Abstract Nonsense

Crushing one theorem at a time

The Group Algebra


Point of post: In this post we discuss the notion of the group algebra on a group.

Motivation

We shall see that it will often be fruitful to consider functions from a finite group G into \mathbb{C} as a perfect setting for our discussion of representation theory. We shall see that there are canonical ways to define representations with this space of functions both as a group (precisely how the group structure works shall be seen below) and as the target space for the group G (the representation space of G). Technically the group algebra is considered to be an adaptation of the free vector space of G over \mathbb{C}. We’ll take a slightly different version here and describe the group algebra as functions as was indicated in the first sentence.

Group Algebra

Let G be a finite group and define the group algebra over G, denoted \mathcal{A}(G), to be the set

 

\mathbb{C}^{G}=\left\{a:G\to \mathbb{C}\right\}

 

with addition and scalar multiplication defined point-wise in the sense that (a+b)(g)=a(g)+b(g) and (\alpha a)(g)=\alpha a(g) where a,b\in\mathcal{A}\left(G\right) and \alpha\in\mathbb{C}. The multiplication on \mathcal{A}\left(G\right) is often called convolution and is denoted a\ast b and given by

 

\displaystyle \left(a\ast b\right)(h)=\sum_{g\in G}a\left(hg^{-1}\right)b(g)

 

and a conjugation given by a^\ast (g)=\overline{a\left(g^{-1}\right)}. We define the mappings \delta_g to be, unimaginatively,

\displaystyle \delta_g:G\to\mathbb{C}:h\mapsto \begin{cases}1 & \mbox{if}\quad g=h\\ 0 & \mbox{if}\quad g\ne h\end{cases}

 

With these concepts in mind we claim that \mathcal{A}(G) is a |G|-dimensional associative unital algebra over \mathbb{C}. More precisely:

Theorem: Let G be a finite group. Then the group algebra \mathcal{A}\left(G\right) with operations of point-wise addition, point-wise scalar multiplication, convolution, and conjugation is an associative unital algebra with identity \delta_e.

Proof: It’s evident that the definitions of point-wise scalar multiplication and addition define a complex vector space structure on \mathcal{A}\left(G\right). Less trivial is to see that convolution interacts with addition and scalar multiplication in such a way to make it into an associative algebra. But, these are just calculations. Namely to prove associativity let a,b,c\in\mathcal{A}\left(G\right) be arbitrary and h\in G. Then:

 

\begin{aligned}\left(a\ast(b\ast c)\right)(h) &= \sum_{g\in G}a\left(hg^{-1}\right)\left(b\ast c\right)(g)\\ &= \sum_{g\in G}a\left(hg^{-1}\right)\left(\sum_{k\in G}b\left(g k^{-1}\right) c(k)\right)\\ &= \sum_{g\in G}\sum_{k\in G}a\left(h g^{-1}\right)b\left(gk^{-1}\right)c\left(k\right)\\ &= \sum_{k\in G}\left(\sum_{g\in G}a\left(hg^{-1}\right)b\left(gk^{-1}\right)\right)c(k)\end{aligned}

but, with equal validity

\begin{aligned}\left(\left(a\ast b\right)\ast c\right)(h) &= \sum_{k\in G}\left(a\ast b\right)\left(hk^{-1}\right)c(k)\\ &= \sum_{k\in G}\left(\sum_{g\in G}a\left(hk^{-1}g^{-1}\right)b(g)\right)c(k)\\ &= \sum_{k\in G}\left(\sum_{g\in G}a\left(h\left(gk\right)^{-1}\right)b\left(gk k^{-1}\right)\right)c(k)\\ &= \sum_{k\in G}\left(\sum_{g\in G}a\left(hg^{-1}\right)b\left(gk^{-1}\right)\right)c(k)\end{aligned}

 

where we’ve used the trick that for each fixed k\in G as g\mapsto gk is a bijection and thus the last equality holds. Since h was arbitrary it follows that \left(\left(a\ast b\right)\ast c\right)=\left(a\ast\left(b\ast c\right)\right). To prove distributivity we note that for any h\in G and a,b,c\in\mathcal{A}\left(G\right) we have that:

 

\begin{aligned}a\ast\left(b+c\right)(h) &= \sum_{g\in G}a\left(h g^{-1}\right)\left(b+c\right)(g)\\ &= \sum_{g\in G}a\left(h g^{-1}\right)\left(b(g)+c(g)\right)\\ &= \sum_{g\in G}\left(a\left(hg^{-1}\right)b(g)+a\left(hg^{-1}\right)c(g)\right)\\ &= \sum_{g\in G}a\left(hg^{-1}\right)b(g)+\sum_{g\in G}a\left(hg^{-1}\right)c(g)\\ &= \left(a\ast b\right)(h)+\left(a\ast c\right)(h)\end{aligned}

 

and since h\in G was arbitrary it follows that a\ast \left(b+c\right)=a\ast b+a\ast c. The scalar distributivity follows easily since for any \alpha\in\mathbb{C}, a,b\in\mathcal{A}\left(G\right), and h\in G we have

\displaystyle \begin{aligned}\left(\left(\alpha a\right)\ast b\right)(h) &=\sum_{g\in G}\left(\alpha a\right)\left(hg^{-1}\right)b(g)\\ &=\sum_{g\in G}\alpha a\left(h g^{-1}\right)b(g)\\ &=\alpha\sum_{g\in G}a\left(h g^{-1}\right)b(g)\\ &= \alpha \left(a\ast b\right)(h)\end{aligned}

 

To see that \delta_e is in fact an identity for convolution we merely note that for any a\in\mathcal{A}\left(G\right) and h\in G we have that

\displaystyle \left(\delta_e\ast a\right)(h)=\sum_{g\in G}\delta_e\left(h g^{-1}\right) a(g)= \delta_e\left(h h^{-1}\right)a(h) =a(h)

 

and since h\in G was arbitrary it follows that \delta_e\ast a=a. It similarly follows that a\ast \delta_e=a. From this we may conclude that \mathcal{A}\left(G\right) is an associative unital algebra with identity \delta_e. \blacksquare

 

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.

 

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January 20, 2011 - Posted by | Algebra, Representation Theory | , , ,

17 Comments »

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