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 \(B\cong\mathcal K(Bp)\). 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 \(\mathop{\mathrm{her}}(e)\). 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 5 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 5 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 \[v^*v=(a-\delta/2)_+,\qquad vv^*\in M_m(\mathop{\mathrm{her}}_B(h)).\] Define \[g(t)=\left(\frac{r(t)}{(t-\delta/2)_+}\right)^{1/2},\] 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 \[\sum_{i=1}^m z_i^*z_i=r(a),\qquad c:=\sum_{i=1}^m z_i z_i^*\in\mathop{\mathrm{her}}_B(h).\] 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 \[ c=\sum_{i=1}^L q((c-(i-1))_+). \tag{10}\] Set \(R=\max(1,\|c\|)\). For every integer \(m\ge1\), \[\big((c-i)_+/R\big)^{1/m}\le q((c-(i-1))_+).\] Indeed, below the spectral threshold \(i\) the left side is zero, and above it the right side is one. Consequently, \[d_\tau((c-i)_+) \le\tau(q((c-(i-1))_+)) \le d_\tau((c-(i-1))_+).\] The last inequality uses Equation (6) and hereditary containment. Summing and using \((c-L)_+=0\) proves Equation (9). 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 \(\mathrm F_{\mathrm{fin}}(A)\) is empty, all functionals are zero or infinite on every nonzero class. Equation (1) then gives \(Nx\le x\), while \(x\le(N+1)x\) is automatic. Take \(u=x\). We henceforth assume \(\mathrm F_{\mathrm{fin}}(A)\ne\varnothing\).

Choose supports with summable relative rank. Choose a real number \(s\) and integers \(M_j\ge1\), \(j\ge1\), such that \[ \frac1{N+1}<s<\frac1N,\qquad E:=\sum_{j=1}^{\infty}\frac1{M_j}<s-\frac1{N+1}. \tag{11}\] Lemma 7 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 \[F_0=0,\qquad F_j(t)=\min\{1,\max\{0,2^jt-1\}\},\qquad r_j=s(F_j-F_{j-1})\quad(j\ge1).\] The functions \(F_j\) increase, and \(r_j\ge0\) vanishes on \([0,2^{-j}]\). Put \(p_j=r_j(a)\). Lemma 9 gives \(c_j\in\mathop{\mathrm{her}}(h_j)_+\) satisfying \[ \tau(c_j)=\tau(p_j)\quad\text{for all extended traces }\tau, \qquad M_j[c_j]\le[e]. \tag{12}\]

Round and sum inside the Cuntz semigroup. Choose integers \(L_j>\|c_j\|\) and put \[ w_j=\sum_{i=1}^{L_j}[(c_j-i)_+],\qquad u=\sup_{k\ge1}\sum_{j=1}^k w_j. \tag{13}\] 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 \[ \varphi(u)=\sum_{j=1}^{\infty}\varphi(w_j) \quad(\varphi\in\mathrm F(\mathrm{Cu}(A))). \tag{14}\]

Fix \(\varphi\in\mathrm F_{\mathrm{fin}}(A)\) and choose its representing trace using Equation (7). Equations (9) and (12) give \[ \varphi(w_j)\le\tau(p_j) \le\varphi(w_j)+\frac{\varphi([e])}{M_j}. \tag{15}\] The finite value \(\varphi([e])\) is the only trace-dependent constant in the error estimate.

Recover the rank of \(a\). Telescoping gives \[\sum_{j=1}^k\tau(p_j)=s\tau(F_k(a)).\] We claim that \[ \tau(F_k(a))\uparrow d_\tau(a)=\varphi(x). \tag{16}\] The values increase and are at most \(d_\tau(a)\) by Equation (6) 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 \(m\ge1\), \[(a-\epsilon)_+^{1/m}\le F_k(a).\] Taking traces and then letting \(m\to\infty\) gives \(d_\tau((a-\epsilon)_+)\le\tau(F_k(a))\). The cut-supremum formula in Equation (5) now proves Equation (16).

Sum Equation (15), then use Equations (14) and (16). In extended nonnegative arithmetic, for every \(\varphi\in\mathrm F_{\mathrm{fin}}(A)\), \[ \varphi(u)\le s\varphi(x) \le\varphi(u)+E\varphi([e]). \tag{17}\] If \(\varphi(x)<\infty\), Equation (8) implies \[(s-E)\varphi(x)\le\varphi(u)\le s\varphi(x).\] Since \(\varphi(x)>0\), Equation (11) yields \[ \frac{\varphi(x)}{N+1}<\varphi(u)<\frac{\varphi(x)}N. \tag{18}\] If instead \(\varphi(x)=\infty\), then \(E\varphi([e])<\infty\) and Equation (17) forces \(\varphi(u)=\infty\). No subtraction of infinite values is used. Since \(\mathrm F_{\mathrm{fin}}(A)\) 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 \(\mathrm F_{\mathrm{fin}}(A)\), the preceding estimates give \[\varphi(Nu)\le\varphi(x),\qquad \varphi(x)\le\varphi((N+1)u).\] 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 \(\mathrm F_{\mathrm{fin}}(A)\) and the infinite-value case excludes \(\varphi(x)=\infty\). Equation (18) makes that inequality strict as well. Applying Equation (1) twice proves \(Nu\le x\le(N+1)u\). ◻

Proof. Theorem 11 gives the full almost-divisibility sandwich, and Lemma 6 gives almost unperforation. ◻

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 \[[T_0]\le m[E],\qquad [T_0]\le m[D].\] By divisibility in (C), choose a positive element \(r\) in a stabilization with \[(m+1)[r]\le[T_0]\le(m+2)[r].\] Almost unperforation, applied with \(m\), gives \([r]\le[E]\) and \([r]\le[D]\). Choose \(\eta>0\) sufficiently small that \[ [(T_0-\delta^2/2)_+]\le(m+2)[(r-2\eta)_+]. \tag{22}\] 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 \(\mathop{\mathrm{her}}((r-\eta)_+)\) with hereditary subalgebras of \(\mathop{\mathrm{her}}(E)\) and \(\mathop{\mathrm{her}}(D)\), 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 \[v=u_D j(r)u_E^*\] 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 \[v^*v=u_Ej(r)^2u_E^*\in\mathop{\mathrm{her}}(E),\qquad vv^*=u_Dj(r)^2u_D^*\in\mathop{\mathrm{her}}(D).\] If \(\|\sigma(w_i)\|\ge\delta\) for some \(i\), then \(\sigma((T_0-\delta^2/2)_+)\ne0\). By (22), 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 20. 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 (20), so failure will provide a representation in which both constructed positives vanish.

