Notes on the Representation Theory of Finite Groups

By David K. Zhang

Under Construction!

Definition: Let $G$ be a group, and let $V$ be a vector space over a field $F.$ A representation of $G$ on $V$ is a group homomorphism $\rho: G \to \GL(V),$ where $\GL(V)$ denotes the group of invertible linear transformations $V \to V.$

We can think of the representation $\rho$ as defining a left action of the group $G$ on the vector space $V,$ given by

\[g \cdot v = \rho(g)(v) = \rho_g v \]

for each $g \in G$ and $v \in V.$ (For brevity of notation, we will often write $\rho_g$ in place of $\rho(g).$) Since $\rho$ is a homomorphism, we must have $\rho_e = \operatorname{Id}_V,$ where $e \in G$ denotes the identity element, and $\rho_{g^{-1}} = (\rho_g)^{-1}$ for all $g \in G.$

Definition: The cardinal number $\dim_F(V)$ is called the degree of the representation $\rho.$

In these notes, we will focus on the special case in which $G$ is a finite group and $V$ is a finite-dimensional vector space over the field $\C$ of complex numbers. This case occurs frequently in physical and chemical applications. TODO: Add some motivation for studying representation theory.

In this special case, we can identify $V$ with the vector space $\C^n$ of $n$-tuples of complex numbers by choosing an ordered basis $\{\vb{v}_1, \dots, \vb{v}_n\}$ of $V$ to identify with the standard basis $\{\vb{e}_1, \dots, \vb{e}_n\}$ of $\C^n.$ The representation $\rho$ thus associates to each group element $g \in G$ an invertible $n \times n$ complex matrix $[\rho_g]$ (which depends implicitly on the choice of basis in $V$).

Definition: The character of the representation $\rho$ is the function $\chi: G \to \C$ defined by

\[\chi(g) = \Tr[\rho_g] = \sum_{i=1}^n [\rho_g]_{ii} \]

for all $g \in G.$ (Recall that the trace of the matrix representation of a linear transformation is independent of the choice of basis.)

1. Group Actions

Definition: Let $G$ be a group, and let $X$ be an arbitrary set. An action of $G$ on $X$ is a function $\alpha: G \times X \to X$ satisfying the following axioms:

  1. $\alpha(g, \alpha(h, x)) = \alpha(gh, x)$ for all $g,h \in G$ and $x \in X.$
  2. $\alpha(e, x) = x$ for all $x \in X,$ where $e$ denotes the identity element of $G.$

Notation: We write $\alpha: G \curvearrowright X$ as a shorthand for the statement $\alpha$ is an action of $G$ on $X.$ When the function $\alpha$ is clear from context, it is common to suppress it notationally by writing $g \cdot x$ or simply $gx$ instead of $\alpha(g, x).$ In this case, we say that $X$ is a $G$-set, and the preceding axioms take the following form:

  1. $g(hx) = (gh)x$ for all $g,h \in G$ and $x \in X.$
  2. $ex = x$ for all $x \in X,$ where $e$ denotes the identity element of $G.$

Group actions arise whenever we want to think of a group $G$ as a collection of transformations acting on some set $X.$ For example, the special orthogonal group $\mathsf{SO}(n, \mathbb{R})$ (which consists of all real orthogonal $n \times n$ matrices with determinant $1$) is naturally viewed as a collection of rotations acting on $\mathbb{R}^n.$ In this case, we have an obvious action $\mathsf{SO}(n, \mathbb{R}) \curvearrowright \mathbb{R}^n$ given by the usual matrix-vector product.

The pervasiveness of group actions in mathematics and mathematical physics is not to be understated. Indeed, the very notion of a group is essentially an algebraic formalization of the idea of a collection of invertible transformations which is closed under composition. Thus, it is common for a group to be understood and characterized in terms of its actions.

This approach can provide significant conceptual advantages over attempting to understand a group through its internal algebraic structure. It is, for example, much easier to think of $\mathsf{SO}(n, \mathbb{R})$ as the group of rigid rotations of Euclidean $n$-space, as opposed to a collection of arrays of real numbers, constrained to satisfy certain algebraic identities and endowed with a mysterious rule of combination.

