Nuclear biduals and extensions

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

There are two different approximation questions here. A finite model of a C*-algebra must approximate each chosen element in norm. A finite model of its bidual is tested by normal functionals. We will connect those questions after constructing the central projection that separates an ideal from its quotient.

The central projection is useful even before approximation enters the argument: it explains how the quotient map acts on the bidual, and why the original algebra need not split. The other construction is a completely positive retraction onto a represented commutant. We build it on a dilation space, including the extra step needed when the coefficient algebra has no unit. These two constructions meet in the nuclearity and extension theorems below.

Prerequisites are Tensor positivity and nuclearity, Finite models of a von Neumann algebra, and Averaging, crossed products, and injectivity. The complete universal predual and bidual product identify the represented bidual with $A^{**}$ and its predual with $A^*$. The normal extension and central quotient prove the image, central kernel and globally normal inverse on the essential representation space. The exact quotient norm and completeness argument and positive two-sided approximate identity supply the nonunital quotient and ideal inputs. Scalar calculus, Hilbert projections, Hahn–Banach, dual-ball compactness and the concrete predual are proved in the scalar and coefficient foundations.

All algebras may be nonunital, and all Hilbert spaces and index sets may be arbitrary. Zero algebras have the unique zero interpretation. Inner products are linear in the second variable.

Read the approximation topology first

Example 3.3. For any set $I$, $c_0(I)^*=\ell^1(I)$ and $c_0(I)^{**}=\ell^\infty(I)$. Coordinate truncations through $\mathbb C^F$, directed by finite subsets $F\subset I$, approximate the identity of $c_0(I)$ in norm. On the bidual the same truncations approximate the identity ultraweakly, because every normal functional is an absolutely summable coordinate family. Both algebras therefore have the appropriate finite models, even when $I$ is uncountable. The maps land in finite commutative algebras, which embed as diagonal matrix algebras with completely positive diagonal retractions.

Proof of the coordinate identifications. Let $\delta_i$ be the function that is one at $i$ and zero elsewhere. For $f\in c_0(I)^*$, put $a_i=f(\delta_i)$. For every finite $F\subset I$, choose scalars $\theta_i$ of modulus one with $\theta_i a_i=|a_i|$, choosing any phase if $a_i=0$. The vector $\sum_{i\in F}\theta_i\delta_i$ has norm at most one, so

Thus $a\in\ell^1(I)$, where the sum is the supremum of its finite subsums. For each positive integer $n$, the set of indices with $|a_i|\geq1/n$ is finite; their union contains the support of $a$. Finite tails therefore tend to zero. Finite-support functions are norm dense in $c_0(I)$: for $x\in c_0(I)$, the set where $|x_i|\geq\varepsilon$ is finite and contains every coordinate needing an error of that size. Hence continuity gives

The upper norm bound follows by absolute summation, and the finite phase tests give the reverse bound. Conversely every absolutely summable family defines this bounded functional.

If $g\in\ell^1(I)^*$, its coordinates $y_i=g(\delta_i)$ satisfy $|y_i|\leq\|g\|$. Density of finite-support families in $\ell^1(I)$ gives $g(a)=\sum_i a_i y_i$; conversely each bounded $y$ defines such a functional. Single-coordinate tests and the absolute-sum bound give $\|g\|=\sup_i|y_i|$. These are the asserted isometric dual identifications. Coordinatewise multiplication on $\ell^\infty(I)$ extends that of $c_0(I)$ and is separately weak* continuous, because multiplication by a fixed bounded family preserves $\ell^1(I)$. The bidual product construction identifies this product with the universal bidual product.

At every matrix size positivity is coordinatewise. Restriction to a finite set, diagonal inclusion into $M_{|F|}$, diagonal compression from $M_{|F|}$, and extension by zero all preserve these positive matrix cones and contract the norm. Their composite is the truncation $P_F$. The empty set uses the zero maps through $M_1$. For $x\in c_0(I)$, $\|x-P_Fx\|\to0$; for $a\in\ell^1(I)$ and bounded $x$, absolute summation gives

This proves both finite-model assertions, including empty and uncountable index sets.

