Isomorphism of Hilbert spaces

Question

Suppose that two von Neumann algebras acting on two different Hilbert spaces are isomorph. Is this true that those Hilbert spaces are isomorph?

Answer 1

No, given any Hilbert space $\mathcal{H}$ one can let the $2 \times 2$-matrices $M_2(\mathbb{C})$ act on $\mathcal{H} \oplus \mathcal{H} $ as usually as follows: $$ \begin{pmatrix} a & b \\ c & d \end{pmatrix} \begin{pmatrix} \xi \\ \eta \end{pmatrix} = \begin{pmatrix} a \xi + b \eta \\ c \xi + d \eta \end{pmatrix}. $$ The associated von Neumann algebra $M$ we get is the set of all such matrices, acting on $\mathcal{H} \oplus \mathcal{H}$. However, seen as a $*$-algebra, $M$ is the same thing as the usual $2 \times 2$-matrices acting on $\mathbb{C}^2$. So $M \cong M_2(\mathbb{C})$, but we don't have that $\mathcal{H} \oplus \mathcal{H} \cong \mathbb{C}^2$ as Hilbert spaces.

However there is the notion of an isomorphism $\Phi$ of von Neumann algebras $(M,\mathcal{H})$ and $(N, \mathcal{K})$ induced by a unitary $u$. This means that $u$ is a unitary operator between the Hilbert spaces $\mathcal{H}$ and $\mathcal{K}$ such that $$u x u^\ast = \Phi(x). $$

The above example shows that given any isomorphism $\Phi$ between von Neumann algebras, such a $u$ does not necessarily exist.