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Positive maps and finite-dimensional approximation · Prerequisite proofs · Sources and terms

Trace-preserving finite models

Written by GPT-6.1 Sol (OpenAI), Ultra, October 2026. Self-checked by the writing AI. New original text: public domain (CC0).

A useful finite record has two jobs: reproduce the ambient trace exactly and represent the chosen unitary observations accurately. We start with the second job and the error allowed when a matrix contraction is replaced by a unitary. We then design a rational record by mixing reconstruction with state preparation, construct the two trace-preserving maps, and check that their composite recovers the observations.

Throughout, MM has a faithful normal tracial state τ\tau, and τm=Tr⁡/m\tau_m=\operatorname{Tr}/m. No factor or separability assumption is needed. Such an algebra is finite: if v∗v=1v^*v=1, traciality gives τ(1−vv∗)=0\tau(1-vv^*)=0, and faithfulness forces vv∗=1vv^*=1. The incoming matrix models are therefore supplied by the direct faithful-trace construction, using its trace realization, trace-retaining finite densities, scalar spatial bound and explicit matrix maps. Its stronger center-valued statements and the continuous-core construction are unnecessary here.

The complete tracial-adjoint lesson supplies finite diagonalization and matrix densities (P03), all CP and Schwarz rules (P04), the trace Hilbert completion (P05), ordinary normal trace adjoints (Lemma 1.1), the weighted Hilbert bound (Theorem 2.1), and the rational-density channel (Theorem 3.1). Their scalar and coefficient-predual inputs are the retained Sections 1–10 and CP01–06. General bounded polar decomposition is proved by the positive-inverse argument T04a; Exercise 9 below proves the finite complement extension. Its closed-range projections and positive inverses are the Hilbert projection and continuous-calculus arguments of retained scalar Sections 8 and 6. For its bounded strong limits, the coefficient-series proof is direct: choose a finite prefix with small summable tail; strong convergence handles the prefix, while the uniform operator bound controls the tail. Thus a bounded strong limit is ultraweak and stays in the given ultraweakly closed algebra. These same precise elementary inputs apply to the direct faithful-trace construction. No trace-preserving conditional expectation or general Radon–Nikodym theorem is required. Unitary budgets for Theorems 2.1 and 3.1 may always be chosen below one. The normal unital model construction below supplies the topology conversion and exact unit correction used when we assemble the trace-preserving maps.

The unitary test uses the state actually induced by reconstruction. The state-preparation construction makes the record faithful and rational. The trace reconstruction and its ordinary adjoint then give both exact trace identities. The worked diagnostics include a density with a zero eigenvalue and a reconstruction that does not preserve matrix trace.

Test unitary observations before designing the record

A reconstruction map determines its own matrix state. The unitary replacement estimate works for that state even when reconstruction does not preserve the normalized matrix trace. This lets us decide the necessary error budget before constructing the finite record.

Lemma 4.1. Let T:Mm→MT:M_m\to M be any ucp map, let u∈Mu\in M be unitary and let y∈Mmy\in M_m be a contraction. If ∥T(y)−u∥2<η\|T(y)-u\|_2<\eta, there is a unitary v∈Mmv\in M_m with

∥T(v)−u∥2<η+2η.(16)\|T(v)-u\|_2<\eta+\sqrt{2\eta}. \tag{16}

Trace preservation of TT is unnecessary.

Proof. Extend the polar partial isometry of the square matrix yy to a unitary vv, so y=v∣y∣y=v|y|. Put φ=τT\varphi=\tau T. Schwarz gives

∥T(v−y)∥22≤φ((v−y)∗(v−y))=φ((1−∣y∣)2)≤1−φ(y∗y)≤1−∥T(y)∥22<2η.(17)\begin{aligned} \|T(v-y)\|_2^2 &\le\varphi((v-y)^*(v-y))\\ &=\varphi((1-|y|)^2) \le1-\varphi(y^*y) \le1-\|T(y)\|_2^2<2\eta. \end{aligned} \tag{17}

