Weak pure infiniteness and O-infinity absorption

Proof. We first prove the bidual description in (i), including its norm closure assertion. One inclusion follows on hDh and then by closure. Conversely, suppose u\in D_+ and e_hu=u. Take increasing continuous positive contractions f_j with f_j(0)=0 and f_j(t)\uparrow1 for t>0. On the compact state space of the unitization, the continuous functions

decrease pointwise to zero. Indeed, the normal extension of each state evaluates the strong limit f_j(h)\uparrow e_h. Dini’s theorem makes this convergence uniform. The positive norm formula and (1-f_j(h))^2\le1-f_j(h) then give \|(1-f_j(h))u\|\to0. Thus u=\lim_j f_j(h)u f_j(h)\in\Her(h). Here f_j(h)\in\Her(h) follows from continuous functional calculus, by first cutting f_j off near zero. For a general element supported on both sides by e_h, apply the positive case to the positive and negative parts of its real and imaginary parts.

If 0\le u\le Kv, compression by 1-e_v gives u^{1/2}(1-e_v)=0. Hence u is supported by e_v and belongs to \Her(v). If u\in\Her(v)_+, take a cutoff approximate identity g_j(v) for \Her(v), with every g_j zero near zero. The quotient function below is continuous when extended by zero at zero, and gives elements

This proves the comparison assertion.

For (ii), the rectangular coefficient z(z^*z+1/j)^{-1/2} compares z^*z to zz^*, since its conjugation of zz^* equals

Apply the same calculation to z^* for the reverse comparison. The inverses lie in unitizations and are multiplied by algebra matrices. Block-diagonal coefficients prove preservation under block sums. For transitivity, first approximate u by x^*vx, then approximate v by y^*wy within the requested error divided by \max\{1,\|x\|^2\}; the coefficient yx compares u to w. A diagonal choice of coefficients proves closure under norm limits in the first variable.

If \phi:D\to E is a *-homomorphism, apply \phi\otimes\id_{\Kc} to a comparison sequence; continuity preserves its norm limit. Applying this in both directions preserves equivalence. For finite d,v, let P_n,P_m be the scalar coordinate projections in the multiplier algebra of D\otimes\Kc. If x_j^*vx_j\to d in the stabilization, then

Conversely, rectangular coefficients embed in the stabilization and give the same convergence there. The multiplier compressions retain algebra membership without a unit, and the argument includes zero matrices.

For (iii), work in the common stabilization D. The closed span of Dv^{1/2}D is \Id_D(v), so finite sums \sum_{i=1}^r x_i v^{1/2}y_i approximate d^{1/2}. For such a sum, define

and put V=\operatorname{diag}(v,\ldots,v)\in M_r(D). Then

The order and rectangular rules already proved give

It follows that z^*z\precsim V. Proper infiniteness, block addition, and induction give V\precsim v for every finite r. Now z^*z\to d, and the norm-limit rule yields d\precsim v. This proof includes v=0.□

Proof. Compress a stabilization comparison coefficient to the first m target and first n source coordinates. These scalar multiplier projections give s\in M_{m,n}(C) with \|d-s^*vs\|<\varepsilon/2. Put z=v^{1/2}s h. Order in the unitization and functional calculus for d give

For the lower bound, use h d h\ge\varepsilon h^2. The two inequalities and Lemma 2.1(i) show \Her(z^*z)=\Her(h). Consequently \Her(|z|)=\Her(h) and e_{z^*z}=e_h. Let z=w|z| be the rectangular polar decomposition in the bidual. Then w^*w=e_h; the factorization through v^{1/2} gives ww^*=e_{zz^*}\le e_v.

For k\in\Her(h)=\Her(|z|), take cutoff functions g_j for |z|, zero near zero, with g_j(|z|)k\to k. Since

norm closure gives wk\in M_{m,n}(C). If k\ge0, then wkw^*=(wk^{1/2})(wk^{1/2})^* belongs to M_m(C). It is supported under e_v, hence belongs to \Her(v). Linear decomposition handles arbitrary k. The identity w^*w=e_h proves multiplicativity and injectivity. It also proves (2.3): functional calculus vanishing at zero commutes with the homomorphism, and the two squares of w(l-\tau)_+^{1/2} are the asserted positives.

For (2.4), normal functional calculus gives e_{wlw^*}=we_lw^*. Thus the image of \Her(l) is supported under e_{wlw^*}, which gives one inclusion by Lemma 2.1(i). A sandwich by the image of l has the form

The inner algebra element is supported on both sides by e_l, so lies in \Her(l). Since conjugation is an injective *-homomorphism on \Her(l), its image is norm closed; taking closures gives the reverse inclusion. Finally, the normal extension of \pi carries (2.1) to the same identities with represented supports. Conjugating back by the represented w^* recovers \pi^{(n)}(k). The two contractive conjugations give both inequalities in (2.5).□

Proof. For (i), choose b\in J_\lambda with \|z-b\|<\varepsilon. Then \|z+J_\lambda\|<\varepsilon, and functional calculus in the quotient gives (z-\varepsilon)_+\in J_\lambda. Since (z-\delta)_+\to z, the particular exhaustion follows. For (ii), apply separation of points by irreducible representations to the quotient D/\Id(v). For (iii), the two support projections coincide. Lemma 2.1(i) gives equality of the hereditary algebras and mutual comparison.□

Proof. Let e=e_{(a-1/8)_+} and H=\Her((a-1/8)_+). Since a is properly infinite, a^{\oplus(k+2)}\precsim a. Apply Lemma 2.2 with cutoff 1/8, and split the resulting row into slots labelled 0,1,\ldots,k,*. They satisfy

Choose a ramp t=t(a)\in H that is zero on [0,1/4] and one on [1/2,1]. It is rich, and its represented norm is one whenever the represented norm of a exceeds 1/2.

We now compute the measurements, before making any choice of S_i. Put F=cyc^*\in C_+. The block calculation behind their definition is as follows. Suppose L\in C^{**} satisfies L^*L=e and has slot coordinates v_0^*L=\gamma_g e, v_i^*L=\sigma_g S_i, and suppose F=LH_0L^* for a positive H_0\in eC^{**}e. Then

Vanishing of the defect in a representation will force this form there, with tH_0=H_0 and with t acting identically on the prescribed supports of S_i. The first measurement therefore recovers H_0, and an inverse on its spectral cutoff recovers each S_i from the second. These are conditional identities used to design the measurements; their verification follows after the graph is built.

Choose a continuous nonnegative function h_{\rm inv} on [0,\infty), zero at zero, equal to \lambda^{-1} for \lambda\ge\alpha_0, and bounded by \alpha_0^{-1}. Define

The sandwiches are (v_it)^*F(v_0t), hence belong to C. They are supported on both sides by e, so they lie in H; the same is true after multiplication by h_{\rm inv}(X). Thus X\ge0 and X,B_i have the asserted algebra membership. The slots, t,c,y determine them without using any S_i.

For the chosen S_i, define in the corner eC^{**}e

Then define in C^{**}

Here S_i\in H, since X\in H. The radicand is at least (1-\gamma_g^2-k\sigma_g^2)e. Its square root belongs to H+\mathbb C e and multiplies H into itself. Hence Lh\in C for every h\in H, term by term in (3.3). The slot identities also give

It follows, by the membership and support argument of Lemma 2.2, that h\mapsto LhL^* is an injective *-homomorphism from H into \Her(a). Indeed, for h\ge0 its value is (Lh^{1/2})(Lh^{1/2})^*\in C, supported under e_a; and L^*L=e gives multiplicativity and recovers h on conjugating back. The same identities give preservation of cutoffs, their Cuntz classes, and represented norms, exactly as in that lemma. Thus p is a rich positive contraction with the stated representation signal. If e=0, all the operators on this corner are zero and these assertions still hold.

It remains to see what vanishing of the defect forces. Work on the essential subspace of a representation in which

and omit the representation symbols. The normal extensions preserve the preceding support identities. For every vector \xi,

Equality first on \ran y and then on its closure shows that c is isometric on e_y\mathcal H. For a positive contraction p, equality \|p\zeta\|=\|\zeta\| implies p\zeta=\zeta, by applying 1-p^2. Consequently ce_y\mathcal H lies in the p=1 subspace, and the positive operator F=cyc^* is supported on that subspace. Since p=LtL^* and L^*L=e, the p=1 subspace lies in the range of L, and its pullback lies in the t=1 subspace. With H_0=L^*FL, we obtain

The last equality uses the isometric action of c on e_y\mathcal H.

Now v_0^*L=\gamma_g e, v_i^*L=\sigma_g S_i, and L^*v_0=\gamma_g e. Substituting these identities into (3.4) yields

Since tX=X, t is the identity on \overline{\ran X}=e_X\mathcal H. Each represented S_i has both supports under e_{(X-\alpha_0)_+}. Therefore tS_i=S_i and S_iXh_{\rm inv}(X)=S_i. The last display gives B_i=S_i, while \|X\|=\|H_0\|>s. This proves (3.2).□

Proof. The contraction assumptions give 0\le d\le y^2\le1. By (3.5), every irreducible representation vanishing on d vanishes on (y-s)_+. The primitive ideal test in Lemma 2.3(ii) therefore gives (y-s)_+\in\Id(d). Set s'=3/8. Apply cutoff exhaustion in Lemma 2.3(i) to (y-s)_+ with cutoff s'-s. After decreasing the resulting parameter if necessary, there is 0<\delta<1 with

Define

The order d\le y^2\le1 gives

Hence z^*z and h_d have the same hereditary algebra, and z^*z\CuEq zz^*=q. It follows that \Id(q)=\Id(h_d). A second application of cutoff exhaustion, to (y-s')_+ with cutoff 1/2-s', gives \delta'>0 such that

Let q_1 be a ramp of q, zero on [0,\delta'/2] and one on [\delta',\infty). It is a positive contraction in \Her(q)\subset\Her(y). If \|\pi(y)\|>1/2, then \pi((y-1/2)_+)\ne0. The ideal inclusion (3.10) forces \pi((q-\delta')_+)\ne0, so \|\pi(q)\|>\delta' and \|\pi(q_1)\|=1.

