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 $p$.



Kernel projections in the universal representation.
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...


