Talk:Von Neumann bicommutant theorem

I think the prove gives here can not work when H is inseaperateable, because the definiton of strong topology.

Problem with proof
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&prime;&prime;. 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 $$\|Xy\|$$ for some y in H, so that for every O in L(H), O is in U if and only if $$\|Oy\|$$ is in V. Since V is open, it contains an open ball of radius d>0 centered at $$\|Xy\|$$.

By choosing h = y, ε = d and repeating the above, we find T in M such that ||Xy - Ty|| < d. Thus $$|\|Xy\| - |Ty\|| < d$$ 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
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)
The problem of finding a $$T$$ in the intersection $$U(h_1, \varepsilon_1)\cap \cdots \cap U(h_n, \varepsilon_n)$$ 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 $$H^{(n)}$$ denote the direct sum of n copies of $$H$$ and let $$ A^{(n)}$$ denote the action of $$A \in L(H)$$ on $$ H^{(n)}$$.

Let $$X, h_i, \varepsilon_i, M$$ be the same as in the current proof. Consider the subspace $$F \subset H^{(n)}$$ which is the closure of $$\{(Sh_1, \ldots, Sh_n): S \in M\}$$. Note that this space is invariant under $$M^{(n)} := \{S^{(n)}: S \in M\}$$ by essentially the same argument as in the proof of the Lemma. We wish to show that $$X^{(n)}(h_1, \ldots, h_n) := (Xh_1, \ldots, Xh_n)$$ is in $$F$$. As in the current proof, we define the projection $$P$$ onto $$F$$. Then the Lemma can be generalised to the statement that $$P \in (M^{(n)})'$$. Indeed, $$M^{(n)}$$ continues to be closed under adjoints in the sense that if $$A \in M$$ then $$ (A^*)^{(n)} \in M^{(n)}$$, 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 $$ (M^{(n)}) = (M)^{(n)}$$ in order for $$X^{(n)}$$ to commute with $$P$$, which is Proposition IX.6.2 and Corollary IX.6.3 of Conway's book and is essentially an exercise in matrix manipulation: given $$S \in (M^{(n)})$$, write $$S = [S_{ij}]$$. Since S commutes with the elementary matrices $$E_{ij}$$ only having 1 in the (i,j) entry (they commute with $$M^{(n)}$$), S is automatically diagonal with equal entries $$C = S_{ii}$$ and therefore is of the form $$C^{(n)}$$ for some $$C \in M$$).

58.168.226.222 (talk) 04:25, 7 July 2020 (UTC)