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.

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theGNS 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