What is the right definition of "real von Neumann algebra"? - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T13:14:13Z http://mathoverflow.net/feeds/question/54898 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/54898/what-is-the-right-definition-of-real-von-neumann-algebra What is the right definition of "real von Neumann algebra"? Paul Siegel 2011-02-09T16:20:22Z 2011-12-16T18:43:37Z <p>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.</p> <p>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?</p> http://mathoverflow.net/questions/54898/what-is-the-right-definition-of-real-von-neumann-algebra/83636#83636 Answer by Jon Bannon for What is the right definition of "real von Neumann algebra"? Jon Bannon 2011-12-16T16:50:58Z 2011-12-16T16:58:43Z <p>Real operator algebras have been studied by Bingren Li and others. Here's a paper of Li on the topic:</p> <p><a href="http://www.kurims.kyoto-u.ac.jp/~kyodo/kokyuroku/contents/pdf/0936-8.pdf" rel="nofollow">http://www.kurims.kyoto-u.ac.jp/~kyodo/kokyuroku/contents/pdf/0936-8.pdf</a></p> <p>I hope this helps!</p> http://mathoverflow.net/questions/54898/what-is-the-right-definition-of-real-von-neumann-algebra/83640#83640 Answer by Manny Reyes for What is the right definition of "real von Neumann algebra"? Manny Reyes 2011-12-16T17:17:22Z 2011-12-16T17:17:22Z <p>Pedersen abstractly characterized von Neumann algebras as AW*-algebras with a separating family of completely additive states. <a href="http://blms.oxfordjournals.org/content/4/2/171.short" rel="nofollow">See here</a>.</p> <p>There's a notion of real AW*-algebra: see Berberian's text <em>Baer</em> $*$-<em>rings</em>, 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.</p> <p>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.</p> <p>(Disclaimer: I'm no expert here, so take my comments with a grain of salt.)</p> http://mathoverflow.net/questions/54898/what-is-the-right-definition-of-real-von-neumann-algebra/83652#83652 Answer by Mkouboi for What is the right definition of "real von Neumann algebra"? Mkouboi 2011-12-16T18:43:37Z 2011-12-16T18:43:37Z <p>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 "<em>a complex $C^\ast$-algebra is the complexification of a real one if and only if it has an involutory ${}^\ast$-antiautomorphism</em>" (here an example by V. Jones of a von Neumann algebra antiautomorphic to itself but without involutory antiautomorphisms: <a href="http://www.mscand.dk/article.php?id=2523" rel="nofollow">http://www.mscand.dk/article.php?id=2523</a>). </p> <p>Conversely, one can study real $C^\ast$-algebras in terms of their complexifications: <em>e.g.</em>, say that $A$ is a real von Neumann algebra if $A\otimes_{\mathbb{R}}\mathbb{C}$ is von Neumann, and so on.</p>