What is the simplest argument which shows that each infinite dimensional von Neumann algebra is not separable (in the norm topology)? It seems that this is a kind of folklore: at least I never saw the proof of this fact.
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3 Answers
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This is proven for instance here, Corollary 1.3.17 p. 26.
I am far from being an expert, however it seems to me that the main ingredient in the proof is the fact that if a von Neumann algebra is infinite dimensional, then it contains an infinite family of non-zero pairwise orthogonal projections (see Proposition 1.3.16).
answered Mar 15, 2016 at 14:21
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4$\begingroup$ It would be hard to find a simpler proof than this. The OP tends not to accept answers, but if it were my question I would accept this one. $\endgroup$ Commented Mar 15, 2016 at 16:00
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1$\begingroup$ Is it any easier to prove that an infinite dimensional $C^*$ algebra contains a self adjoint element that has infinite spectrum? $\endgroup$ Commented Mar 15, 2016 at 16:45
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$\begingroup$ @BillJohnson: the proofs are quite similar. Deriving this result from that one would just take one additional step. $\endgroup$ Commented Mar 15, 2016 at 17:48
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Suppose that we have an infinite-dimensional von Neumann algebra. In particular, this is a C*-algebra so it contains an infinite-dimensional abelian C*-algebra $A$. Consequently, $\overline{A}^{{\rm WOT}}$ is a non-separable abelian von Neumann algebra. No need to invoke the fact that it is isomorphic to $L_\infty(\mu)$ for some measure $\mu$, simply use the Borel functional calculus to build an uncountable, closed discrete subset in $\overline{A}^{{\rm WOT}}$ just like you would do in $L_\infty$.
The fact that every infinite-dimensional C*-algebra contains an infinite-dimensional abelian subalgebra is fairly elementary and can be found in Kadison-Ringrose.
answered Mar 15, 2016 at 20:32
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1$\begingroup$ But proving this fact --- that every infinite dimensional C*-algebra contains an infinite dimensional abelian subalgebra --- is no easier than directly proving that every infinite dimensional von Neumann algebra contains infinitely many pairwise orthogonal projections, which immediately entails that it contains a copy of $l^\infty$. (This is the method used in Francesco's reference.) $\endgroup$ Commented Mar 15, 2016 at 23:03
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$\begingroup$ @NikWeaver I guess the "immediately" implicitly contains some argument about a vN alg being monotone complete? $\endgroup$ Commented Mar 16, 2016 at 0:45
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$\begingroup$ @YemonChoi: maybe just how "immediate" depends on how you define von Neumann algebras, but it should be pretty easy regardless. If you take "von Neumann algebra" to be a WOT-closed unital *-subalgebra of $B(H)$ I think it's pretty immediate ... $\endgroup$ Commented Mar 16, 2016 at 1:46
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I'm not sure if this would qualify as simple but the following argument proves the required result. There is a natural locally convex topology on each von Neumann algebra (the finest such that agrees with the weak operator topology on the unit ball) which has the property that the closed graph theorem holds for every linear mapping into a SEPARABLE Banach space. Hence if the algebra were separable in the norm topology, then the latter would coincide with the former and this can only happen in the finite dimensional case.
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$\begingroup$ Do you know a reference for this closed graph theorem results ? $\endgroup$ Commented Mar 15, 2016 at 14:49
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1$\begingroup$ The commutative case is covered on p. 167 of the book "Saks spaces and applications to functional analysis" which is available online, the reduction to the commutative case follows from standards facts on von Neumann algebras. $\endgroup$– oeirasCommented Mar 15, 2016 at 15:52
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1$\begingroup$ "Standard facts on von Neumann algebras"? $\endgroup$ Commented Mar 15, 2016 at 17:37
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$\begingroup$ That a functional thereon is ultraweakly continuous if its restriction to each commutative subalgebra has the same property. $\endgroup$– oeirasCommented Mar 15, 2016 at 19:20
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$\begingroup$ Let me add some context to the above method. There are many examples of Banach spaces (of bounded continuous, analytic, measurable,..., functions and of operators) with supremum norm and these are, in general, not separable, more precisely, there are results analogous to the one above, namely that separability only occurs under very special conditions. Thus a commutative $C^\star$-algebra with unit is separable iff it is $C(K)$ with $K$ compact, metrisable. The above CGT gives a unified explanation of this fact since these spaces have analogous weaker topologies for which it applies. $\endgroup$– oeirasCommented Mar 15, 2016 at 21:13