Index selection and uniform multiplier control

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.

A double sequence has a row index for an element of $M^\omega$ and a column index for its representative in $M$. Index selection chooses a row slowly while the column follows the ultrafilter. To produce an element of the nontracial multiplier quotient, the rows need common multiplier moduli. Separate membership of every row is insufficient.

We prove the general construction with that uniform control, and then the full finite-trace version, where the control is automatic. An explicit type I example shows why the unrestricted source assertion cannot hold.

Let $M$ have separable predual and a faithful normal state $\varphi$. Use $M^\omega=N_\omega/I_\omega$, $M_\omega=C_\omega/I_\omega$, $E_\omega$, and $\|\cdot\|_\#=\|\cdot\|_{\varphi,\#}$ from the preceding lessons.

Unital subalgebras in this lesson contain the ambient identity. The target state $\varphi^\omega=\varphi\circ E_\omega$ is faithful and normal. The general chosen-family proof gives a normal automorphism of the full quotient for each chosen semi-lift family and preserves its centralizing subalgebra. Neither the diagonal construction nor its kernel statement requires those automorphisms to preserve the chosen state.

1. Why arbitrary rows are not enough

Proposition 1.1. Even for a two-dimensional unital $C^*$-algebra $A\subset\ell^\infty(\mathbb N,M^\omega)$ and the trivial action group, a homomorphism $\Psi:A\to M^\omega$ fixing constant sequences need not exist with

Proof. Take $M=B(\ell^2(\mathbb N_0))$. The type I multiplier theorem identifies $M^\omega$ normally with constant $M$; under that identification $E_\omega=\mathrm{id}$.

Let $w_m$ exchange $\xi_0$ and $\xi_m$ and fix all other basis vectors. Then

The tail projection and off-diagonal terms converge ultraweakly to zero, so

For the off-diagonal terms, each fixed pair of vectors gives a coefficient tending to zero; uniform boundedness extends weak operator convergence to ultraweak convergence on this bounded sequence by trace-class approximation.

View each $w_m$ as a constant element of $M^\omega$, and put $w=(w_m)$, $A=C^*(1,w)$. This is a unital copy of $\mathbb C^2$: $w$ is a self-adjoint unitary different from either scalar sign. Fixing constants requires $\Psi(1)=1$, so any *-homomorphism makes $\Psi(w)$ a self-adjoint unitary. But (1) and $E_\omega=\mathrm{id}$ force $\Psi(w)=1-p_0$, whose square is not $1$. This is impossible. $\square$

The two nonzero spectral projections can be checked without a general dimension assertion. In every row the vectors $(\xi_0+\xi_m)/\sqrt2$ and $(\xi_0-\xi_m)/\sqrt2$ have eigenvalues $+1$ and $-1$, respectively. Thus both $(1+w)/2$ and $(1-w)/2$ are nonzero orthogonal projections with sum one. They span $C^*(1,w)$, giving exactly $\mathbb C^2$.

The weak limit calculation also tests the whole predual. For a fixed pair of $\ell^2$ vectors, each coefficient involving its $m$-th coordinate tends to zero; that treats $p_m$ and both rank-one terms. The uniform operator bound then passes the convergence through any summable normal vector-pair series by a finite-head and bounded-tail estimate. It is therefore ultraweak convergence against every normal functional. In this type I model every quotient element is constant by the full type I theorem, and the weak-limit definition of $E_\omega$ is the identity on those constants. Hence $E_\omega$ is the inverse of that particular normal isomorphism, as used in the contradiction.

This meets the separability and trivial-action hypotheses of Ocneanu's Index Selection Trick in Section 5.5. The obstruction is to the theorem itself, rather than just one choice of diagonal.

2. A uniform row hypothesis

Let $A\subset\ell^\infty(\mathbb N,M^\omega)$ be separable and unital. Write $a=(X_m^a)$. Let $H$ be a countable group of actual semi-lifts acting on $A$ term by term:

Assume $A$ is invariant. The family $\gamma_k$ in the second formula is indexed by the representative column, not the outer row.

