On states and dimensions in Von Neumann algebras

Question

In an exercise, it is given that all the states satisfy normality. Can we prove the algebra to be finite dimensional? Maybe the premise means that it must be separable, and then I can show that it is finite dimensional?

Thanks in advance!

Answer 1

I don't think your idea can work, as normality is not related to the norm-topology.

I also don't see an obvious way to prove the exercise directly from the material in the chapter (not saying it's not possible, just that I don't see it; the fact that $M_*=M^*$ might lead to something).

One way to prove the assertion is to show that every infinite-dimensional von Neumann algebra $M$ has a non-normal state. For this, one shows that in any infinite-dimensional von Neumann algebra there is an increasing net $\{p_j\}$ of projections with $p_j\nearrow 1$ and $p_j\ne 1$ for all $j$. Then the subspace $P=\overline{\operatorname{span}}\{p_j:\ j\}$ does not contain $1$. By Hahn-Banach, there exists $\varphi\in M^*$ with $\varphi(p_j)=0$ for all $j$ and $\varphi(1)=1$.

The problem is that $\varphi$ might not be a state. But Theorem III.4.2.(ii) allows us to write any functional as a linear combination of states. So $\varphi$ is normal, but that's a contradiction. Thus $M$ is finite-dimensional.