Step 1: a matrix corner and two prospective supports. Apply Lemma 17, with seven slots and a cut of \(a\) below \(1/2\), to obtain a c.p.c. order-zero map \(M_7\to\mathop{\mathrm{her}}(a)\) with weight \(d\) satisfying \[ \|\sigma(a)\|\ge\tfrac12\quad\Longrightarrow\quad\|\sigma(d)\|=1. \tag{23}\] Its supporting matrix units identify \[\mathop{\mathrm{her}}(d)=M_7(B),\qquad d=\operatorname{diag}(P),\] 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 \(Q\in\mathop{\mathrm{her}}(b)\) is a positive contraction and \[ \|\sigma(b)\|\ge\tfrac12\quad\Longrightarrow\quad\|\sigma(Q)\|=1. \tag{24}\]

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, \[ U=\begin{pmatrix} \frac12\,1\\[2pt] \frac12 v_0\\[2pt] \frac12 v_1\\[2pt] \bigl(\frac34\,1-\frac14v_0^*v_0-\frac14v_1^*v_1\bigr)^{1/2} \end{pmatrix},\qquad L=UPU^*\in M_7(B),\qquad t=\tfrac12. \tag{25}\] The expression under the square root is at least \(\frac14\,1\), and \(U^*U=1\). Define \[ a'=(L-t)_+,\qquad b'=Q\bigl[Q(c_n^2t-z^*Lz)Q\bigr]_+Q. \tag{26}\] These belong to \(\mathop{\mathrm{her}}(a)\) and \(\mathop{\mathrm{her}}(b)\), 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 18, the compression of \(z^*Lz\) to the support of \(b'\) is at most \(c_n^2t\). Since \(L\ge t p_{a'}\), this implies \[t p_{b'}z^*p_{a'}z p_{b'} \le p_{b'}z^*Lz p_{b'}\le c_n^2t p_{b'},\] which proves (20) 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 \[ Z^*Z\le Q^2, \tag{27}\] 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 18, \[ c_n^2tQ^2\le Z^*UPU^*Z\le tZ^*Z\le tQ^2. \tag{28}\] 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 \[g(x)=\min\{1,\max\{0,8x-2\}\},\qquad G(x)=\min\{1,4x\}.\] They vanish at zero, satisfy \(Gg=g\), and \(g(t)=G(t)=1\). Set \[ X=2g(P)Z_{\mathrm f},\qquad h_0=XX^*,\qquad Y_i=2\operatorname{diag}(G(P))Z_iX^*\quad(i=0,1). \tag{29}\] 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 \(\mathop{\mathrm{her}}(d)\). We have \[ \|X\|\le2,\quad \|h_0\|\le4,\quad \|Y_i\|\le4, \qquad G(P)h_0=h_0=h_0G(P). \tag{30}\] 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 \[r=\tfrac18,\qquad W_j=r^{-1}(Y_0)_j^*\quad(j=1,2),\qquad f=f(h_0),\quad f(x)=\min\{1,\max\{0,4x-1\}\}\quad(x\ge0).\] 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 \[ T=1-VV^*=f^2-uu^*-u c_f-c_f u^*\in B. \tag{31}\] 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 \(\mathbb R\) the continuous functions \[\beta(x)=\min\{1,\max\{0,2x-1\}\},\qquad \chi(x)=\min\{1,\max\{0,4x-1\}\}.\] They vanish at zero, \(\beta\) vanishes on \((-\infty,1/2]\), \(\beta(1)=1\), and \(\chi\beta=\beta\). Put \[ b_T=\beta(T),\qquad B_T=\mathop{\mathrm{her}}_B(b_T),\qquad y=\operatorname{diag}(b_T)Y_1b_T\in M_{3,1}(B_T),\qquad S=\operatorname{diag}(b_T)\in M_3(B_T). \tag{32}\] 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 17 with two slots inside \(\mathop{\mathrm{her}}((yy^*-10^{-3})_+)\), 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 \[ k,k'\in\mathop{\mathrm{her}}((yy^*-10^{-3})_+),\qquad [(yy^*-10^{-2})_+]\le3[(k-0.9)_+]. \tag{33}\] 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 \[\left\|(yy^*-0.01)_+-\sum_{j=1}^3x_j^*(k-0.9)_+x_j\right\|<0.005.\] 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 \[q_j=\left(\frac{(k-0.9)_+}{(k-0.8)_+}\right)^{1/2}x_j,\] where the quotient is defined to be zero when its denominator is zero. This is a continuous function of \(k\). Then \[ \left\|(yy^*-0.01)_+-\sum_{j=1}^3q_j^*(k-0.8)_+q_j\right\|<0.005, \qquad \|q_j\|\le\sqrt{160}<13. \tag{34}\] Indeed the compression sum before replacement is positive and has norm less than \(16\), so each summand has norm less than \(16\). Also \[\frac{(k-0.9)_+}{(k-0.8)_+}\le10(k-0.9)_+.\] 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 \[D'=(1-p')S(1-p')\in M_3(B_T)_+.\] We claim that \[ 3[b_T]=[S]\le[k]+[D']\le2[D']. \tag{35}\] 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 \(p'Sp'\in\mathop{\mathrm{her}}(k)\). For the second inequality it suffices to check that \(k'\in\mathop{\mathrm{her}}(D')\), 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 \[(1-p')c_j(1-p')\le C_jD',\qquad (1-p')c_j(1-p')\longrightarrow k',\] which proves the inclusion and (35).

