Read decompositions from characters

Try

A rotation of three labeled positions gives a representation of C₃ on ℂ³: the generator r sends e₀ to e₁, e₁ to e₂, and e₂ to e₀. Its matrix has trace zero, although the identity has trace three. Can these two trace measurements tell us which smaller representations occur?

Throughout this lesson G is a finite group and representations are finite-dimensional over ℂ. The character of V is the function χV(g)=tr ρV(g). Trace means the sum of diagonal matrix entries; it is independent of basis. An irreducible representation is nonzero and has no proper nonzero invariant subspace. The zero representation has character zero.

Put ζ=exp(2πi/3), so ζ³=1 and 1+ζ+ζ²=0. There are three one-dimensional representations of C₃, obtained by sending r to 1, ζ, or ζ². Read the following as a table of functions, not as three choices of basis for the same representation.

Add the first three rows entry by entry. Their sum is the last row. We will prove why this calculation establishes a decomposition, rather than assuming that equal traces always settle a matrix problem.

Predict

For functions f,h on G define the inner product, linear in its first argument, by ⟨f,h⟩=(1/|G|)Σg f(g) conjugate(h(g)). Predict the inner products of the first three rows. For example, ⟨χ₁,χ₂⟩=(1+ζ²+ζ)/3=0 and ⟨χ₁,χ₁⟩=1.

Does this make every integer-valued function a character? Does knowing χ(e), the dimension, determine a representation? Write down a proposed answer before continuing. The answer to both questions is no: integrality of values is different from integrality of irreducible multiplicities, and representations of equal dimension can have different characters.

Experiment

Construct the vectors vₖ=Σⱼ₌₀² ζ^(−kj)eⱼ for k=0,1,2. Reindex the sum after applying r to check rvₖ=ζᵏvₖ. Their eigenvalues are distinct, so these three nonzero vectors are independent. This gives a direct decomposition of the worked action into the three rows above.

Now change the group and action. Let the generator s of C₂ swap positions 1 and 2, swap positions 3 and 4, and fix position 5 in ℂ⁵. First calculate χ(e) and χ(s). Then solve χ=m₊(1,1)+m₋(1,−1), and exhibit the two invariant parts using coordinate vectors.

Check the changed action

The character is (5,1), so m₊=(5+1)/2=3 and m₋=(5−1)/2=2. The fixed part has basis e₁+e₂, e₃+e₄, e₅. The sign part has basis e₁−e₂, e₃−e₄. These dimensions add to five. The inner product calculation agrees with the direct construction; the proof below makes that agreement general.

Understand

What traces preserve

For h,g∈G, ρ(hgh⁻¹)=ρ(h)ρ(g)ρ(h)⁻¹, so trace invariance under similarity gives χ(hgh⁻¹)=χ(g). Thus a character is a class function, constant on each conjugacy class. A block diagonal matrix shows χ(V⊕W)=χV+χW, and χV(e)=dim V.

The averaged positive Hermitian inner product from the preceding lesson makes every ρ(g) unitary. In an orthonormal basis ρ(g⁻¹)=ρ(g)∗, where ∗ means conjugate transpose. Therefore χV(g⁻¹)=conjugate(χV(g)). This uses finite groups over ℂ, not an arbitrary assertion about traces of inverses.

Averaging on maps counts intertwiners

Let Homℂ(V,W) be the vector space of all linear maps V→W. Give it the action A_g(T)=ρW(g)TρV(g)⁻¹. Multiplication verifies A_g A_h=A_(gh). Its fixed vectors are exactly the intertwiners HomG(V,W), maps satisfying ρW(g)T=TρV(g) for every g.

Define P(T)=(1/|G|)Σg A_g(T). For any h, left multiplication permutes the summands, so A_hP(T)=P(T). Conversely P fixes every intertwiner. Hence P²=P and its image is HomG(V,W). Every vector T splits as P(T)+(T−P(T)), with the two summands in the image and kernel. In a basis adapted to this direct sum, P has identity and zero blocks. Consequently tr P=dim HomG(V,W).

For matrices A on W and B on V, the map T↦ATB has trace (tr A)(tr B). Indeed, in the rectangular matrix-unit basis Eᵢⱼ, the coefficient of Eᵢⱼ in AEᵢⱼB is AᵢᵢBⱼⱼ; summing these coefficients gives the product. Applying this identity to each A_g proves

dim HomG(V,W)=⟨χW,χV⟩.

The order matches our linear-first convention: the target character is first. Schur’s lemma now gives ⟨χᵢ,χⱼ⟩=1 for equivalent irreducibles and 0 for inequivalent ones. For equivalent irreducibles, the intertwiner space is one-dimensional because an isomorphism identifies it with the scalar endomorphisms. These are the row orthogonality relations.

