Proof. Let \(r\ne0\) and \(h=h^*\in rDr\). Choose \(R>\norm{h}+2\) and finite-spectrum self-adjoints \(a_n\in D\) converging in norm to \(h+Rr\). The spectrum of the limit is contained in \(\{0\}\cup(2,\infty)\), so the spectral projections \(f_n=1_{(1,\infty)}(a_n)\) converge in norm to \(r\). For all sufficiently large \(n\), the element \[r f_n+(1-r)(1-f_n)\] is invertible. Its unitary polar part \(u_n\) satisfies \(u_n f_n u_n^*=r\) and \(u_n\to1\) in norm. Since \(a_n\) commutes with \(f_n\), the compression \(r u_n a_n u_n^*r\) has finite spectrum in \(rDr\). Subtracting \(Rr\) gives finite-spectrum approximations to \(h\) in that corner.

For the second assertion, first replace \(\mathcal F\) by its real and imaginary parts. Compress the first self-adjoint test to \(rDr\), approximate it by a finite-spectrum self-adjoint, and partition \(r\) by the latter’s spectral projections. Repeat for each subsequent test in every current corner. An estimate \(\norm{q hq-\lambda q}<\delta\) persists on every subprojection of \(q\). Thus this finite refinement procedure preserves all earlier estimates. Using tolerance less than \(\gamma/2\) for the real and imaginary parts proves (5); zero pieces are discarded. ◻

Proof. First we obtain arbitrarily many mutually orthogonal nonzero positive elements. Choose a maximal abelian unital \(C^*\)-subalgebra \(E\subset A\). Suppose that \(E\) were finite dimensional, with minimal projections \(p_1,\ldots,p_m\) summing to \(1\). Every selfadjoint element of \(p_iAp_i\) commutes with \(E\), so maximality implies \(p_iAp_i=\mathbb Cp_i\). If \(0\ne x\in p_iAp_j\), then \[x^*x=\norm{x}^2p_j,\qquad xx^*=\norm{x}^2p_i.\] Thus \(u=x/\norm{x}\) satisfies \(u^*u=p_j\) and \(uu^*=p_i\). For any \(y\in p_iAp_j\), the element \(u^*y\) is scalar on \(p_j\), and \(y=u(u^*y)\). Consequently each \(p_iAp_j\) has dimension at most one. The decomposition \(A=\sum_{i,j}p_iAp_j\) would make \(A\) finite dimensional, a contradiction. Hence \(E\cong C(X)\) for an infinite compact Hausdorff space \(X\). Given \(k\in\mathbb N\), choose \(k\) distinct points of \(X\), pairwise disjoint neighborhoods, and nonzero positive continuous functions supported in those neighborhoods. Their images give mutually orthogonal nonzero positive elements \(c_1,\ldots,c_k\in A\).

For two nonzero positive elements \(c,d\in A\), simplicity gives \(cAd\ne0\). Indeed, if \(cAd=0\), then \(c\) annihilates the ideal generated by \(d\); this ideal is \(A\), which would force \(c=0\). Choose \(y\in A\) such that \(z=cyd\ne0\). Then \(z^*z\in\overline{dAd}\) and \(zz^*\in\overline{cAc}\). In each tracial von Neumann representation, the polar decomposition of \(z\) gives a partial isometry with initial projection \(\operatorname{supp}(z^*z)\) and final projection \(\operatorname{supp}(zz^*)\). The normal tracial state therefore gives \[\tau(\operatorname{supp}(z^*z))=\tau(\operatorname{supp}(zz^*)) \le \min\{\tau(\operatorname{supp}(c)),\tau(\operatorname{supp}(d))\} \qquad(\tau\in T(A)).\] The elements \(y,z\) are chosen independently of \(\tau\); the support comparison holds in every tracial representation. Start with \(c=c_1\). For \(j=2,\ldots,k\), apply the preceding construction with \(d=c_j\) and replace \(c\) by \(zz^*\). The support trace never increases and is at most the support trace of \(c_j\). The final nonzero positive element \(e=c\) therefore has support trace at most that of each \(c_i\), at every trace. Since the support projections of the \(c_i\) are orthogonal, \[k\,\tau(\operatorname{supp}(e))\le\sum_{i=1}^k\tau(\operatorname{supp}(c_i))\le1 \qquad(\tau\in T(A)).\] Choose \(k\) with \(1/k<\delta\) and rescale \(e\) to have norm one, which does not change its support. It is full because \(A\) is simple. ◻

Proof. The restriction to \(A\) of any trace in \(\operatorname T(D)\) is a tracial state, so \(\operatorname T(A)\ne\varnothing\). Given a positive integer \(n\), apply Lemma 7 to obtain \(b\in A_+\) with \(\norm b=1\) and \[ \rho(b)\le\rho(\operatorname{supp}(b))<\frac1n \qquad(\rho\in\operatorname T(A)). \tag{6}\]

Choose a finite-spectrum self-adjoint \(a\in D\) with \(\norm{a-b}<1/8\), and set \(g=1_{[1/2,\infty)}(a)\). Then \(gbg\ge(3/8)g\), so restriction of any \(\tau\in\operatorname T(D)\) to \(A\) gives \[ \tau(g)\le\frac83\tau(b)\le\frac{8}{3n}. \tag{7}\] We claim that \(g\) is full in \(D\). Otherwise, let \(\pi:D\to D/I\) be the quotient by the proper closed ideal generated by \(g\). The inequalities \(a\ge-1/8\) and \(a(1-g)\le(1/2)(1-g)\) imply \(\norm{\pi(a)}\le1/2\), and hence \(\norm{\pi(b)}\le5/8\). On the other hand, \(\pi|_A\) is a unital homomorphism from the simple algebra \(A\) to a nonzero unital algebra. It is injective and therefore isometric, contradicting \(\norm{b}=1\). Thus \(g\) is full, and choosing \(n\) with \(8/(3n)<\delta\) finishes the proof. ◻

Proof. If \(f=0\), take all \(e_i=0\). We argue by induction on \(m\); for \(m=1\), \(v_1\) itself implements \(f\precsim g_1\). Suppose that \(m>1\). Apply Lemma 6 to choose a finite-spectrum self-adjoint \(a\in fDf\) within \(1/8\) of \(v_1^*v_1\). Set \[e_1=1_{[1/2,\infty)}(a),\qquad r=f-e_1.\] The inequality \(e_1v_1^*v_1e_1\ge(3/8)e_1\) shows that, when \(e_1\ne0\), \[v_1e_1(e_1v_1^*v_1e_1)^{-1/2}\] is a partial isometry with initial projection \(e_1\) and final projection below \(g_1\). On the remaining corner, \[Q:=\sum_{i=2}^m r v_i^*v_i r =r-rv_1^*v_1r\ge\frac38r.\] If \(r=0\), take \(e_2=\cdots=e_m=0\). Otherwise define \(w_i=v_i r Q^{-1/2}\) for \(i=2,\ldots,m\), with the inverse taken in \(rDr\). These elements satisfy \(w_i=g_iw_i r\) and \(\sum_{i=2}^m w_i^*w_i=r\). The induction hypothesis partitions \(r\) as required. ◻

Proof. The case \(f=0\) is immediate. Choose a full projection \(g\) with \(\sup_\tau\tau(g)<\varepsilon\). Fullness provides a finite sum \(\sum_i x_i g y_i\) within distance less than one of \(1\). Write \(X=[x_1\ \cdots\ x_m]\) and let \(Y\) be the column with entries \(gy_i\). Then \(XY\) is invertible, so \(Y\) has a bounded left inverse \(L=(XY)^{-1}X\). In particular, \[1=Y^*L^*LY\le\norm{L}^2Y^*Y,\] and \(Y^*Y\) is invertible. The entries of \(Y(Y^*Y)^{-1/2}f\) form a column as in Lemma 9, with every target projection equal to \(g\). We obtain an orthogonal family with sum \(f\), each member subequivalent to \(g\).

