Cuntz comparison and Jiang–Su absorption

Proof. For the first assertion, choose \|a-r^*br\|<\delta<\epsilon, and a continuous g\ge0, vanishing on [0,\delta], which is one on [\epsilon,\|a\|]. For z=b^{1/2}rg(a),

Thus z^*z and g(a)^2 generate the same hereditary algebra, which contains (a-\epsilon)_+. If w is the polar isometry of z, conjugation by w maps \her(z^*z) into \her(b). Moreover wx\in D for x\in\her(z^*z), by approximation using w|z|=z. Taking v=w(a-\epsilon)_+^{1/2} proves the assertion. Its matrix version gives the claimed hereditary placement.

For the second assertion, approximate a^{1/2} by a finite sum q=\sum_i x_i b y_i so closely that \|a-q^*q\|<\epsilon/2. Write q=XY, where X=(x_i b^{1/2})_i is a row and Y=(b^{1/2}y_i)_i is a column. Then q^*q\le\|X\|^2Y^*Y, while YY^* belongs to the hereditary algebra of H=\operatorname{diag}(b,\ldots,b). Positive domination and equality of the classes of the two squared moduli give

Choose a comparison witness with error less than \epsilon/2 and apply the first assertion to obtain the claimed cut comparison. Alternatively, for a fixed comparison witness against \operatorname{diag}(b,\ldots,b), replacing b by (b-\delta)_+ changes the compression by at most \delta\|r\|^2. This also proves the assertion about an extra target cut.

Let C\subseteq D be hereditary and let (e_i) be an approximate unit of C. Given a,b\in C_+ and a witness r, first choose an index for which the errors in e_i b e_i and e_i a e_i are small. Then e_i r e_i\in C, and its compression is as close to a as desired. This proves the third assertion. The same compression in matrix algebras gives rectangular witnesses. More generally, an element whose two squared moduli belong to the specified hereditary corners lies in their rectangular block, as is seen by multiplying by approximate units on both sides.

For the last assertion choose x with \|u-x^*(y-\beta)_+x\|<\eta, and put

where the quotient is zero when its denominator vanishes. This is a continuous function vanishing near zero, and

The compression and the norm bound follow. If the original comparison is in the stabilization, compression to the given corner first gives x\in D.□

Proof. Suppose first that Equation (1.1) holds and that (l+1)x\le ly. Applying a functional gives \varphi(x)\le\varphi(y), with strict inequality whenever 0<\varphi(y)<\infty. If \varphi(y)=0, then \varphi(x)=0; if \varphi(y)=\infty, the non-strict inequality is automatic. Thus Equation (1.1) gives x\le y.

Conversely, suppose \Cu(A) is almost unperforated, and let x=[a] and y=[b] satisfy the two functional inequalities in Equation (1.1). In B=A\otimes\K, set I=\Ideal_B(b). The map that is zero on I_+ and infinite outside I_+ is a lower semicontinuous extended trace: a sum of positives belongs to an ideal exactly when both summands do, z^*z belongs to it exactly when zz^* does, and the ideal is norm closed. Its rank functional vanishes at y, so the first inequality forces a\in I. Lemma 2.1(2) now gives x\le\sup_{n\ge1}ny. For every functional normalized by \varphi(y)=1, the second inequality gives \varphi(x)<1. The comparison criterion of [15, Proposition 6.2] therefore yields x\le y.□

Proof. Matrix-unit multipliers extract a nonzero matrix entry from any nonzero ideal of B and put it into every matrix position. Simplicity of A then implies simplicity of B. If p\in B were minimal, Bp would be a Hilbert space with inner product x^*y=\langle x,y\rangle p. Left multiplication is faithful by simplicity, and the elements xy^* act as rank-one operators. Since p generates B as an ideal, they span a dense subspace of B, giving B\cong\K(Bp). The multiplier corner 1\otimes e_{11} would consequently identify A with compact operators on a Hilbert space, a contradiction.

There are M nonzero mutually orthogonal positives e_1,\ldots,e_M in \her(e). Indeed, split a nonzero positive element at two distinct nonzero spectral points. If no such points exist, the element is a multiple of a projection, whose corner is not one-dimensional. A shifted nonscalar selfadjoint in that corner has two distinct positive spectral points and can be split instead. Recursion gives the required family while preserving orthogonality.

For nonzero f,g\in B_+, simplicity gives fBg\ne0. If 0\ne v\in fBg, then [v^*v]=[vv^*] is a nonzero class below both [f] and [g]. Iterating this observation gives h\ne0 with [h]\le[e_i] for every i. Therefore M[h]\le\sum_i[e_i]\le[e].□

Proof. Every nonzero b\in B_+ is full. Lemma 2.1 therefore bounds each positive cut of any c\in B_+ by finitely many copies of [b]. If \varphi([b])=0, preservation of increasing suprema gives \varphi([c])=0 for every c. If 0<\varphi([b])<\infty, the same cut comparison gives finiteness on every cut, and the first observation gives positivity on every nonzero class. These alternatives exhaust all functionals.□

Proof. Fullness of h and Lemma 2.1 give [(a-\delta/4)_+]\le m[h] for some finite m\ge1. Apply that lemma in M_m(B), with cut parameter \epsilon=\delta/4 and comparison error less than \delta/4, to the source \operatorname{diag}((a-\delta/4)_+,0,\ldots,0) and target H=\operatorname{diag}(h,\ldots,h). The resulting placement matrix has only its first column nonzero, because all other diagonal entries of its initial squared modulus are zero. Thus there is a column v=(v_i)_{i=1}^m such that

Define

setting it equal to zero where the denominator vanishes. It is continuous and vanishes near zero because r vanishes on [0,\delta]. For z_i=v_i g(a), we have

For the containment, use z_i z_i^*\le\|g\|^2v_i v_i^* and heredity. Additivity and traciality give the asserted identity for every extended trace, with the same c.□

Proof. For q(t)=\min(t,1) on [0,\infty), scalar functional calculus gives the exact identity

Set R=\max(1,\|c\|). For every integer m\ge1,

Indeed, below the spectral threshold i the left side is zero, and above it the right side is one. Consequently,

The last inequality uses Equation (2.3) and hereditary containment. Summing and using (c-L)_+=0 proves Equation (4.1). Only the first rank d_\tau(c) remains as error; the number L of spectral levels does not occur in the error bound.□

Proof. The assertion is immediate for x=0. Otherwise write x=[a], where a\in B_+ has norm one, and put e=(a-1/2)_+. If \Fin(A) is empty, all functionals are zero or infinite on every nonzero class. Equation (1.1) then gives Nx\le x, while x\le(N+1)x is automatic. Take u=x. We henceforth assume \Fin(A)\ne\varnothing.

Choose supports with summable relative rank. Choose a real number s and integers M_j\ge1, j\ge1, such that

Lemma 3.1 supplies nonzero h_j\in B_+ with M_j[h_j]\le[e].

To recover the full rank of a from functions vanishing near zero, choose increasing cutoffs F_j converging to one on (0,1] and transfer their positive increments. Concretely, set

The functions F_j increase, and r_j\ge0 vanishes on [0,2^{-j}]. Put p_j=r_j(a). Lemma 4.1 gives c_j\in\her(h_j)_+ satisfying

Round and sum inside the Cuntz semigroup. Choose integers L_j>\|c_j\| and put

For completeness, this supremum is represented by a positive element of B: choose representatives of w_j, rescale them to norm at most 2^{-j}, and put them in mutually orthogonal stable blocks. Their sum converges in norm. Each finite partial sum is Cuntz below it. Conversely, comparison of all partial sums with one target passes to the norm limit, proving the least-upper-bound property. Therefore

Fix \varphi\in\Fin(A) and choose its representing trace using Equation (2.4). Equations (4.1) and (5.2) give

The finite value \varphi([e]) is the only trace-dependent constant in the error estimate.

Recover the rank of a. Telescoping gives

We claim that

The values increase and are at most d_\tau(a) by Equation (2.3) and hereditary containment. Given 0<\epsilon<1, for sufficiently large k we have F_k(t)=1 on [\epsilon,1]. Thus, simultaneously for all integers m\ge1,

Taking traces and then letting m\to\infty gives d_\tau((a-\epsilon)_+)\le\tau(F_k(a)). The cut-supremum formula in Equation (2.2) now proves Equation (5.6).

Sum Equation (5.5), then use Equations (5.4) and (5.6). In extended nonnegative arithmetic, for every \varphi\in\Fin(A),

If \varphi(x)<\infty, Equation (3.1) implies

Since \varphi(x)>0, Equation (5.1) yields

If instead \varphi(x)=\infty, then E\varphi([e])<\infty and Equation (5.7) forces \varphi(u)=\infty. No subtraction of infinite values is used. Since \Fin(A) is nonempty, these conclusions also prove u\ne0.

Apply comparison with each target. Zero functionals vanish at both x and u, and the everywhere-infinite functionals are infinite at both. For functionals in \Fin(A), the preceding estimates give

The first inequality is strict whenever 0<\varphi(x)<\infty. For the second, if 0<\varphi((N+1)u)<\infty, then the functional belongs to \Fin(A) and the infinite-value case excludes \varphi(x)=\infty. Equation (5.8) makes that inequality strict as well. Applying Equation (1.1) twice proves Nu\le x\le(N+1)u.□

Proof. Theorem 5.1 gives the full almost-divisibility sandwich, and Lemma 2.2 gives almost unperforation.□

Proof. Choose an integer N with 1/N<\eta. There are N mutually orthogonal nonzero positive elements in A: repeatedly split a positive element at two distinct positive spectral values. If a nonzero hereditary corner had no such element, it would contain a minimal projection, and simplicity would identify A with compact operators; unitality would then make A finite dimensional. For any two nonzero positives e,f, simplicity supplies 0\ne v\in eAf. The class [v^*v]=[vv^*] is nonzero and below both [e] and [f]. Iterating over the orthogonal family gives 0\ne h\in A_+ with

Fullness gives finitely many s_i\in A with \sum_i s_i^*hs_i=1. To see the exact equality, choose an invertible finite sum S=\sum_i r_iht_i approximating 1. For the row R=(r_ih^{1/2})_i and column D=(h^{1/2}t_i)_i, the identity S=RD gives S^*S\le\|R\|^2D^*D, so D^*D is invertible. Set s_i=t_i(D^*D)^{-1/2}. The element

therefore satisfies \tau(c)=\tau(p) and d_\tau(c)\le1/N for every \tau\in T(A). Choose an integer L>\|c\| and let w=\bigoplus_{j=1}^L(c-j)_+. Lemma 4.2, applied in the stabilization, gives

This proves the required simultaneous inequalities.□

Proof. The algebra A\otimes\Z and its stabilization absorb \Z. Rørdam’s Theorem 4.5 [26] makes their W-semigroup almost unperforated. Transport through \Cu(\iota) proves (1.2).

For an integer N\ge1, Rørdam’s divisibility lemma [26, Lemma 4.2] supplies z\in\Z_+ such that

For x=[b], tensor these comparisons with b. Comparison is preserved by tensoring with a fixed positive element [26, Lemma 4.1]: tensor the witnesses and use a bounded approximate unit on the other factor. Thus

The middle class is exactly \Cu(\iota)(x). Pulling back the outside class proves (1.3).□

Proof. Let \varphi\in\F(\Cu(A)). By Equation (2.4), it is represented by an extended trace on A\otimes\K. Restricting that trace to the original corner and applying the hypothesis gives

The inequality is strict whenever 0<\varphi([w])<\infty. Since \Cu(A) is almost unperforated, Lemma 2.2 applies and gives [v]\le[w].□

Proof. Choose u=[r] with N[r]\le[b]\le(N+1)[r]. By (2.1), a sufficiently small \eta>0 gives

Lemma 2.1 places the hereditary algebra of \operatorname{diag}((r-\eta)_+,\ldots,(r-\eta)_+), with N diagonal entries, inside \her_B(b). In this hereditary matrix algebra take the cone e_{ij}\mapsto e_{ij}\otimes g(r), where 0\le g\le1, g=0 on [0,\eta], and g=1 on [2\eta,\infty). Transport it by the polar isometry. Every diagonal acts as one on the corresponding transported (r-2\eta)_+, proving both its cut comparison and the norm-one assertion. The off-diagonal entries give equivalence of the orthogonal weights.

For the second assertion do this in each summand, with N=\lceil k/q\rceil and \epsilon=1/2, and restrict to the first k slots. The complete N-slot weight lies in \her(b), so equivalence and additivity of rank give

The retained narrow classes and the cut comparison give

All comparisons take place in the indicated hereditary algebra, by Lemma 2.1. Thus arbitrary bounded traces on that algebra may be used, whether or not they extend boundedly to A. Summing the finitely many summands proves the claim.□

Proof. Put H=s(\lambda-K)s. For a vector \xi in the positive spectral space of H,

The range closure of sH_+^{1/2} is the support space of b. It is contained in the closure of the images s\xi just used. Continuity of the quadratic form proves the compression bound.

