Group Representations

Let G be a group. We say that G acts on a set X (on the left) if there is a set map G × X → X, sending (g, x) to g · x ∈ X, such that 1 · x = x and g · (h · x) = (gh) · x for all x ∈ X and g, h ∈ G.

Now fix a field F. A vector space V over F is called a G-module (or representation of G) if the group G acts on the set V, and if for each g ∈ G there is a linear transformation σ(g) : V → V such that g · x = σ(g)(x) for all x ∈ V. A trivial representation is a representation with g · x = x for all g ∈ G.

Let GL(V) denote the group of linear automorphisms of V; if V = F^{d} then GL(V) = GL_{d}(F). If V is a G-module then σ : G → GL(V) is a group homomorphism. Conversely, any group homomorphism σ : G ⟶ GL(V) makes V into a G-module. Some authors take this as the definition of representation.

A representation V of G is also the same thing as a module over the ring F G. Here the group ring of G is a vector space F G with basis G, made into a ring with the product (∑ α_{i}g_{i})(∑ β_{j}h_{j}) = ∑(α_{i}β_{j})(g_{i}h_{j}), α_{i}, β_{j} ∈ F and g_{i}, h_{j} ∈ G.

A G-map (= homomorphism of G-modules) is a linear transformation f : V → W commuting with the action of G in the sense that f(g · v) = g · f(v). Of course, this is the same thing as a homomorphism of modules over the ring F G.

Permutation representations. Let the set X be a basis of a vector space V. Any action of G on X can be extended linearly into an action of G on V; such a representation is called a permutation representation because G permutes the basis. Each matrix σ(g) consists of 0’s and 1’s. The regular representation is an example: V is the group ring F G, X = G and if v = ∑ a_{i}g_{i} then g · v = ∑ a_{i}(gg_{i}).

1-dimensional representations. A 1-dimensional representation (of G on F) is equivalent to a group map G⟶^{σ}F^{∗}. Since F^{∗} is an abelian group, the commutator subgroup [G, G] must map to 1, so the representation factors through G ⟶ G/[G, G]. If G has n elements each σ(g) must be an n^{th} root of unity, because g^{n} = 1 in G. The absence of n^{th} roots of unity in F can affect the existence of 1-dimensional representations.

Let C_{n} denote the cyclic group of order n, with generator θ. It follows that the 1-dimensional representations of C_{n} (over F) are in 1-1 correspondence with the set of n^{th} roots of unity ζ in F (take σ(θ) = ζ). The group C_{2} has two 1-dimensional representations: the trivial representation and the sign representation (θ · a = −a). The cyclic group C_{3} has three 1-dimensional representations if F = ℂ, but only one if F = ℝ.

Operations. Standard operations on vector spaces (⊕, ⊗, Λ^{∗}, etc.) also induce operations on G-modules. Let V = F^{m} and W = F^{n} be two representations. The direct sum V ⊕W = F^{m+n} is a representation with g ·(v +w) = (g ·v)+(g ·w), and the tensor product V ⊗W = F^{mn} is a representation with g·(v⊗w) = (g·v)⊗(g·w).

Let Λ^{d}V denote the d^{th} exterior product of V, i.e., the vector space of dimension \binom{m}{d} consisting of all alternating d-forms v_{1} ∧⋯∧v_{d} on V. The action of G on Λ^{d}V is given by the formula g · (x_{1} ∧ ⋯ ∧ x_{d}) = (g · x_{1}) ∧ ⋯ ∧ (g · x_{d}). For example, if d = m then under the usual identification of Λ^{m}F^{m} with F the action of g on F is multiplication by det(σ(g)).

Typeset by AMS-TEX

Definition. A nonzero G-module V is called irreducible (= a simple module) if no proper subspace is a G-submodule. V is called completely reducible (= semisimple) if it is a direct sum of irreducible G-modules.

Clearly, every 1-dimensional representation is irreducible. If dim(V) = 2, there is a simple test for irreducibility: V is irreducible if no vector v ≠ 0 in V is an eigenvalue for all of the 2 × 2 matrices σ(g), g ∈ G.

Here is a general test to see if V is irreducible. For every v ≠ 0, does the orbit of v (the set G · v = {g · v, g ∈ G}, which includes 1 · v = v) span V? If so, V is irreducible. If not, the span of G · v is a proper G-submodule.

