Question here about how the projections of a von Neumann algebra $\mathcal{R}$ might be arranged, relative to a projection that is not in $\mathcal{R}$.
Stipulate the following:
$H$ is a Hilbert space of countably infinite dimension;
$B(H)$ is the algebra of bounded linear operators on $H$;
$\mathcal{R} \subseteq B(H)$ is a type II or III von Neumann algebra;
$p \in B(H)$ is a projection NOT in $\mathcal{R}$, and no projection in $\mathcal{R}$ (other than the identity) has a range that contains all of $p$'s range;
We say that one projection "meets" another if the intersection of their ranges is not just $\{ 0 \}$.
Question: Can $p$ meet every nonzero projection in $\mathcal{R}$, yet commute with none of them (other than the identity)?
Note that if $p$, as stipulated above, commutes with all nonzero projections in $\mathcal{R}$, then it meets all of them.* My question is how badly the converse to this can fail.
Added later: I think I have a positive answer in at least one situation, but I would appreciate any more general thoughts on this question (and critiques of the following argument if it's wrong).
Suppose there is a vector $\eta \in H$ that is separating for $\mathcal{R}$. Let $p_\eta$ denote the one-dimensional projection along $\eta$, and let $p_\eta^\perp$ denote its orthogonal complement $(1 - p_\eta)$. $p_\eta$ meets no nonzero projection $q \in \mathcal{R}$ other than the identity, for if $\eta$ were in the range of another $q$, both $q$ and the identity would map $\eta$ to the same vector (namely to $\eta$ itself), violating the definition of "separating." Also $p_\eta$ commutes with no such $q$, since $(p_\eta q) \eta$ will be a scalar multiple of $\eta$ and $(q p_\eta) \eta$ will not. It follows that $p_\eta^\perp$ commutes with no such $q$. Claim: $p_\eta^\perp$ meets each nonzero projection $q \in \mathcal{R}$. This may be obvious but let me prove it. The range of any such $q$ is (since $\mathcal{R}$ is type II or III) an infinite-dimensional subspace of $H$. Let $x, y$ be any linearly independent vectors in this subspace. $x$ can be decomposed as $x = p_\eta x + p_\eta^\perp x$, and $p_\eta x = c_1 \eta$ for some scalar $c_1$. Similarly $y = c_2 \eta + p_\eta^\perp y$ for some $c_2$. So there is a linear combination of these decompositions of $x$ and $y$ such that the $\eta$ terms sum to zero, and the other terms sum to a nonzero vector that is orthogonal to $\eta$. This linear combination is in the range of $q$ and of $p_\eta^\perp$, thus $q$ meets $p_\eta^\perp$.
(Knowing the answer would help me decide whether to investigate these kinds of structures for examples of some order-theoretic properties I've been thinking about. I posted this originally on stackexchange but got no takers there.)
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*Because $p$ can now fail to meet a nonzero $q \in \mathcal{R}$ only by being orthogonal to it, but then $p$'s range would be contained by that of $(1 - q)$, which is a projection in $\mathcal{R}$, contradicting our stipulation about $p$.
