From [Donsing-Pitts-2008, theorem 4.8]:
For $A\subseteq B$ a regular inclusion, with $A$ abelian, and $E:B\to A$ its unique conditional expectation it holds:
The left ideal $$L(E):=\{b\in B:E(b^*b)=0\}$$ is a two-sided ideal $L(E)\trianglelefteq B$.
Can someone clarify the last point for the proof given there (see below) for the above statement?
Sketch of proof from [Donsing-Pitts-2008]:
Let $N(A\subseteq B)$ denote the normalizers.
Evidently it is sufficient to verify this for normalisers, i.e. for $n\in N(A\subseteq B)$ and $b\in B$,
$$E(b^*b)=0\implies E(n^*b^*bn)=0$$
As usual, $n\in N(A\subseteq B)$ induces a partial homeo on characters given by
$$n:\Gamma(A)\to\Gamma(A):\quad n(\alpha)a:=\frac{\alpha(n^*an)}{\alpha(n^*n)}$$
whenever $\alpha\in\Gamma(A)$ satisfies $\alpha(n^*n)\neq0$, which is the domain of the partial homeo.
By the unique extensions property, the induced character extends uniquely to the state
$$\beta:=n(\alpha)\circ E\in SB.$$
This is mentioned in the second paragraph of "4.1 CONTEXT FOR SECTION 4" on page 373.
Now since $E(b^*b)=0$ also indeed
$$\beta(b^*b)=\alpha(n^*E(b^*b)n)=0$$
and from here it is claimed that and hence $\alpha(E(n^*b^*bn))=0$ in this case as well.
Why is that so, can someone elaborate on this last point?
Am I maybe overlooking something?
[Donsing-Pitts-2008]: see https://www.jstor.org/stable/24715826
