$z(M)M$ is atomic

Suppose $M$ is a W*-algebra, $z$ is the supremum of all minimal projections in $M$. It is well-known that $z$ is a central porjection in $M$. Let $M_0=zM$, then $M_0$ is also a W*-algebra. Denote the collection of all minimal projections in a W*-algebra $M$ by $\mathscr P_m^M$.

**Lemma**. $\mathscr P_m^M=\mathscr P_m^{M_0}$.

*Proof*. Since $m=zm$ for any minimal projection $m$ in $M$, $\mathscr P_m^M\subset\mathscr P_m^{M_0}$. Conversely, suppose $m_0$ is a minimal projection in $M_0$, $p\in M$ and $p\leq m_0$, then $zp\leq zm=m$, hence $zp=0$ or $zp=m_0$. Therefore, $m_0p=m_0zp$ equals to $0$ or $m_0$, i.e., $m_0$ is a minimal projection in $M$.

Now, we can denote $\mathscr P_m^M$ and $\mathscr P_m^{M_0}$ by the same notation $\mathscr P_m$.

**Theorem**. $M_0$ is atomic, that is, each projection in $M_0$ dominates a minimal projection.

*Proof*. Suppose $p$ is a non-zero projection in $M_0$, then $p\leq z$, so, there is a minimal projection $m$ such that $pm\neq 0$( otherwise, $z=\sup\{m\in \mathscr P_m\}\leq 1-p$, $pz=0$ ), which is equivalent to $pmp\neq 0$. Since $$pmpMpmp\subset pmMmp=p(\mathbb{C}m) p=\mathbb{C}pmp,$$ $(pmp)^2$ is a scalar multiple of $pmp$. Let $m_p=pmp/\|pmp\|$, then $m_p$ is a projection and $m_pMm_p= \mathbb C m_p$, that is, $m_p$ is a minimal projection. What’s more, $m_p\leq p$ for $m_pp=pm_p=m_p$.

**Theorem**. For any $q$ in $M_0$, we have $q=\sup\{m\in \mathscr P_m|m\leq q\}$.

*Proof*. Denote $\sup\{m\in \mathscr P_m|m\leq q\}$ by $q_0$, then $q_0\leq q$ and $q_0\in M_0$. If $q_0\neq q$, then $q-q_0$ dominates a minimal projection, say $m$. So, $m\leq q$, and thus $m\leq q_0$ by the definition of $q_0$. Therefore, $m(q-q_0)=mq-mq_0=m-m=0$, a contradiction to $m\leq q-q_0$.

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