Almost unperforation now gives \([b_T]\le[D']\). Cut placement supplies a column \(v_1\in M_{3,1}(B_T)\) with \[ v_1^*v_1=(b_T-1/4)_+,\qquad v_1v_1^*\in\mathop{\mathrm{her}}(D'),\qquad \|v_1\|\le1. \tag{36}\] Since \((k-0.8)_+(1-p')=0\), its range condition implies \[ (k-0.8)_+v_1=0. \tag{37}\] We shall also retain the following fixed continuous support identities: \[ (10^{-3}-yy^*)_+k=0,\qquad \chi(T)(v_1)_j=(v_1)_j=(v_1)_j\chi(T)\quad(1\le j\le3). \tag{38}\] The first follows from (33); 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 \[ s=\operatorname{diag}((h_0-1/4)_+)\in M_2(B),\qquad E=(h_0-1/2)_+\in B,\qquad D=s\bigl[s(\tfrac14-p)s\bigr]_+s\in M_2(B). \tag{39}\] Choose an ordered alphabet consisting of all entries of \[ P,h_0,Y_0,Y_1,p,s,E,D,T,b_T,y,k,k',D',v_1,q_1,q_2,q_3 \tag{40}\] 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 21 to the finite list of columns \[ D\mathbf e_jxE,\qquad j=1,2,\quad x\text{ a word of length at most }n, \tag{41}\] including the empty word \(x=1\), with threshold \(1/n\). Here \(\mathbf e_j\) is the \(j\)th scalar basis column. The left and right squared moduli of these columns lie in \(\mathop{\mathrm{her}}(D)\) and \(\mathop{\mathrm{her}}(E)\), respectively. The resulting contraction \(v_0\in M_{2,1}(B)\) satisfies \[ v_0v_0^*\in\mathop{\mathrm{her}}(D),\qquad v_0^*v_0\in\mathop{\mathrm{her}}(E),\qquad v_0^*pv_0\le\tfrac14v_0^*v_0. \tag{42}\] The last assertion is Lemma 18 applied to the pencil defining \(D\). In addition, we have the exact operator identities \[ (1/2-h_0)_+v_0^*=0,\qquad \operatorname{diag}(G(P))v_0=v_0. \tag{43}\] 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 (30). 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 (26). Since they satisfy the required angle bound, the assumed failure gives an irreducible representation \(\sigma_n\) such that \[\max\{\|\sigma_n(a)\|,\|\sigma_n(b)\|\}\ge\tfrac12, \qquad \sigma_n(a')=\sigma_n(b')=0.\] Use the same symbols for the represented operators. The first vanishing gives \(L\le t\), hence \(P=U^*LU\le t\). By (23), the \(a\)-trigger is impossible. Therefore \(\|Q\|=1\) by (24). The two vanishings also give the conditional inequalities (28) 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 (27), (30), and (25). We keep the same notation for their limits.

Equality holds throughout (28). In particular \(Z^*Z=Q^2\) and \(Z^*(t1-UPU^*)Z=0\). Since \[t1-UPU^*=t(1-UU^*)+U(t-P)U^*\] is a sum of positive operators, both positive summands annihilate \(Z\). For \(\xi=U^*Z\), it follows that \[ Z=U\xi,\qquad P\xi=t\xi,\qquad \xi^*\xi=Q^2. \tag{44}\] The first row of \(U\) is \(1/2\), so \(X=2g(P)Z_{\mathrm f}=\xi\). Consequently \[ \|h_0\|=1,\qquad Y_i=\operatorname{diag}(G(P))v_i h_0,\qquad Y_0=v_0h_0. \tag{45}\] The last equality uses the second identity in (43).

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 \[(r^2-R_0)_+p=0,\qquad \|(1-p)(R_0-r^2)_+\|\le(r+1/n)^2-r^2.\] Thus in the limit \[ [p,R_0]=0,\qquad (r^2-R_0)_+p=0,\qquad (1-p)(R_0-r^2)_+=0. \tag{46}\] 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 (45). 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 \[ 0\le p\le\tfrac14\,1. \tag{47}\] The last equation in (46), and invertibility of \(1-p\), now give \(R_0\le r^2\). The other support equation then gives \[ \|Y_0\|\le r,\qquad R_0p=r^2p. \tag{48}\]

Every fixed word \(x\) in the limiting alphabet satisfies \[ D\mathbf e_jxE=0\qquad(j=1,2). \tag{49}\] 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 (41) would give \(\|v_0\|=1\) in the ultraproduct. The first equation in (43) 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 \[\|v_0h_0\|\ge\tfrac12\|v_0\|=\tfrac12.\] This contradicts \(Y_0=v_0h_0\) and (48), 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 \[\mathcal L=\overline{\operatorname{span}} \{xE\eta:x\text{ a word in the alphabet},\ \eta\in\mathcal H\}.\] 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 (49) 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 \[ (1/2-h_0)_+W_j=0\qquad(j=1,2), \tag{50}\] which follows on the full limiting space from \(W_j=r^{-1}h_0(v_0)_j^*\) and (43), and thus restricts to \(\mathcal L\).

By the zero case of Lemma 18, \(D=0\) implies \(s(\frac14-p)s\le0\). Equation (47) 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 (48), this yields \[ R_0s=r^2s. \tag{51}\] Let \(e=1_{(1/4,\infty)}(h_0)\) on \(\mathcal L\). The range of \(s\) is dense in \((e\mathcal L)^{\oplus2}\), so (51) implies \[(W_i e)^*(W_j e)=\delta_{ij}e\qquad(1\le i,j\le2).\] Thus \(W_1,W_2\) restricted to \(e\mathcal L\) are isometries with orthogonal ranges. By (50), 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 \[V^*V=f^2+c_f^2=1.\] Its range-complement projection is exactly \[ 1-VV^*=f^2-uu^*-uc_f-c_fu^*=T. \tag{52}\] Both the initial and final supports of \(u\) are contained in \(e\), and \(f=ef\). As \(c_f\) commutes with \(e\), formula (52) shows \(T=eTe\). In particular \[ Th_0T\ge\tfrac14 T. \tag{53}\] This projection is nonzero. Indeed \(W_2f\ne0\) because \(\|h_0\|=\|f\|=1\) and \(W_2\) is isometric on \(e\mathcal L\), whereas \[V^*W_2f=fW_1^*W_2f+c_fW_2f=0.\] 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 (38) give \[T(v_1)_j=(v_1)_j=(v_1)_jT.\] 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 (32), (36), and (45), we obtain \[ y=v_1Th_0T,\qquad v_1^*v_1=\tfrac34 T,\qquad y^*y=\tfrac34(Th_0T)^2\ge\tfrac{3}{64}T. \tag{54}\] In particular, \(\|yy^*\|\ge3/64\).

The three uniformly bounded witnesses in (34) survive both the ultraproduct and the restriction, giving \[\left\|(yy^*-0.01)_+-\sum_{j=1}^3q_j^*(k-0.8)_+q_j\right\|\le0.005.\] Hence \(q=(k-0.8)_+\) is nonzero: otherwise \(\|yy^*\|\le0.015<3/64\), contradicting (54). On the other hand, (37) and (54) imply \(qy=0\). The first fixed support identity in (38) puts the range of \(k\), and hence of \(q\), in the spectral subspace of \(yy^*\) for \([10^{-3},\infty)\). Therefore \[0=qyy^*q\ge10^{-3}q^2,\] forcing \(q=0\). This contradiction proves the existence of the constant \(c\) and completes the proof. ◻

