The Commutant (Centralizer) of a Set in an Algebra

# The Commutant (Centralizer) of a Set in an Algebra

 Definition: Let $\mathfrak{A}$ be an algebra with unit and let $E \subseteq \mathfrak{A}$. The Commutant (or Centralizer) of $E$ is the set $E^c = \{ a \in \mathfrak{A} : a \cdot e = e \cdot a, \: \forall a \in \mathfrak{A} \}$. The Second Commutant (or Bicommutant) of $E$ is the commutant of $E^c$ and is denoted $E^{cc}$.

In other words, the commutant of $E$ is the set of all points in $\mathfrak{A}$ that commute with every point in $E$.

 Proposition 1: Let $\mathfrak{A}$ be an algebra with unit and let $E \subseteq \mathfrak{A}$. Then the commutant $E^c$ is a subalgebra of $\mathfrak{A}$.
• Proof: To show that $E^c$ is a subalgebra of $\mathfrak{A}$ we only need to show that $E^c$ is closed under addition, closed under scalar multiplication, and closed under vector multiplication.
• Let $x, y \in E^c$. Then $x \cdot e = e \cdot x$ and $y \cdot e = e \cdot y$ for all $e \in E$. Therefore:
(1)
\begin{align} \quad (x + y) \cdot e = x\cdot e + y \cdot e = e \cdot x + e \cdot y = e \cdot (x + y), \quad \forall e \in E \end{align}
• So $(x + y) \in E^c$, showing that $E^c$ is closed under addition.
• Let $x \in E^c$ and let $\alpha \in \mathbf{F}$. Then $x \cdot e = e \cdot x$ for all $e \in E$ and so by the properties of vector multiplication we have that:
(2)
\begin{align} \quad (\alpha x) \cdot e= \alpha (x \cdot e) = \alpha (e \cdot x) = e \cdot (\alpha x) , \quad \forall e \in E \end{align}
• So $\alpha x \in E^c$, showing that $E^c$ is closed under scalar multiplication.
• Let $x, y \in E^c$. Then $x \cdot e = e \cdot x$ and $y \cdot e = e \cdot y$ for all $e \in E$. Therefore by associativity of vector multiplication we have that:
(3)
\begin{align} \quad (x \cdot y) \cdot e = x \cdot (y \cdot e) = x \cdot (e \cdot y) = (x \cdot e) \cdot y = (e \cdot x) \cdot y = e \cdot (x \cdot y), \quad \forall e \in E \end{align}
• So $x \cdot y \in E^c$, showing that $E^c$ is closed under vector multiplication.
• Thus, $E^c$ is a subalgebra of $\mathfrak{A}$. $\blacksquare$
 Proposition 2: Let $\mathfrak{A}$ be an algebra and let $a \in \mathfrak{A}$. Then $\{ a \}^{cc}$ is a subalgebra of $\mathfrak{A}$ containing $a$. Moreover, $\{ a \}^{cc}$ is a commutative subalgebra - that is, vector multiplication is commutative in $\{ a \}^{cc}$.
• Proof: By proposition $1$ we have that $\{ a \}^c$ is a subalgebra of $\mathfrak{A}$. Applying Proposition 1 again shows that $\{ a \}^{cc}$ is a subalgebra of $\{ a \}^c$ and is thus a subalgebra of $\mathfrak{A}$.
• Let $z \in \{ a \}^{cc}$. Then $zy = yz$ for every $y \in \{ a \}^c$. But for every $y \in \{ a \}^c$ we have that $wy = yw$ for every $w \in X$. In particular, $a \in \mathfrak{A}$ and so $ay = ya$ for every $y \in \{ a \}^c$, and so $za = az$. Thus $a \in \{ a \}^{cc}$. $\blacksquare$
 Proposition 3: Let $\mathfrak{A}$ be an algebra with unit and let $a \in \mathrm{Inv}(\mathfrak{A})$. Then $a^{-1} \in \{ a \}^{cc}$.
• Proof: Let $y \in \{ a \}^c$. Then $ay = ya$. Thus:
(4)
\begin{align} \quad ya^{-1} = 1ya^{-1} = a^{-1}aya^{-1} = a^{-1}(ay)a^{-1} = a^{-1}(ya)a^{-1} = a^{-1}y \end{align}
• Since this holds for every $y \in \{ a \}^c$ we see that $a^{-1} \in \{ a \}^{cc}$. $\blacksquare$