Section 18.1-2. In the next 2-3 lectures we will have a lightning introduction to representations of finite groups. For any vector space V over a field F , denote by L(V ) the algebra of all linear operators on V , and by GL(V ) the group of invertible linear operators. Note that if dim(V ) = n, then L(V ) = Matn(F ) is isomorphic to the algebra of n×n matrices over F , and GL(V ) = GLn(F ) to its multiplicative group. Definition 1. A linear representation of a set X in a vector space V is a map φ: X ! L(V ), where L(V ) is the set of linear operators on V , the space V is called the space of the rep- resentation. We will often denote the representation of X by (V; φ) or simply by V if the homomorphism φ is clear from the context. If X has any additional structures, we require that the map φ is a homomorphism. For example, a linear representation of a group G is a homomorphism φ: G ! GL(V ). Definition 2. A morphism of representations φ: X ! L(V ) and : X ! L(U) is a linear map T : V ! U, such that (x) ◦ T = T ◦ φ(x) for all x 2 X. In other words, T makes the following diagram commutative φ(x) V / V T T (x) U / U An invertible morphism of two representation is called an isomorphism, and two representa- tions are called isomorphic (or equivalent) if there exists an isomorphism between them. Example. (1) A representation of a one-element set in a vector space V is simply a linear operator on V . Two such representations are isomorphic if two operators have the same Jordan Normal Form. (2) For any vector space, identity maps id: GL(V ) ! GL(V ) and id: L(V ) ! L(V ) give tautological representations of the group GL(V ) and the algebra L(V ) respectively. (3) For each representation φ: X ! L(V ) one can define its dual representation φ∗ : X ! L(V ∗); φ∗(x)(α)(v) = α(φ(x)(v)) for all x 2 X, v 2 V , and α 2 V ∗. (4) The trivial representation of a group G is a homomorphism φ: G ! GL(V ), where V is a 1-dimensional vector space over the field F , and φ(g) = 1 2 GL(V ) = F × for all g 2 G. n 2 −1 (5) Recall the dihedral group Dn = r; s j r = s = 1; rs = sr . The following map cos (2π=n) − sin (2π=n) 0 1 r 7! ; s 7! sin (2π=n) cos (2π=n) 1 0 defines a 2-dimensional representation φ: Dn ! GL(2; R). 1 2 (6) The following map gives a representation of a symmetric group S4: 00 1 0 01 01 0 0 01 01 0 0 01 B1 0 0 0C B0 0 1 0C B0 1 0 0C (12) 7! B C ; (23) 7! B C ; (34) 7! B C : @0 0 1 0A @0 1 0 0A @0 0 0 1A 0 0 0 1 0 0 0 1 0 0 1 0 (7) There is a natural representations of Sn on the space of polynomials in n variables: (σf)(x1; : : : ; xn) = f(xσ(1); : : : ; xσ(n)): Note, that symmetric polynomials stay invariant under this action. (8) Any Galois group Gal(K=F ) comes with its representation in K over the field F . Definition 3. An invariant subspace of a representation φ: X ! L(V ) is a subspace W ⊂ V invariant under φ(x) for all x 2 X. An invariant subspace gives rise to a subrepresentation φW : X ! L(W ); φW (x) = φ(x)jW ; and a factor representation φV=W : X ! L(V=W ); φV=W (x)(v + W ) = φ(x)(v): Remark 4. If one chooses a basis of the space V of a representation φ: X ! L(V ) in such a way that the first k vectors span an invariant subspace W ⊂ V , then any operator φ(x) is written in matrix form as φ (x) ∗ φ(x) = W 0 φV=W (x) Example. n (1) Let Sn act on F by permuting basis vectors, as in example (4) above. Then the subspace fa; a; : : : ; a) j a 2 F g ⊂ F n is a 1-dimensional invariant subspace. There is a complementary invariant subspace of dimension n − 1 defined by f(a1; : : : ; an) j a1 + ··· + an = 0g : Note that F n is isomorphic to the direct sum of these two subspaces. d (2) For any positive integer d, the subspace Poln of polynomials in n variables of degree d in the space Poln of all polynomials of n variables gives a subrepresentation of Sn. Definition 5. (1) A representation is irreducible (or simple) if it (or rather the underlying vector space) has no invariant subspaces other than 0 and itself, otherwise it is reducible. (2) A representation is indecomposable if it can not be written as a direct sum of two invariant subrepresentations. (3) A representation is completely reducible if it can be written as a direct sum of irre- ducible representations. Example. The representation of a 1-element set given by a 2-dimensional space V = F 2 and an operator a b A = 0 c is reducible, since the subspace he1i ⊂ V is invariant, however it is indecomposable unless b = 0. If b = 0, the representation V is completly reducible: V = he1i ⊕ he2i. Exercise 6. If T : V ! U is a morphism of representations of X, then ker(T ) ⊂ V and im(T ) ⊂ U are invariant subspaces. 