# A primer on group representations

In this post, we will give a brief introduction to the representation theory of general (finite) groups.

Definition: A representation of a group $G$ is a homomorphism $G\to\mathrm{GL}_n(\mathbf{C})$, where $n$ is called the dimension of the representation.

A representation is faithful if the map $G\to\mathrm{GL}_n(\mathbf{C})$ is injective. A representation $G\to\mathrm{GL}_n(\mathbf{C})$ can also be viewed as the action of $G$ on an $n$-dimensional $\mathbf{C}$-vector space $V$ because $\mathrm{Aut}(V)$ is isomorphic to $\mathrm{GL}_n(\mathbf{C})$ via the isomorphism $V\cong \mathbf{C}^n$. However, this requires fixing a basis. This leads to the following definition.

Definition: Let $V$ be a $\mathbf{C}$-vector space and $G$ a group. A representation of $G$ on $V$ is a map $\rho:G\to\mathrm{Aut}(V)$. This is equivalent to an operation $G\times V\to V$ defined via $(g,v)\mapsto \rho(g)(v)$.

An isomorphism of representations $\rho$ and $\rho^\prime$ of $G$ on $V$ and $V^\prime$ respectively is an isomorphism $V\to V^\prime$ that is consistent with the action of $G$ on $V$ and $V^\prime$. We will henceforth denote $\mathrm{Aut}(V)$ by $\mathrm{GL}(V)$. If $\rho:G\to\mathrm{GL}_n(\mathbf{C})$ is a representation of $G$, the character of $\rho$ is $\mathbf{C}$-valued function on $G$ defined as $\chi_\rho(g) = \mathrm{tr}(\rho(g))$. We claim that characters are constant on conjugate classes. Suppose that $g^\prime = hgh^{-1}$. Then $\rho(g^\prime) = \rho(h)\rho(g)\rho(h^{-1})$. Since $\rho(h^{-1}) = \rho(h)^{-1}$, $\rho(g^\prime)$ and $\rho(g)$ are conjugate, they have the same trace.

Consider a character $\chi$ associated to a representation $\rho:G\to\mathrm{GL}(V)$. Then $\chi(1)$ is the dimension of the representation of $G$ since $\rho(1)$ is the identity on $V$, whose trace is the dimension of $V$. Another important statement is the following. Suppose $|G|=k$, and suppose $g\in G$. Then the roots of the characteristic polynomial of $\rho(g)$ are powers $\zeta^n$ of the $k$th roots of unity. Since the trace of a matrix is a sum of its eigenvalues, and if $\omega$ is an eigenvalue of $\rho(g)$, then $\omega^k$ is an eigenvalue of $\rho(g)^k = \rho(g^k)= \rho(1)$, and hence $\omega$ must be a power of $e^{2\pi i/k}$. This also shows that $\rho(g^{-1}) = \overline{\rho(g)}$.

Let $G$ act on a $\mathbf{C}$-vector space $V$. A subspace $W$ of $V$ is $G$-invariant if the action of $G$ on $V$ reduces to an action of $G$ on $W$. If $V=W\oplus Z$, where $W$ and $Z$ are $G$-invariant subspaces, then the representation on $V$ is the direct sum of the representations on $W$ and $Z$. If $\rho$ is the representation on $V$, then the matrix associated to $\rho(g)$ will be in block diagonal form. This leads to the following definition.

Definition: A representation of $G$ on a $\mathbf{C}$-vector space $V$ is irreducible if there is no proper $G$-invariant subspace of $V$.

Let $\rho$ be a non-irreducible (reducible) representation of $G$ on $V$, and consider a proper $G$-invariant subspace $W\subset V$. The basis of $W$ can be extended to $V$, and then the matrix associated to $\rho(g)$ will be in block form, where the upper left matrix acts on the basis of $W$, and the lower right matrix acts on some other representation of $G$. Maschke’s theorem says:

Theorem: If $G$ is a finite group, every representation of $G$ on a nonzero finite-dimensional $\mathbf{C}$-vector space is the direct sum of irreducible representations of $G$.

Proof: The proof is two-part. Say that a representation $\rho:G\to\mathrm{GL}(V)$ on a $\mathbf{C}$-vector space $V$ is unitary if for all $g\in G$, the matrix associated to $\rho(g)$ is unitary, i.e., $\rho(g)^\ast = \rho(g)^{-1}$. Suppose $\rho$ is a reducible unitary representation of $G$ on a Hermitian space $V$. If $W$ is a $G$-invariant subspace of $V$, then $W^\perp$ is also $G$-invariant, because since $\rho$ is unitary, if $v\perp W$, then $\rho(g)(v)\perp \rho(g)(W)$; since $W$ is $G$-invariant, $\rho(g)(v)\in W^\perp$. Since $V=W\oplus W^\perp$, it follows that $\rho$ is the direct sum of the restrictions to $W$ and $W^\perp$. Therefore it follows that every unitary representation on a Hermitian space is an orthogonal sum of irreducible representations.

