The Commutant (Centralizer) of a Set in an Algebra

The Commutant (Centralizer) of a Set in an Algebra

Definition: Let $X$ be an algebra with unit and let $E \subseteq X$. The Commutant (or Centralizer) of $E$ is the set $E^c = \{ x \in X : x \cdot e = e \cdot x, \: \forall e \in E \}$. 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 $X$ that commute with every point in $E$.

Proposition 1: Let $X$ be an algebra with unit and let $E \subseteq X$. Then the commutant $E^c$ is a subalgebra of $X$.
  • Proof: To show that $E^c$ is a subalgebra of $X$ 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 $X$. $\blacksquare$
Proposition 2: Let $X$ be an algebra and let $x \in X$. Then $\{ x \}^{cc}$ is a subalgebra of $X$ containing $x$. Moreover, $\{ x \}^{cc}$ is a commutative subalgebra - that is, vector multiplication is commutative in $\{ x \}^{cc}$.
  • Proof: By proposition $1$ we have that $\{ x \}^c$ is a subalgebra of $X$. Applying proposition 1 again shows that $\{ x \}^{cc}$ is a subalgebra of $\{ x \}^c$ and is thus a subalgebra of $X$.
  • Let $z \in \{ x \}^{cc}$. Then $zy = yz$ for every $y \in \{ x \}^c$. But for every $y \in \{ x \}^c$ we have that $wy = yw$ for every $w \in X$. In particular, $x \in X$ and so $xy = yx$ for every $y \in \{ x \}^c$, and so $zx = xz$. Thus $x \in \{ x \}^{cc}$. $\blacksquare$
Proposition 3: Let $X$ be an algebra with unit and let $x \in \mathrm{Inv}(X)$. Then $x^{-1} \in \{ x \}^{cc}$.
  • Proof: Let $y \in \{ x \}^c$. Then $xy = yx$. Thus:
(4)
\begin{align} \quad yx^{-1} = 1yx^{-1} = x^{-1}xyx^{-1} = x^{-1}(xy)x^{-1} = x^{-1}(yx)x^{-1} = x^{-1}y \end{align}
  • Since this holds for every $y \in \{ x \}^c$ we see that $x^{-1} \in \{ x \}^{cc}$. $\blacksquare$
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