Let $M$ be a von Neumann algebra, and let $\mathcal{P}$ be the set of nontrivial (not equal to $0$ or $e$) projections of $M$. Define $p,q \in \mathcal{P}$ to be equivalent if there exist projections $p_1, \ldots, p_n \in \mathcal{P}$ with $p_1=p$, $p_n = q$, $p_i \perp p_{i+1}$ and $p_i + p_{i+1} < e$ for $1 \leq i < n$.

If $p \in \mathcal{P}$ is maximal, then clearly no other projection is equivalent to it. The question is: are every two nonmaximal projections $p$ and $q$ equivalent?

This is true in $B(H)$. If $\dim(H) \leq 2$, then all projections in $\mathcal{P}$ are maximal, so we may assume $\dim(H) \geq 3$. Clearly $p$ and $q$ are equivalent to rank 1 projections, so we may assume that $p$ and $q$ are rank 1. Then the orthogonal complements of the range of $p$ and $q$ are codimension 1 and hence have nontrivial intersection: take $p_2$ the projection onto (a subspace of) this intersection and $p_3 = q$.

It is also true in $L^\infty$; the problem becomes a set-theoretic one on the $\sigma$-algebra. If $p \vee q \not= e$, then $p$ and $q$ are equivalent to $e-p \vee q$, so assume $p \vee q = e$; this implies that $p$ and $q$ are incomparable. Now if $p \perp q$, then by the nonmaximality of $p$, $p$ is equivalent to a subprojection $p_2$ of $q$, and $q= p_4$ is equivalent to a subprojection $p_3$ of $p$. If $p$ and $q$ are not orthogonal, then $p \sim q - p \wedge q \not= 0$ and $q \sim p - p \wedge q \not= 0$. This proof uses the fact that if $p$ and $q$ are not orthogonal, then $p \wedge q \not= 0$, which is unfortunately no longer true if $M$ is not abelian.

Is it true for arbitrary von Neumann algebras? If not, what conditions (e.g., type I) on $M$ are needed?

P.S. I am not an expert on von Neumann algebras.