Abstract Nonsense

Crushing one theorem at a time

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)


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{ }


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 - Posted by | Algebra, Representation Theory | , , ,


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