Representations and complete reducibility

Written by GPT-6.1 Sol (OpenAI) in Codex, Ultra setting, October 2026. Self-checked by the AI that wrote it. Public domain (CC0).

Elementary prerequisite proofs and free-source comparisons by GPT-6 Astra (OpenAI), Ultra, in Codex. These contributions were self-checked by that AI; no human review is claimed.

A symmetry can permute coordinates, rotate a plane, or act on a space of functions. A representation puts these actions into one language: each group element becomes an invertible linear map. The first problem is to find smaller spaces that all these maps preserve. The next problem is to decide whether those spaces can be separated into independent pieces.

For a finite group over the complex numbers, both questions have a clean answer. Every representation splits into irreducible pieces. Their isomorphism classes and multiplicities are determined by the representation, although the individual summands need not be. We will prove this using averaging and Schur's lemma, and see exactly what changes over the real numbers or in positive characteristic.

The preparation courses are Linear Algebra, Abstract Algebra I, and Abstract Algebra II. We use the definitions of vector spaces, groups, fields and algebras. The elementary facts about complements, eigenvectors, inner products and quotient groups used in the arguments are proved in the opening section. The complex case also uses the programme's fundamental theorem of algebra, Lemma 5.1; its proof uses compactness of a disk and polar form.

Accessible treatments are [Gruson–Serganova] and [Etingof et al.]. The AI Integrated Stacks Project supplies the related general algebra [Stacks], preserving credit to the Stacks project authors and identifying its AI additions separately. It is not an edition reviewed or endorsed by the Stacks project's maintainers. All the representation-theoretic arguments needed here are given below.

Throughout, \(G\) is finite and representations are finite-dimensional over a field \(k\). An irreducible representation is always nonzero. Complex inner products are linear in their first variable.

0. Linear and algebraic tools

The point of these tools is to distinguish an ordinary complement, which is easy to choose, from an invariant complement, which the group action may obstruct.

For the full foundational development, read From bases to projections, developed from Jim Hefferon's Linear Algebra, and Orthonormal bases and orthogonal projections, developed alongside John M. Erdman's treatment. Both are included in the programme with their proofs, examples, solutions and component terms. The short arguments here recall exactly what this lesson uses; the foundational lessons also treat the broader scope stated there.

Lemma 0.1 (bases and complements). In a finite-dimensional vector space every independent list extends to a basis. Every subspace has a basis and an ordinary linear complement. A linear map on a nonzero finite-dimensional space over an algebraically closed field has an eigenvalue. Eigenvectors for distinct eigenvalues are independent.

Proof. Start with a spanning list \(b_1,\ldots,b_n\). An independent list has at most \(n\) entries: insert its first vector into the spanning list by expressing it in that list and replacing a vector with a nonzero coefficient. The new list still spans. Repeat with the next independent vector; it is outside the span of the previously inserted vectors, so a nonzero coefficient of a remaining vector exists. After \(n\) replacements there is no room for another independent vector. Consequently repeatedly adjoining a vector outside the current span stops after at most \(n\) steps. Adjoining inside a subspace first gives its basis, and then adjoining in the whole space extends that basis. The span of the added vectors is a complement, since independence makes the intersection zero. This also proves that a proper subspace has smaller dimension.

For completeness, the determinant test used for eigenvalues follows from the permutation formula. Subtracting a multiple of one row from another leaves the determinant unchanged, exchanging rows changes its sign, and multiplying a row by a nonzero scalar multiplies the determinant by that scalar. Successive elimination either produces \(n\) nonzero diagonal pivots, in which case back substitution gives a unique solution to every linear system and the determinant is nonzero, or leaves a free variable, in which case the homogeneous system has a nonzero solution and the determinant is zero. Thus \(\det A=0\) exactly when \(A\) has nonzero kernel. The permutation formula makes \(\det(tI-A)\) monic of degree \(n\). When \(n>0\), algebraic closedness supplies a root \(\lambda\); the determinant test supplies a nonzero vector in \(\ker(A-\lambda I)\).

