MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Suppose $N \subset M$ are two factors, neither of them Type I, acting on a separable Hilbert space $H$. Let $\pi_1$ be a faithful normal representation of $N$ and $\pi_2$ a faithful normal representation of $M'$. We can consider the von Neumann algebra $N \vee M'$. Suppose that $\pi = \pi_1 \otimes \pi_2$ is a faithful normal representation of $N \vee M'$ (by $\pi(ab') = \pi_1(a) \otimes \pi_2(b')$ for $a \in N, b' \in M'$).

Question: is $\pi$ a spatial isomorphism of von Neumann algebras? And if yes, is there an explicit construction of the unitary implementing the spatial isomorphism (assuming one understands the representations $\pi_1$ and $\pi_2$ well)?

This is the case for example, if I recall correctly, if either $N$ or $M$ is a Type III factor, since in that case $\pi_1(N) \otimes \pi_2(M')$ is of Type III, and it follows that the representation is spatial because isomorphisms between Type III factors are spatial.

The motivation behind this question is that I'm trying to understand the proof of Corollary 1 of D'Antoni & Longo, Interpolation by type $\rm I$ factors and the flip automorphism. Journal of Functional Analysis, 51(3), 361-371 (1983).

share|cite|improve this question
up vote 3 down vote accepted

Okay so I think I have an answer for existence of a spatial isomorphism in the case that the representation $\pi_{1},\pi_{2}$ are the identity representaion on $H.$ . Note that the condition $N\vee M'\cong N\overline{\otimes}M'$ spatially is independent of the way $M$ is represented. Indeed, suppose $M$ is represented on $H$ and $U$ is a unitary which conjugates $N\overline{\otimes}M'$ to $N\vee M'.$ If $K$ is another Hilbert spaces, and $M$ is represented as $M\otimes \Bbb C$ on $H\otimes K,$ then $U\otimes 1$ conjugates $N\overline{\otimes}(M'\overline{\otimes}B(K))$ to $N\vee (M'\overline{\otimes} B(K))$ (here $M'$ means the commutant in $H.$) Similarly, suppose we cut the representation by a projection $p$ in $M'.$ Since $U^{*}pU=1\otimes p$ by assumption, the unitary $U(p\otimes p):pH\otimes pH\to pH$ conjugates $N\overline{\otimes}pM'p$ and $N\vee pM'p.$ By the essential uniques of a normal representation of a Von Neumann Algebra, this proves the claim.

Let us first assume $M$ is type $II.$ Since $M$ is type $II,$ I will assume from here on that $M$ is represented on $L^{2}(M,\tau)$ with $\tau$ a (unique up to scalars) semifinite normal trace.

First off, let's note that $M$ cannot be type $II_{1}$ and this isomorphism be spatial, then we have $N'\cap M\cong (N\overline{\otimes} M')'=N'\overline{\otimes} M.$ Assume $M$ is type $II_{1},$ as noted in my last post, $N'\cap M$ cannot be finite dimensional, and being a subfactor of $M$ (being isomorhpic to $N'\overline{\otimes} M$) it must be a $II_{1}$-factor. But the assumption $N'\cap M$ is infinite dimensional implies that the inclusion $N\subseteq M$ is not finite index, i.e. $\dim_{N}(H)=\infty.$ Then, by definition, we know that $N'$ is infinite. (For more details see Section XIX.2 in Takesaki's Theory of Operator Algebras III).

So we may assume $M$ is type $II_{\infty}.$

So if $M$ is a $II_{\infty}$ factor we may focus on the representation of $M$ on $L^{2}(M,\tau)$ with $\tau$ a fixed semifinite normal trace on $M$ (unique up to scalar multiplication). In this case we claim that this isomorphism is spatial if and only if $N'\cap M$ is infinite. Indeed if $N\vee M'\cong N\overline{\otimes}M'$ spatially then taking commutants implies that $N'\cap M\cong N'\overline{\otimes}M$ spatially, in particular since $M$ is infinite so is $N'\cap M.$

Conversely if $N'\cap M=(N\vee M')'$ is infinite, then since $(N\otimes M')'=N'\otimes M$ is $II_{\infty}$ we have two isomorphic Von Neumann algebras with properly infinite commutants. It is known that two such Von Neumann algebras must be spatially isomorphic (see Theorem V.3.1 in Takesaki's Theory of Operator Algebras I).

If $M$ is type $I$ i.e. isomorphic to $B(H)$ we may take the Hilbert space $M$ is represented on to be $H$ with the canonical action of $M.$ In this case since $M'=\Bbb C,$ it is easy to see that $N\cong N\otimes \Bbb C$ spatially.

Of course, one wouldn't really be able to regard this as explicit, because you would have to know how to write one representation of $M$ in terms of another by tensoring with the trivial representation and cutting by projections.

share|cite|improve this answer
Thanks for your answer! It looks interesting. I don't have my books here, but I'll have a proper look at it after the weekend. – Pieter Naaijkens Apr 30 '11 at 21:02

Potentially a silly question, but do you have an example where neither $M$ nor $N$ is finite type $I$ and one of $M$ and $N$ is not infinite type $I,$ and $N\vee M'\cong N\overline{\otimes} M',$ (by the map you defined) for $N$ and $M$ factors? It's not clear to me that there is a non-trivial case when this actually is an isomorphism, or that there is an example where these algebras are isomorphic by a different map. For example if $N=M$ this is impossible unless $M$ is type I, since $M\vee M'=B(H).$

Also if say $M$ or $N$ is type $II$ you need to have $N$ ``far'' from $M$ for $N\vee M\cong N\overline{\otimes} M,$ for example $N'\cap M$ cannot be finite dimensional (because then $N\vee M$ is type I begin a commutant of a type $I$ von Neumann algebra). So this will rule out finite index inclusions, for example.

share|cite|improve this answer
The condition that there is a spatial isomorphism (of the form in the question) of $N \vee M' \to \pi_1(N) \otimes \pi_2(M')$ for $\pi_i$ faithful normal representations is equivalent to the existance of a Type I factor $F$ such that $N \subset F \subset M$. This is called the split property in the literature, and there are examples from algebraic quantum field theory. I'll see if I can find something more specific. – Pieter Naaijkens Apr 30 '11 at 15:29
Okay. It just wasn't obvious to me that there are nontrivial examples. – Benjamin Hayes Apr 30 '11 at 16:32

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.