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Representation Theory of Finite Groups

Tier: A (Advanced), reaching R. Builds on group theory and linear algebra. The subject turns abstract group structure into concrete linear algebra, and the payoff is that questions about groups become questions about matrices — where the tools are much sharper.

Attempt these from memory before reading.

  • Define a representation of a group GG on a vector space VV.
  • What does it mean for a representation to be irreducible?
  • State Schur’s lemma.
  • What is the character of a representation?
  • How many irreducible representations does a finite group have?

An abstract group is a set with a multiplication table. That table is complete and it is unusable: from the table alone, answering “is this group isomorphic to that one?” or “does this group have a subgroup of order 12?” is genuinely hard.

A representation is a homomorphism ρ:G→GL(V)\rho : G \to GL(V) — every group element becomes an invertible matrix, and the multiplication table becomes matrix multiplication. The point is not that matrices are familiar. The point is that a representation decomposes, and the pieces into which it decomposes are the irreducible representations. Group theory becomes a classification problem about building blocks.

The two results that make the subject work, both due to Frobenius and Schur around 1896–1901:

  • There are finitely many irreducible representations, as many as there are conjugacy classes.
  • Every representation is a direct sum of irreducibles.

So the irreducibles are atoms, and every representation has a unique decomposition into them. That is the same shape as the spectral theorem — a decomposition into indecomposable pieces — and the parallel is exact: for a compact group, the Peter–Weyl theorem is the spectral theorem applied to the regular representation.

Definition (F). A representation of GG on a complex vector space VV is a homomorphism ρ:G→GL(V)\rho : G \to GL(V). It is irreducible if the only GG-invariant subspaces are {0}\{0\} and VV.

Definition (F). The character of ρ\rho is χρ(g)=tr⁡(ρ(g))\chi_\rho(g) = \operatorname{tr}(\rho(g)).

The character is a class function: it is constant on conjugacy classes, because trace is invariant under conjugation. That is why characters are recorded in tables indexed by conjugacy classes — and why the number of irreducibles equals the number of conjugacy classes.

Schur’s lemma, and why everything follows

Section titled “Schur’s lemma, and why everything follows”

Theorem (U, Schur’s Lemma). Let V,WV, W be irreducible representations of GG and φ:V→W\varphi : V \to W a GG-linear map.

  1. If V≇WV \not\cong W then φ=0\varphi = 0.
  2. If V=WV = W then φ\varphi is a scalar multiple of the identity.

Proof idea. ker⁡φ\ker \varphi and im⁡φ\operatorname{im}\varphi are GG-invariant, so by irreducibility each is 00 or everything. For (1), φ\varphi cannot be injective and surjective at once or it would be an isomorphism. For (2), VV is finite-dimensional over C\mathbb{C} so φ\varphi has an eigenvalue λ\lambda; then φ−λI\varphi - \lambda I is GG-linear with non-trivial kernel, so by irreducibility it is zero. ■\blacksquare

Why this lemma carries the subject. Every major theorem — orthogonality of characters, the uniqueness of decomposition, the sum-of-squares formula — is proved by constructing a GG-linear map and applying Schur’s lemma. The proof technique is: build the averaging operator

φ=1∣G∣∑g∈Gρ(g) T ρ(g)−1,\varphi = \frac{1}{|G|} \sum_{g \in G} \rho(g)\, T\, \rho(g)^{-1},

which is GG-linear for any linear TT, then read off what Schur’s lemma says about it.

Worked example (U). For an irreducible VV and any v,w∈Vv, w \in V, apply the averaged rank-one operator T(x)=⟨x,w⟩vT(x) = \langle x, w\rangle v. Schur’s lemma gives φ=λI\varphi = \lambda I, and taking traces computes λ=⟨v,w⟩/dim⁡V\lambda = \langle v, w\rangle / \dim V. Rearranged:

1∣G∣∑g∈Gρii(g)‾ ρjj(g)=δijdim⁡V,\frac{1}{|G|}\sum_{g \in G} \overline{\rho_{ii}(g)}\,\rho_{jj}(g) = \frac{\delta_{ij}}{\dim V},

which is the first orthogonality relation for matrix coefficients. Every orthogonality relation in the subject comes from this one computation.

Characters and the orthogonality relations

Section titled “Characters and the orthogonality relations”

Definition. ⟨χ,ψ⟩=1∣G∣∑gχ(g)ψ(g)‾\langle \chi, \psi \rangle = \frac{1}{|G|}\sum_{g} \chi(g) \overline{\psi(g)}.