To bound the universal-bidual norm, work in the universal nondegenerate representation. For \xi\in e_{h_d}\mathcal H =\mathbf1_{(\delta,\infty)}(d)\mathcal H,

The last inequality uses that y is a contraction. Apply this estimate to vectors in \ran(h_d^{1/2}). The closure of \ran(yh_d^{1/2}) is e_q\mathcal H, so continuity gives

Choose \sqrt{1-\delta}<\tau<1, and take p_1 to be a ramp of p, zero on [0,\tau] and one at 1. Functional calculus gives e_{p_1}\le\tau^{-2}p^2, while e_{q_1}\le e_q. Consequently

The ramp of the rich p is rich and remains in \Her(a). Spectral mapping at its plateau shows that it has norm one in every representation where p has norm one. This proves the first signal in (3.6); the second was proved above. The proof also covers zero inputs.□

Proof. Let f_\eta be the bounded Borel function equal to \lambda^{-1} for \lambda>\eta and zero otherwise. Its norm on the spectrum is at most \eta^{-1}. If \zeta\in e_{(U-\eta)_+}\mathcal H, then

which proves the actual-range inclusion. If \zeta=xQ\xi and Q\zeta=0, then

so \zeta=xQ\xi=0. The assertion concerns the actual range; no closed-range conclusion is being used.□

Proof. We argue by contradiction. If every four-entry choice could be matched in a representation, the fixed elements could be passed to a norm ultraproduct. In an irreducible representation of the resulting image of C, the first pair would give an isometry with nonzero defect, while the last pair would rule out that defect.

All formulas described as fixed below depend only on X,B_i and the displayed constants. Define

Here Q\in\widetilde C_+, whereas x,V,U\in C_+. Let \fin(C) denote the set of finite subsets of C, and order

by the usual order in the first coordinate and inclusion in the second. At \lambda=(m,E), put \eta=1/m, so \eta\to0 along this directed set, and every fixed w\in C eventually belongs to E.

The first two entries. The first pair will force UwV=0 for every fixed w\in C in a limiting image of C. At the index (m,E), we test only the finitely many w\in E. A positive sum lets one norm bound control all of these tests: set

where g_\eta is a ramp that is zero on [0,\eta] and one on [2\eta,\infty). In fact,

Thus a small norm for D_0 makes each A_\eta wV^{1/2} small. We have

Since V is properly infinite, ideal comparison gives D_0\oplus D_0\precsim V. Apply Lemma 2.2 with cutoff \eta, split the carrier into two columns, and multiply these columns by D^{1/2}. Taking adjoints of the resulting two elements gives contractions T_1,T_2\in C with

Their left supports lie under e_D and their right supports under e_V. Both are under e_x\le e_{(X-\alpha_0)_+}; they therefore belong to the required hereditary algebra.

Consider a representation satisfying (4.2) for the entries at this index, and omit its symbols. Then B_1B_1^*=D and Q=(h1-D)_+. The spectral subspace of D for (h,\infty) is killed by Q and lies in e_D\mathcal H\subset e_{(U-\eta)_+}\mathcal H. Lemma 4.1 shows that this subspace is zero. Thus

For the second conclusion, if the norm of D_0 were at least 2\eta, spectral mapping and the plateau of g_\eta would give \|D\|=1>h.

A fixed prospective defect and the last two entries. Choose ramps

so that h_0 vanishes on a neighborhood of [0,\alpha] and equals one on [\beta,\infty), and k_0 vanishes on a neighborhood of [0,\alpha_0] and equals one on [\alpha,\infty). In particular KH=H. Set

The reason for these definitions is the following conditional calculation. In the limiting irreducible representation below, the first pair will give

These relations are not being asserted in C at this stage. They imply b_0r_i=0, and hence

Since r_2^*r_2=H^2\ne0, the isometry v has a nonzero defect projection 1-vv^*. This is the scaling-element construction of Blackadar–Cuntz [3], in the form recalled in [17, Remark 2.4]; the second entry supplies the additional nonzero orthogonal range.

To choose the last two entries before taking any limit, we encode this prospective defect by one fixed element of C. Compression by K puts the formula in the required hereditary algebra, and clipping makes it a positive contraction even before the preceding relations hold. Let \chi(t)=\min\{1,\max\{0,t\}\} on \mathbb R, and define

The compressed selfadjoint element is in C and supported under e_K; since \chi(0)=0, continuous functional calculus in \Her(K) gives

This membership and support statement uses no relation from a representation.

Define the following additional fixed elements:

At the same index \eta, property \propP gives

Apply Lemma 2.2 to this comparison with cutoff \eta. If t_R is a ramp of R, zero on [0,2\eta] and one on [4\eta,\infty), the two carrier columns times t_R give contractions T_3,T_4\in C such that

Their left supports lie under e_{(R-\eta)_+} and their right supports under e_{(R-2\eta)_+}. Both are under e_{x'}\le e_{(X-\alpha_0)_+}. Transport preserves the represented norms, so

In a representation satisfying (4.2), B_3=T_3. Apply Lemma 4.1 to x',Q',R. The spectral subspace of B_3B_3^* above h is killed by Q' and lies under e_{(R-\eta)_+}, so it is zero. Equation (4.11) then gives

Passing only fixed elements to an ultraproduct. Suppose, for a contradiction, that every index in \Lambda admits a representation satisfying (4.2). Choose one, denoted \pi_\lambda, on its nondegenerate essential subspace. There is no size issue in this choice: from any such representation, choose a vector detecting \|\pi_\lambda(X)\|>s and restrict to its reducing cyclic subrepresentation. A set of cyclic representations suffices. Choose an ultrafilter \omega containing all tails of \Lambda, and form the norm ultraproduct

The norm of a class is \lim_{\omega}\|z_\lambda\|. Let \Phi:C\to\mathcal B be \Phi(z)=[(\pi_\lambda(z))], and write B=\Phi(C). The image is a closed C^*-subalgebra. The unital extensions carry the unitized formulas into C^*(B,1_{\mathcal B}). We suppress \Phi on fixed elements from now on.

For any fixed w\in C, eventually w\in E. Positivity of the summands in (4.4) and (4.6) yield

on that tail. Since \|U-(U-\eta)_+\|\le\eta,

It follows that UBV=0 in the image algebra. The other exact relations and bounds at each index give

For HB_i^*=B_i^*, use e_VT_i^*=T_i^* and H=1 on the support of V. Fix also a continuous h_P:[0,1]\to[0,1] with h_P(0)=0 and h_P=1 on [1/2,1]. The left support of T_4 lies under e_{x'}, so we obtain

Every element in (4.14) and (4.15) is the image of a fixed algebra element or of its fixed unitized functional calculus. We have not placed the varying T_i,D,D_0,t_R in B, and we do not require proper infiniteness of B or of \mathcal B. The directed finite sets in (4.13) handle all w\in C without a separability assumption.

A nonzero defect gives the contradiction. Since \|X\|\ge s>\beta, the image V=(X-\beta)_+ is nonzero. Choose an irreducible representation of B in which V\ne0. Extend this nondegenerate representation to C^*(B,1_{\mathcal B})=B+\mathbb C1_{\mathcal B} by sending 1_{\mathcal B} to the identity operator; this remains valid if B has a unit different from 1_{\mathcal B}. Thus the unitized expressions Q,b_0,Q' and the defect below have their stated represented meanings. Again suppress the representation symbols. The closed span of BV\mathcal H is a nonzero reducing subspace, and hence all of \mathcal H. The relation UBV=0 therefore gives U=0. Put D_1=B_1B_1^*. Since D_1\le h1, we have Q=h1-D_1, and

By closure D_1=h1 on e_x\mathcal H. The support of H lies there. Equations (4.7) and (4.14) give

Also HV=V\ne0, so H\ne0 and r_2\ne0.

The conditional calculation preceding (4.8) now applies: v=r_1+b_0 is an isometry with v^*r_2=0. Its defect projection

is nonzero, since Pr_2=r_2\ne0.

We check that this projection is the represented fixed element P_0, a point for which both sides of the cutoff matter. The scalar ramps give KH=H. From Hr_1=r_1 follows Kr_1=r_1, and the definition r_1=h^{-1/2}B_1^*H gives r_1K=r_1. Also K commutes with b_0. Expanding the defect gives

Each term is fixed by multiplication by K on both sides. Therefore KPK=P, and \chi(P)=P because P is a projection. Formula (4.8) yields

Now x'=(P_0-1/2)_+=P/2. Put D_3=B_3B_3^*\le h1; then Q'=h1-D_3. The relation R=0 gives Q'x'=0, so

Equations (4.15) and (4.17) say B_4=PB_4. But B_3^*B_4=0, so D_3B_4=0, whereas (4.18) gives D_3B_4=hB_4. Thus B_4=0, and B_3^*B_3=B_4^*B_4 gives B_3=0. This contradicts (4.18), since h>0 and P\ne0.

The contradiction shows that some index has no representation satisfying (4.2). The four entries chosen at that index prove the proposition.□

Proof. Every positive contraction of C is rich under \propP, since each of its scalar cutoffs is properly infinite. In Proposition 3.1, fix k=4 before the measurements are computed, and choose, for example, \gamma_g=1/2 and \sigma_g=1/(2\sqrt5). Then \gamma_g^2+4\sigma_g^2<1. The graph proposition gives fixed X and B_1,\ldots,B_4 for the inputs a,y,c. Apply Proposition 4.2 to these elements and use its contractions T_i as the graph choices S_i. The resulting rich positive contraction p\in\Her(a) has the required norm-one signal for a.

