Tracial projection methods and uniform property Gamma

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

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 (3.1); 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=\C p_i. If 0\ne x\in p_iAp_j, then

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\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 \supp(z^*z) and final projection \supp(zz^*). The normal tracial state therefore gives

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,

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 \T(D) is a tracial state, so \T(A)\ne\varnothing. Given a positive integer n, apply Lemma 3.2 to obtain b\in A_+ with \norm b=1 and

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\T(D) to A gives

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 3.1 to choose a finite-spectrum self-adjoint a\in fDf within 1/8 of v_1^*v_1. Set

The inequality e_1v_1^*v_1e_1\ge(3/8)e_1 shows that, when e_1\ne0,

is a partial isometry with initial projection e_1 and final projection below g_1. On the remaining corner,

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)<\eps. 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,

and Y^*Y is invertible. The entries of Y(Y^*Y)^{-1/2}f form a column as in Lemma 3.4, 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

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

Since u^*x'u+u^*(h-x')u=y, Lemma 3.4 applied to the column with entries x'u,(h-x')u gives y=k+l and partial isometries \alpha,\beta with

Set \xi=w^*\alpha+l and \zeta=w+\beta. Then

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

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 (3.5). All operations take place inside D and preserve the orthogonal sum f.

Set

For a fixed \tau\in\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

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 (3.4) uniformly in \tau.□

Proof. The assertion is immediate if q=0 or \mathcal F is empty. Choose \gamma>0 with 2\gamma<\eps/2 and use Lemma 3.1 to write q=\sum_{j=1}^m q_j with

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<\eps/2. By Proposition 3.5, 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

□

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

Since yzy\ge0, this implies

If f=0, put t=0. Otherwise the element

is a partial isometry with v^*v=f; set t=vv^*\le1-z. In both cases,

The mass bounds (3.7) follow from (3.10). As f\le y,

Consequently,

which proves (3.8).

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

Trace-norm duality proves (3.9).□

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 [15, 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

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

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 (4.1). 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,

Apply this map to the first inequality in (4.3) 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

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 \normone{b-c}{\tau}.□

Proof. Put \mu=\eta^2 and choose \Phi by Lemma 4.1. In particular, the coefficient list is fixed before r. Choose 0<\sigma<1, then \beta>0, so that

This is possible by choosing \sigma with m\sigma<\mu first.

Scalar matrix approximation. Partition r by Lemma 3.1, using the finite list

Take the compression tolerance smaller than \beta/(2m). For a nonzero partition piece q, set

These are contractions in their respective matrix spaces. Apply any state of qDq entrywise to M,N_x, obtaining scalar matrices G,H_x. Positivity and contractivity of a state at matrix levels give 0\leq G\leq1 and \norm{H_x}\leq1. The compression tolerance implies

Indeed, each unweighted entry differs from its scalar state value by less than twice the compression tolerance, and an m-by-m matrix of entries of norm at most c has norm at most mc.

Construction of the blocks. Rotate the columns of V by a scalar unitary diagonalizing G. Retain the columns corresponding to eigenvalues at least \sigma, forming a row W with k columns and scalar diagonal matrix G_+. The cutoff gives the uniform lower bound needed to normalize the retained row by an inverse square root. If k=0, set b_q=0. Otherwise put

Then U^*U=1_k\otimes q, so b_q is a finite-spectrum positive contraction. Writing W_- for the row of deleted columns, we have \norm{W_-^*W_-}\leq\sigma+\beta, and therefore

The contribution is zero if no columns are deleted. Also

has norm at most \beta, supported on a projection of trace k\tau(q). Consequently, including the case k=0,

We verify explicitly that the compression estimate is independent of the corner and its matrix size. For positive invertible X,Y\geq a1, the resolvent formula for inverse square roots gives

Use Y=G_+\otimes q and a=\sigma/2. If H\otimes q denotes the corresponding retained corner of H_x\otimes q, then \norm H\leq1, and (4.10) gives \norm{W^*xW-H\otimes q}<\beta. Expanding the difference between

and using (4.13) bounds its norm by

The second expression is a scalar matrix over q. The tolerances \sigma,\beta were chosen before the partition of r, so the compression bound is uniform over its pieces.

Write b_q as the finite sum of its nonzero upper spectral projections, weighted by the lengths of their level intervals:

Each p_j is the image of a scalar coordinate projection under U. It therefore has a subrow U_j satisfying (4.5), with q_j=q. Taking matrix corners gives (4.6). Summing (4.12) over the partition gives (4.7).

Averaged commutators. For a finite-spectrum positive contraction b=\sum_s s e_s, include its kernel projection in this spectral resolution. Tracial orthogonality gives, for unitary x,

For the middle inequality, factor |s-t|=|\sqrt s-\sqrt t|(\sqrt s+\sqrt t) and apply Cauchy–Schwarz to the sum. The second squared sum is at most 2\sum_{s,t}(s+t)\normtwo{e_sxe_t}{\tau}^2=4\tau(b). The last inequality is Lemma 4.2.

For b=b_q, equations (4.1), (4.2), and (4.12) give

Thus the contribution of J_q in (4.8) is at most 2\sqrt{3(1+\mu)}\,\eta\tau(q)\leq6\eta\tau(q). Summing over q proves the assertion.□

Proof. Write T=U^*xU, P_e=1_k\otimes e, and choose T_0\in M_k(\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

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

and the fact that the trace of the corner unit is (\Tr\otimes\tau)(P_e)=k\tau(e)=\tau(y). The same estimates hold with e replaced by q.