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 \[\begin{align*} \|[f(H),B]\|_{2,T}^2 &=\sum_{i,j}|f(\alpha_i)-f(\alpha_j)|^2\|e_iBe_j\|_{2,T}^2\\ &\leq\operatorname{Lip}(f)^2 \sum_{i,j}|\alpha_i-\alpha_j|^2\|e_iBe_j\|_{2,T}^2. \end{align*}\] 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 \[H=\begin{pmatrix}0&W\\W^*&0\end{pmatrix},\qquad B=\begin{pmatrix}a&0\\0&m\end{pmatrix},\qquad f(t)=\frac{t}{\sqrt{1+t^2}}.\] 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 29. 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]; it is also part of the stronger Brown–Carrión–White approximation theorem [6]. 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 \[\Phi=\sum_{l=1}^r\lambda_l\beta_l\alpha\] is as close to the identity as prescribed on a given finite set. The integer \(r\) may depend on \(u\) without restriction. Put \[h_l=\beta_l(1),\quad p_l=p_{h_l},\quad \beta_l(x)=h_l\rho_l(x),\quad \mu_l=\rho_l\alpha,\quad\theta_l=h_l\mu_l.\] Here \(\rho_l\) is the supporting homomorphism of the order-zero map [35]; it acts by multipliers of \(\mathop{\mathrm{her}}_D(h_l)\). 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 \[\begin{align*} \left\|\sum_l\lambda_l(a-\theta_l(a))^2\right\| &\leq u^4 &&(a\in F),\tag{74}\\ \left\|a^{1/2}\left(1-\sum_l\lambda_lh_l\right)a^{1/2}\right\| &\leq u^4 &&(a\in\mathcal P). \tag{75}\end{align*}\] For completeness, the Schwarz inequality gives \[0\leq\sum_l\lambda_l(a-\theta_l(a))^2 \leq a^2-a\Phi(a)-\Phi(a)a+\Phi(a^2),\] 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 (75). 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 \([0,1]\): \[g(t)=\begin{cases} 0,&t\leq1-2u,\\ (t-1+2u)/u,&1-2u<t<1-u,\\ 1,&t\geq1-u. \end{cases}\] 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 \[ 1-g_l^4\leq u^{-1}(1-h_l) \tag{76}\] holds by functional calculus. With \[D_l=\sum_{a\in F}p_l(a-\mu_l(a))^2p_l,\qquad B_u=2u^4+8u^2,\] we have, for every \(a\in F\), \[ \sum_l\lambda_l\|(a-\mu_l(a))z_l\|_{2,\tau}^2\leq B_u, \qquad \sum_l\lambda_l\tau(z_lD_l)\leq |F|B_u. \tag{77}\] Indeed, the contribution of \(a-\theta_l(a)\) is bounded by \(u^4\) using (74) 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 \(a\in\mathcal P\), \[\begin{align*} \sum_l\lambda_l\tau(\mu_l(a)g_l^4) &\geq\tau(a)-u^3-\sqrt{B_u},\tag{78}\\ \sum_l\lambda_l\tau(\mu_l(a)z_l) &\leq\tau(a)+\sqrt{B_u}. \tag{79}\end{align*}\] To verify the replacement of \(\mu_l(a)\) by \(a\), for example, \[\left|\sum_l\lambda_l\tau((\mu_l(a)-a)g_l^4)\right| \leq \left(\sum_l\lambda_l\|(\mu_l(a)-a)z_l\|_2^2\right)^{1/2} \left(\sum_l\lambda_l\tau(g_l^8)\right)^{1/2} \leq\sqrt{B_u}.\] The same argument applies to \(z_l\). The lost \(a\)-mass in the first inequality is at most \(u^3\), by (76) and (75); in the second use \(z_l\leq1\).

The first pencil: small supports and summable overlaps. For each \(l\) put \[C_l=\mathop{\mathrm{her}}_D(h_l)\cap\rho_l(E)',\qquad \mathbb E_l=\mathbb E_{\rho_l}.\] 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 \[\begin{align*} S_l&=\sum_{i<l}t_i^2,\qquad K_l=\mathbb E_l\big(u^{-1}D_l+\delta^{-1/2}p_lS_lp_l\big),\\ X_l&=g_l(p_l-K_l)g_l,\qquad b_l=g_l(X_l)_+g_l. \tag{80}\end{align*}\] 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 \(\mathop{\mathrm{her}}_D(h_l)\), 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 \[ 0\leq b_l\leq g_l^2\leq z_l,\qquad b_l\geq g_l^4-g_l^2K_lg_l^2. \tag{81}\] The second inequality follows from \((X_l)_+\geq X_l\). Lemma 18, applied to the pencil in (80), gives \[ p_{b_l}K_lp_{b_l}\leq p_{b_l}. \tag{82}\]

Use the tracial conclusion of Lemma 17 in \(C_l\), which has the form (73), with \(q=\delta\lambda_l\) and positive contraction \(b_l\). It gives a c.p.c. order-zero map \(\psi_l:M_k\to\mathop{\mathrm{her}}_{C_l}(b_l)\) whose weight \(t_l\) satisfies \[ d_\sigma(t_l)\leq\delta\lambda_l d_\sigma(b_l),\qquad d_\sigma((t_l-1/2)_+) \geq\frac{\delta\lambda_l}{3}d_\sigma((b_l-1/2)_+) \tag{83}\] for every bounded positive trace \(\sigma\) on \(C_l\). In particular, on restricting \(\tau\) to \(C_l\), \[ \tau(p_{t_l})\leq\delta\lambda_l,\qquad \sum_l\tau(p_{t_l})\leq\delta,\qquad \tau(S_l)\leq\delta. \tag{84}\] Here ranks are the traces of the corresponding support projections in the bidual. As \(p_{t_l}\leq p_{b_l}\), (82) gives \[\sum_l\tau(t_l^2K_l)\leq\sum_l\tau(t_l^2)\leq\delta.\] 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 \[\begin{align*} \sum_l\|(a-\mu_l(a))t_l\|_{2,\tau}^2 &\leq u\delta &&(a\in F),\tag{85}\\ \sum_l\tau(t_l^2S_l)&\leq\delta^{3/2}. \tag{86}\end{align*}\] These are unweighted sums: the individual support allowances \(\delta\lambda_l\) in (84) 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 \[s_l=1_{(1/2,1]}(t_l),\qquad e_l=1_{(1/2,1]}(b_l).\] Since \(b_l\) is a contraction supported under \(z_l\), scalar functional calculus and (81) give \[ e_l\geq2b_l-z_l\geq2g_l^4-2g_l^2K_lg_l^2-z_l. \tag{87}\] For \(a\in\mathcal P\), the functional \[\sigma_{l,a}(c)=\tau(\mu_l(a)c)\qquad(c\in C_l)\] 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 (83) 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 (87) and \(0\leq\mu_l(a)\leq1\), we obtain \[\begin{align*} \sum_l\tau(\mu_l(a)s_l) &\geq \frac\delta3\sum_l\lambda_l \tau(\mu_l(a)(2g_l^4-z_l)) -\frac{2\delta}{3}\sum_l\lambda_l\tau(g_l^2K_lg_l^2). \tag{88}\end{align*}\] For the last term, remove the expectation against \(g_l^4\in C_l\): \[\begin{align*} \sum_l\lambda_l\tau(g_l^2K_lg_l^2) &=\sum_l\lambda_l\left(u^{-1}\tau(g_l^4D_l) +\delta^{-1/2}\tau(g_l^4S_l)\right)\\ &\leq u^{-1}|F|B_u+\sqrt\delta. \end{align*}\] In the inequality use \(g_l^4\leq z_l\), (77), and \(\tau(S_l)\leq\delta\). Combining this with (78)–(79) proves \[ \sum_l\tau(\mu_l(a)s_l) \geq\frac\delta3\tau(a)-\delta^{3/2}-r_u \qquad(a\in\mathcal P), \tag{89}\] where one may take the explicit error \[r_u=\frac\delta3\left(2u^3+3\sqrt{B_u} +2|F|B_u/u\right)\longrightarrow0.\]