Examples. 1) The regular representation of C_{2} = {1, θ} on the plane is given by θ(x, y) = (y, x). The two vectors (1, ±1) are eigenvectors so this representation is the direct sum F (1, 1) ⊕ F (1, −1).

2) The dihedral group D_{n} (n ≥ 3) is defined as the group of isometries in the plane fixing the regular n-gon; the 2-dimensional representation defining D_{n} is irreducible (as the reflections have different eigenspaces, or because each v and its rotate by 2π/n span the plane).

3) The quaternionic group Q = {±1, ±i, ±j, ±k} has an obvious 4-dimensional representation on the quaternions ℍ. (We take F = ℝ.) If v ≠ 0, I claim that {v, iv, jv, kv} is a basis of ℍ; this shows that ℍ is an irreducible representation of Q. To show this, suppose given a_{i} ∈ ℝ such that a_{1}v +a_{2}(iv)+a_{3}(jv)+a_{4}(kv) = 0. Multiplying on the right by v^{−1} yields a_{1} + a_{2}i + a_{3}j + a_{4}k = 0, so all the a_{i} = 0.

3) The symmetric group S_{4} acts on the regular tetrahedron in ℝ^{3} by permuting the 4 vertices. This extends by linearity to an action of S_{4} on ℝ^{3}, which is irreducible (exercise!). More generally, S_{n} acts on the regular n-simplex in ℝ^{n−1}, giving an irreducible (n − 1)-dimensional representation of S_{n}.

Schur’s Lemma. 1) V is irreducible ⇔ V ≅ (F G)/I for some maximal left ideal.

2) If V, W are irreducible, any nonzero G-map f : V → W is an isomorphism.

3) If V is irreducible, the ring Δ = End_{G}(V) of all G-maps V → V is a division algebra. (A division algebra is an F -algebra in which every non-zero element is a unit.) If F is algebraically closed then Δ = F (multiplication by scalars).

Proof. 1) Any choice of v ≠ 0 in V yields a nonzero G-map F G → V sending 1 to v. Its kernel I is a left ideal, so its image is F G/I. This is a nonzero Gsubmodule of V, and every ideal J containing I yields a submodule J/I of V. If V is irreducible we must have F G/I ≅ V with no ideals J containing I. 2) If W is irreducible and f ≠ 0, f(V) must be W and ker(f) ≠ V. When V is irreducible this forces ker(f) = 0, which means that f is an isomorphism. 3) Any nonzero G-map f : V → V must be an isomorphism by 2), in which case f^{−1} exists and is a G-map. This proves that every nonzero element f is invertible in Δ.

Examples with F = ℝ. The only (finite-dimensional) division algebras over ℝ are ℝ, ℂ and ℍ. (Over F = ℂ the only f.d. division algebra is ℂ itself.)

1) Consider the rotation representation of C_{3} on the plane ℝ^{2}. The 2×2 matrices commuting with this action are products of scaling by r and rotation by α:

( \begin{matrix}r cos(α) & r sin(α) \\ −r sin(α) & r cos(α) \end{matrix} ) = r cos(α) + ir sin(α), i = ( \begin{matrix}0 & 1 \\ −1 & 0 \end{matrix} )

These form a subring Δ of M_{2}(ℝ) isomorphic to ℂ.

2) For the canonical representation of the quaternion group Q on ℍ, we have Δ = ℍ. (Of course!)

Corollary to Schur’s Lemma. If W ⊂ V is a submodule and V is completely reducible, then V = W ⊕ W′ for some complementary submodule W′.

Proof. Write V = ⊕V_{α} with V_{α} irreducible. By Zorn’s lemma, there is a largest family {α_{i}} so that W ∩ ⊕V_{α_{i}} = 0; set W′ = ⊕V_{α_{i}}. If W ⊕ W′ isn’t V, it doesn’t contain some V_{β}; this would imply (W ⊕ W′) ∩ V_{β} = 0, leading to a contradiction.

Remark. The regular representation is never irreducible (unless G = 1). To see this, recall that the norm element of F G is the sum N = ∑ g of every element in G. Since gN = N for all g ∈ G, N generates a 1-dimensional submodule F · N of F G.

The next theorem states that F G is completely reducible (when 1/|G| exists in F), and that it contains every irreducible representation at least once. Therefore F G = F · N ⊕ W′ for some W′. In fact, since N^{2} = |G| · N, the element e = N/|G| is an idempotent of the ring F G and F G = W ⊕ W′ with W′ = F G(1 − e).