We next replace this family by one, say \(s_1,\ldots,s_m\), satisfying \[ s_1\precsim s_2\precsim\cdots\precsim s_m\precsim g. \tag{9}\] This ordering makes the trace weights increase in the same order for every trace, so that one choice of rounding coefficients will work simultaneously for all of them. Here is the pair operation that makes this possible. Suppose that \(x,y\) are orthogonal pieces, both subequivalent to a common projection \(h\). Choose partial isometries \(w,u\) with \[w^*w=x,\quad ww^*=x'\le h, \qquad u^*u=y,\quad uu^*\le h.\] Since \(u^*x'u+u^*(h-x')u=y\), Lemma 9 applied to the column with entries \(x'u,(h-x')u\) gives \(y=k+l\) and partial isometries \(\alpha,\beta\) with \[\alpha^*\alpha=k,\quad \alpha\alpha^*\le x', \qquad \beta^*\beta=l,\quad \beta\beta^*\le h-x'.\] Set \(\xi=w^*\alpha+l\) and \(\zeta=w+\beta\). Then \[\begin{align*} \xi^*\xi&=y,& \xi\xi^*&=w^*\alpha\alpha^*w+l\le x+l,\\ \zeta^*\zeta&=x+l,& \zeta\zeta^*&=x'+\beta\beta^*\le h. \end{align*}\] The cross terms vanish by orthogonality of the initial and final components: \(x\perp l\) and \(x'\perp(h-x')\). Thus replacing \(x,y\) by \(k,x+l\) preserves their sum and orthogonality, and \[k\precsim x\le x+l, \qquad x,y\precsim x+l\precsim h.\] After each adjacent comparison, the new right-hand piece dominates every piece already passed: it contains the previous moving piece. Thus a left-to-right pass with common bound \(g\) leaves a last piece dominating all earlier pieces. Freeze that last piece and repeat on the shorter list, now using it as the common bound. Induction gives (9). All operations take place inside \(D\) and preserve the orthogonal sum \(f\).

Set \[c_i=\lfloor si\rfloor-\lfloor s(i-1)\rfloor\in\{0,1\}, \qquad e=\sum_{i=1}^m c_i s_i.\] For a fixed \(\tau\in\operatorname T(D)\), put \(t_i=\tau(s_i)\) and \(D_j=\lfloor sj\rfloor-sj\). Then \(t_1\le\cdots\le t_m\le\tau(g)\) and \(-1\le D_j\le0\). Summation by parts gives \[\tau(e)-s\tau(f) =D_m t_m+\sum_{i=1}^{m-1}D_i(t_i-t_{i+1}).\] The first term lies in \([-t_m,0]\), while the sum lies in \([0,t_m-t_1]\). It follows that \(|\tau(e)-s\tau(f)|\le t_m\le\tau(g)\), proving (8) uniformly in \(\tau\). ◻

Proof. The assertion is immediate if \(q=0\) or \(\mathcal F\) is empty. Choose \(\gamma>0\) with \(2\gamma<\varepsilon/2\) and use Lemma 6 to write \(q=\sum_{j=1}^m q_j\) with \[\norm{q_j hq_j-\lambda_{j,h}q_j}<\gamma \qquad(h\in\mathcal F)\] for suitable scalars \(\lambda_{j,h}\). These scalars satisfy \(|\lambda_{j,h}|\le\norm{h}+\gamma\). Put \(M=\max_{h\in\mathcal F}\norm{h}+\gamma\) and choose \(\delta>0\) with \(mM\delta<\varepsilon/2\). By Proposition 10, choose \(e_j\le q_j\) so that \(|\tau(e_j)-s\tau(q_j)|<\delta\) for every \(\tau\), and set \(e=\sum_j e_j\). Traciality removes all off-diagonal compressions and gives \[\begin{align*} |\tau(eh)-s\tau(qh)| &\le \gamma\bigl(\tau(e)+s\tau(q)\bigr) +\sum_{j=1}^m |\lambda_{j,h}| |\tau(e_j)-s\tau(q_j)| \\ &\le 2\gamma+mM\delta<\varepsilon. \end{align*}\] ◻

Proof. If \(y=0\), take \(t=0\). Otherwise choose, in \(yDy\), a finite-spectrum self-adjoint \(a\) with \(\norm{a-yzy}<1/8\), and let \(f=1_{(-\infty,1/2)}(a)\). Writing \(d=y-f\), we have \[d(yzy)d\ge\frac38d, \qquad f(yzy)f\le\frac58f.\] Since \(yzy\ge0\), this implies \[ \tau(y-f)\le\frac83\tau(yz), \qquad A_f:=f(1-z)f\ge\frac38f. \tag{14}\] If \(f=0\), put \(t=0\). Otherwise the element \[v=(1-z)f A_f^{-1/2}\] is a partial isometry with \(v^*v=f\); set \(t=vv^*\le1-z\). In both cases, \[\tau(t)=\tau(f),\qquad ftf=f(1-z)f.\] The mass bounds (11) follow from (14). As \(f\le y\), \[\tau(ty)\ge\tau(tf)=\tau(f)-\tau(fz).\] Consequently, \[\begin{align*} \left\|t-y\right\|_{2,\tau}^{2} &=\tau(t)+\tau(y)-2\tau(ty)\\ &\le\tau(y-f)+2\tau(fz) \le\left(\frac83+2\right)\tau(yz) \le5\tau(yz), \end{align*}\] which proves (12).

Finally, decompose \(t-y=t(t-y)+(1-t)(t-y)\). The first summand has left support below \(t\), and the second equals \(-(1-t)y\) and has right support below \(y\). Tracial Cauchy–Schwarz therefore gives \[\left\|t-y\right\|_{1,\tau} \le\bigl(\sqrt{\tau(t)}+\sqrt{\tau(y)}\bigr) \left\|t-y\right\|_{2,\tau} \le2\sqrt{\tau(y)}\,\left\|t-y\right\|_{2,\tau}.\] Trace-norm duality proves (13). ◻

Proof. Let \(E=A\widehat\otimes A\) denote the Banach projective tensor product, with norm \(\norm{\cdot}_\pi\), multiplication map \(m_A(a\otimes b)=ab\), and bimodule actions \(x\cdot(a\otimes b)=xa\otimes b\) and \((a\otimes b)\cdot x=a\otimes bx\). Haagerup’s virtual diagonal theorem (Haagerup 1983, Theorem 3.1) gives a virtual diagonal \(M\in E^{**}\) in the weak-star closed convex hull of \(\{v\otimes v^*:v\in A,\ \norm v\leq1\}\). Its bimodule commutators vanish and \(m_A^{**}(M)=1\). Apply the second adjoint of the bounded linear map \[T:E\longrightarrow E^K\oplus A,\qquad T(d)=\big((x\cdot d-d\cdot x)_{x\in K},m_A(d)\big).\] Its value at \(M\) is \((0,\ldots,0,1)\), in the canonical copy of \(E^K\oplus A\). Weak-star convergence in the bidual restricts to weak convergence in this copy, so this point lies in the weak closure of the image of the finite convex hull. A convex subset of a Banach space has the same weak and norm closures, by Hahn–Banach separation. We may therefore choose a finite convex combination \(d=\sum_l\lambda_l v_l\otimes v_l^*\) such that \[ \norm{x\cdot d-d\cdot x}_\pi<\mu\quad(x\in K),\qquad \norm{1-\sum_l\lambda_l v_lv_l^*}<\mu. \tag{17}\]

Both sums \(\sum_l\lambda_l v_lv_l^*\) and \(\sum_l\lambda_l v_l^*v_l\) are positive contractions. By traciality, the positive element \(c=1-\sum_l\lambda_l v_l^*v_l\) satisfies \(0\leq\tau(c)<\mu\). Since \(\tau(q)-\tau(\Phi(q))=\tau(qc)\), this proves (15). For the centrality estimate, the map \(a\otimes b\mapsto aqb\) has norm at most \(\tau(q)\) from the projective tensor norm to the tracial 1-seminorm: indeed, \[\left\|aqb\right\|_{1,\tau} \leq\left\|aq\right\|_{2,\tau}\left\|qb\right\|_{2,\tau} \leq\norm a\norm b\,\tau(q).\] Apply this map to the first inequality in (17) and multiply by the unitary \(x^*\). ◻

Proof. Work in the tracial von Neumann algebra and put \(d=\sqrt b-\sqrt c\). Traciality and \(b-c=\sqrt b\,d+d\sqrt c\) give \[\tau(\mathop{\mathrm{sgn}}(d)(b-c))=\tau((\sqrt b+\sqrt c)|d|).\] Since \(\sqrt b+\sqrt c\geq d\) and \(\sqrt b+\sqrt c\geq-d\), tracing against the positive and negative parts \(d_+\) and \(d_-\), respectively, bounds the right-hand side below by \(\tau(d_+^2)+\tau(d_-^2)=\tau(d^2)\). Trace-norm duality bounds the left-hand side by \(\left\|b-c\right\|_{1,\tau}\). ◻

Proof. Put \(\mu=\eta^2\) and choose \(\Phi\) by Lemma 13. In particular, the coefficient list is fixed before \(r\). Choose \(0<\sigma<1\), then \(\beta>0\), so that \[ \beta<\sigma/2,\qquad m(\sigma+2\beta)<\mu,\qquad \frac{2\beta}{\sigma} +\frac{(2+\sqrt2)\beta}{\sigma^2}<\eta. \tag{23}\] This is possible by choosing \(\sigma\) with \(m\sigma<\mu\) first.