Put d=y(1-c^*p^2c)y. If a representation \pi had \pi(d)=0 and \|\pi(y)\|>1/4, the graph implication would give \|\pi(X)\|>1/4 and \pi(B_i)=\pi(T_i) for every i. These are exactly the relations excluded by Proposition 4.2. Thus

for every representation. The same p is rich and carries the a-signal, so Lemma 3.2 applies to a,y,c,p and proves the stated conclusions. This argument also covers zero inputs.□

Proof. For an empty index set, D=0 and the zero row has the asserted properties. Otherwise, for each i, put

The positive h_i is properly infinite. Apply Lemma 2.2 to h_i\oplus h_i\precsim h_i with cutoff \delta. If v_i is the resulting rectangular partial isometry, then

The algebra membership follows from transport regularity; no membership of v_i itself is needed. Define

This continuous function vanishes at zero, is bounded by \delta^{-1/2}, and satisfies t^{1/2}\phi_\delta(t)=1 for t>\delta. Consequently

The uniform bound places (s_i) in M_{1,2}(D). Coordinate functional calculus gives (5.1). Its right-hand side converges to d\oplus d in norm as \delta\downarrow0, proving proper infiniteness of d. Scaling covers every positive element of D.□

Proof. If no such bound existed, for each m\ge2 we could choose an algebra C_m satisfying \propP and contractions a_m,y_m,c_m, with the first two positive, such that every pair with the stated membership and simultaneous signals has support norm greater than 1-1/m. Form their bounded product D and the three elements a=(a_m), y=(y_m), c=(c_m)\in D. Lemma 5.1 gives \propP for D, so Proposition 4.3 produces p,q with the signals and \kappa=\|e_pce_q\|<1.

For each m, the coordinate homomorphism D\to C_m preserves hereditary membership, and its normal extension sends the two support projections to the coordinate supports. Thus the coordinate support norm is at most \kappa. If \pi is any representation of C_m, its composition with the coordinate homomorphism is a representation of D, so both signals hold for the coordinate pair as well. Choosing m with 1-1/m>\kappa contradicts its selection. The resulting constant can be increased, if necessary, to lie in (0,1).□

Proof. Fix \theta from Lemma 5.2. On [0,1] set

Both vanish near zero, f(1)=1, and g=1 on [1/2,1]. For a positive contraction h,

Apply Lemma 5.2 once to a_*,b_*,c_*, obtaining p_1,q_1 with support norm at most \theta. Suppose inductively that \|e_{p_k}c_*e_{q_k}\|\le\theta^k. The next normalized input is

Apply the uniform lemma to f(p_k),f(q_k),c^{(k)}. The new supports lie under those of f(p_k) and f(q_k); by (5.3),

The hereditary algebras are nested. In every representation, norm one of p_k implies norm one of f(p_k), by spectral mapping and f(1)=1, and similarly for q_k. Thus the two original signals persist at every step. All positive contractions needed as first inputs are rich under \propP. Choose k with \theta^k<\zeta.□

Proof. Let a_*,b_* be ramps of a,b, zero on [0,\delta] and one on [2\delta,1]. Put M_c=\max\{1,\|c\|\}. Choose a finite iteration in Proposition 5.3 so that its outputs satisfy the signals and

For example, apply the construction to c/M_c and take k with M_c\theta^k<\delta\zeta. This choice precedes the comparison coefficients below.

It remains to realize the diagonal cutoffs through p and q with a coefficient bound depending only on \delta and \rho.

Put d_a=(a-2\delta)_+. If an irreducible representation detects d_a, its norm on a is greater than 2\delta. It then has norm one on a_*, and the signal gives norm one on p. Lemma 2.3(ii) yields d_a\in\Id(p). Since

we have \Id(pap)=\Id(p): one inclusion is immediate, and the displayed order gives the other. The positive pap is properly infinite under \propP, so Lemma 2.1(iii) gives d_a\precsim pap.

Compressing a stabilization comparison coefficient to the scalar source and target corners supplies w\in C with

Set s=pw. Then e_ps=s, s^*as=w^*papw, and

In particular the more general estimate \|pw\|^2\le(1+\rho)/\delta holds without the restriction \rho\le2\delta. The same argument for b,q gives t=qw', e_qt=t, and the identical bounds. Because \|a-d_a\|\le2\delta, these comparisons prove (5.5). Finally

The bound in (5.8) depends on \delta,\rho, not on p,q or the separation accuracy. This is why the preceding choice (5.7) remains valid after the comparison coefficients are found.□

Proof of Theorem 1.1. For arbitrary a,b\in C_+, c\in C, and \varepsilon>0, put M=\max\{1,\|a\|,\|b\|\}. Choose 0<\delta<1/2 and 0<\rho\le2\delta with M(2\delta+\rho)<\varepsilon. Apply Proposition 5.4 to a/M,b/M,c with \zeta=\varepsilon. Multiplying the two diagonal inequalities by M gives all three inequalities in (1.2).

For positive x,y\in C, specialize these inequalities to (a,b,c)=(x^2,y^2,xy). They are precisely (1.1), so C is strongly purely infinite. This uses the unrestricted definition directly. If one instead writes a block matrix Z=\left(\begin{smallmatrix}a&c\\c^*&b\end{smallmatrix}\right), the norm of \operatorname{diag}(s,t)^*Z\operatorname{diag}(s,t)-(a\oplus b) is at most the maximum of the two diagonal errors plus \|s^*ct\|. Hence the same estimates also give diagonal approximation of every such block matrix, in particular of every positive one. The argument includes zero elements and the zero algebra.□

Proof. Take a nonexact C^*-algebra W, whose existence is recorded in [18, Remark (i) following Theorem 1, pp. 163–164]. There are I\triangleleft J such that the canonical surjection

is not injective. The map is surjective because its algebraic image is dense and the range of a C^*-homomorphism is closed. Choose a representative T_0\in W\otimes_{\min}J of a nonzero class in its kernel. Its norm modulo W\otimes_{\min}I is positive, whereas its image in W\otimes_{\min}(J/I) is zero. A sufficiently close finite algebraic approximation to T_0 retains a strict gap between these two norms. After normalizing its nonzero first factors and absorbing their norms into the second factors, we obtain contractions z_j\in W and elements h_j\in J for which

For a tuple \mathbf v=(v_j)_{j=1}^{\ell} in a C^*-algebra D, write

We show that these functions determine both norms in (6.3), with the coefficient algebra and its ideal held fixed.

For arbitrary coefficients d_j\in J, represent the fixed nonzero algebra J faithfully on a Hilbert space \mathcal H. Then

where V:\C^m\to\mathcal H ranges over finite-dimensional isometries. Indeed, also represent D faithfully. The operators on the right are the two-sided compressions in the second Hilbert factor, so their norms are at most the left-hand norm. Finite-rank projections, directed by inclusion, converge strongly to the identity. Their two-sided compressions converge strongly to the tensor sum; evaluation at a fixed vector whose image nearly attains its norm gives the reverse inequality. This applies to arbitrary, including nonselfadjoint, tensor sums.

For the quotient norm, choose a positive contractive approximate identity (f_\lambda) of the fixed ideal I. Put J_D=D\otimes_{\min}I. For T\in D\otimes_{\min}J, define

using the unitized multiplier operations. Elementary tensors and density give

The cutdowns are contractions. Hence, for every fixed K\in J_D,

Taking lower and upper limits and then the infimum over K yields the net limit

The cutdown coefficients lie in J, and no quasicentrality is needed. This argument uses convergence on each fixed element of J_D. A related fixed-element cutdown mechanism appears in [18, proof of Lemma 3, p. 165]; the calculation above establishes the stated formula for general I\triangleleft J and its approximate-identity net. Equations (6.4) and (6.5) show that the displayed quotient norm depends only on the functions \mu_{\mathbf v,m}. Applying the same compression argument in J/I shows the same for \|\sum_j v_j\otimes(d_j+I)\|_{\min}.

If a tuple in an exact algebra had the same functions \mu_{\mathbf z,m} at every matrix level, the two formulas above would transfer this strict gap to that algebra. Exactness makes its canonical quotient map isometric, a contradiction. This proves (i). The conclusion excludes simultaneous equality for all finite matrix tests; it does not assert that a fixed finite list of tests suffices.

For (ii), set W_0=C^*(z_1,\ldots,z_\ell). This algebra is separable, and its inclusion in W preserves matrix norms. It has a faithful representation on a separable Hilbert space. Indeed, choose a dense sequence in the positive unit sphere and, for each member, a state attaining its norm. Their GNS representations have faithful direct sum: for every nonzero positive element, approximation by that sequence shows that some chosen state is positive on it. Choose increasing nonzero finite-rank projections P_r\to1 strongly in that representation, and set

These are contractions. For each fixed m and (A_j), the corresponding two-sided compressions of \sum_j A_j\otimes z_j converge strongly to that operator and have norms at most its norm. Evaluation at a fixed near-norming vector gives the matching lower limit, proving (6.2). The same sequence works for every fixed test. The assertion is pointwise in the test, and uses no multiplicativity of the compressions.□

Proof. Put g=(X-\alpha_0)_+, choose 0<\nu<s-\alpha_0, and put h_g=(g-\nu)_+. Choose a ramp t'=t'(X) that is zero on [0,\alpha_0+\nu] and one on [s,\infty). Then t'\in\Her(h_g), and

The element v=g^{\oplus n}\in M_n(C) is properly infinite by \propF, applied to the scalar positive g. For the dimensions N_r of Lemma 6.2,

The first comparison uses g as a summand of v; the second uses proper infiniteness of v. Lemma 2.2, with cutoff \nu, gives one rectangular carrier

Its conjugation sends M_{N_r}(\Her(h_g)) into M_n(\Her(g))=\Her_{M_n(C)}(v). Use this same carrier for every j, defining

Each matrix has norm at most one, so its entries are contractions in the required hereditary subalgebra.

Suppose that every r admits a representation \pi_r satisfying (6.6) for these entries. Restricting to a reducing cyclic subrepresentation that still detects \|\pi_r(X)\|>s, we may make these choices from a set of representations. Write B_j=[B_{j,pq}]\in M_n(C); these elements are fixed independently of r. At every matrix level m, the normal amplification of the single w_r is isometric on the supported source algebra, by Lemma 2.2. The represented entry equalities and (6.7) therefore give, for every A_j\in M_m(\C),

This equality is exact at each r. Using a different carrier for each j would not give the joint equality.

Define the *-homomorphism

where the numerator is the algebra of bounded sequences and the denominator is its ideal of norm-null sequences. Contractivity of the representations makes each displayed sequence bounded. The image E=\Psi(M_n(C)) is closed and isomorphic to M_n(C)/\ker\Psi. Since M_n(C)\cong M_n\otimes C, nuclearity of matrix algebras and permanence under tensor products make M_n(C) nuclear [2, II.9.4.2 and IV.3.1.1]. Its quotient E is nuclear by [5, Corollary 4], and hence exact by [2, II.9.6.6]. These permanence facts require neither units nor separability. It is this quotient of a nuclear algebra that must be exact; the argument requires no exactness of the ambient \mathcal Q.

