Tensor positivity and nuclearity

Written by GPT-6.1 Sol (OpenAI), Ultra, September 2026. New original text: public domain (CC0).

The minimal tensor product describes two algebras acting on separate Hilbert spaces. The maximal tensor product allows any two commuting actions on one Hilbert space. Nuclearity says that these two ways of coupling systems give the same norm. We will show that this condition is equivalent to approximation by finite-dimensional completely positive models.

The difficult direction is to turn information about all commuting actions into maps with values in the algebra itself. We do this by first encoding operators as normal functionals, then correcting one distinguished functional, and finally using separation to bring the reconstruction maps back into the algebra.

Prerequisites are Completely positive finite models, the retained CF1–10 and CP01–06, and the minimal tensor and universal bidual construction. The density step is proved in Operator density from finite vector tests: its positive matrix approximants have the same norm bound and converge ultraweakly. Its bicommutant and entire contraction-ball arguments also supply the density inputs in the enveloping-algebra construction.

Standard-cone geometry supplies the support, corner, orthogonal-decomposition and norm-estimate arguments used here. Every normal positive functional $\psi$ has a unique cone representative $\xi_\psi$, and

Its commutant support is obtained from its algebra support by the standard conjugation. No faithful normal state on the whole algebra is assumed. The freely accessible primary constructions are Araki, Theorems 4 and 6 and Haagerup, Lemmas 2.6 and 2.10. The natural-cone construction proves endpoint duality and self-duality from bounded multiplication. The cyclic realization proof constructs each positive functional by a convergent modular correction. The geometry component then gives the arbitrary-algebra reduction. We also use Hahn–Banach separation. The dilation and dominated-functional arguments use the freely readable primary papers specified in the references.

Inner products in this lesson are linear in the second variable. Unital algebras in unital assertions are nonzero.

0. Constructing the tensor norms

The representation estimates needed below can be built before using completely positive tensor maps. The scalar inputs are the continuous C*-calculus, positivity, Hilbert completion and forced unitization, the positive scalar extension, and the coordinate proof of compact dual balls. The scalar dilation in the lifting lesson supplies GNS representations; its faithful scalar-state construction supplies a faithful representation. For a nonunital algebra, apply this construction to its forced unitization and restrict. None of these particular constructions uses tensoring or nuclearity.

TN01. Hilbert tensors and a faithful algebraic action

On the algebraic tensor of two Hilbert spaces put

Sesquilinearity makes this independent of the expression for a tensor. Expand its finitely many first coordinates in an orthonormal basis of their span. A tensor then has the form $\sum_k e_k\otimes w_k$, and its squared norm is $\sum_k\|w_k\|^2$. The form is positive definite. Its Hilbert completion defines $L\otimes H$.

For bounded operators $x,y$, orthonormal expansion proves $\|1\otimes y\|\le\|y\|$; interchanging the two coordinates proves $\|x\otimes1\|\le\|x\|$. Thus $\|x\otimes y\|\le\|x\|\|y\|$. Unit vectors approaching the individual norm suprema give the reverse inequality. Testing elementary pairs also gives $(x\otimes y)^*=x^*\otimes y^*$. These formulas define the spatial *-representation of $A\odot B$ associated to two coefficient representations.

If $\pi:A\to B(L)$ and $\rho:B\to B(H)$ are faithful, this algebraic representation is injective. Indeed, express $z=\sum_{i=1}^r a_i\otimes b_i$ with linearly independent $a_i$. The restrictions of vector coefficient functionals $a\mapsto\langle u,\pi(a)v\rangle$ span the dual of $\operatorname{span}\{a_i\}$: their common annihilator is zero, by faithfulness, and finite-dimensional linear algebra gives the spanning assertion. If $(\pi\otimes\rho)(z)=0$, slicing by these coefficients yields $\sum_i f(a_i)\rho(b_i)=0$. Choose coefficients whose restrictions recover each coordinate; then $\rho(b_i)=0$, and faithfulness gives $b_i=0$.

TN02. Scalar states from one faithful representation

We first need only a unital C*-algebra $D$ faithfully and unitally represented on $K$. Its state space is compact for pointwise convergence on $D$: it is the closed subset of the compact dual unit ball given by positivity and $f(1)=1$. Let $C$ be the closed convex hull of its unit-vector states, taken in that state space.