Proof. Write \(T=U^*xU\), \(P_e=1_k\otimes e\), and choose \(T_0\in M_k(\mathbb C)\) with \(\norm{T-T_0\otimes q}<\eta\). Since \(T\) is a contraction, \(\norm{T_0}\leq2\). The map \(a\mapsto UaU^*\) identifies the relevant matrix corner with its image in \(D\), and therefore \[\tau(yx)=(\operatorname{Tr}\otimes\tau)(P_eTP_e),\qquad \tau(|yxy|^2)=(\operatorname{Tr}\otimes\tau)(|P_eTP_e|^2).\] The scalar matrix \(T_0\otimes q\) commutes with \(P_e\), whether or not \(e\) commutes with any entry of \(T\). Thus replacing \(P_eTP_e\) by \(T_0\otimes e\) changes these two traces by at most \(\eta\tau(y)\) and \(3\eta\tau(y)\), respectively. For the latter estimate use \[\norm{R^*R-S^*S}\leq(\norm R+\norm S)\norm{R-S}, \quad R=P_eTP_e,\quad S=T_0\otimes e,\] and the fact that the trace of the corner unit is \((\operatorname{Tr}\otimes\tau)(P_e)=k\tau(e)=\tau(y)\). The same estimates hold with \(e\) replaced by \(q\).

Set \[\alpha=\frac{\operatorname{Tr}(T_0)}{k},\qquad \beta_0=\frac{\operatorname{Tr}(T_0^*T_0)}{k}.\] Then \(|\alpha|\leq2\) and \(0\leq\beta_0\leq4\). The scalar reference traces for \(y\) are \(\alpha\tau(y)\) and \(\beta_0\tau(y)\); for \(p\), they are \(\alpha\tau(p)\) and \(\beta_0\tau(p)\). This proves (30). For the second estimate, the exact identity \[ \left\|[x,f]\right\|_{2,\tau}^2 =2\tau(f)-2\tau(|fxf|^2) \qquad(f\text{ a projection},\ x\text{ unitary}) \tag{32}\] follows by expansion and traciality. Applying it to \(y,p\) and using the squared compression estimates proves (31); the mass-discrepancy coefficient is bounded by \(2+2\beta_0\leq10\). In particular, (32) also accounts for the part of \(x\) that leaves the larger block \(p\). ◻

Proof. The two off-diagonal corners of the commutator are orthogonal in the tracial 2-seminorm, so its square is \[\left\|(1-z)xz\right\|_{2,\tau}^2+\left\|zx(1-z)\right\|_{2,\tau}^2.\] Splitting the \(z\)-corner into the orthogonal \(t_a\)’s writes each term as a sum of squared norms. Since \(1-z\leq1-t_a\), the respective summands are bounded by \(\left\|(1-t_a)xt_a\right\|_{2,\tau}^2\) and \(\left\|t_ax(1-t_a)\right\|_{2,\tau}^2\). ◻

Proof of Theorem 3. Fix a finite \(K\subseteq\mathcal U(A)\). We construct orthogonal projections \(P,S\) such that \(P\) nearly commutes with \(K\), their coefficient traces nearly agree, and \(P+S\) nearly fills the unit in trace. The identity \[2\tau(Px)-\tau(x) =\tau((P-S)x)-\tau((1-P-S)x) \qquad(x\in K)\] then gives the required halving estimate. We obtain \(P,S\) as two labelled sums, sampling equal proportions for the two signs in each round and moving the samples into the unoccupied corner. We use \(H>0\) and \(N\in\mathbb N\), with \(\eta=H/N<1\). First \(H\) is chosen to reduce the unoccupied mass; then \(N\) is increased at fixed \(H\) to reduce the errors.

Proof. The upper bounds follow from the definition of a limit trace. For the reverse bound in (50), choose \(\tau_j\) whose value on \(d_j^*d_j\) is within \(1/j\) of the supremum. The associated limit trace gives equality. For (51), choose approximate maximizers of \(|\bar\tau(d_j)|\). Absolute value commutes with the ultralimit of a bounded complex sequence, so this choice also gives equality. The final assertion is (50) and the definition of \(I_\omega\). ◻

Proof of Theorem 1. Set \(D=B^\omega\). It is a unital real-rank-zero \(C^*\)-algebra by hypothesis and has the tracial states in \(\Lambda_\omega\). The inclusion \(A\subseteq D\) is unital and faithful. Lemma 8 supplies full projections of uniformly arbitrarily small trace over all \(\operatorname T(D)\). Thus Theorem 3 applies.

Choose increasing finite subsets \(F_n\subseteq\mathcal U(A)\) whose union is norm dense in \(\mathcal U(A)\), including \(1\) in each \(F_n\). Theorem 3 gives projections \(P_n\in D\) with \[ \left\|[P_n,x]\right\|_{2,\tau}<1/n,\qquad \left|\tau(P_nx)-\tfrac12\tau(x)\right|<1/n \quad(x\in F_n,\ \tau\in\operatorname T(D)). \tag{52}\] Lift \(P_n\) to positive contractions \((c_{n,j})_j\) in \(B\). Such lifts exist by first taking a self-adjoint lift and then applying the continuous function which clips the real line to \([0,1]\). Since \(P_n\) is a projection, \[\lim_{j\to\omega}\left\|c_{n,j}^2-c_{n,j}\right\|_{2,\operatorname T(A)}=0.\]

For each fixed \(n\), Lemma 18, applied to the commutators and the linear errors in (52), gives an \(\omega\)-large set of coordinates on which \[\begin{align*} \left\|c_{n,j}^2-c_{n,j}\right\|_{2,\operatorname T(A)}&<1/n,\\ \left\|[c_{n,j},x]\right\|_{2,\operatorname T(A)}&<2/n &&(x\in F_n),\\ \sup_{\tau\in\operatorname T(A)} \left|\bar\tau(c_{n,j}x)-\tfrac12\bar\tau(x)\right| &<2/n &&(x\in F_n). \end{align*}\] There are finitely many conditions, so their good coordinate sets have an \(\omega\)-large intersection. Choose one coordinate \(j(n)\) in that intersection and put \(b_n=c_{n,j(n)}\).

The sequence \((b_n)\) consists of positive contractions. Its projection defect, and every eventually included test error, converge to zero in the ordinary sense. Hence \(p=[(b_n)]\in B^\omega\) is a projection, commutes with the tested unitaries, and satisfies \(\lambda(px)=\lambda(x)/2\) on those unitaries for every \(\lambda\in\Lambda_\omega\). Norm density and linear spanning by unitaries extend both conclusions to every \(x\in A\). This construction has used the original \(\omega\); no further ultrapower is taken.

Finally, \(A\) is dense in \(B\) for the designated uniform tracial 2-norm. Indeed, if \(x=[(a_j)]\in C/J=B\), then \[\left\|x-a_m\right\|_{2,\operatorname T(A)}^2 =\sup_{\tau\in\operatorname T(A)} \lim_j\tau((a_j-a_m)^*(a_j-a_m))\longrightarrow0\] by uniform 2-Cauchyness. Given \(x\in B\), choose \(a\in A\) with \(\left\|x-a\right\|_{2,\operatorname T(A)}<\delta\). For every \(\lambda\in\Lambda_\omega\), multiplication by the contraction \(p\) and tracial Cauchy–Schwarz give \[\left\|[p,x]\right\|_{2,\lambda}\leq2\delta,\qquad \left|\lambda(px)-\tfrac12\lambda(x)\right|\leq\tfrac32\delta.\] Let \(\delta\) tend to zero. The first assertion, together with Lemma 18, gives \([p,x]=0\) in \(D\) itself; the second is the required trace identity. Thus the same \(p\) works for the entire constant copy of \(B\), even when \(B\) is not norm separable. ◻

Proof of Corollary 2. Put \(X_B=\{\bar\tau:\tau\in T(A)\}\). By (Carrión et al. 2023, Proposition 3.23(ii)–(iv)), \(X_B\) is a compact face of \(T(B)\) and the designated uniform \(2\)-norm is complete on the unit ball of \(B\). In the terminology of that paper, \((B,X_B)\) is a factorial tracially complete pair. Theorem 1 and (Carrión et al. 2023, Proposition 5.20) give property \(\Gamma\) for this pair, and hence complemented partitions of unity by (Carrión et al. 2023, Theorem 1.4). Evington’s trace theorem (Evington 2025, Theorem B) then gives \(T(B)=X_B\) provided the pair is of type \(\mathrm{II}_1\): each designated GNS von Neumann algebra must have no nonzero type I central summand.

