# Commutativity in a Unital Banach Algebra

Let $A$ be a unital Banach algebra and $S$ a non-empty subset of $A$. The centralizer of $S$ is defined as $$Z(S) \stackrel{\text{def}}{=} \{ a \in A ~|~ \text{ as = sa  for all  s \in S } \}.$$ I need to show that if the elements of $S$ commute with each other, then the elements of $Z(Z(S))$ commute with each other.

I know that $Z(Z(S))$ is a closed sub-algebra of $A$ and that ${\sigma_{A}}(a) = {\sigma_{Z(S)}}(a)$ for any $a \in Z(S)$. Also, $S \subseteq Z(Z(S))$. But I do not see how the commutativity part follows.

Can anybody please give me a hint?

• Could you kindly explain what is meant by ‘$S$ is commuting’? Do you mean ‘$Z(S) = A$’? Commented Feb 11, 2013 at 2:54
• @HaskellCurry I think S commuting means $s_1 s_2=s_2 s_1$ for all $s_1,s_2\in S$ Commented Feb 11, 2013 at 3:01
• Yes, @HaskellCurry it means exactly as i.a.m just said.
– user44349
Commented Feb 11, 2013 at 3:04
• @CB_Student: Thanks for the clarification! I hope you won’t mind if I incorporate this with an edit. Commented Feb 11, 2013 at 3:07
• Not at all ! Thanks for the edit !
– user44349
Commented Feb 11, 2013 at 3:09

Fact: If $S$ and $T$ are subsets of $A$, then $S \subseteq T$ implies $Z(T) \subseteq Z(S)$.

If the elements of $S$ commute with each other, then $S \subseteq Z(S)$. Applying the fact above, we obtain $Z(Z(S)) \subseteq Z(S)$, which then implies that $Z(Z(S)) \subseteq Z(Z(Z(S)))$. Therefore, the elements of $Z(Z(S))$ commute with each other.

• Haskell, Thank you very much. you are brilliant :) Commented Feb 11, 2013 at 3:21
• @HaskellCurry Okay, I get it now. Thank you so much !
– user44349
Commented Feb 11, 2013 at 3:24
• @CB_Student: You’re very much welcome! Commented Feb 11, 2013 at 3:51

This has nothing to do with banach algebras. It holds in every magma, i.e. set equipped with a binary operation. Also, $S \neq \emptyset$ is a useless and unnatural condition, it should be left out.

Observe that $S \subseteq T$ implies $Z(T) \subseteq Z(S)$ $(\star)$. Since $S$ is commutative, we have $S \subseteq Z(S)$. Applying $(\star)$ this we get $Z(Z(S)) \subseteq Z(S)$. Applying $(\star)$ once again, we get $Z(Z(S)) \subseteq Z(Z(Z(S)))$. But this exactly means that $Z(Z(S))$ is commutative.

Edit: Oh, Haskell and I posted the same proof, simultaneously!

• Gosh... We actually did! Commented Feb 11, 2013 at 3:20