If a state $f$ were outside $C$, some neighborhood defined by finitely many real and imaginary evaluation coordinates would miss $C$. The corresponding finite-coordinate image of $C$ is compact and convex and does not contain the coordinate image of $f$. Its closest point to that image exists. The elementary closest-point inequality supplies a real linear functional strictly separating the two images. Since states are Hermitian, this functional is evaluation at one self-adjoint $d\in D$. We would have

The supremum on the right is the spectral upper bound $m$ of the represented self-adjoint operator. Faithfulness and continuous C*-calculus reflect positivity, so $d\le m1$ in $D$. Positivity of $f$ gives $f(d)\le m$, a contradiction. This proves the density assertion with its actual finite-coordinate separation step.

TN03. Every spatial representation satisfies the same bound

Fix a faithful representation $\pi:A\to B(L)$, restricted to its essential subspace, and let $\sigma:A\to B(K)$, $\tau:B\to B(H)$ be any representations. Restrict their tensor action to the two essential subspaces; it is zero on the other summands. We prove

The zero-algebra cases are immediate. Otherwise extend $\pi$ to $A^\dagger$ on $L$ by $\pi^\dagger(a+\lambda1)=\pi(a)+\lambda1_L$, and add the character $\chi(a+\lambda1)=\lambda$. The representation $\widehat\pi=\pi^\dagger\oplus\chi$ on $L\oplus\mathbb C$ is faithful and unital: vanishing of the character first gives $\lambda=0$, after which faithfulness gives $a=0$. This also works when $A$ already has a unit. Extend $\sigma$ by its identity on its essential Hilbert space.

Take $\zeta=\sum_{i=1}^m u_i\otimes v_i\in K\odot H$, with $r=\sum_i\|u_i\|^2>0$. On $M_m(A^\dagger)$ the rule

is positive and has norm $r$. Apply TN02 to $F/r$ in the faithful amplified representation of $\widehat\pi$. It gives a net of finite sums

converging at every matrix argument to $F$. The convex weights have been absorbed into the vectors. Write $\xi_{k\ell i}=(x_{k\ell i},c_{k\ell i})\in L\oplus\mathbb C$, and put

The Gram entries in the last expression are the values of $F_k$ on the individual matrix units times $1$. Hence

For every algebraic tensor $t$, matrix-coordinate convergence gives

Here the character coordinates disappear because $\chi(a)=0$ for $a\in A$. Set $t=z^*z$. The left side is at most $M_k\|(\pi\otimes\tau)(z)\|^2$. Taking limits proves the squared bound on each finite tensor vector $\zeta$; density proves (TN3). A zero $\zeta$, or a decomposition with $r=0$, requires no estimate.

Choose also a faithful $\rho:B\to B(H_0)$ and apply the same argument to the second factor, using the Hilbert flip. We obtain

Consequently any faithful pair defines the same norm $\|z\|_{\min}$. TN01 proves that it is a norm, satisfies the C*-identity, and is a cross norm on elementary tensors. Its completion is $A\otimes_{\min}B$. This also proves isometric inclusion for C*-subalgebras: restrict a faithful spatial pair to the subalgebras and then to their essential spaces. Every spatial representation is contractive for this norm, regardless of faithfulness, degeneracy or Hilbert-space cardinality.

TN04. The norm for commuting actions

For $z=\sum_i a_i\otimes b_i$, define

The coefficient representations are *-representations on one Hilbert space. Their contractivity bounds the supremum by $\sum_i\|a_i\|\|b_i\|$. Spatial pairs are commuting pairs, so TN01 gives nondegeneracy and $\|z\|_{\min}\le\|z\|_{\max}$. Suprema of the representation bounds give submultiplicativity and $\|z^*z\|_{\max}=\|z\|_{\max}^2$. The elementary-tensor upper bound and the spatial lower bound give the cross-norm identity. Completion is a C*-algebra, and each commuting pair extends to its representation on this completion.

Every nondegenerate representation $R$ of this completion has such coefficient representations. Let $(e_i)$, $(f_j)$ be positive contractive approximate identities in $A,B$. The operators $R(a\otimes f_j)$ are uniformly bounded by $\|a\|$. On the dense vectors $R(z)u$, $z\in A\odot B$, they converge, since

in maximal norm, by the finite projective bound above. This defines a strong limit $\pi(a)$. The same reasoning for the adjoints, and multiplication on this dense subspace, proves that $\pi$ is a *-representation. Similarly $R(e_i\otimes b)$ defines $\sigma(b)$. Their actions commute on the dense subspace, and their product there is $R(a\otimes b)$. For a degenerate $R$, make this construction on its essential subspace and extend both actions by zero. Thus the supremum describes precisely the commuting-action representations used later in this lesson.

The identity on algebraic tensors extends to a surjective *-homomorphism $A\otimes_{\max}B\to A\otimes_{\min}B$. Its range is dense and closed. To see closedness, quotient by its closed kernel and use the complete quotient and injective-homomorphism arguments in TN07. The resulting image is complete. We use equality of these two norms for every $B$ as the nuclearity condition.

