proof of von Neumann double commutant theorem

Lemma - Let $H$ be a Hilbert space and $B(H)$ its algebra of bounded operators. Let $\mathcal{N}$ be a *-subalgebra of $B(H)$ that contains the identity operator and is closed in the strong operator topology. If $T\in\mathcal{N}^{\prime\prime}$ , the double commutant of $\mathcal{N}$ , then for each $x\in H$ there is an operator $A\in\mathcal{N}$ such that $\|(A-T)x\|<1$ .

: Let $\overline{\mathcal{N}x}\subseteq H$ be the closure of the subspace $\mathcal{N}x:=\{Sx:S\in\mathcal{N}\}$ . It is clear that $\mathcal{N}x$ is an invariant subspace for $\mathcal{N}$ , hence so is its closure $\overline{\mathcal{N}x}$ (see this entry (http://planetmath.org/InvariantSubspacesForSelfAdjointAlgebrasOfOperators), Proposition 5).

Let $P$ be the orthogonal projection onto $\overline{\mathcal{N}x}$ . Since $\overline{\mathcal{N}x}$ is invariant for $\mathcal{N}$ , we have that $P\in\mathcal{N}^{\prime}$ (see this entry (http://planetmath.org/InvariantSubspacesForSelfAdjointAlgebrasOfOperators), last theorem). Since $\mathcal{N}$ contains the identity operator, we know that $x$ belongs to $\overline{\mathcal{N}x}$ . Hence,

\displaystyle Tx=TPx=PTx

where the last equality comes from the fact that $T\in\mathcal{N}^{\prime\prime}$ and $P\in\mathcal{N}^{\prime}$ . Thus, we see that $Tx\in\overline{\mathcal{N}x}$ , which implies that there exists an $A\in\mathcal{N}$ such that $\|Tx-Ax\|<1$ . $\square$

$\,$

$(1)\Longrightarrow(2)$ Since $\mathcal{M}^{\prime\prime}$ is the commutant of some set, namely it is the commutant of $\mathcal{M}^{\prime}$ , it follows that $\mathcal{M}^{\prime\prime}$ is closed in the weak operator topology (see this entry (http://planetmath.org/CommutantIsAWeakOperatorClosedSubalgebra)). But we are assuming that $\mathcal{M}=\mathcal{M}^{\prime\prime}$ , hence $\mathcal{M}$ is closed in the weak operator topology.

$(2)\Longrightarrow(3)$ This part is obvious since the weak operator topology is weaker than the strong operator topology.

$(3)\Longrightarrow(1)$ Suppose $\mathcal{M}$ is closed in the strong operator topology.

A subset of $B(H)$ is always contained in its double commutant, thus $\mathcal{M}\subseteq\mathcal{M}^{\prime\prime}$ . So it remains to prove the opposite inclusion.

Let $T\in\mathcal{M}^{\prime\prime}$ . We are going to prove that $T$ belongs to the strong operator closure of $\mathcal{M}$ , and since $\mathcal{M}$ is closed under this topology, it will follow that $T\in\mathcal{M}$ .

Recall that the strong operator topology is the topology in $B(H)$ generated by the family of seminorms $\|\cdot\|_{x},x\in H$ defined by $\|S\|_{x}:=\|Sx\|$ . A local base around $T$ , in this topology, consists of sets of the form

\displaystyle V(x_{1},\dots,x_{n};\,\epsilon):=\{S\in B(H):\|(S-T)x_{i}\|\leq% \epsilon,i=1,\dots,n\}\,,\qquad\qquad x_{1},\dots,x_{n}\in H,\epsilon>0

We can however consider $\epsilon$ to be $1$ , since $V(x_{1},\dots,x_{n};\,\epsilon)=V(\epsilon^{-1}x_{1},\dots,\epsilon^{-1}x_{n};1)$ .

For every $x_{1},\dots,x_{n}\in H$ we want to find $A\in\mathcal{M}$ such that $A\in V(x_{1},\dots,x_{n};\,1)$ , i.e. such that $\|(A-T)x_{i}\|<1$ , for each $i$ .

Let $\widetilde{H}$ be the direct sum of Hilbert spaces $\widetilde{H}:=\oplus_{i=1}^{n}H$ . For every $A\in B(H)$ let $\widetilde{A}\in B(\widetilde{H})$ be the direct sum of bounded operators (http://planetmath.org/DirectSumOfBoundedOperatorsOnHilbertSpaces) $\widetilde{A}:=\oplus_{i=1}^{n}A$ , i.e.

\displaystyle\widetilde{A}(y_{1},\dots,y_{n})=(Ay_{1},\dots,Ay_{n})\,,\qquad% \qquad y_{1},\dots,y_{n}\in H

We have that $\mathcal{N}:=\{\widetilde{A}:A\in\mathcal{M}\}$ is a *-algebra of bounded operators in $\widetilde{H}$ .

Claim 1 - $\widetilde{T}\in\mathcal{N}^{\prime\prime}$ .

The algebra $B(\widetilde{H})$ can be canonically identified with the algebra of $n\times n$ matrices with entries in $B(H)$ , and $\mathcal{N}$ corresponds to the diagonal matrices with an element $A\in\mathcal{M}$ in the diagonal. Thus, it is easy to check that $\mathcal{N}^{\prime}$ is precisely the set of matrices whose entries belong to $\mathcal{M}^{\prime}$ .

Since the unit matrices (http://planetmath.org/UnitMatrix) belong to $\mathcal{N}^{\prime}$ , it follows that $\mathcal{N}^{\prime\prime}$ consists solely of diagonal matrices with one element on the diagonal (see this entry (http://planetmath.org/CentralizerOfMatrixUnits)). It is easy to check that $\mathcal{N}^{\prime\prime}$ is precisely the set of diagonal matrices with one element of $\mathcal{M}^{\prime\prime}$ in the diagonal. Hence, we conclude that $\widetilde{T}\in\mathcal{N}^{\prime\prime}$ , and Claim 1 is proved.

Now, we observe that $\mathcal{N}$ is a *-subalgebra of $B(\widetilde{H})$ that contains the identity operator. Since $\mathcal{M}$ is closed in the strong operator topology, it follows easily that $\mathcal{N}$ is also closed in the strong operator topology. Since $\widetilde{T}\in\mathcal{N}^{\prime\prime}$ , Lemma 1 that for each $\widetilde{x}:=(x_{1},\dots,x_{n})\in\widetilde{H}$ there exists an operator $\widetilde{A}\in\mathcal{N}$ such that $\|(\widetilde{A}-\widetilde{T})\widetilde{x}\|<1$ . But this is implies that $\|(A-T)x_{i}\|<1$ for each $1\leq i\leq n$ .

Thus, $T\in V(x_{1},\dots,x_{n};\,1)$ . Hence we conclude that $T$ belongs to the operator closure of $\mathcal{M}$ , but since $\mathcal{M}$ is closed under this topology, $T\in\mathcal{M}$ .

We conclude that $\mathcal{M}^{\prime\prime}=\mathcal{M}$ . $\square$

Title	proof of von Neumann double commutant theorem
Canonical name	ProofOfVonNeumannDoubleCommutantTheorem
Date of creation	2013-03-22 18:40:29
Last modified on	2013-03-22 18:40:29
Owner	asteroid (17536)
Last modified by	asteroid (17536)
Numerical id	7
Author	asteroid (17536)
Entry type	Proof
Classification	msc 46L10
Classification	msc 46H35
Classification	msc 46K05