Exercise 5 gives the estimate that separates ultraweak convergence from norm convergence in this model.

Construct the ideal projection and the quotient map

The dual of a quotient consists of precisely the functionals annihilating its kernel. We first realize that Banach-space fact by a projection in the universal representation.

Proposition 4.1. Let $I$ be a norm-closed two-sided ideal of $A$, and let $q:A\to A/I$ be the quotient map. There is a central projection $p\in A^{**}$ such that

All identifications are normal *-isomorphisms; the first extends the inclusion of $I$, and the second is induced by $q^{**}$.

Proof. Work in the universal representation of $M=A^{**}$. The ultraweak closure $J$ of $I$ is an ideal in $M$: first multiply approximating elements of $I$ by a fixed element of $A$, then use ultraweak density of $A$ and separate ultraweak continuity of multiplication to allow any fixed element of $M$.

Choose a positive contractive approximate identity $(u_i)$ of $I$. It converges strongly to the projection $p$ onto $\overline{IH}$. Indeed it tends to the identity on vectors $a\eta$, $a\in I$, by norm approximation of $a$; it vanishes on the orthogonal complement, since that complement is killed by $I$. Uniform boundedness extends the convergence to all vectors. The subspace $\overline{IH}$ reduces $A$, since $I$ is a two-sided *-ideal, so $p\in M'$. Also $p\in M$, as the bounded strong limit of $(u_i)$. Thus $p$ is central.

Every element of $I$, and hence of $J$, satisfies $x=px$. Conversely $u_i x\in J$ for $x\in M$, and $u_i x\to px$ ultraweakly. Hence $J=pM$. Here is that second-adjoint identification explicitly. Restriction $r:A^*\to I^*$ is onto by Hahn–Banach and gives every functional on $I$ an extension of the same norm. Thus $r^*:I^{**}\to A^{**}$ is a normal isometry. Its range is $(I^\perp)^\perp$: one inclusion is immediate, and an $X$ annihilating $I^\perp=\ker r$ defines a well-defined bounded functional $r(f)\mapsto X(f)$ on $I^*$, with the same norm by those norm-preserving extensions. The weak* annihilator identity gives $(I^\perp)^\perp=J$. Composing the inclusion $I\to A$ with the universal representation of $A$, the full normal-extension construction identifies $r^*$ with its normal star-homomorphic extension, by agreement on the weak* dense copy of $I$. Hence $I^{**}$ is normally and isometrically identified with $J=pM$.

On the dual side $q^*:(A/I)^*\to A^*$ is an isometry with range

Therefore

The middle equality is the weak* annihilator identity proved in the predual annihilator proof: a point outside a weak* closed linear subspace is separated from it by a finite linear combination of the coordinates defining a disjoint basic neighborhood. The map $q^{**}$ is onto: a bounded functional on $I^\perp\subset A^*$ extends to $A^*$ by Hahn–Banach, and that extension is an element of $A^{**}$ with the required image. It is a normal star homomorphism by the normal-extension proof, applied to the quotient followed by its universal representation; agreement on $A$ identifies that extension with $q^{**}$. Restricting to $E=(1-p)M$ gives a bijective normal *-homomorphism $V:E\to N=(A/I)^{**}$. An injective *-homomorphism is isometric. Moreover $V$ maps the closed unit ball onto the closed unit ball: Hahn–Banach gives a lift of any $Y\in N$ with norm $\|Y\|$, and compression by $1-p$ leaves its image unchanged and cannot increase its norm.