The scalar inequality (1−t)2≤1−t2(1-t)^2\le1-t^2 holds for 0≤t≤10\le t\le1. For the final bound, put d=∥T(y)−u∥2<ηd=\|T(y)-u\|_2<\eta. Contractivity and the reverse triangle inequality give 0≤1−∥T(y)∥2≤d0\le1-\|T(y)\|_2\le d; hence 1−∥T(y)∥22≤2d<2η1-\|T(y)\|_2^2\le2d<2\eta. This works for every positive η\eta. The triangle inequality proves (16). □\square

Design the record by mixing state preparations

Lemma 1.1. If T:Mm→MT:M_m\to M is ucp, then for every ε>0\varepsilon>0 there is a ucp map T′T' with ∥T′−T∥<ε\|T'-T\|<\varepsilon such that the density of τT′\tau T' has strictly positive rational eigenvalues. In particular T′T' is faithful on positive elements.

Proof by state preparation. Write

τT(x)=τm(hx),h=∑i=1mλiei,λi≥0,∑iλi=m,(1)\tau T(x)=\tau_m(hx),\qquad h=\sum_{i=1}^m\lambda_i e_i,\qquad \lambda_i\ge0,\quad\sum_i\lambda_i=m, \tag{1}

where the eie_i are rank-one orthogonal projections summing to 11. Choose 0<α<min⁡(1,ε/2)0<\alpha<\min(1,\varepsilon/2) and positive rational numbers rir_i such that

∑iri=m,ri>(1−α)λi.\sum_i r_i=m,\qquad r_i>(1-\alpha)\lambda_i.

Such a choice is possible because the vector ((1−α)λi+α)i((1-\alpha)\lambda_i+\alpha)_i lies in this open simplex and has sum mm. Rational points in the sum-mm hyperplane are dense: approximate the first m−1m-1 coordinates rationally and define the last by subtraction; sufficiently small errors retain all strict inequalities. For m=1m=1, take r1=1r_1=1.

Put g=∑irieig=\sum_i r_i e_i and k=(g−(1−α)h)/αk=(g-(1-\alpha)h)/\alpha. Then k>0k>0 and τm(k)=1\tau_m(k)=1, so ψ(x)=τm(kx)\psi(x)=\tau_m(kx) is a state. The map

T~(x)=(1−α)T(x)+αψ(x)1\widetilde T(x)=(1-\alpha)T(x)+\alpha\psi(x)1

is ucp, has density gg under τ\tau, and satisfies ∥T~−T∥≤2α<ε\|\widetilde T-T\|\le2\alpha<\varepsilon. Thus mixing reconstruction with state preparation produces the same full conclusion, including faithfulness, while changing the actual map and filling zero eigenspaces. The congruence construction below supplies a second realization for the worked diagnostics.

Congruence formulas for the worked diagnostics

Using the diagonalization (1), the same conclusion also has the following direct construction. Choose 0<δ<10<\delta<1. For each positive λi\lambda_i, choose a rational number

(1−δ)λi<qi<λi;(1-\delta)\lambda_i<q_i<\lambda_i;

put qi=0q_i=0 when λi=0\lambda_i=0. Define

b=∑ibiei,bi=qi/λi(λi>0),bi=1(λi=0).(2)b=\sum_i b_i e_i,\qquad b_i=q_i/\lambda_i\quad(\lambda_i>0),\qquad b_i=1\quad(\lambda_i=0). \tag{2}

Thus (1−δ)1≤b≤1(1-\delta)1\le b\le1. Set

T′(x)=T(b1/2xb1/2)+τm(x)T(1−b).(3)T'(x)=T(b^{1/2}xb^{1/2})+\tau_m(x)T(1-b). \tag{3}

Both summands are cp, and their values at 11 add to 11. For ∥x∥≤1\|x\|\le1, contractivity of TT gives

∥T′(x)−T(x)∥≤2∥1−b1/2∥+∥1−b∥≤3δ.(4)\|T'(x)-T(x)\| \le2\|1-b^{1/2}\|+\|1-b\|\le3\delta. \tag{4}

Here ∥1−b1/2∥≤δ\|1-b^{1/2}\|\le\delta follows from 1−t≤1−t1-\sqrt t\le1-t on [0,1][0,1].

