-dimensional irreducible representation of A.

We shall prove unique extension and the exclusion of unintended joins for F\subset Z, and then return to the corner. By Lemma 2.2, it suffices to use \phi_i as the representative of its distinguished equivalence class. More explicitly, conjugating by a unitary of F bijects the sets of state extensions. We first establish uniqueness for these representatives and for every pure state outside their two classes; only afterward will we use the resulting unique lifts.

Consider pure states h,k of F satisfying one of the two tests

Whenever either state belongs to [\phi_i], take it to be \phi_i itself. Choose decreasing positive contractive excision nets x=x_\lambda for h and y=y_\nu for k as in Lemma 2.4; at \phi_i, choose the sequence P_n^i by Lemma 2.5. All limits involving both x and y below are on the product of the two index sets.

We will show that x_\lambda W y_\nu\to0 in norm for every nonempty word W in (3.7). When h=k, this will force all state extensions to agree; for the inequivalent pairs in (4.1), it will keep the lifted representations inequivalent. The stable unitary will then join the designated pair.

A norm estimate for reduced words. We claim that

for every m\geq1, every b_j\in\{s,s^*\}, and every a_j\in F with

By the spanning description (3.7), these words, together with F, span a dense subspace of Z.

Let P_G be the orthogonal projection of the reduced free-product Hilbert D-module K onto the summands starting with a factor in L_G^\circ. For d\in F, the decomposition K=L_F\otimes_D K^{(F)} gives

Indeed, a vector \eta\in P_GK has the form 1\otimes\eta in this decomposition, and

The expectation norm here is the norm in D.

For b\in\{s,s^*\} write b=e_{\ell(b)}be_{r(b)}. Thus (\ell(s),r(s))=(2,1) and (\ell(s^*),r(s^*))=(1,2). For a word in (4.2), put

Each letter of this centered alternating core creates a new leg when applied, from right to left, to a word not starting with G. Therefore

Set d=xa_0e_\ell and d'=e_ra_my. We obtain

Using (4.4) and its adjoint version, it follows that (4.2) holds whenever both

This sufficient criterion does not require the pair of endpoint states to satisfy (4.1).

When an endpoint decays. In the first line of (4.6), increase x^2 to x. The resulting elements are a decreasing net of positive continuous functions on the \ellth convergent sequence in the spectrum of D; they vanish on the other component. At a finite atom Q=P_n^\ell-P_{n+1}^\ell, their pointwise limit is

Here r_h is the support projection in F^{**} and \tau_F is the normalized matrix trace, extended normally to F^{**}. The numerator vanishes by traciality and Lemma 2.7; for example

At the point \infty, the limit is \phi_\ell(a_0^*r_ha_0). It is zero if h\not\sim\phi_\ell, because the normal extension of the \phi_\ell representation annihilates r_h, as in Lemma 2.6. If h=\phi_\ell, the support identity instead gives

Thus the first endpoint decays unless h=\phi_\ell, and it still decays in that case when \phi_\ell(a_0)=0.

The uniform convergence used here is Dini’s theorem for monotone nets [9, Theorem 1.5, p. 306]. We recall its compactness argument. For a fixed \varepsilon>0, the open sets where a chosen function is below \varepsilon cover the compact spectrum. Choose a finite subcover, and then an index dominating its finitely many indices. Monotonicity makes every later function uniformly smaller than \varepsilon. The second line of (4.6) is symmetric: it decays unless k=\phi_r, and also decays in that case when \phi_r(a_m)=0.

Remove the matched endpoints. Call the left endpoint matched when h=\phi_\ell, and the right endpoint matched when k=\phi_r. At each matched endpoint write

Expand the word using these decompositions. In each resulting term, an unmatched endpoint already satisfies its estimate in (4.6), and a matched endpoint with its zero-value coefficient does as well. We treat the scalar terms by moving the corresponding peak through its adjacent letter:

These identities hold in G: the diagonal embedding of D_2 uses \theta^{-1}, so the two projections have the same matrix entry p_n^1 there.

