Maximal Commutative Subsets of an Algebra

# Maximal Commutative Subsets of an Algebra

Recall from the Commutative Subsets of an Algebra page that if $X$ is an algebra and $E \subseteq X$ then $E$ is said to be a commutative subset of $X$ if $xy = yx$ for all $x, y \in E$. The commutant of $E$ is defined to be the set:

(1)\begin{align} \quad E^c = \{ x \in X : ex = xe, \: \forall e \in E \} \end{align}

And the second commutant of $E$ is defined to be $E^{cc} = (E^c)^c$. Lastly we defined the centre of $X$ to be $X^c$.

Definition: Let $X$ be an algebra. A subset $E \subseteq X$ is said to be a Maximal Commutative Subset of $X$ if it is a commutative subset of $X$ and it is not contained in any other different commutative subset of $X$. |

Proposition 1: Let $X$ be an algebra and let $E \subseteq X$. If $E$ is a commutative subset of $X$ then $E$ is contained in some maximal commutative subset of $M$ of $X$. Furthermore, $M$ has the following properties:a) $M$ is a commutative subalgebra of $X$.b) If $X$ is an algebra with unit $1$ then $M$ is a commutative subalgebra of $X$ with unit $1$.c) If $X$ is a normed algebra then $M$ is a closed. |

**Proof:**Let $\mathcal F$ be the collection of all commutative subsets of $X$ containing $E$ as a subset. Then $\mathcal F$ is partially ordered by inclusion. Certainly $\mathcal F \neq \emptyset$ since $E \in \mathcal F$.

- If $\mathcal C \subset \mathcal F$ is a chain, i.e., for every $A, B \in \mathcal C$ we have that either $A \subseteq B$ or $B \subseteq A$ then let $M^* = \bigcup_{A \in \mathcal C} A$. Then $A \subseteq M^*$ for all $A \in \mathcal F$, and so $M^*$ is an upper bound for the chain $\mathcal C$.

- Let $x, y \in M^*$. Then there exists an $A \in \mathcal C$ such that $x, y \in A$ since $\mathcal C$ is a chain. Since $A$ is a commutative subset of $X$ we have that $xy = yx$. Since $xy = yx$ for all $x, y \in M$ we have that $M^*$ is a commutative subset of $X$.

- So every chain in $\mathcal F$ has an upper bound in $\mathcal F$. By Zorn's Lemma, $\mathcal F$ has a maximal commutative subset, call it $M$. Thus every commutative subset is contained in a maximal commutative subset. $\blacksquare$

**Proof of a)**Now since $M$ is a commutative subset we have by one of the propositions on the Commutative Subsets of an Algebra page that $M^{cc}$ is a commutative subalgebra of $X$ and that $M \subseteq M^{cc}$. By maximality of $M$ this implies that $M = M^{cc}$, and so $M$ is a commutative subalgebra of $X$. $\blacksquare$

**Proof of b)**If $X$ is an algebra with unit then $1x = x1$ for all $x \in X$. In particular, $\{ 1 \}$ is a commutative subset of $X$ and is thus contained in $M$, i.e., $1 \in M$. So $M$ is an algebra with unit. $\blacksquare$

**Proof of c)**Let $x, y \in \bar{M}$. Then there exists nets $\{ x(\lambda) \}_{\lambda \in \Lambda}$ and $\{ y(\mu) \}_{\mu \in M}$ such that $x(\lambda) \to x$ and $y(\mu) \to y$. For each $\lambda \in \Lambda$ and $\mu \in M$ we have that $x(\lambda) y(\mu) = y(\mu) x(\lambda)$ since $M$ is a commutative subset of $X$. Thus $\| x(\lambda) y(\mu) - y(\mu) x(\lambda) \| = 0$ for all $\lambda \in \Lambda$ and for all $\mu \in M$. So $xy = yx$. Hence $\bar{M}$ is a commutative subset of $X$. By maximality of $M$ we have that $\bar{M} \subseteq M$ and hence $M = \bar{M}$, i.e., $M$ is closed. $\blacksquare$