If b=0, then H_+^{1/2}s=0. Since H_+ is supported on p_s, and the range of s is dense in its support space, this implies H_+=0, as required.□

Proof. For each j, choose an ultrapower solution of the first j conditions to accuracy 1/j, and choose bounded coordinate representatives. Slightly enlarging each fixed bound once, and truncating representatives, gives bounds independent of j. There is an \omega-large set on which these first j coordinate conditions hold to accuracy 2/j. Replace these sets by decreasing finite intersections, also intersecting the j-th set with \{n\ge j\}. They have empty intersection. At coordinate n, use the solution of the largest applicable level j\le n, with an arbitrary bounded choice when no level applies. The chosen level tends to infinity along \omega; each fixed condition therefore has error tending to zero. The bounds ensure that every unknown represents an element of the norm ultrapower. Positivity or complete positivity can be preserved by choosing positive or completely positive lifts at each level.□

Proof. Set T_0=\operatorname{diag}(w_1^*w_1,\ldots,w_m^*w_m). The hereditary conditions and [w_i^*w_i]=[w_iw_i^*] give

By divisibility in (C), choose a positive element r in a stabilization with

Almost unperforation, applied with m, gives [r]\le[E] and [r]\le[D]. Choose \eta>0 sufficiently small that

This is possible because the class on the left is compactly contained in [T_0], and the classes (m+2)[(r-2\eta)_+] increase to (m+2)[r].

The cut-placement result in the preliminaries supplies partial isometries u_E,u_D which identify \her((r-\eta)_+) with hereditary subalgebras of \her(E) and \her(D), respectively. Their initial projection is the support of (r-\eta)_+. Choose a continuous function j with 0\le j\le1, j=0 on [0,\eta], and j=1 on [2\eta,\infty). Then

belongs to M_{l,1}(B): factor it as (u_Dj(r)^{1/2})(u_Ej(r)^{1/2})^*. Each factor belongs to the ambient stabilized algebra by cut placement, and the product’s left and right supports lie in the prescribed finite matrix corners. It is a contraction, and

If \|\sigma(w_i)\|\ge\delta for some i, then \sigma((T_0-\delta^2/2)_+)\ne0. By (7.3), the source copy u_E(r-2\eta)_+u_E^* is nonzero in this representation: otherwise the comparison would make the represented left side zero. The operator \sigma(v^*v) is the identity on the support of that nonzero copy, since j is one on its spectrum. Thus \|\sigma(v)\|=1.□

Proof of Lemma 7.1. Suppose that no such constant exists. Fix numbers c_n\in(0,1) increasing to 1. For each n, choose an algebra satisfying (C) and data a,b,z for which the conclusion fails with c=c_n. The algebras are allowed to vary; we suppress n until the limiting argument. The construction below always gives (7.1), so failure will provide a representation in which both constructed positives vanish.

Step 1: a matrix corner and two prospective supports. Apply Lemma 6.3, with seven slots and a cut of a below 1/2, to obtain a c.p.c. order-zero map M_7\to\her(a) with weight d satisfying

Its supporting matrix units identify

where B is the hereditary algebra generated by its first diagonal weight and P\in B is a strictly positive contraction. Concretely, in any representation, the support space of d is identified with \mathcal H^{\oplus7}, where B acts nondegenerately on \mathcal H. Choose Q=\min(1,2b) by functional calculus. Then Q\in\her(b) is a positive contraction and

We shall choose contractions v_0\in M_{2,1}(B) and v_1\in M_{3,1}(B). For any such choice set, in multiplier notation,

The expression under the square root is at least \frac14\,1, and U^*U=1. Define

These belong to \her(a) and \her(b), respectively. They are contractions: 0\le L\le1, and the selfadjoint operator inside the positive part in the second expression is bounded above by c_n^2t\,1. By Lemma 6.4, the compression of z^*Lz to the support of b' is at most c_n^2t. Since L\ge t p_{a'}, this implies

which proves (7.1) with c_n, for every choice of the two columns.

Step 2: data chosen before either column. Put Z=p_d zQ and write its rows in groups of sizes 1,2,3,1 as Z_{\mathrm f},Z_0,Z_1,Z_{\mathrm{last}}. These are operators from the original representation space to the corresponding copies of \mathcal H. The hypothesis on z gives

because p_d\le p_a and p_bQ=Q. Thus Z has norm at most one, even though no uniform bound on z itself is assumed.

The next choices address the only way the prospective supports can fail to retain visibility. For any choice of the columns, a representation in which a'=b'=0 would have P\le t and, by the zero case of Lemma 6.4,

As c_n\to1, these inequalities force the limiting range of Z into the range of U, with its U-coordinate in the t-spectral subspace of P. Step 5 proves this alignment precisely. Since the first row of U is 1/2, the first row of Z then recovers that coordinate. The following cutoffs implement this recovery within the hereditary corner, before either column has been chosen.

Fix continuous functions on [0,1] by

They vanish at zero, satisfy Gg=g, and g(t)=G(t)=1. Set

Here h_0\in B_+, Y_0\in M_{2,1}(B), and Y_1\in M_{3,1}(B). For example, their entries are the appropriate entries of sandwiches of zQ^2z^* by g(d) and G(d); these sandwiches belong to \her(d). We have

In particular, G(P) acts as the identity on the support of h_0. None of these data depends on v_0 or v_1.

For later use fix

Thus f vanishes on [0,1/4] and is one on [1/2,\infty). With u=W_1f and c_f=(1-f^2)^{1/2}, put V=u+c_f in \widetilde B and define

This selfadjoint element records the prospective range complement of V. At this stage V need not be an isometry and T need not be positive. Step 7 will produce a reducing subspace in the failure limit on which V is an isometry and T is a nonzero projection. We use the positive spectral part of T now to prepare the second column, which will rule out that possibility. The bounds above give \|W_j\|\le32, \|u\|\le32, and \|T\|\le1089. Fix on \R the continuous functions

They vanish at zero, \beta vanishes on (-\infty,1/2], \beta(1)=1, and \chi\beta=\beta. Put

We have \|y\|\le4, and S is strictly positive in M_3(B_T).

Step 3: the second column and bounded comparison witnesses. Apply Lemma 6.3 with two slots inside \her((yy^*-10^{-3})_+), using the narrower cut (yy^*-10^{-2})_+. The plateau of the slot weights may be chosen equal to one on the transported narrower element. To use the contraction form of that Lemma explicitly, take b=(yy^*-10^{-3})_+/16 and \epsilon=9\cdot10^{-3}/16; positive rescaling preserves all the hereditary algebras and Cuntz classes in question. We obtain equivalent orthogonal positive contractions k,k'\in M_3(B_T) with

If this hereditary algebra is zero, take k=k'=0; the same assertions hold. Choose three matrices x_j\in M_3(B_T) witnessing

There are three matrices because the right side of the comparison is a threefold direct sum; rectangular witnesses can be split into its three blocks. Replace x_j by

where the quotient is defined to be zero when its denominator is zero. This is a continuous function of k. Then

Indeed the compression sum before replacement is positive and has norm less than 16, so each summand has norm less than 16. Also

These give q_j^*q_j\le10x_j^*(k-0.9)_+x_j and the stated bound. The positive gap between the two cuts is what makes these witnesses uniformly bounded.

Choose p'=\min(1,k/0.8) and put

We claim that

For the first inequality, sandwich \frac12\,1\le(p')^2+(1-p')^2 by S^{1/2}. The resulting two summands have the classes of p'Sp' and (1-p')S(1-p'), and p'Sp'\in\her(k). For the second inequality it suffices to check that k'\in\her(D'), since [k]=[k']. Orthogonality gives (1-p')k'(1-p')=k'. Since S is strictly positive, there are positive elements c_j\to k' with c_j\le C_jS for finite constants C_j: one may take c_j=e_jk'e_j with e_j=S(S+1/j)^{-1}. Consequently

which proves the inclusion and (7.16).

Almost unperforation now gives [b_T]\le[D']. Cut placement supplies a column v_1\in M_{3,1}(B_T) with

Since (k-0.8)_+(1-p')=0, its range condition implies

We shall also retain the following fixed continuous support identities:

The first follows from (7.14); the second holds because \chi(T)b_T=b_T and the entries of v_1 belong to B_T. These equations, unlike statements about varying support projections, pass directly to an operator ultraproduct.

Step 4: finite word tests and the first column. Let p be the continuous function of Y_0Y_0^* which is zero up to r^2, linear between r^2 and (r+1/n)^2, and one thereafter. Put

Choose an ordered alphabet consisting of all entries of

and their adjoints. The number of letters is independent of n, as all matrix sizes are fixed. Their norms are bounded independently of the instance: the bounds already given apply, \|s\|,\|E\|\le4, \|D\|\le256, and p,k,k',b_T,D',v_1 are contractions. Thus, for example, 1089 is a common bound on every letter. Neither v_0 nor U is a letter.

Apply Lemma 7.2 to the finite list of columns

including the empty word x=1, with threshold 1/n. Here \mathbf e_j is the jth scalar basis column. The left and right squared moduli of these columns lie in \her(D) and \her(E), respectively. The resulting contraction v_0\in M_{2,1}(B) satisfies

The last assertion is Lemma 6.4 applied to the pencil defining D. In addition, we have the exact operator identities

For the first use the right hereditary condition. For the second, the support of D is contained in that of s, and G(P) fixes the support of h_0 by (7.11). The choice of v_1, all the comparison witnesses, and the whole alphabet precedes the choice of v_0, so the construction is not circular.

Step 5: failure forces alignment in a Hilbert ultraproduct. Make the choices above at coordinate n and define a',b' by (7.7). Since they satisfy the required angle bound, the assumed failure gives an irreducible representation \sigma_n such that

Use the same symbols for the represented operators. The first vanishing gives L\le t, hence P=U^*LU\le t. By (7.4), the a-trigger is impossible. Therefore \|Q\|=1 by (7.5). The two vanishings also give the conditional inequalities (7.9) established in Step 2.

Pass along the fixed free ultrafilter to Hilbert-space ultraproducts of the representation spaces and of their first matrix-coordinate spaces. For clarity, such an ultraproduct is the completion of bounded sequences of vectors modulo those whose norms tend to zero along the ultrafilter, with inner product the ultralimit of the coordinate inner products. Uniformly bounded operator sequences act on these spaces, respect sums, products and adjoints, and have norm equal to the ultralimit of their norms. The last assertion follows by choosing a nearly norm-maximizing unit vector at each coordinate. Fixed continuous functional calculus is preserved, by uniform polynomial approximation on a common spectral interval. All operators used here are uniformly bounded: in addition to the alphabet this includes Q,Z,X,U,v_0, by (7.8), (7.11), and (7.6). We keep the same notation for their limits.

Equality holds throughout (7.9). In particular Z^*Z=Q^2 and Z^*(t1-UPU^*)Z=0. Since

is a sum of positive operators, both positive summands annihilate Z. For \xi=U^*Z, it follows that

The first row of U is 1/2, so X=2g(P)Z_{\mathrm f}=\xi. Consequently

The last equality uses the second identity in (7.24).

Step 6: the moving cutoff makes every word test vanish. Although the function defining p changes with n, p itself is a bounded alphabet variable. At each coordinate it commutes with R_0=Y_0Y_0^* and satisfies

Thus in the limit

The first two equations show that the support of p lies in the spectral subspace of R_0 for [r^2,\infty), hence in the closure of the range of Y_0, and therefore of v_0 by (7.26). On this last closed range the inequality v_0^*pv_0\le\frac14v_0^*v_0 says that the quadratic form of p is at most one quarter of the squared norm. This follows first on vectors v_0\eta, then by density; on the orthogonal complement p is zero. Hence

The last equation in (7.27), and invertibility of 1-p, now give R_0\le r^2. The other support equation then gives

Every fixed word x in the limiting alphabet satisfies

Indeed, if such an operator had norm \delta>0, its coordinate norms would be at least \delta/2 on an ultrafilter-large set. On an ultrafilter-large subset we also have n\ge\operatorname{length}(x) and 1/n\le\delta/2. The transfer condition used in (7.22) would give \|v_0\|=1 in the ultraproduct. The first equation in (7.24) puts the range of v_0^* in the h_0\ge1/2 spectral subspace. Inverting h_0 on that subspace, with inverse norm at most two, gives

This contradicts Y_0=v_0h_0 and (7.29), since r=1/8.