Choose a countable norm-dense unital rational *-algebra $D\subset A$, invariant under $H$. Arrange that $D$ has norm-dense intersections with the closed subalgebras of constant sequences and of sequences all of whose rows belong to $M_\omega$. These subalgebras are separable as closed subspaces of $A$, so their dense generators can be included before taking all countable translates.

Here is the separability argument for both special subalgebras. The map sending $X\in M^\omega$ to its constant sequence is isometric, so its range is norm closed. The set of sequences with every row in $M_\omega$ is a norm-closed *-subalgebra because $M_\omega$ is a norm-closed *-subalgebra of $M^\omega$. Intersecting each range with $A$ gives closed subalgebras of the separable metric space $A$.

Every subspace of a separable metric space is separable: use the countable family of balls with rational radii and centres in one fixed dense set, and choose one point in each ball that meets the subspace. Those chosen points are dense in the subspace. Include the resulting countable dense sets, a countable dense set of $A$, their adjoints and the identity. Take their $H$-orbits and then all finite rational *-words. The result is a countable, norm-dense, invariant unital rational algebra with both required dense intersections. The actual semi-lifts preserve centralizing rows by the chosen-family proof, and termwise actions preserve constant row sequences.

Exhaust the rational algebra by finite sets; these finite sets need not themselves be algebras. At stage $n$, test the first $n$ rational complex scalars in a fixed enumeration. Every sum, product, adjoint, scalar multiple and translated element already belongs to the full algebra and has its representatives chosen, whether or not it lies in that stage's finite set.

For each $a\in D$ choose representatives

For a constant row sequence $X_m^a=X$, use the same representative $u_X(k)$ for every row. For a sequence of centralizing quotient elements, use centralizing representatives in each row.

The norm bound in (5) is available simultaneously for all rows of a fixed $a$. For $C=\|a\|_A>0$, radially clip any row lift $b$ inside the multiplier $C^*$-algebra by

Continuous calculus gives norm at most $C$, and the quotient image is unchanged since the row norm is at most $C$. For zero use the zero lift. Apply the same operation inside the centralizing $C^*$-algebra when the row is centralizing. A constant row sequence uses a single fixed lift in every row; when it is centralizing as well, choose that single lift from the centralizing algebra. Thus the requirements agree on their overlap. Hypothesis (6) is imposed on these chosen representatives.

The additional uniform control is this: for each $a\in D,l\ge1$, there are a positive $\delta_l(a)$, an outer set $R_l(a)\in\omega$, and inner sets $W_l(a,m)\in\omega$ for $m\in R_l(a)$, such that

The modulus is independent of the row. The sets can be made decreasing with $l$ by finite intersections. Separate row membership would only give a modulus depending on both $l$ and $m$; that weaker information does not supply (6).

To make the control data decreasing, replace $\delta_l(a)$ by the minimum of its first $l$ values and $R_l(a)$ by the intersection of its first $l$ sets. For $m$ in that intersection, all first $l$ inner sets are defined; intersect them to obtain the new $W_l(a,m)$. Every intersection here is finite. The new data still satisfy (6) and use a positive threshold independent of the selected row.

Theorem 2.1. Under (4)–(6), there is a unital *-homomorphism $\Psi:A\to M^\omega$ such that:

Injectivity is not asserted for an arbitrary $A$. Its exact kernel will be described below.

3. Selecting rows, then column sets

Let $D_n,H_n$ be increasing finite exhausting sets, with $1\in D_n$, and let $\psi_j$ be norm dense in $M_*$. Put

Each limit exists by bounded ultraweak compactness.

One can construct this limit directly from the full concrete-predual theorem. For fixed $a$, the rule

is a bounded linear functional on $M_*$, of norm at most $\|a\|_A$. Bounded scalar ultralimits exist in a compact complex disc and preserve linear combinations. The identification $(M_*)^*=M$ therefore supplies exactly $L(a)$. This proves existence, uniqueness, linearity and contractivity at the same time. Positivity follows by testing positive rows and positive normal functionals, and $L(1)=1$.

