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Recall that a real C*-algebra is a Banach $\ast$-algebra $A$ over $\mathbb{R}$ which satisfies the standard C* identity and which also has the property that $1 + a^{\ast}a$ is invertible in the unitalization of $A$ for every $a$. This is the "right" definition because the "real Gelfand-Naimark theorem" is true for such algebras: every real C*-algebra is isometrically $\ast$-isomorphic to a norm closed $\ast$-algebra of bounded operators on a real Hilbert space.

Now we turn to von Neumann algebras. A von Neumann algebra is supposed to be a $\ast$-algebra of bounded operators on a (complex) Hilbert space which is closed in the weak topology, or equivalently the strong topology. This can be abstracted to the intrinsic definition of a von Neumann algebra as a C* algebra which is the dual of some (complex) Banach space. My question is: what is the intrinsic definition of a real von Neumann algebra which abstracts the notion of a $\ast$-algebra of bounded operators on a real Hilbert space which is closed in the weak topology or (equivalently?) the strong topology?

Paul Siegel
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    I just glanced back at the proof of the statement that a $\ast$-algebra of bounded operators on a complex Hilbert space is closed in the weak topology iff it is closed in the strong topology iff it is the same as its bicommutant, and I don't see any reason why the proof shouldn't work over a real Hilbert space, but I would be grateful to be corrected. – Paul Siegel Feb 09 '11 at 16:26
  • Why not use something like "A real X is a complex X together with a fixed choice of a complex conjugation such that the complex X is the complexification of the fixed points under conjugation" – Johannes Hahn Mar 11 '14 at 11:26

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Hi, this is intended as a comment on Jon's comment, but I still lack MO reputation to leave comments; sorry for that. I believe what is mentioned in Li's book is wrong; the right statement should be "a complex $C^\ast$-algebra is the complexification of a real one if and only if it has an involutory ${}^\ast$-antiautomorphism" (here an example by V. Jones of a von Neumann algebra antiautomorphic to itself but without involutory antiautomorphisms: http://www.mscand.dk/article.php?id=2523).

Conversely, one can study real $C^\ast$-algebras in terms of their complexifications: e.g., say that $A$ is a real von Neumann algebra if $A\otimes_{\mathbb{R}}\mathbb{C}$ is von Neumann, and so on.

Mkouboi
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Real operator algebras have been studied by Bingren Li and others. Here's a paper of Li on the topic:

http://www.kurims.kyoto-u.ac.jp/~kyodo/kokyuroku/contents/pdf/0936-8.pdf

I hope this helps!

Jon Bannon
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  • To actually answer the question, if the "right" definition is one for which double commutant and Kaplansky density hold, for example, then what you suggest works. Interestingly, it is mentioned in a book of Li that a (complex) operator algebra is the complexification of a real one if and only if the original algebra has a *-antiautomorphism. So not every von Neumann algebra is the complexification of a real one. – Jon Bannon Dec 16 '11 at 18:00
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Pedersen abstractly characterized von Neumann algebras as AW*-algebras with a separating family of completely additive states. See here.

There's a notion of real AW*-algebra: see Berberian's text Baer $*$-rings, exercise 5.14. Since an AW*-algebra is a C*-algebra in which the right annihilator of every set is generated by a single projection, whatever your definition of real C*-algebra is, this condition could easily be added to define a real AW*-algebra.

Thus, one could define a real von Neumann algebra to be a real AW*-algebra with a separating family of completely additive (real-valued?) states.

(Disclaimer: I'm no expert here, so take my comments with a grain of salt.)

Manny Reyes
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  • Of course, this doesn't quite get at the original question, viewing von Neumann algebras as acting on a Hilbert space. – Manny Reyes Dec 16 '11 at 17:28
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A real von Neumann algebra (or real $W^*-$algebra) is a real *-algebra $\mathcal{R}$ of bounded linear operators on a complex Hilbert space containing the identity operator $\bf 1$, which is closed in the weak operator topology and satisfies the condition $\mathcal{R}\cap i\mathcal{R}=\{0\}$.

Rashidkhon
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