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:

\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:

\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:

\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 $x \in \mathrm{Inv}(X)$. Then there exists $x^{-1} \in X$ such that $x \cdot x^{-1} = 1 = x^{-1}x$.

- To show that $x^{-1} \in \{ x \}^{cc}$ we must show that for every $y \in \{ x \}^{c}$ we have that $x^{-1}y = yx^{-1}$.

- Let $y \in \{ x \}^c$. Then $xy = yx$. Thus:

\begin{align} \quad yx^{-1} = 1(yx^{-1)} = x^{-1}(xy)x^{-1} = x^{-1}(yx)x^{-1}= x^{-1}y(xx^{-1}) = x^{-1}y1 = x^{-1}y \end{align}

- Since this holds for every $y \in \{ x \}^c$ we see that $x^{-1} \in \{ x \}^{cc}$. $\blacksquare$