Regularizing the combined row. Form matrices over \(D\) and models in its bidual as follows: \[W=(t_1,\ldots,t_r),\quad D_0=\operatorname{diag}(t_l^2),\quad m(a)=\operatorname{diag}(\mu_l(a)),\quad p=p_{D_0},\] \[R=W(1+W^*W)^{-1/2},\qquad Q_0=R^*R,\qquad G_0=D_0(1+D_0)^{-1}.\] Here \(\mathcal T(p)\leq\delta\) by (84). Each \(t_l\) commutes with \(\mu_l(a)\), so (85) says \[\|aW-Wm(a)\|_{2,\tau}\leq\sqrt{u\delta}=:e_u.\] The diagonal of \(W^*W-D_0\) is zero, and its \((i,j)\) entry off the diagonal is \(t_it_j\). Consequently \[ \|W^*W-D_0\|_{2,\tau}^2 =2\sum_l\sum_{i<l}\tau(t_l^2t_i^2) =2\sum_l\tau(t_l^2S_l)\leq2\delta^{3/2}. \tag{90}\] Lemma 30 gives the first of the following estimates; the second follows by expanding the commutator of \(R^*R\): \[\begin{align*} \|aR-Rm(a)\|_{2,\tau}&\leq e_u, &\|[Q_0,m(a)]\|_{2,\tau}&\leq2e_u. \tag{91}\end{align*}\] 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 \[ \|Q_0-G_0\|_{2,\tau}^2\leq2\delta^{3/2}. \tag{92}\] This regularization has removed every occurrence of \(\|W\|\) from the estimates.

The second pencil in one matrix commutant. Write \(B_0=\mathop{\mathrm{her}}_{M_r(D)}(D_0)\). 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 \[\pi:M_k\longrightarrow M(B_0)\] 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 (92) 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 \[\mathbb E_\pi(X)=\frac1{k^2}\sum_{i,j=0}^{k-1} \pi(u_k^iv_k^j)X\pi(u_k^iv_k^j)^*\] 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, \[ X\leq k^2\mathbb E_\pi(X)\qquad(X\geq0). \tag{93}\] Put \[K=\mathbb E_\pi((Q_0-G_0)^2).\] Then \(K\in B_0\cap\pi(M_k)'\), and \[ \mathcal T(K)\leq2\delta^{3/2},\qquad \|[m(a),K]\|_{2,\tau}\leq4e_u. \tag{94}\] For the commutator use \([m(a),G_0]=0\), \(\|Q_0-G_0\|\leq1\), (91), and the \(2\)-norm contractivity of the expectation, which commutes with multiplication by \(m(a)\).

Choose the fixed piecewise linear cutoff \[g'(t)=\begin{cases} 0,&t\leq1/10,\\ 10t-1,&1/10<t<1/5,\\ 1,&t\geq1/5, \end{cases} \qquad g'=g'(G_0),\] and, with \(L=(20k)^2\), set \[ X'=g'(p-LK)g',\qquad c=g'(X')_+g'. \tag{95}\] As for the first pencil, \(c\in B_0\cap\pi(M_k)'\) and \[ 0\leq c\leq(g')^2\leq p,\qquad c\geq(g')^4-L(g')^2K(g')^2, \qquad p_c\leq1_{(1/10,\infty)}(G_0). \tag{96}\] Lemma 18 gives \(p_cKp_c\leq L^{-1}p_c\). For \(\xi\in p_c\mathcal H\), (93) implies \[\|(Q_0-G_0)\xi\|^2 \leq k^2\langle K\xi,\xi\rangle \leq\frac1{400}\|\xi\|^2.\] On the same space \(G_0\) has compression at least \(1/10\). Therefore \[ c^2\geq cQ_0c\geq\frac1{20}c^2. \tag{97}\] Furthermore \(m(a)\) commutes with \(g'\). Lemma 30, applied to the scalar function \(t\mapsto t_+\) in (95), yields the useful explicit bound \[ \|[m(a),c]\|_{2,\tau} \leq L\|[m(a),K]\|_{2,\tau}\leq4Le_u. \tag{98}\]

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 (96), gives \[s'c^2s'\geq2s'cs'-s'\geq s'-2Ls'Ks'.\] 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 \[\begin{align*} \mathcal T(m(a)c^2) &\geq\mathcal T(m(a)s')-2L\mathcal T(K)\\ &\geq\frac\delta3\tau(a)-(1+4L)\delta^{3/2}-r_u. \tag{99}\end{align*}\] Here we used (89) and (94). 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 (97) implies \(p_d=p_c\), so \(U^*U=p_c\). We claim that the row \[Y=Uc\] belongs to \(M_{1,r}(D)\) and satisfies \[ \sup_{\tau\in T_{\leq1}(D)}\|aY-Ym(a)\|_{2,\tau}=o_u(1) \qquad(a\in F). \tag{100}\] We give uniform estimates for this passage, since \(r\) can grow with \(u\).

For \(\eta>0\) define the algebraic row \[Y_\eta=Rc(d+\eta)^{-1/2}c\in M_{1,r}(D).\] 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 \[ \|Y-Y_\eta\| \leq\sqrt{20}\sup_{t\geq0} \sqrt t\left(1-\sqrt{\frac{t}{t+\eta}}\right) \leq\sqrt{20\eta}. \tag{101}\] 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 \[ \|Y-Y_\eta\|_{2,\tau}\leq\sqrt{20\eta\delta}. \tag{102}\] 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, (91) and (98) imply \[\|[m(a),d]\|_{2,\tau}\leq(8L+2)e_u.\] 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 30. This gives \[\|aY_\eta-Y_\eta m(a)\|_{2,\tau} \leq\left((1+8L)\eta^{-1/2} +(4L+1)\eta^{-3/2}\right)e_u.\] Together with (102), 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 (100) without any restriction on \(r\).

