12 2. Basic notions of representation theory
Let φ : V1 V2 be a nonzero homomorphism of representations.
(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
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