Multiplier ultraproducts and normal embeddings

Written by GPT-6.1 Sol (OpenAI), Ultra. Course exposition: CC0-1.0 to the extent of rights held. Earlier programme and human-source notices retain their own terms.

The central sequence algebra uses sequences that asymptotically commute with normal functionals. A larger construction uses every bounded sequence that preserves the sequences vanishing strong*. This multiplier condition is essential for a nontracial algebra. We will prove that the resulting quotient is a von Neumann algebra, with normal copies of the original algebra and its central sequence algebra.

Let $M\ne0$ have a faithful normal state $\varphi$, and let $\omega$ be a free ultrafilter on $\mathbb N$. Sections 1–4 and 6–7 use only this hypothesis; equivalently, $M$ is countably decomposable. Section 5 assumes separable predual for its central sequence algebra. No factor hypothesis is imposed until the type I example. Inputs are faithful normal state GNS representations, the bounded strong* topology, $C^*$-functional calculus and quotients, the definition of a multiplier by its two multiplication maps, and Kaplansky density.

Write

On bounded subsets of $M$, convergence of the latter to zero is precisely strong* convergence. The full faithful-state topology proof, NT2, uses the normal GNS representation, its separating cyclic vector and the dense vectors from its commutant; it does not require a separable Hilbert space. NT1 proves the normal-functional decompositions, module bounds and Cauchy–Schwarz estimates used below. CQ1 proves the complete C* quotient construction, including its approximate identity and exact norm identity. The bounded spectral and density arguments are the complete CF, SF and BD components linked there.

1. The zero algebra and its multipliers

Put

The bounded strong* description shows that $I_\omega$ is independent of the choice of faithful normal state. It is a norm-closed, self-adjoint algebra: products of two bounded strong*-null sequences vanish strong*, by applying them and their adjoints to fixed vectors. It need not be an ideal in the entire sequence algebra.

For example, in $B(\ell^2(\mathbb N_0))$, take basis vectors $\xi_n$, projections $p_n$ onto $\mathbb C\xi_n$, and operators $a_n\xi_0=\xi_n$, zero on the other basis vectors. Then $p_n\to0$ strong*, but $p_na_n=a_n$ does not tend strongly to zero on $\xi_0$. Thus $a\notin N_\omega$.

Proposition 1.1. $N_\omega$ is a unital $C^*$-algebra containing $I_\omega$ as a closed two-sided ideal. It identifies with the multiplier algebra of $I_\omega$.

Proof. The two ideal-preservation conditions survive products, sums and adjoints. For uniform norm limits, if $a^{(r)}\to a$ and $z\in I_\omega$ is bounded, the products with $a-a^{(r)}$ have operator norms at most $\|a-a^{(r)}\|\sup_n\|z_n\|$. First take the ultralimit for fixed $r$, then let $r\to\infty$. This proves closure and both ideal conditions. The unit sequence belongs to $N_\omega$, and $I_\omega\subset N_\omega$ follows from its own product closure.

To see the multiplier identification, let $q_n$ be the sequence equal to $1$ at coordinate $n$ and zero elsewhere. These central projections belong to $I_\omega$, and $q_nI_\omega$ is the unital coordinate algebra $M$. For an abstract multiplier $m$, the products $mq_n,q_nm$ belong to $I_\omega$. The right-support identity $(mq_n)q_n=mq_n$, and centrality of $q_n$ in $I_\omega$, give $mq_n=q_nmq_n=q_nm$. Thus $mq_n$ specifies an operator $a_n\in M$, with $\|a_n\|\le\|m\|$.

For $z\in I_\omega$, the coordinate of $mz$ is $a_nz_n$, and that of $zm$ is $z_na_n$. Hence $a=(a_n)$ lies in $N_\omega$ and realizes $m$. Conversely every element of $N_\omega$ defines those bounded left and right multiplier maps. Coordinate tests $q_n$ and the operator norm bounds show that the identification is isometric. No assertion that the finite-coordinate projections approximate every element of $I_\omega$ in norm is needed. $\square$

Lemma 1.2. A bounded sequence $a_n$ belongs to $N_\omega$ if and only if, for every $\varepsilon>0$, there are $\delta>0$ and $A\in\omega$ such that

Proof. The displayed uniform condition applies to the coordinates of any bounded $z\in I_\omega$, after rescaling its operator norm bound. Intersect $A$ with the set on which its seminorm is less than $\delta$; this proves both products vanish.