The norm of a class modulo norm-null sequences is the \limsup of its coordinate norms. Subtracting a norm-null sequence gives the lower bound, and subtracting finitely many initial entries gives the upper bound. The same formula holds at each fixed matrix level under the canonical identifications of finite matrices over a product and its norm-null ideal. Since E\subset\mathcal Q is completely isometric, (6.9) and (6.2) give

for every m and every (A_j). Only the fixed elements B_j have been sent into E; the varying choices in (6.8) were used in the coordinatewise equalities. The displayed identity contradicts Lemma 6.2(i). Thus some r admits no such representation, and its entries in (6.8) prove the proposition.□

Proof. Choose \ell from Lemma 6.2, set k=\ell n^2, and choose graph coefficients \gamma_g,\sigma_g>0 with \gamma_g^2+k\sigma_g^2<1. With these choices fixed, apply Proposition 3.1 to obtain its measurements X,B_i. Group the B_i into \ell matrices B_j=[B_{j,pq}]\in M_n(C), and choose the corresponding S_{j,pq} by Proposition 6.3. The graph gives a rich positive contraction p\in\Her(a) with \|\pi(p)\|=1 whenever \|\pi(a)\|>1/2. If d=y(1-c^*p^2c)y vanishes in a representation where \|\pi(y)\|>s, the graph would give all of (6.6), contrary to the choice of the entries. Thus

for every representation. Lemma 3.2 applies to this p, its norm signal, and the displayed implication, and gives the asserted p_1,q_1. The one fixed n controlled both the number of entries and the algebra M_n(C) in the quotient argument; only the finite dimensions N_r varied.□

Proof. We will find n orthogonal targets in \Her(u), each detected by \pi, and one positive element comparing into every target. Transporting n copies of its cutoff will allow \propF to produce richness inside \Her(u).

Put \mathcal H_u=\overline{\pi(u)\mathcal H}. This space is infinite dimensional. Otherwise \pi(u) would have finite rank r>0. Applying the stabilized representation to proper infiniteness of u^{\oplus n} would approximate \pi(u)^{\oplus2n}, of rank 2nr, by positive operators of rank at most nr. On the 2nr-dimensional range of the target, each such approximant has a unit vector in its kernel. Its distance from the target is therefore at least the least positive eigenvalue of \pi(u), contradicting norm approximation.

The action of D=\Her(u) on \mathcal H_u is irreducible. Indeed, \pi(u) has no kernel on its support, so \pi(u)\xi\ne0 for 0\ne\xi\in\mathcal H_u. Irreducibility of \pi gives \overline{\pi(C)\pi(u)\xi}=\mathcal H. Applying \pi(u) and taking closure yields

Thus every nonzero vector is cyclic for the restricted hereditary algebra, which proves irreducibility.

Apply the selfadjoint form of Kadison transitivity [9, Theorem 1] to the irreducible unitization of this action. Choose n+1 orthonormal vectors and prescribe n+1 distinct real eigenvalues. A selfadjoint interpolant has the form h_0+\lambda1, with h_0\in D_{\rm sa}. The corresponding eigenvalues of \pi(h_0) remain distinct, so at most one is zero. Choose n nonzero ones. Nonnegative continuous functions supported in pairwise disjoint neighborhoods of these values, avoiding zero and positive at the chosen values, give orthogonal elements

The functions vanish at zero, so these elements belong to D also when D is nonunital.

There is an element b'\in D_+, still detected by \pi, with b'\precsim b_i for every i. Start with e'=b_1. Given the current e' and the next b_i, choose a vector on which \pi(e') is nonzero. Cyclicity for D and nonzero \pi(b_i) give h\in D such that \pi(z)\ne0, where z=b_ihe'. Then

Replacing e' by z^*z preserves nonzero image and every previous comparison, by transitivity. At the end we obtain b' and

The middle equivalence uses orthogonality, and the last comparison uses \sum_i b_i\in\Her(u).

Choose \tau>0 with 2\tau<\|\pi(b')\|. Transport the n source copies in (6.12), after cutoff \tau, into \Her(u). Let t be a ramp of b', zero on [0,\tau] and one on [2\tau,\infty). It belongs to the common cutoff hereditary algebra and has represented norm one. The sum p of its n orthogonal transported copies is a positive contraction in \Her(u) with \|\pi(p)\|=1. By (2.7), for every \varepsilon\ge0,

The right side is properly infinite by the hypothesis applied to the scalar positive (t-\varepsilon)_+. Thus p is rich.□

Proof. All comparisons remain in C\otimes\Kc, and all uses of shrinking are in C. Since b is properly infinite, b^{\oplus n}\precsim b. Apply Lemma 2.2 with cutoff 1/8 to obtain orthogonal copy maps

Let y_0 be a ramp of b, zero on [0,1/4] and one on [1/2,1], and set a_0=a. For 1\le j\le n we construct positive contractions a_j,y_j such that a_j is rich and belongs to \Her(u), y_j belongs to the common domain of the copy maps, and the following invariants hold:

Once the induction reaches j=n, we will set c_0=a_n+\sum_{i=1}^n\theta_i(y_n). The induction will make the summands orthogonal, and the final use of \propF will prove the sum of the n copies rich without requiring y_n itself to be rich.

Suppose a_{j-1},y_{j-1} have been constructed. At j=1, use the initial pair a_0,y_0; at later stages use the preceding invariants. Apply Proposition 6.4 to a_{j-1},\theta_j(y_{j-1}), and 1\in\widetilde C, and call its outputs p_j,q_j. Thus p_j is rich and belongs to \Her(a_{j-1}), while

The hereditary-image equality (2.4) supplies a positive contraction z_j\in\Her(y_{j-1}) such that

This pullback is an algebra element. Its represented norms equal those of q_j by (2.5).

For 0\le t\le1, put f(t)=(2t-1)_+ and g(t)=\min\{1,2t\}. We next move p_j away from a cutoff of q_j. The support bound ensures that multiplication by 1-g(q_j) is bounded below on the support of p_j. The polar part of (1-g(q_j))p_j will therefore transport p_j, preserving its cutoff Cuntz classes and represented norms. Define

Here T_j=p_j-g(q_j)p_j\in\Her(u). In the universal representation, g(q_j) is supported under e_{q_j}. For \xi\in e_{p_j}\mathcal H, this gives

Since H_j is a contraction, we obtain the order bounds

Let T_j=l_j|T_j| be its polar decomposition in the bidual. The order bounds give \Her(|T_j|)=\Her(p_j) and l_j^*l_j=e_{p_j}. The polar membership argument in the proof of Lemma 2.2, applied now to T_j, gives

It is a positive contraction. The same argument makes each transported cutoff Cuntz equivalent to the corresponding cutoff of p_j, so a_j is rich. The normal extensions of the representations preserve l_j^*l_j=e_{p_j}; conjugating back by l_j^* shows that a_j and p_j have equal norm in every representation.

We check that this change preserves every earlier orthogonality relation. Since gf=f, functional calculus and (6.15) give

and hence \theta_j(y_j)T_j=0. For i<j, the previous orthogonality and p_j\in\Her(a_{j-1}) give \theta_i(y_{j-1})p_j=0, while orthogonality of the copy ranges gives \theta_i(y_{j-1})g(q_j)=0. Thus \theta_i(y_{j-1})T_j=0. Each of these positive elements annihilates \overline{\ran(T_j)}, whose projection is l_jl_j^*, and so annihilates a_j. Because y_j\in\Her(y_{j-1}), the same is true of \theta_i(y_j) for i<j. This proves the orthogonality in (6.14).

The norm signals also persist simultaneously. Wherever \|\pi(a)\|>1/2, the first shrinking step gives \|\pi(p_1)\|=1, hence \|\pi(a_1)\|=1; that signal is above 1/2 and survives every subsequent step. Wherever \|\pi(b)\|>1/2, the initial ramp has \|\pi(y_0)\|=1. If \|\pi(y_{j-1})\|=1, its copy under \theta_j has norm one, shrinking gives \|\pi(q_j)\|=1, and transport gives \|\pi(z_j)\|=1. Since f(1)=1, spectral mapping gives \|\pi(y_j)\|=1.

Finally set

All summands are orthogonal positive contractions in \Her(u). Their sum is a contraction, and in each representation its norm is the maximum of the summand norms. The preceding signals prove (6.13). For every \tau\ge0, orthogonal functional calculus and (2.7) give

The first summand is properly infinite because a_n is rich. The second is properly infinite by \propF, applied to the scalar positive (y_n-\tau)_+. Their block sum is properly infinite, so c_0 is rich. The construction includes zero inputs through zero copy maps and zero cutoffs.□

Proof of Theorem 6.1. Fix u\in C_+, and let \mathcal R_u be the set of rich positive contractions in \Her(u). Let J be the closed ideal generated by all (p-1/2)_+ with p\in\mathcal R_u. We claim that u\in J. Otherwise an irreducible representation of C/J would detect u+J. Its composition with the quotient map is an irreducible representation of C, so Lemma 6.5 gives p\in\mathcal R_u with represented norm one. Then (p-1/2)_+ has nonzero image, contrary to its membership in J.

The ideals generated by finite subfamilies of these cutoffs form an increasing directed family with dense union in J. For \varepsilon>0, Lemma 2.3(i) therefore puts (u-\varepsilon)_+ in the ideal generated by (p_i-1/2)_+ for finitely many p_1,\ldots,p_r\in\mathcal R_u. Starting from the rich zero element, merge them successively by Lemma 6.6. The result is a rich p\in\Her(u) for which every representation satisfies

A norm-one signal survives the next merger because it is above 1/2. The primitive ideal test in Lemma 2.3(ii) now gives (p_i-1/2)_+\in\Id(p) for every i. Consequently

If the cutoff is zero, the zero element already suffices.

