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Let $R \subset S$ be distinct non-type-I von Neumann factors; say two projections $P, Q \in S$ "meet" if they have a common non-null subprojection (i.e. if $P \wedge Q \neq 0$), and call $P$ "$R$-disjoint" if $P$ meets no projection in $R$ except the identity: Must $S$ have a (non-null) $R$-disjoint projection?

For "split inclusions," where there exists a type I factor $N$ such that $R \subset N \subset S$, the answer is yes (there is a *-isomorphism $\phi : N \rightarrow B(K)$ for some Hilbert space $K$; consider a projection in $P \in N$ such that $\phi(P)$ is a one-dimensional projection along a vector $v \in K$ that is separating for $\phi[R]$); so this question is really about non-split inclusions.

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    $\begingroup$ Suppose every projection of $S$ is not disjoint from one of $R$; take the supremum of all the projections in $R$ that are under $S$, and subtract it from the original .... This off-the-top argument seems to yield the result. $\endgroup$
    – David Handelman
    Commented Oct 29, 2019 at 13:54
  • $\begingroup$ Thanks for reply! Can you define "all the projections in $R$ that are under $S$" more explicitly? I fear I was unable to infer the meaning. But I see an argument that may be close to what you're getting at. Let $P \in S \setminus R$ be a projection such that no proj. $Q \in R \setminus \{ 1 \}$ satisfies $Q \geq P$; let $P' \leq P$ be the supremum of all proj's $P \wedge Q$ where $Q$ is a proj. in $R \setminus \{ 1 \}$. Then unless $P' = P$, $(P - P')$ is indeed a non-null $R$-disjoint projection. But could $P' = P$? It would be "weird" in my opinion but I can't prove it's impossible. $\endgroup$
    – Doug McLellan
    Commented Oct 30, 2019 at 0:06
  • $\begingroup$ I think such "weird" projections can exist if $S$ isn't required to be a factor. Let $R$ be a non-type-I factor, let $A$ be any projection $\neq 0, 1$ in the commutant $R'$, and let $S$ be the v. N. algebra $(R \cup \{ A \})''$ generated by them. I think all $S$'s projections will have form $(A \wedge Q_1) \vee ((1-A) \wedge Q_2)$, where $Q_1, Q_2$ are projections in $R$. Now let $P = (A \wedge Q) \vee ((1-A)(1-Q))$ for some $Q \in R \setminus \{0,1\}$. I think $P$ satisfies the conditions to be a "weird" projection in $S \setminus R$ as above (but, to be clear, this $S$ isn't a factor). $\endgroup$
    – Doug McLellan
    Commented Oct 30, 2019 at 0:19

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