Generated von neumann algebra and commutant

Question

Let $(\mathfrak{M}_i, \mathcal{H})_i$ a family of Von neumann algebras acting on the same Hilbert space. The von Neumann algebra

$$\bigcup_{i}\mathfrak{M}_i = \bigcap\{\mathfrak{M} \, : \, \mathfrak{M} \supset \mathfrak{M}_i \, \, \, \forall i\}$$

is called the generated von neumann algebra by $\mathfrak{M}_i$.

I need to prove that $(\bigcup_i \mathfrak{M}_i)' = \bigcap \mathfrak{M}_i'$ and $(\bigcap_i \mathfrak{M}_i)' = \bigcup_i \mathfrak{M}_i'$ (The commutants)

how may I do that?

Martin Argerami · Accepted Answer · 2020-06-20T11:22:04.600

3

Comment: the notation for the von Neumann algebra generated is terrible, as now you have no notation for the union. I will use the union as a union.

Notation: $W^*(S)$ is the von Neumann algebra generated by the set $S$. It is easy to check that $S'=W^*(S)'$: we have $S'=S'''=(S'')'=W^*(S)'$. So you actually only need to do your argument with the union, and not with the algebra generated.

Suppose that $T\in (\bigcup_j M_j)'$. Then, $T\in M_j'$ for any $j$, so $T\in\bigcap_jM_j'$. Conversely, if $T\in \bigcap_jM_j'$ then $T\in M_j'$ for all $j$; so $T\in(\bigcup_jM_j)'$.

Now, using the first equality. $$ W^*\Big(\bigcup_jM_j\Big)=\Big(\bigcup_jM_j\Big)'' =\Big[\big(\bigcup_jM_j\big)'\Big]'=\Big[\bigcap_jM_j'\Big]' $$

Edit: proof that the von Neumann algebra generated by a selfadjoint set $S\subset B(H)$ is $S''$.

Since $S''$ is a von Neumann algebra and it contains $S$, then $W^*(S)\subset S''$. Conversely, since $S\subset W^*(S)$, then $W^*(S)'\subset S'$, so $S''\subset W^*(S)''=W^*(S)$.

edited Jun 20 '20 at 11:22

answered Jun 19 '20 at 17:35

Martin Argerami

205,756

is the union of von neumann algebras a von neumann algebra? – James Arten Jun 19 '20 at 22:16
1

No. In general, a union of subspaces is not a subspace. – Martin Argerami Jun 20 '20 at 00:21
how to prove that both the definitions of generated algebra are equivalent? I mean the bicommutant and the intersection of all von neumann algebras which contain the first one – James Arten Jun 20 '20 at 10:14
It's the first thing you prove after the double commutant theorem. I included two lines in the answer at the end. – Martin Argerami Jun 20 '20 at 11:23
Thank you very much! – James Arten Jun 20 '20 at 11:29
@MartinArgerami Where did you use the "self-adjointness" of $S$ in the proof of the fact that: "von Neumann algebra generated by a selfadjoint set $S$ is $S''$? " It seems the proof works for any subset $S$ of $B(H)$....Isn't that? – sigma Jul 18 '21 at 08:27
@indrajit: No. If $S$ is not selfajoint there is no reason for $S''$ to be a $*$-algebra. – Martin Argerami Jul 18 '21 at 12:04
I understand. Thank you – sigma Jul 18 '21 at 13:15

Generated von neumann algebra and commutant

1 Answers1