Regular Representation

Arun Ram
Department of Mathematics and Statistics
University of Melbourne
Parkville, VIC 3010 Australia
aram@unimelb.edu.au

and

Department of Mathematics
University of Wisconsin, Madison
Madison, WI 53706 USA
ram@math.wisc.edu

Last updates: 22 January 2010

Regular representation

If $A$ is an algebra then $A^{\mathrm{op}}$ is the algebra $A$ except with the opposite multiplication, i.e. $$A^{\mathrm{op}}=\left\{a^{\mathrm{op}}|a\in A\right\}\text{with}{a}_1^{\mathrm{op}}{a}_2^{\mathrm{op}}={\left(a_1{a}_2\right)}^{op}\text{,}\text{for all}{a}_1,a_2\in A\text{.}$$

The left regular representation of $A$ is the vector space $A$ with $A$ action given by left multiplication. Here $A$ is serving as both an algebra and an $A$-module. It is often useful to distinguish the two roles of $A$ and use the notation $\overrightarrow{A}$ for the $A$-module, i.e. $\overrightarrow{A}$ is the vector space $$\overrightarrow{A}=\left\{\overrightarrow{b}|b\in A\right\}\text{with}A\text{-action}a\overrightarrow{b}=\overrightarrow{ab}\text{,}\text{for all}a\in A\text{,}\overrightarrow{b}\in \overrightarrow{A}\text{.}$$

Let $A$ be an algebra and let $\overrightarrow{A}$ be the regular representation of $A$. Then ${\mathrm{End}}_A\left(\overrightarrow{A}\right)\cong A^{\mathrm{op}}$. More precisely $${\mathrm{End}}_A\left(\overrightarrow{A}\right)=\left\{{\phi }_b|b\in A\right\}\text{,}$$ where ${\phi }_b$ is given by $${\phi }_b\left(\overrightarrow{a}\right)=a\overrightarrow{b}\text{,}\text{for all}\overrightarrow{a}\in \overrightarrow{A}\text{.}$$

		Proof.
	Let $\phi \in {\mathrm{End}}_A\left(\overrightarrow{A}\right)$ and let $b\in A$ be such that $\phi \left(\overrightarrow{1}\right)=\overrightarrow{b}$. For all $\overrightarrow{a}\in \overrightarrow{A}$, $$\begin{array}{rcl}\phi \left(\overrightarrow{a}\right)& =& \phi \left(a·\overrightarrow{1}\right)\\ & =& a\phi \left(\overrightarrow{1}\right)\\ & =& a\overrightarrow{b}\text{,}\end{array}$$ and so $\phi ={\phi }_b$. Then ${\mathrm{End}}_A\left(\overrightarrow{A}\right)\cong A^{\mathrm{op}}$ since $$\begin{array}{rcl}\left({\phi }_{b_1}\circ {\phi }_{b_2}\right)\left(\overrightarrow{a}\right)& =& {\phi }_{b_1}\left({\phi }_{b_2}\left(\overrightarrow{a}\right)\right)\\ & =& {\phi }_{b_1}\left(a\overrightarrow{b_2}\right)\\ & =& {\phi }_{b_1}\left(\overrightarrow{a{b}_2}\right)\\ & =& a{b}_2\overrightarrow{b_1}\\ & =& a\overrightarrow{b_2{b}_1}\\ & =& {\phi }_{b_2{b}_1}\left(\overrightarrow{a}\right)\end{array}$$ for all $b_1,b_2\in A$ and $\overrightarrow{a}\in \overrightarrow{A}$. $\square$

Suppose that $A$ is a finite dimensional algebra over an algebraically closed field $𝔽$ such that the regular representation $\overrightarrow{A}$ of $A$ is completely decomposable. Then $A$ is isomorphic to a direct sum of matrix algebras, i.e. $$A\cong \underset{\lambda \in \hat{A}}{⨁}{M}_{d_{\lambda }}\left(\overline{𝔽}\right)\text{,}$$ for some index set $\hat{A}$ and some positive integers $d_{\lambda }$, indexed by the elements of $\hat{A}$.

		Proof.
	If $\overrightarrow{A}$ is completely decomposable, then by the centraliser theorem in the Schur's Lemma page, ${\mathrm{End}}_A\left(\overrightarrow{A}\right)$ is isomorphic to a direct sum of matrix algebras. By proposition 1.1, $$A^{\mathrm{op}}\cong \underset{\lambda \in \hat{A}}{⨁}{M}_{d_{\lambda }}\left(\overline{𝔽}\right)\text{,}$$ for some set $\hat{A}$ and some positive integers $d_{\lambda }$, indexed by the elements of $\hat{A}$. The map $$\begin{array}{rcl}{\left(\underset{\lambda \in \hat{A}}{⨁}{M}_{d_{\lambda }}\left(\overline{𝔽}\right)\right)}^{\mathrm{op}}& ⟶& \underset{\lambda \in \hat{A}}{⨁}{M}_{d_{\lambda }}\left(𝔽\right)\\ a& ⟼& a^t\text{,}\end{array}$$ where $a^t$ is the transpose of the matrix $a$, is an algebra isomorphism. So $A$ is isomorphic to a direct sum of matrix algebras. $\square$

If $A$ is an algebra then the trace $\mathrm{tr}$ of the regular representation is the trace on $$A$$ given by $$\mathrm{tr}\left(a\right)=\mathrm{Tr}\left(\hat{A}\left(a\right)\right)\text{,}\text{for}a\in A\text{,}$$ where $\hat{A}\left(a\right)$ is the linear transformation of $A$ induced by the action of $a$ on $A$ by left multiplication.

Let $A=\underset{\lambda \in \hat{A}}{\oplus }{M}_{d_{\lambda }}\left(𝔽\right)$. The trace of the regular representation is nondegenerate if and only if the integers $d_{\lambda }$ are all nonzero in $\overline{𝔽}$. In characteristic $p$ they could be zero.

		Proof.
	As $A$-modules, the regular representation $$\overrightarrow{A}\cong \underset{\lambda \in \hat{A}}{⨁}{\left(A^{\lambda }\right)}^{\oplus d_{\lambda }}\text{,}$$ where $A^{\lambda }$ is the irreducible $A$-module consisting of column vectors of length $d_{\lambda }$. For $a\in A$ let $A^{\lambda }\left(a\right)$ be the linear transformation of $A^{\lambda }$ induced by the action of $a$. Then the trace $\mathrm{tr}$ of the regular representation is given by $$\mathrm{tr}=\sum _{\lambda \in \hat{A}}{d}_{\lambda }{\chi }^{\lambda }\text{,}\text{where}\begin{array}{rcl}{\chi }_A^{\lambda }:A& ⟶& \overline{𝔽}\\ a& ⟼& \mathrm{Tr}\left(\stackrel{\lambda }{A}\left(a\right)\right)\text{,}\end{array}$$ where ${\chi }_A^{\lambda }$ are the irreducible characters of $A$. Since the $d_{\lambda }$ are all nonzero the trace $\mathrm{tr}$ is nondegenerate. $\square$

Reference

[HA] T. Halverson and A. Ram, Partition algebras, European Journal of Combinatorics 26, (2005), 869-921; arXiv:math/040131v2.

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