Conversely, if (3) fails for some $\varepsilon$, the sets

belong to $\omega$. Otherwise their complements would provide (3). They are decreasing. Let $r(n)$ be the largest $r\le n$ with $n\in B_r$, and zero if none exists. Then $r(n)\to_\omega\infty$. Choose a witnessing $z_n$ for $r(n)>0$, and use zero otherwise. These contractions belong to $I_\omega$, but the sum of the two product seminorms stays at least $\varepsilon$ on an $\omega$-large set. This contradicts $a\in N_\omega$. $\square$

Constant sequences lie in $N_\omega$, because multiplication by a fixed operator preserves bounded strong* convergence on either side.

2. A faithful quotient state

Define

The superscript distinguishes this algebra from the central sequence algebra $M_\omega$.

Lemma 2.1. Formula (5) defines a faithful state on the quotient.

Proof. It is a state on $N_\omega$ by scalar ultralimits and vanishes on $I_\omega$ by Cauchy–Schwarz. It therefore descends to the quotient. If $X\ge0$, lift $X^{1/2}$ to $b\in N_\omega$, and put $h_n=b_n^*b_n$. Then $h$ is positive and bounded, with $\pi(h)=X$. If $\varphi^\omega(X)=0$, then for a bound $h_n\le C1$,

Since $h_n=h_n^*$, this is its symmetric strong* seminorm squared. Thus $h\in I_\omega$ and $X=0$. Faithfulness follows. $\square$

Use its faithful GNS representation, with cyclic vector $\Omega$, and put $\|X\|_{\varphi^\omega}=\varphi^\omega(X^*X)^{1/2}$. For a representative $x\in N_\omega$, this norm equals the ultralimit of $\|x_n\|_\varphi$.

Here the quotient state is used only as a state on a C* algebra; its normality has not yet been asserted. Its GNS construction is direct. Complete the inner-product space with form $\varphi^\omega(X^*Y)$, whose null space is zero by faithfulness. Positivity gives $\|AX\|_{\varphi^\omega}\le\|A\|\|X\|_{\varphi^\omega}$, so left multiplication extends to a unital star representation with cyclic vector $\Omega=1$. It is faithful, since $A\Omega=0$ implies $\varphi^\omega(A^*A)=0$. It is also isometric: for $h=A^*A\ne0$, choose a nonzero continuous spectral cut $f(h)$ supported arbitrarily close to the spectral maximum $\lambda=\|A\|^2$. Then $\|A f(h)\|_{\varphi^\omega}^2\ge(\lambda-\varepsilon)\|f(h)\|_{\varphi^\omega}^2$, and the latter norm is nonzero by faithfulness. This gives the lower operator-norm bound as $\varepsilon$ tends to zero. Thus the represented image is a norm-closed C* algebra, and $\Omega$ separates its elements. Theorem 3.2 proves strong closure and vector-state normality afterward.

3. Completeness on two cyclic vectors

Lemma 3.1. For every fixed $Y\in M^\omega$, the self-adjoint unit ball is complete for

Proof. Select a subsequence $X_r$ of the given Cauchy sequence so that $p_Y(X_r-X_p)<2^{-p-2}$ whenever $r>p$. Choose self-adjoint contraction representatives $a_r(n)\in N_\omega$, using a self-adjoint lift followed by continuous clipping to $[-1,1]$. Fix a multiplier representative $y(n)$ of $Y$.

For every $r$, only finitely many earlier rows $p<r$ need to be tested. Their quotient distances are the ultralimits of the corresponding state seminorms. We can therefore choose decreasing sets $A_r\in\omega$, with $A_r\subseteq\{n\ge r\}$, such that

Start with $A_0=\mathbb N$, and intersect the finitely many sets where these inequalities hold with $A_{r-1}$ and the indicated cofinite set. Put $r(n)=\max\{r\le n:n\in A_r\}$, or zero if the set is empty, and take $a(n)=a_{r(n)}(n)$, with $a_0(n)=0$. Each $a(n)$ is a self-adjoint contraction. For every fixed $p$, the set $A_{p+1}$ makes $r(n)>p$, so (8) gives

This selection uses only finite intersections at each stage; no common intersection of all $A_r$ is required.

To prove that $a$ is a multiplier, let $z=(z(n))\in I_\omega$, with $\|z(n)\|\le K$. Since every fixed row $a_p$ is a multiplier, the triangle inequality and (9) give

Let $p$ increase to infinity. This proves the first null seminorm of $za$. Self-adjointness gives $(za)^*=az^*$, and contractivity gives $\|a(n)z(n)^*\|_\varphi\le\|z(n)^*\|_\varphi$; hence its other null seminorm vanishes as well. Apply the same argument to $z^*$ to obtain $z^*a\in I_\omega$, whose adjoint is $az$. Thus both $za$ and $az$ belong to $I_\omega$, and $a\in N_\omega$.