In the following displays, an untransferred endpoint coefficient means the coefficient chosen in the expansion. For m\ge2, after suppressing its scalar coefficient, a left scalar term becomes

The peak now belongs to the other distinguished state, and b_1 is a bounded outside factor. A right scalar choice similarly moves P_t^{r(b_m)} to P_t^{\ell(b_m)} and leaves b_m as a bounded factor on the right. When both ends are transferred and m\ge3, the resulting expression, again up to its scalar coefficient, is

Suppose at least one core letter remains after the transfers. Each newly exposed endpoint coefficient is one of the original internal a_j. By (4.3), \phi_1(a_j)=\phi_2(a_j)=0: these states evaluate the two components of E_D^F(a_j) at \infty. Thus every new endpoint satisfies its estimate, whether or not its transferred state matches the adjacent letter. Endpoints that were not transferred satisfy their estimates by the preceding expansion. Applying (4.6) to the remaining word therefore proves decay of each such term. This covers no transfers, one transfer when m\ge2, and two transfers when m\ge3, including a remaining core of one letter. The sufficient endpoint criterion requires no assumption that the transferred pair of states satisfies (4.1).

It remains to check the cases in which the core disappears. If m=1, both ends cannot match: the range and source of s or s^* are different, so double matching would be the designated pair excluded in (4.1). For a single left transfer, say h=\phi_i, the remaining compression, up to a scalar and a bounded outside letter, is

The state k is inequivalent to \phi_j. In the equal-state test it equals \phi_i; in the other test, equivalence to \phi_j would again be the excluded designated pair. Hence this expression tends to zero by Lemma 2.6. A single right transfer is symmetric.

Finally let m=2 and transfer at both ends. Both original endpoints matched, so (4.1) forces h=k=\phi_i for some i. Since each letter switches the two matrix indices,

The remaining compression is P_n^j a_1P_t^j, with \phi_j(a_1)=0. If q=\min(n,t), nestedness and excision give

This is convergence on the product of the two sequence indices. Together with cross-excision, whose conclusion also holds on the product net, these cases prove (4.2) in full.

Unique extensions and the exclusion of other joins. In the equal-state test, let \psi be any state of Z extending h. Its GNS vector is fixed by every x_\lambda and y_\nu: a positive contraction x of state value one satisfies \|(1-\pi_\psi(x))\xi_\psi\|^2\leq\psi(1-x)=0. Consequently, for every tested word W,

Its values on F are already prescribed. The dense span above therefore proves uniqueness; existence also follows directly by composing h with the expectation onto F. Conjugacy by unitaries in F now establishes uniqueness for every old pure state, including all states in the two distinguished classes. Their unique extensions are pure by Lemma 2.1.

Next take an inequivalent pair allowed in (4.1). If its unique lifts were equivalent, realize them by unit vectors \xi_h,\xi_k in one irreducible representation \pi of Z. The represented excision elements fix the corresponding vectors, so

For z\in F, the right side tends to zero by cross-excision; for a tested nonempty word it tends to zero by (4.2). Thus the matrix coefficient vanishes on all of Z. This contradicts cyclicity of \xi_k in the irreducible representation. Unitary conjugacy within the old classes extends this exclusion to all representatives. Equivalent old states, on the other hand, have equivalent unique lifts by Lemmas 2.2 and 2.1.

Return to the corner and join the designated pair. States of A=e_1Fe_1 identify with states of F taking value one on e_1, by composition with compression. The same correspondence holds between H(A)=e_1Ze_1 and Z, and it commutes with restriction. Every extension to Z of an e_1-supported state of F is still e_1-supported. Hence this correspondence bijects extension sets and transfers unique extension to A\subset H(A).

Under F=M_2(A) and Z=M_2(H(A)), these supported states have the matrix amplifications of their corner GNS representations, with the representing vector in the first summand. Purity and equivalence of classes are therefore preserved. The matrix-flip unitary of F moves the representative \phi_2 from the second summand to the first, so its class corresponds to [f_2], just as the class of \phi_1 corresponds to [f_1]. The exclusion of all other joins thus descends to the corner.