Define \[V(x)=Uc\pi(x)cU^*=Y\pi(x)Y^*\qquad(x\in M_k).\] 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 (101). 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 \(x,y\in M_k\), \[V(x)V(y)=Uc^4\pi(xy)U^*=0.\] For a selfadjoint \(a\in F\) set \(\Delta_a=aY-Ym(a)\). Since \(m(a)\) commutes with \(\pi(M_k)\) and \(\|Y\|\leq1\), \[[a,V(x)]=\Delta_a\pi(x)Y^*-Y\pi(x)\Delta_a^*,\qquad \|[a,V(x)]\|_{2,\tau}\leq2\|x\|\|\Delta_a\|_{2,\tau}.\] This proves (71) when \(u\) is sufficiently small. Finally \(Y^*Y=c^2\) and \(\|Y\|_{2,\tau}\leq\sqrt\delta\). Rectangular traciality and Cauchy–Schwarz give \[\left|\tau(aV(1))-\mathcal T(m(a)c^2)\right| =\left|\mathcal T(Y^*\Delta_a)\right| \leq\sqrt\delta\,\|\Delta_a\|_{2,\tau}=o_u(1).\] Combining this with (99), and choosing \(u\) to make \(r_u+\sqrt\delta\sup_\tau\|\Delta_a\|_{2,\tau}<\varepsilon\) for the finitely many targets, proves (72). 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 \[F(x,0)\in M_k\otimes1, \qquad F(0,y)\in1\otimes M_k.\] 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 \(\mathbb C^k\otimes\mathbb C^k\), and put \(Q=(1-S)/2\). Since \(S=S^*=S^{-1}\), \(Q\) is a projection, and \[U(t)=\exp(i\pi tQ),\qquad 0\le t\le1,\] is a unitary path from \(1\) to \(S\). For \(z\in M_k\) define \[F_z(x,y)= \begin{cases} (x+y-xy)\,U\bigl(y/(x+y)\bigr)(z\otimes1) U\bigl(y/(x+y)\bigr)^*, &(x,y)\ne(0,0),\\ 0,&(x,y)=(0,0). \end{cases}\] 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 \[\sigma_\omega((z_n))=\lim_{n\to\omega}\sigma_n(z_n)\] on \(D_\omega\), which factors through \(B\) by Cauchy–Schwarz. These choices are allowed to depend on \(n\).

Fix \(0<\delta<1\). Proposition 29, 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 \[ \sigma_\omega(c\psi(1)) \ge\frac{\delta}{3}\sigma_\omega(c)-C_k\delta^{3/2} \tag{107}\] 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 (107) and join the new map to the preceding one using Lemma 31. The new weight commutes with \(r\), and hence, for any trace \(\sigma_\omega\), the defect \[D_j(a)=\sigma_\omega\bigl(a^{1/2}(1-H_j)a^{1/2}\bigr)\] satisfies \[D_{j+1}(a)\le(1-\delta/3)D_j(a)+C_k\delta^{3/2}.\] 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 \[ D_N(a)\le(1-\delta/3)^N+3C_k\sqrt\delta. \tag{108}\] 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 (104)–(105) on an \(\omega\)-large set. Its trace estimates also give (106) 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 (108). 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 \(u,v,x\in A\), \[ d_\tau(|u^*|)=d_\tau(|u|),\quad d_\tau(|xu|),d_\tau(|ux|)\le d_\tau(|u|),\quad d_\tau(|u+v|)\le d_\tau(|u|)+d_\tau(|v|). \tag{110}\] 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 \[u^{[t]}=u\,q_t(|u|),\qquad q_t(\lambda)= \begin{cases}(\lambda-t)_+/\lambda,&\lambda>0,\\0,&\lambda=0, \end{cases}\] 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 (110) 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 (109) 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 \[ [ (|z'_n|-s)_+ ]\le[ (|z_n|-s/2)_+ ], \tag{111}\] 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 (111) 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 \(j\le l\), \[\|[e^{(l)},c_j]\|<1/l,\qquad \|e^{(l)}y_j-y_j\|<1/l.\] 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 \(1\le j\le l\), \[ d_\tau((e^{(l)}_n-1/j)_+)\le l^{-1}d_\tau(b) \quad\text{for all }\tau\text{ with }d_\tau(b)<\infty. \tag{112}\] 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 (112) implies (109). 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 \(I=\mathop{\mathrm{Ideal}}_A(b)\) and choose a sequential positive contractive quasicentral approximate unit \(v_i\) for \(I\) in \(A\). Fix \(t_i>0\) tending to zero, and set \[p_i=(v_i-2t_i)_+,\qquad q_i=\chi_i(v_i),\] 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 \[ [q_i]\le L_i[b]. \tag{116}\] No uniform bound on these integers is required.

Let \(D_i=\mathop{\mathrm{her}}_A(q_i)\). If \(d_\tau(b)<\infty\), the restriction of \(\tau\) to \(D_i\) is a bounded positive tracial functional with \[ \|\tau|_{D_i}\|=d_\tau(q_i)\le L_i d_\tau(b). \tag{117}\] 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 32 inside \(D_i\), with tests \[p_i,\quad q_i x_j q_i\ (j\le i) \quad\text{and their real and imaginary parts},\] and positive target \(p_iap_i\). Choose its accuracy \(\rho_i>0\) after \(L_i\) has been fixed, so that \[ L_i\rho_i\longrightarrow0, \qquad L_i\rho_i^2\longrightarrow0; \tag{118}\] 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 (113), its trace is at most \(C\lambda_i d_\tau(b)\). The other part of the defect is, by traciality, \[\tau\bigl((1-\mu_i(1))^{1/2}p_iap_i (1-\mu_i(1))^{1/2}\bigr) \le L_i\rho_i d_\tau(b).\] Thus \[ \tau\bigl(a^{1/2}(1-\theta_i(1))a^{1/2}\bigr) \le(C\lambda_i+L_i\rho_i)d_\tau(b) \qquad(d_\tau(b)<\infty). \tag{119}\] 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 (117) and functional calculus give \[d_\tau((|z_i|-s)_+) \le s^{-2}\tau(z_i^*z_i) \le s^{-2}L_i\rho_i^2d_\tau(b).\] 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 (118) 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 34 to \(C_0=C^*(A,\theta(M_k))\), and let \(e\) be its absorbing positive contraction. Define \[\phi(z)=(1-e)\theta(z).\] 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 \[ \tau(a^{1/2}j_i a^{1/2}) \le s d_\tau(a)+d_\tau((j_i-s)_+). \tag{120}\] 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 (109) 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 (119) proves (114), 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 (114) 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 \(H_n,h^{(l)}_n\), \[H_n\ge h^{(l)}_n-\delta_{l,n}1, \qquad \delta_{l,n}\to_\omega0.\] 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 (115). 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 22 with \[x=(a-\varepsilon/16)_+, \qquad \gamma=\varepsilon/64.\] 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, \[ \|\sigma(x)\|\ge\gamma \quad\Longrightarrow\quad \|\sigma(xt_n)\|\ge\kappa \qquad\text{for every irreducible }\sigma. \tag{123}\] 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 (123). 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 (123) 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 31.