TN05. Tensor maps and matrix coefficients

For *-homomorphisms, tensor functoriality now follows directly. Compose any commuting target pair with the two coefficient maps to prove maximal contractivity. Compose a spatial target pair and apply TN03 to prove minimal contractivity.

For a cpc map $\phi:A\to C$, the complete dilation argument in finite-model Lemma 4.1 gives $\phi(a)=V^*\Pi(a)V$, $\|V\|\le1$, and, in the commuting case, a second action commuting with $\Pi$. TN03 bounds every spatial dilation by the minimal source norm; TN04 bounds every commuting dilation by the maximal source norm. Compression therefore proves contractivity of $\phi\otimes\operatorname{id}$ for either product. At each matrix level a positive operator matrix remains positive after compression, so these extensions are completely positive. Compose two such maps for tensoring in both variables; rescaling treats arbitrary bounded CP maps. The nonunital dilation used here is the actual bounded approximate-identity construction in that lemma.

The complete matrix-unit unitary in finite-model Lemma 4.2 also identifies

Indeed, a commuting pair becomes $x\otimes1$, $1\otimes\sigma(b)$ on its essential subspace. Its matrix norm is at most the norm in a faithful amplified representation of $B$, by contractivity of the amplified *-homomorphism. The faithful spatial pair gives the reverse bound. TN03 gives the same minimal norm. This includes nonunital $B$ and degenerate pairs.

Finally every bounded scalar functional is a linear combination of four positive bounded functionals, by the state-coefficient decomposition. Their GNS vector functionals give any scalar functional as one vector coefficient of a finite direct sum representation. Explicitly, if $f=\sum_k c_k f_k$ and $f_k(a)=\langle\xi_k,\pi_k(a)\xi_k\rangle$, omit zero coefficients and set

Then $f(a)=\langle u,(\bigoplus_k\pi_k)(a)v\rangle$. For a nonunital positive functional, the positive forced-unit extension and essential cyclic restriction are constructed in Section 7. Tensoring the two coefficient formulas now proves that every algebraic product $f\otimes g$ is minimal-norm bounded. This is the exact scalar-functional fact consumed by Proposition 1.4.

TN06. The split unitization sequence

For either $\nu=\min$ or $\nu=\max$, the inclusion $A\odot B\subset A^\dagger\odot B$ preserves the $\nu$-norm. The minimal assertion is the subalgebra assertion in TN03. For the maximal assertion, extend any commuting $A,B$ pair by sending the new unit to the identity; this extension still commutes with $B$. The reverse inequality follows by restricting any unitized pair. The character and its section consequently give

Both arrows and the section are continuous *-homomorphisms, and $\|s(b)\|=\|b\|$. For a kernel element $x$, choose algebraic $x_k\to x$. The elements $x_k-s((\chi\otimes\operatorname{id})(x_k))$ belong to $A\odot B$ and converge to $x$. The image of $A\otimes_\nu B$ is closed by its isometry, so it is the entire kernel. This proves exactness from the splitting, including the already unital and zero cases. It supplies all tensor details of the unit-removal step in Theorem 3.1.

This proof diagram follows the finite-coordinate approximation in TN03. The scalar character of the forced unitization contributes no value on $A$; dropping its vector coordinates only decreases the nonnegative Gram mass. The bound therefore retains every finite tensor vector and extends by density, with no restriction on the Hilbert spaces. TN04 and TN06 give the two algebra maps shown at the bottom. The proofs are written above; Olesen's free university thesis, Section 1.5, pp.31–34, provides the tensor conventions.

TN07. Quotients, completeness and the closed range

Here are all quotient steps consumed by TN04 and the unitization argument. Use the complete calculus, norm separation, unitization and Hilbert proofs CF1–10, and the positive two-sided approximate identity F09. The same quotient conclusion is proved in F10.

Let $D$ be a C*-algebra and $J$ a closed self-adjoint two-sided ideal. On the algebraic quotient put

This is a norm: zero distance puts $d$ in the closed set $J$, and homogeneity and the triangle inequality follow by choosing representatives within any positive error of each infimum. The quotient map $q$ contracts the norm. Its involution is isometric since $J=J^*$; multiplication is submultiplicative by multiplying two representatives close to their infima. Their product represents the product of the two cosets, because $J$ is a two-sided ideal.

