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Let $A \subseteq \mathcal B(\mathcal H)$ be a unital C*-algebra in its universal representation. The GNS representation $\pi_\mu\colon A \rightarrow \mathcal B(\mathcal H_\mu)$ with base state $\mu$ extends uniquely to a normal $\ast$-homomorphism $\pi_\mu''\colon A'' \to \mathcal B(\mathcal H_\mu)$. Since $A''$ is a von Neumann algebra, there exists a unique projection $p\in A''$ such that $\mathbf{ker} \\ \pi_\mu''\ = A''p$. Is $p$ the least upper bound of operators $a \in A$ such that $\mu(a) = 0$ and $0 \leq a \leq 1$?

EDIT: Jonas provided a simple counterexample. For non-commutative C*-algebras, $\mu(a)=0$ of course does not imply that $\pi_\mu(a)=0$. The naive question is therefore whether $p$ is the least upper bound of operators $a\in A$ such that $0 \leq a \leq 1$ and $\mu(c^\ast a c)=0$ for all $c \in A$. More generally, I would be grateful for any such "intrinsic" characterization of the projection $p$.

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2 Answers 2

up vote 3 down vote accepted

So, it seems like the new question is: Is $\ker\pi_\mu \subseteq A$ dense in $\ker\pi_\mu'' \subseteq A''$. As we're talking about the universal representation, $A''=A^{**}$, the bidual of A.

So, suppose that $\mu$ is a faithful state on A, so $\ker\pi_\mu=\{0\}$. I don't think it's necessary that $\ker \pi_\mu''=\{0\}$: this is equivalent to $\pi_\mu'':A^{**}=A''\rightarrow B(H_\mu)$ being injective, and hence an isomorphism onto its range.

For example, let $G$ be an infinite discrete group, let $A=C^*_r(G)$, and let $\mu$ be the canonical trace on $A$. Then $\pi_\mu(A)'' = VN(G)$ the group von Neumann algebra (as $H_\mu$ can be identified with $\ell^2(G)$), and the predual of $VN(G)$ is $A(G)$ the Fourier algebra. The dual of A is $B(G)$ the Fourier-Stieljtes algebra; as $G$ is not compact, $A(G) \subsetneq B(G)$. Then $A^{**} = B(G)^* = W^*(G)$ (in common notation) and so the map $\pi_\mu'':W^*(G) \rightarrow VN(G)$ is the quotient induced by the adjoint of the inclusion $A(G) \rightarrow B(G)$. In particular, it's not injective.

I could, of course, have misunderstood the question...

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2  
It may be instructive to note that MD's example has nothing to do with (non)commutativity: take $G$ to be the group of integers, so that $C^*(G)=C(T)$; then $\mu$ is just "integration round the circle" and $VN(G)=L^\infty(T)$, and we're observing that the natural map from $C(T)^{**}$* onto $L^\infty(T)$ is not injective. –  Yemon Choi Apr 20 '10 at 18:09
    
(That should have been $C(T)^{**}$ in the last line, obviously) –  Yemon Choi Apr 20 '10 at 19:19
1  
Actually, probably the easiest example is to let $A=C[0,1]$ and to let $\mu$ be Lebesgue measure. Then the observation simply becomes that $C[0,1]^{**}\rightarrow L^\infty[0,1]$ is a proper quotient map, which follows as the inclusion $L^1[0,1]\rightarrow C[0,1]=M[0,1]$ is not a surjection, because not all regular measures on [0,1] are absolutely continuous with respect to Lebesgue measure. –  Matthew Daws Apr 21 '10 at 9:30

I am coming on this problem two years later and trying to remember things I used to know twenty years ago, but my answer (to Jonas' re-stated question) is that $\ker \pi_{\mu}$ is almost never dense in $\ker\pi_{\mu}''$. The universal representation is the direct sum of all the GNS representations and the kernel of $\pi_{\mu}''$ is everything in $A^{**}$ that arises from other GNS representations. So if $\mu$ is a faithful state, $\ker\pi_{\mu}=0$ but $\ker\pi_{\mu}''$ is generally enormous. The projection $p$ is the central cover of the representation $\pi_{\mu}$ and is discussed in the (older?) standard books on C*-algebras.

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