proof of von Neumann double commutant theorem

proof of von Neumann double commutant theorem

Lemma -

Let

H

be a

Hilbert space

and

B

⁢

(

H

)

its algebra of

bounded operators

. Let

𝒩

be a *-

subalgebra

of

B

⁢

(

H

)

that contains the

identity operator

and is closed in the

strong operator topology

. If

T

∈

𝒩

′′

, the

double commutant

of

𝒩

, then for each

x

∈

H

there is an

operator

A

∈

𝒩

such that

∥

(

A

-

T

)

⁢

x

∥

<

1

.

:

Let

𝒩

⁢

x

¯

⊆

H

be the

closure

of the

subspace

𝒩

⁢

x

:=

{

S

⁢

x

:

S

∈

𝒩

}

. It is clear that

𝒩

⁢

x

is an

invariant subspace

for

𝒩

, hence so is its closure

𝒩

⁢

x

¯

(see this entry (

http://planetmath.org/InvariantSubspacesForSelfAdjointAlgebrasOfOperators

),

Proposition

5).

Let

P

be the

orthogonal projection

onto

𝒩

⁢

x

¯

. Since

𝒩

⁢

x

¯

is

invariant

for

𝒩

, we have that

P

∈

𝒩

′

(see this entry (

http://planetmath.org/InvariantSubspacesForSelfAdjointAlgebrasOfOperators

), last

theorem

). Since

𝒩

contains the identity operator, we know that

x

belongs to

𝒩

⁢

x

¯

. Hence,

T

⁢

x

=

T

⁢

P

⁢

x

=

P

⁢

T

⁢

x

where the last

equality

comes from the fact that

T

∈

𝒩

′′

and

P

∈

𝒩

′

. Thus, we see that

T

⁢

x

∈

𝒩

⁢

x

¯

, which implies that there exists an

A

∈

𝒩

such that

∥

T

⁢

x

-

A

⁢

x

∥

<

1

.

□

(

1

)

⟹

(

2

)

Since

ℳ

′′

is the commutant of some set, namely it is the commutant of

ℳ

′

, it follows that

ℳ

′′

is closed in the weak operator topology (see this entry (

http://planetmath.org/CommutantIsAWeakOperatorClosedSubalgebra

)). But we are assuming that

ℳ

=

ℳ

′′

, hence

ℳ

is closed in the weak operator topology.

(

2

)

⟹

(

3

)

This part is obvious since the weak operator topology is weaker than the strong operator topology.

(

3

)

⟹

(

1

)

Suppose

ℳ

is closed in the strong operator topology.

A subset of

B

⁢

(

H

)

is always

contained

in its double commutant, thus

ℳ

⊆

ℳ

′′

. So it remains to prove the opposite

inclusion

.

Let

T

∈

ℳ

′′

. We are going to prove that

T

belongs to the strong operator closure of

ℳ

, and since

ℳ

is

closed under

this

topology

, it will follow that

T

∈

ℳ

.

Recall that the strong operator topology is the topology in

B

⁢

(

H

)

generated by the family of

seminorms

∥

⋅

∥

x

,

x

∈

H

defined by

∥

S

∥

x

:=

∥

S

⁢

x

∥

. A

local base

around

T

, in this topology, consists of sets of the form

V

⁢

(

x

1

,

…

,

x

n

;

ϵ

)

:=

{

S

∈

B

⁢

(

H

)

:

∥

(

S

-

T

)

⁢

x

i

∥

≤

ϵ

,

i

=

1

,

…

,

n

}

,

x

1

,

…

,

x

n

∈

H

,

ϵ

>

0

We can however consider

ϵ

to be

1

, since

V

⁢

(

x

1

,

…

,

x

n

;

ϵ

)

=

V

⁢

(

ϵ

-

1

⁢

x

1

,

…

,

ϵ

-

1

⁢

x

n

;

1

)

.

For every

x

1

,

…

,

x

n

∈

H

we want to find

A

∈

ℳ

such that

A

∈

V

⁢

(

x

1

,

…

,

x

n

;

1

)

, i.e. such that

∥

(

A

-

T

)

⁢

x

i

∥

<

1

, for each

i

.

Let

H

~

be the

direct sum of Hilbert spaces

H

~

:=

⊕

i

=

1

n

H

. For every

A

∈

B

⁢

(

H

)

let

A

~

∈

B

⁢

(

H

~

)

be the

direct sum

of bounded operators (

http://planetmath.org/DirectSumOfBoundedOperatorsOnHilbertSpaces

)

A

~

:=

⊕

i

=

1

n

A

, i.e.

A

~

⁢

(

y

1

,

…

,

y

n

)

=

(

A

⁢

y

1

,

…

,

A

⁢

y

n

)

,

y

1

,

…

,

y

n

∈

H

We have that

𝒩

:=

{

A

~

:

A

∈

ℳ

}

is a *-algebra of bounded operators in

H

~

.

Claim 1 -

T

~

∈

𝒩

′′

.

The algebra

B

⁢

(

H

~

)

can be canonically identified with the algebra of

n

×

n

matrices with entries in

B

⁢

(

H

)

, and

𝒩

corresponds to the

diagonal matrices

with an

element

A

∈

ℳ

in the

diagonal

. Thus, it is easy to check that

𝒩

′

is precisely the set of matrices whose entries belong to

ℳ

′

.

Since the

unit matrices

(

http://planetmath.org/UnitMatrix

) belong to

𝒩

′

, it follows that

𝒩

′′

consists solely of diagonal matrices with one element on the diagonal (see this entry (

http://planetmath.org/CentralizerOfMatrixUnits

)). It is easy to check that

𝒩

′′

is precisely the set of diagonal matrices with one element of

ℳ

′′

in the diagonal. Hence, we conclude that

T

~

∈

𝒩

′′

, and Claim 1 is proved.

Now, we observe that

𝒩

is a *-subalgebra of

B

⁢

(

H

~

)

that contains the identity operator. Since

ℳ

is closed in the strong operator topology, it follows easily that

𝒩

is also closed in the strong operator topology. Since

T

~

∈

𝒩

′′

, Lemma 1 that for each

x

~

:=

(

x

1

,

…

,

x

n

)

∈

H

~

there exists an operator

A

~

∈

𝒩

such that

∥

(

A

~

-

T

~

)

⁢

x

~

∥

<

1

. But this is implies that

∥

(

A

-

T

)

⁢

x

i

∥

<

1

for each

1

≤

i

≤

n

.

Thus,

T

∈

V

⁢

(

x

1

,

…

,

x

n

;

1

)

. Hence we conclude that

T

belongs to the operator closure of

ℳ

, but since

ℳ

is closed under this topology,

T

∈

ℳ

.

We conclude that

ℳ

′′

=

ℳ

.

□

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

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