Question about projections in von Neumann algebras 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.
 A: 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 $q = z^\perp q \sim zp \sim p$.
A: This is a fully worked out answer based mostly on the answer and comments by Jesse Peterson.
If two non-maximal projections commute, then they are equivalent; this follows from the same $L^\infty$-proof as in the question (it is a set-theoretic thing, drawing a Venn diagram makes it clear).
Claim: it suffices to find a non-trivial projection $z \in M$ that commutes with $p$ and $q$ (or equivalently, the von Neumann algebra $W^*(p,q)' \cap M$ is not $\mathbb{C}$).
Proof(claim): In this case $z^\perp$ also commutes with $p$ and $q$, so by the above we are done if either $z$ or $z^\perp$ is non-maximal. Suppose they are both maximal; then they are both minimal as well, and therefore $z \geq pz \in \{0, z\}$ and $z^\perp \geq p z^\perp \in \{0, z^\perp\}$, and so $p = p z + p z^\perp \in \{0,z,z^\perp, e\}$, contradicting the non-maximality of $p \in \mathcal{P}$.
Consider $W^*(p,q)$. If it is not a factor, then $\mathbb{C} \not= W^*(p,q)' \cap W^*(p,q) \subset W^*(p,q)' \cap M$. If it is a factor, then Theorem V.1.41(ii) in Takesaki I shows that either $W^*(p,q)$ is of type $I_2$, or $W^*(p,q)$ is abelian. In the second case $p$ and $q$ commute and we are done, and in the first case $ \mathbb{M}_2(\mathbb{C}) \cong W^*(p,q) \subset M$. 
Let $e_{ij} \in \mathbb{M}_2(\mathbb{C}) \subset M$ be the matrix with $1$ in the $ij$-th position and $0$ elsewhere. Let $N := e_{11} M e_{11}$. Consider the map $\phi \colon M \to \mathbb{M}_2(N)$ given by $\phi(x)_{ij} := e_{1i} x e_{j1}$. Then, using $\sum_{k=1}^2 e_{kk} = e$,
$$ (\phi(x) \phi(y))_{ij} = \sum_{k=1}^2 \phi(x)_{ik} \phi(y)_{kj} = \sum_{k=1}^2 e_{1i} x e_{k1} e_{1k} y e_{j1} = \sum_{k=1}^2 e_{1i} x e_{kk} y e_{j1} = e_{1i} x e_{j1} = \phi(xy)_{ij},   $$
$$ (\phi(x)^*)_{ij} = (\phi(x)_{ji})^* = (e_{1j}x e_{i1})^* = e_{1i} x^* e_{j1} = \phi(x^*)_{ij}. $$
So $\phi$ is a *-homomorphism. Moreover, the map $\theta \colon \mathbb{M}_2(N) \to M$ given by $\theta(a_{ij}) := \sum_{ij=1}^2 e_{i1} a_{ij} e_{1j}$ satisfies
$$ \theta(\phi(x)) = \sum_{ij=1}^2 e_{i1}x e_{j1} e_{1j} = \sum_{ij=1}^2 e_{ii} x e_{jj} = exe = x, $$
$$ \phi(\theta(a_{ij}))_{kl} = e_{1k} \sum_{ij=1}^2 e_{i1} a_{ij} e_{1j} e_{l1} = e_{11} a_{kl} e_{11} = a_{kl}. $$
Hence $\phi$ is invertible and so it is a *-isomorphism. Under this *-isomorphism, $\mathbb{M}_2(\mathbb{C})$ is identified with $\mathbb{M}_2(e_{11} \mathbb{C} e_{11}) \subset \mathbb{M}_2(N)$, since $\phi(e_{kl})_{ij} = e_{1i} e_{kl} e_{j1} = \delta_{ik} \delta_{lj} e_{11} = e_{11} (e_{kl})_{ij} e_{11}$. It is now clear that $W^*(p,q)' \cap M \cong \mathbb{M}_2(\mathbb{C})' \cap \mathbb{M}_2(N)$ is the set of matrices with a single element from $N$ on the main diagonal and $0$ elsewhere, and so it is isomorphic to $N$. $N$ cannot be trivial, otherwise $M \cong \mathbb{M}_2(\mathbb{C})$, contradicting the non-maximality of $p$.  
