As in the title, given $N$, a finite von Neumann algebra with a separable predual, does there always exist a type $II_1$ factor $M$ with a separable predual such that $N \subseteq M$? I've heard that letting $M = N \bar{\ast} R$, the tracial free product of $N$ and the hyperfinite $II_1$ factor $R$, gives a solution, but I am unable to find a proof of this.
Is a finite von Neumann algebra with separable predual always a subalgebra of a type $II_1$ factor with separable predual
operator-algebrasoperator-theoryvon-neumann-algebras
Best Answer
It can be done as you say.
As far as I can tell, the first source (which uses a different technique) is Elliott-Handelman, Math. Scand. 49 (1981), no. 1, 95–98. The requirement of properly infinite commutant is no big deal since it can be achieved by tensoring.
The free-product argument appears for instance in Proposition 8.1 in Bercovici-Collins-Dykema-Li-Timotin, JFA 258 (2010) 1579–1627.