Step 7: a nonzero defect projection on the word space. Let \mathcal H denote the limiting single-coordinate Hilbert space and set

The alphabet includes adjoints, so \mathcal L reduces every letter and the C^*-algebra they generate. Restrict letters to this space, and matrix variables to its amplifications. Equation (7.30) gives D=0 on \mathcal L^{\oplus2}. Moreover E\mathcal H\subseteq\mathcal L, so the restriction retains \|E\|=1/2, hence \|h_0\|=1. We do not assert that v_0 reduces \mathcal L. All further calculations involve only alphabet operators and their continuous functions. The needed range information from v_0 has already become the identity

which follows on the full limiting space from W_j=r^{-1}h_0(v_0)_j^* and (7.24), and thus restricts to \mathcal L.

By the zero case of Lemma 6.4, D=0 implies s(\frac14-p)s\le0. Equation (7.28) gives the reverse inequality, so s(\frac14-p)s=0. Taking a positive square root shows (\frac14-p)s=0, or ps=s/4. Together with (7.29), this yields

Let e=1_{(1/4,\infty)}(h_0) on \mathcal L. The range of s is dense in (e\mathcal L)^{\oplus2}, so (7.32) implies

Thus W_1,W_2 restricted to e\mathcal L are isometries with orthogonal ranges. By (7.31), their ranges lie in the h_0\ge1/2 subspace, where f=1.

It follows that u=W_1f satisfies u^*u=f^2 and c_fu=0, hence also u^*c_f=0. Consequently

Its range-complement projection is exactly

Both the initial and final supports of u are contained in e, and f=ef. As c_f commutes with e, formula (7.33) shows T=eTe. In particular

This projection is nonzero. Indeed W_2f\ne0 because \|h_0\|=\|f\|=1 and W_2 is isometric on e\mathcal L, whereas

Its nonzero range is therefore contained in the range of T.

Step 8: the second column gives the contradiction. Since T is now a projection, b_T=\beta(T)=T and \chi(T)=T. The fixed identities (7.19) give

They hold after restriction because v_1 was included in the alphabet. As G(P) fixes the support of h_0 and T=eTe, it also fixes T; thus \operatorname{diag}(G(P))v_1=v_1. Using (7.13), (7.17), and (7.26), we obtain

In particular, \|yy^*\|\ge3/64.

The three uniformly bounded witnesses in (7.15) survive both the ultraproduct and the restriction, giving

Hence q=(k-0.8)_+ is nonzero: otherwise \|yy^*\|\le0.015<3/64, contradicting (7.35). On the other hand, (7.18) and (7.35) imply qy=0. The first fixed support identity in (7.19) puts the range of k, and hence of q, in the spectral subspace of yy^* for [10^{-3},\infty). Therefore

forcing q=0. This contradiction proves the existence of the constant c and completes the proof.□

Proof. Let I be the closed ideal generated by all the s_\alpha, and let q:A\to A/I be the quotient map. We claim that \|q(x)\|<\gamma. Otherwise, a pure state of the unitization of A/I can be chosen to take the value \|q(x)\| on q(x). For completeness, the states taking that value form a nonempty compact face of the state space: nonemptiness follows by extending the norm-attaining state of the commutative algebra generated by q(x) and the unit. An extreme point of that face is a pure state. Its value on q(x) is positive, so its irreducible representation restricts nontrivially to A/I. Pulling that representation back to A gives an element of \mathcal K(x,\gamma) annihilating every s_\alpha, a contradiction.

The strict quotient-norm inequality permits a finite sum

Indeed, these finite sums are dense in I, and the distance from x to I is strictly less than \gamma. Choose t>0 so small that

If a representation annihilates every cut appearing in this sum, then \|\sigma(x)\|\le\|x-y_t\|<\gamma. This proves the first assertions.

For the last assertion apply what was just proved to the given finite family and put f_t(u)=\min\{u/t,1\} for u\ge0. The function f_t vanishes only at zero, so it preserves supports. Whenever \|\sigma(s_i)\|>t, functional calculus gives \|\sigma(f_t(s_i))\|=1.□

Proof. Set

The space J is a closed left ideal, and H is a hereditary C^*-subalgebra. For the latter assertion, it is a closed selfadjoint subalgebra, and if 0\le a\le b with b\in H_+, then \lambda(a)\le\lambda(b)=0, so \pi(a)\xi=0.

We show that

The inclusion from right to left is immediate. For the converse, let \lambda(y)=0 and put \zeta=\pi(y)\xi, which is orthogonal to \xi. If \zeta=0, then y\in J. Otherwise, on K=\operatorname{span}\{\xi,\zeta\}, define R\xi=\zeta and R\zeta=0. The selfadjoint form of Kadison’s Transitivity Theorem [20] supplies selfadjoint h,g\in D whose restrictions to K are respectively \operatorname{Re}R and \operatorname{Im}R. Thus v=h+ig satisfies

Consequently y-v\in J and v\in J^*, proving (8.2).

Let (e_\alpha) be a positive contractive approximate unit for H and put q_\alpha=1-e_\alpha. Each q_\alpha fixes \xi. For z\in J, the element z^*z belongs to H, and hence

Its adjoint gives \|q_\alpha z^*\|\to0. Applying (8.2) to y-\lambda(y)1 now gives

Choose one index sufficiently late for all the finitely many tests.□

Proof. Apply Lemma 8.3 in \widetilde A to x, all b_i^*b_j, and all r_i(a)^*r_j(a). Their state values are respectively \lambda(x)>\gamma/2, \delta_{ij}, and zero. Choose the excision error at most \gamma/16 for x, and at most \eta^2/(16m) for each of the matrix entries. Put

Here \|v\|\le2, and

The estimate for x therefore gives P_0xP_0\ge(\lambda(x)-\gamma/4)P_0\ge(\gamma/4)P_0. For the matrices, each entry has error at most \eta^2/(4m). An r-by-r operator matrix whose entries have norm at most \delta has norm at most r\delta, by the Cauchy–Schwarz inequality for its row action. Since each matrix here has size at most m, both matrix errors are at most \eta^2/4.

Choose a continuous function f:[0,1]\to[0,1] that is zero on [0,1/2] and has f(1)=1, and set s=f(q)xf(q). This is a positive contraction in A, because A is an ideal in its unitization. Also \lambda(s)=\lambda(x)>0, so \sigma(s)\ne0, and p_s\le P_0. Compressing the preceding inequalities to p_s proves the claims. Further hereditary restriction only decreases the support projection, so preserves all the estimates.□

Proof. If \mathcal K(x,\gamma) is empty, take every t_i=0. Otherwise proceed as follows. Let c<1 be the constant in Lemma 7.1. Before each shrinking operation, apply the last assertion of Lemma 8.2 to the current finite family. This leaves every support unchanged and ensures that each representation in \mathcal K(x,\gamma) gives norm one to at least one of the current seeds.

For one prescribed triple i,j,z, let \beta=\|p_{s_i}zp_{s_j}\|, using the current seeds. If \beta=0, nothing is needed. Otherwise apply Lemma 7.1 to the current seeds s_i,s_j and to z/\beta. Their replacements have supports on which the compression of z has norm at most c\beta. For any \sigma\in\mathcal K(x,\gamma), choose a current seed of represented norm one. If its index is outside \{i,j\}, it survives unchanged. If its index is in \{i,j\}, the representation clause in Lemma 7.1 ensures that at least one replacement is nonzero. Thus the new family still covers \mathcal K(x,\gamma), allowing the next reparametrization.

All new supports are subprojections of their previous supports. Consequently no previously obtained compression bound can increase. For a fixed coefficient z, finitely many repetitions suffice: after r repetitions its bound is at most c^r\|z\|. There are only finitely many prescribed coefficients and pairs. Treating them successively therefore gives (8.8). A final support-preserving reparametrization gives (8.7).□

Proof. If \mathcal K(x,\gamma) is empty, take V=0. Otherwise construct the preceding word model and the seed in Lemma 8.4 for each representation in that set. The seeds cover the set. Lemma 8.2 therefore retains finitely many models, indexed by i=1,\ldots,m. Write n_i for their basis sizes, b_j^{(i)} for their basis representatives, l_j^{(i)} for their levels, and M_a^{(i)} and r_j^{(i)}(a) for their model matrices and residuals. Set N=\sum_i n_i.

Apply Lemma 8.5 with \delta=\eta^2/(2N) to all coefficients

where the residual indices have levels less than d. Let s_i denote the resulting seeds and P_i=p_{s_i}. All the estimates of Lemma 8.4 remain valid. The number of models and their sizes have already been fixed before this cross-compression tolerance is chosen.

By Lemma 6.3, choose c.p.c. order-zero maps \phi_i:M_k\to\her(s_i) such that

In fact it suffices to apply that Lemma to the fixed cut (s_i-1/2)_+. In the universal representation on \mathcal H, consider

The row B is only an operator on these support spaces. Its Gram matrix has (ij,hl)-entry P_i(b_j^{(i)})^*b_l^{(h)}P_h. The diagonal blocks differ from the corresponding identity blocks by at most \eta^2/4. The off-diagonal matrix has N rows and columns with entries bounded by \delta, so has norm at most N\delta=\eta^2/2. In particular,

Let

and let Q be the projection onto the slots of level less than d. The columns of (aB-BM_a)Q are r_j^{(i)}(a)P_i. Its squared norm is the norm of their Gram matrix. The local residual estimates and the same off-diagonal estimate therefore give

For z\in M_k, define operators on \mathcal L by

The map z\mapsto C_z is c.p.c. and order zero: it is the direct sum of the order-zero maps \phi_i, each multiplied by a nonnegative scalar at most one. It is supported on Q, since the taper T vanishes at level d. Put

The last formula shows that V(z)\in A, including when A is nonunital. Row compression gives complete positivity. Furthermore 0\le C_1\le1_{\mathcal L}, so \|V(1)\|\le\lambda_0\|B\|^2\le1; hence V is contractive.

For orthogonal positive contractions z,w\in M_k, we have C_zC_w=0. Thus

and (8.13) gives

This proves (8.9).

We next bound the commutator. Let E_l be the projection onto all slots of level l, summed over the prototypes. By (8.3), the selfadjoint contraction M_a has only its level diagonal and its two adjacent-level diagonals. The operators

each have norm at most one: their separate summands have orthogonal initial spaces and orthogonal final spaces. The level diagonal commutes with T, and

Within each prototype the entries of M_a are scalar and the operator \phi_i(z) is the same in every slot. Thus [M_a,D_z]=0, and

With R_a=aB-BM_a, selfadjointness gives R_a^*=B^*a-M_aB^*, and a direct expansion yields

Because C_z=QC_zQ, (8.14) bounds both outside error terms. Consequently

This proves (8.10); the estimate is uniform over the unit ball of M_k.

Finally fix \sigma\in\mathcal K(x,\gamma). The retained cover gives an index i with \|\sigma(s_i)\|=1, so that h_i=\sigma(\phi_i(1)) has norm one by (8.12). Denote the represented support by P_i^\sigma, and put X=\sigma(x), H=\sigma(V(1)). The unique level-zero slot of the i-th model has representative b_1^{(i)}=1 and taper coefficient one. Every summand of (8.15) at z=1 is positive, so H\ge\lambda_0 h_i. Also

where the first inequality follows by expanding P_i^\sigma X(1-P_i^\sigma)XP_i^\sigma\ge0, and the second uses (8.4). Since h_i=P_i^\sigma h_iP_i^\sigma,

Positive domination and \|X\|\le1 now imply

because \eta<1 gives \lambda_0\ge1/2. This proves (8.11).□

Proof of Proposition 8.1. Choose increasing finite sets F_n of selfadjoint contractions whose union is dense in the selfadjoint unit ball of A. Apply Lemma 8.6 with F=F_n, d=n, and \eta=(n+1)^{-1}, and denote its maps by \nu_n. Complete positivity and contractivity pass to the ultrapower. The error in (8.9) tends to zero, so the resulting map is order zero. The estimate in (8.10), density, and \|\nu_n(z)\|\le\|z\| imply that its range commutes with A. All maps have norm at most one, which supplies the bound required to form their representing sequences. The final estimate in Lemma 8.6 holds for every n and every \sigma\in\mathcal K(x,\gamma), with the same constant \gamma^2/32. This proves the stated simultaneous coordinate quantifiers.□

Proof. For a finite spectral resolution H=\sum_j\alpha_j e_j, the blocks e_iBe_j are orthogonal in L^2(\mathcal M,T). Thus

Approximate an arbitrary H in operator norm by finite spectral step functions of H. The corresponding functions f(H) also converge in norm. Since T(1)<\infty, passage to 2-norm proves the assertion. This passage is made separately in each finite matrix algebra; the resulting inequality has no factor involving its size or T(1).