Since p is properly infinite, ideal comparison with this target in the stabilization, followed by hereditary comparison, yields

The target u is fixed as \varepsilon\downarrow0. Explicitly, choose \varepsilon_j\downarrow0 and, for each j, a comparison coefficient in (6.18) with error less than \varepsilon_j to the doubled cutoff. That cutoff differs from u\oplus u by at most \varepsilon_j. The resulting coefficients therefore compare u\oplus u to u in norm. Thus u is properly infinite. The fixed n was applied only to scalar positives and their scalar cutoffs throughout the proof.□

Proof of Corollary 1.3. Set C=A, the separable nuclear algebra in the corollary, with its stated fixed-amplification hypothesis. Theorem 6.1 gives \propP, and Theorem 1.1 gives strong pure infiniteness. The general nonunital absorption implication is recorded in the paragraph following [13, Definition 3.1]. We spell out the central-sequence interface that includes the nonunital case.

Let C_\omega be the norm ultrapower of C along a free ultrafilter on \N, with C embedded diagonally, and put

Suppose first that C\ne0. Separability makes C \sigma-unital, so F(C) is unital by [10, Corollary 1.10(1)]. The central-sequence embedding statement immediately before [10, Proposition 4.4] gives a unital embedding \Oinf\hookrightarrow F(C): its ambient original algebra is the strongly purely infinite C, and the diagonal copy of C is the required separable nuclear subalgebra of C_\omega. The algebra \Oinf is unital and separable, and Remark 4.6 supplies self-absorption and connectedness of its unitary group. The “in particular” clause of Proposition 4.4(4) therefore gives stable \Oinf-absorption. Since U(\Oinf)=U_0(\Oinf), where U_0 is the identity component, the commutator subgroup of U(\Oinf) lies in U_0(\Oinf); part (5) therefore removes stabilization. Hence C\cong C\otimes_{\min}\Oinf. The Proposition permits C to be nonunital; the unital embedding is into the annihilator quotient F(C). If C=0, the asserted isomorphism is immediate.□

Proof. We first embed the full group algebra Q=C^*(\F_2) into \U. The two free generators are sent to

To see that the resulting homomorphism is injective, take any pair of unitaries S,T on a Hilbert space. Bounded Borel functional calculus gives selfadjoint contractions H,K such that S=\exp(\pi iH) and T=\exp(\pi iK). The contraction (H+iK)/2 evaluates the two displayed unitaries to S,T. Thus every unital representation of Q is a restriction of an evaluation of \U, and the homomorphism is isometric. The logarithms need only belong to the bounded operators on the representation space. Courtney and Sherman give a related embedding with a different normalization in [6, Theorem 6.10].

Nonexactness of C^*(\F_2) was established by Wassermann [23]. Here is an explicit proof for Q. We construct a representation of Q\otimes_{\min}Q that annihilates Q\otimes_{\min}I, where I is the kernel of the regular quotient. Exactness would make it descend to Q\otimes_{\min}(Q/I). A concrete tensor sum, however, has spatial norm in that algebra smaller than the norm the proposed descended representation would assign to it. This will exhibit the required failure of the tensor quotient map. Put G=\F_2, and write u_s for its canonical unitaries in Q. On \ell^2(G) let

There is a state \varphi on Q\otimes_{\min}Q with

For completeness, finite quotients supplying this state can be constructed directly. On the finite word ball of radius m in G, partially defined left translation by each of the two generators is an injection between subsets of equal size, hence extends to a permutation of the ball. Freeness gives a homomorphism q_m from G to the finite permutation group generated by these extensions. The inverse of each extension agrees with inverse left translation whenever both endpoints lie in the ball. If a reduced word has length at most m, applying its permutations to the identity follows its successive suffixes inside the ball and ends at that word. Consequently q_m distinguishes distinct elements of the word ball of radius m.

For the finite group F_m=q_m(G), its commuting left and right regular representations give a representation of C^*(F_m)\otimes_{\min}C^*(F_m): these algebras are finite direct sums of matrix algebras, for which the spatial and maximal tensor norms agree. Pulling its vector state at the identity back to Q\otimes_{\min}Q gives the coefficient \boldsymbol{1}_{q_m(s)=q_m(t)} on u_s\otimes u_t. These coefficients tend to those in (7.3). The group tensors span a dense *-subalgebra, and all the states have norm one, so their limit on that span extends to a state on the spatial completion.

The GNS representation of \varphi is precisely

with cyclic vector \delta_e. Indeed the cyclic vector associated with u_s\otimes u_t corresponds to \delta_{st^{-1}}; the inner products agree because

and left multiplication by u_x\otimes u_y becomes \lambda_x\rho_y. These vectors span \ell^2(G).

Let I=\ker(Q\to C_r^*(G)). The linear unitary R\delta_g=\delta_{g^{-1}} satisfies R\lambda_sR^*=\rho_s. Thus the representation in (7.4) annihilates Q\otimes_{\min}I. Exactness of Q would make it descend to a representation of Q\otimes_{\min}C_r^*(G), in which every u_s\otimes\lambda_s fixes \delta_e.

This contradicts a spatial norm bound. In a faithful unital representation \pi:Q\to B(H), the unitary

satisfies V(1\otimes\lambda_s)V^*=\pi(u_s)\otimes\lambda_s. If a,b are the free generators, the norm of

in this faithful spatial representation is therefore the norm of the adjacency operator on the 4-regular tree. That norm is at most 2\sqrt3<4. To check the bound, orient each edge away from the identity, and for a finitely supported vector \xi bound the contribution of a parent–child edge by

Every nonroot vertex has three children and one parent, and hence total coefficient 2\sqrt3; the root has coefficient 4/\sqrt3\leq2\sqrt3. The quadratic-form bound extends by density and proves the asserted operator norm bound. In the purported descended representation the same sum has eigenvalue 4 on \delta_e, which is impossible for a *-representation. Thus Q is not exact. Its unital inclusion in \U and the inheritance proved above show that \U is not exact either.□

Proof. Contractivity of evaluations proves one inequality. Every unital representation of \U is evaluation at the image of u, so these evaluations also compute the matrix norm on the left. For the other inequality, evaluate at a contraction T\in B(H), and take a unit vector \xi=(\xi_1,\ldots,\xi_r)\in H^r arbitrarily close to attaining the norm of Z(T,T^*). If the entries of Z have degree at most m, let

where the empty word is included. This is a finite-dimensional nonzero space. Set D=P_{H_0}T|_{H_0}, so that D^*=P_{H_0}T^*|_{H_0}. Applying a word successively from the right never changes the relevant vectors under compression, since all its suffix vectors lie in H_0. Hence

and Z(D,D^*)\xi=Z(T,T^*)\xi. Taking the supremum over evaluations and over these vectors proves (7.5). The scalar norm-test formula is proved in [6, Proposition 2.2]; the finite-word compression used here is the argument in the proof of [6, Theorem 3.1]. Including all components of the test vector gives the matrix-valued assertion needed here.□

Proof. If E=0, take C_E=0. Otherwise choose a basis P_1,\ldots,P_k of E, with coordinate functionals f_1,\ldots,f_k, and choose polynomial representatives P_i=\sum_w a_{iw}w. Define

Constants are allowed in the representatives; the empty word has length zero. For Z=\sum_i B_i\otimes P_i\in M_r(E) and scalar unit vectors \alpha,\beta\in\C^r,

It follows that \|B_i\|\leq\|f_i\|\|Z\| for every matrix size r.

Given a finite contraction D, choose \pi,J as in the hypothesis. Both \pi(v) and D are contractions, and telescoping the two intertwining relations gives

for every word w, including words in both letters. More explicitly, the difference for a word A_1\cdots A_m is the sum of m terms, each consisting of contractions on either side of one difference \pi(v)J-JD or \pi(v)^*J-JD^*. Put J_r=1_{\C^r}\otimes J. The coefficient bound and (7.7) yield

Since J_r is an isometry and \pi^{(r)} is contractive,

Taking the supremum supplied by Lemma 7.3 proves (7.6). In particular, C_E depends on the fixed polynomial space and its chosen representatives, but on none of r,d,D,B,v,\pi,J.□

Proof of Proposition 7.1. If the Proposition were false, for every integer j\geq1 there would be a unital exact algebra B_j and a contraction v_j\in B_j satisfying (7.1) with 1/j in place of \gamma. Let \Phi_j:\U\to B_j send u to v_j. Fix a finite-dimensional polynomial space E. Before choosing any matrix amplification, tensor factor, compression, or ideal, choose one integer j=j(E)\geq\max\{1,2C_E\}. Lemma 7.4 then gives

This same j works after tensoring with an arbitrary C^*-algebra C: for every z\in E\odot C,

where \odot denotes the algebraic tensor product and all norms are spatial. The assertion is immediate if C=0. Otherwise, represent C faithfully on a nonzero Hilbert space K, and use faithful representations of \U and B_j for their respective spatial tensor products. Compress the C representation to a nonzero finite-dimensional subspace K_0\subset K. If z=\sum_i P_i\otimes c_i, its compression has coefficients P_{K_0}c_i|_{K_0}\in M_d(\C) and, after interchanging the tensor factors, belongs to M_d(E). Inequality (7.8) bounds its norm by twice the norm of the corresponding compression on the B_j side, which is at most 2\|(\Phi_j\otimes\id_C)z\|. The norms of these compressions have supremum \|z\|: as K_0 increases, the projections 1\otimes P_{K_0} converge strongly to the identity, and their two-sided compressions converge strongly to z. This proves (7.9).