For every \(\tau\in T(A)\), the finite von Neumann algebra \(\pi_\tau(A)''\) has no nonzero type I central summand. Otherwise it would have a nonzero homogeneous summand \(M_n(Z)\) with \(Z\) abelian; composing the representation of \(A\) into this summand with a character of \(Z\) would give a unital finite-dimensional representation of \(A\). Simplicity would make that representation injective, contrary to the infinite dimensionality of \(A\). Thus every such GNS algebra is of type \(\mathrm{II}_1\). Their preservation under completion (Carrión et al. 2023, Proposition 3.23(vi)) makes \((B,X_B)\) a type \(\mathrm{II}_1\) pair, so the cited trace theorem applies. Restriction is therefore a continuous affine bijection between compact Hausdorff trace spaces, and hence a homeomorphism. Finally, for \(\sigma=\bar\tau\), tracial Cauchy–Schwarz gives \(|\sigma(b)|\le\bar\tau(b^*b)^{1/2}\le\|b\|_{2,T(A)}\). ◻

Proof. Take \(e\) from Lemma 7. We first extract an exact finite identity from its fullness. Since the linear span of \(AeA\) is dense in \(A\), choose \(u_i,v_i\in A\), \(1\le i\le n\), so that \[s=\sum_{i=1}^n u_i e v_i,\qquad \norm{1-s}<1.\] Thus \(s\) is invertible. Form the row and column \[U=(u_1e^{1/2},\ldots,u_ne^{1/2}),\qquad V=(e^{1/2}v_1,\ldots,e^{1/2}v_n)^{\mathsf T}.\] Since \(s=UV\), the element \(y=V^*V=\sum_i v_i^*ev_i\) satisfies \[\norm{s^{-1}}^{-2}1\le s^*s =V^*U^*UV\le\norm{U}^2 y.\] In particular \(y\) is invertible. Setting \(x_i=v_i y^{-1/2}\) gives the exact identity \[ \sum_{i=1}^n x_i^* e x_i=1. \tag{55}\]

Define \[c=\sum_{i=1}^n e^{1/2}x_i b x_i^*e^{1/2}\in\operatorname{her}(e)_+.\] Additivity and traciality, followed by Equation (55), give, for every \(\tau\in T\), \[ \tau(c)=\sum_i\tau(bx_i^*ex_i)=\tau(b), \qquad d_\tau(c)\le d_\tau(e)<\delta. \tag{56}\] The norm of \(c\) is finite but may be large. Choose an integer \(L\ge\max\{1,\norm{c}\}\) and put \[v=\bigoplus_{j=1}^{L}(c-j)_+\in M_L(A)_+.\] For \(0\le s\le\norm{c}\), let \(N_L(s)=\sum_{j=1}^L\mathbf 1_{(j,\infty)}(s)\). The elementary counting identity gives \[ 0\le s-N_L(s)\le\mathbf 1_{(0,\infty)}(s). \tag{57}\] For nonintegral positive \(s\) the count is \(\lfloor s\rfloor\); at a positive integer \(s\) it is \(s-1\), so the difference is exactly one. At \(s=0\) the difference is zero. This also checks the upper endpoint when \(\norm{c}=L\).

Apply bounded Borel functional calculus and the normal extension of \(\tau\) to Equation (57). The unnormalized matrix trace adds the support ranks of the direct summands, and hence \[d_\tau(v)=\sum_{j=1}^L\tau\bigl(\mathbf 1_{(j,\infty)}(c)\bigr), \qquad 0\le\tau(c)-d_\tau(v)\le d_\tau(c)<\delta.\] Together with Equation (56), this is Equation (54). ◻

Proof. If \(b=0\), take \(p_0=0\). Otherwise choose \(v\in M_L(A)_+\) as in Proposition 24. Faithfulness of each trace and the support formula give \[0<d_\tau(b),\qquad d_\tau(v)\le\tau(b)\le\norm{b}\,d_\tau(b)<d_\tau(b).\] Strict comparison, applied in \(A\otimes\mathcal K\), gives \(v\precsim b\).

There is a single \(t>0\) such that \[ d_\tau((v-3t)_+)>\tau(b)-2\delta \qquad(\tau\in T). \tag{59}\] Here is the uniformity argument. For fixed \(t>0\), the rank on the left is a lower semicontinuous function of \(\tau\): it is the supremum over \(m\in\mathbb N\) of the continuous functions \[\tau\longmapsto(\tau\otimes\operatorname{Tr}) \bigl(\min\{1,m(v-3t)_+\}\bigr).\] Moreover, these cut ranks increase to \(d_\tau(v)\) as \(t\) decreases to zero. Thus the open sets \[U_t=\{\tau\in T: d_\tau((v-3t)_+)>\tau(b)-2\delta\}\] increase as \(t\) decreases and cover \(T\), by Equation (54). Compactness of \(T\) supplies a finite subcover. The smallest of its cut parameters gives Equation (59).

We next obtain a comparison witness of the required matrix shape. Regard \(b\) as supported in the first matrix corner of \(A\otimes\mathcal K\), and \(v\) as supported in its first \(L\) corners. Let \(P_1\) and \(P_L\) denote the corresponding corner projections. From \(v\precsim b\), choose \(X\in A\otimes\mathcal K\) with \(\norm{X^*bX-v}<t\). The row \(x=P_1XP_L\in M_{1,L}(A)\) satisfies \[w=x^*bx=P_LX^*bXP_L,\qquad \norm{w-v}<t.\] Set \(\alpha=\norm{w-v}\) and \(\gamma=t-\alpha>0\). In \(M_L(A)^{**}\) put \(E=\mathbf 1_{(3t,\infty)}(v)\) and \(F=\mathbf 1_{(t,\infty)}(w)\). The order inequalities \((w-2t)_+\ge w-2t1\) and \(w\ge v-\alpha1\) give \[E(w-2t)_+E \ge E(v-(2t+\alpha)1)E\ge\gamma E.\] Similarly, \[FvF\ge F(w-\alpha1)F\ge\gamma F.\] For a positive element \(a\) in a von Neumann algebra and a projection \(Q\) with \(QaQ\ge\gamma Q\), \(\gamma>0\), the polar decomposition of \(a^{1/2}Q\) has initial projection \(Q\) and final projection at most \(\operatorname{supp}(a)\). Indeed its absolute square is \(QaQ\), which is invertible in the \(Q\)-corner, while its range is contained in the support of \(a\). Applying this observation to the preceding two inequalities and then applying each normal trace gives \[ d_\tau((v-3t)_+)\le d_\tau((w-2t)_+), \qquad d_\tau((w-t)_+)\le d_\tau(v). \tag{60}\] The positive constant \(\gamma\) is independent of \(\tau\).

Set \(a=b^{1/2}x\in M_{1,L}(A)\) and \(w'=aa^*=b^{1/2}xx^*b^{1/2}\in\operatorname{her}(b)_+\). In the bidual of the linking matrix algebra, the polar decomposition \(a=u|a|\) identifies the positive spectral projections of \(a^*a=w\) and \(aa^*=w'\): \[\mathbf 1_{(s,\infty)}(w') =u\mathbf 1_{(s,\infty)}(w)u^* \qquad(s>0).\] Consequently \(w\) and \(w'\) have the same cut ranks, with the unnormalized matrix trace on the \(w\) corner and \(\tau\) on the \(w'\) corner.