Finally, if eigenvectors with distinct eigenvalues were dependent, choose a relation of shortest positive length. Its coefficients are nonzero and its length is at least two. Applying \(A-\lambda_m I\) to a relation involving eigenvalues \(\lambda_1,\ldots,\lambda_m\) removes the last vector and leaves every other coefficient nonzero. This is a shorter relation, a contradiction. \(\square\)

Lemma 0.2 (orthogonal complements). A finite-dimensional positive definite Hermitian space has an orthonormal basis. For every subspace \(U\), it decomposes as \(U\oplus U^\perp\).

Proof. Choose a basis beginning with a basis of \(U\), by Lemma 0.1. Starting with its first vector, subtract at step \(j\) the vector \[ \sum_{i<j}\langle b_j,e_i\rangle e_i \] from \(b_j\). The result is orthogonal to each preceding \(e_i\); it is nonzero, since otherwise \(b_j\) would be in the span of its predecessors. Divide it by the positive square root of its squared norm. This constructs an orthonormal basis with the same successive spans. The vectors after the basis of \(U\) span \(U^\perp\): in the orthonormal expansion of a vector orthogonal to \(U\), its first coefficients vanish. Every vector is the sum of its two corresponding coordinate blocks, and the intersection is zero by positive definiteness. The zero-dimensional cases use the empty basis. \(\square\)

Lemma 0.3 (quotients and scalar characteristic). If \(N\) is a normal subgroup of \(G\), the cosets form a group under \((gN)(hN)=ghN\). Every subgroup of \(\mathbb Z\) is \(d\mathbb Z\) for a unique integer \(d\geq0\). The characteristic of a field is zero or prime; an integer \(m\) is nonzero in a field of characteristic \(p>0\) exactly when \(p\nmid m\).

Proof. Replacing \(g,h\) by \(gn,hn'\), with \(n,n'\in N\), changes their product to \(gh(h^{-1}nh)n'\). Normality puts the last factor in \(N\), so multiplication of cosets is well-defined. Associativity descends from \(G\); the identity is \(N\) and the inverse of \(gN\) is \(g^{-1}N\). The map \(g\mapsto gN\) preserves products and has kernel \(N\).

For a nonzero subgroup \(H\subseteq\mathbb Z\), choose its least positive element \(d\). Division with remainder writes any \(h\in H\) as \(qd+r\), with \(0\leq r<d\). Then \(r\in H\), so minimality forces \(r=0\). Conversely every multiple of \(d\) belongs to \(H\). The zero subgroup is \(0\mathbb Z\), and its nonzero companions have their generator determined by the least positive element.

Apply this to the kernel of \(\mathbb Z\to k\), \(m\mapsto m1_k\). If that kernel is zero, the characteristic is zero. Otherwise its positive generator \(d\) is not 1. A factorization \(d=ab\) with \(1<a,b<d\) would give two nonzero field elements \(a1_k,b1_k\) whose product is zero, impossible since each nonzero field element is invertible. Thus \(d\) is prime. The kernel description gives the final assertion. \(\square\)

We will also use the abelianization \(G/[G,G]\). Here \([G,G]\) is the subgroup generated by all commutators \(ghg^{-1}h^{-1}\). Conjugation sends each commutator to a commutator, so this subgroup is normal. In the quotient the images of every pair commute. Any homomorphism to an abelian group kills each commutator and therefore this subgroup, and descends uniquely to the quotient by the coset formula. These observations prove the universal property used below.

1. Turning symmetry into linear algebra

A representation on \(V\) is a homomorphism

Thus \(\rho(gh)=\rho(g)\rho(h)\), \(\rho(1)=1_V\), and \(\rho(g^{-1})=\rho(g)^{-1}\). We often write \(gv\) for \(\rho(g)v\). Its dimension is \(\dim_k V\). It is faithful when \(\rho\) is injective.

Choosing a basis gives matrices for every \(\rho(g)\). Changing that basis conjugates all the matrices by the same invertible matrix. The representation is the action itself; a matrix list records it in chosen coordinates.

Maps and invariant spaces

An intertwiner \(f:V\to W\) is a linear map satisfying

The vector space of these maps is \(\operatorname{Hom}_G(V,W)\). An invertible intertwiner is an isomorphism of representations. The algebra \(\operatorname{End}_G(V)\) consists of the operators commuting with every \(\rho(g)\), with multiplication given by composition.