Set

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 (5.1). For the second estimate, the exact identity

follows by expansion and traciality. Applying it to y,p and using the squared compression estimates proves (5.2); the mass-discrepancy coefficient is bounded by 2+2\beta_0\leq10. In particular, (5.3) 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

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 \normtwo{(1-t_a)xt_a}{\tau}^2 and \normtwo{t_ax(1-t_a)}{\tau}^2.□

Proof of Theorem 1.3. Fix a finite K\subseteq\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

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\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.

Preparing the averaging maps. When an averaging map acts on the remaining corner, the projections inserted earlier must already approximately commute with its coefficient contractions. Since the map is chosen independently of its input projection, we can arrange these coefficient tests in advance, working backwards through the rounds. Construct finite sets of unitaries

backward. Having chosen K_i, choose the map \Phi_i from Proposition 4.3 for K_i,\eta. If i>1, express each of its coefficient contractions v_l as a linear combination of unitaries with coefficient absolute sum at most two, and include these unitaries, together with K_i, in K_{i-1}.

The induction. We insert finitely many projections in each of N rounds. All inserted projections will be mutually orthogonal and each will carry a label + or -. Let Z_i be their cumulative sum after round i, with Z_0=0. Set

We maintain the bound

In particular, Lemma 5.2 gives the same bound for the squared commutator of any subfamily sum.

In round i, let r=1-Z_{i-1} and choose the weighted family from Proposition 4.3 for \Phi_i,r. For every \tau\in\T(D),

For i=1 the left-hand side is zero. Otherwise each unitary in the chosen decomposition of v_l lies in K_{i-1}, so the induction and Lemma 5.2 give \normtwo{[Z_{i-1},v_l]}{\tau}\leq2\sqrt{E_{i-1}}. Also Z_{i-1}v_lr=[Z_{i-1},v_l]r, proving (5.6).

One round of sampling and insertion. Within this round we suppress i from the block and signed-index notation, and process the finite signed index set a=(j,\pm) in any fixed order. Both indices for j have the target proportion

Let z_a be the occupied projection just before index a, starting with Z_{i-1}. We choose

so that, uniformly over all \tau\in\T(D),

Here j always denotes the block index of a. To make these choices precise, assign positive budgets \eps_a with \sum_a\eps_a\leq\eta^2 before processing the family. At step a, apply Corollary 3.6 with proportion c_a, tests q_j and

and tolerance \eps_a/k_j. Traciality gives

Thus both errors at this step are at most \eps_a. The test h_a, which satisfies 0\leq h_a\leq k_jq_j, is known when it is used. This application may refine q_j internally, but only returns a subprojection e_a; it does not change the fixed weighted family.

Use Lemma 3.7 to replace y_a by a projection t_a\leq1-z_a, retaining its sign, such that

Update the occupied projection to z_a+t_a. At the end of the round this is Z_i.

Mass and overlap. For the estimates in this round, fix \tau\in\T(D), and put

These are the target mass, sampled mass, and overlap with the occupied projection in this round. All subsequent bounds hold for every \tau with the same constants; we suppress \tau in these three quantities. Equations (4.1) and (4.7) give

Using (5.7), we obtain

For the overlap, first use z_a\leq Z_i, then apply the single weighted-operator estimate (4.7):

In the third line we used (5.6), 0\leq\Phi_i(r)\leq1, and \tau(Z_i-Z_{i-1})\leq Y_i, the last following from (5.9). In particular,

Commutators and linear tests. For x\in K_i, sum Lemma 5.1 over this round. Equations (5.7) and (5.10) give

Here the first term in the second estimate is at most 6\eta^2 by (4.8). Passing to the inserted projections, (5.9) implies

Together with K_i\subseteq K_{i-1}, this proves (5.5). The mass left after the round satisfies

by (5.10) and the first inequality in (5.9). All sums in this argument are controlled by weighted projection masses or energies. No bound on the raw sum \sum_jw_j, or on the number of blocks, has been used.

Completing the packing. Let P,S be the sums of all inserted projections labelled + and -, respectively. Then P+S=Z_N, and for x\in K, \tau\in\T(D),

The first follows from Lemma 5.2. For the second, iterate (5.16), use (1-\eta)^N\leq e^{-H}, and sum (5.12). For the third, the + and - targets c_a\tau(p_jx) cancel for every block in every round. Their total sampling error is at most 7N\eta^2, by (5.13). The remaining displacement error is, by Lemma 3.7 and Cauchy–Schwarz over all signed indices,

Solving (5.4) gives

Since x is a contraction, |\tau((1-Z_N)x)|\leq\tau(1-Z_N). Equations (5.18) and (5.19) therefore yield

Choose H large enough to make the first term small. With this H fixed, let N increase through integers and put \eta=H/N. Then every other term in (5.20), and \sqrt{E_N} in (5.17), tends to zero. This proves the two desired uniform estimates.□

Proof. The upper bounds follow from the definition of a limit trace. For the reverse bound in (6.1), choose \tau_j whose value on d_j^*d_j is within 1/j of the supremum. The associated limit trace gives equality. For (6.2), 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 (6.1) and the definition of I_\omega.□

Proof of Theorem 1.1. 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 3.3 supplies full projections of uniformly arbitrarily small trace over all \T(D). Thus Theorem 1.3 applies.

