Bicommutant of self-adjoint subset of an involutive algebra

Question

1.1.9. Let $A$ be an involutive algebra and $M$ a self-adjoint subset of $A$... If the elements of $M$ commute pairwise, then $M\subset M'$, so that $M'\supset M''$ and $M''$ is commutative...

^{Source: Dixmier's}

Here $M'$ and $M''$ are the commutant and bicommutant of $M$. $M\subset M'$ is pretty straightforward, because each element in $M$ commutes with $M$, so each of the elements in $M$ must belong to the commutant $M'$. But how to comprehend $M'\supset M''$? And why would $M''$ be commutative?

score 1 · Accepted Answer · 2018-09-17T15:43:34.747

1

In general, if $A \subset B$, then $B' \subset A'$. Indeed, if $x \in B'$ then $xb = bx$ for every $b \in B$. Since $A \subset B$, we have in particular, that $xa = ax$ for every $a \in A$. Hence $x \in A'$.

Now, $M \subset M'$, hence $M'' \subset M'$ and finally $M'' \subset M'''$. So $M''$ is contained in its commutant, and hence commutative.

edited Sep 17 '18 at 15:43

answered Sep 17 '18 at 08:01

I understand now why $B'\subset A'$, but where along the way was $M=M''$ proved? – Frenzy Li Sep 17 '18 at 14:28
I was too quick. The answer is now corrected. – Sep 17 '18 at 15:43
Wow, using the tricommutant (?) here is quite neat. Thanks! – Frenzy Li Sep 17 '18 at 17:28

Bicommutant of self-adjoint subset of an involutive algebra

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