Theorem (U, orthogonality). The irreducible characters are orthonormal: ⟨χi,χj⟩=δij\langle \chi_i, \chi_j \rangle = \delta_{ij}. Every character is a non-negative integer combination of irreducible characters, and the multiplicity of χi\chi_i in VV is ⟨χV,χi⟩\langle \chi_V, \chi_i \rangle.

Worked example (U, the character table of S3S_3). S3S_3 has three conjugacy classes: ee, transpositions (3 of them), 3-cycles (2 of them). So there are three irreducibles. The trivial and sign representations give two; the third is found from 12+12+d32=∣S3∣=61^2 + 1^2 + d_3^2 = |S_3| = 6, so d3=2d_3 = 2:

ee(12)(123)
trivial111
sign1−1-11
standard20−1-1

The standard representation is the action on C3\mathbb{C}^3 modulo the diagonal, and it is where the symmetry of the three roots of a cubic lives — which is exactly how Galois theory connects to representation theory.

Exercise. Verify the orthogonality relations hold for this table, and compute ⟨χreg,χi⟩\langle \chi_{\text{reg}}, \chi_i \rangle for each irreducible, where χreg\chi_{\text{reg}} is the character of the regular representation.

The regular representation and the sum-of-squares theorem

Section titled “The regular representation and the sum-of-squares theorem”

Theorem (U). The regular representation of GG decomposes as

C[G]≅⨁iVi⊕di,\mathbb{C}[G] \cong \bigoplus_i V_i^{\oplus d_i},

with di=dim⁡Vid_i = \dim V_i. Taking dimensions:

∣G∣=∑idi2.|G| = \sum_i d_i^2.

Worked example (U). For S3S_3: 12+12+22=61^2 + 1^2 + 2^2 = 6 ✓. For the quaternion group Q8Q_8: four one-dimensional irreducibles and one two-dimensional, giving 1+1+1+1+4=81 + 1 + 1 + 1 + 4 = 8 ✓.

The sum-of-squares theorem is a hard constraint. It immediately rules out candidate character tables before any representation is constructed, and it is the first thing to check whenever a table is being built.

Two proofs. The first derives character orthogonality from the matrix-coefficient relation computed above; the second derives both the regular-representation decomposition and the sum-of-squares formula from it. Neither needs a new idea — both are Schur’s lemma applied to a different averaged operator, which is the point the page has been making.

Theorem (character orthogonality). For irreducible χi,χj\chi_i, \chi_j,

⟨χi,χj⟩=1∣G∣∑gχi(g)χj(g)‾=δij.\langle \chi_i, \chi_j \rangle = \frac{1}{|G|}\sum_g \chi_i(g)\overline{\chi_j(g)} = \delta_{ij}.

Proof. Sum the matrix-coefficient relation 1∣G∣∑gρi(g)ab‾ ρj(g)cd=δijδacδbddi\frac{1}{|G|}\sum_g \overline{\rho_i(g)_{ab}}\,\rho_j(g)_{cd} = \frac{\delta_{ij}\delta_{ac}\delta_{bd}}{d_i} over the diagonal indices a=ca = c and b=db = d. The left side becomes 1∣G∣∑gχi(g)‾ χj(g)\frac{1}{|G|}\sum_g \overline{\chi_i(g)}\,\chi_j(g), since a character is the trace and the trace is the sum of diagonal entries. The right side becomes δijdi∑a,bδab=δijdidi=δij\frac{\delta_{ij}}{d_i}\sum_{a,b}\delta_{ab} = \frac{\delta_{ij} d_i}{d_i} = \delta_{ij}. ■\blacksquare

One line, and it is worth being clear about why: a character is a sum of matrix coefficients, and the matrix-coefficient relation is an orthogonality statement for each coefficient separately, so summing the relation over diagonal indices is exactly the operation “take the inner product of the traces”. The converse direction also holds and is useful: any relation among characters gives a relation among representations, because ⟨χV,χi⟩\langle \chi_V, \chi_i\rangle is the multiplicity of ViV_i in VV.

Theorem (regular representation). C[G]≅⨁iVi⊕di\mathbb{C}[G] \cong \bigoplus_i V_i^{\oplus d_i}, and consequently ∣G∣=∑idi2|G| = \sum_i d_i^2.