Since bb commutes with hh, the new density is

h′=hb+τm(h(1−b))1=∑i(qi+c)ei,c=1−1m∑iqi>0.(5)h'=hb+\tau_m(h(1-b))1 =\sum_i(q_i+c)e_i,\qquad c=1-\frac1m\sum_iq_i>0. \tag{5}

The number cc is rational, and so are all qi+cq_i+c. Strict positivity also fills every original zero eigenspace. The density remains normalized, because τm(h′)=1\tau_m(h')=1. Choose 3δ<ε3\delta<\varepsilon. If x≥0x\ge0 and T′(x)=0T'(x)=0, then τm(h′x)=0\tau_m(h'x)=0; invertibility of h′h' gives x=0x=0. □\square

The correction in (3) changes the actual cp map as well as its density. Simply replacing hh by a rational matrix would not specify a compatible positive reconstruction map.

From a weighted reconstruction to two trace-preserving maps R takes the density state phi to ordinary matrix trace, L takes that trace back, and T equals T2 composed with L. Its ordinary adjoint S equals T sharp, and both traces are preserved exactly. The displayed error budgets refer to the two distinct approximation stages. Rational recording supplies an ordinary matrix trace Mₘ, φ = τT₂positive rational density h M_q, τ_qordinary normalized trace M, τoriginal faithful trace R L T = T₂L τ_qR = φ; φL = τ_q; τT = τ_q First budget: ‖T(R(x_k)) − u_k‖₂ < η + 2√η (Theorem 2.1) S = T♯ : M → M_q; τ_qS = τ If ‖T(y_k) − u_k‖₂ < δ, then ‖TS(u_k) − u_k‖₂ < √(2δ) + δ Second budget: Theorem 3.1, the complex pairing estimate (14), and Schwarz

Figure. The upper loop changes the state on the finite record, while the right arrow composes its return with the perturbed reconstruction. The lower arrow is the actual ordinary trace adjoint. The two error labels use different parameters: choose the first small enough to obtain the second stage's desired δ\delta. The formulas assert approximation of the selected unitaries and exact trace identities; they do not assert that the finite maps are algebra embeddings.

Normal unital models with an explicit trace-norm budget

The direct faithful-trace theorem gives normal cpc maps Ai:M→MmiA_i:M\to M_{m_i} and cpc maps Bi:Mmi→MB_i:M_{m_i}\to M for which Φi=BiAi\Phi_i=B_iA_i converges pointwise ultraweakly to the identity. Here cpc means completely positive and contractive. For a cp map the norm is its unit-value norm, so 0≤Ai(1)≤10\le A_i(1)\le1, 0≤Bi(1)≤10\le B_i(1)\le1, and 0≤Φi(1)≤10\le\Phi_i(1)\le1.

Schwarz also holds for these subunital maps. Apply complete positivity to the Gram block [x∗xx∗x1]\left[\begin{smallmatrix}x^*x&x^*\\x&1\end{smallmatrix}\right], then add the positive lower-right block 1−Φi(1)1-\Phi_i(1). Sandwiching by [1−Φi(x)]\left[\begin{smallmatrix}1\\-\Phi_i(x)\end{smallmatrix}\right] gives Φi(x)∗Φi(x)≤Φi(x∗x)\Phi_i(x)^*\Phi_i(x)\le\Phi_i(x^*x). Consequently

∥Φi(x)−x∥22≤τ(Φi(x∗x))−2Re⁡τ(x∗Φi(x))+τ(x∗x)⟶0.(N1) \|\Phi_i(x)-x\|_2^2 \le \tau(\Phi_i(x^*x))-2\operatorname{Re}\tau(x^*\Phi_i(x))+\tau(x^*x) \longrightarrow0. \tag{N1}

Both tests on the right are normal: fixed multiplication preserves the coefficient predual by P02 of the tracial-adjoint lesson. Thus ultraweak convergence of these contractions gives the required 22-norm convergence directly.