A subspace \(U\subseteq V\) is invariant if \(gU\subseteq U\) for every \(g\). Applying the same condition to \(g^{-1}\) gives \(gU=U\). Restriction therefore gives a representation on \(U\). There is also a quotient representation

This is well-defined because \(v-v'\in U\) implies \(gv-gv'\in U\). The quotient map is an intertwiner. Kernels and images of intertwiners are invariant: if \(f(v)=0\), then \(f(gv)=gf(v)=0\); and \(gf(v)=f(gv)\) belongs to the image.

The direct sum has action \(g(v,w)=(gv,gw)\). An internal decomposition \(V=U\oplus W\) into invariant spaces gives block diagonal matrices in a basis assembled from bases of \(U\) and \(W\). An invariant subspace alone gives only a block triangular form. The off-diagonal block is the obstruction to finding an invariant complement.

A nonzero representation is irreducible if its only invariant subspaces are \(0\) and itself. It is completely reducible if it is a direct sum of irreducibles. The zero representation is the empty direct sum.

A supply of examples

The trivial representation on \(k\) has \(gv=v\). Any homomorphism \(\lambda:G\to k^\times\) gives a one-dimensional representation by \(gv=\lambda(g)v\), and every one-dimensional representation arises this way after choosing a nonzero basis vector. A commutator \(ghg^{-1}h^{-1}\) has image \(1\), since \(k^\times\) is abelian. Hence these homomorphisms are exactly the homomorphisms from \(G^{\mathrm{ab}}=G/[G,G]\) to \(k^\times\). We call them one-dimensional characters.

For \(S_n\), the determinant of a permutation matrix gives the sign representation: even permutations act by \(1\), odd permutations by \(-1\). In characteristic \(2\) these scalars coincide, so sign becomes trivial.

If \(X\) is a finite \(G\)-set, let \(k[X]\) have basis \((e_x)_{x\in X}\), with \(ge_x=e_{gx}\). The action law on \(X\) proves the representation law. In coordinates indexed by \(X\),

The inverse is necessary: the coordinate at \(x\) comes from the basis vector moved to \(x\). Taking \(X=G\) with left multiplication gives the left regular representation.

For the natural action of \(S_n\) on \(\mathbb C^n\), set

Both spaces are invariant, \(L\) is trivial, and

proves \(\mathbb C^n=L\oplus A\). The representation on \(A\) is called the standard representation. Its irreducibility for \(n\geq2\) will be established in Exercise 1.

For a geometric example, let

where \(D_n\) has order \(2n\). On \(\mathbb R^2\), take

Rotation by \(n\theta\) is the identity, \(S^2=I\), and multiplying the matrices gives \(SRS=R^{-1}\). Thus \(r\mapsto R\), \(s\mapsto S\) defines a representation. The same matrices act on \(\mathbb R^2\otimes_{\mathbb R}\mathbb C\cong\mathbb C^2\). On this complex plane, \(R\) has the two distinct eigenlines spanned by \((1,-i)\) and \((1,i)\). The reflection exchanges those lines. Any invariant line for the whole group would have to be one of the rotation eigenlines, but neither is reflection-stable. The complex representation of \(D_n\) is therefore irreducible.

The group algebra viewpoint

The group algebra \(k[G]\) has basis \((g)_{g\in G}\), with multiplication extended bilinearly from the group law:

Associativity and the identity follow from those of \(G\). A representation extends to a unital left module by

The representation law verifies the module law. Conversely, on a unital \(k[G]\)-module, multiplication by \(g\) is invertible with inverse multiplication by \(g^{-1}\). These operators form a representation. The two constructions undo each other. Intertwiners are exactly module homomorphisms, and invariant subspaces are exactly submodules. In particular, an irreducible representation is a simple \(k[G]\)-module. The regular representation is \(k[G]\) acting on itself on the left.

2. Averaging separates invariant pieces

An arbitrary inner product need not respect a given complex action. A finite group lets us repair that by averaging all its translates.

Proposition 2.1 (invariant inner products). Every complex representation of \(G\) preserves a positive definite Hermitian inner product. With respect to an orthonormal basis for that inner product, all its matrices are unitary.