Proof. The character of the regular representation is

χreg(g)={∣G∣g=e0g≠e,\chi_{\mathrm{reg}}(g) = \begin{cases} |G| & g = e \\ 0 & g \neq e, \end{cases}

because left multiplication by gg permutes GG with no fixed points unless g=eg = e. Decompose C[G]=⨁iVi⊕mi\mathbb{C}[G] = \bigoplus_i V_i^{\oplus m_i}, so χreg=∑imiχi\chi_{\mathrm{reg}} = \sum_i m_i \chi_i. Taking the inner product with χj\chi_j:

mj=⟨χreg,χj⟩=1∣G∣∑gχreg(g)χj(g)‾=1∣G∣∣G∣ χj(e)‾=χj(e)=dj.m_j = \langle \chi_{\mathrm{reg}}, \chi_j \rangle = \frac{1}{|G|}\sum_g \chi_{\mathrm{reg}}(g)\overline{\chi_j(g)} = \frac{1}{|G|}|G|\,\overline{\chi_j(e)} = \chi_j(e) = d_j.

Every multiplicity equals the dimension. Taking dimensions of both sides, ∣G∣=∑imidi=∑idi2|G| = \sum_i m_i d_i = \sum_i d_i^2. ■\blacksquare

The computation is again one line, and again it is Schur’s lemma doing the work — character orthogonality is Schur’s lemma summed over the group. What makes the result striking is that a statement with no visible input from analysis, the arithmetic identity ∣G∣=∑di2|G| = \sum d_i^2, comes out of an orthogonality relation. The number of conjugacy classes equals the number of irreducible representations, and summing the squares of the dimensions exhausts the group.

ClaimTrue?Counterexample
Characters determine the representationyes, over C\mathbb{C}false over R\mathbb{R}: the two complex conjugate 2-dimensional irreducibles of Q8Q_8 have the same real character
Irreducible implies one-dimensionalnothe standard representation of S3S_3 is 2-dimensional and irreducible
χ(g)\chi(g) determines ggnoχ(123)=χ(132)=−1\chi(123) = \chi(132) = -1 in S3S_3: characters are class functions
Subrepresentations always splitno, over R\mathbb{R} or Z\mathbb{Z}over C\mathbb{C} they do (complete reducibility); over fields of characteristic dividing $

The last row is the research frontier. Modular representation theory studies representations over fields whose characteristic divides ∣G∣|G|, where Maschke’s theorem fails, representations need not be semisimple, and the classification becomes vastly harder. The decomposition matrix relating ordinary and modular irreducibles is the central object, and Brauer theory is its framework.

  • Modular representation theory. Over a field of characteristic pp dividing ∣G∣|G|, complete reducibility fails. The subject becomes the study of indecomposable modules, Brauer characters, and decomposition matrices. The McKay correspondence connects the irreducible representations of a finite subgroup of SL(2,C)SL(2,\mathbb{C}) to the affine Dynkin diagrams; a bridge from representation theory to algebraic geometry that is still generating work.
  • Character degrees. The set of irreducible character degrees strongly constrains the group: GG is solvable iff every character degree is coprime to its prime factors… the precise statement is the Itô–Michler theorem, and the converse directions of several such results are recent.
  • Categorification. Khovanov homology categorifies the Jones polynomial; geometric representation theory categorifies the entire character theory. This is where representation theory meets topology.
ObjectLooks likeIs actually
χV=χW\chi_V = \chi_W implies V≅WV \cong Wtrue over C\mathbb{C}false over R\mathbb{R}
$d_i \midG$
Every group of order p2p^2 is abelianyesand hence has only 1-dimensional irreducibles
Irreducible characters are linearly independentyesthat is why the character table has no repeated rows
  • State Schur’s lemma in both parts.
  • Why are characters class functions?
  • State the sum-of-squares theorem and apply it to Q8Q_8.
  • What does Maschke’s theorem say, and where does it fail?
  • What is the McKay correspondence?

Mix these with material from Abstract Algebra, Linear Algebra, Spectral Theory and Measure Theory.

  1. (Group theory.) Show that every group of order p2p^2 is abelian, and deduce that all its irreducibles are 1-dimensional.
  2. (Linear algebra.) Prove that a finite group of unitary matrices is diagonalisable, using averaging over the group.
  3. (Spectral theory.) Show that the left-regular representation of a compact group extends to a decomposition of L2(G)L^2(G), and connect to the Peter–Weyl theorem.
  4. (Measure theory.) For a compact group with Haar measure, show the matrix coefficients of inequivalent irreducibles are orthogonal in L2(G)L^2(G).
  5. (Research.) Show that the number of irreducible representations of a finite abelian group equals its order, and reconcile this with the sum-of-squares theorem.