On proof of weak operator closed ness of kernel of the representation

Question

Let $\pi:L^{\infty}(X,\mu)\mapsto B(\mathcal{H})$ be a representation of an abelian von Neumann algebra. Where $\mu$ is a probability measure and $\mathcal{H}$ is a seperable Hilbert space. Prove that $\text{ ker }\pi$ is weak operator closed.

Answer 1

It's not clear what you are asking. There are two possibilities:

• You are requiring $\pi$ to be wot continuous. In that case, $\ker\pi=\pi^{-1}(\{0\})$ is closed.

• You are not requiring $\pi$ to be wot continuous. In that case the assertion is not true. Take $X=\mathbb N$ with the counting measure. Fix a free ultrafilter $\omega$, and let $\pi:\ell^\infty(\mathbb N)\to\mathbb C$ be $\pi(x)=\lim_{n\to\omega}x_n$. Then $\pi$ is a representation, and $\ker\pi\supset c_0$. But $c_0$ is wot dense in $\ell^\infty(\mathbb N)$.