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.



Question about projections in von Neumann algebras
This is true. Since $p, q \in \mathcal P(M)$ are non-trivial and not maximal, then under your definition they are equivalent to any non-trivial subprojection. Also, we have $M \not= \mathbb M\_2(\mathbb C)$, and so there exists a non-trivial projection $z \in \mathcal P(M)$ commuting with $p$ and $q$, such that $zp \not= 0$ (see, e.g., Theorem 1.41 in Volume 1 of Takesaki's books).


Then we can consider separate cases:


If $q = p^\perp$, then $q \sim p\_0 \sim p$, where $p\_0$ is any projection $0 \not= p\_0 \lneq p^\perp$.


If $z = q$ then $q \sim zp \sim p$.


If $z^\perp = q$ and $z \not= p$, then $q \sim zp \sim p$.


If $zq \not= 0$ and $z \not= q$, then $q \sim zq \sim z^\perp \sim zp \sim p$.


If $zq = 0$, $z^\perp \not= q$, then $q = z^\perp q \sim zp \sim p$.