Choose continuous \(g:[0,\infty)\to[0,1]\) which is zero on \([0,t]\) and one on \([2t,\infty)\), and put \(p_0=g(w')\). This is a positive contraction in \(\operatorname{her}(b)\) since \(g(0)=0\). Scalar functional calculus gives \[\mathbf 1_{(2t,\infty)}\le g\le\mathbf 1_{(t,\infty)}, \qquad 0\le g-g^2\le \mathbf 1_{(t,\infty)}-\mathbf 1_{(2t,\infty)}.\] Taking traces and using the equality of cut ranks yields \[\begin{split} d_\tau((w-2t)_+)&\le\tau(p_0)\le d_\tau((w-t)_+),\\ \tau(p_0-p_0^2)&\le d_\tau((w-t)_+)-d_\tau((w-2t)_+). \end{split}\] Equations (54), (59) and (60) now imply Equation (58). All ranks here are finite, so the displayed differences are well defined. The argument also allows \(v=0\); its rank approximation then gives \(\tau(b)<\delta\) at every trace, and \(x=0\) is a permissible witness. ◻

Proof. Choose positive contractive lifts \((h_\ell)_{\ell\in\mathbb N}\) of \(h\). For \(b_\ell=(h_\ell-\eta)_+\) we have \(\norm{b_\ell}\le1-\eta<1\). Apply Lemma 25 with \(\delta=1/\ell\) to obtain positive contractions \(p_\ell\in\operatorname{her}(b_\ell)\) satisfying \[\sup_{\tau\in T}|\tau(p_\ell)-\tau(b_\ell)|\le2/\ell, \qquad \sup_{\tau\in T}\tau(p_\ell-p_\ell^2)\le2/\ell.\] Let \(p\) be the class of \((p_\ell)\) in \(D\). Since \(p_\ell\) is a positive contraction, \[0\le(p_\ell-p_\ell^2)^2\le p_\ell-p_\ell^2.\] Its projection defect therefore has uniform tracial \(2\)-norm tending to zero, so \(p=p^*=p^2\) in \(D\). For an arbitrary limit trace \(\tau=\lim_\omega\tau_\ell\) on represented sequences, the first estimate gives \[\tau(p)=\lim_{\ell\to\omega}\tau_\ell(b_\ell) =\tau((h-\eta)_+).\]

For the support assertion, functional calculus in \(A^{**}\) gives \[\operatorname{supp}(b_\ell)=\mathbf 1_{(\eta,1]}(h_\ell), \qquad k_\eta(h_\ell)\operatorname{supp}(b_\ell)=\operatorname{supp}(b_\ell).\] Every element of \(\operatorname{her}(b_\ell)\) is supported on both sides by this projection. Hence \(k_\eta(h_\ell)p_\ell=p_\ell\) exactly at every coordinate. Passing to the quotient and taking adjoints proves Equation (62).

Finally represent \(D^{**}\) faithfully on a Hilbert space. The identity \((1-k_\eta(h))p=0\) says that the range of \(p\) lies in the kernel of \(1-k_\eta(h)\). Its kernel projection is \[\mathbf 1_{\{1\}}(k_\eta(h)) =\mathbf 1_{[\eta,1]}(h).\] Thus \(p\) has the asserted order bound in \(D^{**}\). This argument uses an exact identity in \(D\) and does not require any normal trace on \(D^{**}\) to be faithful. ◻

Proof. Apply Proposition 26 to \(h=q\) and \(\eta=1/2\). It gives a projection \(p\le q\) with \(\tau(p)=\tau(q)/2\) for all \(\tau\in\Lambda\). Then \(r=2p-q\) belongs to \(qDq\), is selfadjoint, and satisfies \(r^2=4p-4p+q=q\) and \(\tau(r)=0\). This includes \(q=0\). ◻

Proof. It suffices to treat \(0<\varepsilon,\delta<1\). We first consider a single selfadjoint contraction \(x\). Choose continuous functions \(f_1,\ldots,f_m:[-1,1]\to[0,1]\) and \(\lambda_1,\ldots,\lambda_m\in[-1,1]\) such that \[\sum_{j=1}^m f_j=1, \qquad \operatorname{supp}(f_j)\subseteq \{t\in[-1,1]:|t-\lambda_j|\le\varepsilon\}.\] For example, piecewise linear hat functions on a sufficiently fine finite subdivision give such a partition of unity.

We will repeatedly apply these functions inside the current residual corner and remove a projection supported where the corresponding function is nonzero. At any trace where a run starts with residual trace greater than \(\delta\), a full pass through all the functions will remove a fixed positive trace mass. To prove this uniformly, we first estimate how the traces of the functional-calculus values change when the corner shrinks.

If \(v\in D\) is a projection, write \(f^v(vxv)\) for functional calculus in the corner \(vDv\), whose unit is \(v\); the value in the zero corner is defined to be zero. In particular, for a polynomial \(P(t)=\sum_{k=0}^d a_kt^k\), \[ P^v(vxv)=a_0v+\sum_{k=1}^d a_k(vxv)^k. \tag{64}\]

We give an explicit uniform continuity estimate for this changing corner functional calculus. Suppose \(v\le u\) are projections and fix \(\tau\in\Lambda\). Put \(t=\tau(u-v)\). Since \(u-v\) is a projection, \[\norm{u-v}_{2,\tau}=\sqrt t, \qquad \norm{uxu-vxv}_{2,\tau}\le2\sqrt t.\] The second inequality follows by writing the difference as \((u-v)xu+vx(u-v)\) and using the tracial \(2\)-norm multiplication inequalities. The usual telescoping identity, with all powers of \(uxu\) and \(vxv\) contractive, gives \[\norm{(uxu)^k-(vxv)^k}_{2,\tau}\le 2k\sqrt t \qquad(k\ge1).\] Consequently, by \(|\tau(a)|\le\norm{a}_{2,\tau}\) and Equation (64), \[\left|\tau\bigl(P^u(uxu)\bigr) -\tau\bigl(P^v(vxv)\bigr)\right| \le C_P\sqrt t, \qquad C_P=|a_0|+2\sum_{k=1}^d k|a_k|.\] The term \(|a_0|\) records the change of corner units. If \(f\in C([-1,1])\) and \(\norm{f-P}_\infty\le a\), functional calculus in each corner and \(\tau(u),\tau(v)\le1\) yield \[ \left|\tau\bigl(f^u(uxu)\bigr) -\tau\bigl(f^v(vxv)\bigr)\right| \le 2a+C_P\sqrt{\tau(u-v)}. \tag{65}\] This estimate does not depend on \(x\), the two projections, or the chosen trace.

For each \(j\) choose a polynomial \(P_j\) with \(\norm{f_j-P_j}_\infty\le\delta/(16m)\), and set \[C=\max\{1,C_{P_1},\ldots,C_{P_m}\}, \qquad \alpha=\frac{\delta}{4m}, \qquad \kappa=\min\left\{\frac{\delta}{4m}, \left(\frac{\delta}{8mC}\right)^2\right\}>0.\] Thus whenever \(v\le u\) and \(\tau(u-v)<\kappa\), Equation  (65) gives, for every \(j\), \[ \left|\tau\bigl(f_j^u(uxu)\bigr) -\tau\bigl(f_j^v(vxv)\bigr)\right| \le\frac{\delta}{4m}. \tag{66}\]

Starting with an arbitrary residual projection \(u\), perform one run through the functions as follows. Put \(v_0=u\) and, successively, let \[y_j=f_j^{v_{j-1}}(v_{j-1}xv_{j-1})\in v_{j-1}Dv_{j-1} \qquad(1\le j\le m).\] Apply Proposition 26 to \(y_j\) with cutoff \(\alpha\) to obtain a projection \(a_j\in D\) satisfying \[a_j\le\mathbf 1_{[\alpha,1]}(y_j) \le\operatorname{supp}(y_j)\le v_{j-1}, \qquad \tau(a_j)=\tau((y_j-\alpha)_+) \ge\tau(y_j)-\alpha \quad(\tau\in\Lambda),\] and put \(v_j=v_{j-1}-a_j\). The trace estimate uses the stronger scalar bound \(0\le y_j-(y_j-\alpha)_+\le\alpha v_{j-1}\). The projections removed during the run are mutually orthogonal. In the bidual corner, \(\operatorname{supp}(y_j)\) lies under the spectral projection of \(v_{j-1}xv_{j-1}\) for \([\lambda_j-\varepsilon,\lambda_j+\varepsilon]\cap[-1,1]\). Compressing the corresponding spectral order bounds by \(a_j\) gives \[-\varepsilon a_j\le a_jxa_j-\lambda_ja_j \le\varepsilon a_j, \qquad \norm{a_jxa_j-\lambda_ja_j}\le\varepsilon.\] These are norm estimates in \(D\) and do not require \(a_j\) to commute with \(x\).

For every run and every \(\tau\in\Lambda\) we claim that \[ \tau(u)>\delta\quad\Longrightarrow\quad \tau(u-v_m)\ge\kappa. \tag{67}\] Indeed, suppose \(\tau(u)>\delta\) but \(\tau(u-v_m)<\kappa\). The partition identity in the initial corner implies \[\sum_{j=1}^m\tau\bigl(f_j^u(uxu)\bigr)=\tau(u)>\delta,\] so one index \(j\) has \(\tau(f_j^u(uxu))>\delta/m\). Because \(v_{j-1}\le u\) and \(\tau(u-v_{j-1})<\kappa\), Equation (66) applies at that step. Therefore \[\tau(a_j)\ge\tau(y_j)-\alpha >\frac{\delta}{m}-\frac{\delta}{4m}-\frac{\delta}{4m} =\frac{\delta}{2m}>\kappa,\] contradicting \(a_j\le u-v_m\). The index used to prove progress may depend on the trace, but every run processes all \(m\) indices and the constant \(\kappa\) is fixed beforehand.

