# Abstract Von Neumann Algebras

I have just read this question Is a von Neumann algebra just a C*-algebra which is generated by its projections? and am wondering about Robert Israel's answer when he says that a subalgebra of $C(X)$ is not a Von Neumann algebra. Some abstract definition must be being used, but he also says weak closure (by which I assume he means weak operator closure) and that only exists in $B(H)$. When someone says that a C* algebra $A$ is a Von Neumann algebra, is he saying that $A$ is *-isomorphic (i.e. a bijective map preserving the 3 algebraic operations, is an isometry, and preserves ) to a (concrete) Von Neumann algebra? ( Subalgebra of $B(H)$ equal to its bicommutant) That doesn't seem like what people would mean since such an isomorphism would say nothing about how $A$ lies within $B(H)$ topologically. On the other hand, $A$ isn't known to lie in some external space so one can't hope for a big isomorphism that happens to take $A$ to a von Neumann algebra, and the bigger space containing $A$ to $B(H)$. I am similarly confused about what "weak topology" means.

I have also heard of a predual characterization of Von Neumann algebras, although I've never heard the equivalence of this notion and mine formally stated nor proved. If that's what's going on here, could someone please direct me to a sufficiently precise description of what everything means?

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My understanding of what's going on: a von Neumann algebra is a C*-algebra with a predual. A choice of predual induces a weak topology; I am not sure whether this topology depends on the choice of predual or not. –  Qiaochu Yuan Sep 13 '12 at 21:46
Yeah I'm not sure either, because one of the requirements for a Von Neumann Algebra (VNA) is that it be weakly closed in $B(H)$ and I don't see a way any notion of the word "is" in your comment could be construed to capture this. Isomorphisms, weak to weak operator homeomorphism, and so on, captures everything intrinsic, but being weakly closed in $B(H)$ seems decidedly extrinsic. –  Jeff Sep 13 '12 at 22:12
There is no "choice of predual", AFAIK. The predual of a von Neumann algebra is unique. See Sakai, "C*-algebras and W*-algebras", sec. 1.13. books.google.ca/books?id=DZ5JvQaIz8QC –  Robert Israel Sep 13 '12 at 22:55
I gave you two references to Sakai's theorem in this question of yours. –  t.b. Sep 14 '12 at 6:53
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