At stage $n$, choose a row $r(n)\ge n$ in all $R_l(a)$ for $a\in D_n,l\le n$, and satisfying

These are finitely many outer ultrafilter conditions, so their intersection is nonempty.

Having fixed that row, choose decreasing inner sets $V_n\in\omega$, inside $\{k\ge n\}$ and all $W_l(a,r(n))$ for $a\in D_n,l\le n$. Require on $V_n$:

1. Every addition, adjoint, multiplication and rational scalar relation on the finite data has symmetric seminorm error less than $1/n$.
2. For $a\in D_n,j\le n$,

3. For $a\in D_n,\gamma\in H_n$,

4. For a centralizing-row sequence $a\in D_n$ and $j\le n$,

Lemma 3.1. These inner sets exist.

Proof. Every algebraic relation holds in the fixed row quotient, so its representative error is in $I_\omega$. Equation (10) uses its ultraweak-limit expectation. For (11), the two representatives give the same row element $\gamma(X_{r(n)}^a)=X_{r(n)}^{\gamma a}$, by (4). Equation (12) uses the centralizing representative of that row. Each finite test therefore holds on an inner $\omega$-large set. Intersect these with the finitely many uniform-modulus sets, the previous $V_{n-1}$, and the cofinite set. $\square$

For the action test, both representatives in (11) belong to the multiplier algebra and represent the same element in that one fixed row. Their difference is consequently in $I_\omega$. At this stage the row is already fixed, while the family acts at the actual column $k$. The relation and expectation tests likewise use only a finite list of errors in that same row. Explicitly, intersect their inner large sets with

This is large, and choosing it as $V_n$ gives nestedness and empty total intersection. Outer largeness was used to choose $r(n)$; inner largeness is used only after that row choice. No countable intersection of large sets is taken.

The moduli in this intersection are the fixed $\delta_l(a)$ of (6). A new modulus shrinking with $r(n)$ would not give the subsequent multiplier proof.

4. The diagonal and its full extension

Set $V_0=\mathbb N$, and let $s(k)$ be the largest $n\ge1$ with $k\in V_n$, or zero if none. Since $V_n\subset\{k\ge n\}$, the level is finite; since every $V_n$ is large, $s(k)\to_\omega\infty$. Define

using $r(0)=1$ on the initial band.

Lemma 4.1. Each $y_a$ belongs to $N_\omega$.

Proof. Fix $a,l$. On an $\omega$-large set the level $n=s(k)$ is high enough that $a\in D_n,l\le n$. Its row $r(n)$ lies in $R_l(a)$, and its actual column $k\in V_n$ lies in $W_l(a,r(n))$. Thus (6) controls both products at that coordinate with the same fixed $\delta_l(a)$.

For a contraction sequence $z_k\in I_\omega$, the condition $\|z_k\|_\#<\delta_l(a)$ holds on another large set. Both product seminorms have ultralimit at most $1/l$. Let $l\to\infty$, and rescale bounded null sequences. This proves both multiplier conditions. $\square$

At a fixed coordinate $k$, membership in $V_n$ forces $n\le k$, so the set of possible positive levels is a finite initial segment. Its maximum exists, and on $V_n\setminus V_{n+1}$ the level is exactly $n$. For any fixed threshold $N$, the large set $V_N$ has $s(k)\ge N$. This proves the required ultrafilter divergence, including the harmless level-zero band.

The multiplier proof quantifies over every bounded null input chosen after the diagonal. Divide it by one common operator bound. For fixed $a,l$, the set where the level includes $a,l$ is large; on it the row and column satisfy both control-set conditions in (6). Intersect with the large set where the rescaled input is smaller than the fixed $\delta_l(a)$. The product seminorms then have ultralimit at most $1/l$. The same two products are considered for every $l$, so both limits are zero.

The relation tests make $\Psi_0$ a unital rational *-homomorphism, and (5) gives $\|\Psi_0(a)\|\le\|a\|_A$. It extends by norm continuity to a complex unital *-homomorphism on $A$.

