# 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 $$.

*:* 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,

$Tx=TPx=PTx$ |

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 $$. $\mathrm{\square}$

$$

$(1)\u27f9(2)$ 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.

$(2)\u27f9(3)$ This part is obvious since the weak operator topology is weaker than the strong operator topology.

$(3)\u27f9(1)$ 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^{} $\parallel \cdot {\parallel}_{x},x\in H$ defined by ${\parallel S\parallel}_{x}:=\parallel Sx\parallel $. A local base around $T$, in this topology, consists of sets of the form

$V({x}_{1},\mathrm{\dots},{x}_{n};\u03f5):=\{S\in B(H):\parallel (S-T){x}_{i}\parallel \le \u03f5,i=1,\mathrm{\dots},n\},{x}_{1},\mathrm{\dots},{x}_{n}\in H,\u03f5>0$ |

We can however consider $\u03f5$ to be $1$, since $V({x}_{1},\mathrm{\dots},{x}_{n};\u03f5)=V({\u03f5}^{-1}{x}_{1},\mathrm{\dots},{\u03f5}^{-1}{x}_{n};1)$.

For every ${x}_{1},\mathrm{\dots},{x}_{n}\in H$ we want to find $A\in \mathcal{M}$ such that $A\in V({x}_{1},\mathrm{\dots},{x}_{n};\mathrm{\hspace{0.17em}1})$, i.e. such that $$, for each $i$.

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

$\stackrel{~}{A}({y}_{1},\mathrm{\dots},{y}_{n})=(A{y}_{1},\mathrm{\dots},A{y}_{n}),{y}_{1},\mathrm{\dots},{y}_{n}\in H$ |

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

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

The algebra $B(\stackrel{~}{H})$ 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 $\stackrel{~}{T}\in {\mathcal{N}}^{\prime \prime}$, and Claim 1 is proved.

Now, we observe that $\mathcal{N}$ is a *-subalgebra of $B(\stackrel{~}{H})$ 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 $\stackrel{~}{T}\in {\mathcal{N}}^{\prime \prime}$, Lemma 1 that for each $\stackrel{~}{x}:=({x}_{1},\mathrm{\dots},{x}_{n})\in \stackrel{~}{H}$ there exists an operator $\stackrel{~}{A}\in \mathcal{N}$ such that $$. But this is implies that $$ for each $1\le i\le n$.

Thus, $T\in V({x}_{1},\mathrm{\dots},{x}_{n};\mathrm{\hspace{0.17em}1})$. 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 ${\mathcal{M}}^{\prime \prime}=\mathcal{M}$. $\mathrm{\square}$

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 |