commutant is a weak operator closed subalgebra

Let $H$ be a Hilbert space and $B(H)$ the algebra of bounded operators in $H$. Recall that the commutant of a subset $ℱ\subset B(H)$ is the set of all bounded operators that commute with those of $ℱ$, i.e.

${ℱ}^{\prime }:=\{T\in B(H):TS=ST,\forall S\in ℱ\}.$

- If $ℱ\subset B(H)$, then ${ℱ}^{\prime }$ is a subalgebra of $B(H)$ that contains the identity operator and is closed in the weak operator topology.

: It is clear that ${ℱ}^{\prime }$ contains the identity operator, since it commutes with all operators in $B(H)$ and in particular with those of $ℱ$.

Let us now see that ${ℱ}^{\prime }$ is a subalgebra of $B(H)$. Let $T_1,T_2\in {ℱ}^{\prime }$ and $\lambda \in ℂ$. We have that, for all $S\in ℱ$,

$S(T_1+T_2)=S{T}_1+S{T}_2=T_{1}S+T_{2}S=(T_1+T_2)S$

$S(\lambda T_1)=\lambda S{T}_1=\lambda T_{1}S$

$S(T_1{T}_2)=T_{1}S{T}_2=T_1{T}_{2}S$

thus, $T_1+T_2$, $\lambda T_1$ and $T_1{T}_2$ all belong to ${ℱ}^{\prime }$, and therefore ${ℱ}^{\prime }$ is a subalgebra of $B(H)$.

It remains to see that ${ℱ}^{\prime }$ is weak operator closed. Suppose $(T_i)$ is a net in ${ℱ}^{\prime }$ that converges to $T$ in the weak operator topology. Then, for all $x,y\in H$ we have that $⟨T_{i}x,y⟩\to ⟨Tx,y⟩$. Thus, for all $S\in ℱ$, we have

$⟨(TS-ST)x,y⟩$ $=$ $⟨TSx,y⟩-⟨Tx,S^{*}y⟩$

$=$ $lim\left(⟨T_{i}Sx,y⟩-⟨T_{i}x,S^{*}y⟩\right)$

$=$ $lim⟨(T_{i}S-S{T}_i)x,y⟩$

$=$ $lim⟨(T_{i}S-T_{i}S)x,y⟩$

$=$ $0$

Hence, $TS-ST=0$, so that $T\in {ℱ}^{\prime }$. We conclude that ${ℱ}^{\prime }$ is closed in the weak operator topology. $\mathrm{\square }$

Title	commutant is a weak operator closed subalgebra
Canonical name	CommutantIsAWeakOperatorClosedSubalgebra
Date of creation	2013-03-22 18:39:32
Last modified on	2013-03-22 18:39:32
Owner	asteroid (17536)
Last modified by	asteroid (17536)
Numerical id	7
Author	asteroid (17536)
Entry type	Theorem
Classification	msc 46L10