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Over an algebraically closed field, every endomorphism of an irreducible representation is scalar

Statement

Let $k$ be an algebraically closed field, let $G$ be a group, and let $V$ be an irreducible representation of $G$ over $k$ . Then every $T \in {End}_{G} (V)$ is of the form $T = λ {id}_{V}$ for some $λ \in k$ .

Facts & Assumptions

Given: An algebraically closed field $k$ , a group $G$ , an irreducible representation $V$ of $G$ over $k$ , and an endomorphism $T \in {End}_{G} (V)$ .

[L1]

Every finite-dimensional endomorphism over an algebraically closed field is triangularisable (Every finite-dimensional endomorphism over an algebraically closed field is triangularisable).

[L2]

For an irreducible representation, every nonzero $G$ -endomorphism is an isomorphism and ${End}_{G} (V)$ is a division ring (Schur's lemma for irreducible representations: a nonzero intertwiner is an isomorphism, and ${End}_{G} (V)$ is a division ring).

Proof

technique · direct

1.1

The representation $V$ is nonzero because it is irreducible, so [L1] applies to $T$ and yields a basis in which $T$ is upper triangular. Let $λ$ be one diagonal entry. Then $T - λ {id}_{V}$ is upper triangular with a zero diagonal entry, hence is not invertible.

L1given

2.1

For every $g \in G$ , the identity map commutes with the action of $g$ , so $T - λ {id}_{V}$ is still a $G$ -endomorphism. By [L2], a nonzero $G$ -endomorphism of an irreducible representation must be invertible. Therefore the noninvertible endomorphism $T - λ {id}_{V}$ is zero.

step 1.1L2givenalgebra

3.1

Hence $T = λ {id}_{V}$ , as required.

step 2.1∎

Depends on

Used by

Over ℂ, a cyclic group of order n has exactly n irreducible representations up to equivalence, represented by the characters g↦ λ with λⁿ=1 Example

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Sources

Pavel Etingof et al., Introduction to Representation Theory, Corollary 1.17 (standard reference, not scraped)