The hypothesis (that 1/|G| exists in F) fails only when F has characteristic p > 0 and p divides |G|. In this case, the regular representation is never completely reducible, because F · N ⊂ F G has no complement. (If F G = F · N ⊕ W′ then some nonzero multiple of N must be idempotent, which is impossible because N^{2} = |G| · N = 0.)

Maschke’s Theorem. If G is a finite group and \frac{1}{|G|} ∈ F, then:

1) Every representation of G is completely reducible.

2) There are only a finite number s of irreducible representations V_{i} (up to isomorphism), with V_{i} occuring n_{i} ≥ 1 times in the regular representation of G on F G.

We will write Δ_{i} for the division algebra End_{G}(V_{i}), and set d_{i} = dim_{F} (Δ_{i}).

3) The i^{th} irreducible representation has dimension d_{i}n_{i}. Hence

|G| = ∑_{i=1}^{s} d_{i}n_{i}^{2}.

4) F G is the product of the s matrix rings M_{n_{i}}(Δ_{i}). The projection F G → M_{n_{i}}(Δ_{i}) allows us to identify V_{i} with the M_{n_{i}}(Δ_{i})-module Δ_{i}^{n_{i}}.

Some explanation of parts 3) and 4) is in order. The matrix ring M_{n_{i}}(Δ_{i}) is the direct sum of its n_{i} columns, each being the irreducible representation V_{i} = Δ_{i}^{n_{i}} of dimension d_{i}n_{i} over F. Summing over i = 1, ..., s yields the decomposition of the |G|-dimensional representation F G into its irreducible components.

Corollary (Complex representations). Suppose that F = ℂ. Then

|G| = ∑ n_{i}^{2}, where n_{i} = dim(V_{i}).

If G is abelian then there are exactly |G| irreducible representations, all of them 1dimensional. Every representation is a direct sum of 1-dimensional representations.

Indeed, we must have Δ_{i} = ℂ and F G = M_{n_{1}}(ℂ)×⋯×M_{n_{s}}(ℂ). If G is abelian, F G is a commutative ring; this forces n_{1} = ⋯ = n_{s} = 1, or F G = Π_{i=1}^{s}ℂ.

Proposition. Let c denote the number of conjugacy classes of elements of G. If F = ℂ then there are c irreducible representations of G. In general, the center E_{i} of Δ_{i} is a finite field extension of F, and c = ∑_{i=1}^{s} dim_{F} (E_{i}) ≥ s.

The connection to Maschke’s theorem comes from the observation that the center of F G is ∏ E_{i}. Write C_{1}, …, C_{c} for the conjugacy classes of G. The c elements z_{j} = ∑{g ∈ C_{j}} are central elements of F G, and form a basis for the center of F G.

Examples. 1) If F = ℂ then C_{3} has three irreducible 1-dimensional representations. If F = ℝ then C_{3} has only two irreducible representations: the trivial representation V_{1} = ℝ and the rotation representation on the plane V_{2} = ℝ^{2}.

2) The dihedral group D_{2} = C_{2} × C_{2} is abelian, so it has 4 one-dimensional representations–even over ℝ. The regular representation F D_{2} is the sum of these 4 representations. Finding the irreducible representations of D_{3} and D_{5} is an exercise.

3) The dihedral group D_{4} has 8 elements, and D_{4}/[D_{4}, D_{4}] is C_{2} × C_{2}. Thus it has exactly 4 one-dimensional representations. We have already observed that D_{4} has a 2-dimensional irreducible representation V as its “birth certificate”. Since 8 = 4 · 1 + 2^{2}, this accounts for all the irreducible representations of D_{4}.

4) The quaternionic group Q has 8 elements and 5 conjugacy classes. Since Q/[Q, Q] = Q/{±1} = C_{2}×C_{2}, there are exactly 4 one-dimensional representations. Counting (8 = 4 + 4) shows there is exactly one other irreducible representation V_{5}, of dimension 2 or 4 depending on F. If F = ℝ, then V_{5} is the 4-dimensional representation of Q on ℍ; if F = ℂ then V_{5} is the 2-dimensional representation of Q on ℍ ≅ ℂ^{2} (and n_{5} = 2).