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, \[ \|\sigma(a)\|>3\varepsilon/32 \quad\Longrightarrow\quad \|\sigma(a^{1/2}g(h_n^\nu)^2a^{1/2})\|>\varepsilon/20. \tag{124}\] 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 (123). Since \[\|\sigma_n((a-\varepsilon/16)_+)\| >3\varepsilon/32-\varepsilon/16=\varepsilon/32>\gamma,\] that trigger applies, even if the represented norms tend to the boundary \(3\varepsilon/32\).

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 (123) gives \[\| (\bar a-\varepsilon/16)_+\bar t\|\ge\kappa.\] Consequently the joint spectral part where \(\bar a>\varepsilon/16\) and \(\bar t>\kappa/2\) is nonzero. Indeed the product is zero where \(\bar a\le\varepsilon/16\) 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 \[\|\bar a^{1/2}g(\bar H)^2\bar a^{1/2}\| \ge\varepsilon/16>\varepsilon/20.\] 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 (124) 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 (124) hold exactly. Include \(\{n\ge l\}\) in the \(l\)th set, and use its largest admissible stage at coordinate \(n\). Lemma 19 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 (124) holds, every irreducible representation annihilating \((y_n^\nu-\varepsilon/20)_+\) annihilates \((a-3\varepsilon/32)_+\). Indeed annihilation of the former is precisely the condition \(\|\sigma(y_n^\nu)\|\le\varepsilon/20\), which implies \(\|\sigma(a)\|\le3\varepsilon/32\). Primitive ideal separation gives \[(a-3\varepsilon/32)_+\in\mathop{\mathrm{Ideal}}_A((y_n^\nu-\varepsilon/20)_+).\] The additional positive cut \(\varepsilon/8-3\varepsilon/32=\varepsilon/32\) now gives a finite integer \(L_{a,\varepsilon,n}\) in (122), 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 36 for this \(k\). Fix its coordinate maps \(\nu_n\), their weights, and every integer \(L_{a,\varepsilon,n}\) before making any subsequent choices. They will not be reindexed in this proof.

For a test \((a,\varepsilon)\) retain \(b=(a-\varepsilon/8)_+\) and put \(d=(a-\varepsilon/4)_+\). Then \(d\in\mathop{\mathrm{Ideal}}_A(b)\) and \([d]\le[b]\), so Proposition 35 applies to \((d,b)\) with relative rank constant one. Let \(E_r\) be the good sets in (122) for the enumerated tests. Form decreasing \(\omega\)-large sets \[G_l=\{n\ge l\}\cap\bigcap_{r\le l}E_r,\] 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 35 for the first \(l(n)\) pairs \((d_r,b_r)\), with the individually prescribed accuracies \[\eta_n/L_{a_r,\varepsilon_r,n}.\] 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, \[ \tau\bigl(d^{1/2}(1-h_n^\theta)d^{1/2}\bigr) \le\frac{\eta_n}{L_{a,\varepsilon,n}}d_\tau(b) \qquad(d_\tau(b)<\infty). \tag{127}\] 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: \[ H_n=1-(1-h_n^\theta)^{1/2}(1-h_n^\nu) (1-h_n^\theta)^{1/2}. \tag{128}\] 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 \[ u_n=(1-f(H_n))^{1/2}a(1-f(H_n))^{1/2}. \tag{129}\] 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 (125) 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 \[ y^\nu=ag(h^\nu)^2\le ag(H)^2. \tag{130}\] 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 \[z_i=ag(H)^2\varrho(e_{ii})\quad(1\le i\le k), \qquad z_1=y^k(a).\] 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 (130) commute, so scalar functional calculus, splitting a cut of an orthogonal sum, and enlarging its individual cuts give \[ [(y^\nu-\varepsilon/22)_+] \le k[(y^k(a)-\varepsilon/30)_+] \quad\text{in }\mathrm{Cu}(A_\omega). \tag{131}\]

Here the cut margins yield an exact inequality before any multiplication by a varying integer. Put \(\Delta=\varepsilon/20-\varepsilon/22>0\). Choose one rectangular comparison witness for (131) 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, \[\left\|(y_n^\nu-\varepsilon/22)_+ -X_n^*\operatorname{diag}\bigl((y_n-\varepsilon/30)_+, \ldots,(y_n-\varepsilon/30)_+\bigr)X_n \right\|<\Delta.\] Norm comparison at that coordinate gives \[[(y_n^\nu-\varepsilon/20)_+] \le k[(y_n-\varepsilon/30)_+] \le k[(y_n-\varepsilon/32)_+].\] Intersect with the good set for (122). Only now multiply the exact Cuntz inequality by \(L_n=L_{a,\varepsilon,n}\) to obtain \[ [b]\le kL_n[(y_n-\varepsilon/32)_+]. \tag{132}\] 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 \[ d_\tau((u_n-\varepsilon)_+) \le\tfrac12 d_\tau((y_n-\varepsilon/32)_+) \qquad\text{for every }\tau \tag{133}\] on an \(\omega\)-large set. Put \(c_n=(1-f(H_n))^{1/2}\) and \(v_n=c_ndc_n\). Since \(\|a-d\|\le\varepsilon/4\), one has \(\|u_n-v_n\|\le\varepsilon/4\). Comparing \(u_n\) with \((v_n-\varepsilon/2)_+\) gives \[[(u_n-\varepsilon)_+]\le[(v_n-\varepsilon/2)_+], \qquad d_\tau((u_n-\varepsilon)_+)\le(2/\varepsilon)\tau(v_n).\] For the first inequality the norm error is at most \(\varepsilon/4+\varepsilon/2<\varepsilon\); the second is the elementary spectral trace bound. From the scalar inequality \(f(t)\ge t\) and (128), \[f(H_n)\ge H_n\ge h_n^\theta.\] This uses functional calculus on \(H_n\) itself and positive domination of \(H_n\); it makes no operator-monotonicity assertion about \(f\). Traciality and (127) consequently give, whenever \(d_\tau(b)<\infty\), \[ d_\tau((u_n-\varepsilon)_+) \le\frac{2}{\varepsilon} \tau\bigl(d^{1/2}(1-f(H_n))d^{1/2}\bigr) \le\frac{2\eta_n}{\varepsilon L_n}d_\tau(b). \tag{134}\]

