Let $\mathcal{M}$ be a von Neumann algebra and $\varphi$, a state on $\mathcal{M}$. Let $I=\{a\in \mathcal{M}: \varphi(a^*a)=0\}$.
It follows from Cauchy-Schwartz inequality that $I$ is a left ideal. If we further assume that $\varphi$ is normal, then $I$ will be $\sigma$-weakly closed. I was wondering what would still be true if we don’t assume normality? For example, can we find a corner of $\mathcal{M}$ on which $\varphi$ is faithful?
I think yes and I give my reasons below.
We let $J$ be the $\sigma$-weak closure of $I$, i.e., $J=\overline{I}^{\text{$\sigma$-weak}}$. Then, there exists a projection $q\in\mathcal{M}$ such that $J=Mq$. Let $p=1-q$. I want to show that $\varphi|_{p\mathcal{M}p}$ is faithful.
Towards this end, let $x\in p\mathcal{M}p$ be such that $\varphi(x^*x)=0$. We can write $x=p\tilde{x}p$ for some $\tilde{x}\in\mathcal{M}$. Moreover, $x\in J$ and hence, can be written as $yq$ for some $y\in \mathcal{M}$.
Therefore, $xp=yqp=0$. However, $xp=p\tilde{x}pp=p\tilde{x}p=x$. This establishes the claim.
Please let me know if this is fine. Thanks for your help!!
As s.harp mentioned, the problem is that $q$ can be the identity. For example, take $\mathcal M=B(H)$ and $\varphi$ any non-normal state that is zero on $K(H)$ (these are common once you have the right perspective). So $I\supseteq K(H)$, which means $I=K(H)$ (see this answer). Then $\overline{K(H)}=B(H)$ and so $q=1$.