For completeness, take a Cauchy sequence of cosets and select a subsequence $(z_k)$ with $\|z_{k+1}-z_k\|_q<2^{-k}$. Choose a representative $d_1$ of $z_1$ and, for each difference, a representative $v_k$ with $\|v_k\|<2^{1-k}$. The norm series $d=d_1+\sum_{k\ge1}v_k$ converges in $D$, by CF1. Its partial sums represent $z_{k+1}$, and their tails tend to zero in quotient norm by contractivity of $q$. Hence $z_k\to q(d)$. The original Cauchy sequence has the same limit: use its Cauchy bound and a sufficiently late member of that subsequence. Thus no completeness theorem for a quotient is an additional assumption.

Let $(e_i)$ be the positive contractive two-sided approximate identity of $J$ from F09, computing $1-e_i$ in the forced unitization. For every $d\in D$, $d(1-e_i)\in D$ represents $q(d)$. Therefore its norm is at least $\|q(d)\|_q$. For any $j\in J$,

The second term tends to zero and $\|1-e_i\|\le1$. Infimizing the first term gives

The same argument with compression on both sides gives, for every $c\in D$,

Indeed the compression still represents $q(c)$, while its difference from the compression of $c+j$ has norm at most $\|(1-e_i)j(1-e_i)\|\to0$. Apply (TN7b) to $c=d^*d$ and use the C*-identity in the unitization:

Consequently the quotient, with precisely its distance norm, is a C*-algebra. This includes $J=0$, $J=D$ and $D=0$, using respectively the constant zero approximate identity or the usual identity of the ideal.

For clarity, an injective star homomorphism $\phi:D\to E$ is isometric. Its forced unital extension is a unital star homomorphism; preservation of inverses gives spectrum inclusion. The normal spectral-radius formula and the C*-identity give contractivity, including for the original nonunital algebras. If $h\in D_+$ lost norm, choose a continuous nonnegative function vanishing on $[0,\|\phi(h)\|]$, vanishing at zero, and nonzero at $\|h\|$. The isometric calculus gives a nonzero $f(h)\in D$, while polynomial approximation and spectrum inclusion give $\phi(f(h))=f(\phi(h))=0$, contradicting injectivity. Thus norms of positive elements are preserved, and applying this to $d^*d$ preserves every norm.

For an arbitrary star homomorphism $\phi$, its closed kernel is self-adjoint, and the induced map $D/\ker\phi\to E$ is injective and therefore isometric. Its range is complete by the quotient argument and hence closed in $E$: a convergent sequence in the range is Cauchy there and its range limit must equal its limit in $E$. A star homomorphism with dense range is consequently onto. These are the precise closed-range and quotient facts used in TN04.

Figure. The two compressions retain the same quotient cosets; their norm limits prove the C*-identity of the complete distance quotient. Equations (TN7a)–(TN7c) supply the exact proof. Each $e_i$ is a positive contraction in the ideal; the quotient-norm infimum need not be attained.

1. Positive bilinear forms and the dual order

The Banach dual $B^*$ has a matrix order: a matrix $[f_{ij}]$ is positive when

is a positive functional on $M_n(B)$. The transpose in the corresponding matrix pairing is important.

Lemma 1.1. A positive functional $\omega$ on $A\otimes_{\max}B$ determines a completely positive map

Conversely, every bounded completely positive map $\theta:A\to B^*$ determines a positive functional on $A\otimes_{\max}B$. The two constructions are inverse and preserve norms.

Proof. Suppose first that $\omega\ge0$. If $[a_{ij}]\ge0$ and $[b_{ij}]\ge0$, write $a_{ij}=\sum_kx_{ki}^*x_{kj}$ and $b_{ij}=\sum_ly_{li}^*y_{lj}$. Then

Evaluating $\omega$ gives positivity of $[\theta_\omega(a_{ij})]$ in the dual order. Also $\|\theta_\omega\|\le\|\omega\|$ by the tensor cross norm.

Conversely define $\omega(a\otimes b)=\theta(a)(b)$ algebraically. For $z=\sum_i a_i\otimes b_i$, complete positivity applied to $[a_i^*a_j]$ and $[b_i^*b_j]$ gives $\omega(z^*z)\ge0$.

The GNS construction for this positive algebraic functional has commuting left actions of $A$ and $B$. They are bounded: for $a\in A$, the difference $\|a\|^2z^*z-z^*(a^*a\otimes1)z$ has nonnegative $\omega$-value by the same matrix test, and likewise for $b\in B$. The expressions with $1$ denote left multiplier actions when an algebra is nonunital; their differences are finite algebraic tensors and the positive matrix test still applies. Thus the actions extend to commuting representations, giving a representation $\pi$ of the maximal tensor product.