3 Corollary 7. If V and U are irreducible representations of X, any morphism T : V ! U is either 0 or an isomorphism. Proof. Since V is irreducible, the subrepresentation ker(T ) is either 0, in which case T is injective, or coincides with the whole V , in which case T = 0. Since U is irreducible, we must now have im(T ) = U unless T = 0, which forces T to be an isomorphism. Corollary 8 (Schur's lemma). Any endomorphism T : V ! V of an irreducible represen- tation V of X over an algebraically closed field is a scalar, that is T = λ · Id for some λ 2 F . Proof. Let T 2 EndX (V ) = HomX (V; V ) be an endomorphism of V . Then for any λ 2 F , so is T −λ·Id. Choosing λ to be an eigenvalue of T (which we can always do over an algebraically closed field) we see that ker(T − λ · Id) is a subrepresentation of X in V . Therefore, since V is irreducible, we must have T = λ · Id. Corollary 9. Let φ: X ! L(V ) and : X ! L(U) be a pair of irreducible representations over an algebraically closed field. Then any morphisms T;S : V ! U differ by a scalar factor. Proof. If one of the two morphisms is zero, the statement is obvious. Otherwise, both mor- phisms are isomorphisms, hence ST −1 is an endomorphism of an irreducible representation −1 U which implies ST = λ Id. Corollary 10. Any irreducible representation of an abelian group over an algebraically closed field is 1-dimensional. Proof. If G is abelian and φ: G ! GL(V ) is a representation of G, operators φ(g) commute for all g 2 G, hence any of them can be viewed as an endomorphism of V . If V is irre- ducible, By Schur's lemma, all of them must be scalar, therefore any subspace is invariant and representation V can be irreducible if and only if dim(V ) = 1. 2 Example. The 2-dimensional plane R is an irreducible representation of the abelian group SO(2; R) of plane rotations cos θ − sin θ G = θ 2 R : sin θ cos θ However, over the field of complex numbers, we see that cos θ − sin θ eiθ 0 ∼ : sin θ cos θ 0 e−iθ Proposition 11. Every subrepresentation and factor representation of a completely reducible representation is completely reducible. Proof. Let V be a completely reducible representation, and U be its subrepresentation. 0 0 Then for any U1 ⊂ U, there exists an invarinat subspace V ⊂ V such that V ' U1 ⊕ V . 0 Setting U2 = V \ U we see that U2 is invariant and U ' U1 ⊕ U2. Now, let π : V ! V=W be a canonical projection to a factor representation, and U1 ⊂ V=W −1 be the subrepresentation. Then V1 = π (U1) is an invariant subspace of V , which con- tains W . Now, we have V = V1 ⊕ V2 for some invariant subspace V2, which implies that V=W ' U1 ⊕ U2 where U2 = π(V2) is an invariant subspace in V=W . Proposition 12. If a representation φ: X ! L(V ) is completely reducible, then V can be written as a direct sum of minimal (nonzero) irreducible subspaces. Conversely, if V can be written as a sum of minimal invariant subspaces V = V1 + ··· + Vn, then V is completely reducible. 4 Proof. The first statement is immediate: choose any minimal invariant subspace V1 ⊂ V and write V = V1 ⊕V2, continue the same process for V2. Now, let V = V1 +···+Vn for some minimal invariant subspaces Vi ⊂ V , and U ⊂ V be any invariant subspace. Then U \ Vj is either 0 or coincides with Vj for any j = 1; : : : ; n. Setting 0 X U = Vj j:U\Vj =? 0 we get V = U ⊕ U . Definition 13. A sum of representations φ: X ! L(V ) and : X ! L(U) is a representa- tion φ + : X ! V ⊕ U; (φ + )(x) = (φ(x); (x)): Remark 14.