For the row assertion use

This function has Lipschitz constant one, and f(H)=\left(\begin{smallmatrix}0&R\\R^*&0\end{smallmatrix}\right). The squares of the two commutator norms in the preceding inequality are respectively 2\|aR-Rm\|_{2,T}^2 and 2\|aW-Wm\|_{2,T}^2.□

Proof of Proposition 9.1. We will combine small matrix cones chosen in the commutants of the nuclear approximation maps. Their supports are chosen to make the model and overlap errors summable, after which a regularized row gives a contraction. A second support compression makes this row bounded below on the retained support. The polar isometry of the compressed row then transports an exact matrix cone to D while retaining the tracial mass needed for the stated estimate.

Fix F,\mathcal P,k,\delta. Our approximation parameter is 0<u<1/4. The notation o_u(1) below always means a quantity tending to zero uniformly over T_{\leq1}(D) with this fixed data. It has no dependence on the number of approximation summands.

The convex approximation and its weighted errors. The hereditary algebra D is nuclear. The convex order-zero return approximation is due to Hirshberg, Kirchberg, and White [18, Theorem 1.4]; it is also part of the stronger Brown–Carrión–White approximation theorem [6, Theorem 3.1]. This supplies a finite-dimensional unital algebra E, one c.p.c. map \alpha:D\to E, positive numbers \lambda_1,\ldots,\lambda_r with \sum_l\lambda_l=1, and c.p.c. order-zero maps \beta_l:E\to D, such that

is as close to the identity as prescribed on a given finite set. The integer r may depend on u without restriction. Put

Here \rho_l is the supporting homomorphism of the order-zero map [35]; it acts by multipliers of \her_D(h_l). In particular h_l commutes with \mu_l(D).

By approximating F, its squares, and a sufficiently late positive contractive approximate-unit element e\in D, we arrange

For completeness, the Schwarz inequality gives

whose norm is at most 3v if both relevant approximation errors are at most v. For the second estimate choose e so that \|a^{1/2}(1-e)a^{1/2}\|\leq u^4/2 for all targets, and approximate e within u^4/2. Since \theta_l(e)\leq h_l, this proves (9.5). All appearances of 1 in this proof are allowed in the bidual; no unit of D is assumed.

Choose the following explicit continuous cutoff on [0,1]:

Set g_l=g(h_l) and z_l=1_{(1-2u,1]}(h_l). Thus p_{g_l}\leq z_l\leq p_l, and the scalar inequality

holds by functional calculus. With

we have, for every a\in F,

Indeed, the contribution of a-\theta_l(a) is bounded by u^4 using (9.4) and the trace inequality \tau(z_lxz_l)\leq\tau(x) for x\geq0. On z_l, \|(\theta_l(a)-\mu_l(a))z_l\|\leq2u. Apply \|x+y\|_2^2\leq2\|x\|_2^2+2\|y\|_2^2 and \sum_l\lambda_l\tau(z_l)\leq1.

Weighted Cauchy–Schwarz now gives, for a\in\mathcal P,

To verify the replacement of \mu_l(a) by a, for example,

The same argument applies to z_l. The lost a-mass in the first inequality is at most u^3, by (9.6) and (9.5); in the second use z_l\leq1.

The first pencil: small supports and summable overlaps. For each l put

Construct maps \psi_l:M_k\to C_l inductively, writing t_l=\psi_l(1). Once the earlier maps have been chosen, we penalize both the model error and overlap with their weights by setting

These formulas define b_l\in C_l. To check membership before taking positive parts, sandwich each term of D_l by g_l. Terms involving a\in D then lie in \her_D(h_l), and terms involving \mu_l(a) lie there by multiplier action. The same is true of g_lS_lg_l. Averaging preserves this hereditary algebra and commutes with multiplication by g_l. Hence X_l belongs to it and to the commutant, as does b_l.

The inequalities X_l\leq p_l and X_l\geq-(\|K_l\|+1)p_l imply 0\leq(X_l)_+\leq p_l by scalar functional calculus. Therefore

The second inequality follows from (X_l)_+\geq X_l. Lemma 6.4, applied to the pencil in (9.10), gives

Use the tracial conclusion of Lemma 6.3 in C_l, which has the form (9.3), with q=\delta\lambda_l and positive contraction b_l. It gives a c.p.c. order-zero map \psi_l:M_k\to\her_{C_l}(b_l) whose weight t_l satisfies

for every bounded positive trace \sigma on C_l. In particular, on restricting \tau to C_l,

Here ranks are the traces of the corresponding support projections in the bidual. As p_{t_l}\leq p_{b_l}, (9.12) gives

We may remove \mathbb E_l inside these traces because t_l^2\in C_l. Both summands defining K_l are positive. It follows that

These are unweighted sums: the individual support allowances \delta\lambda_l in (9.14) have absorbed the convex coefficients. Their total is \delta, so increasing the number of approximation summands does not enlarge either error bound.

We next obtain a lower bound for their high spectral cuts. Let

Since b_l is a contraction supported under z_l, scalar functional calculus and (9.11) give

For a\in\mathcal P, the functional

is a bounded positive trace. Indeed \mu_l(a) is a positive contraction commuting with C_l, so positivity and the trace identity follow from those of \tau. Normality shows that d_{\sigma_{l,a}}(t_l)=\tau(\mu_l(a)p_{t_l}), and gives the same formula for every positive spectral cut. Thus (9.13) applies to this single weighted trace and yields \tau(\mu_l(a)s_l)\geq(\delta\lambda_l/3)\tau(\mu_l(a)e_l). Using (9.17) and 0\leq\mu_l(a)\leq1, we obtain

For the last term, remove the expectation against g_l^4\in C_l:

In the inequality use g_l^4\leq z_l, (9.7), and \tau(S_l)\leq\delta. Combining this with (9.8)–(9.9) proves

where one may take the explicit error

Regularizing the combined row. Form matrices over D and models in its bidual as follows:

Here \mathcal T(p)\leq\delta by (9.14). Each t_l commutes with \mu_l(a), so (9.15) says

The diagonal of W^*W-D_0 is zero, and its (i,j) entry off the diagonal is t_it_j. Consequently

Lemma 9.2 gives the first of the following estimates; the second follows by expanding the commutator of R^*R:

Finally, Q_0=1-(1+W^*W)^{-1} and G_0=1-(1+D_0)^{-1}. The resolvent identity and the contractivity of both resolvents show that

This regularization has removed every occurrence of \|W\| from the estimates.

The second pencil in one matrix commutant. Write B_0=\her_{M_r(D)}(D_0). Both W^*W and D_0 belong to B_0: for the former, use its entries t_it_j and an approximate unit obtained by functional calculus of D_0. Hence Q_0,G_0\in B_0. Let

be the diagonal of the supporting homomorphisms of \psi_l, whose unit in the bidual is p. Its values commute with \operatorname{diag}(t_l) and with m(a). To see the latter assertion, recover each supporting homomorphism by bounded strong limits from t_l and \psi_l(M_k), all of which commute with \mu_l(a). The multiplier assertion follows because its products with \operatorname{diag}(t_l) belong to M_r(D); the same observation shows that m(a) acts by multipliers on B_0 and commutes with p.

The row R is contractive, but its Gram operator Q_0 need not commute with \pi(M_k), so R\pi(\cdot)R^* need not have order zero. We use the small overlap cost in (9.22) to find a \pi(M_k)-invariant support on which R is bounded below. The polar isometry of the compressed row will then transport the matrix action while the estimates below control the lost tracial mass.

Let u_k,v_k\in M_k be the clock and shift unitaries. The map

is the trace-preserving expectation onto the commutant of \pi(M_k) in the corner with unit p. The usual matrix-unit calculation shows that the average kills the nonconstant matrix coefficients, and fixes the commutant. Since the identity conjugation is one of its k^2 positive summands,

Put

Then K\in B_0\cap\pi(M_k)', and

For the commutator use [m(a),G_0]=0, \|Q_0-G_0\|\leq1, (9.21), and the 2-norm contractivity of the expectation, which commutes with multiplication by m(a).

Choose the fixed piecewise linear cutoff

and, with L=(20k)^2, set

As for the first pencil, c\in B_0\cap\pi(M_k)' and

Lemma 6.4 gives p_cKp_c\leq L^{-1}p_c. For \xi\in p_c\mathcal H, (9.23) implies

On the same space G_0 has compression at least 1/10. Therefore

Furthermore m(a) commutes with g'. Lemma 9.2, applied to the scalar function t\mapsto t_+ in (9.25), yields the useful explicit bound

There remains a lower bound on the mass surviving this second pencil. Set s'=\operatorname{diag}(s_l). On s_l one has t_l>1/2, so t_l^2/(1+t_l^2)>1/5, and hence g's'=s'. The scalar inequality c^2\geq2c-p, followed by (9.26), gives

For a\in\mathcal P, m(a) is a positive contraction commuting with s'. Decomposing the trace along s' and p-s' shows that \mathcal T(m(a)c^2)\geq\mathcal T(m(a)s'c^2s'); the omitted diagonal contribution is nonnegative. Thus

Here we used (9.19) and (9.24). The argument only uses the commutation of m(a) with s'; it does not require c to commute with s'.

Polar correction and exact order zero. Put d=cQ_0c and let U be the polar partial isometry of the row Rc in the bidual. Equation (9.27) implies p_d=p_c, so U^*U=p_c. We claim that the row

belongs to M_{1,r}(D) and satisfies

We give uniform estimates for this passage, since r can grow with u.

For \eta>0 define the algebraic row

The domination c^2\leq20d implies a factorization c=d^{1/2}x with \|x\|\leq\sqrt{20}. One elementary proof is to define x^* initially on the range of d^{1/2} by x^*(d^{1/2}\xi)=c\xi; the stated domination makes this well-defined and bounded, and it extends by zero on the orthogonal complement. Using Rc=Ud^{1/2} now gives

For the last scalar bound split at t=\eta: below it use \sqrt t, and above it use 1-\sqrt{t/(t+\eta)}\leq\eta/t. In particular the norm limit Y lies in the rectangular algebra. Both Y and Y_\eta have right support under p, so

This uses \mathcal T(p)\leq\delta, rather than a bound on the trace of the full matrix unit.

At fixed \eta the intertwining error for Y_\eta tends to zero uniformly with u. To spell this out, (9.21) and (9.28) imply

On [0,1] the function (t+\eta)^{-1/2} has norm at most \eta^{-1/2} and Lipschitz constant at most \tfrac12\eta^{-3/2}. Expand aY_\eta-Y_\eta m(a) through its four factors and apply Lemma 9.2. This gives

Together with (9.32), this bounds the error for Y by the displayed right-hand side plus 2\sqrt{20\eta\delta}. First choose \eta small and then u small. This proves (9.30) without any restriction on r.

Define

These elements lie in D: the rows Y_\eta\pi(x) belong to M_{1,r}(D) by multiplier action, and converge in norm to Y\pi(x) by (9.31). The map is completely positive and contractive, since V(1)=Uc^2U^*\leq1. It is exactly order zero. Indeed c commutes with \pi(M_k), U^*U=p_c, and for orthogonal positive x,y\in M_k,

For a selfadjoint a\in F set \Delta_a=aY-Ym(a). Since m(a) commutes with \pi(M_k) and \|Y\|\leq1,

This proves (9.1) when u is sufficiently small. Finally Y^*Y=c^2 and \|Y\|_{2,\tau}\leq\sqrt\delta. Rectangular traciality and Cauchy–Schwarz give

Combining this with (9.29), and choosing u to make r_u+\sqrt\delta\sup_\tau\|\Delta_a\|_{2,\tau}<\varepsilon for the finitely many targets, proves (9.2). All choices were made for the entire family T_{\leq1}(D) at once, completing the proof.□

Proof. Let \mathcal C_k=C_0((0,1],M_k). Its unitization is the algebra of continuous M_k-valued functions on [0,1] whose value at zero is scalar. Realize \widetilde{\mathcal C}_k\otimes\widetilde{\mathcal C}_k as the algebra of continuous M_k\otimes M_k-valued functions F on the square satisfying

These conditions also make F(0,0) scalar. Here is a direct verification that these conditions characterize the tensor product. The tensor product contains every scalar continuous function on the square, and its evaluation at each point is the entire indicated fiber: use scalar functions vanishing at zero to realize matrix entries on either nonzero coordinate. Given a continuous function satisfying the conditions, choose, at each point, an element of the tensor product having the same value there. Continuity gives a neighborhood on which the error is less than a prescribed tolerance. A finite scalar partition of unity subordinate to these neighborhoods gives an approximation in the tensor product. Closedness proves the assertion.

Let S be the unitary that interchanges the two factors of \C^k\otimes\C^k, and put Q=(1-S)/2. Since S=S^*=S^{-1}, Q is a projection, and

is a unitary path from 1 to S. For z\in M_k define

At y=0 this is xz\otimes1, and at x=0 it is y1\otimes z. At the origin continuity follows from \|F_z(x,y)\|\le(x+y)\|z\|; no limit of the unitary itself is needed. Thus F_z belongs to the tensor product. Pointwise it is a positive scalar contraction times a *-homomorphism, so z\mapsto F_z is completely positive, contractive, and order zero. Its weight is (x+y-xy)1.