For unital algebras the GNS vector is $[1\otimes1]$, and the functional has norm $\omega(1\otimes1)=\theta(1)(1)$. Its vector formula gives $|\theta(a)(b)|\le\theta(1)(1)\|a\|\|b\|$, while testing the two units gives the reverse bound. Hence this norm is $\|\theta\|$.

For completeness, the cyclic vector in the nonunital case can be constructed without assuming the boundedness of $\omega$ on the tensor completion. Take positive contractive approximate identities $(e_i)$ in $A$, $(f_j)$ in $B$. In the algebraic GNS space,

Choose a weakly convergent subnet of these bounded vectors, with limit $\xi$. For each finite algebraic tensor $z$,

in the GNS norm. Indeed, expand the squared norm of the difference into finitely many values $\theta(a)(b)$; multiplication by the approximate identities makes their coefficients converge in the original C*-norms, and $|\theta(a)(b)|\le\|\theta\|\|a\|\|b\|$. Hence $\pi(z)\xi=[z]$. The same coefficient convergence gives

This vector functional is bounded on the maximal completion with norm at most $\|\xi\|^2\le\|\theta\|$. The opposite norm inequality follows from the tensor cross norm, as in the first paragraph of the proof. Thus the constructions preserve norms in the nonunital case as well. $\square$

The finite-rank vector forms in a spatial representation have an especially useful factorization.

Lemma 1.2. Let $A$ act on $L$, let $B$ act on $H$, and take $\zeta=\sum_{i=1}^n u_i\otimes v_i$. For $\omega_\zeta(a\otimes b)=\langle\zeta,(a\otimes b)\zeta\rangle$, the associated map factors completely positively through $M_n$.

Proof. Define

The first map is compression by the operator $e_i\mapsto u_i$, hence is completely positive. The matrix of functionals $[\langle v_i,(\,\cdot\,)v_j\rangle]$ is positive in the dual order: on a positive matrix $[b_{ij}]$, its value is $\sum_{i,j}\langle v_i,b_{ij}v_j\rangle\ge0$.

Here is the dual matrix criterion used to conclude that $T$ is completely positive. If $F=[T(E_{ij})]$ is positive, take $X=[x_{pq}]\in M_m(M_n)_+$ and $Y=[b_{pq}]\in M_m(B)_+$. Write the positive scalar matrix $X$, indexed by $(p,i),(q,j)$, as a sum of rank-one matrices $\overline{u_{pi}}u_{qj}$. For each term, the pairing of $[T(x_{pq})]$ with $Y$ is the pairing of $F$ with the positive scalar compression $U^*YU$. It is nonnegative. Thus $T$ is completely positive. Necessity follows by applying $T$ to the positive matrix $[E_{ij}]$.

Expanding the vector functional gives $\theta_{\omega_\zeta}=TS$. If $B$ is a concretely represented von Neumann algebra, $T$ takes values in its predual. $\square$

Finite sums of these factorizations still factor through one matrix algebra, by diagonal blocks. For a positive functional $\omega$ of norm at most $c$, the composite map has norm at most $c$, even when the separate maps in a chosen vector decomposition do not have norm at most one.

Lemma 1.3. In any faithful representation of a C*-algebra $D$, the convex hull of its vector states is weak* dense in its state space. For a spatial representation $D=A\otimes_{\min}B$, vectors that are finite sums of elementary tensors suffice.

Proof. If $D=0$, its state space is empty and the assertion is vacuous. For a faithful representation $\pi:D\to B(K)$, first restrict to the essential subspace $K_{\mathrm{ess}}=\overline{\pi(D)K}$. The representation is zero on its orthogonal complement, so this restriction remains faithful and is nondegenerate. A contractive approximate identity acts strongly as the identity there; hence every unit vector in $K_{\mathrm{ess}}$ defines a functional of norm one, and therefore a state. All unit vectors and spectral suprema below are taken in this essential representation.

If a state were outside the weak* closed convex hull, Hahn–Banach would separate it by a self-adjoint $d\in D$. But the supremum of $\langle\xi,d\xi\rangle$, over unit vectors, is the largest point of the spectrum of $d$ in a faithful representation. No state can take a larger value. Applying this to $-d$ also rules out separation on the other side.

For the spatial representation, restrict the two coefficient representations to $L_{\mathrm{ess}}=\overline{A L}$ and $H_{\mathrm{ess}}=\overline{B H}$. The essential subspace of their tensor representation is $L_{\mathrm{ess}}\otimes H_{\mathrm{ess}}$; the other tensor summands carry the zero representation. Finite sums of elementary tensors from these essential spaces are norm-dense there. Normalizing a nearby nonzero vector approximates its vector state in norm, so does not change the weak* closed convex hull. $\square$

Proposition 1.4. A positive functional $\Omega$ on $A\otimes_{\max}B$, of norm at most one, is minimal-norm bounded if and only if its associated map $A\to B^*$ is in the pointwise weak* closure of the finite-rank completely positive contractions.

