The element $z \in M$ commutes with every element of $M$

Question

Let $M$ be a von Neumann algebra on a Hilbert space $H$. Let $p$ be a projection in $M$. Now consider the set $P(M)$ as the set of all projections in $M$, that is, $$P(M):=\{x \in M: x =x^2=x^*\}.$$ Then given $p$ belongs to $P(M)$. Now we say that $q_1\sim q_2$ for any elements $q_1,q_2 \in P(M)$ if there exists an element $v \in M$ such that $q_1=v^*v$ and $q_2=vv^*$. Now consider the set $\cal U$ as, $$\cal U=\{q \in P(M) : q \sim p\}.$$ Let $K=\displaystyle\overline{\bigcup_{q\in \cal U} q(H)}$, that is, $K$ is the closure of union of the image of $q$ for all $q \in \cal U$. Let $z\in P(M)$ such that $z(H)=K$, that is $z \in M$ be a projection from $H$ to $K$. Now I have to show that $zx=xz$ for all $x \in M$.
Now notice that $p$ is an element of $\cal U$, since $p=p^*p=pp^*$. This gives us $p\le z$. Also, any element of a unital $C^*$-algebra can be written as linear span of its unitaries. So, if we show that $zu=uz$ for all unitary elements $u$ in $M$, then we are done. Please help me to solve this. Thank you.

Answer 1

What you said last is the key. If $u$ is any unitary in $M$, and $q\in\mathcal U$, then $uqu^*\in\mathcal U$ (with partial isometry $pu^*$). This shows that $q\in\mathcal U$ if and only if $uqu^*\in\mathcal U$ for all unitary $u$ in $M$. It follows that $$ K=\overline{\bigcup_{q\in\mathcal U}uqu^*H}=\overline{\bigcup_{q\in\mathcal U}uqH}=u\,\overline{\bigcup_{q\in\mathcal U}qH}=uK. $$ Then for any $h\in H$ we have $uzh=zuzh$, which gives $$ uz=zuz $$ for every $u\in M$ unitary. In particular $$ u^*z=zu^*z=(zuz)^*=(uz)^*=zu^*. $$ Applying the above to $u^*$ we get that $uz=zu$ for any unitary in $M$.