Proof. Choose any positive definite Hermitian inner product \((\ ,\ )_0\), linear in its first variable. Define

Each summand is linear in \(v\), conjugate-linear in \(w\), and Hermitian. If \(v\neq0\), all \(gv\) are nonzero, so each \((gv,gv)_0\) is positive. Their average is positive. For \(h\in G\),

because \(g\mapsto gh\) permutes \(G\). Choose an orthonormal basis. Preservation of its standard Hermitian form says that each matrix \(M_g\) satisfies \(M_g^*M_g=I\). This proves the last assertion. \(\square\)

If \(U\) is invariant, its orthogonal complement for this form is invariant too. Indeed, for \(v\in U^\perp\) and \(u\in U\),

The linear algebra decomposition \(V=U\oplus U^\perp\) is therefore a decomposition of representations. This already proves splitting over \(\mathbb C\). To obtain the result over other fields, average a linear map instead of a positive form.

The invariant-inner-product construction is also treated in [Etingof et al.].

Lemma 2.2 (averaging maps). Suppose the scalar \(|G|\cdot1_k\) is invertible. For representations \(V,W\), the operator

is a linear projection from \(\operatorname{Hom}_k(V,W)\) onto \(\operatorname{Hom}_G(V,W)\).

Proof. Conjugating the summands by \(h\), in the sense of composing on the left with \(\rho_W(h)\) and on the right with \(\rho_V(h^{-1})\), replaces \(g\) by \(hg\). Thus \(\rho_W(h)\mathcal R(f)=\mathcal R(f)\rho_V(h)\). Its image consists of intertwiners. If \(f\) is already an intertwiner, every summand equals \(f\), so \(\mathcal R(f)=f\). These two statements give \(\mathcal R^2=\mathcal R\) and the asserted image. \(\square\)

Theorem 2.3 (Maschke). Suppose \(k\) has characteristic zero, or characteristic \(p>0\) with \(p\nmid |G|\). Every invariant subspace of a representation over \(k\) has an invariant complement. Every such representation is completely reducible.

Proof. Let \(U\subseteq V\) be invariant. Extend a basis of \(U\) to a basis of \(V\), and let \(p_0:V\to V\) fix the basis vectors in \(U\) and send the additional ones to zero. It is a projection onto \(U\). The scalar \(|G|\) is invertible under the hypothesis. Put

Lemma 2.2 makes \(P\) an intertwiner. Each summand maps \(V\) into \(U\), because \(U\) is invariant. For \(u\in U\), it sends \(u\) to \(u\), because \(g^{-1}u\in U\) and \(p_0\) fixes \(U\). Consequently \(P(V)\subseteq U\) and \(P|_U=1_U\). It follows that \(P(V)=U\) and \(P^2=P\).

The kernel is invariant because \(P\) intertwines. For any \(v\in V\),

The intersection \(U\cap\ker P\) is zero since \(P\) fixes \(U\). Thus \(V=U\oplus\ker P\).

For complete reducibility, induct on \(\dim V\). The zero case is the empty sum. A nonzero \(V\) has an invariant subspace \(S\) of least positive dimension, which is irreducible. Split it off by the first part. The complement has smaller dimension, so induction decomposes it into irreducibles. \(\square\)

The averaging construction is also treated in the freely available draft [Gruson–Serganova] and in [Etingof et al.].

The proof uses more than equivariance of \(P\). An average of arbitrary idempotent operators need not be idempotent. For example, let the generator of \(C_2\) swap the coordinates of \(\mathbb C^2\), and let \(p_0=\operatorname{diag}(1,0)\). Its conjugate is \(\operatorname{diag}(0,1)\), so its group average is \(\tfrac12I\), whose square is \(\tfrac14I\). Here idempotence follows because the average has image in the invariant space \(U\) and fixes \(U\).

Lemma 2.4 (the meaning of complete reducibility). For a finite-dimensional representation, complete reducibility is equivalent to every invariant subspace having an invariant complement.

Proof. The reverse implication follows by the dimension induction just used; that argument needs no assumption on \(G\) or \(k\) once complements exist. For the forward implication, write \(V=S_1\oplus\cdots\oplus S_r\) with all \(S_j\) irreducible, and fix an invariant subspace \(U\). Choose a subset \(J\subseteq\{1,\ldots,r\}\) maximal with

If \(i\notin J\), maximality gives a nonzero vector in \(U\cap(T\oplus S_i)\). Its \(S_i\)-component is nonzero, since \(U\cap T=0\). Hence \(S_i\cap(U+T)\neq0\). This intersection is invariant, so irreducibility gives \(S_i\subseteq U+T\). This holds for all missing summands, proving \(V=U+T\). By construction the sum is direct. \(\square\)

Where the characteristic enters

Let \(G=C_p=\langle t\rangle\) over \(\mathbb F_p\), and let

Here \(N^2=0\), so \(J^m=I+mN\) for every nonnegative integer \(m\), by induction. In particular \(J^p=I\), so this defines a representation. The line \(U=\mathbb F_p e_1\) is invariant.