Now start at \(q\) and perform \(K=\lfloor1/\kappa\rfloor+1\) runs, each on the preceding residual projection. Collect all removed projections as \(p_1,\ldots,p_L\), discarding zero projections if desired, and let \(w=q-\sum_s p_s\) be the final residual. If some \(\tau\in\Lambda\) satisfied \(\tau(w)>\delta\), the residual at the start of every run would have trace greater than \(\delta\) at this same \(\tau\). Equation (67) would imply \[\tau(q-w)\ge K\kappa>1,\] which is impossible. Thus \(\tau(w)\le\delta\) simultaneously for all \(\tau\in\Lambda\). This is a fixed finite construction; it uses no compactness of \(\Lambda\). Each removed projection retains its scalar compression estimate, proving the one-coefficient case.

For the finite family, begin with the single block \(q\). At stage \(j=1,\ldots,n\), suppose there are \(L_{j-1}\) current blocks. Apply the one-coefficient case for \(x_j\) inside each of them, with compression tolerance \(\varepsilon\) and trace-loss tolerance \[\frac{\delta}{n\max\{1,L_{j-1}\}}.\] Replace the old blocks by their new subprojections. The total trace lost at this stage is at most \(\delta/n\), at every trace. All prior compression bounds survive: if \(p\le p_0\) and \(\norm{p_0x_kp_0-\mu p_0}\le\varepsilon\), then \[px_kp-\mu p=p(p_0x_kp_0-\mu p_0)p, \qquad \norm{px_kp-\mu p}\le\varepsilon.\] There are finitely many blocks at every stage. Summing the trace losses over the \(n\) stages proves Equation (63). For an empty coefficient family, simply take the block \(q\). ◻

Proof. The assertion is immediate if \(q=0\) or \(H\) is empty. Otherwise put \(M=\max\{1,\max_{b\in H}\norm{b}\}\) and choose \(0<\theta\le\min\{1/4,\zeta/(8M)\}\). Apply Lemma 28, with both tolerances \(\theta\), to the real and imaginary parts of every \(b/M\), \(b\in H\). It gives orthogonal projections \(p_s\le q\), a residual \(w=q-\sum_s p_s\), and complex scalars \(\nu_{s,b}\) such that \[\norm{p_sbp_s-\nu_{s,b}p_s}\le2M\theta, \qquad \tau(w)\le\theta\quad(\tau\in\Lambda).\] By Corollary 27, choose \(r_s=r_s^*\in p_sDp_s\) with \(r_s^2=p_s\) and \(\tau(r_s)=0\) for every \(\tau\in\Lambda\). Set \[r=w+\sum_s r_s.\] Orthogonality of the supporting projections gives \(r=r^*\in qDq\) and \(r^2=w+\sum_s p_s=q\). For a trace \(\tau\), its restriction to \(p_sDp_s\) has norm \(\tau(p_s)\). Traciality therefore gives \[\begin{split} |\tau(br_s)| &=\left|\tau\bigl((p_sbp_s-\nu_{s,b}p_s)r_s\bigr)\right| \le2M\theta\,\tau(p_s),\\ |\tau(bw)|&=|\tau(wbw)|\le M\tau(w). \end{split}\] Summing these estimates, using \(\sum_s\tau(p_s)\le1\), yields \[|\tau(br)|\le2M\theta\sum_s\tau(p_s)+M\theta \le3M\theta\le\frac{3\zeta}{8}<\zeta.\] The error is weighted by the traces of the blocks, so their number does not enter the bound. ◻

Proof. Adjoin \(1_D\) to \(E\). This preserves finite dimensionality: if \(e\) is the unit of \(E\), it adds only the scalar summand \(\mathbb C(1_D-e)\) when \(e\ne1_D\) (and if \(E=0\), take \(e=0\)). It therefore suffices to work with the resulting unital finite-dimensional algebra. Choose matrix units \[\{e^{(\ell)}_{ij}:1\le i,j\le n_\ell\},\qquad 1\le\ell\le a, \qquad \sum_{\ell=1}^a\sum_{i=1}^{n_\ell}e^{(\ell)}_{ii}=1_D.\]

We first impose finitely many coefficient tests. Fix a finite \(H\subset S\) and \(\xi>0\), and write \(N_0=\sum_\ell n_\ell\). For each block apply Lemma 29 in the corner \(e^{(\ell)}_{11}De^{(\ell)}_{11}\) to the finite family \[H_\ell= \{e^{(\ell)}_{1i}s e^{(\ell)}_{i1}: s\in H,\ 1\le i\le n_\ell\},\] with tolerance \(\xi/N_0\). Denote the resulting corner symmetry by \(u_\ell\), and put \[R=\sum_{\ell=1}^a\sum_{i=1}^{n_\ell} e^{(\ell)}_{i1}u_\ell e^{(\ell)}_{1i}.\] Each summand is selfadjoint, and the matrix-unit relations give \[R^2=\sum_{\ell,i}e^{(\ell)}_{ii}=1_D, \qquad Re^{(\ell)}_{jk} =e^{(\ell)}_{j1}u_\ell e^{(\ell)}_{1k} =e^{(\ell)}_{jk}R.\] Terms from distinct matrix blocks have orthogonal supports. Thus \(R\) is a selfadjoint unitary commuting with \(E\). For every \(s\in H\) and \(\tau\in\Lambda\), cyclicity gives \[\tau(sR)=\sum_{\ell=1}^a\sum_{i=1}^{n_\ell} \tau\bigl(e^{(\ell)}_{1i}s e^{(\ell)}_{i1}u_\ell\bigr), \qquad |\tau(sR)|<N_0\frac{\xi}{N_0}=\xi.\] The complementary scalar block in the nonunital case is included in these formulas, so the square of \(R\) is the full unit of \(D\).

Choose a norm-dense sequence \((s_k)_{k\ge1}\) in the unit ball of \(S\). Apply Lemma 21 to one variable in the selfadjoint unit ball of \(D\), with the tests \[\norm{r^2-1_D}_{2,\Lambda}=0, \qquad \norm{[r,e^{(\ell)}_{ij}]}_{2,\Lambda}=0, \qquad \sup_{\tau\in\Lambda}|\tau(s_kr)|=0\quad(k\ge1).\] Every finite list is satisfiable to arbitrary precision by the unitaries \(R\) just constructed; the square and commutation relations already hold exactly for those trial solutions. The diagonal lemma provides \(r=r^*\in D\) satisfying every test. Since the uniform tracial \(2\)-norm detects zero in \(D\), we have \(r^2=1_D\) and \([r,e^{(\ell)}_{ij}]=0\) in \(D\). Thus \(r\) is a selfadjoint unitary in \(E'\). Finally, \[|\tau((s-s_k)r)|\le\norm{s-s_k}\norm{r}\] extends the trace identities from the dense sequence to the unit ball of \(S\), and scaling gives Equation (69) for all \(s\in S\). ◻

Proof. Choose an increasing sequential contractive approximate unit \((u_n)\). The continuous functions \(\sigma\mapsto\sigma(u_n)\) increase to \(1\) on compact \(T\), so Dini’s theorem gives uniform convergence. In a tracial von Neumann representation, for any bounded sequence \((a_n)\), \[\|(1-u_n)a_n\|_{2,\sigma}^2 \le\|a_n\|^2\sigma((1-u_n)^2) \le\|a_n\|^2(1-\sigma(u_n)).\] The same estimate holds on the right. Thus \((u_n)\) represents \(1_D\). Its value under every limit trace is \(1\). Uniform convergence on the constant approximate units also shows that the restriction of each limit trace to \(A\) has norm one. The norm identity follows by choosing a trace at each coordinate within an arbitrarily small error of the supremum. Its last assertion is the definition of the quotient. ◻

Proof. First choose a nonzero \(b\in A_+\) with \(d_\sigma(b)<\xi\) for all \(\sigma\in T\). An extreme point of the compact convex set \(T\) has a factorial tracial GNS representation. This factor cannot be finite dimensional: the representation is faithful, and would make \(A\) finite dimensional. It is therefore a \(\mathrm{II}_1\) factor and contains unital matrices of every size. By (95) and order-zero lifting, there is a nonzero order-zero map \(M_l\to A\) for every \(l\). The support projections of its diagonal images are orthogonal and equivalent in every tracial representation. Their sum has trace at most one, so each diagonal has dimension at most \(1/l\). Taking \(l>1/\xi\) gives \(b\).

