Tracial absorption in constant matrix products

In a long tensor block, basis words with the same letter counts have the same state density. Pairing those words gives square-zero matrix units which exactly centralize the product state and occupy almost all of it. A complete matrix-unit repair turns these partial units into commuting full blocks. This proves tracial absorption and the implication from strong lambda-stability to strong stability.

1. The products and the conclusion

Fix $d\ge2$ and numbers

p_i>0,\qquad \sum_{i=1}^d p_i=1.

Let $h=\operatorname{diag}(p_1,\ldots,p_d)$, let $\rho(a)=\operatorname{Tr}(ha)$ on $M_d(\mathbb C)$, and form the specified infinite product

P_h=\bigotimes_{n=1}^{\infty}(M_d(\mathbb C),\rho). \tag{CTA.1}

Section 4 constructs its faithful product representation. Write

R=\bigotimes_{n=1}^{\infty}(M_2(\mathbb C),\operatorname{tr}_2).

Constant-product absorption theorem. There is a normal isomorphism with normal inverse

P_h\cong P_h\bar\otimes R. \tag{CTA.2}

No type classification of $P_h$ is needed for this conclusion.

More generally, for any nonzero von Neumann algebra $B$,

B\cong B\bar\otimes P_h \quad\Longrightarrow\quad B\cong B\bar\otimes R. \tag{CTA.3}

In particular, with

h_\lambda=\frac1{1+\lambda} \begin{pmatrix}\lambda&0\\0&1\end{pmatrix}, \qquad 0<\lambda<1,

the product $P_{h_\lambda}$ is precisely the specified $R_\lambda$ of definition preceding LA01. Thus strong $\lambda$-stability implies tracial-product absorption.

The proof first establishes the following ordinary matrix criterion.

Approximate-halving criterion. If $A\ne0$ has separable predual, the following are equivalent:

$A\cong A\bar\otimes R$ normally, with normal inverse.
$A$ has a bounded centralizing sequence of square-zero partial isometries $v_n$ with $v_n^*v_n+v_nv_n^*=1$.
$A$ has centralizing contractions $x_n$ such that
$x_n^2\longrightarrow0,\qquad x_n^*x_n+x_nx_n^*\longrightarrow1 \quad\hbox{strongly*}. \tag{CTA.4}$

Here centralizing means $\|\varphi x_n-x_n\varphi\|\to0$ for every $\varphi\in A_*$, with the module conventions of ECA02. All strongly* limits below concern uniformly bounded operators and the intrinsic sigma-strong* topology.

We use the previously proved estimates for centralizing sequences and homogeneous rounding, ECA02–04, the faithful-state continuity and predual density arguments, CDS02 and CDS05–06, and the complete summable tensor-splitting theorem, TS01–06. Sections 2–3 give the ordinary, non-equivariant repair in full, so the criterion does not require a circle action.

2. From approximate units to exact full units

Assume condition 3. The bounded centralizing sequences form a *-algebra. Strongly* null bounded sequences centralize, and they may be multiplied on either side by a bounded centralizing sequence while remaining strongly* null. These are the fully proved estimates of ECA02. Thus equality modulo such a null sequence is compatible with the algebraic operations on centralizing sequences; write it as $\sim$.

Set $a_n=x_n^*x_n$, $b_n=x_nx_n^*$. From (CTA.4),

a_n+b_n\sim1,\qquad a_nb_n\sim b_na_n\sim0,\qquad a_n^2\sim a_n,\quad b_n^2\sim b_n. \tag{CTA.5}

For example $a_nb_n=x_n^*x_n^2x_n^*\sim0$, and

a_n^2=x_n^*b_nx_n \sim x_n^*(1-a_n)x_n =a_n-(x_n^*)^2x_n^2\sim a_n.

Every moving multiplier here is centralizing.

Put

p_n=1_{[1/2,1]}(a_n),\qquad z_n=(1-p_n)x_np_n. \tag{CTA.6}

The scalar inequality

|1_{[1/2,1]}(t)-t|\le2(t-t^2),\qquad0\le t\le1,

proved in ECA03, gives $p_n-a_n\to0$ strongly*. Hence $p_n$ is centralizing. The relations $a_nx_n\sim0$ and $x_na_n\sim x_n$ give $z_n\sim x_n$.

