Isomorphism between von Neumann algebras

Question

Let $R_{1}$ and $R_{2}$ be two von Neumann a algebras with wot dense sub algebras $U_{1}$ and $U_{2}$. Suppose $\varphi$ is a * isomorphism from $U_{1}$ onto $U_{2}$. Is there always an isomorphism $\phi$ between $R_{1}$ and $R_{2}$ which is an extension of $\varphi$?

Answer 1

The answer is no.

Let $F_2$ be the free group on two generators, then the reduced group $C^\ast$-algebra $C^\ast_r(F_2)$ is wot dense in the group von Neumann algebra $L(F_2) \subseteq \mathbb B(L^2(F_2))$. As $C^\ast_r(F_2)$ is separable, simple, and not type I, there is a faithful irreducible representation $\pi \colon C^\ast_r(F_2) \to \mathbb B(L^2(F_2))$, so the image $A:=\pi(C^\ast_r(F_2))$ is $\ast$-isomorphic to $C^\ast_r(F_2)$, and $A'' = \mathbb B(L^2(F_2)) \not \cong L(F_2)$.

Answer 2

This is not even true when $U_1=U_2$. For instance if you take $U_1=U_2=UHF(2^\infty)$, then by taking the von Neumann algebra obtained via GNS with the trace, you get $\mathcal R_1$ the hyperfinite II$_1$-factor. While by using an appropriate state $\psi_\lambda$, you can get $\mathcal R_2$ to be one Power's III$_\lambda$ factors.