Choose increasing finite subsets F_n\subseteq\U(A) whose union is norm dense in \U(A), including 1 in each F_n. Theorem 1.3 gives projections P_n\in D with

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,

For each fixed n, Lemma 6.1, applied to the commutators and the linear errors in (6.3), gives an \omega-large set of coordinates on which

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

by uniform 2-Cauchyness. Given x\in B, choose a\in A with \normtwo{x-a}{\T(A)}<\delta. For every \lambda\in\Lambda_\omega, multiplication by the contraction p and tracial Cauchy–Schwarz give

Let \delta tend to zero. The first assertion, together with Lemma 6.1, 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 1.2. Put X_B=\{\bar\tau:\tau\in T(A)\}. By [2, 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 1.1 and [2, Proposition 5.20] give property \Gamma for this pair, and hence complemented partitions of unity by [2, Theorem 1.4]. Evington’s trace theorem [11, 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 [2, 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. For every \tau\in T, the map [a]\mapsto d_\tau(a) is a functional on \Cu(A), by [10, Proposition 4.2], applied to the lower semicontinuous trace \tau\otimes\Tr. Only this direction of the trace–functional correspondence is needed.

Let u,v\in(A\otimes_{\min}\K)_+ satisfy the functional inequalities. If v=0, any \tau\in T gives d_\tau(u)=0. The faithful stabilized trace then gives u=0, so u\precsim v. Assume v\ne0. Faithfulness implies d_\tau(v)>0 for every \tau\in T. Thus

Fix s>0. There is a positive element c\in M_k(A) in a finite matrix corner such that \norm{u-c}<s/2: compress u by a sufficiently large finite-rank corner projection. In the tracial von Neumann representation for any \tau\in T, put P=1_{(s,\infty)}(u). Then

The polar decomposition of c^{1/2}P therefore has initial projection P and final projection dominated by \supp(c). Taking the normal semifinite trace shows that

This finite bound uses the same k for every \tau. Consequently, if d_\tau(v)=\infty, then d_\tau((u-s)_+)<d_\tau(v) as well. For the remaining traces, monotonicity gives

Definition 7.1 now yields (u-s)_+\precsim v for every s>0. For each n, choose a comparison witness y_n such that

Since \norm{u-(u-1/n)_+}\le1/n, these witnesses prove u\precsim v.□

Proof. First, elements in each prescribed domain have lifts in the corresponding coordinate domains with the same radius. For a selfadjoint element, symmetrize an arbitrary lift and apply the continuous clipping function to [-R_i,R_i]. For a positive element clip to [0,R_i] instead. For a general element with radius R_i>0, if a is any lift, replace it by a f(a^*a), where

This continuous functional calculus has norm at most R_i and does not change the quotient element. Radius zero has only the zero lift.

Only countably many parameters occur in the polynomials. Fix one bounded representative sequence for each of them, once and for all. No multiplicative compatibility among these chosen lifts is required. Given an admissible lifted tuple, let t_{k,l} be the coordinate test obtained by evaluating the lifted polynomial in A and replacing \Lambda by T. Equation (7.1) and evaluation of polynomials in the quotient give

Each of these coordinate sequences is bounded, since its polynomial involves only finitely many bounded parameters and variables in fixed balls.

For each n, choose an admissible trial tuple x^{(n)} with t_k(x^{(n)})<c_k+1/(2n) for k\le n, and choose lifts x^{(n)}_{i,l} in the prescribed coordinate domains. The set

belongs to \omega. Put

These are decreasing members of \omega with empty intersection. For each l that belongs to at least one F_n, let r(l)=\max\{n:l\in F_n\}, which is finite and at most l, and set x_{i,l}=x^{(r(l))}_{i,l}. At the remaining coordinates set x_{i,l}=0.

The resulting sequences remain in their fixed balls. If n\ge k and l\in F_n, then r(l)\ge n and l\in E_{r(l)}, so

Taking the ultralimit, and then letting n\to\infty, proves the claim. The last assertion follows from Equation (7.1).□

Proof. Take e from Lemma 3.2. 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

Thus s is invertible. Form the row and column

Since s=UV, the element y=V^*V=\sum_i v_i^*ev_i satisfies

In particular y is invertible. Setting x_i=v_i y^{-1/2} gives the exact identity

Define

Additivity and traciality, followed by Equation (8.2), give, for every \tau\in T,

The norm of c is finite but may be large. Choose an integer L\ge\max\{1,\norm{c}\} and put

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

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 (8.4). The unnormalized matrix trace adds the support ranks of the direct summands, and hence

Together with Equation (8.3), this is Equation (8.1).□

Proof. If b=0, take p_0=0. Otherwise choose v\in M_L(A)_+ as in Proposition 8.1. Faithfulness of each trace and the support formula give

Strict comparison, applied in A\otimes\K, gives v\precsim b.

There is a single t>0 such that

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\N of the continuous functions

Moreover, these cut ranks increase to d_\tau(v) as t decreases to zero. Thus the open sets

increase as t decreases and cover T, by Equation (8.1). Compactness of T supplies a finite subcover. The smallest of its cut parameters gives Equation (8.6).