Now it suffices to prove that if $\rho:G\to\mathrm{GL}(V)$ is a representation of a finite group on a $\mathbf{C}$-vector space $V$, then there is a $G$-invariant positive-definite Hermitian form on $V$ (making it a Hermitian space). Consider some positive-definite Hermitian form $[,]$ on $V$, and define

$\displaystyle\langle v,w \rangle = \frac{1}{|G|}\sum_{g\in G}[gv,gw].$

Since $G$ is finite, this is clearly $G$-invariant. Checking that this new form is positive-definite and Hermitian is easy, and this proves the required result. QED.

We have Schur’s lemma:

Theorem: If $\rho,\rho^\prime$ are irreducible representations of $G$ on $V$ and $V^\prime$, and if $T:V\to V^\prime$ is a $G$-invariant linear transformation, then either $T$ is an isomorphism or $T=0$. In addition, if $T:V\to V$ is a $G$-invariant linear transformation then $T = cI$ for some scalar $c$.

Proof: To prove the first statement, consider the subspace $\ker T$. Clearly this is $G$-invariant, so $\ker T$ is either $V$ or $\{0\}$. If $T\neq 0$, then $\ker T$ cannot be $V$, so $\ker T=\{0\}$, i.e., $T$ is injective. In this case, because $\mathrm{im}(T)$ is also $G$-invariant, it is either $V^\prime$ or $\{0\}$. Since $T\neq 0$, the image of $T$ is $V^\prime$, and $T$ is an isomorphism. To prove the second statement, let $\zeta$ be an eigenvalue of $T$. Then $\ker(T-\zeta I)$ is not zero since it contains an eigenvector of $T$. Therefore, $T-\zeta I$ is not an isomorphism, and so $T-\zeta I = 0$, i.e., $T=\zeta I$. QED.

There are a number of important consequences. For example, suppose that $G$ is abelian, and consider an irreducible representation $\rho:G\to\mathrm{GL}(V)$. Because $G$ is abelian, $\rho(g)\rho(h)v = \rho(h)\rho(g)v$, and therefore, for every $g\in G$, by Schur’s lemma, $\rho(g)$ must be scalar multiplication. In this case, if $V$ is not one-dimensional, then every subspace of $V$ is a $G$-invariant subspace – but this contradicts irreducibility! Therefore, every irreducible representation of an abelian group $G$ is one-dimensional. We will see another proof below of this when the abelian group is finite.

Definition: Let $\chi,\chi^\prime$ be characters of $G$. There is a Hermitian product defined via

$\displaystyle\langle \chi,\chi^\prime \rangle = \frac{1}{|G|}\sum_g\chi^\prime(g)\overline{\chi(g)}.$

Because $\chi$ is constant on conjugacy classes it is clear that if $C_j$ denotes the conjugacy classes of $G$ and $g_j$ are representatives of $C_j$, then

$\displaystyle\langle \chi,\chi^\prime \rangle = \frac{1}{|G|}\sum |C_j|\chi^\prime(g_j)\overline{\chi(g_j)} = \sum\frac{1}{|C_G(g_j)|}\chi^\prime(g_j)\overline{\chi(g_j)}.$

The main theorem of representation theory (without proof) is the following.

Theorem: Suppose $G$ is a finite group. Then:

1. The irreducible characters of $G$ form an orthonormal basis of class functions (functions that are constant on conjugacy classes) of $G$. If $\rho,\rho^\prime$ are irreducible representations of $G$, then $\langle \chi,\chi^\prime \rangle = 1$ if $\rho$ and $\rho^\prime$ are isomorphic, and is $0$ otherwise.
2. The number of isomorphism classes of irreducible representations is finite, and is the equal to the number of conjugacy classes of the group.
3. Suppose $\rho_i$ represent the irreducible representations of $G$. Then each $\mathrm{dim}(\rho_i)$ divides $|G|$, and: $|G| = \sum \mathrm{dim}(\rho_i)^2$.

There are many corollaries of this. For example, we get another proof that representations of abelian groups are one-dimensional, in the case when the group is finite. If $G$ is finite abelian and $\chi:G\to\mathbf{C}^\times$ is an irreducible character of $G$, then $\chi$ must be one-dimensional. To see this, note that since $G$ is finite abelian, the number of conjugacy classes of $G$ is $|G|$, which is also the number of irreducible characters of $G$. Since $|G| = \sum^{|G|} \mathrm{dim}(\chi_i)^2$, each $\mathrm{dim}(\chi_i)$ must be $1$. We will conclude with the following result.

Corollary: Representations of a finite group $G$ are isomorphic if and only if their characters are equal.

Proof. Since any representation $\rho$ can be written as $\bigoplus_i m_i\rho_i$, so the associated character $\chi$ can be written as $\sum m_i\chi_i$. Clearly $m_i = \langle \chi,\chi_i \rangle$ by the orthogonal projection formula, completing the proof.

Best,
SKD