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If $\mathcal{R} \subseteq \mathcal{S}$ are factors acting on Hilbert space $H$, with $\mathcal{S}$ type I and $\mathcal{R}$ not type I, and $P$ is a maximal projection in $\mathcal{S}$ (meaning the identity operator is the only projection in $\mathcal{S}$ with larger range), can there exist a nonzero projection $Q \in \mathcal{R}'$ (the commutant of $\mathcal{R}$) such that $P \wedge Q = 0$ (meaning their ranges are disjoint except for the null vector)?

No such $Q$ can exist in $\mathcal{S}$ or in $\mathcal{S'}$ -- see arguments below -- but I don't see how/whether these arguments can be extended to show that no such $Q$ can exist in all of $\mathcal{R}'$. Am I missing something obvious?

No such $Q$ exists in $\mathcal{S}$. Suppose $Q \in \mathcal{R}' \cap \mathcal{S}$ is a counterexample. Note first that $\mathcal{R}' \cap \mathcal{S}$ ($\mathcal{R}$'s relative commutant in $\mathcal{S}$) is a subfactor of $\mathcal{S}$ having the same type as $\mathcal{R}$. Since $\mathcal{S}$ is a type I factor, there is a *-isomorphism $\phi$ from $\mathcal{S}$ to the algebra $B(K)$ of all bounded operators on some Hilbert space $K$. $\phi(P)$'s range will be a subspace of $K$ with co-dimension 1, and $\phi(Q)$'s range will have infinite dimension (since it is a member of a non-type-I subfactor of $B(K)$), and a subspace always has nontrivial intersection with another subspace if its dimension is higher than the other's co-dimension. So $\phi(P) \wedge \phi(Q) \neq 0$, and by isomorphism $P \wedge Q \neq 0$.

(Note this answers the question in the case that $\mathcal{S}$ is the whole algebra $B(H)$.)

Furthermore, no such $Q$ exists in $\mathcal{S}'$, since every nonzero projection in a von Neumann factor meets every nonzero projection in that factor's commutant.

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