We next obtain a comparison witness of the required matrix shape. Regard b as supported in the first matrix corner of A\otimes\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\K with \norm{X^*bX-v}<t. The row x=P_1XP_L\in M_{1,L}(A) satisfies

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

Similarly,

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 \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

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\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':

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 \Her(b) since g(0)=0. Scalar functional calculus gives

Taking traces and using the equality of cut ranks yields

Equations (8.1), (8.6) and (8.7) now imply Equation (8.5). 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\N} of h. For b_\ell=(h_\ell-\eta)_+ we have \norm{b_\ell}\le1-\eta<1. Apply Lemma 8.2 with \delta=1/\ell to obtain positive contractions p_\ell\in\Her(b_\ell) satisfying

Let p be the class of (p_\ell) in D. Since p_\ell is a positive contraction,

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

For the support assertion, functional calculus in A^{**} gives

Every element of \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 (8.9).

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

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 8.3 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

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,

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,

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

Consequently, by |\tau(a)|\le\norm{a}_{2,\tau} and Equation (9.2),

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

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

Thus whenever v\le u and \tau(u-v)<\kappa, Equation  (9.3) gives, for every j,

Starting with an arbitrary residual projection u, perform one run through the functions as follows. Put v_0=u and, successively, let

Apply Proposition 8.3 to y_j with cutoff \alpha to obtain a projection a_j\in D satisfying

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, \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

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

Indeed, suppose \tau(u)>\delta but \tau(u-v_m)<\kappa. The partition identity in the initial corner implies

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 (9.4) applies at that step. Therefore

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 (9.5) would imply

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

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

There are finitely many blocks at every stage. Summing the trace losses over the n stages proves Equation (9.1). 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 9.1, 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

By Corollary 8.4, 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

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

Summing these estimates, using \sum_s\tau(p_s)\le1, yields

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 \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

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 9.2 in the corner e^{(\ell)}_{11}De^{(\ell)}_{11} to the finite family

with tolerance \xi/N_0. Denote the resulting corner symmetry by u_\ell, and put

Each summand is selfadjoint, and the matrix-unit relations give

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

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 7.3 to one variable in the selfadjoint unit ball of D, with the tests

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,

extends the trace identities from the dense sequence to the unit ball of S, and scaling gives Equation (9.7) for all s\in S.□

Proof. We first construct a supporting homomorphism \pi in D^{**}, then use the supplied projections to transport its matrix units into D. The final estimates measure the total trace lost in this transport, independently of the matrix sizes. The supporting-homomorphism step is the finite-dimensional case of the order-zero structure theorem [37, Theorem 3.3]; we include the details to specify the support used below. Put q=\supp(h)\in D^{**}. For every projection e\in F, the positive elements \phi(e) and \phi(1_F-e) are orthogonal. Consequently \phi(e) commutes with their sum h. Projections linearly span F, so h commutes with \phi(F). For t>0 define

This map is completely positive and contractive, since 0\le\pi_t(1_F)=h(h+t1)^{-1}\le1. If e is a projection, orthogonality of \phi(e) and \phi(1_F-e) gives

Because projections span F, \pi_t has a pointwise strong limit \pi:F\to D^{**}. Positivity at every matrix level passes to this limit. Thus \pi is completely positive, \pi(1_F)=q, and \pi takes every projection to a projection. Its range lies in qD^{**}q: indeed, for positive x\in F one has 0\le\pi(x)\le\norm{x}q, and then one uses linearity. Viewed as a unital completely positive map into that corner, \pi is multiplicative. To recall the multiplicative-domain argument, equality in the Schwarz inequality at a projection e forces \pi(ex)=\pi(e)\pi(x) and \pi(xe)=\pi(x)\pi(e) for every x; this follows by applying Schwarz to e+zx for all z\in\C and comparing the linear terms. Applying this to all projections and using their linear span proves multiplicativity. Finally,

so h\pi(x)=\phi(x). The commutation with h passes to the strong limit, and hence

The zero map is included, with q=0.

Write F=\bigoplus_{\alpha}M_{n_\alpha}, and choose matrix units e^{(\alpha)}_{st}. Work first in one block and omit its superscript. Set a=\phi(e_{11}). Proposition 8.3 supplies a projection p\in D with

The final support inequality follows from \pi(e_{11})=\supp(\phi(e_{11})). On [0,1] define

Both functions are continuous, and k(s)=s g(s). The spectral containment of p gives k(a)p=pk(a)=p. In particular the elements

are in D, even though the individual \pi(e_{s1}) were constructed only in D^{**}. The equality uses \pi(e_{s1})a=\phi(e_{s1}), which follows from Equation (10.2). Define

Since p\le\pi(e_{11}), multiplication gives

For different blocks the products vanish, because their supporting matrix units are orthogonal. Thus these formulas extend to a *-homomorphism \rho:F\to D. Direct multiplication by the matrix units also shows that, for P=\rho(1_F),

This construction uses the support containment of p; it does not require p to commute with a, nor P to commute with h.

Fix \tau\in\Lambda. In all expressions on D^{**} use its canonical normal tracial extension, still denoted by \tau. Faithfulness of this extension is not needed. Scalar functional calculus gives, in each block,

Traciality and the partial isometries \pi(e_{s1}) imply that \phi(e_{ss}), \rho(e_{ss}), and \pi(e_{ss}) have the same traces as a, p, and \pi(e_{11}), respectively. Summing over all diagonal entries of all blocks therefore gives

The total loss is measured by the trace of q, so no matrix-size factor appears.