Suppose now that we are given a group action $\alpha: G \curvearrowright X.$ We can consider, for each $g \in G$, the function $\alpha_g: X \to X$ obtained by fixing the first argument of $\alpha$ at $g$ and allowing the second to vary. Explicitly, we define

\[\alpha_g(x) = \alpha(g, x) \]

for all $g \in G$ and $x \in X.$ Observe that each of the functions $\alpha_g$ is a bijection, since by axioms 1 and 2, $\alpha_g$ has an inverse given by $\alpha_{g^{-1}}.$ This means that each $\alpha_g$ is a simply a permutation of the elements of $X.$ Indeed, we can equivalently think of the group action $\alpha$ as an embedding of $G$ into the symmetric group $\operatorname{Sym}(X)$ of permutations of $X,$ i.e., a group homomorphism $G \to \operatorname{Sym}(X)$ given by $g \mapsto \alpha_g.$

1.1. Orbits and Stabilizers

At this point, there are two natural questions we may wish to ask regarding the behavior of the action of a group $G$ on a particular element $x \in X$.

  1. What elements of $X$ can $x$ be transformed into under the action of $G$?
  2. Which elements of $g$ effect a transformation that leaves $x$ unchanged?

To explore these questions, we make the following definitions.

Definition: If $X$ is a $G$-set and $x \in X$, then the orbit of $x$ under $G$ is the set

\[Gx \coloneqq \{ gx : g \in G \}. \]

The stabilizer of $x$ under $G$ is the set

\[G_x \coloneqq \{ g \in G : gx = x \}. \]

(These notations are unfortunately very similar, particularly in handwriting.)

It is easy to see that the stabilizer $G_x$ of any element $x \in X$ is a subgroup of $G$. Indeed, if two permutations $\alpha_g$ and $\alpha_h$ leave $x$ unchanged, then clearly their composition $\alpha_g \circ \alpha_h = \alpha_{gh}$ also leaves $x$ unchanged, and the inverse $(\alpha_g)^{-1} = \alpha_{g^{-1}}$ also leaves $x$ unchanged. For this reason, $G_x$ is often called the stabilizer subgroup or isotropy subgroup of $G$ with respect to $x$.

It is also easy to see that the orbits of any two elements $x,y \in X$ are either disjoint or identical; they cannot partially overlap. Indeed, if the intersection $Gx \cap Gy$ is nonempty, then there exist $h,k \in G$ such that $hx = ky$. It follows that we can write $x = h^{-1}ky$, and hence that

\[Gx = \{ gx : g \in G \} = \{ gh^{-1}ky : g \in G \} = \{ gy : g \in G \} = Gy. \]

Thus, we see that the collection of all orbits of elements of $X$, denoted by

\[X/G \coloneqq \{ Gx : x \in X \}, \]

forms a partition of the set $X.$

Having introduced orbits and stabilizers side-by-side, we hope that the reader is beginning to suspect some sort of connection between these two notions. Indeed, such connections are provided by the following two results:

Theorem 1. (Orbit-Stabilizer Theorem)
TODO: Write up the Orbit-Stabilizer Theorem and its proof.

Theorem 2. (Burnside's Lemma)
TODO: Write up Burnside's Lemma and its proof.

2. Vector Spaces

Definition: Let $F$ be a field. A vector space over $F$ consists of a group $(V, +)$ together with a binary operation $\mu: F \times V \to V$ called scalar multiplication, here written as $(f, \vv) \mapsto f \cdot \vv,$ that satisfies the following axioms:

  1. $1 \cdot \vv = \vv$ for all $\vv \in V,$ where $1$ denotes the multiplicative identity element of $F.$
  2. $(ab) \cdot \vv = a \cdot (b \cdot \vv)$ for all $a,b \in F$ and $\vv \in V.$
  3. $(a + b) \cdot \vv = (a \cdot \vv) + (b \cdot \vv)$ for all $a,b \in F$ and $\vv \in V.$
  4. $a \cdot (\vv + \vw) = (a \cdot \vv) + (a \cdot \vw)$ for all $a \in F$ and $\vv, \vw \in V.$

Notation: Here, we adopt the convention (common among physicists and engineers) that elements of $V,$ referred to as vectors, are denoted by upright bold lowercase letters, while elements of $F,$ referred to as scalars, are denoted by italic lowercase letters. Accordingly, we refer to the group operation $+$ of $V$ as vector addition, and we will write $0$ and $\vo$ for the additive identity elements of $F$ and $V$ respectively.