Here is a direct proof that the inverse is normal. The concrete predual of $E$ is the quotient of $M_*$ by the annihilator of $E$, with $E=(E_*)^*$ isometrically, by the complete concrete-predual quotient proof. The preadjoint $V_*:N_*\to E_*$ is an isometry because $V$ maps the unit ball onto the unit ball, so its range is norm closed. If $X\in E=(E_*)^*$ annihilates that range, then $\rho(VX)=0$ for every $\rho\in N_*$, whence $VX=0$ and $X=0$. Hahn–Banach makes the range norm dense: otherwise it would supply a nonzero bounded functional annihilating its closed range. Thus $V_*$ is a surjective isometry, its inverse is bounded, and $V^{-1}=(V_*^{-1})^*$ is weak* continuous. This proves normality of the inverse, including zero summands. Combining the two central summands proves the decomposition. $\square$

The splitting takes place in $A^{**}$. It does not assert that $A$ is a direct sum of its ideal and quotient.

Construct a retraction on a dilation space

Proposition 2.1. If $A$ is nuclear, every state GNS commutant $C=\pi_\varphi(A)'$ is injective.

Proof. Write $\pi:A\to B(H)$ for the state GNS representation and put $C=\pi(A)'$. Only nondegeneracy of $\pi$ is needed in this proof. The state case has $H\ne0$. Nuclearity and the tensor-norm construction give a *-homomorphism

Indeed the two commuting representations first give a homomorphism for the maximal norm, and nuclearity identifies that norm with the minimal norm. Spatial tensor injectivity embeds $D=C\otimes_{\min}A$ in $L=B(H)\otimes_{\min}A$.

Use forced unitizations, with the same newly adjoined unit for $D^\dagger\subset L^\dagger$, even if these algebras already have units. The map $\rho^\dagger(d+\lambda1)=\rho(d)+\lambda1_H$ is a unital *-homomorphism. The full Arveson extension proof extends it to a ucp map $\Phi:L^\dagger\to B(H)$. The full unital dilation construction supplies a representation $\sigma:L^\dagger\to B(K)$ and an isometry $V:H\to K$ such that

For every $d\in D$, the fact that $\Phi$ agrees with a homomorphism on $D$ gives

For completeness, the squared norm of $\sigma(d)V\eta-V\rho(d)\eta$ is the quadratic form of $\Phi(d^*d)-\Phi(d)^*\Phi(d)=0$. Apply the same computation to $d^*$ and take adjoints to get the second identity.

The operator $b\otimes1_A$ need not be an element of $L$. We now construct its action without assuming that $A$ has a unit. Let $(u_i)$ be a positive contractive approximate identity of $A$, put $h_i=1_H\otimes u_i$, and set $K_0=\overline{\sigma(L)K}$. The operators $\sigma(h_i)$ converge strongly to the projection $P$ onto $K_0$. On vectors $\sigma(l)\zeta$ this follows from $h_i l\to l$ in norm; on $K_0^\perp$ every $\sigma(l)$ vanishes. The common contraction bound extends convergence to all of $K$. Since

strongly, $V^*PV=1_H$. Consequently $\|(1-P)V\eta\|^2=0$, so $VH\subset K_0$. The space $K_0$ reduces $\sigma$, and we henceforth restrict the dilation to it.

For $b\in B(H)$, the uniformly bounded operators $\sigma(b\otimes u_i)|_{K_0}$ converge strongly to an operator $\tau(b)$. To check this, use the dense vectors $\sigma(c\otimes a)\zeta$, with $c\in B(H)$, $a\in A$:

The bound $\|\sigma(b\otimes u_i)\|\le\|b\|$ makes the limit well defined on all of $K_0$, independent of the chosen approximate identity. This formula on dense vectors proves linearity and multiplication for $\tau$. Applying it also to $b^*$, and taking vector coefficients, proves $\tau(b)^*=\tau(b^*)$. The case $b=1_H$ gives $\tau(1_H)=1_{K_0}$. Thus $\tau:B(H)\to B(K_0)$ is a unital *-representation. On the same dense vectors, $\tau(b)$ commutes with $\sigma(1_H\otimes a)$ for every $a\in A$.