The cone homomorphisms associated to \phi_1 and \phi_2 extend to commuting unital homomorphisms from their unitizations. Nuclearity of the cones lets their product be evaluated on the spatial tensor product. The functions F_z vanish at (0,0). The ideal of functions vanishing there is generated by \mathcal C_k\otimes1 and 1\otimes\mathcal C_k: quotienting by those two ideals leaves only the scalar evaluation at the origin. Its image consequently lies in C^*(\phi_1(M_k),\phi_2(M_k)), even when B is nonunital. Evaluation of F_z gives the asserted map and its weight. Finally \phi(1)-h_1=(1-h_1)h_2\ge0, and similarly for h_2. The last assertion follows from containment of the range in the generated algebra.□

Proof. The sequences that vanish in the uniform 2-seminorm define a closed two-sided ideal J_2 of D_\omega. Indeed the uniform 2-seminorm is at most the norm, and \|uz\|_2,\|zu\|_2\le\|u\|\|z\|_2 for every bounded trace; the triangle inequality and the same inequalities establish ideality and closedness. Write B=D_\omega/J_2. Every choice \sigma_n\in T_{\leq1}(D) gives a bounded trace

on D_\omega, which factors through B by Cauchy–Schwarz. These choices are allowed to depend on n.

Fix 0<\delta<1. Proposition 9.1, applied at coordinate n with its auxiliary error tending to zero, has the following consequence. Given finitely many positive contractions in B and finitely many elements with which to commute, there is a c.p.c. order-zero map \psi:M_k\to B commuting with those elements and satisfying

for all the chosen targets and all the traces just described. To see the quantifiers, choose positive contractive lifts of the finitely many targets and bounded lifts of the tests, and apply that proposition separately at every coordinate, including real and imaginary parts of the tests, rescaled to contractions, with the requested errors tightened by their fixed norms. The coordinate errors tend to zero uniformly over T_{\leq1}(D). The coordinate maps are order zero, and their commutator sequences belong to J_2. This proves the assertion. The lifts exist by completely positive lifting for the finite-dimensional domain when a map, rather than a single element, is being lifted.

Start with joined weight H_0=0. Having constructed the preceding joined map, let r=1-H_j and prescribe the targets r^{1/2}ar^{1/2} for a\in\mathcal P. Prescribe commutation with the preceding map and with the fixed tests. Apply (10.5) and join the new map to the preceding one using Lemma 10.1. The new weight commutes with r, and hence, for any trace \sigma_\omega, the defect

satisfies

This follows by subtracting \sigma_\omega(r^{1/2}ar^{1/2}\psi(1)) from D_j(a); traciality and commutation of \psi(1) with r justify the subtraction without any commutation assumption on a. Since D_0(a)\le1, induction gives

Choose \delta so that the second term is less than \rho/4, and then a finite N so that the first term is less than \rho/4.

Lift the resulting c.p.c. map from B to coordinate c.p.c. maps into D. Its exact order-zero and commutation identities in B give (10.2)–(10.3) on an \omega-large set. Its trace estimates also give (10.4) on an \omega-large set. Otherwise choose, at every failing coordinate, a violating tracial contraction and one of the finitely many targets; pass to an \omega-large set where the target is fixed. Their ultralimit would contradict the strict bound \rho/2 from (10.6). Intersect these finitely many good sets and choose a coordinate. This proves all the required uniform estimates.□

Proof. We record the rank facts used in verifying ideality. For u,v,x\in A,

The first follows from Cuntz equivalence of the two squared moduli. For the second use u^*x^*xu\le\|x\|^2u^*u, and then the first identity for right multiplication. For the last use (u+v)^*(u+v)\le2(u^*u+v^*v) and the Cuntz inequality for a sum of positive elements.

For t>0 the polar cut

belongs to A, has modulus (|u|-t)_+, and satisfies \|u-u^{[t]}\|\le t. Consequently a fixed positive cut of a sum or product can be Cuntz compared to the modulus of a sum or product of sufficiently small polar cuts. In detail, choose t so small that the norm difference of the two expressions makes the difference of their moduli less than the desired cut. This is possible uniformly on the bounded sets at issue, by uniform continuity of z\mapsto|z|. Norm comparison and (10.8) then bound that cut rank by the sum of the ranks of the input polar cuts, or by one such rank for multiplication and adjoints. Apply (10.7) with error \eta/2 for each of the two summands, and with error \eta for a product or adjoint. Finite intersection of the good sets proves closure under addition, adjoints, and left and right multiplication by bounded sequences. Scalar multiplication follows in the same way.

Here are the perturbation details, including the zero-rank case. If (z_n) and (z'_n) represent the same element, then, for a fixed s>0, on an \omega-large set \||z'_n|-|z_n|\|<s/2. Therefore

by comparing |z'_n| with (|z_n|-s/2)_+ in norm, at error strictly less than s. This proves representative independence. If z^{(l)}\in J_b converges in the norm of A_\omega to z, fix a cut s>0 and choose l so that bounded lifts of z^{(l)} and z are close enough on an \omega-large set to give (10.9) with z^{(l)} on the right. Apply its predicate at cut s/2 and arbitrary error \eta. This proves norm closedness. When d_\tau(b)=0, every rank comparison in these arguments has an exactly zero right side on the common good set. No assertion about the uncut rank of a norm-small perturbation is being used.

Put I_0=J_b\cap C_0. Choose dense sequences (c_j) in the unit ball of C_0 and (y_j) in the unit ball of I_0. A quasicentral approximate unit for I_0 in C_0 supplies positive contractions e^{(l)}\in I_0 such that, for j\le l,

Choose positive contractive lifts e^{(l)}_n and bounded lifts of the c_j,y_j. For each l, there is an \omega-large set G_l on which these finitely many norm estimates hold with 2/l in place of 1/l and, simultaneously, for 1\le j\le l,

Each displayed inequality is one uniform coordinate predicate, so only finitely many predicates are intersected at this stage. In particular, this construction does not enumerate traces.

Replace G_l by the decreasing sets E_l=\{n\ge l\}\cap\bigcap_{r\le l}G_r. For n\in E_1 let l(n) be the largest l\le n with n\in E_l, and set e_n=e^{(l(n))}_n; put e_n=0 elsewhere. Then l(n)\to_\omega\infty. The norm tests imply that e=(e_n) commutes with C_0 and absorbs I_0. Given s,\eta>0, choose j with 1/j<s and then require l(n)\ge j and 1/l(n)\le\eta. Equation (10.10) implies (10.7). For zero-rank traces it continues to assert exact zero at each selected coordinate. Thus e\in J_b, completing the proof.□

Proof. The relative rank bound will control both the approximate-unit tail in the hereditary construction and the trace mass lost when we remove the ideal J_b. Write I=\Ideal_A(b) and choose a sequential positive contractive quasicentral approximate unit v_i for I in A. Fix t_i>0 tending to zero, and set

where \chi_i is zero on [0,t_i], increases linearly to one on [t_i,2t_i], and is one thereafter. We may assume 2t_i<1. Then p_iq_i=p_i, \|p_i-v_i\|\le2t_i, and (p_i) is still a quasicentral approximate unit. Moreover [q_i]\le[(v_i-t_i/2)_+]. Compact containment and v_i\in I give an integer L_i\ge1 with

No uniform bound on these integers is required.

Let D_i=\her_A(q_i). If d_\tau(b)<\infty, the restriction of \tau to D_i is a bounded positive tracial functional with

Indeed every positive contraction of D_i is supported by p_{q_i}, so its trace is at most d_\tau(q_i); an approximate unit obtained from q_i shows equality. Equivalently, use q_i^{1/m} and lower semicontinuity. Finiteness of the bound turns the additive tracial map on positives into a bounded positive linear functional on D_i. For d_\tau(b)=0 that functional is zero. For a positive finite d_\tau(b), dividing it by L_i d_\tau(b) gives an element of T_{\leq1}(D_i).

Choose a dense sequence (x_j) in the unit ball of A, and apply Lemma 10.2 inside D_i, with tests

and positive target p_iap_i. Choose its accuracy \rho_i>0 after L_i has been fixed, so that

for example \rho_i=1/(i(1+L_i)) works. Denote the resulting maps by \mu_i:M_k\to D_i, and put \theta_i(z)=p_i\mu_i(z)p_i. These are c.p.c. maps into A.

Their trace defects have the required relative bound already before correction. Put \lambda_i=\|a^{1/2}(1-p_i^2)a^{1/2}\|, so \lambda_i\to0. The positive element defining \lambda_i is supported by p_a; using (10.11), its trace is at most C\lambda_i d_\tau(b). The other part of the defect is, by traciality,

Thus

At zero rank both contributions are exactly zero.

Let \mu,\theta be the c.p.c. maps represented by the coordinate maps. Their relevant relations first hold modulo J_b. In fact, for each tested 2-norm defect z_i, normalization in (10.15) and functional calculus give

The estimate holds for every fixed s>0; at zero rank it holds because all the coordinates at issue belong to I and the restricted trace is zero. Equation (10.16) proves that these defect sequences belong to J_b. Consequently, modulo J_b, the map \mu is order zero and commutes with p=(p_i) and each (q_i x_jq_i). Also [p_i,q_ix_jq_i]=q_i[p_i,x_j]q_i\to0 in norm. It follows that \theta=p\mu(\cdot)p is order zero modulo J_b and commutes there with (q_ix_jq_i). It commutes there with the constant x_j as well: using q_ip_i=p_i, the difference between the two commutators has norm at most 2\|[p_i,x_j]\|. For example p_i\mu_i(z)p_i x_j(1-q_i) =p_i\mu_i(z)[p_i,x_j](1-q_i) for \|z\|\le1; the other term is its analogous left-sided expression. Density gives the assertion for all of A.

Apply Lemma 10.4 to C_0=C^*(A,\theta(M_k)), and let e be its absorbing positive contraction. Define

Since e commutes with the range, this is c.p.c. Every order-zero polynomial defect of \theta belongs to J_b\cap C_0 and is annihilated by 1-e. Multiplication by (1-e)^2 therefore proves the exact order-zero identities for \phi. Likewise [\phi(z),x]=(1-e)[\theta(z),x]=0 for x\in A. Thus \phi:M_k\to A_\omega\cap A' has the required norm relations.

To check the trace loss, the removed weight j=e\theta(1) is a positive contraction in J_b. Choose positive contractive representatives j_i and h_i of j and h=\phi(1). Then \theta_i(1)-h_i-j_i\to_\omega0 in norm. For s>0, functional calculus gives j_i\le s1+(j_i-s)_+, whence

Here the second term bounds the trace of a compression of the positive contraction (j_i-s)_+, and its rank is unchanged by a positive rescaling if necessary. Given a desired relative error, first choose s so that sC is small, then use (10.7) at that fixed cut with a sufficiently small error. Norm-null differences of the lifts contribute at most their norm times d_\tau(a)\le C d_\tau(b). Combining these observations with (10.17) proves (10.12), uniformly for positive finite d_\tau(b). If d_\tau(b)=0, then d_\tau(a)=0 and every positive contraction h_i has exactly zero defect against a. This also proves the assertion for every choice of positive contractive representative of h.

For completeness, the passage to several pairs permits arbitrary finite coordinate tests. Construct the map for the first pair. For the next pair use the preceding construction and reindex its coordinate lifts as follows. At coordinate n the matrix-unit values of all already fixed maps are a finite list of ordinary elements of A. Norm centrality of the new map allows an index at which its commutators with that list and with the first n fixed dense tests are less than 1/n. Intersect this set of indices with the sets imposing its order-zero errors and its uniform trace predicate for its fixed pair, to any desired accuracy tending to zero. A finite intersection is \omega-large and hence nonempty. The selected sequence still has (10.12) and now commutes with the earlier maps. Repeat for the finite list of pairs and join.

The joined weight H dominates every original weight h^{(l)}. This domination also preserves the uniform trace estimates at coordinates. Indeed choose positive lifts D^{(l)}_n of H-h^{(l)}; for any positive contractive lifts H_n,h^{(l)}_n,

Their trace-defect difference against a_l is therefore at most C_l\delta_{l,n}d_\tau(b_l) when that rank is finite; at zero rank both defects vanish. Lift the joined c.p.c. map and select one coordinate in the finite intersection imposing all desired norm relations, all prescribed commutators, and all relative trace bounds with strict slack. This gives (10.13). The finite set F, and each \eta_l, are arbitrary at this final step.□

Proof. First consider finitely many tests. For one of them apply Proposition 8.1 with

It gives a norm-central order-zero map with positive contractive weight representatives t_n and a constant 0<\kappa\le1 such that, on a common \omega-large set,

If no representation satisfies the trigger, one can take the zero map and any 0<\kappa\le1; the assertions below then remain vacuous on the relevant representations. Choose a continuous positive contraction \chi on [0,1], zero at zero and equal to one on [\kappa/2,1], and reparametrize the order-zero map by \chi.

