Would a supersymmetric theory of von Neumann algebras be useful? - MathOverflow

Would a supersymmetric theory of von Neumann algebras be useful?

2010-10-25T23:26:43Z

While looking over the first chapter of

1) Quantum Fields and Strings: A Course For Mathematicians (P. Deligne, P. Etingof, D.S. Freed, L. Jeffrey, D. Kazhdan, J. Morgan, D.R. Morrison and E. Witten, eds.,), 2 vols., American Mathematical Society, Providence, 1999.

I wondered whether there would be any use to developing a theory of super-von Neumann algebras, mimicking the usual theory. Not knowing whether or not this would be a sterile or trivial exercise, I never tried.

I have always wondered, though:

Would there be any benefit in developing a theory of super-von Neumann algebras, and if so what would the benefit likely be? Particularly, could such a theory tell us anything useful about ordinary von Neumann algebras we otherwise couldn't easily obtain?

Perhaps this is a trivial question, but I'm curious if anyone with broader knowledge can shed some light on this.

Of course, the dream is that looking at something like this would miraculously unveil something cool like a canonical time-evolution on $II_{1}$-factors.

(This is another candidate for the 'dumb question' tag!)

Answer by Greg Kuperberg for Would a supersymmetric theory of von Neumann algebras be useful?

2010-10-26T09:23:58Z

A von Neumann algebra is an associative Banach algebra over $\mathbb{C}$, which also has an anti-linear anti-involution * such that $||a^*a|| = ||a||^2$, and which also has a predual as a Banach space. In context, you can think of it as a non-commutative algebra with a certain semisimple-like property and certain fairly strong analytic closure properties.

Now, you can have a non-commutative superalgebra, but this is a somewhat thin combination, because the associativity axiom of an algebra (and in fact every axiom for a von Neumann algebra) does not use the switching map $v \otimes w \mapsto w \otimes v$ or its superized version $v \otimes w \mapsto (-1)^{(\deg v)(\deg w)} w \otimes v$. A supercommutative algebra is not usually a commutative algebra, a Lie superalgebra is not usually a Lie algebra, and a Hopf superalgebra is not usually a Hopf algebra; all of these objects have axioms that use the switching map. But an associative superalgebra is an associative algebra and a von Neumann superalgebra is a von Neumann algebra.

On the other hand, in quantum physics one is often interested in a classical limit which is commutative, or in the supersymmetry context, supercommutative. It is an interesting fact that you can make a commutative von Neumann algebra, which is then a model of classical probability. But you can't make a nontrivially supercommutative von Neumann algebra, because it doesn't have the semisimple-like properties of a von Neumann algebra. However, von Neumann algebra axioms really are necessary for the quantum probability model. So the conventional thing to do is to embed the supercommutative algebra that exists in a theory such as supersymmetry in a von Neumann algebra, even though it is not a von Neumann subalgebra. Or, you could say that supersymmetry (if you accept it) and quantum probability are two ultimately different reasons that classical probability has to be changed. Supersymmetry can be viewed as more of a geometric reason than a probabilistic reason.

Answer by Marcel Bischoff for Would a supersymmetric theory of von Neumann algebras be useful?

2010-11-10T11:21:31Z

Maybe a dump answer. There is the notion of $\mathbb{Z}_2$-graded von Neumann algebra acting on a $\mathbb{Z}_2$ graded Hilbert space. There is also the notion of a graded commutant etc. and it is used in the context fermionic nets, CAR algebra, supersymmetric conformal nets etc. So it looks for me more like "trivial exercise".