If $z_n=s_n|z_n|$ is the polar decomposition, define

q_n=1_{[1/2,1]}(z_n^*z_n),\qquad w_n=s_nq_n. \tag{CTA.7}

Then $w_n^*w_n=q_n\le p_n$ and $w_nw_n^*\le1-p_n$; thus $w_n$ is a square-zero partial isometry. To record both error seminorms, set

g(t)= \begin{cases} t,&0\le t<1/2,\\ (1-\sqrt t)^2,&1/2\le t\le1. \end{cases}

Polar calculus gives

\begin{aligned} (w_n-z_n)^*(w_n-z_n)&=g(z_n^*z_n),\\ (w_n-z_n)(w_n-z_n)^*&=g(z_nz_n^*). \end{aligned} \tag{CTA.8}

The scalar bound $0\le g(t)\le2(t-t^2)$, including $g(0)=0$, is proved in ECA03. Since both arguments are projections modulo strongly* null sequences, (CTA.8) proves $w_n-z_n\to0$ strongly*. Consequently

w_n\sim x_n,\qquad w_n^*w_n+w_nw_n^*\longrightarrow1. \tag{CTA.9}

The sequence $w_n$ is centralizing.

We next prove that every nonzero projection $f\in A$ contains a nonzero square-zero partial isometry. The contractions $y_n=fx_nf$ centralize $fAf$: extend $\chi\in(fAf)_*$ normally by $\widetilde\chi(a)=\chi(faf)$, and restrict $[x_n,\widetilde\chi]$ to the corner. This is exactly $[y_n,\chi]$. Centrality of $x_n$ also gives $[x_n,f]\to0$ strongly*, by ECA02. Using its small-error multiplication rule,

y_n^2\longrightarrow0,\qquad y_n^*y_n+y_ny_n^*\longrightarrow f \quad\hbox{in }fAf. \tag{CTA.10}

For example $y_n^2-fx_n^2f=fx_n(fx_n-x_nf)f$; the other products are expanded in the same way. Applying (CTA.5)–(CTA.9) inside $fAf$ gives partial isometries whose support sums tend to $f$. A normal positive functional nonzero on $f$ shows that some one of them is nonzero.

In fact $f$ can be filled exactly. Order the square-zero partial isometries in $fAf$, including zero, by orthogonal extension:

v\preceq v+r,\qquad r\in(f-v^*v-vv^*)A(f-v^*v-vv^*).

Along a chain, initial and final projections increase. The identity

\|(v_j-v_i)\eta\|^2 =\langle(v_j^*v_j-v_i^*v_i)\eta,\eta\rangle \quad(v_i\preceq v_j)

and its adjoint version give a bounded strong* chain limit. Products pass to that limit, so it is still a square-zero partial isometry and an upper bound. Zorn's lemma gives a maximal one. If its support sum were less than $f$, the preceding corner construction would give a nonzero orthogonal extension. Therefore there is $r\in fAf$ with

r^2=0,\qquad r^*r+rr^*=f. \tag{CTA.11}

Apply (CTA.11) to $f_n=1-w_n^*w_n-w_nw_n^*$, taking $r_n=0$ when $f_n=0$, and put $v_n=w_n+r_n$. These are full square-zero partial isometries. Both $r_n^*r_n$ and $r_nr_n^*$ lie below $f_n\to0$, so $r_n\to0$ strongly*. Hence

v_n-x_n\longrightarrow0\quad\hbox{strongly*}, \tag{CTA.12}

and the predual small-error estimate in ECA02 proves that $v_n$ is centralizing. This establishes condition 2 from condition 3. No global commutator estimate was imposed on units in an intermediate fixed corner.

3. Exact commutation and the absorption criterion

To turn the units into a tensor factor, they must be chosen to commute with previous blocks. Let $D\subset A$ be a unital full matrix algebra of size $q$, with matrix units $e_{ij}$, and let $C=D'\cap A$. The faithful normal tracial expectation from TS01 is

E_D(a)=\frac1q\sum_{i,j=1}^q e_{ji}ae_{ij}. \tag{CTA.13}

For a centralizing sequence $x_n$ satisfying (CTA.4),

E_D(x_n)-x_n =\frac1q\sum_{i,j}e_{ji}[x_n,e_{ij}] \longrightarrow0\quad\hbox{strongly*}. \tag{CTA.14}

The equality uses $q^{-1}\sum_{i,j}e_{ji}e_{ij}=1$. Fixed-operator commutation follows from ECA02, and the sum is finite.