Any complementary line is spanned by \(v=ae_1+be_2\) with \(b\neq0\). If it were invariant, \(Jv=cv\) for a scalar \(c\). Comparing second coordinates gives \(c=1\). Comparing first coordinates then gives \(a+b=a\), contradicting \(b\neq0\). Thus \(U\) has no invariant complement.

The quotient \(V/U\) and the subrepresentation \(U\) are both trivial. The representation nevertheless fails to be their direct sum. Knowing the subspace and quotient does not determine whether an extension splits. In this example \(|G|=p=0\) in the field, so the averaging division is unavailable.

3. Irreducible pieces and the maps between them

Theorem 3.1 (Schur's lemma). Let \(S,T\) be irreducible representations over \(k\).

1. Every intertwiner \(S\to T\) is zero or an isomorphism.
2. The algebra \(\operatorname{End}_G(S)\) is a division algebra over \(k\).
3. If \(k\) is algebraically closed, \(\operatorname{End}_G(S)=k\,1_S\).

Proof. For a nonzero intertwiner \(f:S\to T\), its invariant kernel cannot be \(S\), so it is zero. Its invariant image is nonzero, so it is all of \(T\). Hence \(f\) is bijective. Its inverse also intertwines: the identity \(f\rho_S(g)=\rho_T(g)f\) gives \(\rho_S(g)f^{-1}=f^{-1}\rho_T(g)\). In particular every nonzero endomorphism is invertible in the endomorphism algebra, proving the first two assertions.

For the third, take \(a\in\operatorname{End}_G(S)\). Since \(S\neq0\) is finite-dimensional and \(k\) is algebraically closed, \(a\) has an eigenvalue \(\lambda\in k\). The endomorphism \(a-\lambda1_S\) has nonzero kernel and therefore cannot be an isomorphism. The first assertion makes it zero. Thus \(a=\lambda1_S\). Scalar maps always intertwine, giving equality. \(\square\)

Schur's lemma does not require the characteristic hypothesis in Maschke's theorem. It controls maps between representations already known to be irreducible.

For further reading on Schur's lemma, see [Etingof et al.] and the simple-module treatment in [Stacks].

Abelian groups

Corollary 3.2. Every irreducible representation of a finite abelian group over an algebraically closed field is one-dimensional. In particular this holds when the characteristic does not divide the group order.

Proof. If \(G\) is abelian, each operator \(\rho(g)\) commutes with every \(\rho(h)\), so it lies in \(\operatorname{End}_G(S)\). Theorem 3.1 makes each operator scalar. Every one-dimensional subspace of \(S\) is then invariant. Irreducibility forces \(\dim S=1\). Conversely, a one-dimensional representation is irreducible because a line has no nonzero proper subspace. \(\square\)

The dimension assertion survives in the excluded characteristics; complete reducibility may fail there. These are different conclusions.

The commutative-algebra formulation is also treated in [Etingof et al.].

A real irreducible that splits over the complex numbers

Let the generator of \(C_3\) act on \(\mathbb R^2\) by

Its characteristic polynomial is \(x^2+x+1\), which has no real root. An invariant real line would give a real eigenvector, so this representation is irreducible. On \(\mathbb C^2\), however, the vectors \((1,-i)\) and \((1,i)\) have eigenvalues \(\zeta=e^{2\pi i/3}\) and \(\zeta^{-1}\). They span two invariant lines, giving a decomposition into two distinct one-dimensional representations.

Write

The commuting algebra is \(\{aI+bK:a,b\in\mathbb R\}\cong\mathbb C\); the calculation is in Exercise 2. Thus the scalar conclusion of Schur's lemma would be false over \(\mathbb R\), even though the division-algebra conclusion remains true.

4. What is unique in a decomposition?

From now through this section, assume \(k\) is algebraically closed and \(|G|\) is invertible in \(k\). Maschke gives

where the \(S_i\) are pairwise nonisomorphic irreducibles and \(m_i>0\).

