# Abstract Nonsense

## Left Regular Representation

Point of post: In this post we shall discuss the notion of the left and right regular representations on the group algebra $\mathcal{A}\left(G\right)$

Motivation

We saw in our last post that $\ast$-representations of the group algebra $\mathcal{A}\left(G\right)$ on a pre-Hilbert space $\mathscr{V}$ are, in a sense, one-to-one. In this post we give a specific example of a $\ast$-representation of $\mathcal{A}\left(G\right)$ on itself which will then induce a representation of $G$ on $\mathcal{A}\left(G\right)$. This representation will have a canonical form once we parse it out. This representation will be called (and one very similar to it) will be called the left regular representation of $G$ on $\mathcal{A}\left(G\right)$.

Canonical Inner Product on $\mathcal{A}\left(G\right)$

As it stands we have yet to put an inner product on the group algebra $\mathcal{A}\left(G\right)$ of some finite group $G$. We begin by defining the canonical inner product on $\mathcal{A}\left(G\right)$. It’s defined by

$\text{ }$

$\displaystyle \left\langle a,b\right\rangle=\frac{1}{|G|}\sum_{g\in G}a(g)\overline{b(g)}$

$\text{ }$

The first order of business is to verify that this is, in fact, an inner product on $\mathcal{A}\left(G\right)$.

Theorem: Let $G$ be a finite group and $\mathcal{A}\left(G\right)$ the group algebra over $G$. Then,

$\text{ }$

$\displaystyle \langle\cdot,\cdot\rangle:\mathcal{A}\left(G\right)\times\mathcal{A}\left(G\right)\to\mathbb{C}:(u,v)\mapsto \frac{1}{|G|}\sum_{g\in G}a(g)\overline{b(g)}$

$\text{ }$

is an inner product on $\mathcal{A}\left(G\right)$.

Proof: To see that $\langle\cdot,\cdot\rangle$ is linear in the first entry let $a,b,c\in\mathcal{A}\left(G\right)$ and $\alpha,\beta\in\mathbb{C}$. Then,

$\text{ }$

\displaystyle \begin{aligned}\left\langle \alpha a+\beta b,c\right\rangle &=\frac{1}{|G|}\sum_{g\in G}\left(\alpha a(g)+\beta b(g)\right)\overline{c(g)}\\ &= \frac{\alpha}{|G|}\sum_{g\in G}a(g)\overline{c(g)}+\frac{\beta}{|G|}\sum_{g\in G}b(g)\overline{c(g)}\\ &= \alpha\langle a,c\rangle+\beta\langle b,c\rangle\end{aligned}

$\text{ }$

and so it’s linear in the first entry. It’s conjugate symmetric since

$\text{ }$

\displaystyle \begin{aligned}\overline{\langle a,b\rangle} &=\overline{\frac{1}{|G|}\sum_{g\in G}a(g)\overline{b(g)}}\\ &= \frac{1}{\overline{|G|}}\sum_{g\in G}\overline{a(g)}\overline{\overline{b(g)}}\\ &= \frac{1}{|G|}\sum_{g\in G}\overline{a(g)}b(g)\\ &= \langle b,a\rangle\end{aligned}

$\text{ }$

and lastly it’s positive-definite since

$\text{ }$

$\displaystyle \langle a,a\rangle=\frac{1}{|G|}\sum_{g\in G}a(g)\overline{a(g)}=\frac{1}{|G|}\sum_{g\in G}|a(g)|^2\geqslant 0$

and evidently from this it’s clear that $\langle a,a\rangle=0$ if and only if $a=\bold{0}$. The conclusion follows. $\blacksquare$

One last thing to notice about this inner product is that it makes $\{\sqrt{|G|}\delta_g\}_{g\in G}$ an orthonormal basis for $\mathcal{A}\left(G\right)$. Indeed, we already know it’s a basis and if $g\ne g'$ then

$\text{ }$

\displaystyle \begin{aligned}\langle \sqrt{|G|}\delta_g,\sqrt{|G|}\delta_{g'}\rangle &= \sum_{h\in G}\delta_g(h)\overline{\delta_{g'}(h)}\\ &=\delta_g(g)\delta_{g'}(g)+\delta_g(g')\delta_{g'}(g')\\ &= 0\end{aligned}

$\text{ }$

and if $g=g'$ then

$\text{ }$

\displaystyle \begin{aligned}\left\langle \sqrt{|G|}\delta_g,\sqrt{|G|}\delta_g\right\rangle &= \sum_{h\in G}\delta_g(h)\overline{\delta_g(h)}\\ &= \delta_g(g)\delta_g(g)\\ &= 1\end{aligned}

$\text{ }$

References:

1.Simon, Barry. Representations of Finite and Compact Groups. Providence, RI: American Mathematical Society, 1996. Print.

2. Serre, Jean Pierre. Linear Representations of Finite Groups. New York: Springer-Verlag, 1977. Print

January 23, 2011 -

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