We may now pass (9) to the quotient: $p_Y(\pi(a)-X_p)\le2^{-p}$. The selected subsequence converges in the self-adjoint unit ball, and its original Cauchy sequence has the same limit. This proves completeness. $\square$

Theorem 3.2. $M^\omega$ is a von Neumann algebra in this GNS representation, and $\varphi^\omega$ is faithful normal.

Proof. Suppose $X_i$ is a strong-operator Cauchy net of self-adjoint contractions. For every $Y\in M^\omega$, it is Cauchy for (7), since that expression tests the vectors $\Omega,Y\Omega$. Metric completeness applies to this net: choose increasing indices $i_r$ beyond a $2^{-r}$-Cauchy tail for that metric. The sequence $X_{i_r}$ has a limit by Lemma 3.1, and the triangle inequality between the sequence and each sufficiently late net element proves convergence of the whole net. No countable cofinal subset of its directed index set is assumed. Thus there is an element $X_Y$ with

All the $X_Y\Omega$ are the same limit. The state is faithful, so equality on $\Omega$ implies equality of these algebra elements. Denote their common value by $X$. Equation (11) for all $Y$, density of $M^\omega\Omega$, and uniform boundedness give $X_i\to X$ strongly.

Thus the self-adjoint unit ball is strongly closed. Kaplansky density makes it strongly dense in the self-adjoint unit ball of the bicommutant. These balls therefore agree, and real and imaginary parts show that the whole represented algebra equals its bicommutant. The state is its vector state, hence normal, and remains faithful by Lemma 2.1. $\square$

4. The normal copy of the original algebra

The constant map

is a unital injective *-homomorphism: a constant sequence can belong to $I_\omega$ only if its operator is zero. Its state restriction is $\varphi^\omega\circ\iota=\varphi$. It is normal. If $0\le a_i\uparrow a$, the supremum $Z=\sup_i\iota(a_i)$ in $M^\omega$ satisfies $Z\le\iota(a)$, and

Faithfulness gives $Z=\iota(a)$. We henceforth identify $M$ with this normal copy.

5. The normal centralizing subalgebra

Now assume $M$ has separable predual, but allow nonfactor algebras. Let $C_\omega$ consist of bounded sequences with $\|[x_n,\psi]\|\to_\omega0$ for every $\psi\in M_*$, using the functional convention of the preceding central sequence lessons. Every element of $I_\omega$ belongs to $C_\omega$, by the normal-functional Cauchy–Schwarz estimate.

Also $C_\omega\subset N_\omega$. To verify the potentially troublesome right multiplication, take $\|a_n\|\le C$ and $\|z_n\|\le K$, with $a$ centralizing and $z\in I_\omega$. Moving $a_n^*$ past the state gives

The second line is Cauchy–Schwarz applied to the positive bounded factors. The other right/left adjoint estimate is identical with $z_nz_n^*$ and $[a_n,\varphi]$. The two remaining seminorms are bounded directly by $C\|z_n\|_\varphi$ or $C\|z_n^*\|_\varphi$.

Theorem 5.1. The inclusion induces a normal embedding of the finite von Neumann algebra

Its faithful normal trace is $\tau_\omega=\varphi^\omega|_{M_\omega}$. In the factor case this is the previously defined scalar-limit trace.

Proof. The kernel of inclusion on $C_\omega$ is exactly $I_\omega$, so the quotient embeds isometrically. Its state is faithful by Lemma 2.1, and is tracial because

Consequently its $2$-norm is the ultralimit of the symmetric seminorm (1).

Here is the completeness argument without assuming a scalar weak limit. Choose a norm-dense sequence of normal states $\psi_j$. For a fast $2$-Cauchy subsequence $X_r$ of quotient contractions, choose centralizing contraction representatives $x_r(k)$, with $\|X_r-X_p\|_2<2^{-p}$ for $r>p$. Choose nested sets $A_r\in\omega$, inside $\{k\ge r\}$, on which

Each finite collection is available by the ultralimit trace identity and centralizing. Set $r(k)=\max\{r\le k:k\in A_r\}$, or zero when empty, and $x(k)=x_{r(k)}(k)$, using $x_0=0$. Since $r(k)\to_\omega\infty$, the first line and norm density make $x$ centralizing. The second line gives $\|\pi(x)-X_p\|_2\le2^{-p}$. The unit ball is therefore $2$-complete.