Exercises. 1) Consider the rotation representation of C_{3} on the complex plane ℂ^{2}. Write this as the direct sum of two 1-dimensional representations over F = ℂ.

2) Provide details for the sketch given above that the 2-dimensional representation of D_{n} is irreducible when n ≥ 3.

3) Describe all irreducible representations of D_{3} and D_{5} over ℝ and over ℂ. Hint: Find two actions of D_{5} on the regular pentagon.

4) Prove that the 3-dimensional representation of S_{4} arising from the action on the regular tetrahedron is irreducible.

5) Determine all irreducible complex representations of the alternating group A_{4} (12 elements). Hint. Use the fact that [A_{4}, A_{4}] has 4 elements to write down all group maps A_{4} → ℂ^{∗}. Then let G act on the set X = {(12)(34), (13)(24), (14)(23)} of elements of A_{4} by conjugation, and prove that F X is irreducible.

6) If V is irreducible and W is any 1-dimensional representation of G, show that the tensor product V ⊗ W is also an irreducible representation of G.

Young Tableaux. Let S_{n} denote the symmetric group on n elements. The number of conjugacy classes of S_{n} equals the number of unordered partitions of n; the unordered partition λ = {r_{1}, ..., r_{h}} corresponds to the congugacy class of (1, …, r_{1}) … (n+1−r_{h}, …, n). Since the order of the r_{i} doesn’t matter, we always assume that r_{1} ≥ r_{2} ≥ ... ≥ r_{h}. Each partition λ determines an arrangement of n empty boxes into h rows, the i^{th} row has r_{i} boxes; such an arrangement is called a Young Tableau of shape λ and size n. The corresponding irreducible representation S^{λ} of S_{n} is sometimes called the Specht module of λ. We shall write f^{λ} for dim(S^{λ}).

If we fill in the boxes of a Young tableau of shape λ with the numbers 1, ..., n we get a Young diagram D. We call D standard if a) the entries in every row are increasing, and b) the entries in every column are increasing. The number of standard Young diagrams of shape λ equals f^{λ} = dim S^{λ}.

There is a simple product formula for f^{λ}, called the hook formula. If (i, j) is a box in a Young Tableau, the corresponding hooklength h_{ij} is the number of boxes in the “hook” {(i, k), k ≥ j} ∪ {(k, j), k ≥ i} with vertex (i, j). The hook formula says that

dim(S^{λ}) = f^{λ} = \frac{n!}{∏_{(i,j)∈λ} h_{ij}} .

If R (resp. C) denotes the subgroup of S_{n} consisting of permutations which merely permute the entries in the rows (resp. in the columns) of D, then the Specht module may be described as S^{λ} = (F S_{n})f_{D} ⊆ F S_{n}, where f_{D} ∈ F S_{n} is the sum

f_{D} = ∑_{\begin{matrix}τ∈C \\ σ∈R \end{matrix}}(−1)^{τ} τσ.

Representations of S_{4}. The only partitions of n = 4 are {1, 1, 1, 1, }, {2, 1, 1}, {2, 2}, {3, 1} and {4}, corresponding to the 5 Young tableau of size 4. Therefore there are exactly 5 irreducible representations of S_{4}. The only way to add up to 24 using five squares is 24 = 1+1+4+9+9, so S_{4} has two irreducible 3-dimensional representations (corresponding to two actions of S_{4} on the regular tetrahedron), one irreducible 2-dimensional representation (S_{4} acts on the triangle in the plane by S_{4} → D_{3} ⊂ GL_{2}(F)) and two 1-dimensional representations (the trivial representation and the sign representation). Of course, the dimensions of these representations can also be found by the hook formula.

Characters of finite groups. For simplicity, we concentrate on representations of a finite group G over ℂ. The character χ_{V} of a representation V = ℂ^{n} is defined to be the set map χ_{V} : G → ℂ sending g to the trace of the matrix σ(g). This map is independent of the choice of basis for V, since the trace is independent of this choice. This also shows that if V and W are isomorphic representations then χ_{V} = χ_{W}. Note that χ_{V} determines the dimension of V, because χ_{V} (1) = trace(I) = dim(V). We will see that in fact χ_{V} completely determines V (over F = ℂ).

Examples. 1) Let V be the 2-dimensional rotation representation of the cyclic group C_{n}. Then χ_{V} (θ^{k}) = 2 cos(2πk/n) for all k.

