agreseza

2022-01-04

If A is a commutative C*-subalgebra of linear bounded operator space B(H) on some Hilbert space H, so is the double commutant A. It follows from A is dense in A and the multiplication is continuous on each factor respectively, respect to the strong operator topology.
1) Can the assertion A is commutative be verified in an algebraic way?
2) Or, a more general question: If A is a commutative subalgebra of an algebra B, is the double commutant A of A also commutative?

Elois Puryear

Expert

If A is commutative, you have
${A}^{\prime }\subset A{}^{″}$
Then
${A}^{\prime }\supset A{}^{″}$
And then
$A{}^{″}\subset A{}^{‴}$
So A'' is contained in its commutant and is thus commutative.
The above reasoning shows that all even higher commutants of A are commutative.

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