For a strong Cauchy net in this represented unit ball, convergence on the faithful vector $\Omega$ makes it $2$-Cauchy. Completeness gives a limit in $M_\omega$. Trace equality makes both its first and adjoint state seminorms converge, so the bounded faithful-state strong* criterion in $M^\omega$ gives strong convergence to that limit. Kaplansky density now proves that this unit ball, and hence the subalgebra, is strongly closed. It is a von Neumann subalgebra of $M^\omega$, with the normal restriction of its faithful state. This proves normality of the embedding and finiteness. $\square$

6. Automorphisms and convergent implementing families

Theorem 6.1. A fixed normal automorphism $\beta$ acts term by term on $M^\omega$. More generally, if $\beta_n\to\beta$ in the $u$-topology, the sequence action induces a normal automorphism

It restricts to $\beta$ on the constant copy of $M$. At separable-predual generality it also preserves $M_\omega$.

Proof. For bounded $z\in I_\omega$,

The adjoint seminorm has the same estimate. Inverses converge in the $u$-topology as well. For any fixed $\psi\in M_*$, precomposition by $\beta_n$ is isometric, so

The last limit tests the fixed normal functional $\psi\circ\beta^{-1}$. This needs neither a metric nor a separable Banach space. Thus the termwise sequence automorphism preserves $I_\omega$ in both directions. It preserves its multiplier algebra, so induces a *-automorphism of the quotient. A bijective *-homomorphism between von Neumann algebras is normal: its order inverse preserves the least-upper-bound property of every increasing bounded positive net.

For fixed $a\in M$, $u$-convergence gives $\beta_n(a)\to\beta(a)$ strong*. Expand each state square norm of their difference; multiplicativity makes the square term $\beta_n(a^*a)$, and the cross terms involve fixed normal functionals applied to $\beta_n(a)$. Their limits cancel. The adjoint expansion is the same. This proves the constant restriction.

For a centralizing $x_n$, the predual identity gives

The inverse family gives equality of the invariant subalgebra. $\square$

A constant automorphism family is called liftable; a convergent variable family is semi-liftable. Formula (18) keeps the family in its notation: the limit $\beta$ alone does not determine that automorphism. The next lesson gives the exact counterexample.

If $\operatorname{Ad}(u_n)\to\alpha$, the unitaries $u_n$ belong to $N_\omega$. Indeed, the only nonautomatic square seminorms of their products with $z_n\in I_\omega$ are tested by $\varphi\circ\operatorname{Ad}(u_n)$ or its inverse; these converge in predual norm to the corresponding fixed normal states of $\alpha$. The estimate (19) applies. Thus $U=\pi(u_n)$ is a unitary and

This realizes an approximately inner automorphism of $M$ by an actual inner automorphism in the larger algebra.

7. Dense representatives and type I factors

Lemma 7.1. If $A_0\subset M$ is an ultraweakly dense *-subalgebra, every $X\in M^\omega$ has a bounded multiplier representative $b_n\in A_0$.

Proof. Start with a multiplier representative $a_n$, uniformly bounded by $C$. If $C=0$ use zero. Otherwise Kaplansky density for the norm closure of $A_0$, followed by norm approximation from $A_0$, gives $b_n\in A_0$ with $\|b_n\|\le2C$ and $\|b_n-a_n\|_{\varphi,\#}<1/n$. The difference belongs to $I_\omega\subset N_\omega$; hence $b$ is a multiplier and $\pi(b)=X$. No assumption that $A_0$ contains the unit is needed. $\square$

Theorem 7.2. If $M=B(H)$ is a countably decomposable type I factor, then every multiplier sequence converges strong* along $\omega$. The constant embedding is a normal isomorphism $M\cong M^\omega$.

Proof. The finite-dimensional case follows from compactness of bounded operator balls. In infinite dimension $H$ is separable. Indeed, the rank-one projections from any orthonormal basis have strictly positive values under a faithful normal state, and every finite sum of those values is at most one. For each positive integer $r$, at most $r$ of them can have value at least $1/r$; their union exhausts the basis. Choose a basis $\xi_j$ and the faithful diagonal state

Changing the faithful state does not change (2). Put $P_m$ for the first $m$ basis projection and $q_m=1-P_m$. Then $\|q_m\|_{\varphi_0,\#}\to0$.

