Kernel projections in the universal representation. 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 $p$.
 A: 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...
A: 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.