Put ai=Ai(1)a_i=A_i(1), bi=Bi(1)b_i=B_i(1), and define

Aiu(x)=Ai(x)+τ(x)(1−ai),Biu(y)=Bi(y)+τmi(y)(1−bi).(N2) A_i^u(x)=A_i(x)+\tau(x)(1-a_i),\qquad B_i^u(y)=B_i(y)+\tau_{m_i}(y)(1-b_i). \tag{N2}

A positive scalar functional is completely positive: a scalar compression of a positive block has nonnegative functional value. Multiplying the resulting positive scalar matrix by a fixed positive operator is again positive, by its square-root sandwich. Both corrections are therefore cp, their unit values fill the exact defects, and both maps are ucp. The first correction is normal because τ\tau is normal and has fixed finite-dimensional range. The second has finite-dimensional domain.

Write ci=1−Φi(1)c_i=1-\Phi_i(1). Since 0≤ai≤10\le a_i\le1,

0≤Bi(1−ai)=bi−Φi(1)≤ci,0≤1−bi≤ci. 0\le B_i(1-a_i)=b_i-\Phi_i(1)\le c_i, \qquad0\le1-b_i\le c_i.

Every positive contraction d≤cid\le c_i has ∥d∥22=τ(d2)≤τ(d)≤τ(ci)\|d\|_2^2=\tau(d^2)\le\tau(d)\le\tau(c_i). Expanding the composite of (N2), using ∣τ(x)∣≤∥x∥|\tau(x)|\le\|x\| and ∣τmi(Aiu(x))∣≤∥x∥|\tau_{m_i}(A_i^u(x))|\le\|x\|, therefore gives

∥BiuAiu(x)−Φi(x)∥2≤2∥x∥τ(ci)⟶0.(N3) \|B_i^uA_i^u(x)-\Phi_i(x)\|_2 \le 2\|x\|\sqrt{\tau(c_i)}\longrightarrow0. \tag{N3}

The last limit uses the normal trace test at 11. Combining (N1)–(N3), one sufficiently late index meets any finite list and any prescribed positive error. These are exactly the normal ucp recording maps and ucp reconstruction maps used in Theorem 2.1, with the original injectivity hypothesis intact. The index set is a net; no countable approximation assumption entered.

Reconstruct the observations with the exact trace

Theorem 2.1. Suppose MM is injective. Given unitaries u1,…,un∈Mu_1,\ldots,u_n\in M and ε>0\varepsilon>0, there are a ucp map T:Mq→MT:M_q\to M and contractions yk∈Mqy_k\in M_q such that

τT=τq,∥T(yk)−uk∥2<ε.(6)\tau T=\tau_q,\qquad \|T(y_k)-u_k\|_2<\varepsilon. \tag{6}

Proof. Fix a small η>0\eta>0. Semidiscreteness gives ucp maps S1:M→MmS_1:M\to M_m, T1:Mm→MT_1:M_m\to M with ∥T1S1(uk)−uk∥2<η/2\|T_1S_1(u_k)-u_k\|_2<\eta/2. Put xk=S1(uk)x_k=S_1(u_k), so ∥xk∥≤1\|x_k\|\le1. Lemma 1.1 gives a faithful ucp T2T_2 with rational strictly positive density hh and ∥T2−T1∥<η/2\|T_2-T_1\|<\eta/2. Therefore

∥T2(xk)−uk∥2<η.(7)\|T_2(x_k)-u_k\|_2<\eta. \tag{7}

Write φ=τT2=τm(h ⋅)\varphi=\tau T_2=\tau_m(h\,\cdot). The improved weighted Hilbert bound in the preceding lesson implies, for every contraction xx,

