separating vector

Question

Let $\mathcal{A}=vN(\mathbb F_2)$ be the group von Neumann algebra of the free group on two generators.

Is it possible for the algebra $\mathcal{M}=M_2(\mathcal{A})$ to have a separating vector?

Answer 1

The group von Neumann algebra for $\mathbb{F}_2$ (which is typically denoted by $L(\mathbb{F}_2)$, since it comes from the left regular representation) is a $II_1$ factor, which implies that it has a unique faithful normal tracial state $\tau$.

On $M_2(L(\mathbb{F}_2))$, define $$\tilde{\tau}(x) = \frac{1}{2}\sum_{i = 1, 2} \tau(x_{i, i})$$ (like the usual normalized trace on $M_2(\mathbb{C})$, but applying the $L(\mathbb{F}_2)$-trace along the diagonal). This will again be a faithful normal tracial state on $M_2(L(\mathbb{F}_2))$. You can see Chapter V, Proposition 2.14 of Takesaki's Theory of Operator Algebras for the proof (where you think of $M_2(L(\mathbb{F}_2))$ as $L(\mathbb{F}_2) \otimes M_2(\mathbb{C})$), but you could also check for yourself by hand.

The point is that $M_2(L(\mathbb{F}_2))$ admits a faithful normal tracial state, and the distinguished unit vector $\xi_{\tilde{\tau}}$ from its GNS representation will be separating, since $\tilde{\tau}$ is faithful.

So yes, you could have a cyclic and separating vector for $M_2(L(\mathbb{F}_2))$, but like MaoWao pointed out, it depends on what representation/space you're working with.