2) The character of the regular representation V = ℂG is easy to work out. The matrix σ(g) consists of 0’s and 1’s, and the (i, i) entry is 1 exactly when g · g_{i} = g_{i} in G. This never happens when g ≠ 1, meaning that all diagonal entries are 0, and so the trace is 0. In conclusion, if g ≠ 1 then χ_{ℂG}(g) = 0.

3) The character of a 1-dimensional representation V is χ_{V} (g) = σ(g), simply because the trace of a 1 × 1 matrix (a) is a. These characters are not so interesting.

The characters χ_{1}, ..., χ_{s} of the irreducible representations V_{1}, ..., V_{s} are called the “irreducible” characters. Every character χ_{V} is a linear combination of the irreducible characters in the vector space ℂ^{G} of all set maps G → ℂ. To see this, write V as the sum V = V_{i_{1}} ⊕ ⋯ ⊕ V_{i_{r}} of irreducible representations. This puts the matrices σ(g) in block diagonal form, and we have χ_{V} (g) = χ_{i_{1}}(g) + ⋯ + χ_{i_{r}}(g) for all g ∈ G. In particular, the character of the regular representation is ∑ n_{i}χ_{i}, from which we deduce that for every g ≠ 1 we have ∑ n_{i}χ_{i}(g) = 0. (And of course ∑ n_{i}χ_{i}(1) = ∑ n_{i}^{2} = |G|.)

A class function on G is a function ϕ : G → ℂ which is constant on conjugacy classes, i.e., if g′ = hgh^{−1} then ϕ(g′) = ϕ(g). The characters χ_{V} are examples of class functions, because χ_{V} (g) and χ_{V} (g′) are the traces of the matrices σ(g) and σ(g′) = P σ(g)P^{−1}, P = σ(h). We choose representatives g_{1}, …, g_{c} of the c conjugacy classes of G; any class function (including χ_{V}) is then completely determined by its values on these c elements. Thus the class functions form a c-dimensional vector subspace of ℂ^{G}. We are going to show that the irreducible characters χ_{i} of G form a basis of this vector space. Since they belong to this subspace, and there are s = c of them, it suffices to show that they are linearly independent. This follows from the following result, whose proof we omit.

Orthogonality Relations. The irreducible characters are orthogonal (with respect to the usual hermitian inner product) in the vector space ℂ^{G}:

⟨χ_{i}|χ_{j}⟩ = ∑_{g∈G} χ_{i}(g)^{∗}χ_{j}(g) = \{ \begin{matrix}|G| & if i = j \\ 0 & if i ≠ j. \end{matrix}

Corollary. The irreducible characters {χ_{1}, ..., χ_{c}} form a basis for the vector space of class functions on G.

Thus every class function ϕ uniquely determines complex numbers a_{1}, ..., a_{c} such that ϕ(g) = ∑ a_{i}χ_{i}(g) for all g ∈ G. In fact, a_{i} = \frac{1}{|G|}⟨ϕ|χ_{i}⟩. In particular, the character χ_{V} of any representation V uniquely determines integers m_{i} such that

V ≅ \overbrace{V_{1} ⊕ … V_{1}}^{m_{1}} ⊕ \overbrace{V_{2} ⊕ … V_{2}}^{m_{2}} ⊕ ⋯ ⊕ \overbrace{V_{c} ⊕ … V_{c}}^{m_{c}} .

Character tables. The complex numbers χ_{i}(g_{j}) assemble to form a c × c matrix, called the character table of G. The above results state that the character table tells us almost everything about all representations of G. The Orthogonality Relations imply that the columns are linearly independent (being orthogonal). The character table is not quite a unitary matrix; it satisfies the relation A^{∗t}A = |G| · I instead.

• The character tables of C_{2} and C_{3} are ( \begin{matrix}1 & 1 \\ 1 & −1 \end{matrix} ) and ( \begin{matrix}1 & 1 & 1 \\ 1 & ω & ω^{2} \\ 1 & ω^{2} & ω \end{matrix} ), ω = e^{2πi/3}.

• For S_{3} there are 3 conjugacy classes, represented by: {1, (12), (123)}. The character table for S_{3} is

( \begin{matrix}1 & 1 & 1 \\ 1 & 1 & −1 \\ 2 & −1 & 0 \end{matrix} ) .