Is the centre of a dense sub algebra $A$ of the Von Neumann algebra $M$ is trivial ($Z_A \subset A$) then shall we conclude $Z_M(M)$ is trivial hence $M$ is factor?
Remember that we see $Z_A(A) \subset A$ within that subalgebra not in whole $M$.
Is the centre of a dense sub algebra $A$ of the Von Neumann algebra $M$ is trivial ($Z_A \subset A$) then shall we conclude $Z_M(M)$ is trivial hence $M$ is factor?
Remember that we see $Z_A(A) \subset A$ within that subalgebra not in whole $M$.
Copyright © 2021 JogjaFile Inc.
The spirit of the situation is that taking a weak operator closure can make lots of things appear. Like when you take the double commutant of a projectionless algebra and you get a von Neumann algebra with more projections than you can wish for.
Here we can exploit the fact that most commonly C$^*$-algebras admit very distinct representations. For instance take $A$ to be the UHF$(2^\infty)$ algebra. This algebra is unital and simple, and so it's center is trivial. Doing GNS for the trace we get a representation $\pi:A\to B (H)$ such that $\pi(A)''$ is the hyperfinite II$_1$-factor. But you can choose to instead do GNS with respect to a well-chosen state, and now you get $\sigma:A\to B (H)$ such that $\sigma(A)''$ is a Powers' Factor, which is a type III$_\lambda$-factor.
Note that because $A$ is simple, all representations are faithful.
What we can do next is consider $\pi\oplus\sigma:A\to B (H\oplus H)$. Because these representations map into factors they don't have subrepresentations, which forces them to be disjoint as said factors are non-isomorphic. By standard considerations in representation theory of C$^*$-algebras, we have that $$ (\pi\oplus\sigma)(A)''=\pi(A)''\oplus\sigma(A)'', $$ with non-trivial center. So we have a separable C$^*$-algebra with trivial centre that is dense in a separable (as a von Neumann algebra) von Neumann algebra with non-trivial centre.
We can play the same game with all $\lambda\in(0,1)$ at the same time, and that way we can get a representation $\rho$ such that the centre of $\rho(A)''$ is isomorphic to $\ell^\infty[0,1]$.
And all we did above can be done with most of the canonical separable, simple, unital C$^*$-algebras, like the Cuntz algebras or the reduced free group algebras.