Every fixed rational relation occurs at all sufficiently high levels. For example, its multiplication error is the representative

with symmetric seminorm below $1/s(k)$ on a large set. This is null. Adjoint, addition and each rational scalar relation are treated in the same way. The constant identity is fixed, and the coordinate norm bound gives contractivity. Extension from rational to complex scalars follows by approximating the scalar and using this norm bound; bounded products and adjoints pass through norm limits. Since $D$ is norm dense, this gives the asserted unital *-homomorphism on all of $A$.

Combining (9) and (10) gives scalar expectation errors below $2/s(k)$, hence the first line of (7) on $D$. The map $L:A\to M$ is linear and contractive: these properties hold coordinatewise for the expectations and survive scalar ultralimits. Thus both sides extend by norm continuity to $A$.

More explicitly, for a fixed $a,\psi_j$, write $n=s(k)$ and split

Both summands have absolute value below $1/n$ on the large set of sufficiently high levels. The expression inside the functional is bounded in norm by $2\|a\|_A$. Approximation of any normal functional by a fixed $\psi_j$ therefore gives zero scalar ultralimit for every element of $M_*$. The weak-limit definition of the canonical expectation identifies that limit with $E_\omega(\Psi_0(a))$, proving the first clause. Both sides are norm continuous on $A$, so this clause extends to every input.

Constant sequences are fixed on $D$ because their row representatives were chosen identical, making (13) exactly $u_X(k)$. Norm density in the constant subalgebra extends this conclusion. Equation (12) gives centralizing images on the dense centralizing-row subalgebra; $M_\omega$ is norm closed, so the entire centralizing-row subalgebra has its image there.

For a centralizing-row input in $D$, every fixed $\psi_j$-commutator has norm at most $1/s(k)$ at sufficiently high levels. For arbitrary $\psi\in M_*$, use

The ultralimit and norm density prove that the full representative is centralizing. Norm density of the special intersection and norm closedness of the centralizing quotient then give that clause on its entire subalgebra. The constant clause uses the separately norm-dense constant intersection and the exact identical-row representative. These are norm-density extensions for a $C^*$-domain.

Finally (11) compares $\gamma_k(y_a(k))$ with $y_{\gamma a}(k)$ at the actual column $k$. It gives equivariance on $D$, and norm continuity gives it on $A$. These arguments prove Theorem 2.1.

Define the state

The expectation identity implies $\varphi^\omega\Psi=\rho$. Faithfulness of $\varphi^\omega$ then gives the exact formula

The functional $\rho$ is a state: it is the composition of the positive unital maps $L$ and $\varphi$, or equivalently of the constructed unital *-homomorphism with the target state. The kernel calculation also proves that this particular square-null set is a closed two-sided ideal. If its element $a$ has zero image, then $\Psi(bac)=\Psi(b)\Psi(a)\Psi(c)=0$ for all $b,c$, and continuity treats norm limits. No injectivity of $\Psi$ is needed to obtain the formula.

Indeed $\rho(a^*a)=\varphi^\omega(\Psi(a)^*\Psi(a))$ vanishes precisely when $\Psi(a)=0$. In particular a faithful $\rho$ makes this selection injective.

5. The full finite-trace version

Corollary 5.1. Suppose $M$ has separable predual and a faithful normal tracial state $\tau$. For every separable unital $A\subset\ell^\infty(\mathbb N,M^\omega)$ invariant under a countable semi-lift group acting as in (4), a map satisfying all of (7) exists. No factor hypothesis is needed.

Proof. Use $\varphi=\tau$. Every bounded sequence in $M$ is an inner multiplier. Choose contractive row representatives as in (5); for constants use identical representatives, and for centralizing rows choose centralizing ones. For a contraction $z$,

Thus (6) holds with $R_l(a)=W_l(a,m)=\mathbb N$ and, for example, $\delta_l(a)=(4l(1+\|a\|_A))^{-1}$. All countable algebra, action and intersection data can be chosen as above, so Theorem 2.1 applies. $\square$

Here (15) becomes

Consequently the selection embeds the $C^*$-quotient by this trace-null ideal into $M^\omega$, preserving its trace. In particular it fixes every constant $X\in M^\omega$ whose constant sequence belongs to $A$. This is the full tracial diagonal-selection phenomenon, including nonfactor finite algebras.