∥[h1/2,x]∥2,τm2=φ(x∗x)+φ(xx∗)−2τm(h1/2x∗h1/2x)≤2−2∥T2(x)∥22.(8)\begin{aligned} \|[h^{1/2},x]\|_{2,\tau_m}^2 &=\varphi(x^*x)+\varphi(xx^*) -2\tau_m(h^{1/2}x^*h^{1/2}x)\\ &\le2-2\|T_2(x)\|_2^2. \end{aligned} \tag{8}

The two cross traces agree by cyclicity; they are real and nonnegative. From (7), ∥T2(xk)∥2>1−η\|T_2(x_k)\|_2>1-\eta and ∥T2(xk)∥2≤1\|T_2(x_k)\|_2\le1. In particular

∥[h1/2,xk]∥2,τm2<4η.(9)\|[h^{1/2},x_k]\|_{2,\tau_m}^2<4\eta. \tag{9}

Apply the rational density theorem to obtain ucp maps R:Mm→MqR:M_m\to M_q and L:Mq→MmL:M_q\to M_m satisfying

τqR=φ,φL=τq,∥LR(xk)−xk∥φ#<2η.(10)\tau_qR=\varphi,\qquad\varphi L=\tau_q,\qquad \|LR(x_k)-x_k\|_\varphi^\#<2\sqrt\eta. \tag{10}

For arbitrary aa, Schwarz and traciality of the target give

∥T2(a)∥22≤12φ(a∗a+aa∗)=(∥a∥φ#)2.(11)\|T_2(a)\|_2^2 \le\tfrac12\varphi(a^*a+aa^*)=(\|a\|_\varphi^\#)^2. \tag{11}

Set yk=R(xk)y_k=R(x_k) and T=T2LT=T_2L. These are contractions and a ucp map. Equations (7), (10) and (11) give

∥T(yk)−uk∥2<η+2η,τT=φL=τq.(12)\|T(y_k)-u_k\|_2<\eta+2\sqrt\eta, \qquad \tau T=\varphi L=\tau_q. \tag{12}

Choosing η+2η<ε\eta+2\sqrt\eta<\varepsilon proves the result. □\square

One common error budget η\eta controls the perturbed reconstruction throughout (7)–(12). The original perturbation and the semidiscrete error have already been added in (7).

Use the adjoint to preserve both traces

Theorem 3.1. For every finite list a1,…,aN∈Ma_1,\ldots,a_N\in M in an injective faithfully tracial algebra and every ε>0\varepsilon>0, there are ucp maps

M→SMq→TMM\xrightarrow{S}M_q\xrightarrow{T}M

such that SS is normal and

τqS=τ,τT=τq,∥TS(aj)−aj∥2<ε.(13)\tau_qS=\tau,\qquad \tau T=\tau_q,\qquad \|TS(a_j)-a_j\|_2<\varepsilon. \tag{13}

Proof. First treat a finite list of unitaries. Choose T,ykT,y_k from Theorem 2.1 with error less than η\eta. Its ordinary trace adjoint S=T♯S=T^\sharp is normal and cp by the preceding lesson. Since τT=τq\tau T=\tau_q, its density is 11, so S(1)=1S(1)=1. Pairing with 11 gives τqS=τ\tau_qS=\tau. Both maps are L2L^2 contractions by Schwarz and their trace identities.

The adjoint identity gives the complex estimate

∣τq(S(uk)yk∗)−1∣=∣τ(uk(T(yk)−uk)∗)∣<η.(14)\left|\tau_q(S(u_k)y_k^*)-1\right| =\left|\tau(u_k(T(y_k)-u_k)^*)\right|<\eta. \tag{14}

Consequently its real part exceeds 1−η1-\eta. As both ∥S(uk)∥2\|S(u_k)\|_2 and ∥yk∥2\|y_k\|_2 are at most 11,

∥S(uk)−yk∥22<2η,∥TS(uk)−uk∥2<2η+η.(15)\|S(u_k)-y_k\|_2^2<2\eta, \qquad \|TS(u_k)-u_k\|_2<\sqrt{2\eta}+\eta. \tag{15}

Choose the small budget before applying Theorem 2.1.

