1.6. Lecture 6

1.6.1. The group algebra and the regular representation

Let $G$ be a finite group. Recall (cf. Definition 1.1.8) that we defined the free vector space ${\mathbb{C}}(G)$ to be the vector space of formal sums

\sum_{g\in G}z_{g}g,

where all $z_{g}\in{\mathbb{C}}$ , with vector addition and scalar multiplication given by

\left(\sum_{g\in G}z_{g}g\right)+\left(\sum_{g\in G}w_{g}g\right):=\left(\sum_% {g\in G}(w_{g}+z_{g})g\right),\quad\lambda\left(\sum_{g\in G}z_{g}g\right):=% \sum_{g\in G}\left(\lambda z_{g}\right)g.

The group multiplication lets us define a multiplication operation on ${\mathbb{C}}(G)$ :

\left(\sum_{g\in G}z_{g}g\right)\left(\sum_{h\in G}w_{h}h\right):=\sum_{g,h\in G% }z_{g}w_{h}gh=\sum_{g\in G}\left(\sum_{h\in G}z_{gh^{-1}}w_{h}\right)g.

Definition 1.6.1.

Let $G$ be a finite group. The free vector space ${\mathbb{C}}(G)$ equipped with the multiplication rule defined above is called the group algebra $\mathbb{C}G$ (or ${\mathbb{C}}[G]$ ) of $G$ .

Example 1.6.2.

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

	$\displaystyle{\bf x}\cdot{\bf y}$	$\displaystyle=\big{(}e-(12)\big{)}\big{(}2(23)+(123)\big{)}$
		$\displaystyle=2(23)+(123)-2(12)(23)-(12)(123)$
		$\displaystyle=2(23)+(123)-2(123)-(23)$
		$\displaystyle=(23)-(123).$

Remark 1.6.3.

A vector space $V$ over a field $k$ with a multiplication rule “ $\cdot$ ” $:V\times V\rightarrow V$ that satisfies

(i)

${\bf v}\cdot({\bf u}+{\bf w})={\bf v}\cdot{\bf u}+{\bf v}\cdot{\bf w}$
(ii)

$({\bf u}+{\bf w})\cdot{\bf v}={\bf u}\cdot{\bf v}+{\bf w}\cdot{\bf v}$
(iii)

$(\alpha{\bf v})\cdot(\beta{\bf u})=(\alpha\beta){\bf v}\cdot{\bf u}$

for all ${\bf v},{\bf u},{\bf w}\in V$ and $\alpha,\beta\in k$ is called an algebra. Note that one does not require $\cdot$ to be associative however we will only consider algebras with an associative product. Examples of algebras include ${\mathbb{R}}^{3}$ with the cross product, the space of all polynomials $k[X]$ with coefficients in $k$ with multiplication being pointwise multiplication of polynomials, and $\operatorname{Hom}(V)$ for any vector space $V$ with multiplication being composition of linear maps.

The formal sums $g=1g\in{\mathbb{C}}G$ form a basis of ${\mathbb{C}}G$ , hence $\operatorname{dim}{\mathbb{C}}G=|G|$ .

Any representation $(\rho,V)$ of $G$ gives rise to an algebra homomorphism from ${\mathbb{C}}G$ to $\operatorname{Hom}(V)$ , which we also denote as $\rho$ :

\rho\left(\sum_{g\in G}z_{g}g\right):=\sum_{g\in G}z_{g}\rho(g).

Remark 1.6.4.

For many this is the definition of representation of a finite group that they use, in other words an algebra homomorphism from the group algebra to a vector space. This is because this definition can be extended to general algebra, not just group algebras (see Problem 1.6.8).

Recall also that we have the permutation representation of $G$ on ${\mathbb{C}}(X)$ (Definition 1.1.9); in this case we normally denote this by $(\lambda,{\mathbb{C}}G)$ , and call it the regular representation, i.e.

\lambda(g)\left(\sum_{h\in G}z_{h}h\right):=\sum_{h\in G}z_{h}gh=\sum_{h\in G}% z_{g^{-1}h}h.

Theorem 1.6.5.

Let $(\pi,V)$ be any representation of $G$ . Then there is an isomorphism of vector spaces

\operatorname{Hom}_{G}(\mathbb{C}G,V)\cong V.

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

The proof of this is short, but can be 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.

Proof.

If $T:\mathbb{C}G\rightarrow V$ is a $G$ -homomorphism, define

\Phi(T):=T(e)\in V.

Conversely, given ${\bf v}\in V$ , let $\Psi({\bf v})\in\operatorname{Hom}_{G}(\mathbb{C}G,V)$ be the linear map defined by

\Psi({\bf v})\left(\sum_{g\in G}z_{g}g\right)=\sum_{g\in G}z_{g}\pi(g){\bf v};

you can check that $\Psi({\bf v})$ is indeed a $G$ -homomorphism (see Exercise 1.6.2).

We claim that the maps $\Phi$ and $\Psi$ are linear maps between $V$ and $\operatorname{Hom}_{G}(\mathbb{C}G,V)$ that are 2-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({\bf v}))={\bf v}

and

\Psi(\Phi(T))=T

for all ${\bf v}\in V$ , $T\in\operatorname{Hom}_{G}(\operatorname{C{}}G,V)$ .

