Commutative Subsets of an Algebra

# Commutative Subsets of an Algebra

 Definition: Let $\mathfrak{A}$ be an algebra. A subset $E \subseteq \mathfrak{A}$ is said to be a Commutative Subset of $\mathfrak{A}$ if $xy = yx$ for all $x, y \in E$.
 Definition: Let $\mathfrak{A}$ be an algebra and let $E \subseteq \mathfrak{A}$. The Commutant of $E$ is the set $E^c = \{ a \in \mathfrak{A} : ea = ae \: \forall e \in E \}$. The Second Commutant of $E$ is the set $E^{cc} := (E^c)^c = \{ a \in \mathfrak{A} : e'a = ae' \: \forall e' \in E \}$. The Centre of $\mathfrak{A}$ is the commutant of $\mathfrak{A}$, i.e., $\mathfrak{A}^c$.
 Proposition 1: Let $\mathfrak{A}$ be an algebra and let $E \subseteq \mathfrak{A}$. Then $E^c$ is a subalgebra of $\mathfrak{A}$. Moreover, if $\mathfrak{A}$ is an algebra with unit $1$ then $E^c$ is a subalgebra with unit $1$. _
• Proof: Let $x, y \in E^c$. Then $ex = xe$ and $ey = ye$ for all $e \in E$. So $(x + y)e = xe + ye = ex + ey = e(x + y)$ for all $e \in E$, so $x + y \in E^c$ whenever $x, y \in E^c$, showing that $E^c$ is closed under addition.
• Let $\alpha \in \mathbf{F}$ and let $x \in E^c$. Then $ex = xe$ for all $e \in E$. So $e(\alpha x) = \alpha(ex) = \alpha(xe) = (\alpha x)e$ for all $e \in E$. Therefore $\alpha x \in E^c$ whenever $\alpha \in \mathbf{F}$ and $x \in E^c$, showing that $E^c$ is closed under scalar multiplication.
• Let $x, y \in E^c$. Then $ex = xe$ and $ey = ye$ for all $e \in E$ again. So $e(xy) = (ex)y = (xe)y = x(ey) = x(ye) = (xy)e$ for all $e \in E$. Therefore $xy \in E^c$ whenever $x, y \in E^c$, showing that $E^c$ is closed under multiplication.
• So $E^c$ is a subalgebra of $\mathfrak{A}$.
• If $\mathfrak{A}$ is an algebra with unit $1$ then by definition, $a1 = a = 1a$ for all $a \in \mathfrak{A}$. In particular, $e1 = e = 1e$ for all $e \in E$ and so $1 \in E^c$. So if $\mathfrak{A}$ is an algebra with unit $1$ then $E^c$ is a subalgebra with unit $1$. $\blacksquare$
 Proposition 2: Let $\mathfrak{A}$ be an algebra and let $E \subseteq \mathfrak{A}$. Then $E$ is a commutative subset of $\mathfrak{A}$ if and only if $E \subseteq E^c$.
• Proof: $\Rightarrow$ Suppose that $E$ is a commutative subset of $\mathfrak{A}$. Let $x \in E$. Then $xy = yx$ for all $y \in E$. So $x \in E^c$ showing that $E \subseteq E^c$.
• $\Leftarrow$ Conversely, suppose that $E \subseteq E^c$. Let $x, y \in E$. Then $x, y \in E^c$. Since $x \in E^c$ and $y \in E$ we have that $xy = yx$. Since this holds true for all $x, y \in E$ we see that $E$ is a commutative subset of $\mathfrak{A}$.
 Proposition 3: Let $\mathfrak{A}$ be an algebra and let $E, F \subseteq \mathfrak{A}$. If $E \subseteq F$ then $F^c \subseteq E^c$.
• Proof: Suppose that $E \subseteq F$. Let $x \in F^c$. Then $xf = fx$ for all $f \in E$. Since $E \subseteq F$ this tells us that $xe = ex$ for all $e \in E$. So $x \in E^c$. Thus $F^c \subseteq E^c$. $\blacksquare$
 Proposition 4: Let $\mathfrak{A}$ be an algebra and let $E \subseteq \mathfrak{A}$. If $E$ is a commutative subset of $\mathfrak{A}$ then $E^{cc}$ is a commutative subalgebra of $\mathfrak{A}$ and $E \subseteq E^{cc} \subseteq E^c$;
• Proof: By Proposition 1, $E^c$ is a subalgebra of $\mathfrak{A}$. Applying this proposition again and we see that $E^{cc}$ is a subalgebra of $E^c$, and so $E^{cc}$ is a subalgebra of $\mathfrak{A}$.
• Now since $E$ is a commutative subset of $\mathfrak{A}$ we have by Proposition 2 that $E \subseteq E^c$. So by Proposition 3 we have that $E^{cc} \subseteq E^c$.
• Now let $x \in E$. Let $y \in E^c$. Then $ey = ye$ for all $e \in E$. In particular, $xy = yx$ for all $y \in E^c$. Thus $x \in E^{cc}$ and hence $E \subseteq E^{cc}$. Combining the above two set inclusions gives us that $E \subseteq E^{cc} \subseteq E$.
• Lastly, let $x, y \in E^{cc}$. Then from above since $E^{cc} \subseteq E^c$ we have that $x, y \in E^c$. So for $x \in E^{cc}$ and $y \in E^c$ we have by definition of $x \in E^{cc}$ that $xy = yx$. So $E^{cc}$ is a commutative subalgebra of $\mathfrak{A}$. $\blacksquare$
 Proposition 5: Let $\mathfrak{A}$ be an algebra. Then the centre of $\mathfrak{A}$ is a commutative subalgebra of $X$.
• Proof: By Proposition 1, $\mathfrak{A}^c$ is a subalgebra of $\mathfrak{A}$. Let $x, y \in \mathfrak{A}^c$. Since $x \in \mathfrak{A}^c$ by definition we have that $xz = zx$ for all $z \in \mathfrak{A}$. In particular, $xy = yx$ since $y \in \mathfrak{A}$. So $xy = yx$ for all $x, y \in \mathfrak{A}^c$ and thus $\mathfrak{A}^c$ is a commutative subset of $\mathfrak{A}$. By Proposition 4, $\mathfrak{A}^c$ is a commutative subalgebra of $\mathfrak{A}$. $\blacksquare$