The same equivalence holds with completely positive matrix factorizations whose composites have norm at most one in place of all finite-rank completely positive contractions.

Proof. If $\Omega$ is minimal-norm bounded, Lemmas 1.2 and 1.3 approximate it by positive finite-rank vector forms, with norms at most one. Their maps are finite-rank completely positive contractions and converge pointwise weak*.

Conversely, a finite-rank bounded map $A\to B^*$ corresponds to an algebraic sum $\sum_k f_k\otimes g_k$, with $f_k\in A^*$, $g_k\in B^*$. Each such product functional is bounded on the minimal tensor product: realize the two functionals as vector coefficients of their universal representations and take their tensor product. Therefore the sum is minimal-norm bounded. If its associated map is completely positive, Lemma 1.1 gives positivity on all algebraic squares. Continuity then gives positivity on the minimal completion. Its positive-functional norm equals the norm of the associated map, so is at most one.

The weak* closed unit ball of the minimal dual, pulled back to the maximal tensor product, is closed under convergence on elementary tensors when the norms are uniformly bounded. Hence the limit functional is also minimal-norm bounded. $\square$

For the additional statement, the forward approximations from Lemma 1.2 already have completely positive matrix factorizations. The reverse implication holds because every such composite is a finite-rank completely positive map.

Proposition 1.5 (Convexity of matrix factorizations). For a target $D$ that is a C*-algebra or a dual $B^*$, the completely positive maps $A\to D$ factoring through matrices form a convex cone. Their intersection with the unit ball is convex, even when only the composite norms are bounded.

Proof. For finitely many factorizations $T_jS_j$ and coefficients $t_j\ge0$, use the recording map $S(a)=\bigoplus_j S_j(a)$ and reconstruction map $T((x_j)_j)=\sum_jt_jT_j(x_j)$. Both are completely positive in the inherited matrix orders. Block diagonal inclusion and block compression replace the middle direct sum by one full matrix algebra. Their composite is $\sum_jt_jT_jS_j$. If the $t_j$'s sum to one and the original composites have norms at most one, the triangle inequality bounds the new composite norm by one. No bounds on the separate arrows are needed for this assertion. $\square$

Proposition 1.6 (Normal-valued approximation). Let $N$ be a von Neumann algebra. Suppose $\Omega$ is a positive functional of norm at most one on $A\otimes_{\min}N$ and its associated map $\theta:A\to N^*$ has range in $N_*$. Then $\theta$ is in the pointwise norm closure of completely positive matrix factorizations $A\to M_n\to N_*$ whose composites have norm at most one.

Proof. Use a faithful spatial representation with $N$ normally represented. Lemmas 1.2 and 1.3 give finite-tensor vector approximations with reconstruction values in $N_*$; their convex combinations also factor through matrices by Proposition 1.5. Their composite norms are at most one by Lemma 1.1. Their convergence against every element of $N$ is precisely weak convergence in $N_*$, since both the approximants and the limit take values in that predual. For each finite list from $A$, apply Hahn–Banach to the convex image in the finite product of copies of $N_*$. Its weak and norm closures agree, giving the assertion. A positive functional of norm less than one is handled by scaling the state approximations; the zero case is immediate. No unit or separability of $A$ is required. $\square$

2. Keep a marginal exactly fixed

Choose a weight-constructed standard representation $N\subset B(H)$. The marginal-correction statement concerns the abstract algebra and its predual, so this choice adds no hypothesis on $N$. The commutant standard representation uses the opposite right Hilbert algebra and has the same cone. In translating the programme construction’s first-variable inner product to the second-variable convention here, put $[u,v]=\langle v,u\rangle$; the norm, conjugation, cone and support projections are unchanged.

Let $N\subset B(H)$ be a von Neumann algebra in standard form. For $\psi\in N_*^+$, write $\xi_\psi$ for its cone vector and $q_\psi\in N'$ for the projection onto $\overline{N\xi_\psi}$.

We recall an elementary consequence of the GNS dominated-functional theorem. The map

is a complete order isomorphism onto the linear span of the positive normal functionals bounded above by a scalar multiple of $\psi$. Its inverse is not claimed bounded on an arbitrary normed subspace. On positive functionals $f\le c\psi$, the inverse satisfies $0\le R_\psi^{-1}(f)\le cq_\psi$.

