Fast reindexing with liftable actions

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.

Reindexing lets a separable part of an ultraproduct move farther out along its representatives. The finite tests must preserve multiplication, the multiplier condition and the desired action. This lesson proves the fast construction with hypotheses that survive the semi-lift counterexample.

Let $M$ have separable predual and a faithful normal state $\varphi$. It need not be a factor or finite. Use the multiplier ultraproduct $M^\omega=N_\omega/I_\omega$, its normal centralizing subalgebra $M_\omega$, and the canonical expectation $E_\omega:M^\omega\to M$. Write $\|\cdot\|_\#$ for the symmetric seminorm associated to $\varphi$, and $\|\cdot\|_{\varphi^\omega,\#}$ for its quotient version. On bounded subsets these describe strong* convergence. Subalgebras here share the ambient identity, as in the unital target and constant-fixing clauses. The complete faithful-state topology and coefficient proofs, bounded density, and positive order-normal to predual theorem NF1–4 provide the normal extension inputs. The general centralizing-algebra proof applies without a factor assumption.

1. The corrected theorem

Theorem 1.1. Let $P,Q\subset M^\omega$ be von Neumann subalgebras with separable predual. Let $G$ be a countable group of liftable automorphisms of $M^\omega$, meaning constant lifts $\theta^\omega$ of normal automorphisms $\theta$ of $M$. Suppose $G$ leaves $P$ invariant. There is a normal unital injective *-homomorphism $\Phi:P\to M^\omega$ such that

In particular $E_\omega\Phi=E_\omega|_P$, and $\Phi$ preserves the faithful normal state $\varphi^\omega|_P$.

The expectation identity keeps the nonfactor and nontracial cases. For a factor and $x\in P\cap M_\omega$, it implies

For $a\in Q\cap M_\omega$ as well, (2) is trace factorization on the finite centralizing algebra. Full scalar factorization for every $x\in P$ would contradict fixed constants, even for $M=R$.

We prove all assertions, including membership of the reindexed representatives in $N_\omega$. An arbitrary reindexing of a multiplier sequence need not preserve that membership.

2. Countable data and bounded representatives

Replace $P,Q$ by $P\vee M,Q\vee M$, and eventually restrict the constructed map back to the original $P$. These joins still have separable predual. Indeed each original algebra, and $M$, has a countable set generating it as a von Neumann algebra. Their union generates each join. In the faithful normal state GNS representation of that join, countably many rational *-words applied to the cyclic vector have dense span, by strong density. The representation Hilbert space is separable, so its represented algebra has separable predual. The enlarged $P$ remains $G$-invariant because constant lifts preserve $M$.

Here are the predual details behind the countability assertion and Exercise 1. For a separable Hilbert space $H$, complete its finite-rank operators for the norm $\operatorname{Tr}|t|$; this is the trace-class space $S_1(H)$. For a finite-rank operator, finite-dimensional singular-value decomposition and Cauchy–Schwarz show that the infimum of $\sum_j\|\xi_j\|\|\eta_j\|$ over its rank-one decompositions equals that norm. The functional $b\mapsto\operatorname{Tr}(tb)$ has the same norm: the upper bound follows term by term, and the adjoint polar partial isometry attains the trace norm. Thus the coefficient map extends isometrically to the completion, into the norm-closed normal predual of $B(H)$. Every normal coefficient series gives a trace-norm Cauchy series of rank-one operators, so this map is onto. A completion element also gives its operator-norm limit, since $\|t\|\le\operatorname{Tr}|t|$. If this limit is zero, its normal functional vanishes on every finite-rank matrix unit and then on every bounded operator by bounded strong density and normality, proving that the completion identifies with actual trace-class operators.

Finite rational rank-one sums from a countable dense set of vectors are dense in this trace norm, so $S_1(H)$ is separable. Restriction of the full coefficient series to any represented von Neumann subalgebra is onto its concrete predual: that predual is exactly the quotient coefficient space furnished by the full predual theorem. Hence it is a separable quotient. A unital subalgebra of a separable-predual algebra is represented faithfully and normally on the latter's separable faithful-state GNS space and has the same quotient conclusion. This also treats both intersections used to choose the countable data below.

Choose countable unital *-algebras over $\mathbb Q(i)$,

ultraweakly dense in their ambient algebras. Arrange that $\mathcal A\cap M$ and $\mathcal B\cap M$ are dense in $M$, and that $\mathcal A\cap M_\omega$ and $\mathcal B\cap M_\omega$ are dense in their respective intersections. Include countable strong* dense sets from the unit balls of those intersections before generating the algebras. Include all $G$-translates of the generators for $\mathcal A$; countability of $G$ makes $\mathcal A$ countable and invariant.

