Ultraweak topology closed hereditary subalgebra induce an ultraweak topology closed left ideal? I am trying to learn about center-valued trace carefully recently, and I met a question right now.
We suppose that $M$ is a von Neumann algebra, and $\varphi$ is a normal positive linear functional, then we know that $\ker \varphi$ is a subspace of $M$ and it is also $\sigma-WOT$ closed. Then we claim that $I=\{x:x^*x\in \ker \varphi \}$ is also $\sigma-WOT$ closed.
I am kind of confused about the claim, how do we know that I is $\sigma-WOT$ closed? I think we may need to use joint continuity for multiplication of $\sigma-WOT$. However, it is not true in general.
We know the fact that every closed hereditary subalgebra will induce a closed left ideal in norm topology, is it still true for ultraweak topology? I am not sure about that.
Any help will be truly grateful!
 A: The kernel of $\phi$ is not an ideal unless $\phi$ is multiplicative. Since $\phi$ is normal, $\ker\phi$ is a $\sigma$-wot closed subspace of $M$. It is $I$ that is a left-ideal.
Since $\phi$ is normal, it has a support projection $p\in M$. We have $\phi=p\phi=p\phi p$, and $p\phi$ is faithful on $pMp$. Hence
\begin{align}
x\in I
&\iff \phi(x^*x)=0 \iff \phi(px^*xp)=0\\[0.3cm]
&\iff px^*xp=0\iff xp=0\\[0.3cm]
&\iff x=x(1-p). 
\end{align}
Hence $I=M(1-p)$. Now it is straighforward to check that $I$ is $\sigma$-wot closed.

The support of $\phi$: the support $p$ of $\phi$ on $M$ is
$$
p=1-\bigvee\{z\in\mathcal P(M):\ \phi(z)=0\}.
$$
From the normality of $\phi$ we deduce that $\phi(p)=1$. It follows that, for any $x\in M$,
$$
|\phi(x(1-p))|^2\leq\phi(x^*x)\phi(1-p)=0.
$$
Hence $\phi(x)=\phi(xp)$ for all $x$. Since $\phi$ preservers adjoints (or doing an analog of the inequality above) we get that $\phi(x)=\phi(px)$. Hence $\phi(x)=\phi(pxp)$ for all $x$.
Finally, $\phi$ is faithful on $pMp$. Indeed, suppose that  $x=pxp$ with $x\geq0$ and $\phi(x)=0$. If $x\ne0$, there exists $\lambda>0$ and a spectral projection $q\leq p$ such that $\lambda q\leq  xq$. Then $$
0\leq\lambda\phi(q)\leq\lambda\phi(xq)=\lambda \phi(x^{1/2}qx^{1/2})\leq\lambda\phi(x)=0. 
$$
Hence $\phi(q)=0$, which implies that $q\leq 1-p$. Hence $q=0$ and so $x=0$, showing that $\phi$ is faithful on $pMp$.