The target trace used here is defined on the full multiplier quotient. Both products of two bounded representatives have equal $\tau$-values at every coordinate, so their ultralimits agree; this proves that $\tau^\omega$ is tracial on all of $M^\omega$. It is the already established faithful normal quotient state for the chosen trace. Taking the outer ultralimit makes $\rho$ tracial as well. The common kernel in (17) is closed and two-sided by (15). The map $a+\ker\Psi\mapsto\Psi(a)$ is therefore a well-defined unital injective *-homomorphism, and its trace agrees with the descended $\rho$. A zero positive square under that descended trace has zero image under the faithful target trace and hence is zero. This proves the faithful trace and exact quotient embedding asserted for the finite nonfactor case.

The general hypothesis (6) fails intrinsically in Proposition 1.1. Use the diagonal state (13) of the ordinary multiplier lesson. Its $\|p_m\|_\#^2=\lambda_m$ tends to zero, whereas

For any representatives of the constant row element $w_m$, taking the inner ultralimit preserves these fixed-row product seminorms, since multiplication by $p_m$ is fixed within that row. Thus no common small-input modulus can hold for unbounded $m$ in an outer ultrafilter set. Merely requiring each individual row to be a multiplier leaves precisely this gap.

The obstruction also persists for every choice of dense rational algebra in the swap example. Such an algebra must contain a nonscalar element $a=\alpha1+\beta w$ with $\beta\ne0$, since the scalar subalgebra is closed and proper in $\mathbb C^2$. In row $m\ge1$, the diagonal state gives the exact identity

The mixed matrix entries have zero value under the diagonal state, which verifies the displayed equality. For any alternative lift of that row, the difference from the constant row is inner-null, and multiplication by its fixed $p_m$ keeps it null. The output norm thus has the same inner ultralimit, at least $|\beta|/2$.

Choose $l$ with $1/l<|\beta|/2$. On the outer large set $R_l(a)$, choose $m$ so large that $\|p_m\|_\#<\delta_l(a)$; this is possible since a free-filter large set meets every cofinite set. Condition (6) would bound this output by $1/l$ on the inner large set $W_l(a,m)$, contradicting that inner ultralimit. This proves intrinsic failure of the common-modulus hypothesis for every dense rational algebra and every representative choice in the example.

6. Exercises with complete solutions

Exercise 1. Verify the swap formula (2).

Solution. On $\xi_0$, its first three terms give zero and the term $|\xi_m\rangle\langle\xi_0|$ gives $\xi_m$. On $\xi_m$ it gives $\xi_0$, and on every other basis vector it gives that same vector. Thus it is the stated self-adjoint permutation unitary, with square $1$.

Exercise 2. Why is the ultraweak limit in (3) not unitary?

Solution. It annihilates $\xi_0$ and fixes the orthogonal complement, so it is the proper projection $1-p_0$. Its square is itself, which differs from $1$. A unital homomorphism cannot send the unitary $w$ to it.

Exercise 3. Explain why $A=C^*(1,w)$ is already a separable counterexample.

Solution. The relation $w=w^*,w^2=1$ makes its continuous functional calculus a quotient of $\mathbb C^2$. Both spectral projections $(1\pm w)/2$ are nonzero because every swap has both signs in its spectrum. Hence the algebra is exactly $\mathbb C^2$, finite-dimensional and separable.

Exercise 4. Distinguish the row and column actions in (4).

Solution. The outer action sends row $X_m$ to the same actual automorphism $\gamma(X_m)$. Within its representative, that action is implemented by $\gamma_k$ at column $k$. Equation (11) therefore uses $\gamma_k$, with the row $r(n)$ held fixed. Replacing it by $\gamma_{r(n)}$ would apply a different rule.