Do this for each of the finitely many tests, choosing commuting maps by successive reindexing. Here reindexing preserves the representation quantifier in (11.3). More explicitly, at coordinate n, previously chosen matrix-unit lifts form a finite list of elements of A. Choose an index of the next map at which its norm commutators with those elements and the first n dense tests are small, its order-zero defects are small, and its own uniform implication (11.3) holds. Each condition holds on an \omega-large set, and a finite intersection is nonempty. The implication is simultaneous over all representations at the selected index, so it is retained without choosing representations during reindexing. Join the reparametrized maps using Lemma 10.1.

Let \nu denote this finite join for the moment, and choose c.p.c. coordinate lifts with weights h_n^\nu. We claim that, on an \omega-large set, simultaneously in all irreducible representations,

Suppose instead that its failure set belongs to \omega, and choose a failing representation \sigma_n at each such coordinate. Intersect with the good set for (11.3). Since

that trigger applies, even if the represented norms tend to the boundary 3\eps/32.

Take the Hilbert space ultraproduct of these representations, and write \bar a,\bar t,\bar H for the resulting positive operators representing a, the old unmodified weight for this test, and the joined weight. Norm centrality gives [\bar a,\bar t]=0, and (11.3) gives

Consequently the joint spectral part where \bar a>\eps/16 and \bar t>\kappa/2 is nonzero. Indeed the product is zero where \bar a\le\eps/16 and has norm at most \kappa/2 where \bar t\le\kappa/2. On the indicated nonzero part the reparametrized weight \chi(\bar t) is one. The joined weight dominates this weight and commutes with it and with \bar a; being a contraction, it too is one there. It follows that

This contradicts the upper bound at the failing coordinates. Arbitrary positive weight lifts of the joined map give the same limiting operators, so no particular lifting convention was needed for this contradiction. This proves (11.4) for each test in the finite group.

For each l perform this construction for the first l tests. Take c.p.c. coordinate lifts of the resulting maps. Choose a decreasing sequence of \omega-large sets on which the first l norm centrality tests and matrix-unit order-zero identities hold to accuracy tending to zero, and all the first l implications (11.4) hold exactly. Include \{n\ge l\} in the lth set, and use its largest admissible stage at coordinate n. Lemma 6.5 applies, since all these maps and weights are contractions. The diagonal defines a single norm-central order-zero \nu, with c.p.c. lifts, and preserves each of the implications on an \omega-large set.

At a coordinate where (11.4) holds, every irreducible representation annihilating (y_n^\nu-\eps/20)_+ annihilates (a-3\eps/32)_+. Indeed annihilation of the former is precisely the condition \|\sigma(y_n^\nu)\|\le\eps/20, which implies \|\sigma(a)\|\le3\eps/32. Primitive ideal separation gives

The additional positive cut \eps/8-3\eps/32=\eps/32 now gives a finite integer L_{a,\eps,n} in (11.2), by the compact-cut ideal comparison from the preliminaries. Choose the integer to be at least one and put it equal to one outside the good set. This makes all constants specified at all coordinates and proves the proposition.□

Proof. Apply Proposition 11.1 for this k. Fix its coordinate maps \nu_n, their weights, and every integer L_{a,\eps,n} before making any subsequent choices. They will not be reindexed in this proof.

For a test (a,\eps) retain b=(a-\eps/8)_+ and put d=(a-\eps/4)_+. Then d\in\Ideal_A(b) and [d]\le[b], so Proposition 10.5 applies to (d,b) with relative rank constant one. Let E_r be the good sets in (11.2) for the enumerated tests. Form decreasing \omega-large sets

and let l(n) be the largest l\le n with n\in G_l, or zero if there is none. Thus l(n)\to_\omega\infty, and all ideal bounds for the first l(n) tests are valid at coordinate n. Choose positive numbers \eta_n\to0, for example \eta_n=1/n.

At coordinate n use the finite-coordinate conclusion of Proposition 10.5 for the first l(n) pairs (d_r,b_r), with the individually prescribed accuracies

Include the first n members of a fixed dense family in A, and all matrix-unit values \nu_n(e_{ij}), in the commutator tests. Ask for norm commutators and norm order-zero defects less than 1/n. Denote the resulting c.p.c. map by \theta_n, and its weight by h_n^\theta. Where l(n)=0 one may use the zero map. Every finite problem here has already specified its finite integers; its requested trace accuracy is allowed to be arbitrarily small. Consequently the maps represent a norm-central order-zero map \theta commuting with \nu, and, for each fixed test, on an \omega-large set,

At zero rank this asserts exact zero; alternatively it follows directly from [d]\le[b].

Join \theta and \nu to obtain \zeta^k. Use the following specific positive contractive representatives of its weight:

They belong to A: equivalently, they are h_n^\theta+(1-h_n^\theta)^{1/2}h_n^\nu(1-h_n^\theta)^{1/2}. They lie between h_n^\theta and 1. Since the two weights commute in the ultrapower, they represent the joined weight H. For u^k(a) use the positive contractive representatives

For y^k(a) choose positive contractive representatives y_n; for example, if B_n are positive contractive representatives of g(\zeta^k)(e_{11}), one can take y_n=B_naB_n. All products in (11.5) are positive because the matrix map is norm-central.

We first convert the norm visibility into an exact coordinate Cuntz inequality. In A_\omega the old weight h^\nu and H commute, h^\nu\le H, and both commute with a. Monotonicity of g in this commutative functional calculus gives

Write \varrho for the supporting homomorphism of \zeta^k. Since a commutes with the range of \zeta^k, it also commutes with the supporting matrix units: these are the strong limits of \zeta^k(e_{ij})(H+t)^{-1} as t\downarrow0 in the bidual. The right side is the orthogonal sum of

The products are in A_\omega. All z_i are Cuntz equivalent, as are their corresponding positive cuts. Indeed a^{1/2}g(H)\varrho(e_{ij})\in A_\omega has its two squared moduli equal to z_j,z_i; taking polar cuts gives the same fact for (z_j-t)_+ and (z_i-t)_+. The two positives in (11.10) commute, so scalar functional calculus, splitting a cut of an orthogonal sum, and enlarging its individual cuts give

Here the cut margins yield an exact inequality before any multiplication by a varying integer. Put \Delta=\eps/20-\eps/22>0. Choose one rectangular comparison witness for (11.11) in M_{k,1}(A_\omega) with norm error less than \Delta/2. Such rectangular witnesses suffice because the source is in one corner and the target is the diagonal of k elements of A_\omega. Choose bounded coordinate lifts X_n of this witness. Since functional calculus commutes with the quotient, on an \omega-large set,

Norm comparison at that coordinate gives

Intersect with the good set for (11.2). Only now multiply the exact Cuntz inequality by L_n=L_{a,\eps,n} to obtain

The bound on X_n may depend on the fixed k and the test. No norm error is multiplied by L_n in this argument.

We next prove the coordinate rank gap

on an \omega-large set. Put c_n=(1-f(H_n))^{1/2} and v_n=c_ndc_n. Since \|a-d\|\le\eps/4, one has \|u_n-v_n\|\le\eps/4. Comparing u_n with (v_n-\eps/2)_+ gives

For the first inequality the norm error is at most \eps/4+\eps/2<\eps; the second is the elementary spectral trace bound. From the scalar inequality f(t)\ge t and (11.8),

This uses functional calculus on H_n itself and positive domination of H_n; it makes no operator-monotonicity assertion about f. Traciality and (11.7) consequently give, whenever d_\tau(b)<\infty,

This estimate handles every possible trace value. Set R_n=d_\tau((y_n-\eps/32)_+). If R_n<\infty, the exact inequality (11.12) gives d_\tau(b)\le kL_nR_n<\infty, and therefore

For the fixed k and test, eventually 2k\eta_n/\eps\le1/2. If R_n=0, the same argument first forces d_\tau(b)=0 and then gives exactly zero on the left. If R_n=\infty, inequality (11.13) is automatic in [0,\infty]. All good sets used in these alternatives were chosen uniformly over the trace families; thus their intersection proves (11.13) simultaneously for every lower semicontinuous extended trace.

Apply Lemma 6.2 in A at each coordinate of this common good set. It gives

Choose x_n\in A with

These witnesses need not themselves be bounded. Define the continuous scalar function

and set r_n=q_s(y_n)x_n on the good set and r_n=0 elsewhere. Then

The norm estimate follows from q_s(t)^2\le s^{-1}(t-s)_+. Thus these representatives have the fixed bound \|r_n\|^2\le64/\eps, and their ultrapower element has norm squared at most 32/\eps. The compression errors tend to zero along \omega, proving (11.6). Neither bound depends on k or on any L_n.□

Proof. For fixed m\ge1, apply Proposition 11.2 with k=pm and identify M_{pm}=M_p\otimes M_m. Define

These are norm-central contractions, and P_m is positive. The matrix-unit relations for the supporting homomorphism give

The scalar identity fg=g gives \alpha_m(e_{11})s_j^{(m)}=s_j^{(m)}. The weight of \alpha_m is f(H^{pm}). Under the indicated identification the first diagonal slot of M_{pm} is e_{11}^{(p)}\otimes e_{11}^{(m)}, so Proposition 11.2 supplies, for every test,

The witness norm bounds are independent of m.

We spell out the diagonalization because the number of slots grows. For every m choose c.p.c. coordinate lifts \alpha_{m,n}, contractive lifts s_{j,m,n} for 1\le j\le m, and positive contractive lifts P_{m,n}=s_{1,m,n}. Choose lifts of each comparison witness bounded by \sqrt{64/\eps}; the sharper ultrapower bound in Proposition 11.2 leaves room for this choice. Put h_{m,n}=\alpha_{m,n}(1). At stage l use m=l and require:

1. 1.

   the matrix-unit order-zero identities of \alpha_{l,n}, and its commutators with the first l dense tests of A, to have error less than 1/l;

2. 2.

   the centrality and the two algebraic relations in (11.15) for all slot indices at most l to have error less than 1/l;

3. 3.

   for the first l comparison tests, the equations with positive coordinate expressions

   \bigl((1-h_{l,n})^{1/2}a(1-h_{l,n})^{1/2}-\eps\bigr)_+ \quad\text{and}\quad r_{a,\eps,l,n}^*P_{l,n}aP_{l,n}r_{a,\eps,l,n}

   to have error less than 1/l.

For fixed l all these finitely many estimates hold on an \omega-large set, because their equations are exact in the ultrapower. The positive coordinate expressions represent the desired products since the weights and slots commute with A in the ultrapower.

Intersect the first l such good sets with \{n\ge l\} to obtain decreasing E_l\in\omega, and take m(n) to be the largest l\le n with n\in E_l (zero if none). At coordinate n use the row indexed by m(n). For a slot or witness whose index is larger than m(n), use zero at that coordinate. Every fixed variable is eventually present along \omega, since m(n)\to_\omega\infty. All maps and slots are contractions, and the witness for a fixed test has its individual bound \sqrt{64/\eps} independently of the row. These are precisely the boundedness hypotheses of Lemma 6.5. The selected maps therefore define a c.p.c. norm-central order-zero \alpha, and the selected slots and witnesses define elements with (11.15)–(11.16). No uniform trace estimate as m grows was needed: each fixed matrix size had already produced exact equations with the stated individual norm bounds.□

Proof. Let 0\ne x\in J_+. There is a pure state \lambda of the unitization D^{\sim} such that b=(\lambda\otimes\id_C)(x)\ne0. Indeed, product positive functionals detect nonzero positive elements of a spatial tensor product. Fixing the second functional gives a nonzero positive affine function on the state space of D^{\sim}, which is nonzero at some extreme point. The resulting pure state does not annihilate D. Its GNS representation is therefore nondegenerate on D; write \xi for its unit cyclic vector.

Lemma 8.3 gives, on any prescribed finite set of coefficients in D^{\sim}, positive contractions g\in D^{\sim} with

Here \pi_\lambda(g)\xi=\xi, since g is a positive contraction. Choose once and for all a positive contraction u\in D such that \|\pi_\lambda(u)\xi-\xi\|<1/8. Approximation of x by finite algebraic sums and excision of their first coefficients show that

The element used to approximate it is (ug\otimes1)x(gu\otimes1), which belongs to J; all products with 1 here are multiplier products. Moreover,

Thus the first factor retains a fixed norm lower bound as the excision accuracy improves.

