Build the full S₃ character table and decompose a new action

Try

Let S₃ permute three coordinate axes of C³: a permutation g sends the coordinate vector eᵢ to e indexed by g(i). The vector c=(1,1,1) is fixed by every permutation. The plane W consisting of vectors whose coordinates sum to zero is also invariant. Every vector x decomposes uniquely as its coordinate mean times c plus a vector in W.

Choose v₁=(1,−1,0) and v₂=(0,1,−1) as a basis of W. Our task is to describe all irreducible complex representations of S₃, then decompose an unfamiliar nine-dimensional action. We will compute matrices, prove irreducibility, derive characters and recover actual invariant subspaces; a list of character rows alone is not the whole explanation.

Predict

Write r=(123) and s=(12), with products applied right to left. Predict r(v₁), r(v₂), s(v₁) and s(v₂). Does an invariant plane necessarily contain an invariant line? A single complex matrix has an eigenvector, but its eigenline need not be preserved by the other group matrices.

Use invariant complements, Schur’s lemma and character orthogonality and completeness as prerequisites when needed. All representations here are finite-dimensional over C; the standard Hermitian product is linear in its first argument.

Experiment

Compute each image in coordinates relative to v₁,v₂. The coordinate columns give the matrices R=ρ(r) and S=ρ(s).

Input	Image under r	Image under s
v₁	v₂	−v₁
v₂	−v₁−v₂	v₁+v₂

Thus R=[[0,−1],[1,−1]] and S=[[−1,1],[0,1]]. Multiply them to verify R³=I, S²=I and SRS=R⁻¹=[[−1,1],[−1,0]]. These relations agree with the permutations. The construction by restricting the actual coordinate action already proves that this is a representation; the relation checks are an independent way to catch a sign or composition error.

The basis is not orthonormal. Its Gram matrix is B=[[2,−1],[−1,2]], so for coordinate columns x,y the inner product is xᵀB conjugate(y). Check RᵀBR=B and SᵀBS=B. These real matrices preserve the metric B even though RᵀR is not I. Changing a basis changes the matrix of the inner product as well as the matrices of the action.

Replace v₂ by v₁+v₂. The change-of-basis matrix is P=[[1,1],[0,1]], whose columns are the new basis written in the old one. Predict the new matrices as P⁻¹RP and P⁻¹SP and the new Gram matrix as PᵀBP. The characters must stay unchanged because trace is invariant under conjugation.

Understand

Prove the plane representation irreducible

The characteristic polynomial of R is X²+X+1. Its two distinct eigenvalues are ω and ω², where ω³=1 and ω≠1. Any invariant line for the whole group would be an eigenline of R. But R S=S R⁻¹ implies that if Rv=λv, then R(Sv)=λ⁻¹Sv. Hence S exchanges the two R-eigenlines. Neither is invariant under S. There is no nonzero proper invariant subspace of the two-dimensional W, proving irreducibility. This is a direct argument, independent of the character table we are about to derive.

Derive and exhaust the character table

The three conjugacy classes are identity, the three transpositions, and the two three-cycles. We already have three representations: the trivial line, the sign line, and the plane W. The first two are irreducible because they are one-dimensional; their characters differ on transpositions. The plane is irreducible by the preceding proof and differs in dimension.

For a permutation representation the trace is the number of fixed coordinate basis vectors. On C³ this gives the character (3,1,0). Since C³ is the direct sum of its trivial constant line and W, subtract (1,1,1) to obtain the standard character (2,0,−1). This also agrees with tr(I)=2, tr(S)=0 and tr(R)=−1.

Representation	Identity, size 1	Transpositions, size 3	Three-cycles, size 2
Trivial, denoted 1	1	1	1
Sign, denoted ε	1	−1	1
Standard plane W	2	0	−1

Weighted row inner products use the class sizes: ⟨χ,ψ⟩ = (χ(1)conj(ψ(1)) + 3χ(s)conj(ψ(s)) + 2χ(r)conj(ψ(r)))/6. For example ⟨χ_W,χ_W⟩=(4+0+2)/6=1, and ⟨χ_W,1⟩=(2+0−2)/6=0. Omitting class sizes produces wrong multiplicities.

The sum of squared dimensions is 1²+1²+2²=6=|S₃|. The regular-representation theorem says the squared dimensions of all inequivalent irreducibles sum to six. Every additional irreducible would contribute a positive square, which is impossible. Thus this list is exhaustive. We have not inferred completeness merely from finding three attractive examples.

