Talk:Von Neumann bicommutant theorem
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I think the prove gives here can not work when H is inseaperateable, because the definiton of strong topology.
Problem with proof
[edit]The end of the proof that ii)=> iii) doen not work, since we showed that for every h there is a T, not that for h1... hn there is a single T.
I will correct the proof to the following, unless I missed something and you'll enlighten me:
Let X be in M′′. We now show that it is in the strong operator topology closure of M. For every neighborhood U of X that is open in the strong operator topology, it is the preimage of V, an open neighborhood of for some y in H, so that for every O in L(H), O is in U if and only if is in V. Since V is open, it contains an open ball of radius d>0 centered at .
By choosing h = y, ε = d and repeating the above, we find T in M such that ||Xy - Ty|| < d. Thus and T is in U. Thus in every neighborhood U of X that is open in the strong operator topology there is a member of M, and so X is in the strong operator topology closure of M. Dan Gluck (talk) 18:39, 1 May 2014 (UTC)
- I have made the necessary correction and few other necessary adjustments. Dan Gluck (talk) 11:46, 3 May 2014 (UTC)
Location of a correct proof
[edit]The end of the proof still seems wrong. As pointed out in the "clarification needed" : "This part is incomplete since we must intersect a finite number of these subbasic open sets (September 2015)."
I found a correct proof in the online notes by Vaughan Jones entitled "Von Neumann Algebras", 2015, section 3.2, p. 12. It is quite different -- short but tricky since it involves tensor products (explicitly, you work with matrices whose entries are matrices).
2001:171C:2E60:D7E1:FCD3:E5C6:10B1:1DED (talk) 13:48, 24 April 2020 (UTC)
Clarification to (iii)
[edit]The problem of finding a in the intersection can be solved by taking the approach in Conway's "Functional Analysis" (Section IX.6, specifically Proposition IX.5.3 and Theorem IX.6.4) which is very similar to the current proof. I have attempted to give a direct argument here (without the notation of the lattice of invariant subspaces involved).
Some notation: let denote the direct sum of n copies of and let denote the action of on .
Let be the same as in the current proof. Consider the subspace which is the closure of . Note that this space is invariant under by essentially the same argument as in the proof of the Lemma. We wish to show that is in . As in the current proof, we define the projection onto . Then the Lemma can be generalised to the statement that . Indeed, continues to be closed under adjoints in the sense that if then , so the same argument as in the current proof applies.
From here, the argument after the Lemma is the same (the only thing to check is that in order for to commute with , which is Proposition IX.6.2 and Corollary IX.6.3 of Conway's book and is essentially an exercise in matrix manipulation: given , write . Since S commutes with the elementary matrices only having 1 in the (i,j) entry (they commute with ), S is automatically diagonal with equal entries and therefore is of the form for some ).