For a general finite list, use an exact finite linear combination of unitaries for each element. Indeed a self-adjoint contraction bb is (v+v∗)/2(v+v^*)/2, where v=b+i(1−b2)1/2v=b+i(1-b^2)^{1/2} is unitary. Applying this to the real and imaginary parts, with their norm bounds, gives at most four unitary summands for any element. Collect their finite list and take a unitary error smaller than ε\varepsilon divided by the maximum sum of absolute coefficients. Zero elements need no tests. Linearity proves (13). □\square

In (14) the difference of the complex pairing from 11 is controlled directly. A lower bound on its absolute value alone would allow an incorrect phase and would not justify (15).

Corollary 4.2. Given finitely many unitaries in an injective faithfully tracial MM, they can be approximated in 22-norm by T(vk)T(v_k), where T:Mm→MT:M_m\to M is ucp, τT=τm\tau T=\tau_m, and every vkv_k is unitary. This follows from Theorem 2.1 and Lemma 4.1 with a sufficiently small initial budget. Their reconstruction images still need to be moved into an actual finite-dimensional subalgebra; the next lessons do that in a factor.

Worked diagnostics and complete solutions

Exercise 1. In Lemma 1.1 take h=diag⁡(2,0)h=\operatorname{diag}(2,0), q1=19/10q_1=19/10, q2=0q_2=0. Compute b,c,h′b,c,h'.

Solution. Equations (2) and (5) give b=diag⁡(19/20,1)b=\operatorname{diag}(19/20,1), c=1/20c=1/20, and h′=diag⁡(39/20,1/20)h'=\operatorname{diag}(39/20,1/20). Its normalized trace is 11, its eigenvalues are positive rationals, and its previously zero eigenvalue is now positive. For any δ>1/20\delta>1/20 the required strict rational inequalities hold.

Exercise 2. Explain why the scalar correction in (3) is completely positive.

Solution. A positive scalar functional is cp. At any matrix level, [τm(xij)]≥0[\tau_m(x_{ij})]\ge0 when [xij]≥0[x_{ij}]\ge0. Tensoring this scalar positive matrix with the positive operator T(1−b)T(1-b) gives the positive block matrix [τm(xij)T(1−b)][\tau_m(x_{ij})T(1-b)]. This is exactly the amplification of the correction.

Exercise 3. Check that τm(h′)=1\tau_m(h')=1 even when hh has a kernel.

Solution. Equation (5) gives τm(h′)=m−1∑iqi+c=1\tau_m(h')=m^{-1}\sum_iq_i+c=1. The coordinates with zero eigenvalues contribute qi=0q_i=0, and the positive constant cc occurs in every coordinate. No division by a zero eigenvalue occurs in (2).

Exercise 4. Derive the cross term in (8) from the square of the commutator.

Solution. Expand (h1/2x−xh1/2)∗(h1/2x−xh1/2)(h^{1/2}x-xh^{1/2})^*(h^{1/2}x-xh^{1/2}) and take τm\tau_m. The first two terms become τm(hxx∗)\tau_m(hxx^*) and τm(hx∗x)\tau_m(hx^*x). Cyclicity makes both negative cross terms τm(h1/2x∗h1/2x)\tau_m(h^{1/2}x^*h^{1/2}x). This equals ∥h1/4xh1/4∥2,τm2\|h^{1/4}xh^{1/4}\|_{2,\tau_m}^2, so it is real and nonnegative.

Exercise 5. Give an explicit positive η\eta making the bound in (12) smaller than a prescribed ε>0\varepsilon>0.

Solution. Take η=min⁡(ε/4,ε2/64,1/4)\eta=\min(\varepsilon/4,\varepsilon^2/64,1/4). Then η≤ε/4\eta\le\varepsilon/4, and 2η≤ε/42\sqrt\eta\le\varepsilon/4. Their sum is at most ε/2<ε\varepsilon/2<\varepsilon. The extra bound η≤1/4\eta\le1/4 keeps the near-unitary lower bound positive.

Exercise 6. Why do the two trace identities in (13) imply L2L^2 contractivity?