Define

Compression of a unital *-representation by an isometry is ucp. The intertwining identities for $1_H\otimes a\in D$ now show

Hence $E(b)\in C$. If $c\in C$, apply the intertwining identity instead to $c\otimes u_i\in D$. Strong convergence gives

Therefore $E(c)=c$. This is the required ucp retraction onto $C$, and proves injectivity. No normality of $E$ is asserted or needed. $\square$

Assemble cyclic information by finite central covers

Lemma 1.1. Suppose that $\pi_\varphi(A)''$ is injective for every state $\varphi$ on $A$. Then $A^{**}$ is injective.

Proof. The zero algebra has zero bidual and needs no construction. Otherwise put $M=A^{**}$. For each state $\varphi$, its normal representation extension has a central kernel $M(1-z_\varphi)$, and identifies $Mz_\varphi$ normally with $\pi_\varphi(A)''$. Thus each $Mz_\varphi$ is injective. Moreover

An element killed by all these extensions acts as zero in every state GNS summand of the faithful universal representation. Hence the intersection of their kernels is zero, which proves the displayed equality.

For a finite set $F=\{\varphi_1,\ldots,\varphi_n\}$, let $z_F=\bigvee_{\varphi\in F}z_\varphi$. Although the projections may overlap, they admit the finite partition

These are orthogonal central projections with sum $z_F$, and $p_k\le z_{\varphi_k}$. Central corners and finite products of injective algebras are injective, so $Mz_F$ is injective. This conclusion is independent of the enumeration of $F$.

Represent $M$ faithfully and normally on $H$. For each $F$, choose a ucp retraction $B(z_FH)\to Mz_F$, and compose with compression $b\mapsto z_Fbz_F$. Regard its values as elements of $M$, and call the resulting cpc map $E_F:B(H)\to M$. For $x\in M$,

The finite sets are directed by inclusion, and $z_F\uparrow1$ strongly, hence ultraweakly, by the bounded monotone convergence theorem. The compactness of the product of ultraweakly compact balls gives a subnet of $(E_F)$ converging ultraweakly at each $b\in B(H)$. Its limit $E$ is linear and completely positive: each matrix quadratic form is a finite sum of vector coefficients and preserves positivity under the limit. Its norm is at most one, $E(1)=1$, and $E(x)=x$ for all $x\in M$. Thus $E$ is a ucp retraction onto $M$. This uses finite overlapping covers rather than a maximal disjoint family of cyclic summands, and works for arbitrary state sets and Hilbert spaces. $\square$

The complete normal-extension and central-quotient proofs supply the central kernels used here. C02 proves the arbitrary projection supremum and strong convergence. The product compactness and predual quotient arguments are in the scalar foundations and coefficient foundations. For commutant permanence use the complete normal-representation corner comparison and amplification, opposite and commutant proofs. Their standard pair is supplied by the arbitrary-algebra construction; the full natural-cone construction provides a standard representation as well. These arguments let the hypothesis be stated using injectivity of each $\pi_\varphi(A)'$.

Convert the two approximation questions

Theorem 3.1. For any C*-algebra $A$,

Proof. If $A=0$, the zero maps through $M_1$ give its identity, and its zero bidual is injective. Now assume $A\ne0$. Starting with nuclear $A$, Proposition 2.1 constructs a retraction onto every state GNS commutant. Commutant permanence gives injectivity of the corresponding represented von Neumann algebras; the finite-cover construction of Lemma 1.1 then gives injectivity of $A^{**}$.