Observe that axioms 1 and 2 are precisely the conditions for $\mu$ to define a group action $\mu: F^* \curvearrowright V,$ where $F^*$ denotes the multiplicative group of nonzero elements of $F.$ Thus, the scalar multiplication $\mu$ can be thought of as a “multiplicative field action” of $F$ on $V,$ which is required to distribute over addition in $F$ and $V$ (as stipulated in axioms 3 and 4).

Note that while we are writing the group operation of $V$ additively, we have not assumed that $V$ is abelian! Many textbook authors unnecessarily make this assumption, when in fact we can prove from axioms 1-4 that $V$ must be abelian.

Theorem: Let $V$ be a vector space over a field $F.$ Then $(V, +)$ is an abelian group, where $+$ denotes vector addition.

Proof: Let $\vv, \vw \in V.$ Observe that

\[(\vv + \vw) + (\vv + \vw) = (\vv + \vv) + (\vw + \vw) \]

since by axioms 1, 3, and 4, both the LHS and RHS are equal to the vector $(1 + 1) \cdot (\vv + \vw).$ By associativity, this implies

\[\vv + (\vw + \vv) + \vw = \vv + (\vv + \vw) + \vw \]

and by cancelling $\vv$ on the left and $\vw$ on the right, we obtain $\vw + \vv = \vv + \vw,$ as desired. QED

This shows that the requirement of a “multiplicative field action” in the definition of a vector space is quite restrictive no non-abelian group admits one!

Examples of vector spaces:

  1. Any field $F$ is trivially a vector space over itself. Simply take vector addition and scalar multiplication to be field addition and multiplication in $F$. The vector space axioms then follow from the field axioms.

  2. More generally, for any $n \in \N$ there is a natural vector space structure on $F^n$, the set of all $n$-tuples of elements of $F$, with vector addition and scalar multiplication defined by

    \[\mqty[x_1 \\ \vdots \\ x_n] + \mqty[y_1 \\ \vdots \\ y_n] \coloneqq \mqty[x_1 + y_1 \\ \vdots \\ x_n + y_n] \qquad c \cdot \mqty[x_1 \\ \vdots \\ x_n] \coloneqq \mqty[cx_1 \\ \vdots \\ cx_n]_{\text{\normalsize .}} \]

    We will later see that this is an example of a vector space of dimension $n,$ coinciding with the intuition that there are $n$ independent “degrees of freedom” given by modifying each component separately. The most familiar examples of vector spaces (as typically seen in undergraduate linear algebra) are obtained by taking $F = \R$ or $F = \C.$

    Note that for $n = 0$, the vector space $F^0$ obtained from this construction is simply the singleton set $\{(\ )\}$, where $(\ )$ is the empty tuple. This is called the trivial vector space over $F,$ and is the unique vector space having dimension $0$ and cardinality $1.$

  3. Even more generally, let $X$ be any set, and consider the set of all functions $\{f: X \to F\}$ endowed with pointwise addition and scalar multiplication. This is naturally a vector space in the same way as the previous example; indeed, the previous example is just the special case of a finite indexing set $X = \{1, \dots, n\}$.

    Note that this vector space does not, in general, have dimension equal to the cardinality of $X!$ This may seem counterintuitive, since a function $f: X \to F$ has one “degree of freedom” for each element of $X$. This is a subtlety in the way that dimension is defined for vector spaces, which we will return to later.

Definition: Let $V$ and $W$ be vector spaces over a common field $F$. A function $f: V \to W$ is called a linear transformation, linear map, or vector space homomorphism if it satisfies the following conditions:

  1. $f(\vv + \vw) = f(\vv) + f(\vw)$ for all $\vv, \vw \in V$.
  2. $f(c\vv) = cf(\vv)$ for all $\vv \in V$ and $c \in F$.

2.1. Vector Spaces over Subfields

Suppose $V$ is a vector space over a field $F$, and let $K$ be a subfield of $F$. Then by restricting the scalar multiplication function $\mu: F \times V \to V$ to $\mu|_K: K \times V \to V$, we obtain a vector space structure of $V$ over the smaller field $K$.

