Regular Representation

## Regular representation

If $A$ is an algebra then ${A}^{\mathrm{op}}$ is the algebra $A$ except with the opposite multiplication, i.e. $A op = a op | a ∈ A with a 1 op a 2 op = a 1 a 2 op , for all a 1 a 2 ∈ A .$

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 $\stackrel{\to }{A}$ for the $A$-module, i.e. $\stackrel{\to }{A}$ is the vector space $A → = b → | b ∈ A with A -action a b → = a b → , for all a ∈ A , b → ∈ A → .$

Let $A$ be an algebra and let $\stackrel{\to }{A}$ be the regular representation of $A$. Then ${\mathrm{End}}_{A}\left(\stackrel{\to }{A}\right)\cong {A}^{\mathrm{op}}$. More precisely $End A A → = φ b | b ∈ A ,$ where ${\phi }_{b}$ is given by $φ b a → = a b → , for all a → ∈ A → .$

 Proof. Let $\phi \in {\mathrm{End}}_{A}\left(\stackrel{\to }{A}\right)$ and let $b\in A$ be such that $\phi \left(\stackrel{\to }{1}\right)=\stackrel{\to }{b}$. For all $\stackrel{\to }{a}\in \stackrel{\to }{A}$, $φ a → = φ a · 1 → = a φ 1 → = a b → ,$ and so $\phi ={\phi }_{b}$. Then ${\mathrm{End}}_{A}\left(\stackrel{\to }{A}\right)\cong {A}^{\mathrm{op}}$ since $φ b 1 ∘ φ b 2 a → = φ b 1 φ b 2 a → = φ b 1 a b 2 → = φ b 1 a b 2 → = a b 2 b 1 → = a b 2 b 1 → = φ b 2 b 1 a →$ for all ${b}_{1},{b}_{2}\in A$ and $\stackrel{\to }{a}\in \stackrel{\to }{A}$. $\square$

 Proof. If $\stackrel{\to }{A}$ is completely decomposable, then by the centraliser theorem in the Schur's Lemma page, ${\mathrm{End}}_{A}\left(\stackrel{\to }{A}\right)$ is isomorphic to a direct sum of matrix algebras. By proposition 1.1, $A op ≅ ⨁ λ ∈ A ˆ M d λ 𝔽 ‾ ,$ for some set $\stackrel{ˆ}{A}$ and some positive integers ${d}_{\lambda }$, indexed by the elements of $\stackrel{ˆ}{A}$. The map $⨁ λ ∈ A ˆ M d λ 𝔽 ‾ op ⟶ ⨁ λ ∈ A ˆ M d λ 𝔽 a ⟼ a t ,$ 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 $tr a = Tr A ˆ a , for a ∈ A ,$ where $\stackrel{ˆ}{A}\left(a\right)$ is the linear transformation of $A$ induced by the action of $a$ on $A$ by left multiplication.

 Proof. As $A$-modules, the regular representation $A → ≅ ⨁ λ ∈ A ˆ A λ ⊕ d λ ,$ 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 $tr = ∑ λ ∈ A ˆ d λ χ λ , where χ A λ : A ⟶ 𝔽 ‾ a ⟼ Tr A λ a ,$ 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.