Thus $E_D(x_n)$ is centralizing on the whole of $A_*$ and still satisfies (CTA.4). It also centralizes $C_*$: a normal functional on $C$ extends by composition with $E_D$, and bimodularity identifies the restricted commutator. Apply Section 2 inside $C$. Its spectral formulas also give the same strong* estimates in $A$, because the original error relations and centrality hold there. The unused projections tend strongly* to zero in $A$, so the arbitrary fillers in their corners are small there as well. Consequently $C$ contains a sequence of full halving units which remains centralizing on all of $A_*$.

Choose normal states $(\psi_j)$ norm dense in the state space of $A$. Their complex linear span is norm dense in $A_*$, by CP07's positive decomposition. If a faithful state is needed to express a bounded strong* limit by one seminorm, $\sum_j2^{-j}\psi_j$ provides it: a positive element killed by that sum is killed by every $\psi_j$, hence by every normal state by norm density, and is zero. Normality follows from norm completeness of the predual in CP06.

Inductively construct commuting full $M_2$ subalgebras $D_k$, with $v_k=e_{12}^{(k)}$, such that

\|\psi_jv_k-v_k\psi_j\|\le2^{-k} \qquad(1\le j\le k). \tag{CTA.15}

At each step the previously chosen blocks generate a full finite matrix algebra, by TS03. Apply the preceding relative-commutant construction and choose a sufficiently late unit for the finitely many estimates. This gives exact commutation of different blocks.

For every fixed $j$, the defects in (CTA.15) are summable; the omitted finite initial segment is finite. TS06 at parameter $1$ gives the bound

\|\psi_j-\psi_jE_{D_k}\| \le3\|\psi_jv_k-v_k\psi_j\|. \tag{CTA.16}

The complete tail-expectation and spatial-splitting proof of TS01–05 therefore gives

A\cong R\bar\otimes Q,\qquad Q=\left(\bigvee_kD_k\right)'\cap A, \tag{CTA.17}

by normal multiplication with normal inverse. This proves generation and the whole spatial tensor product, not just its finite matrix moments.

The interleaving unitary of ECA07 identifies $R\bar\otimes R$ normally with $R$. Tensor flips and (CTA.17) now give

A\bar\otimes R \cong R\bar\otimes Q\bar\otimes R \cong R\bar\otimes R\bar\otimes Q \cong R\bar\otimes Q \cong A. \tag{CTA.18}

All these maps have normal inverses.

This proves 3 implies 1 in the criterion. Conversely, in $A\bar\otimes R$ the sequence $1\otimes e_{12}^{(n)}$ consists of full halving units and is centralizing. ECA08 proves centrality in $R$; the normal product-functional density argument of CDS06 proves it in $A\bar\otimes R$. Transfer by the assumed normal isomorphism gives 1 implies 2. Condition 2 immediately implies 3. The approximate-halving criterion is proved.

4. The faithful product representation

We now construct the particular algebra in (CTA.1). On one leg use the Hilbert space of $d$-by-$d$ matrices with Hilbert–Schmidt norm and reference vector

\Omega_h=h^{1/2}.

Its norm is one. Left multiplication represents $M_d$. Right multiplication is bounded as well:

\|xa\|_2^2=\operatorname{Tr}(a^*x^*xa) \le\|a\|^2\operatorname{Tr}(x^*x).

It commutes with every left multiplication. Since $h^{1/2}$ is invertible, the vectors obtained from $\Omega_h$ by either all left or all right multiplications span the entire finite-dimensional Hilbert space.

Form the Hilbert space completion of finite-support tensor vectors with reference $\Omega_h$ in the omitted legs. Let $\Omega$ be the product reference vector, and let $P_h$ be the von Neumann algebra generated by the finite left matrix algebras. Both their cyclic vectors and the cyclic vectors of the commuting finite right algebras are dense. Thus $\Omega$ is cyclic for $P_h$, and is separating: if $X\Omega=0$, then $Xb'\Omega=0$ for every finite right operator $b'$, and density gives $X=0$.

It follows that

\Phi(X)=\langle X\Omega,\Omega\rangle \tag{CTA.19}

is a faithful normal state. Its restriction to the first $m$ legs is the matrix state with density $h^{\otimes m}$. The Hilbert space is separable: in each leg choose a finite orthonormal basis containing $\Omega_h$; the finite-support product basis is countable. CP04–06 therefore gives separable predual for $P_h$.

If $h=I/d$, the local states are normalized traces. Traciality extends to the entire algebra: first keep one factor in a finite head and extend the other by normality, and then extend the first factor by normality. In particular the $d=2$ case is the faithful tracial product defining $R$. No classification of the general $P_h$ has entered this representation argument.