(i)

Let ${\bf v}\in V$ . Then

$\Phi(\Psi({\bf v}))=\Psi({\bf v})(e)=\pi(e){\bf v}={\bf v},$

as required.
(ii)

Let $T\in\operatorname{Hom}_{G}(\mathbb{C}G,V)$ and define $S=\Psi(\Phi(T))\in\operatorname{Hom}_{G}({\mathbb{C}}G,V)$ . We wish to show that $S=T$ . To do this, we evaluate $S$ at the basis vectors $g\in{\mathbb{C}}G$ . Since $S$ and $T$ are linear, if they agree on a basis, then they are the same:

$S(g)=\Psi\big{(}\Phi(T)\big{)}(g)=\pi(g)\Phi(T)=\pi(g)T(e)=T(\lambda(g)e)=T(g),$

since $T$ is a $G$ -homomorphism. ∎

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 1.6.6.

(i)

Every irreducible representation $(\rho,W_{\rho})$ of $G$ is a constituent of the regular representation with multiplicity $\operatorname{dim}\rho$ . In other words,

$(\lambda,{\mathbb{C}}G)\cong\bigoplus_{\rho\in\mathrm{Irr}(G)}(\rho,W_{\rho})^% {\operatorname{dim}\rho}$

(here $(\rho,W_{\rho})^{\operatorname{dim}\rho}=(\rho,W_{\rho})\oplus(\rho,W_{\rho})% \oplus\ldots\oplus(\rho,W_{\rho})$ ; $\operatorname{dim}\rho$ times).
(ii)

(Sum of squares formula.) We have

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

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

Proof.

(i)

By Maschke’s theorem (or more precisely Theorem 1.5.12), we can decompose ${\mathbb{C}}G$ as a direct sum of irreducibles. By Lemma 1.5.14, an isomorphism class of an irreducible representations $\rho$ appears in the decomposition $\operatorname{dim}\operatorname{Hom}_{G}(\mathbb{C}G,\rho)$ times. By Theorem 1.6.5, $\operatorname{dim}\operatorname{Hom}_{G}(\mathbb{C}G,\rho)=\operatorname{dim}\rho$ .
(ii)

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

1.6.2. Exercises

Problem 37. Verify that the sum of squares formula holds for dihedral groups.

Problem 38. Fill in the following details from the proof of Theorem 1.6.5:

(a)

Verify that $\Phi:\operatorname{Hom}_{G}({\mathbb{C}}G,V)\rightarrow V$ is linear.
(b)

Verify that $\Psi:V\rightarrow\operatorname{Hom}({\mathbb{C}}G,V)$ is linear.
(c)

Show that for every ${\bf v}\in V$ , $\Psi({\bf v})$ is a $G$ -homomorphism, that is

$\Psi({\bf v})\lambda(g)=\pi(g)\Psi({\bf v})$

(hence $\Psi({\bf v})\in\operatorname{Hom}_{G}({\mathbb{C}}G,V)$ ).

Problem 39. Decompose $(\lambda,{\mathbb{C}}(D_{3}))$ into irreducible subrepresentations.

Problem 40. Show the following:

Theorem 1.6.7.

If every irreducible representation of a finite group $G$ is 1-dimensional, then $G$ is Abelian.

Hint: Consider the representation $(\lambda,{\mathbb{C}}G)$ . Show that $\ker\lambda=\{e\}$ , and thus $G$ is isomorphic to $\lambda(G)<\operatorname{GL}({\mathbb{C}}G)$ . Use Theorem 1.6.6 to show that there is a basis of ${\mathbb{C}}G$ for which the matrices of all $\lambda(g)$ are diagonal, hence $\lambda(G)$ is Abelian.

Remark: Combining this with Proposition 1.3.1 and Theorem 1.4.3 gives the following nice result:

Theorem 1.6.8.

A finite group $G$ is Abelian if and only if all its irreducible representations are 1-dimensional.

Problem 41. Let $A$ be an algebra over ${\mathbb{C}}$ . An $A$ -module, or algebra representation of $A$ is a pair $(\pi,V)$ , where $V$ is a ${\mathbb{C}}$ -vector space and $\pi\in\operatorname{Hom}(A,\operatorname{Hom}(V))$ is such that

\pi({\bf a}{\bf b})=\pi({\bf a})\pi({\bf b})\qquad\forall\,{\bf a},{\bf b}\in A.

In an analogous manner to group representations, a subspace $W\subseteq V$ is said to define a submodule $(\pi,W)$ of $(\pi,V)$ if $\pi({\bf a}){\bf w}\in W$ for all ${\bf a}\in A$ , ${\bf w}\in W$ , and $(\pi,V)$ is said to be irreducible if it has no non-trivial submodules.

(a)

Show Schur’s lemma for irreducible algebra-modules: if $(\pi,V)$ is a finite dimensional irreducible algebra module and $T\in\operatorname{Hom}(V)$ is such that

$\pi({\bf a})T=T\pi({\bf a})$

for all ${\bf a}\in A$ , then $T\in{\mathbb{C}}\,{\,\mathrm{Id}}$ .
(b)

Show that $(\pi,V)$ is an irreducible representation of a group $G$ if and only if $(\pi,V)$ is an irreducible ${\mathbb{C}}G$ -module.

Problem 42. Let $A$ be an algebra, and define $Z(A)=\{{\bf a}\in A\,|\,{\bf a}{\bf b}={\bf b}{\bf a}\;\forall{\bf b}\in A\}$ . This set is called the centre of $A$ .

(a)

Show that $Z(A)$ is a subalgebra of $A$ .
(b)

Compute $Z({\mathbb{C}}S_{3})$ .
(c)

Show that if $(\pi,V)$ is an irreducible $A$ -module, then there exists a homomorphism of commutative algebras $\psi:Z(A)\rightarrow{\mathbb{C}}$ such that

$\pi({\bf z})=\psi({\bf z}){\,\mathrm{Id}}$

for all ${\bf z}\in Z(A)$ .