Topology of state space in von Neumann algebras

What are the sufficient conditions for a von Neumann algebra to have a first countable set of states with respect to the weak * operator topology?

• Just to clarify: do you want the set of all states, or just the normal ones? – Yemon Choi Sep 27 '18 at 8:22
• The set of states of a separable C* algebra is first countable in the weak * topology. This criterion does not apply to infinite dimensional von Neumann algebras. Is there any other sufficient condition? – val 72 Sep 27 '18 at 8:45
• I wonder if there is a criterion for all states. But I think a restriction to normal states will do. – val 72 Sep 27 '18 at 8:47

Every von Neumann algebra is a C$$^*$$-algebra. So the usual theorem that a C$$^*$$-algebra $$A$$ is (norm) separable iff its state space is first countable in the weak-* topology (i.e. the topology $$\sigma(A^*,A)$$) applies. As von Neumann algebras are norm separable iff they are finite-dimensional, we conclude that the state space of a von Neumann algebra $$A$$ is (weak-*) first countable iff $$A$$ is finite-dimensional.

On reflection, the proof I thought of for first countability didn't work. I thought that first-countable compact Hausdorff spaces were second countable, but there are counterexamples, such as $$[0,1]^2$$ with the order topology arising from the lexicographic order (Steen & Seebach, Counterexamples in Topology, Counterexample 48).
So I will instead show it for second countability. Unaccept this answer if you think it's no longer adequate with that restriction. As all von Neumann algebras are unital, we can restrict to the case of a unital C$$^*$$-algebra $$A$$. Suppose that the state space $$X$$ is second countable. As $$A$$ is unital, $$X$$ is also compact, so by Urysohn's metrization theorem, $$X$$ is metrizable. Therefore the C$$^*$$-algebra $$C(X)$$ is separable (see Conway's Functional Analysis, chapter V, Theorem 6.6). By a theorem of Kadison, the map $$\zeta : A \rightarrow C(X)$$ defined by $$\zeta(a)(\phi) = \phi(x)$$ embeds $$A$$ as a closed subspace (but not subalgebra) of $$C(X)$$. Specifically, the image of $$\zeta$$ is the continuous affine functions (see Lemma 2.5 of Kadison's A Representation Theory for Commutative Topological Algebra. This only does it for the self-adjoint elements and real-valued functions, but each element $$a$$ of $$A$$ can be uniquely expressed as $$a = a_r + ia_i$$ where $$a_r,a_i$$ are self-adjoint, and this is enough to extend it to all elements of $$A$$ and complex-valued functions). As a subspace of a separable metrizable space is separable, this shows that $$A$$ is separable.
• @DougMcLellan The predual $A_*$ of a W$^*$-algebra $A$ embeds in the dual space $A^*$ by the double dual embedding, and this embedding is an isomorphism onto the span of the normal states, so I will consider the normal state space $X$ as a convex subset of $A_*$, and then the weak-* topology $\sigma(A^*,A)$ on normal states becomes the weak topology $\sigma(A_*,A)$. If there is a countable $\sigma(A_*,A)$-dense subset $S \subseteq X$, then $X$ is its $\sigma(A_*,A)$-closed convex hull, and therefore its norm-closed convex hull, and so $X$ is norm separable, so $A_*$ is norm separable. – Robert Furber Sep 4 at 19:16