5. Pairing basis words of equal density

In a block of $m$ legs, index the standard matrix basis by words

a=(a_1,\ldots,a_m)\in\{1,\ldots,d\}^m.

Let $k_i(a)$ be the number of occurrences of $i$. The density of the basis word is

p(a)=\prod_{i=1}^d p_i^{\,k_i(a)}. \tag{CTA.20}

Partition the words by their count vectors $(k_1,\ldots,k_d)$. Within each class, order the words lexicographically and pair consecutive words. For a pair $(a,b)$, use the matrix unit $E_{ab}=|a\rangle\langle b|$, always with the first word as the range and the second as the source. Put

x^{(m)}=\sum_{\text{chosen pairs }(a,b)}E_{ab}. \tag{CTA.21}

All words in different pairs are disjoint, so

(x^{(m)})^2=0,\qquad (x^{(m)})^*x^{(m)}+x^{(m)}(x^{(m)})^*=1-f_m, \tag{CTA.22}

where $f_m$ is the projection onto the unpaired words. Thus $x^{(m)}$ is a contraction and a partial isometry; it may be zero for a small block.

For every pair, $p(a)=p(b)$. Hence

h^{\otimes m}x^{(m)}=x^{(m)}h^{\otimes m}. \tag{CTA.23}

This is exact state centrality in that block. No approximation of the density has been made.

There is at most one unpaired word in each count class. Every coordinate of a count vector lies in $\{0,\ldots,m\}$, so there are at most $(m+1)^d$ such classes. Put

r=\max_i p_i<1.

Every word has density at most $r^m$, and therefore

0\le\rho^{\otimes m}(f_m)\le(m+1)^d r^m \longrightarrow0. \tag{CTA.24}

To see the last limit explicitly, the ratio of consecutive upper bounds is $r((m+2)/(m+1))^d$, which tends to $r<1$. It is eventually bounded by a number strictly less than one, so the remaining tail decreases geometrically. For $d=2$ there are exactly $m+1$ possible counts, giving the sharper bound $(m+1)r^m$.

The same-density condition in (CTA.23) is the reason for partitioning by count vectors. Pairing arbitrary words would keep the square-zero relation but could destroy state centrality.

6. Tail units centralize the entire predual

Place $x^{(n)}$ in legs $n+1,\ldots,2n$ of $P_h$, with identity in the other legs, and call the resulting operator $x_n$. Let $f_n$ denote its unused projection. Then

x_n^2=0,\qquad x_n^*x_n+x_nx_n^*=1-f_n,\qquad \Phi(f_n)\le(n+1)^d r^n\longrightarrow0. \tag{CTA.25}

Since $f_n$ is a projection and $\Phi$ is faithful, $\|f_n\|_\Phi^\#=\sqrt{2\Phi(f_n)}\to0$. CDS02's faithful-state argument makes this strong* convergence on the bounded set of projections.

Each $x_n$ centralizes $\Phi$ exactly. To check it on a finite local element, take a finite head containing both that element and the block carrying $x_n$. The state is the corresponding finite product matrix state, and its density commutes with $x_n$ by (CTA.23). The trace identity then gives $\Phi(x_na)=\Phi(ax_n)$. Both sides are normal functionals of $a$, so equality extends from all finite heads to every $a\in P_h$.

Let $\mathcal L$ be the union of the finite head matrix algebras. Every fixed $a,b\in\mathcal L$ commutes with $x_n$ once $n$ exceeds the head containing them. The module identity

[x_n,a\Phi b] =[x_n,a]\Phi b+a[x_n,\Phi]b+a\Phi[x_n,b] \tag{CTA.26}

then gives zero for all sufficiently large $n$.

The span of these functionals $a\Phi b$ is norm dense in $(P_h)_*$. Indeed $\mathcal L\Omega$ is dense in the product GNS Hilbert space by construction. The complete coefficient-series description in CP04–06, truncation of one series, and approximation of its finitely many coefficient vectors give precisely that norm density, as in CDS05. Since $\|x_n\|\le1$, the commutator maps on the predual have norm at most $2$. Approximation therefore extends the eventual-zero assertion on the dense family to

\|\eta x_n-x_n\eta\|\longrightarrow0 \qquad(\eta\in(P_h)_*). \tag{CTA.27}

Thus $x_n$ satisfies condition 3 of the criterion. The algebra has separable predual by Section 4. Sections 2–3 give (CTA.2).

The original tail blocks need not commute with one another: the intervals $\{n+1,\ldots,2n\}$ can overlap. Exact commutation of the final matrix blocks is supplied by the relative-commutant induction of Section 3.

