# 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

In a sense you have to distinguish between and abstract and a concrete von Neumann algebra. But, as with C$^*$-algebras, since you can always represent them as concrete, the distinction is not that important.

But it is true that a von Neumann algebra can be represented on a Hilbert space in such a way that it is not equal to its double commutant. For instance you take a II$_1$ factor and consider an irreducible representation: its image will be dense in some (nontrivial) $B(K)$ but of course it cannot be the whole thing. It is hard to imagine that these representations are of any use, so you always represent your vN algebra in a way that suits you best.

So the meaningful question is whether you can tell intrinsically if a C$^*$-algebra is (or, better said, can be represented as) a von Neumann algebra. This is what you would say is an abstract definition of a von Neumann algebra.

Sakai's characterization is in a sense too abstract. Because it is explicitly known what the predual should be: the normal functionals. So a C$^*$-algebra is isomorphic to a concrete von Neumann algebra precisely when the normal functionals separate points.

When people say "weak topology" in the context of von Neumann algebras they are usually referring to the topology induced by the normal functionals, which is the weak$^*$ topology when the algebra is seen as the dual of the normal functionals. In a concrete von Neumann algebra, this is the ultraweak topology; it agrees with the weak operator topology on bounded sets.

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I recently study the book "$C^*$- algebras and their automorphisms". In chapter 3 theorem3.9.8 you can see some sufficient condition in which a $C^*$ algebra be a von Neumann algebra.

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