Try
Permuting three sensor labels moves vectors as well as labels. Write e₁,e₂,e₃ for the coordinate vectors of V=C³. For a permutation g in S₃, define ρ(g)eⱼ equal to the coordinate vector indexed by g(j) and extend linearly. Thus r=(1 2 3) sends (x,y,z) to (z,x,y), while s=(1 2) sends it to (y,x,z). Products of permutations act right to left, so ρ(gh)=ρ(g)ρ(h).
Try applying each map to (1,1,1) and (1,−1,0). The first vector stays fixed. The second moves, but its coordinate sum stays zero. Predict whether every vector has a unique decomposition into a constant vector and a sum-zero vector.
Predict
A finite-dimensional complex representation of a finite group G is a homomorphism ρ:G→GL(V), where GL(V) denotes the invertible complex-linear maps on V. In particular ρ(e)=I and ρ(g⁻¹)=ρ(g)⁻¹. A subspace W is invariant when ρ(g)W⊆W for every g; inverses then give equality. A representation is irreducible when V is nonzero and its only invariant subspaces are 0 and V. The zero representation space is allowed, but is not called irreducible.
Two representations ρ on V and σ on U are equivalent if there is an invertible linear map T:V→U with Tρ(g)=σ(g)T for every g. This changes coordinates, not the group element. Individual matrices being similar separately is insufficient: the same T must work for all g.
An ordinary vector-space complement W′ with V=W⊕W′ need not be invariant. For our constant line L=C(1,1,1), the plane E of vectors with third coordinate zero is a complement. Yet r sends e₂ in E to e₃ outside E. Predict how a geometry preserved by the whole group could select a better complement.
Experiment
Use L and H=❴(x,y,z):x+y+z=0❵. They are invariant, their intersection is zero, and every v=(x,y,z) splits as m(1,1,1)+(v−m(1,1,1)), where m=(x+y+z)/3. The second summand lies in H. This proves V=L⊕H, including uniqueness. It does not yet prove H irreducible.
In the basis u=e₁−e₂, v=e₂−e₃ of H, record columns as images of basis vectors. Calculate one column before opening the table.
| Map | Image of u | Image of v | Matrix on H, given by rows |
|---|
| r | v | −u−v | (0,−1); (1,−1) |
| s | −u | u+v | (−1,1); (0,1) |
| Identity | u | v | (1,0); (0,1) |
Multiply these two small matrices to verify R³=I, S²=I and SRS=R⁻¹. Their basis is not orthonormal: the ordinary inner product gives Gram entries 2,−1;−1,2. The basis changes the coordinate description of lengths, although permutation of the three original coordinates preserves length. Change the vector to (4,−2,1): its constant component is (1,1,1), and its H component is (3,−3,0)=3u. Under r those components become (1,1,1) and 3v.
Understand
We now construct preserved geometry for every finite-group complex representation, not just permutation matrices. Recall the linear-first complex inner product and orthogonal complements. Choose any positive-definite Hermitian product h₀, linear in its first argument and conjugate-linear in its second. Define
h(x,y)=(1/|G|) Σ over g∈G of h₀(ρ(g)x,ρ(g)y).
The sum is finite. Linearity, conjugate-linearity and h(y,x)=conjugate(h(x,y)) follow term by term. If x≠0, the identity summand h₀(x,x) is positive and all other diagonal summands are nonnegative. Therefore h(x,x)>0. For k∈G, h(ρ(k)x,ρ(k)y) uses ρ(gk) in its summands; right multiplication g↦gk permutes G. Hence it equals h(x,y). This average is an inner product for which all group operators are unitary. It is not a claim that their matrices were unitary in the original metric.
Let W be invariant and define W⊥ using this new h. For x∈W⊥ and w∈W,
h(ρ(k)x,w)=h(x,ρ(k⁻¹)w)=0,
because ρ(k⁻¹)w is in W. Thus W⊥ is invariant. Finite-dimensional positive-definite orthogonal decomposition gives V=W⊕W⊥. Equivalently, the projection onto W along W⊥ commutes with every ρ(k): both summands are preserved. This is the invariant-complement step in Maschke’s theorem over C.
To finish the theorem, induct on d=dim V. At d=0 the direct sum is empty. For d>0, if V is irreducible we are done. Otherwise choose a nonzero proper invariant W. Its invariant orthogonal complement has dimension d−dim W, and both dimensions are strictly below d. Apply the induction hypothesis to both restrictions and combine the finite decompositions. Every finite-dimensional complex representation is therefore a direct sum of nonzero irreducible subrepresentations. This proves existence, not uniqueness of the summands as actual subspaces or of their multiplicities; later arguments address those distinctions.
Why the characteristic restriction matters: a complete failure example
Let p be any prime and work over Fₚ, the field of p elements. Let a generator of the cyclic group Cₚ act on Fₚ² by U with rows (1,1),(0,1). Write U=I+N, where N²=0. Induction gives Uʲ=I+jN for every nonnegative integer j. Thus Uᵖ=I in characteristic p, so this defines a representation. It is nontrivial, and indeed has order p since jN≠0 for 1≤j<p.
The line W=Fₚe₁ is invariant. Any complementary line has a spanning vector (a,b) with b≠0. If that line were invariant, U(a,b)=(a+b,b) would equal λ(a,b). The second coordinate forces λ=1; the first then forces b=0, a contradiction. No invariant complement exists. This argument also works when p=2: −1=1 changes no step, and b is still nonzero. Division by |Cₚ| is impossible in Fₚ; the complex averaging proof cannot be imported into this setting. Failure here does not mean every representation in characteristic p fails to split.
Apply
Change the complex example to the order-two action on C² generated by A with rows (1,2),(0,−1). Verify A²=I. The line W=Ce₁ is invariant, but Ce₂ is not. Starting from h₀(x,y)=x₁ conjugate(y₁)+x₂ conjugate(y₂), compute the averaged h and find W⊥. Predict whether its spanning vector has positive or negative first coordinate.
Compare your averaged metric and complement
Since G has two elements, h=(h₀+h₀(A·,A·))/2. In linear-first coordinates h(x,y)=xᵀG conjugate(y), the real symmetric matrix G has rows (1,1),(1,3). Therefore h(x,e₁)=x₁+x₂, and W⊥=C(−1,1). Indeed A(−1,1)=(1,−1), the negative of that vector. Its squared h-length is 2; the metric is positive since h(x,x)=|x₁+x₂|²+2|x₂|². The columns e₁ and (−1,1) give one simultaneous basis in which the identity and A become diagonal. This particular action has two one-dimensional invariant summands; it does not assert that every finite group representation splits into lines.
Explain exactly where finite dimension and finiteness of G entered the general proof. Then repair the claim “a subspace complement always inherits an action”: replace an arbitrary complement by the orthogonal complement in an averaged invariant metric. Finally compare the F₂ example above with A over C. A numerical similarity of displayed entries cannot identify matrices over different fields.
Continue
Return to group actions for the action law and to invariant subspaces and blocks for the linear decomposition language. Here an action preserves the vector-space operations as well as the set of vectors.
Continue with intertwiners and Schur’s lemma. An invariant complement proves that decomposition exists. Intertwiners will explain which irreducible pieces are equivalent and why a complex irreducible endomorphism is scalar. These proof tasks are public practice requiring mathematical review, not automatic certification of understanding.
