A question about a projection in a W$^*$-algebra

Question

Let $M$ be a W$^*$-algebra algebra over a Hilbert space $\mathcal H$. Let $N$ be a W$^*$-subalgebra of $M$ and $\phi:M \to N$ be a normal positive linear map from $M$ onto $N$. Let us consider the set $\mathcal K=\overline{\bigcup\left\{q(\mathcal H):q \text{ is a projection in $M$ such that } \phi(q)=0\right\}}.$ Let us now define a projection $p$ as $p(\mathcal H)=\mathcal K$, that is, $p$ is a projection from $\mathcal H$ onto $\mathcal K$. Now I want to know whether $\phi(p)=0$?
Also if for any unitary $u \in N$ we have $\phi(u^*xu)=u^*\phi(x)u,~\forall x \in M$, I want to prove that $p\in N'$, the commutant of $N$. Please help me with some hint or answer to solve these two problems. Thanks in advance.

Answer 1

You have by definition that $$ p=\bigvee\{q\in M:\ \phi(q)=0\}. $$ This net is monotone, so by normality you have $\phi(p)=\sup\{\phi(q):\ \phi(q)=0\}=0$.

If you have $\phi(uxu^*)=u\phi(x)u^*$ for all $x\in M$, in particular you have that $\phi(q)=0$ if and only if $\phi(uqu^*)=0$. So, for a fixed $u$, we have the equality $$ \{q\in M:\ \phi(q)=0\}=\{uqu^*\in M:\ \phi(q)=0\}. $$ From this we get that $p=upu^*$. We can write this as $pu=up$. As the unitaries of $N$ span $N$, we obtain $py=yp$ for all $y\in N$, which says that $p\in N'$.