Hyperfinite factors and increasing fatorization of states

If a factor $$R$$ contains a matrix algebra $$M\subset R$$ (i.e., a $$M$$ is a type $$I_n$$ factor), then $$R \cong M \otimes M^c$$ where $$M^c=R\cap M'$$ is the relative commutant. Each state $$\omega$$ on $$R$$ restricts to states $$\omega\rvert_M$$ on $$M$$ and $$\omega\rvert_{M^c}$$ on $$M^c$$. The product state $$\omega^\times$$ defined on $$R$$ by linear continuation of $$\omega^\times(xy) = \omega\rvert_M(x)\omega\rvert_{M^c}(y)$$, $$x\in M$$, $$y\in M^c$$, is called the (normal) factorization of $$\omega$$ relative to $$M$$ [Kadison–Ringrose 2, 11.4.13].

Let $$R$$ be a hyperfinite factor with dense increasing sequence finite type I factors $$M_1\subset M_2\subset \dotsb \subset R$$ and let $$\omega$$ be a normal state on $$R$$. Denote by $$\omega_n^\times$$ the normal factorization of $$\omega$$ relative to $$M_n$$. Does it follow for all normal states that the $$\omega_n^\times$$ approximate $$\omega$$ in norm? I.e. does $$\lVert \omega - \omega^\times_n \rVert \to 0$$?

Some remarks: It is clear that the sequence converges weakly. Very similar statements are proved in section 11.4 of Kadison–Ringrose 2.

Since the answer to this question depend on the nature of $$R$$ and the choice of the sequence $$M_n$$, let's say that $$R$$ satisfies property $$(P)$$ w.r.t. the filtration $$M_n$$ if $$\|\omega - \omega_n^\times\| \to 0$$ for every normal state $$\omega$$ on $$R$$ where $$\omega_n^\times$$ is defined as in the question.

Below I prove the following results.

Proposition 1. If $$R$$ is the hyperfinite II$$_1$$ factor, then property $$(P)$$ holds w.r.t. any filtration $$M_n$$.

Recall that a factor is said to be ITPFI if it can be written as the infinite tensor product of a sequence of matrix algebras equipped with faithful states. By a result of Krieger (see [A. Connes and E.J. Woods, Approximately transitive flows and ITPFI factors. Ergodic Theory Dynam. Systems 5 (1985), 203-236] for a systematic account), not all injective factors are ITPFI.

Proposition 2. If $$R$$ is an injective factor that is not ITPFI, there is no filtration for which property (P) holds.

Proposition 3. If $$R$$ is an ITPFI factor, then property (P) holds w.r.t. some filtrations, but not necessarily w.r.t. all filtrations.

Proof of Proposition 1. Note that for all normal states $$\omega$$ and $$\mu$$ on $$R$$, we have that $$\|\omega_n^\times -\mu_n^\times\| \leq 2 \|\omega - \mu\|$$ for all $$n$$. Denote by $$E_n$$ the unique trace preserving conditional expectation of $$R$$ onto $$M_n$$. Denote by $$S_0$$ the set of normal states on $$R$$ that are of the form $$\eta \circ E_n$$ for some $$n$$ and some state $$\eta$$ on $$M_n$$. Then $$S_0$$ is norm dense in the set of all normal states on $$R$$. When $$\mu\in S_0$$, we have that $$\mu_n^\times = \mu$$ for $$n$$ large enough. In combination with the inequality above, it follows that $$\|\omega - \omega_n^{\times}\| \to 0$$.

Proof of Proposition 2. Assume that $$R$$ satisfies property (P) w.r.t. some filtration $$M_n$$. I prove that $$R$$ must be an ITPFI factor. Choose a faithful normal state $$\omega$$ on $$R$$. Replacing $$M_n$$ by a subsequence, we may assume that $$\|\omega_n^\times - \omega_{n-1}^\times\| < 2^{-n}$$ for all $$n$$.

Denote $$A_n = M_n \cap M_{n-1}'$$, so that we can view $$M_n$$ as the tensor product of the matrix algebras $$A_1 \otimes \cdots \otimes A_n$$. Define the faithful state $$\mu_n$$ on $$A_n$$ by restricting $$\omega$$ to $$A_n$$. Define the faithful state $$\gamma_n$$ on $$M_n$$ by $$\gamma_n = \mu_1 \otimes \cdots \otimes \mu_n$$. Finally define the faithful normal states $$\eta_n$$ on $$R$$ by $$\eta_n(xy) = \gamma_n(x) \omega(y)$$ for all $$x \in M_n$$ and $$y \in R \cap M_n'$$.

