double commutant of the left representation of $C^*$ algebra

Question

If $A$ is a unital $C^*$ algebra with a trace $\tau$,$\lambda:A\to B(L^2(A,\tau))$ is the left representation.where $L^2(A,\tau)$ is the GNS space.Let $\lambda(A)^{"}$ be the double commutant of $\lambda(A)$,Suppose there is a central projetion $p$ in the center of $\lambda(A)^{"} $.Is the map $A\to p\lambda(A)^{"} $ surjective?

Answer 1

If you mean the map $a \mapsto p\lambda(a)$, then no, the map won't be surjective in general. For example, take $A$ to be $C(\mathbb{T})$, continuous functions on the complex unit circle, with $\tau$ integration against the Haar measure (i.e. normalized Lebesgue). Then $\lambda$ just becomes the representation of pointwise multiplication of $f \in C(\mathbb{T})$ on $L^2(\mathbb{T})$.

For any nonzero projection $p \in L^\infty(\mathbb{T}) = \lambda(A)''$, the map $a \mapsto p\lambda(a)$ isn't surjective. One way of seeing this is that $p\lambda(A)''$ is a von Neumann algebra, so it's generated by its projections, but the only projections in $p\lambda(A)$ are $0$ and $p$.