My favorite model of quantum probability is by von Neumann algebras, i.e., a quantum measurable space is a von Neumann algebra and a quantum distribution is a normal state. Then, one important new phenomenon in quantum probability is the existence of bosons and fermions. Another important new phenomenon, of which bosons and fermions are a special case, is that every automorphism of the von Neumann algebra $L(H)$, the bounded operators on a Hilbert space $H$, is inner. This is a quantum Noether's theorem that says that if the algebra of observables is $L(H)$, then every symmetry can also be used as an observable. The connection between these two phenomena is that if you have two physical systems with the same algebra $L(H)$, then the symmetry $x \otimes y \mapsto y \otimes x$ of $L(H) \otimes L(H)$ is inner if you use the von-Neumann-completed tensor product. (Clearly, since it is directly given by an operator on $H \otimes H$.) So, you can measure whether the joint system is in a bosonic or fermionic state and (in this abstracted, simplified model) it has to be one of the two.


So my question is, which von Neumann algebras $M$ have the property that $x \otimes y \mapsto y \otimes x$ is an inner automorphism of $M \otimes M$? I thought that I had learned that $L(H)$ is the only von Neumann algebra for which every automorphism is inner, but from the comments it seems that that is not true. What about this automorphism in particular?



Which von Neumann algebras have inner permutation of tensor factors?
Here's an argument showing that in the ${\rm II}\_1$ case the flip automorphism is never inner. 


Let $M$ be a type ${\rm II}\_1$ factor and $\tau$ its trace, so that $M \subset L^2(M, \tau)$, and $M$ acts standardly on $L^2(M, \tau)$. Suppose that the flip automorphism is implemented by a unitary $U \in \mathcal U(M \overline \otimes M)$. I'll reach a contradiction by showing that $U$ is orthogonal to every vector in the dense subspace of $L^2(M \overline \otimes M, \tau \otimes \tau)$ spanned by vectors of the form $v \otimes w$ where $v, w \in \mathcal U(M)$.


Fix $v, w \in \mathcal U(M)$, and $\varepsilon > 0$. Take $n \in \mathbb N$ such that $2^{-n} < \varepsilon$, and take a partition of unity $\{ p\_k \}\_{k = 1}^{2^n} \subset M$ such that each $p\_k$ is a projection of trace $2^{-n}$. Then 
$$
| \langle U, v \otimes w \rangle | = \left| \sum\_{k = 1}^{2^n} (\tau \otimes \tau) ((p\_k \otimes 1)(v^\* \otimes w^\*) U (p\_k \otimes 1) ) \right|
$$
$$
\leq \sum\_{k = 1}^{2^n} | (\tau \otimes \tau) ((v^\* \otimes w^\*) U (p\_k \otimes vp\_kv^\*) ) |
$$
$$
\leq \sum\_{k = 1}^{2^n} (\tau \otimes \tau)(p\_k \otimes vp\_kv^\*)
$$
$$
= \sum\_{k = 1}^{2^n} \tau(p\_k)^2 = 2^{-n} < \varepsilon.
$$


**Update**: I've recently come across the 1975 paper of Sakai "Automorphisms and tensor products of operator algebras" where he proves that the flip automorphism for a von Neumann algebra $M$ is inner if and only if $M$ is a type ${\rm I}$ factor. His proof is roughly as follows:


For the type ${\rm II}\_1$ case he proceeds as I did above by showing that the unitary $U$ would have to be orthogonal to every vector of the form $v \otimes w$. His argument for this is not as direct as the one above, but the argument I gave above is based on techniques of Popa which came later.


For the type ${\rm II}\_\infty$ case he writes $M$ as $N \overline \otimes \mathcal B(\mathcal H)$ where $N$ is a type ${\rm II}\_1$ factor and then shows with a simple argument that if the flip automorphism is inner on $M$ then it must also be inner on $N$ which it cannot be by the arguments above.


For the type ${\rm III}$ case he first writes $M$ as $N \overline \otimes \mathcal B(\mathcal H)$ where $N$ is type ${\rm III}$ and countably decomposable. Next he shows that if the flip is inner on $N$ then $N$ has trivial outer automorphism group. Indeed, if $\sigma$ denotes the flip and $\rho \in {\rm Aut}(N)$ then since $\sigma$ is inner, and ${\rm Inn}(M)$ is a normal subgroup, we must have $\tilde \rho = \sigma (\rho^{-1} \otimes {\rm id}) \sigma (\rho \otimes {\rm id})$ is also inner. Restricting $\tilde \rho$ to $N \otimes \mathbb C$ we then see then that there is a unitary $V \in N \overline \otimes N$ such that $(a \otimes 1)V = V(\rho(a) \otimes 1)$ for all $a \in N$. If we then consider the normal conditional expectation $E$ from $N \overline \otimes N$ to $N \otimes \mathbb C$, then there exists some operator $x \in \mathbb C \otimes N$ such that $E(Vx) \not= 0$, and we then have $a E(Vx) = E(Vx) \rho(a)$ for all $a \in N$. By conjugating this formula it then follows easily that $E(Vx)E(Vx)^\* \in \mathcal Z(N) = \mathbb C$ and also $E(Vx)^\*E(Vx) \in \mathbb C$, hence $E(Vx)$ is a non-zero scalar multiple of a unitary showing that $\rho$ is inner. Tomita-Takesaki theory though gives continuum many outer automorphism of $N$, a contradiction.