This estimate handles every possible trace value. Set \(R_n=d_\tau((y_n-\varepsilon/32)_+)\). If \(R_n<\infty\), the exact inequality (132) gives \(d_\tau(b)\le kL_nR_n<\infty\), and therefore \[d_\tau((u_n-\varepsilon)_+) \le (2k\eta_n/\varepsilon)R_n.\] For the fixed \(k\) and test, eventually \(2k\eta_n/\varepsilon\le1/2\). 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 (133) is automatic in \([0,\infty]\). All good sets used in these alternatives were chosen uniformly over the trace families; thus their intersection proves (133) simultaneously for every lower semicontinuous extended trace.

Apply Lemma 16 in \(A\) at each coordinate of this common good set. It gives \[(u_n-\varepsilon)_+\precsim (y_n-s)_+, \qquad s=\varepsilon/32.\] Choose \(x_n\in A\) with \[\|x_n^*(y_n-s)_+x_n-(u_n-\varepsilon)_+\|<1/n.\] These witnesses need not themselves be bounded. Define the continuous scalar function \[q_s(t)= \begin{cases}\bigl((t-s)_+/t\bigr)^{1/2},&t>0,\\0,&t=0, \end{cases}\] and set \(r_n=q_s(y_n)x_n\) on the good set and \(r_n=0\) elsewhere. Then \[r_n^*y_nr_n=x_n^*(y_n-s)_+x_n, \qquad \|r_n\|^2\le s^{-1}\|x_n^*(y_n-s)_+x_n\| \le(32/\varepsilon)(1+1/n).\] The norm estimate follows from \(q_s(t)^2\le s^{-1}(t-s)_+\). Thus these representatives have the fixed bound \(\|r_n\|^2\le64/\varepsilon\), and their ultrapower element has norm squared at most \(32/\varepsilon\). The compression errors tend to zero along \(\omega\), proving (126). Neither bound depends on \(k\) or on any \(L_n\). ◻

Proof. For fixed \(m\ge1\), apply Proposition 37 with \(k=pm\) and identify \(M_{pm}=M_p\otimes M_m\). Define \[\alpha_m(z)=f(\zeta^{pm})(z\otimes1_m),\qquad s_j^{(m)}=g(\zeta^{pm})(e_{11}^{(p)}\otimes e_{j1}^{(m)}) \quad(1\le j\le m), \qquad P_m=s_1^{(m)}.\] These are norm-central contractions, and \(P_m\) is positive. The matrix-unit relations for the supporting homomorphism give \[(s_i^{(m)})^*s_j^{(m)}=\delta_{ij}P_m^2.\] 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 37 supplies, for every test, \[((1-\alpha_m(1))a-\varepsilon)_+ =(r_{a,\varepsilon}^{(m)})^*(P_m^2a)r_{a,\varepsilon}^{(m)}.\] 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 \(\sqrt{64/\varepsilon}\); the sharper ultrapower bound in Proposition 37 leaves room for this choice. Put \(h_{m,n}=\alpha_{m,n}(1)\). At stage \(l\) use \(m=l\) and require:

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. the centrality and the two algebraic relations in (135) for all slot indices at most \(l\) to have error less than \(1/l\);

3. for the first \(l\) comparison tests, the equations with positive coordinate expressions \[\bigl((1-h_{l,n})^{1/2}a(1-h_{l,n})^{1/2}-\varepsilon\bigr)_+ \quad\text{and}\quad r_{a,\varepsilon,l,n}^*P_{l,n}aP_{l,n}r_{a,\varepsilon,l,n}\] 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 \(\sqrt{64/\varepsilon}\) independently of the row. These are precisely the boundedness hypotheses of Lemma 19. The selected maps therefore define a c.p.c. norm-central order-zero \(\alpha\), and the selected slots and witnesses define elements with (135)–(136). 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 of Theorem 2. Fix the free ultrafilter in the theorem. Proposition 46, through the support, norm-cone, tracial-cone, comparison, and packing constructions above, supplies a unital \(*\)-homomorphism \(\mathcal Z\to F_\omega(A)\). Proposition 49 then gives \(A\cong A\otimes_{\min}\mathcal Z\). ◻

Proof of Theorem 1. The zero algebra, if admitted by the convention for simplicity, is immediate. Nuclearity implies exactness, so Corollary 12 shows that \(A\) satisfies condition (C). Theorem 2 then gives \(A\cong A\otimes_{\min}\mathcal Z\). ◻

Proof of Corollary 3. Again the zero algebra is immediate. Lemma 15 derives (C) from the canonical induced map \(\mathrm{Cu}(\iota)\), using that \(\mathrm{Cu}(\iota)([b])=[b\otimes1_\mathcal Z]\). Theorem 2 then proves the assertion. ◻

Proof of Corollary 4. The forward implication follows from functoriality of \(\mathrm{Cu}\) under \(*\)-isomorphisms. For the converse, let \(\Phi:\mathrm{Cu}(A)\to\mathrm{Cu}(A\otimes_{\min}\mathcal Z)\) be an abstract isomorphism. The case \(A=0\) is immediate, so put \(D=A\otimes_{\min}\mathcal Z\) and suppose \(A\ne0\). Then \(D\) is nonzero, separable, and nuclear.

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

Approximately unitarily equivalent homomorphisms induce the same map on \(\mathrm{Cu}\). Indeed, after stabilization, norm approximation by unitary conjugates gives comparison of every positive cut of either image with the other image, by Lemma 5(1); approximate units place the comparison witnesses in the algebra when the conjugating unitaries are multipliers. Taking the suprema of cuts in Equation (4) gives equality of the two Cuntz classes. Thus \(\mathrm{Cu}(j_D)=\mathrm{Cu}(\mathop{\mathrm{id}}_A\otimes\varphi)\) is an isomorphism. Lemma 15, applied to \(D\), gives condition (C) on \(\mathrm{Cu}(D)\). Both almost unperforation and the full almost-divisibility sandwich are intrinsic ordered-semigroup properties, so \(\Phi\) transports them to \(\mathrm{Cu}(A)\). Theorem 2 now gives \(A\cong A\otimes_{\min}\mathcal Z\). ◻