12 2. Basic notions of representation theory

Let φ : V1 → V2 be a nonzero homomorphism of representations.

Then:

(i) If V1 is irreducible, φ is injective.

(ii) If V2 is irreducible, φ is surjective.

Thus, if both V1 and V2 are irreducible, φ is an isomorphism.

Proof. (i) The kernel K of φ is a subrepresentation of V1. Since

φ = 0, this subrepresentation cannot be V1. So by irreducibility of V1

we have K = 0.

(ii) The image I of φ is a subrepresentation of V2. Since φ = 0,

this subrepresentation cannot be 0. So by irreducibility of V2 we have

I = V2.

Corollary 2.3.10 (Schur’s lemma for algebraically closed fields). Let

V be a finite dimensional irreducible representation of an algebra A

over an algebraically closed field k, and let φ : V → V be an inter-

twining operator. Then φ = λ · Id for some λ ∈ k (a scalar operator).

Remark 2.3.11. Note that this corollary is false over the field of

real numbers: it suﬃces to take A = C (regarded as an R-algebra)

and V = A.

Proof. Let λ be an eigenvalue of φ (a root of the characteristic poly-

nomial of φ). It exists since k is an algebraically closed field. Then

the operator φ−λ Id is an intertwining operator V → V , which is not

an isomorphism (since its determinant is zero). Thus by Proposition

2.3.9 this operator is zero, hence the result.

Corollary 2.3.12. Let A be a commutative algebra. Then every

irreducible finite dimensional representation V of A is 1-dimensional.

Remark 2.3.13. Note that a 1-dimensional representation of any

algebra is automatically irreducible.

Proof. Let V be irreducible. For any element a ∈ A, the operator

ρ(a) : V → V is an intertwining operator. Indeed,

ρ(a)ρ(b)v = ρ(ab)v = ρ(ba)v = ρ(b)ρ(a)v

(the second equality is true since the algebra is commutative). Thus,

by Schur’s lemma, ρ(a) is a scalar operator for any a ∈ A. Hence