Simplicity now gives \(c\in\operatorname{her}(b)_+\) satisfying \(|\sigma(c)-\sigma(g)|<\xi\) on \(T\), without a norm bound on \(c\). Here are details of this trace transfer. Approximate \(g^{1/2}\) in the ideal generated by \(b\) by \(UD_bV\), with \(U\) a row, \(V\) a column, and \(D_b=\operatorname{diag}(b,\ldots,b)\), so that \((UD_bV)^*(UD_bV)\) is within \(\xi\) of \(g\). Set \[C=(D_bU^*UD_b)^{1/2}\in M_N(\operatorname{her}(b)),\qquad v=CV.\] Each entry \(v_j\) belongs to \(\overline{bA}\). Thus \(c=\sum_jv_jv_j^*\in\operatorname{her}(b)_+\), and traciality gives \(\sigma(c)=\sigma(v^*v)\), proving the assertion.

For an integer \(N\ge\|c\|\), put \[f_j(s)=\min(1,\max(0,s-j+1)),\qquad k=\bigoplus_{j=1}^N f_j(c).\] On the spectrum of \(c\), the functions sum to \(s\), while \[0\le\sum_{j=1}^N1_{\{f_j(s)>0\}}-s\le1_{\{s>0\}}.\] Hence \(\sigma(k)=\sigma(c)\) and \(d_\sigma(k)-\sigma(k)\le d_\sigma(c)\le d_\sigma(b)<\xi\). ◻

Proof. We first compare finite-matrix approximations to the desired trace with a spectral cut of the target. A small auxiliary matrix corner supplies the strict rank inequality needed for comparison. We then discard that corner and obtain elements of \(A\) whose trace and projection defects vanish along \(\omega\).

Take positive contraction lifts \(h_n,z_n,a_n\). Set \[\alpha_n=\sup_{\sigma\in T}(\sigma(a_n)-\sigma(z_n))_+, \qquad \beta_n=\|(1-h_n)z_n\|_{2,T}.\] Both tend to zero along \(\omega\). Choose \(0<\delta_n<1\) tending to zero slowly enough that \(\beta_n/\delta_n\to_\omega0\), and put \[\gamma_n=\alpha_n+\beta_n/\delta_n+\delta_n, \qquad H_n=(h_n-1+\delta_n)_+.\] In a tracial representation, the support \(Q\) of \(H_n\) satisfies \(\delta_n(1-Q)\le1-h_n\). Since \(0\le z_n\le1\), tracial Cauchy–Schwarz gives \[\sigma(z_n)\le\sigma(Q)+\sigma((1-Q)z_n) \le d_\sigma(H_n)+\beta_n/\delta_n.\] Consequently \(\sigma(a_n)\le d_\sigma(H_n)+\gamma_n\).

Apply Lemma 41 with \(g=a_n\) and \(\xi=\gamma_n\) to obtain \(k_n\). Also choose a positive contraction \(u\in A\) with \(\inf_T\sigma(u)>7/8\), and apply that lemma to \(8\gamma_nu\) with the same tolerance. Its output \(c_n\) satisfies \[5\gamma_n\le d_\sigma(c_n)\le10\gamma_n.\] Thus, in disjoint finite matrix blocks, \[ d_\sigma(k_n)\le d_\sigma(H_n)+3\gamma_n <d_\sigma(H_n\oplus c_n)\qquad(\sigma\in T). \tag{97}\] The target is nonzero, and all these ranks are finite at each coordinate. Every densely finite trace is a multiple of a state in \(T\); hence the comparison hypothesis gives \(k_n\precsim H_n\oplus c_n\).

We use the cutdown in a form that also controls its trace. Suppress \(n\), and write \(b=H_n\oplus c_n\). Choose \(x\) with \(\|x^*bx-k\|<1/4\) and set \(d=(k-1/2)_+^{1/2}\), \(y=b^{1/2}xd\). Then \(|y|^2\ge d^2/4\). For the polar decomposition \(y=v|y|\) in the bidual, the support of \(d\) lies under \(v^*v\), and \[\|(|y|+\epsilon)^{-1}d\|\le2, \qquad \|vd-y(|y|+\epsilon)^{-1}d\|\le2\epsilon.\] The first inequality follows by conjugating \(d^2\le4|y|^2\) with \((|y|+\epsilon)^{-1}\) and taking norms. The second proves that \(vd\) belongs to the closed right ideal supported by \(b\). Therefore \[w=\sqrt2vd,\qquad w^*w=2(k-1/2)_+, \qquad r=ww^*\in\operatorname{her}(b),\qquad 0\le r\le1.\] For \(f(s)=2(s-1/2)_+\) on \([0,1]\), \[0\le s-f(s)\le1_{\{s>0\}}-s, \qquad 0\le f(s)-f(s)^2\le1_{\{s>0\}}-f(s).\] Traciality transfers these estimates from \(f(k_n)=w^*w\) to \(r_n=ww^*\). By (96), both \(|\sigma(r_n)-\sigma(a_n)|\) and \(\sigma(r_n-r_n^2)\) tend uniformly to zero along \(\omega\).

Compress \(r_n\) to its \(H_n\) matrix slot, and identify the resulting positive contraction \(p_n\) with an element of \(A\). The complementary corner is supported by \(c_n\), so this removes at most \(10\gamma_n\) of trace. If \(P\) is the first-slot projection, the defect identity is \[Pr_nP-(Pr_nP)^2 =P(r_n-r_n^2)P+Pr_n(1-P)r_nP.\] The last term has trace at most \(d_\sigma(c_n)\), since \(r_n^2\le r_n\). In particular, both \(|\sigma(p_n)-\sigma(a_n)|\) and \(\sigma(p_n-p_n^2)\) are at most \(12\gamma_n\), uniformly on \(T\). Moreover \(p_n\in\operatorname{her}(H_n)\), so \(\|(1-h_n)p_n\|\le\delta_n\). The class \(p=[(p_n)]\) is a projection: its positive contraction defects have vanishing trace, hence vanishing uniform \(2\)-norm. It has all the required properties. ◻

Proof. If \(q=0\), take \(r=0\); if \(Y=0\), apply Proposition 42 to \((h,z,a)=(q,q,tq)\). Scaling \(Y\) and the tolerance reduces to \(0\le Y\le q\). All functions below are evaluated at \(Y\) in the unital corner \(qDq\).

Choose a large integer \(m\). Let \(F_0=0\), \(F_m=1\); for \(1\le j<m\) let \(F_j\) be \(1\) up to \((j-1/2)/m\), decrease linearly to \(0\) at \(j/m\), and vanish thereafter. Let \(G_j\) be \(1\) up to \(j/m\), decrease to \(0\) by \((j+1/3)/m\), and vanish thereafter for \(j<m\); put \(G_m=1\). These functions satisfy \[G_jF_j=F_j,\qquad F_{j+1}G_j=G_j.\] Construct orthogonal projections \(r_j\) recursively with \[G_jr_j=r_j,\qquad \tau(r_j)=t\tau(F_j-F_{j-1}).\] Indeed, if \(R=\sum_{i<j}r_i\), then \(F_jR=G_jR=R\) and \(\tau(R)=t\tau(F_{j-1})\). Apply Proposition 42 to \[h=G_j-R,\qquad z=F_j-R,\qquad a=t(F_j-F_{j-1}).\] They are positive contractions, \(hz=z\), and \(\tau(z)-\tau(a)=(1-t)\tau(F_j)\ge0\). The selected projection is orthogonal to \(R\), lies under \(q\), and is fixed by \(G_j\).