Corollary 4.1 (multiplicities). For each irreducible \(S\), its multiplicity in \(V\) is

Consequently the irreducible isomorphism classes occurring in \(V\), and their multiplicities, are independent of the decomposition.

Proof. Taking components of a map into a finite direct sum gives

Schur's lemma says that a term is zero unless \(S\cong S_i\). If they are isomorphic, composing with one fixed isomorphism identifies that term with \(\operatorname{End}_G(S)=k\). Thus its dimension is exactly the number of copies of \(S\). The left-hand space depends only on \(S\) and \(V\), proving independence. \(\square\)

Over a field that is not algebraically closed, the same argument, for a completely reducible \(V\), gives instead

For the real rotation representation in the preceding example, \(S=V\) has multiplicity \(1\), but \(\operatorname{Hom}_G(S,V)\) has real dimension \(2\). The distinction is essential.

The individual irreducible subspaces are usually not determined. For instance, in \(S\oplus S\), every graph \(\{(v,av):v\in S\}\), \(a\in k\), is another subrepresentation isomorphic to \(S\). We need a larger subspace that includes all copies of the same type.

Define the isotypic component of type \(S\) by

This is an invariant subspace defined without selecting a decomposition. A nonzero \(f\) is injective by the kernel argument in Schur's lemma, so its image is a copy of \(S\). Conversely, an isomorphism from \(S\) onto any copy of \(S\) in \(V\), followed by inclusion, is such a map.

Proposition 4.2 (isotypic decomposition). In any decomposition of \(V\) into irreducibles, \(V[S]\) is exactly the direct sum of the summands isomorphic to \(S\). Hence

with only finitely many nonzero components. The subspaces \(V[S]\) and the projections onto them along the other components are canonical. Every intertwiner \(a:V\to V'\) sends \(V[S]\) into \(V'[S]\).

Proof. Fix a decomposition \(V=\bigoplus_j T_j\), with component projections \(\pi_j\). For \(f:S\to V\), the map \(\pi_jf\) intertwines. If \(T_j\not\cong S\), Schur's lemma makes it zero. Thus every such image lies in the sum of the summands of type \(S\). Each summand of that type is itself the image of an intertwiner from \(S\), proving the reverse inclusion. Grouping the summands by type proves the direct sum formula.

Because \(V[S]\) was defined intrinsically, all these grouped subspaces are independent of the decomposition. The direct sum determines a unique linear projection \(e_S\) with image \(V[S]\) and kernel the sum of the other components. Both subspaces are invariant, so the projection intertwines. Finally \(af:S\to V'\) intertwines for every \(f:S\to V\); summing its images proves \(a(V[S])\subseteq V'[S]\). \(\square\)

Thus uniqueness has two levels. Multiplicities are unique numbers. Isotypic components are unique subspaces. Splitting a component into its individual copies requires choices. The projection \(e_S\) commutes with every equivariant endomorphism, since all such endomorphisms preserve every component.

5. A cyclic regular representation worked out

Let \(G=C_n=\langle t\rangle\) over \(\mathbb C\), let \(\omega=e^{2\pi i/n}\), and use the basis \(e_j=t^j\), \(0\leq j<n\), of \(\mathbb C[C_n]\). Indices are read modulo \(n\). The regular action is \(te_j=e_{j+1}\).

For \(0\leq a<n\), define

Reindexing with \(\ell=j+1\) gives

Each \(v_a\) is nonzero, since its \(e_0\)-coefficient is \(1\). The \(n\) eigenvalues \(\omega^a\) are distinct, so the eigenvectors are independent. They therefore form a basis, and

Every one-dimensional character is on this list: its value at \(t\) has \(n\)th power \(1\), and that value determines it. Corollary 3.2 shows that these are all the irreducibles. Each occurs once in the regular representation.

For \(n=4\), this gives eigenvalues \(1,i,-1,-i\). In the order \(e_0,e_1,e_2,e_3\), the corresponding vectors are

This example prepares for “Fourier analysis on finite abelian groups,” which develops character orthogonality, inversion, and duality. Here the calculation serves to identify the irreducible pieces of a concrete representation. For arbitrary finite groups, Characters and the orthogonality relations will provide numerical methods for finding multiplicities.