Exhaust these data by increasing finite sets

and let $\psi_j$ be norm dense in $M_*$. The finite sets need not themselves be algebras: all resulting sums, products and translates already have representatives indexed by the whole countable algebra.

For each $x\in\mathcal A$, choose a representative $u_x(k)\in N_\omega$, uniformly bounded by $\|x\|$. Such a lift exists by continuous radial clipping of any lift: for $C=\|x\|>0$, replace a lift $b$ by $b f(b^*b)$, where $f(t)=\min(1,C/\sqrt t)$, with $f(0)=1$. Its image is unchanged and its norm is at most $C$. Choose $u_0=0$. For $x\in M$ choose the constant representative. For $x\in M_\omega$ choose a representative in $C_\omega$, using the same quotient lifting inside that $C^*$-algebra.

Likewise choose bounded multiplier representatives $v_a(k)$ for $a\in\mathcal B$, and use constants for $a\in M$. Exact algebraic coherence of the representatives is unnecessary. For example

Addition and rational complex scalar multiplication have the same null-error property.

3. Multiplier moduli and the finite selection

For every $x\in\mathcal A$ and positive integer $l$, the multiplier criterion gives $\delta_l(x)>0$ and $W_l(x)\in\omega$ such that

Shrinking the moduli and taking finite intersections lets us assume they decrease with $l$. No countable intersection is used.

At outer index $n$, choose an inner index $p(n)\ge n$ satisfying the following finite list:

1. $p(n)\in W_l(x)$ for $x\in\mathcal A_n,\ l\le n$.
2. Each addition, adjoint, multiplication and rational scalar relation for operands in $\mathcal A_n$ has symmetric seminorm error below $1/n$ at $p(n)$. For scalars, test the first $n$ elements of a fixed enumeration of $\mathbb Q(i)$.
3. For $x\in\mathcal A_n\cap M_\omega,\ j\le n$, $\|[u_x(p(n)),\psi_j]\|<1/n$.
4. For $x\in\mathcal A_n\cap M_\omega,\ a\in\mathcal B_n$, $\|[u_x(p(n)),v_a(n)]\|_\#<1/n$.
5. For $x\in\mathcal A_n,\ a\in\mathcal B_n,\ j\le n$,

6. For $x\in\mathcal A_n,\ \theta^\omega\in G_n$,

Lemma 3.1. Such a choice exists for every $n$.

Proof. Every condition specifies an $\omega$-large set of admissible inner indices. The first uses (6); the second uses null quotient relations such as (5). The third uses the centralizing representative.

For the fourth, $v_a(n)$ is a fixed element of $M$ while choosing the inner index. Centralizing sequences commute strong* along $\omega$ with every fixed element of $M$, by the normal-functional centrality criterion proved in the ordinary central sequence lesson. Thus the fourth condition is also available.

For the fifth, $u_x(k)$ converges ultraweakly along $\omega$ to $E_\omega(x)$; at this fixed $n$, the functional $y\mapsto\psi_j(v_a(n)y)$ is normal and fixed. For the sixth, the difference represents zero because $\theta^\omega$ is the constant lift of $\theta$. It therefore lies in $I_\omega$.

Intersect these finitely many sets with the cofinite set $\{k\ge n\}$. A free ultrafilter contains the resulting nonempty infinite set, so a choice is possible. $\square$

The normal functional in (7) varies when $n$ changes, but it is fixed during each individual selection. This is the reason fast selection handles its moving coefficient $v_a(n)$.

4. The map on the countable algebra

Set

Lemma 4.1. Every $w_x$ belongs to $N_\omega$. For $x\in\mathcal A\cap M_\omega$, it is an ordinary centralizing sequence.

Proof. Fix $x,l$. Once $n$ is large enough that $x\in\mathcal A_n$ and $n\ge l$, the selected index lies in $W_l(x)$. Thus (6), with its fixed positive $\delta_l(x)$, holds for every sufficiently large outer index. If $z_n$ is a contraction sequence in $I_\omega$, its seminorm is below this fixed modulus on an $\omega$-large set. Both products with $w_x(n)$ have seminorm below $1/l$ there. Let $l\to\infty$, and rescale to handle any bounded null sequence. This proves both multiplier conditions.

For centralizing $x$, condition 3 gives ordinary convergence of its commutators with each $\psi_j$. The bound $\|[w_x(n),\rho]\|\le2\|x\|\|\rho\|$ and norm density give ordinary convergence for every $\rho\in M_*$. $\square$

The finite relation tests make $\Phi_0$ a unital *-homomorphism over $\mathbb Q(i)$. Its operator bound is

Taking $a=1$ in (7), and using norm density of the normal functionals, gives

It follows that $\varphi^\omega\Phi_0=\varphi^\omega|_{\mathcal A}$. Applying this to $x^*x$ and $xx^*$ gives the exact isometry

In particular the map is injective on $\mathcal A$.

