proof of von Neumann double commutant theorem
Lemma - Let be a Hilbert space and its algebra of bounded operators. Let be a *-subalgebra of that contains the identity operator and is closed in the strong operator topology. If , the double commutant of , then for each there is an operator such that .
: Let be the closure of the subspace . It is clear that is an invariant subspace for , hence so is its closure (see this entry (http://planetmath.org/InvariantSubspacesForSelfAdjointAlgebrasOfOperators), Proposition 5).
Let be the orthogonal projection onto . Since is invariant for , we have that (see this entry (http://planetmath.org/InvariantSubspacesForSelfAdjointAlgebrasOfOperators), last theorem). Since contains the identity operator, we know that belongs to . Hence,
where the last equality comes from the fact that and . Thus, we see that , which implies that there exists an such that .
Since is the commutant of some set, namely it is the commutant of , it follows that is closed in the weak operator topology (see this entry (http://planetmath.org/CommutantIsAWeakOperatorClosedSubalgebra)). But we are assuming that , hence is closed in the weak operator topology.
This part is obvious since the weak operator topology is weaker than the strong operator topology.
Suppose is closed in the strong operator topology.
We can however consider to be , since .
For every we want to find such that , i.e. such that , for each .
We have that is a *-algebra of bounded operators in .
Claim 1 - .
The algebra can be canonically identified with the algebra of matrices with entries in , and corresponds to the diagonal matrices with an element in the diagonal. Thus, it is easy to check that is precisely the set of matrices whose entries belong to .
Since the unit matrices (http://planetmath.org/UnitMatrix) belong to , it follows that consists solely of diagonal matrices with one element on the diagonal (see this entry (http://planetmath.org/CentralizerOfMatrixUnits)). It is easy to check that is precisely the set of diagonal matrices with one element of in the diagonal. Hence, we conclude that , and Claim 1 is proved.
Now, we observe that is a *-subalgebra of that contains the identity operator. Since is closed in the strong operator topology, it follows easily that is also closed in the strong operator topology. Since , Lemma 1 that for each there exists an operator such that . But this is implies that for each .
Thus, . Hence we conclude that belongs to the operator closure of , but since is closed under this topology, .
We conclude that .
|Title||proof of von Neumann double commutant theorem|
|Date of creation||2013-03-22 18:40:29|
|Last modified on||2013-03-22 18:40:29|
|Last modified by||asteroid (17536)|