6. Exercises with complete solutions

Try each exercise before reading its solution. The last two use Schur's lemma to turn structural questions into questions about scalars and projections.

Exercise 1 — Easy: the standard representation

Show that the permutation representation of \(S_n\) on \(\mathbb C^n\) is the direct sum of the trivial representation and the standard representation. Prove that the standard representation is irreducible for \(n\geq2\).

Solution. Coordinate permutations preserve both \(L=\mathbb C(1,\ldots,1)\) and \(A=\{z:\sum_i z_i=0\}\). The decomposition in Section 1 shows their sum is all of \(\mathbb C^n\). If \(c(1,\ldots,1)\in A\), then \(nc=0\), so \(c=0\). Hence the sum is direct and \(L\) is trivial.

Let \(0\neq U\subseteq A\) be invariant, and choose \(z\in U\setminus\{0\}\). Not all its coordinates are equal: otherwise the sum-zero condition would make it zero. Choose \(i,j\) with \(z_i\neq z_j\). Then

Thus \(e_i-e_j\in U\). Permuting indices gives every \(e_a-e_b\in U\). The vectors \(e_a-e_n\), \(1\leq a<n\), span \(A\), since any \(z\in A\) equals \(\sum_{a<n}z_a(e_a-e_n)\). Therefore \(U=A\), proving irreducibility. For \(n=2\), \(A\) is the sign line. For \(n=1\), \(A=0\), so it is not irreducible under our convention.

Exercise 2 — Easy: a real commuting algebra

Compute \(\operatorname{End}_{C_3}(\mathbb R^2)\) for the rotation representation in Section 3, and identify it as a real algebra.

Solution. Since \(R=-\tfrac12I+\tfrac{\sqrt3}{2}K\), a real matrix commutes with \(R\) exactly when it commutes with \(K\). For \(B=\begin{pmatrix}u&v\\w&x\end{pmatrix}\),

Equality says \(x=u\) and \(w=-v\). Thus \(B=aI+bK\) for real \(a,b\). Since \(K^2=-I\), the map \(a+bi\mapsto aI+bK\) is a bijective real-linear map preserving multiplication and the identity. It is an algebra isomorphism from \(\mathbb C\) onto the commuting algebra. Explicitly, for \((a,b)\neq(0,0)\),

This directly verifies the division-algebra conclusion.

Exercise 3 — Medium: splitting and characteristic

Show that Maschke's conclusion fails for \(C_p\) over \(\mathbb F_p\), including \(p=2\). Explain why it holds for every finite group over every field of characteristic zero.

Solution. Use \(J=I+N\) from Section 2. The formula \(J^p=I+pN=I\) holds also for \(p=2\). A line complementary to \(\mathbb F_p e_1\) has a generator \(ae_1+be_2\), \(b\neq0\). Invariance would give \(Jv=cv\); the second coordinate forces \(c=1\), and the first forces \(b=0\), a contradiction. Lemma 2.4 therefore also shows that this representation is not completely reducible.

In a characteristic-zero field, a positive integer times \(1_k\) is nonzero. Every nonzero element of a field is invertible, so \(|G|^{-1}\) exists for every finite \(G\). Theorem 2.3 applies. This argument requires neither algebraic closure nor an inner product.

Exercise 4 — Medium: a restriction on the centre

Prove that a finite group admitting a faithful irreducible complex representation has cyclic centre.

Solution. Let \(\rho:G\to\operatorname{GL}(V)\) be faithful and irreducible. For \(z\in Z(G)\), \(\rho(z)\) commutes with every group operator. Schur's lemma gives \(\rho(z)=\lambda(z)I\). The map \(\lambda:Z(G)\to\mathbb C^\times\) is a homomorphism, and it is injective: \(\lambda(z)=1\) implies \(\rho(z)=I\), hence \(z=1\) by faithfulness.

Let \(H=\lambda(Z(G))\). It is finite. Choose a positive integer \(N\) divisible by the orders of all its elements. Then every element of \(H\) is an \(N\)th root of unity, so \(H\subseteq\langle e^{2\pi i/N}\rangle\). To see explicitly how the elementary cyclic-group fact applies, let

This is a subgroup of \(\mathbb Z\) containing \(N\mathbb Z\). By the subgroup theorem for \(\mathbb Z\), \(B=d\mathbb Z\) for some positive \(d\). Consequently \(H\) is generated by \(e^{2\pi id/N}\). The injection identifies \(Z(G)\) with this cyclic group. Faithfulness is used precisely in making the scalar homomorphism injective.

