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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.

But note that "double commutant" and "commutative" are both algebraic terms, I want to ask:

  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?

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If $A$ is commutative, you have $$A\subset A'.$$ Then $$ A'\supset A''.$$ 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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  • $\begingroup$ You mean all the even order? $\endgroup$
    – C. Ding
    Aug 1 '18 at 23:56
  • $\begingroup$ Yes, of course. Thanks. $\endgroup$ Aug 2 '18 at 1:36

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