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′′. 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)