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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?

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$B(H)$ is not the only vNa with only inner automorphisms. There are some spectacular results on II$_1$ factors with no outer automorphisms following from "super-rigidity" theorems - see e.g. arxiv.org/abs/math/0605456 and the references therein. –  Ollie Margetts Dec 9 '12 at 17:00
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For a simpler example, consider $M_n({\bf C}) \oplus M_k({\bf C})$ with $n \neq k$. Any automorphism has to take each summand to itself, and hence is implemented by a unitary in each summand. –  Nik Weaver Dec 9 '12 at 17:36
    
Huh. The remark at the end was plainly misstated then; as Nik points out, I overlooked the obvious. Maybe I had in mind only factors, but Ollie's remark shoots that down as well. –  Greg Kuperberg Dec 9 '12 at 17:56
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Connes showed that if the flip automorphism of a $II_1$ factor is a pointwise limit of inner automorphisms then it must be the hyperfinite $II_1$ factor. –  Leonel Robert Dec 9 '12 at 17:59
    
@Leonel: Is there anything known about this for the hyperfinite $III_1$-factor? –  Ulrich Pennig Dec 9 '12 at 18:15

2 Answers 2

up vote 17 down vote accepted

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.

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So, I guess, while there are now known to be $\mathrm{II}_1$ factors that have no outer automorphisms, none of them are tensor squares of other $\mathrm{II}_1$ factors. –  Greg Kuperberg Dec 10 '12 at 1:06
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In fact, many of the examples of ${\rm II}_1$ factors without ourter automorphisms can also be shown to be prime, i.e., they are not isomorphic to any tensor product of other ${\rm II}_1$ factors. –  Jesse Peterson Dec 10 '12 at 1:17
    
Thanks a lot! You bagged the question! –  Greg Kuperberg Oct 12 '13 at 17:01

Not sure if this should be an answer or a comment. If $M$ has nontrivial center then the flip automorphism is not inner: if $p$ is a nontrivial central projection then $p \otimes (1-p) \mapsto (1-p) \otimes p$, but $p\otimes(1-p)$ times anything in $M\otimes M$ lies under itself.

On the positive side, I think that the hyperfinite $II_1$ factor implements its own flip automorphism, since this can be done on the approximating matrix algebras and you can take a weak* limit. Possibly this sort of argument works for any factor?

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Hi Nik. It can count as a partial answer. Indeed I was starting to wonder about exactly both of these comments and I was not sure if they were true. –  Greg Kuperberg Dec 9 '12 at 18:00
    
Probably it works for any hyperfinite factor (?). –  Ulrich Pennig Dec 9 '12 at 18:02
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For the hyperfinte ${\rm II}_1$ factor the flip is approximately inner which can be seen by restricting to finite dimensional subalgebras, but the unitaries you get won't converge. –  Jesse Peterson Dec 10 '12 at 0:51

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