To see that the designated join actually occurs, use the stable unitary relation. The state \widetilde f_2\circ\Ad v takes value one on every p_n^1, so its restriction to A is f_1. Unique extension, now proved, gives

This proves the asserted equivalence criterion. If A is simple, Lemma 2.10 applies to the faithful expectation and the unique-extension property, proving that H(A) is simple.□

Proof. At a successor, unique extensions compose, and the intermediate lift is pure by Lemma 2.1. At a limit, two extensions agree on each earlier algebra, hence on their dense union. Existence follows from state extension. Transfinite induction gives the first assertion. For the second, a unitary implementing equivalence in the original algebra also conjugates the unique lifts; use Lemma 2.2.□

Proof. Otherwise realize h,k by unit vectors \xi,\eta in one irreducible representation \pi of B_\lambda. For each earlier \alpha in the given tail, the cyclic subspaces

are reducing and carry inequivalent irreducible representations of B_\alpha. The orthogonal projection onto N_\alpha, restricted to M_\alpha, is an intertwiner, so it is zero. Thus \langle\pi(a)\xi,\eta\rangle=0 for every a\in B_\alpha. Continuity gives the same equality for every a\in B_\lambda, contradicting cyclicity of \xi.□

Proof. Simplicity and faithfulness follow from the permanence facts above applied to the matrix stages. Separability and infinite dimension are immediate from those stages. A state taking value one on p_n must restrict to the vector state of the first basis vector on M_{2^n}. These vector states are compatible, so they determine exactly one state on the dense union and its completion. This proves peaking and purity. The trace formula is the normalized matrix trace formula.

A nonzero finite-dimensional irreducible representation of an algebra containing A_0 restricts to a unital representation of A_0. Its kernel is zero by simplicity, which would embed the infinite-dimensional algebra A_0 in a finite-dimensional algebra.□

Proof. Let k\in T be active. Both sequences for its bridge lie in A[T\cap k], by (5.1). The original expectations onto their generated abelian algebras restrict to this smaller base, since their ranges already lie in it. The restriction property Corollary 3.3 identifies the algebra obtained by adjoining v_k with that reduced HNN extension.

The sequences peak to pure states of the smaller base by Lemma 2.3, applied to its inclusion in X_k. That same lemma gives each of these particular states its unique extension to X_k. If the two smaller states were equivalent, Lemma 2.2 and uniqueness would make their extensions equivalent, contrary to the choice of the original active step. The smaller base contains A_0, so Lemma 5.3 excludes finite-dimensional irreducibles. All hypotheses of Theorem 4.1 therefore hold in the inherited iteration.

Continuity follows because a polynomial uses only finitely many indices. Simplicity follows by induction from Lemma 2.10 and the limit observation above. Faithfulness follows either from the restricted trace on X_\delta or from the same permanence argument. Finally, a countable set of stable unitaries together with the separable algebra A_0 generates a separable algebra.□

Proof. We will insert the omitted indices, in their original order, after A[S]. Recall that the ordinal j is the set of indices below j, whereas j+1 also contains the index j itself. Fix an active j\notin S, and for j+1\le\beta\le\delta set

These index sets are dependency-closed. Initially, O_{j+1}=X_j and R_{j+1}=X_{j+1}; afterward both chains perform precisely the inherited steps with indices in S.

The omitted pair stays inequivalent on O_\beta. Both (p_n) and (q_n^j) peak on every O_\beta by Lemma 2.3, since they peak in X_\delta. Their states are initially inequivalent. Suppose k>j is an active member of S, and they are still inequivalent on O_k. The q_n^k-sequence also lies in O_k and peaks there. Each of these three particular states has a unique extension to X_k, by Lemma 2.3; this does not use the theorem being proved.