Now fix any ideal I\subset C, with quotient map q:C\to C/I and a positive contractive approximate unit (e_\lambda) of I. For z\in E\odot C, every z(1\otimes(1-e_\lambda)) still belongs to E\odot C. Consequently (7.9) applies to the entire net with the single index j(E). Formula (7.2) and exactness of B_j give

Finite sums of polynomial tensors are dense in \U\otimes_{\min}C. Although j(E) may vary with the polynomial space, the resulting constant 2 does not. The last inequality therefore extends by continuity to every z\in\U\otimes_{\min}C. It says that the canonical surjective homomorphism

has zero kernel: if \kappa(z+\U\otimes_{\min}I)=0, the quotient class has norm zero. An injective *-homomorphism of C^*-algebras is isometric, so the fixed factor 2 suffices to prove that \kappa is an isomorphism. Since C,I were arbitrary, \U would be exact, contrary to Lemma 7.2.□

Proof. In the common stabilization,

Thus x^{\oplus k} belongs to the ideal generated by h^{\oplus n}. The proper-target ideal comparison in Lemma 2.1 gives (7.10). This argument takes place in the stabilization and therefore covers every finite rectangular corner and nonunital C. It also covers h=0, when x=0.□

Proof. If f(x)=0, take R=0. Otherwise, choose \delta>0 so that f vanishes on [0,\delta], and choose 0<\varepsilon<\delta. Then f(x)^{1/2}\in\Her((x-\varepsilon)_+). Apply Lemma 2.2 to x\precsim h at this cutoff, and let w\in M_{s,m}(C^{**}) be its rectangular partial isometry. Its initial projection is the support of (x-\varepsilon)_+. The membership and support conclusions of that lemma therefore give

The norm identity follows from \|R\|^2=\|R^*R\|. Only the product R, rather than w itself, is asserted to belong to the rectangular algebra.□

Proof. Finite matrix exactness follows, for every C^*-algebra D_0, from the identifications

which preserve the ideal kernels in the definition of exactness. For the unitization step, let A_0 be a nonunital exact algebra, let I\triangleleft D_0 be a closed two-sided ideal, and write q_0\colon D_0\to D_0/I. The scalar character \chi\colon\widetilde A_0\to\C gives, for every z\in\widetilde A_0\otimes_{\min}D_0, the bounded decomposition

Indeed, both projections are bounded, and the decomposition holds first on algebraic tensors and then on their completion. If (\id\otimes q_0)(z)=0, applying \chi\otimes\id gives d\in I. The isometric inclusion

then shows (\id\otimes q_0)(z_0)=0. Exactness of A_0 gives z_0\in A_0\otimes_{\min}I, hence z\in\widetilde A_0\otimes_{\min}I. If M_n(A) is already unital, this second step is unnecessary.□

Proof. Write K=\{P\in\Prim(C):\|c_P\|\geq\alpha\}. If K is empty, y=0 satisfies all conclusions. Suppose henceforth that K is nonempty. The contradiction will have three levels of choices. First we fix comparison rows and a single contraction in M_n(C). Each finite matrix contraction then determines an admissible perturbation. If it fails, a primitive quotient witnessing that failure supplies an approximate reducing copy for that fixed contraction. All these copies therefore come from one element of the same exact algebra \widetilde{M_n(C)}.

Fixed comparison rows. Choose a continuous function e\colon[0,\infty)\to[0,1] that vanishes near zero and is one on [\alpha/2,\infty). Apply Lemma 7.6 to the last two comparisons in (7.13). This gives rows R,S\in M_{1,n}(C) with

The relation p^*ph=h, together with its adjoint, shows that left and right multiplication by p^*p are the identity on \Her(h). Hence (1-p^*p)RR^*(1-p^*p)=0, which implies (1-p^*p)R=0. The same argument for S gives

At this point fix the single element

The rows R,S and the element V will not depend on any finite matrix model. Fix now also the global contraction

The inverse is evaluated in the unitization, so its product with V remains in M_n(C).

Reducing copies of arbitrary finite contractions. Let d\geq1 and let D\in M_d(\C) be a contraction. Put c_1=(c-\alpha/2)_+. Condition (F_n) and Lemma 7.5, applied to the scalar element c_1, give

Choose a continuous \varphi\colon[0,\infty)\to[0,1], vanishing near zero and equal to one on [\alpha/4,\infty), and put e_1(c)=\varphi(c_1). Lemma 7.6 yields L\in M_{n,d}(C) such that

Because e(c)c_1=c_1, multiplication by E is the identity on \Her(c_1^{\oplus n}). The range-square argument used for (7.16) gives EL=L. Define

Scalar matrix multiplication here requires no unit in C.

For each P\in K, choose an irreducible representation \pi_P\colon C\to B(H_P) with kernel P, and use its matrix amplifications. Identify C/P with the image of the induced faithful representation on H_P, and make the same identification at rectangular matrix levels. The spectral subspace 1_{(3\alpha/4,\infty)}(\pi_P(c))H_P is nonzero, since \|c_P\|\geq\alpha. Choose a unit vector \xi_P in that subspace, so that \pi_P(e_1(c))\xi_P=\xi_P. Define

Equation (7.18) proves J_P^*J_P=1_d. Both D and D^* preserve \C^d\otimes\xi_P, and the same equation gives

Thus one copy reduces both w_P and its adjoint. No eigenvector at the norm of c_P, and no continuous choice of the vectors \xi_P, is required.

Admissible tests and the two optional conclusions. For the chosen d,D, take

If h,k\in\mathcal D, then RR^*,SS^*\in\mathcal D. For a positive contractive approximate unit (a_\lambda) of \mathcal D, the identity

and its analogue for S give a_\lambda RwS^*a_\lambda\to RwS^* in norm. Each approximant belongs to \mathcal D C\mathcal D\subset\mathcal D, so y_0\in\mathcal D. This uses the range squares of the rows and makes no assertion that their individual entries belong to \mathcal D.

If p,q\geq0 and pq=0, then p^2R=R implies pR=R: multiply (1-p^2)R=0 by (1+p)^{-1} in the unitization. Likewise qS=S. It follows that pS=qR=0 and S^*R=0. Consequently y_0^2=0 and py_0^*q=0. The positive elements y_0y_0^* and y_0^*y_0 are orthogonal, whence

We may therefore use y=y_0+y_0^* when selfadjointness is requested, and y=y_0 otherwise. In both cases pyq=py_0q; hereditary membership is preserved if it is also requested.

One global contraction and the uniform gap. Suppose that no admissible y satisfies (7.14). For each d,D, choose a witness P\in K for the test just constructed. Then

Multiplying by the contractive outer factors R^*p^* and q^*S, and using (7.16), gives

Although P may depend on d,D, the element V is the one fixed in (7.17).

Functional calculus commutes with quotient maps for the fixed global clipping \widehat V. Hence in each witness quotient

The last strict inequality follows from (7.21) and \|w_P\|\leq1. The clipping identity follows by squaring the absolute value of V_P-V_P\max\{1,|V_P|\}^{-1} and applying scalar functional calculus. Combining (7.21), (7.22), and (7.20) gives

By Lemma 7.7, B=\widetilde{M_n(C)} is a unital exact algebra. The amplified representations \pi_P^{(n)} extend unitally to B. Although the witnessing P varies with d,D, the contraction \widehat V and the exact algebra B remain fixed. Thus \widehat V\in B has, for every d,D, a unital representation and an isometry satisfying both inequalities in (7.23). This contradicts Proposition 7.1 and proves the Proposition with both optional conclusions and the same \sigma.□

Proof. Fix an irreducible representation \pi:C\to B(H_\pi) with kernel P. First, no nonzero \pi(a), a\in C_+, has finite rank. Indeed, if its rank were d\in\N, applying the amplified representation to a^{\oplus 2n}\precsim a^{\oplus n} would express an operator of rank 2nd as a norm limit of operators of rank at most nd. This is impossible, since the set of operators of rank at most a fixed finite integer is norm closed.

Put H_h=\overline{\pi(h)H_\pi}. We claim that the restriction of \pi(\Her(h)) to H_h is strongly dense in B(H_h). Irreducibility and the bicommutant theorem [2, I.9.1.1] give strong density of \pi(C) in B(H_\pi). Consequently the strong closure of \pi(\Her(h)) contains \pi(h)B(H_\pi)\pi(h). For t>0, let E_t=1_{[t,\infty)}(\pi(h)). Every operator E_tTE_t, with T\in B(H_h) extended by zero on H_h^\perp, has the form \pi(h)S\pi(h): the inverse of \pi(h) on E_tH_\pi is bounded. As t\downarrow0, E_tTE_t\to T strongly on H_h. This proves the claim.

The space H_h is infinite dimensional by the finite-rank obstruction. Choose orthonormal vectors \xi_1,\ldots,\xi_m in H_h and distinct numbers \lambda_1,\ldots,\lambda_m\in(0,1). Let T\in B(H_h) be the positive contraction with T\xi_j=\lambda_j\xi_j and zero on their orthogonal complement. Choose nonnegative continuous functions f_j on \R, with disjoint compact supports in (0,1) and f_j(\lambda_j)=1. Kaplansky density [2, I.9.1.3] supplies a bounded net of selfadjoint elements b_\lambda of the represented hereditary algebra converging strongly to T. Continuous functional calculus on bounded selfadjoint sets is strongly continuous; this follows by uniform polynomial approximation on a common bounded interval. Hence f_j(b_\lambda)\xi_j\to\xi_j. One b_\lambda makes all these m vectors nonzero. Lift it to a selfadjoint b\in\Her(h). The elements f_j(b) lie in \Her(h), are pairwise orthogonal, and are nonzero at P. The functions vanish at zero, so this argument also applies to a nonunital hereditary algebra. This proves (i).

For two positives u,v nonzero at P, strong density of \pi(C) gives x\in C with \pi(uxv)\ne0. For example, choose a vector on which \pi(v) is nonzero and a rank-one operator carrying its image to a vector not annihilated by \pi(u); approximate that operator strongly on the image under \pi(v). With z=uxv, the positive d=zz^* satisfies

Here and below \CuEq denotes Cuntz equivalence, and the comparison rules are those of Lemma 2.1. For a finite list of positives nonzero at P, repeating this operation produces one positive, nonzero at P, subequivalent to every element of the list. At each step the new element lies in the hereditary algebra of the preceding one, so all previous comparisons and any prescribed hereditary membership persist.

