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?
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1$\begingroup$ Just to clarify: do you want the set of all states, or just the normal ones? $\endgroup$– Yemon ChoiCommented Sep 27, 2018 at 8:22
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$\begingroup$ 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? $\endgroup$– val 72Commented Sep 27, 2018 at 8:45
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$\begingroup$ I wonder if there is a criterion for all states. But I think a restriction to normal states will do. $\endgroup$– val 72Commented Sep 27, 2018 at 8:47
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1 Answer
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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.
Added in edit
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.
For the statement that von Neumann algebras are finite-dimensional iff they are norm separable, I have found that this has been asked and answered before here. It is proved in Corollary 1.3.17 of this thesis.
answered Sep 28, 2018 at 23:35
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$\begingroup$ Thank you very much! I would like a reference for the 'usual' theorem, if possible, and a proof sketch for the last the equivalence of norm separabilith and finite-dimensionality in von Neumann algebras. So far I knew only of the following implications: a) for C* algebras norm separability implies first countablity of the state spaceof the von Neumann algebras b) for von Neumann algebras (weak*-) first countability implies finite dimensionality. I was not aware of the inverse implications. $\endgroup$– val 72Commented Sep 29, 2018 at 8:20
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$\begingroup$ Thank you so much for the time and trouble to provide a detailed proof. I am going to uncheck the question to leave open the possibitlity of an answer, since you have proven the implication from second countability to separability (not from first countability). I was wondering if you could indicate a reference (or provide a proof) from the finite dimensionality of a von Neumann algebra to its separability; the corollary shows the one-way implication, not equivalence. I hope it is not too much that I ask. Thanks in advance. $\endgroup$– val 72Commented Sep 30, 2018 at 11:17
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$\begingroup$ @val72 Every topological vector space over the complex numbers is a topological vector space over the reals, and this doubles the dimension, so preserves finiteness of the dimension. For every finite-dimensional topological vector space over the reals, we can pick a finite basis, and the rational linear combinations of that basis form a countable dense subset (by continuity of addition and multiplication by reals). $\endgroup$ Commented Sep 30, 2018 at 22:44
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$\begingroup$ Does anyone know of results on the restriction of this question to normal states, suggested by @yemon-choi? Specifically, conditions under which the normal state space of a von Neumann algebra is second-countable or separable in the weak-* topology? $\endgroup$ Commented Sep 3, 2019 at 15:59
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1$\begingroup$ @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. $\endgroup$ Commented Sep 4, 2019 at 19:16