In X_k, the q_n^j-state is equivalent to the p_n-state, because step j has already been performed. The q_n^k-state is inequivalent to that state, because step k is active. Equivalence on O_k would pass to these particular unique extensions. Consequently, on O_k the q_n^k-state is equivalent to neither the p_n-state nor the q_n^j-state. The k-bridge joins only the classes of its designated pair, by Theorem 4.1. It therefore does not join the two states of the omitted j-bridge. At limits their inequivalence persists by Lemma 5.2. Transfinite induction proves the claim.

The omitted adjunction commutes past the inherited steps. Write H_j for the reduced HNN adjunction with the fixed data (p_n),(q_n^j). We claim, respecting the trace and named generators,

This is true initially. At an active k\in S above j, the data for both adjunctions lie in O_k. The expectation from R_k=H_j(O_k) onto O_k, followed by either k-data expectation, is a trace-preserving expectation onto that data algebra. By Lemma 2.8, it agrees with the expectation supplied there by the corresponding peaking sequence. The same observation applies with j and k interchanged. Thus the reordering isomorphism Proposition 3.4 applies and gives

proving the next instance of (5.3). Other successor steps are identities. Figure 1 displays the two orders at an active index k\in S.

At a limit, the isomorphisms already constructed agree on overlaps because they fix the named generators. The expectations from the O_\beta onto the two fixed abelian algebras generated by (p_n) and (q_n^j) also agree on overlaps, by Lemma 2.8. Applying Lemma 3.5 to these fixed data identifies the closure of the preceding H_j(O_\beta) with H_j of the limit base, proving (5.3) there.

At \beta=\delta we have O_\delta=A[S\cup j] and R_\delta=A[S\cup(j+1)]. The two sequences still peak to inequivalent states on O_\delta, by the first part of the proof. Thus Theorem 4.1 applies to (5.3), giving (U) for

If j\in S or j is inactive, this inclusion is the identity. The continuous chain \bigl(A[S\cup j]\bigr)_{j\le\delta} starts at A[S] and ends at X_\delta; continuity follows because a polynomial involves only finitely many indices outside S. Now Lemma 5.1 proves the theorem.□

Proof. Construct increasing countable dependency-closed sets S_n, starting with S_0=\varnothing. Choose a countable dense subset D_n of A[S_n]. For each a\in D_n and each integer m\ge1, Lemma 2.4 supplies a positive contraction x_{a,m} in X_\delta satisfying

Each witness has countable support. Adjoin all these supports to S_n and close under dependencies to obtain S_{n+1}.

Set S=\bigcup_n S_n and B=A[S]. Suppose h|_B=t h_1+(1-t)h_2 for states h_1,h_2 and 0<t<1. Each h_i takes value one on every witness x_{a,m}. A positive contraction of state value one fixes the state’s GNS vector, so h_i(a)=h_i(x_{a,m}a x_{a,m}) and h_i(x_{a,m}^2)=1. Hence

Letting m tend to infinity shows that both states agree with h on every D_n. Their union is dense in B, so h|_B is pure. Its unique extension is h by Theorem 5.5.□

Proof. There are at most

countable subsets of \delta. For each one, the corresponding algebra A[S] is separable and has at most \mathfrak c states: values on a countable dense subset determine a state. By Lemma 5.6, every pure state of X_\delta is the unique extension of one such restriction. There are at most \mathfrak c\cdot\mathfrak c=\mathfrak c possibilities.□

Proof of Theorem 1.1. Set \kappa=\mathfrak c^+. Every ordinal below \kappa has cardinality at most \mathfrak c, and every countable subset of \kappa is bounded: otherwise \kappa would be a countable union of sets of cardinality at most \mathfrak c, a contradiction.

Fix the schedule first. Before constructing any algebras, choose an injection

To do so, enumerate \kappa\times\kappa in order type \kappa. When a pair (\beta,\eta) is encountered, choose an unused ordinal at least \beta. Fewer than \kappa ordinals have been used, whereas the tail above \beta has cardinality \kappa. This defines the entire schedule without referring to any states.