Here is the complete order argument, including a possibly nonfaithful $\psi$. If $\psi=0$, both spaces are zero. Otherwise abbreviate $q=q_\psi$ and $\xi=\xi_\psi$. For $0\le f\le c\psi$, Cauchy–Schwarz for $f$ gives

Thus the form $f(x^*y)$ is well defined and bounded on the dense subspace $N\xi\subset qH$. Its representing operator satisfies

For $a,x,y\in N$, the form identity gives

Density shows that $z_f$ commutes with the restricted $N$-action. Extending it by zero on $(1-q)H$ gives an element of $qN'q$. Taking $x=1$ shows $R_\psi(z_f)=f$.

Conversely, if $0\le z\le cq$ in $qN'q$, then

is normal and positive, and is at most $c\psi$. Any element of the corner is a linear combination of four positive elements bounded by scalar multiples of $q$. This proves the claimed range. If $R_\psi(z)=0$, commutation gives

Density proves injectivity, and hence also uniqueness of each $z_f$.

To check every matrix level, take $Z=[z_{ij}]\in M_m(qN'q)_+$ and $Y=[y_{ij}]\in M_m(N)_+$. On $qH$, factor the two matrices as $Z=U^*U$ and $qYq=V^*V$. Their entries commute across the two algebras, so

Its value at $\xi$ is $\sum_{i,j}R_\psi(z_{ij})(y_{ij})$. Thus $R_\psi$ is completely positive in the stated dual matrix order.

For the inverse, let $[f_{ij}]$ be positive in that inherited dual order and put $z_{ij}=R_\psi^{-1}(f_{ij})$. For arbitrary $a_1,\ldots,a_m\in N$,

because $[a_i^*a_j]$ is positive. The tuples $(a_i\xi)_i$ are dense in $(qH)^m$, so $[z_{ij}]\ge0$. This proves complete positivity of the inverse without asserting any uniform inverse norm bound. It also explains why restricting to the actual support, rather than assuming a faithful state, is sufficient.

The exact norm-additive decomposition of a self-adjoint normal functional is proved in NJ01–NJ06. That construction uses only the retained CF1–10 scalar and Hilbert arguments and CP01–06 concrete predual, together with the scalar positive-functional norm identity proved in MG01. Its normal positive summands remain in the predual; it does not use a bidual or a cone. The decomposition of a vector in a real self-dual cone is a separate statement.

Lemma 2.1 (Marginal correction). Let $A$ be unital, and suppose a completely positive map $\theta:A\to N_*$ is in the pointwise norm closure of completely positive matrix factorizations. Put $\psi=\theta(1)$. For every finite $F\subset A$ and $\varepsilon>0$, there is a completely positive factorization $\theta_3=TS$ through a matrix algebra such that

The recording map $S$ may be taken unital, and $\|\theta_3\|=\|\theta\|$.

Proof. The case $\psi=0$ gives $\theta=0$; take a state as the recording map to $\mathbb C$ and zero as the reconstruction map. Otherwise choose a factorization $\theta_1$ with errors less than $\delta$ on $F\cup\{1\}$, where $\delta$ will tend to zero. Put $\psi_1=\theta_1(1)$, and take the Jordan decomposition

For a state $\omega$ on $A$, set $\theta_2(a)=\theta_1(a)+\omega(a)f$. This still factors through matrices: add a scalar block to the previous factorization. Its marginal $\chi=\theta_2(1)=\psi_1+f$ satisfies $\chi\ge\psi$ and $\|\chi-\psi\|=\|g\|<\delta$. Also

Normalize its recording map to be unital. Explicitly, if $\theta_2=T_0S_0$ and $h=S_0(1)$, invert $h^{1/2}$ on its finite-dimensional support and add a state times the complementary projection. This gives a ucp $S$ with $S_0(a)=h^{1/2}S(a)h^{1/2}$. Here are the support details of this normalization. Let $p$ be the support of the positive finite matrix $h$, and let $d$ be its inverse square root on $p\mathbb C^n$, extended by zero. For $a\in A$, the matrix

is positive, by the scalar Cauchy–Schwarz bound on its quadratic form. Applying $(S_0)_2$ and testing a vector in $\ker h$ in either diagonal coordinate forces both $S_0(a)$ and its adjoint to vanish on that kernel: arbitrary scalar phases in the other coordinate make every mixed coefficient zero. Hence $S_0(a)=pS_0(a)p$. The explicit formula

is completely positive, has $S(1)=p+(1-p)=1_n$, and satisfies the asserted identity. The scalar state map is completely positive by its positive matrix quadratic-form test. If the original recording map is normal and a normal state is used, both displayed terms are normal as well. Thus the support normalization used in the semidiscrete converse preserves normality. A nonzero concrete von Neumann algebra has a normal state by taking a nonzero vector coefficient and dividing by its value at the unit.