Let $a\in N_\omega$, and let $A$ be its bounded ultraweak $\omega$-limit. Given a fixed $j$, Lemma 1.2, applied to $q_m$ for sufficiently large $m$, makes both

arbitrarily small on an $\omega$-large set. For the first, the nonadjoint seminorm of $q_ma_n$ includes the square in (23) with coefficient $2^{-j-1}/2$; for the second use the adjoint seminorm of $a_nq_m$.

In the finite-dimensional range of $P_m$, the vectors $P_ma_n\xi_j$ converge in norm along $\omega$ to $P_mA\xi_j$, by ultraweak convergence of their finitely many coefficients. Choose $m$ also large enough that $q_mA\xi_j$ is small. Equation (23) then proves $a_n\xi_j\to_\omega A\xi_j$. The same argument for adjoints gives $a_n^*\xi_j\to_\omega A^*\xi_j$. Uniform boundedness extends both conclusions from the basis span to every vector. Thus $a_n-A\in I_\omega$, and every quotient element is constant. Normality and injectivity of the constant embedding were proved in Section 4. $\square$

8. Exercises with complete solutions

Exercise 1. Verify that the null space in (2) is self-adjoint and product closed.

Solution. Adjointing interchanges the two state seminorms. In a faithful normal representation, if $x_n,y_n\to_\omega0$ strong* with norm bounds $C,D$, then $\|x_ny_n\xi\|\le C\|y_n\xi\|\to0$, and $\|(x_ny_n)^*\xi\|\le D\|x_n^*\xi\|\to0$. Thus the products vanish strong* as well.

Exercise 2. In the rank-one example in Section 1, compute the symmetric seminorm of $p_na_n$ for the state (22).

Solution. Since $p_na_n=a_n$, $a_n^*a_n=p_0$ and $a_na_n^*=p_n$. Its squared symmetric seminorm is $(2^{-1}+2^{-n-1})/2$, which tends to $1/4$, rather than zero. Meanwhile $\|p_n\|_{\varphi_0,\#}^2=2^{-n-1}\to0$. This proves both the nonideal phenomenon and failure of the multiplier condition.

Exercise 3. Why does the diagonal witness in Lemma 1.2 avoid a countable intersection of ultrafilter sets?

Solution. For each fixed $r$, $B_r\cap\{n\ge r\}$ is an $\omega$-large set on which $r(n)\ge r$. These separate threshold bounds give $r(n)\to_\omega\infty$. A common intersection over all $r$ is neither required nor asserted.

Exercise 4. Explain why multipliers commute with the coordinate projections $q_n$.

Solution. The element $mq_n$ lies in $I_\omega$ and has right support $q_n$, since $(mq_n)q_n=mq_n$. Centrality of $q_n$ within $I_\omega$ makes it also left-supported there. The same argument applies to $q_nm$, and both equal $q_nmq_n$. Their coordinate value therefore lies in the unital coordinate algebra $M$.

Exercise 5. Prove faithfulness of the quotient state on a positive element without assuming the state is tracial.

Solution. Lift the positive quotient element as $h=b^*b$ in the multiplier algebra. If its state is zero, the positive operator inequality $h_n^2\le Ch_n$ gives vanishing of $\varphi(h_n^2)$. Self-adjointness makes this both strong* seminorms, so $h\in I_\omega$. The quotient element is zero. No cyclic trace interchange occurs.

Exercise 6. Verify that changing a representative outside an $\omega$-large set leaves both the quotient and multiplier membership unchanged.

Solution. The bounded difference is zero on that set, so its symmetric seminorm has ultralimit zero. It belongs to $I_\omega$, which is an ideal inside $N_\omega$ and itself lies in $N_\omega$. Adding it preserves multiplier membership and the quotient image.

Exercise 7. Explain why self-adjointness is used in the multiplier step of Lemma 3.1.

Solution. The diagonal representative is a self-adjoint contraction at every coordinate. Its one-sided state distance from each fixed multiplier row has ultralimit at most $2^{-p}$. For a bounded null sequence $z$, (10) therefore makes the first seminorm of $za$ vanish as $p$ increases. Self-adjointness gives $(za)^*=az^*$, whose first seminorm is at most $\|z^*\|_\varphi$. Applying this same pair of estimates to $z^*$ makes $z^*a$ null in both seminorms; its adjoint $az$ is consequently null as well. Thus both products belong to $I_\omega$. A one-sided estimate for a general nonsymmetric representative would not give these adjoint identities or both multiplier conditions.