Solution. Schwarz gives T(x)∗T(x)≤T(x∗x)T(x)^*T(x)\le T(x^*x). Applying τ\tau gives ∥T(x)∥22≤τm(x∗x)\|T(x)\|_2^2\le\tau_m(x^*x). The same argument for SS, applying τm\tau_m and using τmS=τ\tau_mS=\tau, proves its contraction inequality. Bounded elements are dense in their trace Hilbert spaces, so both maps extend contractively.

Exercise 7. Why is ∣z∣>1−η|z|>1-\eta insufficient for (15), whereas ∣z−1∣<η|z-1|<\eta suffices?

Solution. The number z=−1z=-1 has absolute value 11, but real part −1-1; a squared-distance formula involving 2Re⁡z2\operatorname{Re}z would then have the wrong bound. If ∣z−1∣<η|z-1|<\eta, then Re⁡z>1−η\operatorname{Re}z>1-\eta, which is the estimate actually used in (15).

Exercise 8. Show that the state in (17) need not be τm\tau_m.

Solution. Take M=CM=\mathbb C, m=2m=2, and T(x)=x11T(x)=x_{11}. This is ucp, but τT(e11)=1\tau T(e_{11})=1 while τ2(e11)=1/2\tau_2(e_{11})=1/2. The proof therefore uses φ=τT\varphi=\tau T rather than the normalized matrix trace. Schwarz and the scalar functional calculus inequality remain valid for this state.

Exercise 9. Explain how to extend the polar partial isometry of a square matrix to a unitary.

Solution. Its initial and final projections have the same rank, equal to the rank of the matrix. Their complements therefore have the same dimension. Choose an isometric bijection between those complements and add it to the polar partial isometry. The orthogonal initial and final partitions make the sum unitary, and the extra term annihilates ∣y∣|y|, so y=v∣y∣y=v|y| remains true.

Exercise 10. Suppose the error in Theorem 3.1 is at most γ\gamma on all unitary summands of a=∑jαjuja=\sum_j\alpha_j u_j. Bound its error on aa.

Solution. Linearity and the triangle inequality give ∥TS(a)−a∥2≤γ∑j∣αj∣\|TS(a)-a\|_2\le\gamma\sum_j|\alpha_j|. A finite list has a finite maximum of these coefficient sums. Choosing its unitary budget below the desired error divided by that maximum gives the claimed general finite-set approximation. If every element is zero, any unital trace-preserving scalar model suffices.

References and proof scope

The rational perturbation and trace-preserving factorization were developed by Uffe Haagerup in A new proof of the equivalence of injectivity and hyperfiniteness for factors on a separable Hilbert space, Journal of Functional Analysis 62 (1985), 160–201. A freely accessible method comparison is Xiaoyan Zhou and Junsheng Fang, A note on relative amenability of finite von Neumann algebras, Journal of Operator Theory 81:1 (2019), 107–132: Lemmas 3.5–3.6 and the scalar specialization of Claim 1 in the proof of Theorem 3.2, printed pp.120–127. The source's later relative-algebra hypotheses are not imported. The finite injectivity input is the direct local faithful-trace proof, and the exact complex estimate (14) supplies the needed real part without inferring its sign from an absolute-value lower bound.

The state-preparation proof gives the primary rational perturbation with bound 2α2\alpha. Every faithfully tracial injective algebra is retained; factoriality and separable predual enter the later subalgebra conclusions. These finite maps prove no AFD conclusion by themselves. The new normalization argument (N1)–(N3) proves the incoming topology and unit correction with its own explicit trace budget.

For optional reading on trace-preserving expectations, Matthew Daws's Conditional Expectations, Section 4, Theorem 4.1 (source label thm:main), Proof 1, gives an accessible Hilbert-space compression proof: the trace on the ambient algebra is normal, semifinite and faithful, and its restriction to the subalgebra must also be semifinite. Editable author source at commit a2d5477. The repository notice licenses those notes under CC BY-NC-SA 4.0. This is supplementary reading; no expectation theorem is an input of this lesson's finite-model proofs.