Now \(r=\sum_jr_j\) has trace \(t\tau(q)\). The spectral support conditions give \[r_jYr_j\le\frac{j+1/3}{m}r_j,\] whereas \(F_j-F_{j-1}\) is supported at values at least \((j-3/2)/m\). The difference of these thresholds is \(11/(6m)<2/m\). Summing therefore gives \[\tau(rY)\le t\sum_j\frac{j+1/3}{m}\tau(F_j-F_{j-1}) \le t\bigl(\tau(Y)+(2/m)\tau(q)\bigr).\] Take \(2/m<\varepsilon\). ◻

Proof. For a matrix summand of size \(l\) with matrix units \(e_{ij}\), apply Lemma 43 in \(e_{11}De_{11}\) to the observable \(l^{-1}\sum_j e_{1j}Ye_{j1}\). Repeat the resulting projection along the diagonal, using the matrix units. The repeated projection commutes with the summand and splits its traces proportionally. Its cost is \(l\) times the selected corner cost. Adding over the orthogonal summands gives (99), since their units sum to \(1_D\). ◻

Proof. Choose a dense sequence of self-adjoint contractions in \(E\) and represent each by self-adjoint contractions in \(A\). At coordinate \(n\), apply Proposition 45 to the first \(n\) entries of these representatives, with \(\rho=1/n\). The resulting sequence gives a positive contraction in \(D\) whose commutators with the dense sequence have zero uniform \(2\)-norm. They are therefore zero in \(D\). Passing to the limit traces proves the inequalities on the dense sequence, and norm continuity extends them to the entire self-adjoint unit ball. ◻

Proof. For an integer \(m\ge1\), put \(\lambda=1-c^{1/m}\). Successively choose \(e_1,\ldots,e_m\) from Lemma 48, enlarging \(E\) to contain the previous choices at each step. They commute, and \[P_j=\prod_{i=1}^j(1-e_i),\qquad P_0=1,\] are positive contractions commuting with \(E\). For a self-adjoint contraction \(x\in E\), the element \(P_{j-1}x\) is a permitted test at step \(j\). The resulting recurrence gives \[|\tau(P_mx)-c\tau(x)|\le m\lambda^{3/2}.\] For commuting positive contractions \(a,b\), \[ab-(ab)^2=b(a-a^2)+a^2(b-b^2)\le(a-a^2)+(b-b^2).\] Thus \(\tau(P_m-P_m^2)\le m\lambda^{3/2}\) as well. Since \(\lambda\le(-\log c)/m\), both errors tend to zero.

We spell out the diagonal passage because \(E\) need not consist of constant sequences. Fix representatives for a dense sequence of self-adjoint contractions of \(E\). For each \(k\), choose \(m\) large enough and a positive contraction lift \(P_n^{(k)}\) so that on an \(\omega\)-large set \(G_k\), the first \(k\) commutators and trace moments, and the uniform \(2\)-norm projection defect, are all within \(1/k\) of their targets. Such a set exists by the uniform limit-trace estimates and exact commutation in \(D\); use \(\tau((P_m-P_m^2)^2)\le\tau(P_m-P_m^2)\) for the defect. Put \[F_k=\{n:n\ge k\}\cap G_1\cap\cdots\cap G_k.\] These sets decrease, belong to \(\omega\), and have empty intersection. At coordinate \(n\in F_1\), use \(P_n^{(k(n))}\) for the largest \(k(n)\) with \(n\in F_{k(n)}\); use zero off \(F_1\). Then \(k(n)\to_\omega\infty\). The resulting class is a projection commuting with \(E\) and satisfying (107) on the dense sequence. Continuity and complex linearity complete the proof. ◻

Proof. Start with the separable unital algebra \(C^*(A,1_D)\). At step \(j=1,\ldots,n-1\), adjoin the previously selected projections, apply Proposition 49 with \(c=1/(n-j+1)\), and multiply its projection by the remaining complement \(q\). The factors commute, so this gives a projection \(f_j\le q\). Apply (107) to \(qa\) in the enlarged algebra. Inductively \(\tau(qa)=(n-j+1)\tau(a)/n\), so \(\tau(f_ja)=\tau(a)/n\). Take the final complement for \(f_n\). These are exactly the projection identities in the definition of uniform property \(\Gamma\) (Castillejos et al. 2022, Definition 2.1). ◻

Proof. The approximate-unit sequence defines \(u\in A_\omega\cap A'\) with \(ua=a=au\) for constant \(a\in A\). For \(y\in A_\omega\cap A'\), \((uy-y)a=(ua-a)y=0\), and likewise on the left. Thus \(u\) is the unit modulo the annihilator.

If \(y\) annihilates \(A\), choose a bounded representative \((y_n)\) of norm at most \(M\). For each fixed \(k\), \(\|y_nu_k\|\to_\omega0\), while \[\|y_n(1-u_k)\|_{2,T}^2 \le M^2\sup_{\sigma\in T}(1-\sigma(u_k))\longrightarrow0 \quad(k\to\infty).\] Taking these limits in order proves \(y\in J\).

For surjectivity, let \(v\in A_\omega\) lift a member of \(D\cap A'\). Its commutators with \(A\) lie in \(J\). A quasicentral approximate unit for this ideal gives positive contractions \(d^{(k)}\in J\) that commute with the first \(k\) elements of a dense sequence in \(A\) to within \(1/k\), and satisfy \(d^{(k)}[v,a]=[v,a]\) to that accuracy on those elements. Choose positive contraction representatives. On an \(\omega\)-large set their uniform \(2\)-norms are below \(1/k\), and their coordinate norm errors are below \(2/k\). The decreasing-set selection used in Proposition 49 now gives \(d\in J\cap A'\) satisfying \(d[v,a]=[v,a]\) for every \(a\in A\). Hence \(v-dv\) is a central lift of the prescribed element. This is also the central-surjectivity statement of (Castillejos et al. 2022, Lemma 1.2).

Apply matrix order-zero lifting to the resulting quotient map and (108) to obtain \(\psi\), and compose with the quotient to \(F_\omega(A)\) to obtain \(\varphi\). The image of \(\psi(1)\) in \(D\) is \(1_D\). ◻

Proof. Choose \(\epsilon>0\) such that \(b=(a-\epsilon)_+\ne0\). Every member of \(T\) is faithful, so compactness gives \[m=\min_{\sigma\in T}\sigma(b)>0.\] Since \(\nu(a)<\infty\), (109) gives an \(\omega\)-large set on which \(\nu_n(b)\le C\) for some finite \(C\). A lower-semicontinuous trace \(\eta\) finite on a nonzero positive element of a simple algebra is necessarily densely finite. Indeed, the set \(N_\eta=\{x:\eta(x^*x)<\infty\}\) is linear by \((x+y)^*(x+y)\le2x^*x+2y^*y\). Traciality and positivity give \[\eta((xy)^*(xy))=\eta(xyy^*x^*)\le\|y\|^2\eta(x^*x), \qquad \eta((yx)^*(yx))\le\|y\|^2\eta(x^*x).\] Thus \(N_\eta\) is a nonzero two-sided ideal and is norm dense by simplicity. Approximating square roots shows that finite-trace positives are dense. Therefore, on this set, our trace hypothesis gives \(\nu_n=r_n\sigma_n\) with \(\sigma_n\in T\) and \(0\le r_n\le C/m\). The zero trace is included by \(r_n=0\), and the trace infinite on every nonzero positive element cannot occur there. Extend the scalars and states boundedly off this set.

For every bounded positive representative \(z_n\) and \(\delta>0\), \[0\le r_n\sigma_n(z_n)-r_n\sigma_n((z_n-\delta)_+) \le(C/m)\delta.\] The regularization in (109) can therefore be removed, and \[\nu=(\lim_\omega r_n)(\lim_\omega\sigma_n).\] This is a finite, possibly zero multiple of a trace through \(D\). For \(z=\psi(1)\) from Lemma 51, the image of \(z\) in \(D\) is \(1_D\). Consequently \(\nu_a(\varphi(1))=\nu(az)=\nu(a)\) for every permitted test. The formula is independent of the representative modulo the annihilator, since that ideal annihilates \(a\). ◻

Proof of Theorem 5. Theorem 50 and (108), followed by Lemmas 51 and 52, give tracially large order-zero maps \(M_n\to F_\omega(A)\) for every \(n\). This is uniform McDuffness in the nonunital sense of (Castillejos et al. 2023, Definition 2.4).

Our comparison hypothesis also makes \(\mathop{\mathrm{Cu}}(A)\) almost unperforated. Indeed, if \((k+1)[x]\le k[y]\) with \(y\ne0\), then additivity and monotonicity of dimension functions give \[d_\tau(x)\le\frac{k}{k+1}<1 \qquad\text{whenever }d_\tau(y)=1.\] The stated comparison gives \([x]\le[y]\). The case \(y=0\) is immediate. This uses precisely the specified finite-target-rank tests, including the case that this family is empty. Dimension-function monotonicity follows from cutdown under subequivalence, as in the proof of Proposition 42, followed by increasing the supports of the cutdowns.

For simple algebras, almost unperforation is the strict comparison formulation used in (Castillejos et al. 2023, sec. 2.3). Its nonunital absorption criterion (Castillejos et al. 2023, Proposition 5.4) states that a simple, separable, nuclear algebra with this comparison and nontrivial lower-semicontinuous traces is \(\mathcal Z\)-stable if it is uniformly McDuff. All these hypotheses have now been verified. Hence \(A\cong A\otimes\mathcal Z\). ◻