Exercise 8. In Theorem 3.2, why do all the limits $X_Y$ agree?

Solution. They give the same vector limit on $\Omega$, because the first term of every $p_Y$ is the same norm. Thus $(X_Y-X_Z)\Omega=0$. Faithfulness of the state implies $\varphi^\omega((X_Y-X_Z)^*(X_Y-X_Z))=0$ only when $X_Y=X_Z$. This produces one algebra element controlling every dense cyclic vector.

Exercise 9. Prove normality of the constant embedding by increasing positive nets.

Solution. If $a_i\uparrow a$, the image supremum $Z$ is bounded by $\iota(a)$. Normality of $\varphi$ and the vector-state normality of $\varphi^\omega$ give zero state on the positive difference, as in (13). Faithfulness makes that difference zero, so the embedding preserves every such supremum.

Exercise 10. When $M$ has a faithful normal tracial state, show that $N_\omega=\ell^\infty(\mathbb N,M)$.

Solution. Use that trace to define the unchanged ideal. For bounded $a_n$, the tracial $2$-norm inequalities $\|a_nz_n\|_2,\|z_na_n\|_2\le\sup_n\|a_n\|\|z_n\|_2$ show preservation on both sides. Thus every bounded sequence is a multiplier. The ideal restriction is needed in the general nontracial construction.

Exercise 11. Derive the predual error term in (20).

Solution. Replace $\psi\circ\beta_n$ by the fixed functional $\psi\circ\beta$ inside $[x_n,\cdot]$. The uniform estimate $\|[x_n,\rho]\|\le2\|x_n\|\|\rho\|$ bounds the difference by the displayed second term. Normalizing by $\beta_n^{-1}$ does not change its norm. The fixed commutator and the predual error both tend to zero.

Exercise 12. Why is the implementing unitary in (21) allowed to have a nonscalar quotient even though the induced map has a prescribed limit on $M$?

Solution. Membership in $N_\omega$ means preservation of strong*-null sequences, rather than scalar-triviality. Its quotient unitary can contain additional asymptotic information. Equation (21) specifies its action on the constant copy of $M$; it does not determine its action on all varying quotient elements. The tensor-tail counterexample in the next lesson exhibits this distinction.

Exercise 13. Produce bounded representatives from a dense *-subalgebra that does not contain $1$.

Solution. Let $A_0$ be the finite-rank operators in $B(\ell^2)$, and let $a_n$ represent $X$ with bound $C>0$. Kaplansky approximation in their norm closure, followed by finite-rank norm approximation, gives $b_n\in A_0$, bounded by $2C$, with symmetric state error less than $1/n$. Their difference lies in $I_\omega$, so they represent the same element and remain multipliers. For the unit itself one can simply use increasing finite-rank projections.

Exercise 14. Identify the role of finite-dimensional compactness in Theorem 7.2.

Solution. The multiplier condition makes the tails in (23) uniformly small on an $\omega$-large set. Within a fixed finite-dimensional range, weak coefficient convergence is norm convergence. Adding the small tails gives norm convergence of each full column and row. Ultraweak convergence alone, without the multiplier tail control, would not give this conclusion.

References

Adrian Ocneanu's freely accessible Actions of discrete amenable groups on factors, thesis, Chapter 5, Sections 5.1–5.2, printed pp.47–52 (PDF pp.61–66), gives the normalizing quotient, two-vector completeness, normal subalgebras and convergent automorphism families. Its completeness proof modifies representatives and takes an operator limit at each coordinate. Here finite simultaneous row tests select one diagonal representative, and the limiting multiplier estimates establish both null seminorms before passing to the quotient. Sections 1–4 and 6–7 require a given faithful normal state; the centralizing subalgebra in Section 5 explicitly retains separable predual. The normality proof handles arbitrary strong-Cauchy nets, and a semi-liftable automorphism keeps its chosen family.

Hiroshi Ando and Uffe Haagerup, Ultraproducts of von Neumann algebras, v3, Lemma 3.13, gives the simultaneous-vector diagonal method comparison. Lemmas 3.8 and 3.10, Propositions 3.14–3.15 and Theorem 3.7 give a separate support-corner realization for sigma-finite algebras. Their Theorem 3.9 quotes the left-ideal/hereditary correspondence from another text; no such correspondence or standard-form corner construction is a premise of the direct completeness argument here. Full faithful-state topology, quotient functional calculus and bounded density proofs are supplied in the programme. Human-source rights remain separate from the course dedication.