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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!)

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  • $\begingroup$ I know next to nothing about von Neumann algebras, but I've heard that people sometimes use them to describe quantum field theories. I guess your question may amount to asking if non-super von Neumann algebras can adequately describe supersymmetric quantum field theories (and perhaps, whether people who study von Neumann algebras would find such theories interesting). $\endgroup$
    – S. Carnahan
    Oct 26, 2010 at 5:30
  • $\begingroup$ This link might be useful for those (like me) who forget what the "super" bit means: secure.wikimedia.org/wikipedia/en/wiki/Super_algebra $\endgroup$ Oct 26, 2010 at 8:39
  • $\begingroup$ @Scott: I wish I were being so sophisticated. This is really a duh question. I have heard Kawahigashi mention super von Neumann algebras in connection with is work (w/ Roberto Longo) on QFTs, but I don't know what's going on enough to ask a truly intelligent question about it! $\endgroup$
    – Jon Bannon
    Oct 26, 2010 at 15:10

2 Answers 2

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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.

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  • $\begingroup$ Disclaimer: Witten would be making a much more geometric and field-theoretic point than whether or not the Atiyah-Singer index theorem can be expressed with von Neumann algebras. So, while I think that my answer is correct, I don't know how relevant it is to what Witten was doing. $\endgroup$ Oct 26, 2010 at 11:40
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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".

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  • $\begingroup$ I don't see this as a dump answer. This was my feeling, too, Michael. I wondered if there is some surprise that one would see down the line that is not immediately evident. $\endgroup$
    – Jon Bannon
    Nov 10, 2010 at 20:57
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    $\begingroup$ @JonBannon: I've used this concept in my thesis (see here pages 10 and 11) $\endgroup$ Jan 26, 2014 at 16:50
  • $\begingroup$ Cool! It's nice to have something on this. $\endgroup$
    – Jon Bannon
    Jan 26, 2014 at 18:51

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