In the reverse direction, let $M=A^{**}$ be injective. The full continuous-core construction, faithful-trace finite matrix models, finite-trace corner passage, and normal coefficient-compression transfer prove the injectivity-to-semidiscreteness implication used here. The core uses the real-line modular action of a faithful normal semifinite weight; the algebra itself need not have a faithful normal state. Its weight, spectral and coefficient inputs are proved in the earlier scalar and weight foundations, with the complete balanced-weight graph and cocycle arguments. All normal tests and arbitrary projection families are retained. The injective/semidiscrete equivalence in the averaging lesson produces cpc predual factorizations

approximating the identity at each predual element in norm. The identification $M_*=A^*$ preserves both the norm and the dual matrix orders. These factorizations therefore satisfy the pointwise weak* hypothesis of Corollary 4.2 in the tensor-positivity lesson, which proves nuclearity of $A$. The norm on $M_n^*$ is the trace norm. Thus the passage from normal tests back to norm approximation uses predual approximation and the proved tensor corollary. $\square$

Corollary 3.2. If $A$ is nuclear, the von Neumann algebra generated by every nondegenerate representation of $A$ is injective and semidiscrete. Conversely, this property for the universal representation, or injectivity for every state GNS representation, implies nuclearity.

Proof. A nondegenerate represented closure is normally isomorphic to a central summand of $A^{**}$. Central corners preserve injectivity, and injectivity gives semidiscreteness. The universal representation realizes the whole bidual. The last assertion follows from Lemma 1.1 and Theorem 3.1. $\square$

For a degenerate representation, the represented closure is understood on its essential Hilbert space. Taking the bicommutant in an ambient space with a zero representation summand adds an unrelated identity there; the normal extension theorem is applied to the essential space.

Apply the projection to an extension

Theorem 4.2. For a closed two-sided ideal $I\subseteq A$,

Proof. Write the normal splitting from Proposition 4.1 as $A^{**}=pA^{**}\oplus(1-p)A^{**}$. If $A$ is nuclear, Theorem 3.1 makes its bidual injective. Its two central corners are injective, so their identifications with $I^{**}$ and $(A/I)^{**}$, followed by Theorem 3.1, make $I$ and $A/I$ nuclear. Conversely, nuclearity of those two algebras makes both bidual summands injective. Their product is injective, and the normal splitting then makes $A^{**}$ injective. Theorem 3.1 returns nuclearity of $A$. $\square$

Example 4.3. Let $A=C([0,1])$, and $I=\{f:f(0)=0\}\cong C_0((0,1])$. Evaluation at zero gives $A/I\cong\mathbb C$. The ideal and quotient are nuclear by the commutative finite-model construction, so the extension theorem gives nuclearity of $A$. The central projection selecting $I^{**}$ belongs to $A^{**}$ and not to $A$; the next exercise section verifies this directly.

The finite models in this example. For $n\geq1$, put $t_j=j/n$, $0\leq j\leq n$. Let $h_j$ be the continuous piecewise linear function that is one at $t_j$, zero at all other grid points, and linear on each interval between consecutive grid points. Then $h_j\geq0$, $\sum_jh_j=1$, and $h_j(t)>0$ only when $|t-t_j|<1/n$. Define

The recording map is a unital star homomorphism into diagonal matrices. The reconstruction map is ucp: for a positive matrix $[X_{ab}]$ over $M_{n+1}$, its pointwise image is the positive matrix $\sum_jh_j(t)[(X_{ab})_{jj}]$, a convex combination of positive scalar compressions. Its unit value is one; the complete unit-value norm formula gives contractivity. Uniform continuity yields

For $I=\{f:f(0)=0\}$, omit the coordinate $j=0$ in both maps. Each remaining $h_j$ vanishes at zero, so reconstruction takes values in $I$; its unit-value function in the ambient algebra is $1-h_0\leq1$, and the same matrix calculation gives complete positivity and contractivity. Since the omitted sample equals $f(0)=0$, the composite and its error estimate are unchanged. The quotient $\mathbb C$ uses the identity through $M_1$. The full finite-model/nuclearity equivalence in Theorem 3.1 of the tensor lesson therefore proves all three nuclearity assertions directly.

Figure. The plotted triangle is exactly $h_2$ for $n=4$; all five hats together give the partition of unity. The maps and the error bound are the ones proved above. Restricting to $I$ removes the zero sample while preserving the composite because $f(0)=0$.