The remaining tests already imply, on the countable data,

For the middle equality, (7) makes the representative error ultraweakly null, while $v_a(n)E_\omega(x)$ has ultraweak ultralimit $E_\omega(a)E_\omega(x)$. All products are legitimate multiplier products by Lemma 4.1. The first equality follows from condition 4, and the last from (8). Constants in $\mathcal A\cap M$ are fixed exactly.

5. Normal extension and all assertions

Lemma 5.1. A norm-contractive unital *-homomorphism on an ultraweakly dense countable rational *-algebra, preserving a faithful normal state as above, extends uniquely to a normal injective *-homomorphism on its von Neumann closure.

Proof. First extend by norm continuity to the $C^*$-closure $A=\overline{\mathcal A}^{\,\|\cdot\|}$. Continuity with respect to rational complex scalars gives complex linearity. Multiplication, adjoints, the state identity and (12) survive norm limits.

For $x\in P$, Kaplansky density supplies a bounded net $x_i\in A$ converging strong* to $x$, with $\|x_i\|\le\|x\|$. The symmetric state seminorm makes it Cauchy; (12) makes its image Cauchy in the same seminorm, with the same uniform operator bound. In the von Neumann algebra $M^\omega$, a bounded strong*-Cauchy net has a strong* limit. Explicitly, in its faithful normal representation, each vector image of the net and each adjoint-vector image is Cauchy. Their limits define bounded operators $T,S$ with the common norm bound. Passing to the limit in their inner products gives $S=T^*$. Strong closedness of $M^\omega$ puts $T$ in that algebra. Cauchy convergence of the symmetric faithful-state seminorm gives this full strong* Cauchy property: first on the cyclic separating vector and its dense commutant orbit, by commutation, then on all vectors by the uniform bound. Finite heads and summable tails of vector tests give the amplified topology as well. This is an argument for the entire net. The faithful normal state criterion therefore gives an image limit $\Phi(x)$.

If two nets approximate $x$, their differences have vanishing symmetric seminorm, so (12) makes the image limits equal. The construction is thus well-defined and agrees with the original map. Using simultaneous bounded approximating nets for two elements, joint strong* continuity of multiplication on bounded sets and continuity of adjoints give multiplicativity, linearity and the *-identity. The operator bound and the state identity survive. Faithfulness of the state makes $\Phi$ injective.

For normality, let $0\le x_i\uparrow x$ in $P$. In $M^\omega$, put $Y=\sup_i\Phi(x_i)\le\Phi(x)$. Preservation and normality of the state give

Its faithfulness gives $Y=\Phi(x)$. Thus $\Phi$ preserves increasing positive suprema. Composing it with a positive normal functional of the target gives a bounded order-normal positive functional of $P$; NF1–4 puts that functional in $P_*$. The four-positive-functional decomposition proves ultraweak continuity for every normal coefficient. Thus $\Phi$ is normal on its whole domain. Uniqueness follows from ultraweak density and normality. $\square$

Apply this lemma to (9). The constant restriction extends from $\mathcal A\cap M$ to $M$. For the second assertion of (1), use bounded strong* density of $\mathcal A\cap M_\omega$ in $P\cap M_\omega$. Its images are centralizing by Lemma 4.1 and commute with the ultraweakly dense algebra $\mathcal B$. The normal subalgebra $M_\omega$ and the relative commutant $Q'\cap M^\omega$ are both strongly closed, so the full image has the required inclusion.

The expectation equality in (13) extends separately: first in $x$, for each fixed $a\in\mathcal B$, because $x\mapsto E_\omega(a\Phi(x))$ and $x\mapsto E_\omega(a)E_\omega(x)$ are normal linear maps. Then extend in $a$, for each fixed $x\in P$, by the same normality. This uses no joint ultraweak continuity of multiplication. Equivariance extends by normality of $\Phi$ and the constant lifts. These arguments prove every part of Theorem 1.1.

Restricting from the enlarged $P$ to the original algebra proves the originally stated version. The countable group hypothesis governs the finite action tests; separable predual governs the countable algebra and functional tests.

6. Exercises with complete solutions

Exercise 1. Why does adjoining $M$ preserve separable predual here?

Solution. The two algebras have countable generating sets, so their join has one. Its faithful normal state GNS space is the closure of the span of countably many rational *-words applied to its cyclic vector. Kaplansky density makes that span dense. Hence this space is separable. The represented algebra has predual a quotient of the trace-class operators on that space, and therefore separable predual.

Exercise 2. Why can arbitrary bounded representatives be used instead of an exactly linear choice of lifts?