Multiplicities and the regular action

Maschke’s theorem decomposes any V as a direct sum of irreducibles. Additivity and row orthogonality give mᵢ=⟨χV,χᵢ⟩ for the multiplicity of the i-th irreducible. Thus these numbers are nonnegative integers and do not depend on the chosen decomposition. Equal characters imply equivalent representations. Also ⟨χV,χV⟩=Σᵢmᵢ², so a nonzero representation is irreducible exactly when its character has squared norm one.

We must still show that there are only finitely many irreducibles and that they account for all class functions. The group algebra ℂ[G] is the vector space with basis e_g and multiplication e_g e_h=e_(gh), extended bilinearly. Its left regular representation sends e_h to e_(gh). Thus χreg(e)=|G| and χreg(g)=0 for g≠e: a nonidentity left translation fixes no basis element.

Let W be any irreducible representation and choose w≠0. The map e_g↦ρW(g)w, extended linearly, is equivariant. Its image is nonzero and invariant, hence is W. Its kernel has an invariant complement by Maschke, so W occurs in the regular representation. Decompose this finite-dimensional regular representation once: every irreducible must be equivalent to one of its finitely many constituents. Label the distinct classes V₁,…,Vᵣ and put dᵢ=dim Vᵢ. Then

mᵢ(reg)=⟨χreg,χᵢ⟩=dᵢ, and |G|=Σᵢdᵢ².

For C₃ the three displayed one-dimensional representations already contribute 1+1+1=3. There can be no others. This conclusion follows from the regular action, not merely from having found three examples.

Why the character table is square

Let c be the number of conjugacy classes. Class functions form a c-dimensional vector space, with indicator functions of classes as a basis. The r orthonormal irreducible characters give r≤c. To prove the reverse inequality, consider the center Z of ℂ[G].

Write z=Σg a_g e_g. Conjugating z by e_h permutes its coefficients according to g↦hgh⁻¹. Therefore z is central exactly when its coefficients are constant on conjugacy classes. The sums of basis elements over individual classes form a basis of Z, so dim Z=c.

Extend each representation linearly to ℂ[G]. A central z commutes with its action, so Schur’s lemma says that z acts on Vᵢ by a scalar λᵢ(z). The linear map z↦(λ₁(z),…,λᵣ(z)) from Z to ℂʳ is injective. To see this, if all scalars vanish, z acts as zero on every irreducible constituent of the regular representation, hence as zero on all of ℂ[G]. But left multiplication by z sends e_e to z itself. Thus z=0. This is where faithfulness of the regular action matters; an arbitrary representation would not suffice.

Injectivity gives c≤r. Therefore r=c, and the irreducible characters form an orthonormal basis of all class functions. This proof has not assumed a decomposition theorem for the group algebra beyond Maschke’s theorem for its regular representation.

Orthogonality of columns

Let Cⱼ be the conjugacy classes with representatives gⱼ. Form the square matrix U with entries Uᵢⱼ=√(|Cⱼ|/|G|)χᵢ(gⱼ). Row orthogonality states UU∗=I. A square matrix with a right inverse is invertible, so U∗U=I as well. Reading its column entries, then conjugating if needed, gives

Σᵢ χᵢ(gⱼ) conjugate(χᵢ(gₖ)) = 0 if j≠k, and = |G|/|Cⱼ| if j=k.

The orbit-stabilizer theorem for conjugation identifies |G|/|Cⱼ| with the order of the centralizer C_G(gⱼ), the subgroup commuting with gⱼ. Consequently columns for nonconjugate elements are orthogonal; for conjugate elements their common squared norm is the centralizer order. The class-size weights in the rows are essential. Unweighted row sums generally give the wrong answer.

Apply

For C₄ generated by r, its four one-dimensional characters have χₖ(rʲ)=i^(kj), k=0,1,2,3. Their dimensions already exhaust |C₄|=4. A proposed character has values (2,1,0,1) at (e,r,r²,r³). Calculate all four multiplicities, and decide whether such a representation exists. Do not accept it simply because all four displayed values are nonnegative integers.

Check the multiplicity obstruction

The multiplicities are (1,1/2,0,1/2). For example the k=1 coefficient is (2−i+i)/4=1/2. Multiplicities must be integers, so this function is not a character. By contrast, values (2,0,−2,0) have multiplicities (0,1,0,1), realized by the generator diag(i,−i). A class function can have negative values and still be a character.

The character method decides equivalence and multiplicities for finite-dimensional complex representations of a finite group. It does not, by itself, find a useful basis of each invariant summand. Nor does this proof establish analogous results in a field whose characteristic divides |G|: division by |G| and complete reducibility were essential steps.

Continue

Return to averaging and invariant complements for the decomposition mechanism, or intertwiners and Schur’s lemma for the scalar-action argument. Next, the complete S₃ example uses these proofs to assemble and verify a nonabelian character table.