The integration state in Exercise 4 uses the complete scalar integral and $L^2$ construction. Its GNS norm is exactly $\|f\|_2^2=\int_0^1|f|^2$. Continuous functions are dense in this $L^2$ space by the compact-cutoff approximation in that construction, so its GNS representation is multiplication on $L^2([0,1])$. Multiplication by an interval indicator is a bounded projection commuting with every continuous multiplication operator. The nonzero normalized interval indicator belongs to its range, proving its norm is one, as used in the exercise.

Order laboratory: domination and ambient norm

The retraction proof above uses a dilation. A different useful question is how a dominated functional determines an operator in a cyclic commutant. The following construction is retained in full for Exercises 3 and 4; it explains why an order inverse need not be an ambient Banach-space inverse.

Fix a state $\varphi$, its nondegenerate GNS representation $(\pi,H,\xi)$, and $C=\pi(A)'$. The GNS dominated-functional theorem gives the complete order isomorphism

where $D_\varphi$ is the linear span of positive functionals dominated by scalar multiples of $\varphi$. It sends $1$ to $\varphi$, and

The inverse need not be bounded for the ambient norm of $A^*$. The distinction between these two bounds is the subject of Exercises 3 and 4.

The full order proof is the cyclic construction in Section 2 of Tensor positivity and nuclearity, now applied to $\pi(A)\xi$. Namely, for $0\le f\le r\varphi$, define

Cauchy–Schwarz and domination bound this form by $r\|\pi(a)\xi\|\|\pi(b)\xi\|$, so it represents a unique $0\le z_f\le r1$. The form identity with $a^*cb$, for $c\in A$, shows that $z_f$ commutes with $\pi(A)$, hence belongs to $C$. If $A$ is nonunital, use a positive contractive approximate identity $(u_i)$: $\pi(u_i)\xi\to\xi$ and $f(u_i a)\to f(a)$ give

Commutation identifies all other form coefficients from this functional, proving injectivity. Conversely each $0\le z\le r1$ gives the dominated positive vector functional $R_\varphi(z)\le r\varphi$; decomposing an arbitrary element of $C$ into four positive parts gives exactly the claimed linear range. At matrix level, positive matrices over the two commuting algebras pair positively, as proved in the tensor lesson. For the inverse, positivity of $[f_{ij}]$ gives

Density of $\pi(A)\xi$ proves $[z_{ij}]\ge0$. Thus both directions preserve every matrix cone even in the nonunital case, while no ambient inverse norm bound has been introduced.

The central support used by a cyclic restriction

For a projection $e\in M'$, let $w$ project onto $\overline{M'eH}$. This closed space reduces $M'$. It also reduces $M$: for $x\in M$ and $y\in M'$, the vector $xye\eta=yex\eta$ stays in $M'eH$, and the same calculation applies to $x^*$. Consequently $w$ belongs to both commutants and is central. It majorizes $e$. If $z$ is any central projection majorizing $e$, then $M'eH\subset zH$; hence $w\leq z$. This proves directly that $w=c_{M'}(e)$, as in the complete central-support construction C04.

The restriction $T$ of $M$ to $eH$ has kernel $M(1-w)$. If $xe=0$, then $xy e\eta=yxe\eta=0$ for every $y\in M'$, so $xw=0$ by density; the converse follows from $e\leq w$. Restriction is normal: a series-vector functional on $B(eH)$ remains such a functional when its vectors are included in $H$. On $Mw$, it is an injective star homomorphism and hence isometric by TN07. Multiplication by $w$ does not change its value, so the image of the unit ball of $M$ is the image of the unit ball of $Mw$. Normality and dual-ball compactness make this image ultraweakly compact and closed. Bounded strong density ODF07 for the represented algebra then puts the entire unit ball of its closure in that image. Thus the restriction maps $Mw$ isometrically onto the represented closure. The complete preadjoint argument in the central quotient and globally normal inverse, applied to this corner, proves that its inverse is normal as well: its preadjoint is an isometry with closed range, and Hahn–Banach and injectivity make that range all of the corner predual.

