# Isomorphism between von Neumann algebras

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 $$\varphi$$?

• What if $U_2=R_2$ but $U_1$ is a proper subalgebra of $R_1$? Then you obviously cannot extend $\varphi$ to an injective function. Or am I missing something? Commented Jun 28, 2018 at 10:11

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 $A'' = \mathbb B(L^2(F_2)) \not \cong L(F_2)$.
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.