Denoting by $$\omega_n^c$$ the restriction of $$\omega$$ to $$M_n^c = R \cap M_n'$$, we may view $$\eta_n$$ as the tensor product $$\mu_1 \otimes \cdots \otimes \mu_n \otimes \omega_n^c$$. So we get that $$\|\eta_n - \eta_{n-1}\| = \|(\mu_n \otimes \omega_n^c - \omega_{n-1}^c)|_{M_{n-1}^c}\|$$. On $$M_{n-1}^c$$ the state $$\mu_n \otimes \omega_n^c$$ coincides with $$\omega_n^\times$$ and on $$M_{n-1}^c$$, the state $$\omega_{n-1}^c$$ coincides with $$\omega_{n-1}^\times$$. We thus conclude that $$\|\eta_n-\eta_{n-1}\| \leq \|\omega_n^\times - \omega_{n-1}^\times\| < 2^{-n}$$. It follows that $$(\eta_n)$$ is a Cauchy sequence in $$R_*$$. Denote by $$\eta$$ the normal state on $$R$$ such that $$\|\eta-\eta_n\| \to 0$$.

For all $$i \geq n$$, the restriction of $$\eta_i$$ to $$M_n$$ equals $$\gamma_n$$. So the restriction of $$\eta$$ to $$M_n$$ equals $$\gamma_n$$. Denote by $$(N,\mu)$$ the infinite tensor product of the matrix algebras $$(A_n,\mu_n)$$. The identity map on all $$M_n$$ then extends to a unitary operator $$U : L^2(N,\mu) \to L^2(M,\eta)$$. We have the faithful normal GNS representations $$\pi_\mu : N \to B(L^2(N,\mu))$$ and $$\pi_\eta : M \to B(L^2(M,\eta))$$. By construction, $$U \pi_\mu(x) U^* = \pi_\eta(x)$$ for all $$x \in M_n$$ and all $$n$$. It follows that $$U \pi_\mu(N) U^* = \pi_\eta(M)$$, so that $$N \cong M$$. So $$M$$ is an ITPFI factor.

Proof of Proposition 3. When $$R$$ is the infinite tensor product of the sequence of matrix algebras $$A_n$$ with faithful states $$\mu_n$$, we denote by $$\mu$$ the infinite product state on $$R$$ and we can take the filtration $$M_n = A_1 \otimes \cdots \otimes A_n$$. For every $$n$$, there is a unique $$\mu$$-preserving conditional expectation $$E_n : R \to M_n$$. From here the proof is identical to the proof of proposition 1.

As a counterexample, let $$\mu_0$$ be a faithful state on $$A_0 = M_2(\mathbb{C}) \otimes M_2(\mathbb{C})$$ that cannot be written as the tensor product of two states. Then also the norm distance of $$\mu_0$$ to the set of tensor product states is strictly positive, say larger than $$\delta > 0$$. Define $$R$$ as the infinite tensor product of copies of $$(A_0,\mu_0)$$, with infinite product state $$\mu$$. Define the filtration $$M_n$$ by \begin{align} & M_{2n} = \underbrace{A_0 \otimes \cdots \otimes A_0}_{\text{n times}} \; ,\\ & M_{2n+1} = \underbrace{A_0 \otimes \cdots \otimes A_0}_{\text{n times}} \otimes M_2(\mathbb{C}) \; . \end{align} So by definition, $$M_n$$ is a tensor product of $$n$$ copies of $$M_2(\mathbb{C})$$. I prove that $$R$$ does not satisfy property (P) w.r.t. this filtration $$M_n$$. Indeed, the restriction of $$\mu - \mu_{2n+1}^\times$$ to the $$n+1$$ copy of $$A_0$$ equals $$\mu_0 - \mu_1 \otimes \mu_2$$ for some states $$\mu_1$$ and $$\mu_2$$ on $$M_2(\mathbb{C})$$. Thus, $$\|\mu - \mu_{2n+1}^\times\| > \delta$$ for all $$n$$, so that property (P) does not hold w.r.t. the filtration $$M_n$$.

• Thanks for your answer Stefaan! I'm a bit puzzled by the "unique trace preserving conditional expectation" that you mention. I only understand this if the factor $R$ is also finite.
– Lau
Oct 19, 2023 at 4:54
• @Lauritz. I see. In my conventions (but not generally agreed upon), hyperfinite means "finite and approximately finite dimensional (AFD)". I will update my answer, because for certain type III factors the property does not hold. Oct 19, 2023 at 9:08