In Exercise 2, $\overline{A\xi}$ reduces $M$ by bounded strong density, contains $\xi$, and lies in $zH$; its projection is a nonzero $e\in M'$ with $e\leq z$. The preceding least-central-projection argument gives $0\ne w\leq z$, without requiring $e\in M$.

5. Exercises with solutions

Exercise 1. Let $A=\mathbb C$, and take the direct sum of two copies of its scalar representation on $\mathbb C$. Compute the commutant of this sum and compare it with the product of the two separate commutants.

Solution. The summed representation is $\lambda\mapsto\lambda1_2$, whose commutant is all of $M_2$. The product of the separate commutants is the diagonal algebra $\mathbb C\oplus\mathbb C$. Off-diagonal matrix units intertwine the identical summands. This is why Lemma 1.1 uses central supports rather than a product assertion about universal commutants.

Exercise 2. Assume, as in Lemma 1.1, that every state GNS represented closure of $A$ is injective. Let $M=A^{**}$ act universally, let $z\in M$ be a nonzero central projection, choose a unit vector $\xi\in zH$, and let $e\in M'$ be the projection onto $\overline{A\xi}$. Verify that $w=c_{M'}(e)\le z$, although $e$ need not belong to $M$.

Solution. The projection $z$ is central in both $M$ and $M'$, and $e\le z$. The central support in $M'$ is the least central projection of that algebra majorizing $e$, so it is at most $z$. Since $e\ne0$, also $w\ne0$. If the cyclic represented closure is injective, its normal extension has kernel $M(1-w)$: an operator commuting with $M'$ and vanishing on $eH$ vanishes on $\overline{M' eH}=wH$. The normal-extension theorem then identifies that closure with the nonzero central summand $Mw\subset zM$.

Exercise 3. Suppose a positive map $\Theta:B\to A^*$ has $\Theta(1)=\varphi$, where $B$ is unital and $\varphi$ is a state. Explain why $R_\varphi^{-1}\Theta$ is defined on all of $B$. If $\Theta$ is completely positive, show that this composition has norm one.

Solution. For $b\ge0$, the inequality $b\le\|b\|1$ gives $0\le\Theta(b)\le\|b\|\varphi$, placing $\Theta(b)$ in $D_\varphi$. Positive elements span $B$, so every image belongs to that linear space. The complete order inverse makes the composition completely positive, and its value at the unit is $R_\varphi^{-1}(\varphi)=1$. A unital completely positive map has norm one. No ambient norm bound for $R_\varphi^{-1}$ is needed.

Exercise 4. For $A=C([0,1])$ with the integration state $\varphi(f)=\int_0^1f(t)\,dt$, let $c_n$ be multiplication by $1_{[0,1/n]}$ in the GNS commutant. Compute $\|c_n\|$ and $\|R_\varphi(c_n)\|$. What does this show about the complete order inverse in the order laboratory?

Solution. The multiplication projection has norm one. Its positive functional is $f\mapsto\int_0^{1/n}f(t)\,dt$, with norm $1/n$, attained at the unit. Thus $R_\varphi^{-1}$ is unbounded in the inherited Banach norm of its range. The composition in Exercise 3 remains bounded because it is completely positive and unital, with the required order bounds. Proposition 2.1 constructs its retraction directly by dilation.

Exercise 5. In Example 3.3, prove ultraweak convergence of finite-coordinate truncations on $\ell^\infty(I)$. Explain why their convergence on the unit is usually not norm convergence.

Solution. For $x\in\ell^\infty(I)$, $a\in\ell^1(I)$, and $P_Fx=1_Fx$,

Every absolutely summable family has finite tails as small as desired, giving the directed convergence. If $I$ is infinite, $\|1-P_F1\|_\infty=1$ for every finite $F$. Semidiscrete approximation is tested by normal functionals rather than the operator norm of the unit error.

Exercise 6. Verify surjectivity of $q^{**}$ in Proposition 4.1 directly from $q^*$, including its norm control.