We use the following elementary cut observation. If a,b are positive and \|a\otimes b+J\|<t^2, then

To see this, represent (D\otimes C)/J faithfully and nondegenerately. The commuting multiplier operators representing a\otimes1 and 1\otimes b have product of norm less than t^2. Their joint spectrum has no point with both coordinates larger than t. Joint continuous functional calculus proves (12.1). Now fix 0<t<\min\{1/2,\|b\|\} and choose g so that the preceding distance is less than t^2. Both cuts in (ug^2u-t)_+\otimes(b-t)_+ are nonzero, and their tensor product belongs to J.□

Proof. Let J_0 be the ideal generated by those tensors. If J_0\ne J, primitive ideal separation gives a nondegenerate irreducible representation \pi with \pi(J_0)=0 and \pi(J)\ne0. Let \pi_D and \pi_C be its commuting leg representations, and put I=\ker\pi_D, L=\ker\pi_C. Nuclearity of D and of its quotient D/I [11, Corollary 4], together with the quotient identity for the maximal tensor product, gives

For completeness, the maximal quotient identity follows by the universal property of commuting representations: a representation annihilating the indicated two leg ideals is exactly a pair factoring through D/I and C/L. The nuclearity of the first leg identifies both maximal products in this statement with spatial products.

The representation induced by \pi on the target of q is faithful. Here is the needed verification. Write M=\pi_D(D)'' and N=\pi_C(C)''. The center of M commutes with both legs, hence with the irreducible representation \pi; it consists of scalars. Thus M is a factor, and the same holds for N. If nonzero positive elements x\in M, y\in N had xy=0, their support projections e\in M, f\in N\subset M' would satisfy ef=0. Since f commutes with M, it annihilates every vev^*, v\in\mathcal U(M). The join of these projections is the central support of e, which is 1 in the factor M. This would give f=0, a contradiction. The leg representations of D/I and C/L are faithful, so the induced tensor representation kills no nonzero positive elementary tensor. Lemma 12.1 applied to its kernel proves faithfulness.

The nonzero ideal q(J) therefore contains a nonzero positive tensor \bar a\otimes\bar b. Choose positive lifts a\in D, b\in C. Since a\otimes b\in J+\ker q, compression by (1-e)\otimes(1-f), where e and f run through positive contractive approximate units of I and L, gives

Indeed these compressions send each term of a finite sum from I\otimes C+D\otimes L to zero in norm, and they preserve J. The quotient images of a_e,b_f remain \bar a,\bar b. Fix 0<t<\min\{\|\bar a\|,\|\bar b\|\}. Once the displayed distance is less than t^2, (12.1) yields (a_e-t)_+\otimes(b_f-t)_+\in J_0 with nonzero image under q. Faithfulness of the induced representation contradicts \pi(J_0)=0.□

Proof. Let \mathcal Q be the cone of maps a\mapsto\sum_{j=1}^N b_j^*wa b_j, where b_j\in E. It is convex: sums concatenate the lists, and multiplication by a nonnegative scalar is absorbed into the b_j. It suffices first to place \Phi in its point-weak closure.

Fix a_1,\ldots,a_m\in D and \ell_1,\ldots,\ell_m\in E^*. We may represent all these functionals in one nondegenerate cyclic representation \sigma:E\to\mathcal B(H), with cyclic vector \xi, in the form

To verify this, decompose the real and imaginary parts of each \ell_i into differences of positive functionals. A positive multiple of their sum dominates all the positive summands. In its GNS representation a dominated positive functional gives a bounded positive form on vectors \sigma(b)\xi; its representing operator commutes with \sigma(E), as is seen by moving \sigma(x) between the two arguments of the form. Linear combinations give the c_i in (12.4). Extend \sigma unitally to E^{\sim} when necessary.

Put C=C^*(1,c_1,\ldots,c_m)\subset\sigma(E)'. The commuting representations define

continuity for the spatial tensor norm follows from nuclearity of D. For q=v,w, let p_q be the support projection of \sigma(q) and let \pi_q be the restriction to p_qH. This is a reducing subspace for \pi: each q commutes with D, and each element of C commutes with E^{\sim}.

We claim

Suppose a\otimes c\in\ker\pi_w. Commutation with the weight and positivity give

Since c^*c commutes with all of \sigma(E), this identity persists after multiplying \sigma(wa^*a) on either side by elements of \sigma(E) and taking closed linear spans. The ideal hypothesis va^*a\in\Ideal_E(wa^*a) therefore gives \sigma(va^*a)c^*c=0. This says \pi(a\otimes c)\sigma(v)^{1/2}=0, and, by density of the range of \sigma(v)^{1/2} in p_vH, gives a\otimes c\in\ker\pi_v. Lemma 12.2 says that the elementary tensors in \ker\pi_w generate that kernel, and proves (12.5).

Let K=\ker\pi_w. The vectors

are dense in p_wH: cyclicity first gives density before application of \sigma(w)^{1/2}, and its range is dense in its support space. Finite sums of their vector functionals have weak-* closure equal to the positive functional cone of (D\otimes C)/K. Here is the separation argument. A selfadjoint element of this quotient that is nonnegative on all these functionals has a positive representing operator on p_wH, by the density of the \eta_b. Faithfulness of the quotient representation makes the element positive. Thus the polar of the cone of these vector functionals is exactly the positive cone of the quotient. The bipolar form of Hahn–Banach separation identifies its weak-* closure with all positive functionals on that quotient. This argument also applies to a nonunital quotient.

The positive functional

factors through this quotient by (12.5). Hence, simultaneously on the finitely many tensors a_i\otimes c_i, it is approximated by finite sums of vector functionals for \eta_{b_j}. On these tensors the identities are

They prove the required point-weak closure, with the same finite list of coefficients for all the tests. Finally, for a fixed finite list a_1,\ldots,a_m, the set

is convex. Its weak and norm closures coincide by Hahn–Banach separation. This proves (12.3).□

Proof. Fix the map and slots from Proposition 11.3. Thus P=s_1\ge0, all s_j are central contractions,

and, for every element a of the countable dense family \mathcal A of positive contractions fixed in Section 11, closed under positive rational cuts, and every rational \eps>0, there is a bounded element r_{a,\eps} with

Set E=A_\omega, v=1-h, w=P^2, and define \Phi(a)=va, \Psi(a)=wa. Both maps are completely positive and contractive, and \Psi has order zero. Since A is nuclear, composing finite-dimensional completely positive approximations to \id_A with either map proves that both maps are nuclear.

Letting \eps\downarrow0 in the last comparison gives \Phi(d)\in\Ideal_E(\Psi(d)) for d\in\mathcal A. We verify the extension to an arbitrary positive contraction a\in A. Choose d_n\in\mathcal A and positive rational \delta_n\downarrow0 with \|d_n-a\|<\delta_n. The cuts b_n=(d_n-\delta_n)_+ belong to \mathcal A\cap\Ideal_A(a), by positive-element perturbation, and b_n\to a in norm. Centrality of w implies

The same assertion follows for the closed ideal by approximation with finite sums. Consequently \Psi(b_n)\in\Ideal_E(\Psi(a)), and

Taking the norm limit proves the ideal hypothesis for a; scaling proves it for every positive element.

Apply Lemma 12.3. Any resulting sum packs into a single element z=\sum_{j=1}^N s_jv_j\in E, because, for a\in A,

There is no uniform bound here on z or on the individual v_j, and the coefficients need not commute with A. A compression by one approximate-unit element will supply both the bound needed for diagonalization and, through simultaneous tests, control of the commutators.

Fix a finite set F of contractions in A, and let \eta,\delta>0. Choose a positive contraction t\in A from an approximate unit so that, for all a\in F,

Approximate \Phi to within \delta, using one sum in (12.9), on

Put S'=tz. The t^2 test gives

Also \alpha(e_{11})S'=S' exactly, since \alpha(e_{11}) commutes with t and fixes all the s_j. Writing \kappa=\delta+\eta, the three tests give

For the first line, multiply \|S'^*S'-vt^2\|<\delta by a; the other two lines follow directly from their respective tests. Taking the adjoint of (12.12) controls S'^*a^*S' as well. Now expand the full commutator square:

Its four terms are within \kappa of, respectively, va^*a, -va^*a, -va^*a, and va^*a, because v commutes with A. Therefore

Take increasing finite sets dense in the unit ball of A and choose \eta,\delta\downarrow0, always with \delta<1. Each resulting S' has norm at most \sqrt2 by (12.10), and may be represented by a sequence bounded by 2. Fix representatives of the finitely many matrix entries of \alpha throughout. The countable diagonalization of Lemma 6.5, applied to these fixed representatives and the norm tests (12.11), (12.14), and \alpha(e_{11})S'=S', gives a single element S of norm at most 2. It commutes with every element of A, retains the fixed support equation, and satisfies (12.6) on a dense set, hence on all of A by continuity. Only the uniformly bounded S' are diagonalized at this last step; the compression coefficients and their unbounded sums are not retained as variables.□

Proof. The model and its generators. Decompose the first matrix leg of M_{p+1}\otimes M_p into blocks of sizes p and 1, and put

The upper diagonal block has size p^2 and the lower one size p. At each t the map a\mapsto A_a(t) is completely positive, contractive, and order zero. Direct matrix multiplication gives

and

For the last identity, T^*T is (1-t) times the bottom block projection, and the ith summand is (1-t) times the ith top block projection.

Let

and define the boundary algebra

We claim that A_a and T generate B_p. First A_1+T^*T=1, and (13.5) gives the coordinate scalar function t1. Thus their generated algebra contains C([0,1])1. At t=1 the generators have image j(M_p). At 0\le t<1, the nonzero link T(t), its adjoint, and the top matrix action yield all matrix units of M_{p+1}\otimes1_p: the links A_{e_{i1}}(t)T(t) give the ith top-to-bottom units, and their products give all top-to-top units and the bottom projection. At t=0 these are exactly the prescribed fiber. When 0<t<1, compression of A_a(t) to the bottom block and division by the scalar t gives the full bottom copy of M_p. Multiplying by the preceding matrix units gives all of M_{p+1}\otimes M_p in that fiber.

Fiber generation and the scalar functions prove the claim in norm. More explicitly, for f\in B_p and \eps>0, at each t choose an element g_t of the generated algebra whose value at t agrees with f(t), or is within \eps/2 of it. Continuity gives an open neighborhood on which \|g_t-f\|<\eps. A finite subcover and a subordinate scalar partition of unity produce an element of the generated algebra within \eps of f on the whole interval.

The representation j of M_p has multiplicity p+1, so some unitary V\in M_{p+1}\otimes M_p conjugates j(M_p) onto 1_{p+1}\otimes M_p. The finite-dimensional unitary group is path connected; choose a unitary path V(t) from 1 to V. Conjugation by this path changes only the endpoint presentation: the boundary algebra becomes the one with fibers M_{p+1}\otimes1_p at 0 and 1_{p+1}\otimes M_p at 1. Reversing the interval and interchanging the tensor legs identifies it with I(p,p+1).

Arbitrary representations of the relations. We have identified the algebra generated by the model. To obtain the asserted map into B, we now show that every pair (\alpha,s) in the statement gives a representation of that algebra. Represent B faithfully and unitally on a Hilbert space H. The order-zero structure theorem gives a supporting representation \rho:M_p\to\mathcal B(H), with unit the support projection of h=\alpha(1), such that \alpha(a)=h\rho(a)=\rho(a)h. In the polar decomposition

the initial space of u is H_s=1_{[0,1)}(h)H. The equation \alpha(e_{11})s=s implies that \alpha(e_{11}) is the identity on the closed range of s. Since 0\le\alpha(e_{11})\le h\le1, both h and \rho(e_{11}) are the identity on this range. Thus u is an isometry from H_s into 1_{\{1\}}(h)\rho(e_{11})H.

For 1\le i\le p the subspaces \rho(e_{i1})uH_s are mutually orthogonal. Their direct sum H_t reduces \rho and is identified with \C^p\otimes H_s by the unitary sending e_i\otimes\zeta to \rho(e_{i1})u\zeta. On H_t the weight is 1 and the representation is a\otimes1_{H_s}. Since H_t is contained in the h=1 space, it is orthogonal to H_s. The orthogonal complement

reduces both generators: s and s^* vanish there, and \alpha acts there as the unital matrix representation \rho|_{H_r}. This summand is therefore obtained by evaluating B_p at 1.