The three-letter permutation representation is 1⊕W, with actual subspaces already exhibited. The regular representation has character (6,0,0) and decomposition 1⊕ε⊕W⊕W. The two copies of W are two invariant summands of the regular space, not a new four-dimensional irreducible.

Change the action: all nine ordered pairs

Let U have basis eᵢⱼ for all 1≤i,j≤3, including diagonal pairs. Define g(eᵢⱼ)=e indexed by g(i),g(j). An ordered pair is fixed exactly when both coordinates are fixed. If g fixes k letters, it fixes k² ordered pairs, so the character of U is (9,1,0).

Take its weighted inner product with each row. The multiplicities are (9+3)/6=2 for the trivial representation, (9−3)/6=1 for sign, and (18+0)/6=3 for W. Therefore U is equivalent to 2·1⊕ε⊕3·W. The dimensions check: 2+1+3·2=9. Character arithmetic predicts the result; now verify it using the vectors.

The diagonal pairs span an invariant three-dimensional space D equivalent to the original permutation representation. The off-diagonal pairs span an invariant six-dimensional space O. They have disjoint basis sets, so U=D⊕O. The action on off-diagonal pairs is free and transitive: there is exactly one permutation sending (1,2) to any specified pair of different letters, since the remaining image is forced. Thus O is a regular representation. This proves the same decomposition independently by orbit structure.

Locate the summands explicitly

In D the sum of e₁₁,e₂₂,e₃₃ spans the trivial line, and Σxᵢeᵢᵢ with Σxᵢ=0 gives a copy of W. In O the sum of all six basis vectors gives a second trivial line. The vector z=Σg∈S₃ sign(g)e indexed by g(1),g(2) is nonzero because its six basis vectors are distinct; reindexing shows h(z)=sign(h)z. It spans the sign line.

For x∈W define A(x)=Σi≠j xᵢeᵢⱼ and C(x)=Σi≠j xⱼeᵢⱼ. Both maps are intertwiners: applying a permutation relabels the same coordinate and pair indices. They are injective because a nonzero coordinate appears in an off-diagonal coefficient. Their images intersect only in zero. Indeed A(x)+C(y)=0 implies xᵢ+yⱼ=0 whenever i≠j. For each fixed i compare the other two j values; this forces all coordinates of y equal. Since y sums to zero over C, y=0, and then x=0. Thus their sum is a direct sum of two copies of W, of dimension four.

To justify that the two displayed lines add directly to those four dimensions, use Schur’s lemma. On A(W)⊕C(W), the coordinate projections onto its two irreducible W summands are intertwiners. A trivial or sign line contained there would project nontrivially into W, contradicting inequivalence. More generally an intersection with the sum of the trivial and sign lines, if nonzero, contains an irreducible summand by complete reducibility; it would be one of those lines and yield the same contradiction. Hence O is the direct sum of the two lines and A(W),C(W), whose dimensions total six. This makes all nine dimensions in U’s character decomposition concrete.

Apply

Try a different action: S₃ acts on the three unordered two-element subsets of its letters. Compute its character and decomposition. Construct an explicit equivariant bijection to the three letters. Then examine the candidate class function (3,0,0): is it the character of any complex representation of S₃? Check multiplicities instead of guessing from its positive value at identity.

Check the subset action and the rejected character

Send a two-element subset to its unique missing letter. This is a bijection and commutes with every permutation, so the subset representation is equivalent to the three-letter permutation representation. Its character is (3,1,0), and it decomposes as 1⊕W. The candidate (3,0,0) has inner product 3/6=1/2 with the trivial character, which cannot be a nonnegative integer multiplicity. Therefore it is not a character. A class function need not come from a representation.

For the changed basis in Experiment, P⁻¹RP=[[-1,−1],[1,0]], P⁻¹SP=[[-1,−1],[0,1]], and PᵀBP=[[2,1],[1,2]]. Traces are still −1 and zero. The new basis describes the same plane action with a different coordinate metric.

Continue

Rebuild the table from matrices and conjugacy classes, then decompose the ordered-pair action without consulting either worked method. Explain why matching characters proves equivalence, why class sizes enter the formula and how actual invariant subspaces realize the multiplicities. Use the saved tasks and mixed group assessment for synthesis. This is a finite-group complex-representation foundation; induction, modular representations and broader algebra projects remain further work.