Timeline for Definition of a von Neumann algebra

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Sep 30, 2012 at 10:22	comment	added	jbc		@Dmitri Pavlov. would like to suggest the following. Consider all finite or infinite partitions of unity on $ A $. Now take the closed subspace of $A'$ consisting of those functionals which behave well over all of these partitions. I claim (tentatively) that the dual of the latter does what you want. If the family of partitions is finite then the condition is vacuous and you simply get the double dual. If $A$ is already a von Neumann algebra, you get the normal functonals, i.e., the predual. If A is equal to the dual, then it is, of course, a von Neumann algebra.
Dec 25, 2010 at 12:02	history	edited	Dmitri Pavlov	CC BY-SA 2.5	added 296 characters in body
Dec 25, 2010 at 2:03	comment	added	Dmitri Pavlov		@Yemon: Yes, the ultraweak topology on A is the σ(A,A)-topology. I now see a potential problem with this approach: If the embedding A→A* is not ultraweakly continuous, then we cannot say that the ultraweak topology pulls back to itself.
Dec 25, 2010 at 0:15	comment	added	Yemon Choi		But unless I have misunderstood terminology: isn't the ultraweak topology on $A^{}$ the $\sigma(A^{},A^)$-topology, so the restriction of this topology to (the emebdded copy of) $A$ would be the $\sigma(A,A^)$-topology, a.k.a. the weak topology on $A$?
Dec 24, 2010 at 23:31	comment	added	Dmitri Pavlov		@Yemon: I mean that we pull back the ultrweak topology on A along the morphism A→A. In other words, we restrict the ultraweak topology on A** to the embedded copy of A.
Dec 24, 2010 at 23:11	comment	added	Yemon Choi		I'm being slow here, but when you say "pull back the ultraweak topology on $A^{**}$ to $A$" do you mean something different from "restriction to the embedded copy of $A$"?
Dec 24, 2010 at 22:45	history	answered	Dmitri Pavlov	CC BY-SA 2.5