Apply (i) with m=n separately to h and k, obtaining orthogonal lists h_1,\ldots,h_n and k_1,\ldots,k_n. Take a common element c for the resulting list of 2n positives. Then

and the same argument gives c^{\oplus n}\precsim k. The asserted choice in \mathcal D follows from the hereditary membership just observed.□

Proof. In the state space of the unitization \widetilde B, the set

is nonempty: evaluation at the largest spectral value of b on C^*(1,b) extends to a state of \widetilde B by the positive Hahn–Banach extension theorem [2, II.6.3.1]. It is a weak-star compact face, since every state has value at most \|b\| on b. By the Krein–Milman theorem it has an extreme point, which is therefore a pure state of \widetilde B; see [2, II.6.3.2–3]. Let \widetilde\pi be its irreducible GNS representation [2, II.6.4.8]. It does not annihilate B, because its value on b is positive. The closure of \widetilde\pi(B)H_{\widetilde\pi} is a nonzero reducing subspace for \widetilde\pi(\widetilde B), hence the whole space. If an operator commutes with \widetilde\pi(B), then its commutator with \widetilde\pi(x) annihilates \widetilde\pi(B)H_{\widetilde\pi} for every x\in\widetilde B, since xB\subset B. It therefore commutes with \widetilde\pi(\widetilde B). Thus the restriction \pi=\widetilde\pi|_B is nondegenerate and irreducible. The vector state gives \|\pi(b)\|\ge\|b\|, and contractivity gives equality.□

Proof. The first set is \{P:(c-\alpha)_+\notin P\}, which is open in the hull-kernel topology. For the second assertion, extend any relative open cover to ambient open sets. Write those sets as

where each J_\lambda is a closed ideal of C. If \|c+I\|\ge\alpha, Lemma 8.2 applied in C/I gives an irreducible representation attaining that norm. Its pullback has a primitive kernel containing I, belongs to K(c,\alpha), and lies outside every \mathcal O_\lambda. This contradicts the covering property. Thus \|c+I\|<\alpha; if C/I=0, the same inequality is immediate. Choose d\in I with \|c-d\|<\alpha, and then approximate d by an element d_F\in\sum_{\lambda\in F}J_\lambda for a finite index set F, closely enough that \|c-d_F\|<\alpha still holds. Any primitive ideal containing every J_\lambda, \lambda\in F, has \|c_P\|<\alpha. Hence the finitely many \mathcal O_\lambda, \lambda\in F, cover K(c,\alpha). The proof uses quasi-compactness only and does not assert that K(c,\alpha) is closed.□

Proof. If K_0 is empty, choose U=0 and the empty family. Otherwise, Lemma 8.1 associates to every original choice a positive c nonzero at its specified point, with c^{\oplus n}\precsim h,k. Choose 0<\alpha<\|c_P\|. The resulting open sets \Omega(c,\alpha) cover K_0, so choose a finite subcover \Omega_i=\Omega(c_i,\alpha_i), 1\le i\le N.

Set \theta=\sigma/4, \eta_i=\theta^{i-1}, and U_0=0. At stage i, apply Proposition 7.8 with T=p_iU_{i-1}q_i, scale \eta_i, and the selected data. The identities p_ih_i=h_i, q_ik_i=k_i imply p_i^2h_i=h_i, q_i^2k_i=k_i. The simultaneous hereditary and selfadjoint options give a selfadjoint contraction y_i\in\mathcal D such that U_i=U_{i-1}+\eta_i y_i satisfies

Let U=U_N. Later perturbations change the ith corner by at most

Consequently

In particular each selected corner is nonzero throughout its entire open set \Omega_i, proving the assertion.□

Proof. If K_0=\varnothing, take x=0. Assume henceforth that K_0 is nonempty. We first construct a selfadjoint element that is nonscalar on every relevant hereditary support space. Coupling its separated spectral windows and deleting the nearby spectral blocks will produce a selfadjoint element whose positive and negative parts are both nonzero at every point of K_0. A final coupling of those two orthogonal parts gives the required square-zero element.

First coupling: a nonscalar selfadjoint element. At each P\in K_0, choose h\in\mathcal D_+ with h_P\ne0. Lemma 8.1(i), with m=2, provides two orthogonal positives in \Her(h)\subset\mathcal D nonzero at P. Apply the preceding cutoff construction to these two positives and use the resulting four inputs in Lemma 8.4 to obtain U=U^*\in\mathcal D. For an irreducible representation \pi with kernel P, put H_{\mathcal D,\pi}=\overline{\pi(\mathcal D)H_\pi}. This space reduces \pi(\mathcal D), which acts as zero on its orthogonal complement. At P one selected pUq is nonzero. Since p,q\in\mathcal D and pq=0, this implies

In particular, nonscalarity is obtained on the support space of \mathcal D, without using its zero action on the complement.

Second coupling: separated spectral windows. At a specified P\in K_0, the selfadjoint operator in (8.2) has two distinct spectral values. Choose disjoint closed intervals I,J\subset\R at positive distance, and continuous nonnegative functions F,G supported in I,J, respectively, such that both F(\pi(U))|_{H_{\mathcal D,\pi}} and G(\pi(U))|_{H_{\mathcal D,\pi}} are nonzero. The functions may be chosen on all of \R with compact support. If a window contains zero, its function need not vanish at zero. We evaluate scalar functions in \mathcal D+\C1 inside a fixed unitization \widetilde C of C, and take spectral projections in \widetilde C^{**}. Let (d_\lambda) be a positive contractive approximate identity of \mathcal D. The positive elements

belong to \mathcal D. Their images on H_{\mathcal D,\pi} converge strongly to F(\pi(U))^2 and G(\pi(U))^2, respectively, so a choice of middle elements makes both images nonzero. These positives are supported, globally in the bidual, on 1_I(U) and 1_J(U).

The same cutoff construction gives the corresponding p,q,h,k, retaining these spectral supports. Apply Lemma 8.4 again to obtain V=V^*\in\mathcal D and finitely many selected pairs (p_i,q_i), supported in windows (I_i,J_i), such that

Put \rho=\min_i\dist(I_i,J_i)>0. The positive elements p_i,q_i need not commute with U; only the displayed support identities will be used.

Spectral deletion produces both signs. We retain the corners in (8.3) while removing all sufficiently narrow diagonal spectral blocks. A positive operator with all those diagonal blocks zero must vanish; the same holds for a negative operator. Since the retained corners prevent vanishing, this will force both signs in every quotient over K_0.

Fix a compact interval [-M,M] containing the spectrum of U in the unitization. Take a finite continuous nonnegative partition of unity f_1,\ldots,f_s on this interval, whose supports S_j have diameter less than \rho/8. Define

The summation uses ordered pairs, and its condition is symmetric in j,k, so H=H^*. Although a factor f_j(U) may have a scalar part, it lies in \mathcal D+\C1; the middle factor V\in\mathcal D implies H\in\mathcal D.

For each selected pair (p_i,q_i) we have

Indeed, expand p_iVq_i using \sum_jf_j(U)=1 on both sides. For a summand p_if_j(U)Vf_k(U)q_i to be nonzero, the supports S_j,S_k must meet I_i,J_i, respectively, by (8.3). Their distances then satisfy

Every potentially nonzero summand is therefore retained in (8.4). This argument requires no commutation of p_i,q_i with U. Equations (8.3) and (8.5) show that H_P\ne0 for every P\in K_0.

Let B be any Borel interval of length less than \rho/4. The two supports in a retained pair cannot both meet B, since their distance exceeds \rho/2. Hence

Partition [-M,M] into finitely many disjoint Borel intervals B_j of length less than \rho/4, assigning endpoints so that their spectral projections sum to the identity. In any unital extension of a representation, the resulting projections E_j thus satisfy E_j\pi(H)E_j=0 and \sum_jE_j=1. If \pi(H)\ge0, then

for every j,\xi, whence \pi(H)=0. Applying the same argument to -\pi(H) excludes a nonzero negative semidefinite image. The spectral projection calculations can equivalently be made in the bidual and passed through the normal extensions of the representations. We conclude that

Both positive parts belong to \mathcal D and are orthogonal.

Third coupling: the square-zero product. At every P\in K_0, apply the same cutoff construction to H_+,H_- to obtain p,h\in\Her(H_+) and q,k\in\Her(H_-) for Lemma 8.4. The contractions p,q are orthogonal. The resulting U'=U'^*\in\mathcal D has a selected nonzero corner at every point of K_0. Put

If x_P=0, then (pU'q)_P=0 for every p\in\Her(H_+) and q\in\Her(H_-): approximate them in norm by elements H_+aH_+ and H_-bH_-, and use

This contradicts the selected nonzero corner at P. Thus x_P\ne0 throughout K_0, while x^2=H_+U'(H_-H_+)U'H_-=0. All constructed elements lie in \mathcal D; spectral projections have only been used to verify their supports.□

Proof of Theorem 1.2. Fix a\in C_+ and r>0. The set K_0=K(a,r) is quasi-compact by Lemma 8.3. If it is empty, every primitive image of (a-r)_+ is zero, so (a-r)_+=0, and ((a-r)_+)^{\oplus2}\precsim a is immediate. Suppose that K_0 is nonempty. Proposition 8.5 applied to \Her(a) gives x\in\Her(a) with x^2=0 and x_P\ne0 throughout K_0. Set z_1=x^*x. Its Cuntz-equivalent positive xx^* is orthogonal to it, since (x^*x)(xx^*)=x^*x^2x^*=0. Therefore

Repeat the construction inside \Her(z_1), and then inside successive hereditary subalgebras, always over the same K_0. It yields z_j\in C_+ nonzero throughout K_0 and z_{j+1}^{\oplus2}\precsim z_j. Finite direct sums and transitivity give

Choose a finite j\ge1 with 2^j\ge n. Dropping extra summands gives z_j^{\oplus n}\precsim a. If (z_j)_P=0, then P\notin K_0, so \|a_P\|<r and ((a-r)_+)_P=0. The primitive ideal test in Lemma 2.3 therefore gives (a-r)_+\in\Id(z_j). Condition (F_n) makes z_j^{\oplus n} properly infinite because z_j is a scalar positive element. Lemma 7.5, used with two copies on the left, then implies

Only scalar positives z_j\in C have been tested against (F_n). Finally, \|a-(a-r)_+\|\le r. As r\downarrow0, norm-limit closure of comparison with the fixed target a gives

This includes a=0; when n=1, it also follows directly from the hypothesis. Therefore C satisfies (P). Theorem 1.1 now gives strong pure infiniteness, completing the proof.□

Proof. Property (P) makes h properly infinite, so Lemma 7.5 applies with multiplicity one. If h=0, the ideal is zero and the assertion is immediate. For exact C, property (P) is precisely the amplification hypothesis of Proposition 7.8 with n=1 for every scalar positive element. Its constant was fixed independently of n, so neither the constant nor the hereditary and selfadjoint options change.□

Proof. Write the polar decomposition in C^{**} as

All three positive elements a,b,u are contractions. Choose continuous functions on [0,1] to separate the small singular values of ab from a spectral corner on which an inverse square root is available. The small values will give small cross products; on the remaining corner we will preserve exact cancellation while perturbing the two sides separately. Use the following functions:

• •

  0\le L\le1, L=1 on [0,5\ell], and L=0 on [6\ell,1]; put N=1-L;

• •

  0\le g\le1, g=0 on [0,\ell], and g=1 on [2\ell,1];

• •

  0\le R\le1, R=0 on [0,\ell/2], and R(s)=\sqrt{\ell/s} for s\ge\ell.

