Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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 $p$.

share|improve this question
add comment

2 Answers

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...

share|improve this answer
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
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
add comment

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.

share|improve this answer
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.