Potentially a silly question, but do you have an example where neither $H\ne \Bbb C$ M$nor$N$is finite type$I$and one of$M$and$N$is not infinite type$I,$and$N\vee M'\cong N\overline{\otimes} M',$(by the map you defined) even non-spatially (for$N$and$M$factors)factors? It's not clear to me that there is a non-trivial case when this actually is an isomorphism, or that there is an example where these algebras are isomorphic by a different map. For example if$N=M$this is impossible unless$M$is type I, since$M\vee M'=B(H).$Also if say$M$or$N$is type$II$you need to have$N$far'' from$M$for$N\vee M\cong N\overline{\otimes} M,$for example$N'\cap M$cannot be finite dimensional (because then$N\vee M$is type I begin a commutant of a type$I$von Neumann algebra). So this will rule out finite index inclusions, for example. Post Undeleted by Benjamin Hayes 2 deleted 722 characters in body I don't have Potentially a full answer to your silly question, but note that it is impossible to have$N$be type$II_{1},$because in this case we cannot have$N\vee M'\cong N\overline{\otimes} M$even non-spatially. In fact$N\vee M'\subseteq N\vee N'= B(H),$by the double commutant theorem. So at least this gives some reduction, for example if$M$is type$II_{1},$then$N$must be finite-dimensional. (Although I really don't know how to handle the$II_{\infty}$or type$I$case off the top of my head). Also, for rather silly reasons$H$of course must be infinite dimensional, or$H=\Bbb C$(although in the case$H=\Bbb C$there is nothing to prove), for this isomorphism to be spatial. However this example leads me to a clarification question: do you have an example where$H\ne \Bbb C$and$N\vee M'\cong N\overline{\otimes} M',$(by the map you defined) even non-spatially (for general$N$and$M$factors)? It's not clear to me that there is a non-trivial case when this actually is an isomorphism. For example if$N=M$this is impossible unless$N$or$M$is type I, since$M\vee M'=B(H).\$