For a contraction c\in F, Equation (10.2) and commutation with \pi(c) give

Here q-h is a positive contraction and q-P is a projection; both commute with \pi(c). The squared triangle inequality now yields Equation (10.1).□

Proof. Apply Theorem 10.1 to the finite set

with error \eta. In the finite-family notation following that theorem, put T_i=\phi_i\psi_i and T_0=\sum_i\beta_iT_i. Thus \norm{T_0(x)-x}<\eta on this set. Regard the outgoing maps as taking values in D and write h_i=\phi_i(1_{F_i}). Fix one trace \tau\in\Lambda throughout the following weighted estimates. Positivity and contractivity give T_i(1_A)\le h_i\le1, and hence

For a\in E, the Schwarz inequality for the completely positive contraction T_i gives T_i(a)^2\le T_i(a^2). Expansion of the squares and traciality imply

In the last inequality each trace term is bounded by the norm of its argument, and \norm{a}\le1.

Apply Lemma 10.2, with the same \eta, to each \phi_i, obtaining \rho_i:F_i\to D. Since \psi_i(a) is a contraction, Equations (10.1) and (10.6) give

A second squared triangle inequality and Equation (10.7) yield

Choose the r_i successively. At step i apply Proposition 9.3 with

The algebra S_i is unital and norm separable, while E_i is finite dimensional and may be nonunital. We obtain

Because r_i commutes with \rho_i\psi_i(a) and multiplication by a unitary preserves the tracial 2-seminorm,

Thus Equation (10.8) gives

Every weighted estimate has been obtained at the same fixed \tau; their constants are independent of that trace.

Set z=\sum_i\sqrt{\beta_i}\,r_i. We spell out the cancellation that converts Equation (10.10) into a bound for z. If a word contains r_j exactly once and all its other letters belong to S_j, cyclicity writes its trace as \tau(sr_j) for some s\in S_j. It vanishes by Equation (10.9). For i<j, each of the four terms in [r_i,a]^*[r_j,a] has this form, since a,r_i\in S_j. The reverse mixed term is its adjoint and also has zero trace. Therefore

For any b\in A, the same argument gives \tau(br_i)=0 and \tau(br_ir_j)=0 for i\ne j. The diagonal terms use r_i^2=1 and \sum_i\beta_i=1, proving \tau(bz)=0 and \tau(bz^2)=\tau(b). These are coefficient identities, and do not require the different r_i to commute.

For the fourth moment write z_i=\sum_{j\le i}\sqrt{\beta_j}\,r_j, with z_0=0. At step i put X=z_{i-1}, r=r_i, and c=\sqrt{\beta_i}. The full noncommutative expansion, grouped only after applying the trace, is

Indeed, the four words with one r are cyclic permutations of X^3r. The four words with three r reduce, by cyclicity and r^2=1, to Xr. Of the six words with two r’s, four have trace \tau(X^2), and two are alternating. The terms with one or three r vanish by Equation (10.9). Tracial Cauchy–Schwarz gives

The already proved second-moment cancellation, applied to the partial sum, gives \tau(X^2)=\sum_{j<i}\beta_j. Consequently

Summing over i proves the claimed bound, including its sharper form:

Finally, \norm{z}\le\sum_i\sqrt{\beta_i}\le\sqrt M, which explains why this construction alone does not give a common norm bound as the approximation data vary.□

Proof. Fix \tau and form the Hilbert space L^2(D,\tau) by quotienting out the zero vectors of the tracial seminorm and completing. Left and right multiplication by the bounded selfadjoint z define commuting bounded selfadjoint operators L_z and R_z on this space. Continuous functional calculus gives g_R(L_z)=L_d and g_R(R_z)=R_d. Let E_z be their joint spectral measure, and let \mu_a be the finite positive measure \mu_a(U)=\langle E_z(U)a,a\rangle. The scalar 1-Lipschitz inequality for g_R gives

This proves Equation (10.12) in tracial L^2.

For real s, the scalar inequalities

follow by considering |s|\le R and |s|>R. Functional calculus therefore bounds the traces of |z-d| and |z^2-d^2| by the first two right-hand sides of Equation (10.13). For completeness, the trace inequality that permits arbitrary contractive coefficients is

To see it, take the polar decomposition x=u|x| in D^{**} and use the normal tracial extension of \tau. With y=|x|^{1/2} and c=bu, traciality and Cauchy–Schwarz give

This proves Equation (10.13) without a commutation assumption on b. Combining it with Equation (10.5) proves Equation (10.14).□

Proof. Choose 0<\eta<1 so that \sqrt{72\eta}<\sigma/2. Proposition 10.3 and Lemma 10.4 give \bar d=\bar d^*\in D with \norm{\bar d}\le R satisfying Equation (10.14). Choose a selfadjoint lift of \bar d to \ell^\infty(\N,A) and apply g_R at every coordinate. The resulting lift (d_l) remains a lift of \bar d and satisfies d_l=d_l^* and \norm{d_l}\le R for every l.