Solution. Identify $(A/I)^*$ isometrically with $I^\perp$ through $q^*$. An element $Y\in(A/I)^{**}$ becomes a bounded functional on $I^\perp$ of norm $\|Y\|$. Hahn–Banach extends it to $X\in(A^*)^*=A^{**}$ with the same norm. The adjoint definition then gives $q^{**}X=Y$. Multiplying $X$ by $1-p$ leaves its image unchanged and cannot increase its norm.

Exercise 7. In Example 4.3 use $u_n(t)=\min(1,nt)$ as an approximate identity for $I$. Determine the values of the central support projection $p\in A^{**}$ under all point-evaluation states. Prove $p\notin A$.

Solution. Uniform convergence $u_nf\to f$ for $f\in I$ follows by making $f$ small near zero and using $u_n=1$ away from zero for large $n$. Hence $u_n\to p$ strongly in the universal representation. The canonical normal extensions of evaluation at $t$ take value zero on $p$ when $t=0$ and value one when $t>0$. If $p$ came from $A$, that continuous function would have these point values, which is impossible at zero. Thus the bidual splitting uses a projection absent from the original C*-algebra.

Exercise 8. Show that the nuclearity–bidual theorem does not justify passing nuclearity to an arbitrary C*-subalgebra merely by inclusion. Identify the unsupported step in that attempted argument.

Solution. An inclusion $B\subseteq A$ induces an inclusion $B^{**}\subseteq A^{**}$. Injectivity of the larger algebra does not itself supply a ucp retraction onto this smaller von Neumann algebra. The permanence results proved here apply to corners, products, suitable directed limits and the explicit averaged fixed-point algebras. An arbitrary von Neumann subalgebra has not been shown to be one of those. Thus Theorem 3.1 cannot establish general subalgebra permanence by that argument; an additional extension or retraction mechanism would be required.

References

Nathanial P. Brown, The symbiosis of C*- and W*-algebras, arXiv:0812.1763v1 (9 December 2008). Proposition 2.3.6, printed p.7, gives injectivity of the commutant of a nuclear represented algebra by an extension into the commuting range. Proposition 3.2.1, Theorem 3.2.2 and Corollary 3.2.3, printed p.10, describe the nuclearity–bidual equivalence and the ideal/quotient extension consequence. Brown leaves injectivity implying semidiscreteness to further von Neumann algebra theory; that implication here uses the full construction in Averaging, crossed products, and injectivity. Proposition 2.1 constructs the represented retraction by a tensor dilation and proves the approximate-identity step needed to define the first tensor factor on the essential dilation space. Lemma 1.1 assembles the state kernels through finite overlapping central covers and a pointwise ultraweak limit. Brown's shared-unit tensor extension and standard-form conjugation are distinct from these full local constructions.

William B. Arveson, Subalgebras of C*-algebras, Acta Mathematica 123 (1969), 141–224. Theorem 1.2.3 proves the extension method behind Brown's commuting-range construction. Lemma 1.4.1 and Theorem 1.4.2 prove the dilation-commutant order correspondence. The full extension and unital dilation providers are linked in Proposition 2.1, which then supplies the nonunital tensor-factor construction. The order laboratory supplies the scalar cyclic form, both matrix directions and the precise domination bound.

• Bruce Blackadar, Operator Algebras, author-hosted edition dated 8 February 2017. III.1.1.13, printed pp.238–239 (PDF pp.246–247), describes central ideals; III.5.2.9–11, printed p.339 (PDF p.347), describes representation extensions, second adjoints of homomorphisms and the ideal bidual splitting. The full ideal approximate-unit, duality and normal-inverse arguments are given in the linked proofs and Proposition 4.1 above.
• [Bidual foundations] The complete state-coefficient predual and bidual product, normal extension, and central quotient with normal inverse.
• [GNS order] The full exact-marginal cyclic order construction and the order laboratory above, including both matrix directions and domination bounds.
• [Central support] The full C04 central-support construction and the cyclic-restriction proof above.

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