Minimal projections Assume $M$ is a W*-algebra  such that the set of  minimal projections is not empty. Let $z(M)$ be the supremum of all minimal projections in $M$. It is well-known that $z(M)$ is a central projection. 
Let us consider the W*-algebra $M_0=z(M)M$. Let $q$ be  a projection in $M_0$. 
Does the following statement hold? 
$$q=\textrm{sup}\{e: e ~\textrm{is a minimal projection}\}$$ 
 A: This answer is some details support and reorganization for Martin's.
First, it is easy to verify that the collection of all minimal projections in $M$ coincides with that in $M_0$. We denote them by a single notation $\mathscr P_m$.
Lemma. $M_0$ 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 $p\leq z(M)$, so, there is a minimal projection $m$ such that $pm\neq 0$( otherwise, $z(M)=\sup\{m\in \mathscr P_m\}\leq 1-p$, $pz(M)=0$ ), 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 by the above lemma, 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 $m\leq q-q_0$.
A: I assume that what you mean is whether $q=r $, where $r=\sup\{e:\ e\leq q \text { and }e \text { minimal}\} $.
If $q\ne r $, then $q-r $ is a nonzero projection. If $f\leq q-r $ is a minimal projection, then $f\leq q $ and $r+f \geq r $, a contradiction. If $q-r $ dominates no minimal projection, then $z-(q-r) $ is the supremum of all mimimal projections, a contradiction. So $q=r $.