Linear interpolation on the unspecified intervals gives such functions. A function letter without an argument denotes evaluation at u. L selects the small singular values, g confines the perturbations away from zero, and R normalizes u on that region. Functional calculus and the polar decomposition give

The second identity uses N(0)=0; the first remains valid with L(0)=1. Moreover

where both quotient functions are extended continuously by zero at s=0, since R and N vanish near zero. We use the unitization for scalar terms throughout this proof.

The four columns and their products. We first specify conditions under which four columns have the product estimates we need. Suppose that orthogonal positive contractions e,f\in C and invertibles Z,W\in\widetilde C satisfy

for some z,w\in C. For such a choice, define

These elements belong to C: the only polar factor is vR\in C, and all unitized factors occur multiplied by elements of C. For any such admissible choice, the proposed outputs are

The norm of each positive output dominates the norms of its two columns in every primitive quotient. At the end, a row polar decomposition will bound the product of these outputs by the four cross products estimated next.

Put E_R=RuR-\ell. Since E_Rg=0 and f=gf, we have

In particular, no commutation of W with g is needed. Using v^*ab=u and retaining the indicated order of factors gives

The other three products obey

Indeed,

Here \|uL(u)\|\le6\ell. For 0\le s\le\ell, \ell^{-1/2}R(s)sL(s)\le\sqrt\ell; for s\ge\ell the same expression equals \sqrt{s}L(s)\le\sqrt{6\ell}. Together with (9.4), these scalar estimates prove (9.8). We now construct e,f,Z,W so that, in each prescribed primitive quotient, at least one of the two relevant columns also has a fixed lower norm bound.

Two orthogonal cores. Set h_0=(u-2\ell)_+. Since h_0\oplus h_0\precsim h_0, Lemma 7.6 supplies a row [r_1,r_2]\in M_{1,2}(C) such that

where \theta\colon[0,\infty)\to[0,1] is continuous, vanishes near zero, and equals one on [2\ell,\infty). Define

They are orthogonal positive contractions in \Her(h_0); each belongs to this hereditary algebra because it is dominated by their sum. In any primitive quotient with \|u_P\|\ge4\ell, \|(h_0)_P\|\ge2\ell, and hence

This includes equality at the threshold 4\ell by the chosen plateau.

Put

and let e=\chi(e_0) and f=\chi(f_0), where \chi\colon[0,1]\to[0,1] is continuous, vanishes near zero, and is one on [1/2,1]. Then

For the last two identities, gh_0=h_0 implies that multiplication by g is the identity on \Her(h_0). By (9.10), both (k_e)_P and (k_f)_P are nonzero whenever \|u_P\|\ge4\ell.

The quotient regions requiring a high column. Define

We first check why a lower bound on the region \|(d_i)_P\|\ge\tau will recover any signal missed by the corresponding low column. Suppose \|a_P\|\ge r and \|(X_0)_P\|<r/2. Since a=aL(u_*)+aN(u_*), we have \|(aN(u_*))_P\|>r/2. The identity N(u_*)=vNv^* gives \|(avN)_P\|>r/2. On [0,1], the scalar inequality R^2N^2\ge\ell N^2 holds: where N is nonzero, s>5\ell and R(s)^2=\ell/s\ge\ell. Consequently

and therefore

It follows that \|(d_1)_P\|>2\tau. Similarly, if \|b_P\|\ge r and \|(Y_0)_P\|<r/2, then \|(bN)_P\|>r/2. The order inequality bR^2N^2b\ge\ell bN^2b gives \|(d_2)_P\|>2\tau. Thus every prescribed quotient either has a low column of norm at least r/2 or belongs to the corresponding perturbation region.

Comparison inputs and normalized left factors. We now verify the inputs to Proposition 7.8 on these two regions. If (d_i)_P\ne0, then C_iN=C_igN implies \|(C_ig)_P\|>2\tau, so (h_i)_P\ne0. The same assumption implies N_P\ne0, and therefore \|u_P\|>5\ell. Consequently (k_e)_P,(k_f)_P\ne0. The primitive ideal test and Corollary 8.6 give

Let D_i=C_ig, and use continuous functional calculus in the unitization to set

Thus \|F_i\|\le\tau^{-1} and p_i\in C. The functional-calculus intertwining identity

follows first for polynomials in D_iD_i^* and D_i^*D_i, and then by uniform approximation. In particular,

The use of |D_i^*| in the left factor F_i is essential to this identity.

Two compatible inverse perturbations. Fix

Apply Proposition 7.8, with n=1, to

at scale \eta. Its hypotheses follow from (9.11), (9.13), and (9.15). For the resulting contraction y\in C, put

Since \|Z_1-1\|\le\eta<1, the element Z_1 is invertible; define W_1=(Z_1^*)^{-1}. Since ge=e, the conclusion of the Proposition becomes

Next apply the same Proposition at scale \eta' to

The element T is in C; the Proposition imposes no bound on its norm. With the resulting contraction y'\in C, set

Since gf=f, the second conclusion is

We record the inverse estimates, including the factorizations needed to preserve (9.16). For a_0\in C with \|a_0\|<1, the Neumann series gives

The series of middle factors converges in C with norm at most \|a_0\|/(1-\|a_0\|). This formula does not require g to be a projection. Applying it to Z_1^* gives

Thus W=1+gwg for w=w_1+\eta'y'\in C, with

Applying (9.18) to W^*=1+gw^*g now proves that W is invertible. Define Z=(W^*)^{-1}. The same series gives

In particular, all conditions in (9.4) hold.

The second perturbation changes the inverse used in the first column. To keep the first lower bound independent of \ell, we must estimate this change without the potentially large factor \|F_1\|. The inverse difference has the exact order

As Zg=g(1+zg^2), its effect on the first normalized column is bounded by

Since 3\eta'/2\le\sigma\eta/2, (9.16) remains bounded below by \sigma\eta/2 with Z in place of Z_1. Using \|F_i\|\le\tau^{-1} in each primitive quotient, and \tau\ell^{-1/2}=r/8, we obtain

The cancellation \tau\ell^{-1/2}=r/8 is what removes \ell from both lower bounds.

Assembling the two positive outputs.

To check the hereditary membership of Q_a,Q_b, (9.5) and the bounds on Z,W give

By Lemma 2.1, the first two summands belong to \Her(a) and the last two to \Her(b). Thus Q_a\in\Her(a) and Q_b\in\Her(b). The earlier coverage reduction and (9.21), together with domination of the column norms by the output norms, prove (9.1) with

Here m\le r/2 and \eta'\le\eta/2, so this single lower bound applies to all cases. It is independent of \ell.

Finally put U_a=[X_0,X] and U_b=[Y_0,Y] in M_{1,2}(C). Their range squares are Q_a^2 and Q_b^2. The rectangular polar decompositions in the bidual have the form U_a=Q_av_a and U_b=Q_bv_b, where v_av_a^* and v_bv_b^* are the support projections of Q_a and Q_b, respectively. Hence both identities

hold. Since partial isometries have norm at most one, they imply

by (9.7) and (9.8). This proves (9.2).

The construction also covers zero elements and empty norm regions. If h_0=0, the row in (9.9) may be zero. Empty perturbation targets permit zero perturbations, and the coverage argument then assigns every prescribed quotient to a low column. All inverse operations have either the positive floor \tau or the convergent series (9.18).□

Proof. Choose a comparison implementer v\in C with

Such an implementer lies in C by compression to the two scalar matrix corners. On [0,\|Q\|] define the continuous function

Then (Qx)^*(Qx)=v^*(Q^2-m^2/4)_+v and \|Qx\|^2<1+\delta<2, which in particular gives \|Qx\|\le2. Moreover the scalar inequality

holds because j vanishes on [0,m/2]. Functional calculus followed by multiplication by v^* and v gives

Thus \|x\|\le(2/m)\|Qx\|\le4/m, independently of the norm of the original comparison implementer v.□

Proof. Fix the constant \sigma in Proposition 7.8 with n=1. Let a,b\in C_+ be contractions and let \varepsilon>0. Choose 0<r<1 with r^2<\varepsilon/3 and put m=r\sigma^2/1024. Next choose 0<\ell<1/10 so small that

Take Q_a,Q_b from Lemma 9.1. If (a^2-r^2)_+ has nonzero image at P, then \|a_P\|>r, so \|(Q_a)_P\|\ge m. Thus (Q_a^2-m^2/4)_+ has nonzero image there. The primitive ideal test and Corollary 8.6 imply

Finally choose 0<\delta<1 such that

Apply Lemma 9.2 to the comparisons in (9.24). We obtain x_a,x_b\in C with

and

Consequently

Since Q_a\in\Her(a)=\overline{aCa}, Q_ax_a\in\overline{aC}; similarly Q_bx_b\in\overline{bC}. Choose s,t\in C such that

For elements U,V of norm at most 2, replacing them by elements within \delta changes U^*U or U^*V by at most (4+\delta)\delta, by expanding the difference and using the triangle inequality. Also \|a^2-(a^2-r^2)_+\|,\|b^2-(b^2-r^2)_+\|\le r^2. Therefore

by (9.23) and (9.25). The choice order is r, then m, then \ell, then \delta; the lower bound m has no dependence on the later parameters.

For arbitrary positives a,b, put M=\max\{1,\|a\|,\|b\|\}. Apply the preceding result to a/M,b/M with tolerance \varepsilon/M^2. Multiplication of all three inequalities by M^2 gives the required inequalities for a,b with the same s,t\in C. Zero elements, including the zero algebra, are included. These are precisely the three inequalities in (1.1), so C is strongly purely infinite.□