Solution. Every rational algebraic relation holds in the quotient, so the discrepancy of any chosen bounded lifts is in $I_\omega$. At each stage only finitely many such discrepancies are tested. Making their symmetric seminorms less than $1/n$ at the selected coordinate makes the reindexed discrepancy ordinarily null. The quotient map then respects the relation exactly.

Exercise 3. What is the role of the fixed modulus $\delta_l(x)$?

Solution. It controls multiplication for all contraction inputs whose seminorm is small, rather than just a finite selected set of inputs. For every sufficiently large outer index it applies to $u_x(p(n))$. Every null input sequence eventually satisfies the fixed modulus on an ultrafilter set. This proves the full multiplier condition; finite algebraic tests alone would not.

Exercise 4. Why is it legitimate to test a moving representative $v_a(n)$ in condition 4?

Solution. At the selection for a given $n$, $v_a(n)$ is a single fixed operator of $M$. The centralizing inner sequence commutes strong* with that operator as its inner index tends along $\omega$. Thus an $\omega$-large set satisfies this one test. The selected inner index can change at the next stage, allowing a different fixed operator there.

Exercise 5. Explain why no countable intersection of ultrafilter sets is needed.

Solution. At stage $n$, only finite sets of elements, group actions, functionals, moduli and algebraic relations are used. Their finite intersection is in the ultrafilter and is nonempty. Every fixed datum is included at all sufficiently large stages, so its errors vanish ordinarily along the outer index. This sequence of finite choices supplies the countable conclusion.

Exercise 6. Derive the symmetric state isometry (12).

Solution. Multiplicativity and state preservation give $\varphi^\omega(\Phi_0(x)^*\Phi_0(x))=\varphi^\omega(x^*x)$. Applying the same identity to $xx^*$ gives the adjoint square. Averaging the two equalities and taking square roots gives (12). This supplies the bounded strong* continuity required for the extension.

Exercise 7. Why is the extension multiplicative although strong convergence is not jointly continuous without bounds?

Solution. The Kaplansky approximants and their images have uniform norm bounds. On such bounded sets, products converge strong*, by splitting $a_i b_i-ab=a_i(b_i-b)+(a_i-a)b$, and applying the adjoint version too. The same estimates hold for the image nets, so the products of their limits agree with the limit of the products.

Exercise 8. Explain the separate extension of the expectation identity.

Solution. For each fixed $a$, left multiplication by $a$, $\Phi$ and $E_\omega$ are normal linear maps. Thus the equality extends ultraweakly in $x$. After doing so, fix $x$; right multiplication by the fixed $\Phi(x)$ and multiplication by the fixed $E_\omega(x)$ are normal in $a$. This extends to all $a\in Q$ without asserting a jointly ultraweakly continuous product.

Exercise 9. Why does the equivariance test fail for a general semi-lift represented by $\theta_k\to\theta$?

Solution. The representative of its action on $x$ is $\theta_k(u_x(k))$. After reindexing, equivariance would compare this with $\theta_n(u_x(p(n)))$, whose automorphism index is the outer $n$, rather than the selected inner $p(n)$. Convergence on fixed elements does not control this varying input. The tensor-tail example supplies an exact contradiction to imposing all those equivariance requirements together with relative commutation.

Exercise 10. Give the correct factor trace conclusion and explain the failure for arbitrary constants.

Solution. For a factor and centralizing $x$, $E_\omega(x)=\tau_\omega(x)1$; applying $\varphi$ to (1) gives (2). If $a$ is centralizing too, this is trace factorization on $M_\omega$. For a constant trace-zero self-adjoint unitary $x=a$, fixing constants instead gives $\tau^\omega(a\Phi(x))=1$, whereas the product of scalar traces is zero. The expectation identity correctly gives $E_\omega(a)E_\omega(x)=a^2=1$.

References

Adrian Ocneanu, Actions of discrete amenable groups on factors, thesis, Chapter 5, Section 5.3, Fast Reindexation Trick, printed pp.53–54 (PDF pp.67–68), gives the classical finite-test selection with a countable family of constant lifts. Its ordered product formula uses the canonical expectation, with the input in the map's domain and the prescribed coefficient in the other algebra. Its normal-extension step is brief.

The complete construction here supplies countability of the joined algebras and their intersections, norm-controlled lifts, every rational algebraic relation, fixed multiplier moduli and both ideal-preservation estimates. A faithful-state isometry gives the full bounded-net extension; positive suprema and normal coefficients prove whole-domain normality. The expectation identities extend separately in their two variables. The theorem retains general separable-predual algebras, while the preceding tensor-tail obstruction explains the restriction to constant lifts. Human-source rights remain separate from the course dedication.