Commutant(Algebra)

Let $R$ be a ring with unity 1, and $E$ be a $R$-module.

Definition

The commutant $R^{\prime }$ of $R$ is the ring $R^{\prime }=R^{\prime }(E)=En{d}_R(E)$.

And we can iteratate as double commutant(bicommutant) $R^{\prime \prime }=R^{\prime }(R^{\prime }(E))=En{d}_{R^{\prime }}(E)$.

From the definition, we can consider $E$ as a $R^{\prime }$-mod as $\phi \cdot x=\phi (x)$ for $\phi \in R^{\prime }$ and $x\in E$. For each $\alpha \in R$, there is a $R$-mod homomorphism $f_{\alpha }:E\to E$ such that $e\mapsto \alpha e$. Then for $\phi \in R^{\prime }$, $\phi \circ f_{\alpha }=f_{\alpha }\circ \phi (e)$ so that $R^{\prime }$ and $R$ are commutes each other.¹

Similarly, for $f\in R^{\prime \prime }$ and $\phi \in R^{\prime }$, $f\circ \phi =\phi \circ f$.

Example

• Let $R=\mathbb{R}$, $E={\mathbb{R}}^n$. Then $R^{\prime }=En{d}_R(E)=Ma{t}_n(\mathbb{R})$. Also, $R^{\prime \prime }=En{d}_{Ma{t}_n(\mathbb{R})}({\mathbb{R}}^n)=R\cdot I{\cong }_{Ring}R$.

• $V$ be a vector space, $V^{\ast }=Hom(V,\mathbb{R})$. $V^{\ast \ast }{\cong }_{\text{Nat’l Vect Isom}}\mathbb{R}$ where $ev:V\to V^{\ast \ast }$ such that $v\mapsto \left(\phi \mapsto \phi (v)\right)$. $ev$ is 1-1 to $v\in V$ and onto if $dimV<\infty$.

• Evaluation map $ev:R\to R^{\prime \prime }$ such that $r\mapsto (e\mapsto re)$. Define $f_r:E\to E$ such that $f_r(e)=re=ev(r)(e)$.

Ring homomorphism

Claim

1. $f_r\in R^{\prime \prime }$.
2. $ev$ in Example is a ring homomorphism.

Proof

1. $f_r(\phi (e))=r\phi (e)=\phi (re)=\phi (f_r(e))$.

2. $ev(r+r^{\prime })=ev(r)+ev(r^{\prime })$, $ev(1)=1$, and $ev(r)\cdot ev(r^{\prime })(e)=ev(r)(r^{\prime }e)=r{r}^{\prime }e=ev(r{r}^{\prime })(e)$.

Footnotes

1. I guess this is the reason that called ‘commut’ant. ↩