Construct the algebras in ordinal order. Build a continuous bridge iteration (X_\beta)_{\beta\le\kappa} starting from the algebra of Lemma 5.3. For each \beta<\kappa, once X_\beta has been constructed, choose a list

containing every pure state of X_\beta, with repetitions allowed. This is possible by Corollary 5.7, since |\beta|\le\mathfrak c for \beta<\kappa.

At stage j<\kappa, let f be the state of X_j peaked by (p_n). If j is outside the range of \sigma, make the identity step. Otherwise write j=\sigma(\beta,\eta) and let g be the unique lift of h_{\beta,\eta} to X_j. The list for \beta is already available because \beta\le j; when \beta=j, it is chosen before the successor step. If g\sim f, again make the identity step.

Suppose g\not\sim f. By Lemma 5.6, choose a countable dependency-closed S\subset j such that g|_{A[S]} is pure and uniquely determines g on X_j. The restriction f|_{A[S]} is pure by Lemma 2.3. The algebra A[S] is separable, simple and unital by Lemma 5.4. Applying Theorem 6.1, choose an automorphism \gamma of A[S] with

If a state on A[S] takes value one on every q_n^j, composing it with \gamma gives the unique state taking value one on every p_n. Thus (q_n^j) peaks to g|_{A[S]}. By Theorem 5.5, it peaks to g on X_j as well.

The trace requirement deserves a separate check. For each n, \gamma carries the unital matrix stage M_{2^n}\subset A_0 onto a unital matrix subalgebra of A[S], and carries its minimal projection p_n to q_n^j. The ambient trace restricts to the unique normalized trace on this matrix algebra. Consequently,

This argument does not require an automorphism to preserve the specified trace on all of A[S].

We can therefore perform the bridge of Theorem 4.1 with (p_n),(q_n^j), and set I_j=S. At an identity step set I_j=\varnothing. At limit stages take the norm completion of the union. Each successor satisfies (U) and has a faithful trace-preserving expectation onto its base. The results of Section 5 therefore apply throughout the recursion.

All choices are set-sized. At every stage the algebra is generated by at most \kappa elements, so its density is at most \kappa and its underlying cardinality at most \lambda=\kappa^{\aleph_0}. Fix a set of cardinality \lambda. Algebra structures on its subsets, together with their norms, traces, embeddings, states, automorphisms and lists, form a set of codes. Well-order that set to make the required choices; recode successor algebras with their embeddings and use norm inductive limits. Ordinary choice and transfinite recursion in ZFC suffice.

Every final pure state is treated. Let A=X_\kappa and let h be any pure state of A. By Lemma 5.6, there is a countable dependency-closed S\subset\kappa such that h|_{A[S]} is pure and has unique extension h to A. Boundedness of S gives \beta<\kappa with A[S]\subset X_\beta. The state h|_{X_\beta} is the unique extension of h|_{A[S]} to X_\beta: any other extension could itself be extended to A, contradicting uniqueness there. Hence h|_{X_\beta} is pure.

Choose \eta with h_{\beta,\eta}=h|_{X_\beta} and put j=\sigma(\beta,\eta). Its lift to X_j is h|_{X_j}. At stage j this lift either already belongs to the p_n-class or is joined to it by the bridge. By Lemma 5.1, the equivalence persists to A. Thus all pure states of A have unitarily equivalent GNS representations.

Every nonzero irreducible representation \pi of the unital algebra A is unital: the range of \pi(1) is a nonzero closed invariant subspace. Every nonzero vector is cyclic by irreducibility, so each unit vector defines a pure state whose GNS representation is equivalent to \pi. The p_n-state supplies one such representation. Therefore A has exactly one nonzero irreducible representation up to unitary equivalence.

Finally, A is simple and has a faithful tracial state by Lemma 5.4 and the limit permanence facts. It is unital and contains the infinite-dimensional algebra A_0. An algebra K(H) is unital only when H is finite-dimensional: the identity on an infinite-dimensional Hilbert space is not compact, as an infinite orthonormal sequence has no norm-convergent subsequence. The finite-dimensional case, including H=0, cannot be isomorphic to A. This proves the theorem.□