Replace $T_0$ by $T_2(x)=T_0(h^{1/2}xh^{1/2})$. Then $T_2(1)=\chi$.

By the dominated-functional theorem,

is ucp. Since $\psi\le\chi$, its support in $N$ is below the support of $\chi$. Standard conjugation gives $q_\psi\le q_\chi$, so $q_\chi\xi_\psi=\xi_\psi$.

Define

This is completely positive and normal-functional-valued, and $T(1)=\psi$. Put $\theta_3=TS$.

For $z=B(S(a))$, contractivity gives $\|z\|\le\|a\|$. Comparing its vector functionals at $\xi_\chi$ and $\xi_\psi$ gives

The cone estimate bounds the last factor by $\sqrt\delta$, and $\|\xi_\chi\|^2=\|\chi\|\le\|\psi\|+\delta$. The right side tends to zero uniformly on the finite $F$. Choose $\delta$ small enough. Finally, for completely positive maps from the unital $A$ into $N_*$, Lemma 1.1 gives the norm as the norm of the unit value. Both maps have unit value $\psi$, so their norms agree. In particular contractivity of the composite is retained when $\theta$ is contractive. $\square$

Only finitely many normal functionals are involved at each approximation. The argument needs no countable decomposition of $N$.

3. Nuclearity produces finite models

Theorem 3.1 (Choi–Effros–Kirchberg). For every C*-algebra $A$, the following are equivalent:

1. $A$ is nuclear.
2. The identity of $A$ has cpc factorizations through matrix algebras converging pointwise in norm.

If $A$ is unital, both factor maps may be chosen unital. If $A$ is separable, the approximations may be a sequence; in general they are a net.

Proof. The implication from 2 to 1 and the assertion about sequences were proved in Completely positive finite models. We prove 1 to 2 first for unital $A$, including the normalization of both factor maps.

The universal predual and bidual product identify every positive $f\in A^*$ with its positive normal evaluation functional on $A^{**}$, preserving its norm. Use NC00–NC08 and CG01–CG10 for the standard representation and its cone, CG07–CG08 for the complete dominated-functional inverse, and ODF06 for the exact positive-matrix density used below. The reverse implication uses the complete commuting and nonunital dilations in finite-model Lemma 4.1, its matrix-unit proof, and Theorem 4.3. Propositions 1.1–1.2 supply the net and separable-sequence assertions.

Encode finitely many tests. Let $a_1,\ldots,a_r\in A$ and let $f_1,\ldots,f_s$ be positive functionals on $A$. Put $w=\sum_j f_j$, viewed as a normal functional on $M=A^{**}$. Represent $M$ in standard form, and write $\xi=\xi_w$. Set $N=M'$. The formula

defines a positive functional on $A\otimes_{\max}N$, because the two actions commute. Nuclearity makes it a positive functional on $A\otimes_{\min}N$.

Approximate the functional by matrix factorizations. Represent $A$ faithfully on a Hilbert space $L$, keeping the given normal representation of $N$ on $H$. Lemmas 1.2 and 1.3 approximate $\Omega$ weak* by positive finite-tensor vector forms and their convex combinations. Their associated maps $A\to N_*$ factor completely positively through matrices. The limit is

The convergence is pointwise weak in $N_*$, since testing against every $y\in N$ is precisely its weak topology. Convexity and Hahn–Banach, applied in $(N_*)^r$ for each finite set, turn this into pointwise norm approximation.

Recover an operator approximation. Apply Lemma 2.1, with exact marginal $\theta(1)=\psi=\omega_\xi|_N$. Take $S:A\to M_n$ ucp and $T:M_n\to N_*$ completely positive with $T(1)=\psi$ and $TS$ close to $\theta$ on the chosen elements.

Let $e\in M$ project onto $\overline{N\xi}$. The dominated-functional inverse turns $T$ into a ucp map $T':M_n\to eMe$, where the unit of the corner is $e$. Viewed in $M$, $T'$ is cpc. Each $f_j\le w$ has a GNS Radon–Nikodym contraction $b_j\in M'=N$ with

To obtain this contraction explicitly, apply the preceding dominated-form proof to $f_j\le w$: it gives $0\le d_j\le1$ in $M'$, and take $b_j=d_j^{1/2}$. Also $e\xi=\xi$, and $e$ commutes with every $b_j$. Thus $eb_j\xi=b_j\xi$. The definition of the dominated inverse says exactly that $TS(a)(y)=\langle\xi,yT'S(a)\xi\rangle$. Consequently