By Equation (7.1) and the trace-test identity in Lemma 7.3,

There are only finitely many tests, indexed by E, H, and the one fourth-moment test. For each test the desired bound in Equation (10.16) holds on a set in \omega, using the strict margin \sigma/2 for commutators and the positive margin \sigma for moments. Their finite intersection belongs to \omega and is nonempty. Any coordinate in it supplies the required d.□

Proof. Choose a norm-dense sequence (c_k)_{k\geq1} in the unit ball of C and a norm-dense sequence (x_k)_{k\geq1} in its selfadjoint unit ball. Choose lifts (c_{k,l})_{l\geq1} consisting of contractions in A and lifts (x_{k,l})_{l\geq1} consisting of selfadjoint contractions in A. These lifts are chosen separately for each element; no multiplicative lifting of C is needed.

At coordinate l, use Proposition 10.5 with the finite sets

with tolerance 1/l, and with the same fixed number R at every coordinate. Obtain d_l=d_l^*\in A, \norm{d_l}\leq R, such that for k\leq l,

The local proposition permits arbitrary finite subsets of A, so their dependence on l causes no change in its hypotheses. The common bound R makes (d_l)_l a bounded sequence. Let d be its image in D.

For a fixed k, the displayed estimates hold for all l\geq k. Taking ultralimits, using Equation (7.1) for the commutators and the definition of limit traces for the moment tests, gives

for every \tau\in\Lambda. Since the uniform tracial 2-norm detects zero in D, [d,x_k]=0. Norm continuity and linearity give d\in C'. For completeness, the relevant continuity estimates are

They extend the two moment bounds from (c_k)_k to the entire unit ball of C, uniformly over \Lambda. Nuclearity was used in obtaining the local assertion in A; this transfer places no nuclearity hypothesis on C.□

Proof. Fix N. At the jth step, apply Proposition 11.1 with R=N to C_{N,j}. This algebra is unital and norm separable, since A is separable and only finitely many elements have been adjoined. The resulting d_{N,j} commutes exactly with A and its predecessors and satisfies Equation (11.2). This constructs each row separately.

We prove the asserted convergence quantitatively. Fix t\in\R, \tau\in\Lambda, and a contraction a\in A. Define

At step j, the coefficient b=aP_{j-1} belongs to C_{N,j} and has norm at most one. Write d=d_{N,j}. Scalar Taylor expansion with its integral remainder gives

Functional calculus, followed by the tracial inequality |\tau(by)|\leq\tau(|y|) for a contraction b, therefore yields

The last inequality is Hölder’s inequality for the probability measure associated with d and the tracial state \tau, since \tau(|d|^3)\leq\tau(d^4)^{3/4}. Inserting the two conditional moment bounds from Equation (11.2) proves, for every 1\leq j\leq N,

Every bound here holds for all \tau\in\Lambda and all contractions a\in A, with the same right-hand side.

For fixed t, one has |q_N|\leq1 once N\geq t^2/4. Telescoping the recurrence gives

Since the elements in a row commute, P_N=e^{its_N}. Consequently

This proves Equation (11.3). The coefficients aP_{j-1} are exactly why moment bounds against the whole earlier algebra were required. No independence of the variables or equality of their distributions has been assumed.□

Proof. Suppose the conclusion fails. Then \norm{f}_\infty>0, and there are \delta>0, integers N_k\to\infty, and traces \tau_k\in\Lambda such that

The positive functional x\mapsto\tau_k(a^{1/2}xa^{1/2}) has norm \tau_k(a) and agrees with x\mapsto\tau_k(ax) by traciality. Thus the left side of Equation (11.7) is at most 2\norm{f}_\infty\tau_k(a), and in particular

On the compact spectrum of s_{N_k}, the formula

defines a positive unital functional. Its representing probability measure, viewed as a measure \mu_k on \R, satisfies

for every bounded continuous h on \R. By Equation (11.3), for every fixed t\in\R,

The classical Lévy continuity theorem for characteristic functions [9, Theorem 1.1] now implies \mu_k\to\gamma weakly: the limiting characteristic function is that of a probability measure and is continuous at zero. Hence \int f\,d\mu_k\to\int f\,d\gamma. Multiplying by 0<\tau_k(a)\leq1 contradicts Equation (11.7). This argument permits the testing trace to vary with k, which proves the asserted uniformity. It also accommodates the varying spectra of s_{N_k}; no common spectral bound is needed for the continuity theorem.□

Proof. For m=1, take p_1=1. Suppose m\geq2, and choose the Gaussian quantiles

so that the consecutive intervals I_i between q_{i-1} and q_i each have \gamma-measure 1/m. The assignment of the finite endpoints is immaterial because \gamma has no atoms.

Fix a finite set H of positive contractions in A and a tolerance \varepsilon>0. Choose \theta>0 so small that

Choose pairwise disjoint open intervals about the finitely many quantiles q_1,\ldots,q_{m-1} whose union U has \gamma(U)<\theta. Interpolating linearly across these intervals gives continuous functions f_1,\ldots,f_m:\R\to[0,1] with

At a transition around q_i, only f_i and f_{i+1} vary, with one decreasing from one to zero and the other its complement. In particular,

The second estimate uses (u-u^2)^2\leq1/16 for 0\leq u\leq1, and the integrand vanishes outside U.

Keep these functions fixed. Apply Lemma 11.3 to every pair (a,f_i) with a\in H, and to (1,(f_i-f_i^2)^2) for each i. For all sufficiently large N, all of these finitely many convergence errors are less than \theta. Set v_i=f_i(s_N). These are positive contractions in D\cap A' and satisfy \sum_i v_i=1. Equation (11.9) gives

Thus arbitrarily accurate finite sets of trace-splitting and projection tests can be met by m positive contractions that already commute with A and sum to one.