7. Strong lambda-stability and dual absorption

Let $B\cong B\bar\otimes P_h$ be a normal isomorphism with normal inverse. Tensor it with $\mathrm{id}_R$, reassociate, use (CTA.2) on the $P_h$ leg, and return through the original isomorphism:

B\bar\otimes R \cong B\bar\otimes P_h\bar\otimes R \cong B\bar\otimes P_h \cong B. \tag{CTA.28}

Normality of tensor isomorphisms and their inverses follows from the explicit spatial comparison in TG03; reassociation is a Hilbert-space tensor unitary. This proves (CTA.3) for arbitrary $B$, without a separable-predual assumption on $B$.

Set $d=2$ and use the two eigenvalues of $h_\lambda$ in Section 1. Equation (CTA.1) is then the actual product defining $R_\lambda$, so

R_\lambda\cong R_\lambda\bar\otimes R,\qquad B\cong B\bar\otimes R_\lambda \ \Longrightarrow\ B\cong B\bar\otimes R. \tag{CTA.29}

This is the claimed passage from strong $\lambda$-stability to strong stability.

There is now a direct application to the remaining dual-action step. Suppose $B$ has separable predual, $\varphi$ is faithful normal, $\sigma_T^\varphi$ is approximately inner, and $B\cong B\bar\otimes R_\lambda$ for one $0<\lambda<1$. By (CTA.29) $B$ absorbs $R$. The complete state-transport and central-sequence theorem CDS01–07 therefore proves, for the positive-character dual action,

\left(B\rtimes_{\sigma_T^\varphi}\mathbb Z,\gamma\right) \cong \left((B\rtimes_{\sigma_T^\varphi}\mathbb Z)\bar\otimes R, \gamma\otimes\beta\right). \tag{CTA.30}

Both directions are normal and intertwine the actions exactly.

For the AFD type III₁ uniqueness argument, this closes the passage from an established strong-$\lambda$-stability statement to tracial stability and then to dual absorption. The separate theorem asserting strong $\lambda$-stability for every such factor, and the action-classification steps, still require their own proofs.

8. Exact finite checks and boundary cases

Two binary legs. Take $p_1=1/3$, $p_2=2/3$. In the ordered basis $11,12,21,22$, the block density is

H=\operatorname{diag}(1/9,2/9,2/9,4/9).

The only nontrivial count class has words $12,21$, so the construction gives

x=|12\rangle\langle21|,\qquad x^2=0,\qquad Hx=xH.

Its two support projections have total state mass $4/9$; the unused mass is $5/9$. The construction supplies approximate fullness only as the block length grows, not in every small block.

Four binary legs. The count-class sizes are $1,4,6,4,1$. All but the two extreme classes can be paired completely. Therefore at the same probabilities the unused mass is exactly

(1/3)^4+(2/3)^4=\frac{17}{81}. \tag{CTA.31}

These counts also give seven disjoint matrix pairs and fourteen occupied words out of sixteen.

A wrong pairing changes the state. In the two-leg example, pair $11$ with $22$ instead and set $y=|11\rangle\langle22|$. It is still square-zero, but evaluation of its functional commutator at $y^*$ gives

\Phi(yy^*)-\Phi(y^*y)=\frac19-\frac49=-\frac13. \tag{CTA.32}

Thus its state commutator norm is at least $1/3$. Small or square-zero matrix units alone do not guarantee the predual property in the criterion.

One-dimensional and zero-weight boundaries. For $d=1$, every block has one word, all words remain unpaired, and the unused mass is one. The assumption $d\ge2$ is substantive. If a density is concentrated on one letter, that pure word also remains unpaired with mass one at every length; moreover the reference state is no longer faithful on the full matrix algebra. The argument uses both strict positivity of the leg weights for the faithful-state topology and $r<1$ for (CTA.24).

Fullness versus centrality. The estimates (CTA.25) and (CTA.27) supply different inputs. The first controls how much of the unit is missing; the second controls every normal functional. Section 2 repairs exact fullness, while Section 3 creates exact mutual commutation and proves the normal tensor decomposition. Keeping these steps separate prevents replacing a spatial absorption theorem by a finite-block approximation.

Source

Masamichi Takesaki, Theory of Operator Algebras III, Chapter XVIII, Section 4, proof of Theorem 4.16(i), printed p.396, uses the implication from strong $\lambda$-stability to strong stability. The proof above establishes it directly for the specified constant product, and for all constant faithful finite matrix products, through equal-density word pairing and the complete approximate-halving criterion.