On H_t\oplus H_s\cong\C^{p+1}\otimes H_s the generators are precisely

where h_s and \rho_s denote the source restrictions, with \rho_s=0 on \ker h_s. Split H_s as H_0\oplus H_{\mathrm{mid}} according to the spectral sets \{0\} and (0,1). On \C^{p+1}\otimes H_0, (13.6) is the amplification of evaluation at 0: its bottom matrix action is zero and its link has coefficient 1. On H_{\mathrm{mid}} the representation \rho_s is unital and commutes with h_s. Continuous functional calculus therefore gives a unital representation

Amplifying by M_{p+1} and restricting to B_p sends (13.2)–(13.3) to (13.6) on this summand. Together with the two endpoint representations this gives a unital representation of B_p on all of H, sending its generators to \alpha(a) and s. Since these generators generate B_p, the image lies in the represented copy of B. The identification of B_p with I(p,p+1) proves the assertion.□

Proof. The congruences p\equiv0\pmod r and p\equiv-1\pmod s have a simultaneous solution, and adding a sufficiently large multiple of rs makes it at least 2. Write p=mr, p+1=ns, and use the unital amplifications

Their tensor product, applied pointwise, sends the 0 boundary M_r\otimes1_s into M_p\otimes1_{p+1} and the 1 boundary 1_r\otimes M_s into 1_p\otimes M_{p+1}.□

Proof. Choose a nuclear approximation D\xrightarrow{\theta}M_k\xrightarrow{\eta}D, with both maps completely positive and contractive, whose composition is within \eps of the identity on G. The matrix-valued extension theorem of Arveson [4, Theorem 1.2.3] extends \theta to a c.p.c. map \widetilde\theta:L\to M_k. Then R=\eta\widetilde\theta has the stated properties. Applying the approximation also to 1_D makes R(1_L) as close to 1_D as desired; an exactly unital retraction is unnecessary.□

Proof. Realize \Z as the closure of an increasing sequence of its injectively included prime dimension-drop stages D_l. By Proposition 13.1 and Lemma 13.2, every D_l has a unital homomorphism \chi_l:D_l\to F_\omega(A).

Let \mathcal Z_0 be a countable dense *-subalgebra over \mathbb Q+i\mathbb Q containing 1, and exhaust it by finite sets. For any finite set of contractions F\subset\mathcal Z_0 and any \eps>0, approximate F\cup F^2\cup\{1\} by one stage, and apply Lemma 13.3 to those stage approximants. With the errors chosen sufficiently small, the resulting c.p.c. map R:\Z\to D_l, viewed in \Z, satisfies

Consequently, for x,y\in F,

Here the norm calculation takes place in \Z, with D_l included isometrically. Composing with \chi_l produces c.p.c. maps \beta_n:\Z\to F_\omega(A) such that

for every x,y\in\mathcal Z_0. Including successively larger scaled tests handles elements outside the unit ball.

We now turn these approximate quotient relations into exact ones. Put C_A=A_\omega\cap A', and let q:C_A\to F_\omega(A) be the quotient map. By the nuclear completely positive lifting theorem of Choi–Effros [10, Theorem 3.10 and Corollary 3.11], each \beta_n has a c.p.c. lift L_n:\Z\to C_A. Regarding L_n as a map into A_\omega and applying the same theorem to \ell^\infty(\N,A)\to A_\omega gives a c.p.c. lift

Each coordinate map L_{n,k}:\Z\to A is completely positive and contractive, so all its values on each fixed test have a bound independent of n and k.

The relevant quotient norm errors are multiplied errors. If d\in C_A and a\in A, then

Indeed, replacing d by d+j with j\in\Ann(A) does not change either product, and taking the infimum over such j gives the inequalities. Apply this to

where e represents the unit as above. Equations (13.7) and (13.8) give errors tending to zero in A_\omega after multiplication by each fixed a\in A. Also [L_n(x),a]=0 exactly for every n,x,a.

Choose a countable dense set \{a_j\} in A, and diagonalize over \mathcal Z_0, its product tests, and these a_j. More explicitly, choose increasing row indices m_l so that the first l multiplied defects in row m_l have ultrapower norm less than 2^{-l-1}. Their coordinate errors, together with the first l commutator errors, are less than 2^{-l} on some set E_l\in\omega; replacing these sets by their successive finite intersections with \{k\ge l\} makes them decreasing. Let l(k) be the largest l\le k with k\in E_l, or zero when no such level exists. Put \gamma_k=L_{m_{l(k)},k} if l(k)>0, and \gamma_k=0 otherwise. Then l(k)\to_\omega\infty, and

defines a c.p.c. map \Gamma:\Z\to A_\omega\cap A'. The maps \gamma_k are linear and completely positive at every coordinate; no separate lifting of individual elements is involved. The diagonal conditions say

All the displayed defects are central relative to A, so they annihilate it on both sides. Passing through q therefore gives multiplicativity and unitality on \mathcal Z_0. Continuity extends these identities to \Z and proves the proposition.□

Proof. We first check connectedness for a prime block I(r,s), with \gcd(r,s)=1. A unitary is a continuous path in U(rs) whose endpoints belong to

Both endpoint groups are connected. Attach a path in H_0 from 1 to the initial value and a path in H_1 from the final value to 1. Attaching these paths gradually, while reparametrizing the original path, is a homotopy through paths satisfying the same endpoint conditions. The result is a based loop in U(rs). Changing the attached H_0 path by an endpoint loop changes its determinant winding by an arbitrary multiple of s; changing the H_1 path changes it by an arbitrary multiple of r. Bezout’s identity therefore permits choices for which the total winding is zero. The determinant identifies \pi_1(U(rs)) with \mathbb Z, so this loop contracts with its basepoint fixed. That contraction retains endpoint value 1, and hence remains in the prime block. Thus every unitary of I(r,s) is connected to 1.

For a unitary in the injective unital inductive limit \Z, choose a sufficiently close element of a stage. Its two unitary defects are small, so its polar part is a unitary in the same stage and is still close to the original unitary. Two unitaries of distance less than 1 are joined by the logarithm path for their quotient. Connectedness of the stage unitary group proves connectedness of \mathcal U(\Z). Strong self-absorption gives an isomorphism \Z\otimes\Z\cong\Z, proving the same assertion for the tensor square.

Finally, a unitary v connected to 1 in D/J is a finite product of exponentials of selfadjoints. To see this, subdivide a continuous unitary path finely enough that consecutive values differ by less than 1, and take a selfadjoint logarithm of each consecutive quotient. Lift these finitely many selfadjoints through q and multiply their exponentials in D. The resulting unitary lifts v.□

Proof. Write D=A^{\sim}\otimes\Z and A_1=A\otimes1, and form the unital algebra and its annihilator quotient

First-factor sequences give a homomorphism F_\omega(A)\to Q: centrality and annihilation are both preserved by tensoring with 1. This homomorphism is unital, since the approximate-unit representative e\otimes1 differs from 1_{D_\omega} by an annihilator of A_1. Its composition with \varphi is a unital map \varphi_1:\Z\to Q and commutes with the constant second-factor map \varphi_2(z)=1\otimes z+\Ann(A_1). Because \Z is nuclear, these two maps give a unital homomorphism

The approximately inner half-flip of the strongly self-absorbing algebra \Z [33, Proposition 1.5] says that, for a finite set H\subset\Z and \delta>0, there is a unitary V\in\Z\otimes\Z with

By Lemma 13.5, V is connected to 1. Thus \Theta(V) is connected to 1 in Q, and the same lemma lifts it to a unitary W\in C_D.

This unitary has a representing sequence of actual unitaries in D. Indeed choose bounded representatives x_k. Since W is unitary, both \|x_k^*x_k-1\| and \|x_kx_k^*-1\| tend to zero along \omega. On a good set both are less than 1/2, and the polar parts u_k=x_k(x_k^*x_k)^{-1/2} are unitaries with \|u_k-x_k\|\to_\omega0. Put u_k=1 on the remaining coordinates. Then W=[(u_k)], and W\in C_D gives

For each z\in H choose a lift c_z\in A_\omega\cap A' of \varphi(z); a single c.p.c. lift of \varphi is available by Choi–Effros lifting, so these lifts may satisfy \|c_z\|\le\|z\|. In C_D the difference

has quotient norm less than \delta by (13.10). The analogue of (13.8) gives

Since W commutes with a\otimes1 in the ultrapower, this is the concrete norm estimate

Here (ac_z)\otimes1 is represented by a first-factor sequence (ac_{z,k})\otimes1, with ac_{z,k}\in A. Thus no estimate in an annihilator quotient is being used as an unmultiplied estimate in a norm ultrapower.

For finitely many elementary tensors, choose one coordinate where (13.11) and (13.12) all have the desired accuracy. This gives (13.9) for those tensors. For a general finite G, approximate each of its elements by a finite sum of elementary tensors, include all the corresponding second factors in H, and choose \delta small enough for the finite sums of the bounds \delta\|a\|. Unitary conjugation preserves the initial approximation errors, so the same conclusion holds for G.□

Proof. The zero algebra is immediate, so assume A\ne0. Set B=A\otimes\Z and let \iota(a)=a\otimes1. The algebra A^{\sim}\otimes\Z acts by multipliers on its ideal B, so all the unitaries in Lemma 13.6 act as automorphisms of B by conjugation.

Choose sequences (a_j) and (b_j) dense in A and B, respectively, and put \eps_n=2^{-n}. We construct products of multiplier unitaries U_n and embeddings \iota_n=\Ad(U_n)\circ\iota as follows. Start with U_0=1. Before step n, let F_n be a finite subset of A containing a_1,\ldots,a_n and all elements c_{k,j} chosen at earlier steps k<n, 1\le j\le k. Apply Lemma 13.6 with these domain tests and the finite target set

Obtain u_n with errors less than \eps_n, and put U_n=U_{n-1}u_n. The commutator estimate gives

and the range estimate gives elements c_{n,j}\in A with

Store these finitely many preimages in every subsequent F_m. They are legitimate finite norm tests: each has finite norm, indeed \|c_{n,j}\|<\|b_j\|+\eps_n, because \iota_n is isometric.

For each dense domain element a_j, equation (13.13) holds at every step n\ge j. Summability of (\eps_n) gives convergence of \iota_n(a_j). As all \iota_n are isometric, approximation by the dense set extends this to point-norm convergence on A. The limit \Phi:A\to B is a homomorphism by continuity of multiplication and adjoints, and it is isometric since \|\Phi(a)\|=\lim_n\|\iota_n(a)\|=\|a\|.

Fix j and take any n\ge j. The stored preimage c_{n,j} belongs to every F_m with m>n, so

Letting n\to\infty shows that every b_j is in the closure of \Phi(A). The range of an isometric homomorphism is closed, and the b_j are dense, so \Phi(A)=B. This proves the desired isomorphism.□

Proof of Theorem 1.2. Fix the free ultrafilter in the theorem. Proposition 13.4, through the support, norm-cone, tracial-cone, comparison, and packing constructions above, supplies a unital *-homomorphism \Z\to F_\omega(A). Proposition 13.7 then gives A\cong A\otimes_{\min}\Z.□

Proof of Theorem 1.1. The zero algebra, if admitted by the convention for simplicity, is immediate. Nuclearity implies exactness, so Corollary 5.2 shows that A satisfies condition (C). Theorem 1.2 then gives A\cong A\otimes_{\min}\Z.□

Proof of Corollary 1.3. Again the zero algebra is immediate. Lemma 6.1 derives (C) from the canonical induced map \Cu(\iota), using that \Cu(\iota)([b])=[b\otimes1_\Z]. Theorem 1.2 then proves the assertion.□

Proof of Corollary 1.4. The forward implication follows from functoriality of \Cu under *-isomorphisms. For the converse, let \Phi:\Cu(A)\to\Cu(A\otimes_{\min}\Z) be an abstract isomorphism. The case A=0 is immediate, so put D=A\otimes_{\min}\Z and suppose A\ne0. Then D is nonzero, separable, and nuclear.

Strong self-absorption of \Z supplies an isomorphism \varphi:\Z\to\Z\otimes\Z approximately unitarily equivalent to z\mapsto z\otimes1_\Z [33, Definition 1.3(iv) and Example 1.14(iv)]. Consequently \id_A\otimes\varphi:D\to D\otimes\Z is an isomorphism approximately unitarily equivalent to the first-factor map j_D(d)=d\otimes1_\Z. This also holds when A is nonunital: if u_n\in\Z\otimes\Z are the implementing unitaries, then 1_{M(A)}\otimes u_n\in M(A\otimes\Z\otimes\Z) implement the equivalence, first on elementary tensors and then on all of D by density.

Approximately unitarily equivalent homomorphisms induce the same map on \Cu. Indeed, after stabilization, norm approximation by unitary conjugates gives comparison of every positive cut of either image with the other image, by Lemma 2.1(1); approximate units place the comparison witnesses in the algebra when the conjugating unitaries are multipliers. Taking the suprema of cuts in Equation (2.1) gives equality of the two Cuntz classes. Thus \Cu(j_D)=\Cu(\id_A\otimes\varphi) is an isomorphism. Lemma 6.1, applied to D, gives condition (C) on \Cu(D). Both almost unperforation and the full almost-divisibility sandwich are intrinsic ordered-semigroup properties, so \Phi transports them to \Cu(A). Theorem 1.2 now gives A\cong A\otimes_{\min}\Z.□