Choose a countable norm-dense family (a_k)_{k\geq1} in the positive unit ball of A. Apply Lemma 7.3 to the m variables constrained to be positive contractions, with the countable tests

The preceding finite construction verifies the hypotheses of that lemma. In this application the diagonal variables are only the m positive contractions v_i; the elements s_N are not diagonal variables. In particular all required lifts lie in fixed norm balls. The sum-to-one equation is included among the tests, regardless of which positive contractive lifts of a trial tuple are chosen.

The resulting positive contractions p_i are projections, commute with A, and sum to one, because the uniform tracial 2-norm detects zero in D and commutation extends by norm density and linearity. They are pairwise orthogonal: for each i,

and positivity forces p_jp_i=0 for j\ne i. The trace equations extend from the dense family to every positive contraction, uniformly over \Lambda, and then by scaling and linearity to every element of A. This proves Equation (11.8), which is uniform property \Gamma as in Definition 7.4.□

Proof of Theorem 1.4. If T(A)\ne\varnothing, Theorem 11.4 gives uniform property \Gamma. Lemma 7.2 converts the assumed stabilized comparison to the convention used in Theorem 7.5. That theorem therefore gives A\cong A\otimes_{\min}\Zalg.

Suppose now that T(A)=\varnothing. The comparison convention in the hypothesis says that every positive element of the stabilization is Cuntz subequivalent to every nonzero positive element. We show directly that every nonzero hereditary subalgebra of A contains an infinite projection.

Given 0\ne b\in A_+, the stipulated comparison gives 1\precsim b. The comparison witnesses can be taken in A: view 1 and b in the corner determined by e=1_A\otimes e_{11}, and compress every stabilized witness z_n to x_n=ez_ne. Then

which is x_n^*b x_n\to1 after identifying this corner with A. Choose x\in A for which t=x^*b x is invertible and put

Then w^*w=1, while

is a projection Murray–von Neumann equivalent to 1.

There are orthogonal nonzero positive elements b,c\in A: take nonnegative continuous functions with disjoint supports near two distinct spectral points of a nonscalar selfadjoint element. For this b, the projection p\in\Her(b) above annihilates c, so p\ne1. Thus 1 is an infinite projection. For any nonzero hereditary subalgebra E\subset A, choose 0\ne h\in E_+ and repeat the construction to obtain a projection p_h\in\Her(h) \subset E equivalent to 1. It is infinite inside E: an isometry w_h with w_h^*w_h=1 and w_hw_h^*=p_h identifies A with p_hAp_h\subset E, and carries a proper subprojection equivalent to 1 to a proper subprojection equivalent to p_h. We have proved the hereditary-subalgebra definition of pure infiniteness for the simple algebra A.

Kirchberg–Phillips’ absorption theorem [20, Theorem 3.14] now gives A\otimes_{\min}\Oinf\cong A. The algebra \Oinf is strongly self-absorbing [33, Examples 1.14(ii)], and Winter’s theorem gives \Oinf\otimes_{\min}\Zalg\cong\Oinf [35, Theorem 3.1]. Associativity of the minimal tensor product therefore yields

The absorption theorems used here require no Universal Coefficient Theorem hypothesis. This completes both trace-space cases.□

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),

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 (12.1) 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\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

Each entry v_j belongs to \overline{bA}. Thus c=\sum_jv_jv_j^*\in\Her(b)_+, and traciality gives \sigma(c)=\sigma(v^*v), proving the assertion.

For an integer N\ge\|c\|, put

On the spectrum of c, the functions sum to s, while

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

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

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

Consequently \sigma(a_n)\le d_\sigma(H_n)+\gamma_n.

Apply Lemma 13.1 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

Thus, in disjoint finite matrix blocks,

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

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

For f(s)=2(s-1/2)_+ on [0,1],

Traciality transfers these estimates from f(k_n)=w^*w to r_n=ww^*. By (13.1), 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

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\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 13.2 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

Construct orthogonal projections r_j recursively with

Indeed, if R=\sum_{i<j}r_i, then F_jR=G_jR=R and \tau(R)=t\tau(F_{j-1}). Apply Proposition 13.2 to

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

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

Take 2/m<\varepsilon.□

Proof. For a matrix summand of size l with matrix units e_{ij}, apply Lemma 13.3 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 (13.4), since their units sum to 1_D.□

Proof. Consider all finite-dimensional unital E^0, order-zero maps \theta:E^0\to A, and self-adjoint contractions b_x\in E^0. Their error functions

are nonnegative continuous affine functions on T. At each fixed \sigma, hyperfiniteness of M_\sigma supplies a unital finite-dimensional subalgebra approximating the tuple in 2-norm by self-adjoint contractions. Lift its inclusion through (12.1) and choose a coordinate. This makes L(\sigma) arbitrarily small, including its unit-defect term.

For every \kappa>0, a finite convex combination of these functions has maximum less than \kappa. Otherwise separate their convex hull from the open convex set \{f\in C(T,\R):\max_Tf<\kappa\}. The separating functional is positive, since this set is closed under adding arbitrary negative functions. Normalize it to a probability measure \mu. Its barycenter \sigma_\mu belongs to compact convex T, and affinity gives L(\sigma_\mu)=\int L\,d\mu\ge\kappa for every candidate L, a contradiction. Denote the resulting data by D_i^0,\theta_i,b_{i,x},w_i.