Definition: Let $V$ be a vector space over $F,$ and let $S$ be a subset of $V.$ A linear combination of elements of $S$ is a vector of the form

\[c_1 \cdot \vv_1 + c_2 \cdot \vv_2 + \cdots + c_n \cdot \vv_n \]

where $n$ is a natural number, $c_1, \dots, c_n \in F,$ and $\vv_1, \dots, \vv_n \in S.$ If $n = 0,$ then we define the empty linear combination to have the value $\vo \in V.$ The span of $S$ is the set of all linear combinations of elements of $S.$

If we wish to be specific about the field $F$ from which the coefficients $c_i$ are drawn, we can refer to linear combinations as $F$-linear combinations. For example, we might wish to take $\R$-linear combinations in a vector space over $\C.$ Note that our convention for the empty linear combination implies that the span of the empty set is the singleton $\{\vo\}.$

Definition: Let $V$ be a vector space over $F.$ A subset $S$ of $V$ is said to be linearly independent if no linear combination of elements of $S$ equals $\vo$ except the trivial linear combination. In other words, for every $n \in \mathbb{N}$ and every pair of sequences $c_1, \dots, c_n \in F$ and $\vv_1, \dots, \vv_n \in S,$ we have

\[c_1 \cdot \vv_1 + \cdots + c_n \cdot \vv_n = \vo \]

only when $c_1 = \cdots = c_n = 0.$ Otherwise, if some nontrivial linear combination gives $\vo,$ $S$ is said to be linearly dependent.

Note that any subset of $V$ containing $\vo$ is automatically linearly dependent. Moreover, the empty subset of any vector space is linearly independent, since the condition above is vacuously satisfied.

Definition: Let $V$ be a vector space over $F.$ A basis of $V$ is a subset $S$ of $V$ which is linearly independent and whose span is equal to $V.$

Lemma: Let $V$ be a vector space over $F.$ If a subset $S$ of $V$ is linearly independent, then for any vector $\vv \in V$ lying outside the span of $S,$ $S \cup \{\vv\}$ is linearly independent.

Proof: by contraposition. Suppose $S \cup \{\vv\}$ is linearly dependent. Then we can pick $n \in \mathbb{N},$ $c_1, \dots, c_n, c_{n+1} \in F,$ and $\vv_1, \dots, \vv_n \in V$ such that

\[c_1 \cdot \vv_1 + \cdots + c_n \cdot \vv_n + c_{n+1} \vv = \vo. \]

If $c_{n+1} = 0,$ then we have

\[c_1 \cdot \vv_1 + \cdots + c_n \cdot \vv_n + \vo = \vo \]

showing that $S$ is linearly dependent. Otherwise, if $c_{n+1} \ne 0,$ then we can write

\[-\frac{c_1}{c_{n+1}} \cdot \vv_1 - \cdots - -\frac{c_n}{c_{n+1}} \cdot \vv_n = \vv \]

showing that $V$ lies in the span of $S.$ QED

Theorem: Every vector space has a basis.

Proof: Let $V$ be a vector space over a field $F,$ and consider the poset $P$ of linearly independent subsets of $V$ ordered by inclusion. We will proceed by applying Zorn's lemma to $P,$ so we first verify that the hypotheses of Zorn's lemma hold. $P$ is clearly nonempty, since the empty set is vacuously a linearly independent subset of $V.$

Suppose that $\{C_i : i \in I\}$ is a chain in $P,$ indexed by some index set $I.$ Let $\mathscr{C} = \bigcup_{i \in I} C_i.$ Clearly, $\mathscr{C}$ is an upper bound of $\{C_i : i \in I\}$; we claim that it is also linearly independent. Indeed, if $\mathscr{C}$ were linearly dependent, then we could find a nontrivial linear combination of elements of $\mathscr{C}$ equal to zero. But the vectors $\vv_1, \dots, \vv_n \in \mathscr{C}$ involved in such a linear combination would all have to be present in $C_j$ for some $j \in I,$ since there are only a finite number of such vectors. Thus, $C_j$ would be linearly dependent, contradicting the assumption that $\{C_i : i \in I\}$ is a chain in $P.$

Zorn's lemma now guarantees the existence of a maximal element $B$ of $P.$ We claim that $B$ must be a basis of $V.$ Indeed, if the span of $B$ were not equal to $V,$ then by the previous lemma, we could enlarge $B$ by adding a vector $\vv \in V$ lying outside of its span. This would contradict the maximality of $B$; hence, $B$ is a basis of $V.$ QED

Lemma: If a vector space has a finite basis, then all of its bases are finite and have the same size.


Theorem: All bases of a given vector space have the same cardinality.


Definition: The dimension of a vector space is the cardinality of one (and hence all) of its bases.