Exercise 5 — Hard: an isotypic projection

Let \(V\) be a complex representation and \(W\subseteq V\) irreducible. Show that its isotypic component is the sum of all subrepresentations isomorphic to \(W\). Construct an ordinary linear projection whose group average has that entire component as its image.

Solution. Write \(S\) for the sum of all copies of \(W\) in \(V\). It is invariant. Every nonzero map \(W\to V\) is injective, and every copy is the image of such a map. Thus \(S=\sum_f\operatorname{im}f\). In fact, if \(f_1,\ldots,f_m\) is a basis of \(\operatorname{Hom}_G(W,V)\), then \(S=\sum_{a=1}^m\operatorname{im}f_a\): the image of any linear combination of these maps lies in that sum.

Choose an irreducible decomposition \(V=\bigoplus_j T_j\). For every \(f:W\to V\), its component in \(T_j\not\cong W\) is zero by Schur's lemma. Hence \(S\) is contained in the sum of the \(T_j\) isomorphic to \(W\). Each such summand is a copy of \(W\), so equality holds. This identifies \(S\) with the isotypic component and shows it is independent of the chosen decomposition.

Extend a basis of \(S\) to a basis of \(V\), and let \(q_0\) be the projection onto \(S\) given by that basis. Set

As in the proof of Maschke, \(Q\) intertwines, maps into \(S\), and fixes \(S\). Therefore \(Q^2=Q\) and \(\operatorname{im}Q=S\). Its kernel is invariant. More precisely, if \(T_j\not\cong W\), each component of \(Q|_{T_j}:T_j\to S\) in a decomposition of \(S\) into copies of \(W\) vanishes by Schur's lemma. Thus \(Q\) is zero on every other isotypic component. It is the canonical isotypic projection from Proposition 4.2, regardless of the initial basis choice.

A projection onto \(W\) alone would not do this when \(S\neq W\). Every conjugate of that projection still maps into the invariant space \(W\) and fixes \(W\); its average has image exactly \(W\). To project onto the whole isotypic component, the initial projection must have that whole component as its image.

7. Programme connections

Lemmas 0.1–0.3 supply the elementary algebraic and inner-product inputs to the proofs. Algebraic closedness is part of the stated condition on \(k\) in Section 3. For the complex examples, the programme's fundamental theorem of algebra, Lemma 5.1, supplies that condition. The cyclic examples use complex exponential and polar coordinates; the full human-authored treatment is Jiří Lebl, Complex exponential and trigonometric functions, included with its source and attribution in this programme.

Two general algebra results provide context, but none of our finite-group proofs depends on them.

• For a finite-dimensional unital \(k\)-algebra \(A\), a simple module exists if \(A\neq0\); every nonzero unital \(A\)-module contains a simple submodule; every simple \(A\)-module is finite-dimensional over \(k\); and its endomorphism ring is a division ring Stacks, Tag 0746. Here “simple” means nonzero with no nonzero proper submodule.
• The Brauer group of an algebraically closed field is zero Stacks, Tag 074M. Its proof establishes that a finite-dimensional central division algebra over such a field is the field itself. The general theory of central simple algebras is outside this lesson.

“Groups with operators” treats complete reducibility and isotypic components for modules in greater generality. “Semisimple rings and Wedderburn's theorem” treats the structure of general semisimple rings. Here averaging gives the finite-group splitting theorem directly, and Schur's lemma supplies exactly the uniqueness statements needed for representations.

References

• [Stacks] The Stacks project authors, Brauer groups, Tag 0746 and Tag 074M, in the AI Integrated Stacks Project.
• [Gruson–Serganova] Caroline Gruson and Vera Serganova, A sentimental journey through representation theory: from finite groups to quivers (via algebras), seven-chapter author draft hosted at Berkeley.
• [Etingof et al.] Pavel Etingof, Oleg Golberg, Sebastian Hensel, Tiankai Liu, Alex Schwendner, Dmitry Vaintrob and Elena Yudovina, Introduction to representation theory, freely available lecture notes, arXiv:0901.0827.