We next replace each \theta=\theta_i, now viewed in D, by a homomorphism on a projection, controlling the loss of unit trace by the original unit defect. Write \theta=h\pi, and put f(s)=(2s-1)_+, g(s)=\min(1,2s). For the e_{11} corner of a matrix summand, apply Proposition 13.2 to

The support condition holds because gf=f, and the trace inequality is equality. We obtain a projection v with

It lies under \pi(e_{11}). The elements \pi(e_{j1})v belong to D, because

Repeating v along each matrix diagonal and summing over the summands gives a projection Q\in D commuting with \pi(D_i^0), with Q\pi(d)\in D for d\in D_i^0. Traciality and repetition give

Set D_i=\C(1-Q)+Q\pi(D_i^0) and x^{(i)}=Q\pi(b_{i,x}). This is a unital finite-dimensional subalgebra; zero image summands are simply absent. No commutation of Q with h is needed. For a contraction b\in D_i^0, tracial L^2 contractivity and the triangle inequality give

since 1-f(s)\le2(1-s). With m=|\mathcal F| and C_0=(1+\sqrt2)^2, this implies

The weighted sum is at most (2+2mC_0)\kappa. Take \kappa sufficiently small. Limit traces restrict to T by Lemma 12.1, so the bound holds on \Lambda.□

Proof. Use the unnormalized matrix trace and set

For l(r)=\min(r^2,1), the first diagonal corner of l(V) is e. The function l is globally 2-Lipschitz. Left and right multiplication by V are commuting self-adjoint operators on tracial L^2, so their joint spectral calculus gives

The squared norm on the right before the factor 4 is 2\sum_i\|[p_i,x]\|_{2,\tau}^2. Compressing the left side to the first corner proves (14.3). This argument uses scalar Lipschitzness on L^2, not operator-norm Lipschitzness.

Let S be the support of s in the tracial von Neumann representation. The row W is supported on the right by \operatorname{diag}(p_1,\ldots,p_N), so its polar decomposition gives \tau(S)\le\lambda. Since \tau(s^2)=\lambda+2R,

Both s-e and e-e^2 are positive, supported under S, and bounded above by |s-S|. Cauchy–Schwarz gives

which proves (14.4).□

Proof of Proposition 14.1. We first work in D. Choose a large parameter K>1, and apply Proposition 14.2 with \Delta=K^{-2}. Recursively use Corollary 13.4 to choose p_i\in D\cap D_i' with

and tolerance K^{-1}. In particular, \tau(p_i)=t_i and \tau(p_ix^{(i)})=t_i\tau(x^{(i)}). For each \tau\in \Lambda, write

Both are nonnegative, and the single selected cost controls both:

In particular H_\tau\le\lambda^2/(2K)+2\lambda/K^2, uniformly in \tau and independently of the number of models.

Since [p_i,x^{(i)}]=0, for x\in\mathcal F,

The positive functional z\mapsto\tau(p_i z) has norm t_i, whence

Scalar Cauchy–Schwarz and (14.2) therefore give, for s=s_N,

Set e_D=\min(s,1). Lemma 14.3 gives

Replacing s by e_D in (14.6) costs at most \tau(s-e_D). By (14.5), the upper bounds for the trace and projection errors tend to \lambda^{3/2} as K\to\infty, while the commutator bound tends to zero. Choose K giving the desired inequalities with slack strictly smaller than \rho. A positive contraction lift of e_D and one good coordinate then give (14.1) in A.□

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 14.1 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 15.1, enlarging E to contain the previous choices at each step. They commute, and

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

For commuting positive contractions a,b,

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

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 (15.2) 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 15.2 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 (15.2) 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 [3, Definition 2.1].□

Proof. The approximate-unit sequence defines u\in\Aom\cap A' with ua=a=au for constant a\in A. For y\in\Aom\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

Taking these limits in order proves y\in J.

For surjectivity, let v\in\Aom 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 15.2 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 [3, Lemma 1.2].

Apply matrix order-zero lifting to the resulting quotient map and (16.1) to obtain \psi, and compose with the quotient to \Fw 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

Since \nu(a)<\infty, (16.2) 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

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,

The regularization in (16.2) can therefore be removed, and

This is a finite, possibly zero multiple of a trace through D. For z=\psi(1) from Lemma 16.1, 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 1.5. Theorem 15.3 and (16.1), followed by Lemmas 16.1 and 16.2, give tracially large order-zero maps M_n\to\Fw for every n. This is uniform McDuffness in the nonunital sense of [5, Definition 2.4].

Our comparison hypothesis also makes \Cu(A) almost unperforated. Indeed, if (k+1)[x]\le k[y] with y\ne0, then additivity and monotonicity of dimension functions give

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 13.2, followed by increasing the supports of the cutdowns.

For simple algebras, almost unperforation is the strict comparison formulation used in [5, Section 2.3]. Its nonunital absorption criterion [5, Proposition 5.4] states that a simple, separable, nuclear algebra with this comparison and nontrivial lower-semicontinuous traces is \Z-stable if it is uniformly McDuff. All these hypotheses have now been verified. Hence A\cong A\otimes\Z.□