Exercise 5. Why do the row choices satisfy (9)?

Solution. For each fixed $a,j$, the scalar $\psi_j(E_\omega(X_m^a))$ has ultralimit $\psi_j(L(a))$. Its error below $1/n$ is therefore an outer large-set condition. At stage $n$ only finitely many such conditions and row-modulus sets are intersected, together with a cofinite set.

Exercise 6. Why is the uniform modulus independent of the selected row essential?

Solution. A null input has small seminorm relative to every fixed positive threshold on a large set. It need not be small relative to thresholds that shrink with its coordinate. Lemma 4.1 needs one fixed $\delta_l(a)$ while its selected rows vary. A row-dependent threshold cannot justify that step; the swaps in (18) exhibit its failure.

Exercise 7. Prove centralizing membership of the selected image.

Solution. For a sequence of centralizing rows, (12) bounds every fixed $\psi_j$-commutator by $1/s(k)$ on a large set. Levels tend to infinity along $\omega$. Uniform operator bounds and norm density extend the convergence to every normal functional. This is exactly membership in $C_\omega$.

Exercise 8. Check constant fixing for an element of $M^\omega$, whose inner representative may vary.

Solution. If every row equals $X$, choose $x_m(k)=u_X(k)$ independently of $m$. The selected diagonal in (13) is then $u_X(k)$ at every coordinate, even though that representative varies with $k$. Its quotient is precisely $X$.

Exercise 9. Derive the kernel formula (15).

Solution. Multiplicativity and the state identity give $\rho(a^*a)=\varphi^\omega(\Psi(a)^*\Psi(a))$. If $\Psi(a)=0$, the value is zero. Conversely faithfulness of the target state makes a positive square with zero state vanish, so $\Psi(a)=0$. Both implications prove the formula.

Exercise 10. Establish the uniform row modulus in the finite-trace case.

Solution. The trace $2$-norm satisfies $\|bz\|_2,\|zb\|_2\le\|b\|\|z\|_2$. The representatives have a common bound $\|a\|_A$, so their sum is at most $2\|a\|_A\|z\|_2$, regardless of row or column. The positive modulus in Corollary 5.1 therefore works on all indices.

Exercise 11. Why is injectivity inappropriate for arbitrary sequence algebras?

Solution. A nonzero sequence can have row $2$-norm tending to zero along the outer filter. For instance a sequence supported at just one row is nonzero in $\ell^\infty$ but trace-null for a free filter. Equation (17) forces the selection to annihilate it. The correct injective object is the quotient by that ideal.

Exercise 12. Show that a different choice of representatives cannot repair the swap counterexample's common-modulus failure.

Solution. For each fixed row $m$, an alternative representative differs from the constant $w_m$ by an inner null sequence. Multiplication by the fixed $p_m$ preserves that nullity, so its inner product seminorm has ultralimit equal to the value in (18). For unbounded $m$, the input $\|p_m\|_\#$ is arbitrarily small but the output remains at least $1/2$. This contradicts any fixed modulus on an outer large set.

References

Adrian Ocneanu, Actions of discrete amenable groups on factors, thesis, Chapter 5, Section 5.5, printed pp.56–58 (PDF pp.70–72), gives the classical row–column Index Selection Trick. Its stated domain contains rows in the full multiplier quotient. The row choice, finite relation and action tests, and band-selected representatives provide the comparison method. Its unrestricted nontracial conclusion is contradicted by the two-dimensional type I example of Proposition 1.1, which satisfies the source's separability and trivial-action hypotheses.

The source's selected-column control set belongs to an individual selected row. It does not supply a positive small-input threshold common to moving rows. Hypothesis (6) supplies exactly that control here; the complete construction proves both multiplier conditions before taking the quotient, every norm-continuous extension, and the exact state-square kernel. In the finite-trace case the common threshold is automatic, giving the full nonfactor trace-null quotient embedding. The swap calculation proves failure of the added hypothesis for every dense rational algebra and every choice of its row representatives.