1 Representation theory of finite groups 1.5 Maschke’s theorem and complete reducibility 2 Character theory

1.6 The group ring and the regular representation

Definition 1.53.

Let $G$ be a finite group. The group ring $\mathbb{C}[G]$ has as elements formal linear combinations

\sum_{g\in G}a_{g}[g]

with $a_{g}\in\mathbb{C},$ multiplied according to the ‘rule forced by $[g][h]=[gh]$ ’, that is,

\sum_{g\in G}a_{g}[g]\sum_{g\in G}b_{g}[g]=\sum_{g,h\in G}a_{g}b_{h}[gh].

This is a noncommutative ring (if $G$ is not abelian).

To elaborate, the symbols $[g]$ for $g\in G$ are basis vectors for a vector space of dimension $\dim G$ , which we call the group ring.

Example 1.54.

Let $x=[e]-[(12)]$ and $y=2[(23)]+[(123)]$ be elements of $\mathbb{C}[S_{3}]$ . Then

	$\displaystyle xy$	$\displaystyle=2[e][(23)]-2[(12)][(23)]+[e][(123)]-[(12)][(123)]$
		$\displaystyle=2[(23)]-2[(123)]+[(123)]-[(23)]$
		$\displaystyle=[(23)]-[(123)].$

If $(\rho,V)$ is a representation of $G$ , then we can ’multiply’ any element of $V$ by any element of $\mathbb{C}[G]$ :

(\sum_{g\in G}a_{g}[g])v=\sum_{g\in G}a_{g}\rho(g)v.

The group ring is a vector space of dimension $\dim(G)$ . It has a linear left action of $G$ :

g\cdot\sum a_{h}[h]=\sum a_{h}[gh].

Therefore we have a representation of $G$ on $\mathbb{C}[G]$ called the left regular representation.

In fact, the regular representation is just the permutation representation for the action of $G$ on itself by left multiplication. If we adopt the ’functional point of view’ mentioned before, we get another way of thinking about this.

Definition 1.55.

Let

\mathbb{C}^{G}=\{f:G\rightarrow\mathbb{C}\}

be the space of functions from $G\rightarrow\mathbb{C}$ . Define a representation $\rho$ of $G$ on $\mathbb{C}^{G}$ by

(\rho(g)f)(h)=f(g^{-1}h).

Lemma 1.56.

The representation $\mathbb{C}^{G}$ is isomorphic to the regular representation $\mathbb{C}[G]$ .

This is simply the discussion in Remark 1.9 applied to this partiuclar case.

When we refer to the ’regular representation’, then, we mean either $\mathbb{C}[G]$ or $\mathbb{C}^{G}$ according to which is most convenient.

Theorem 1.57.

Let $V$ be any representation of $G$ . Then there is an isomorphism of vector spaces

\operatorname{Hom}_{G}(\mathbb{C}[G],V)\rightarrow V.

Equivalently, $\dim\operatorname{Hom}_{G}(\mathbb{C}[G],V)=\dim V$ .

Proof.

This is short, but difficult to wrap your head around. The idea is to provide a recipe to turn a $G$ -homomorphism $\mathbb{C}[G]\rightarrow V$ into an element of $V,$ and a recipe to turn an element of $V$ into a $G$ -homomorphism $\mathbb{C}[G]\rightarrow V,$ and check that these recipes are inverse to each other.

If $f:\mathbb{C}[G]\rightarrow V$ is a $G$ -homomorphism, let

\phi(f)=f([e])\in V

Conversely, if $v\in V,$ let $\psi(v)\in\operatorname{Hom}_{G}(\mathbb{C}[G],V)$ be given by

\psi(v)(\sum a_{g}[g])=\sum a_{g}[g]v;

you can check that $\psi(v)$ is indeed a $G$ -homomorphism.

Then I claim that the maps $\phi$ and $\psi$ are linear maps between $V$ and $\operatorname{Hom}_{G}(\mathbb{C}[G],V)$ that are two-sided inverses of each other, so these vector spaces are isomorphic. It is clear that they are linear maps. We must check that

\phi(\psi(v))=v

and

\psi(\phi(f))=f

for all $v\in V,$ $f\in\operatorname{Hom}_{G}(\mathbb{C}[G],V)$ .

•

Let $v\in V$ . Then

$\phi(\psi(v))=\psi(v)([e])=[e]v=v$

as required.
•

Let $f\in\operatorname{Hom}_{G}(\mathbb{C}[G],V)$ and let $f^{\prime}=\psi(\phi(f))$ . Then, for $g\in G,$

$f^{\prime}([g])=[g]\phi(f)=[g]f([e])=f([g][e])=f([g])$

since $f$ is a $G$ -homomorphism. As $f,f^{\prime}$ are linear, this implies $f=f^{\prime}$ as required. ∎

This has a beautiful consequence: the sum of the squares of the dimensions of the irreducible representations is equal to the order of the group. We write $\mathrm{Irr}(G)$ for the set of isomorphism classes of irreducible representations of $G$ .

Theorem 1.58.

1.

Every irreducible representation $\rho$ of $G$ is a constituent of the regular representation with multiplicity $\dim\rho$ . In other words,

$V\cong\bigoplus_{\rho\in\mathrm{Irr}(G)}\rho^{\dim\rho}.$
2.

(Sum of squares formula.) We have

$\sum_{\rho\in\mathrm{Irr}(G)}\dim(\rho)^{2}=|G|$

where the sum runs over the isomorphism classes of irreducible representations of $G$ . In particular, $\mathrm{Irr}(G)$ is finite.

Proof.

1.

By Maschke’s theorem, we can decompose $\mathbb{C}[G]$ as a direct sum of irreducibles in which each isomorphism class $\rho$ of irreducible representations appears $\dim\operatorname{Hom}_{G}(\mathbb{C}[G],\rho)=\dim(\rho)$ times.
2.

Immediate from equating dimensions on both sides of the first part and noting that $\dim\mathbb{C}[G]=|G|$ . ∎

Exercise 1.59.

Verify the sum of squares formula for dihedral groups.

Note that this gives another approach to the classification of representations of dihedral groups. First, write down all the irreducible representations and check that they are non-isomorphic. Then, by the sum of squares formula, you have found everything!