Take the 2-minute tour ×

Take a separable Hilbert space $H$ and consider the $C^{\ast}$-algebra $A:=B(H)$. Forget for a moment that this algebra is of this sort and do the GNS construction for $A$. You will get a much larger Hilbert space $\mathfrak{H}$ with the property that $A$ can be embedded in $B(\mathfrak{H})$ but this huge Hilbert space is not needed. Now start with an abstract $C^{\ast}$-algebra. The question is: under which assumption $A$ can be embedded in $B(H)$ for separable Hilbert space? I'm also interested in the same question for $W^{\ast}$-algebras. I would be also happy if I could see the proof/reference for the proof. I'm pretty convinced that the answer should be known.

share|improve this question
The "huge" Hilbert space is the space HS(H) of Hilbert-Schmidt operators on H. If H is separable, then so is HS(H). The same is true for an arbitrary von Neumann algebra: A can be embedded into B(H) for some separable H if and only if the GNS construction of A is separable. –  Dmitri Pavlov Jul 23 '12 at 0:39
In the classical GNS construction this huge Hilbert space is a direct sum over all possible states and in the case of $A=B(H)$ it is a Hilbert space of dimension $2^c$, so much larger than initial one. In the case of von Neumann algebras I've meant not the standard embedding "good" for $C^{\ast}$ purposes but an embedding $A$ in $B(H)$ such that $A$ is also closed in the weak operator topology. –  truebaran Jul 23 '12 at 0:54
@Dmitri: I assume truebaran meant doing GNS over states on A, which in the case B(H) is not the same as doing GNS for the "standard" (nsf) tracial weight. I am slightly confused as to what "the" GNS construction for an arbitrary vN alg is supposed to be if one neither specifies a choice of nsf weight, nor decides to take a big direct product over a separating family of such weights –  Yemon Choi Jul 23 '12 at 1:00
@Yemon: If A is a von Neumann algebra, then the GNS construction of A is simply its L^2-space (aka the standard form), which is completely independent of the choice of a faithful semifinite weight. For any faithful semifinite weight m the space GNS(A,m) is canonically isomorphic to L^2(A). What I said about separability is true in this context. –  Dmitri Pavlov Jul 23 '12 at 1:52
@truebaran: You are referring to the Gelfand-Neumark theorem, not the GNS construction. The GNS construction establishes a bijection between cyclic representations and positive functionals. –  Dmitri Pavlov Jul 23 '12 at 10:03

Your Answer



Browse other questions tagged or ask your own question.