PropositionStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedaudited 2026-09-22

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Finite-dimensional modules decompose into weight spaces

Statement

Assume the Axiom of Choice. Let $g$ be a finite-dimensional complex semisimple Lie algebra with Cartan subalgebra $h$ , and let $V$ be a finite-dimensional representation of $g$ (Representations of Lie algebras). Then $V$ is the direct sum of its weight spaces for $h$ (Weight and weight space): $V = ⨁_{μ \in h^{*}} V_{μ},$ and only finitely many of the spaces $V_{μ}$ are nonzero.

Facts & Assumptions

Given: The Axiom of Choice, such $g, h$ and a finite-dimensional representation $V$ .

[A1]

The Axiom of Choice is assumed; it enters through the root-space theory supplying [L1] and [L4], whose contracts carry the assumption (The Axiom of Choice).

[L1]

For every root $α$ of $(g, h)$ there are $e_{α} \in g_{α}$ and $f_{α} \in g_{- α}$ with $[e_{α}, f_{α}] = h_{α}$ , $[h_{α}, e_{α}] = 2 e_{α}$ and $[h_{α}, f_{α}] = - 2 f_{α}$ , so that $span {e_{α}, f_{α}, h_{α}}$ is a copy of ${s l}_{2}$ (The root ${s l}_{2}$ triple).

[L2]

For a nonzero finite-dimensional module over ${s l}_{2}$ , the operator $h$ acts diagonalisably with integer eigenvalues (Finite-dimensional representations of ${s l}_{2}$ ).

[L3]

A family of diagonalisable endomorphisms of a finite-dimensional vector space is simultaneously diagonalisable if and only if its members commute pairwise (A family of diagonalisable endomorphisms of a finite-dimensional space is simultaneously diagonalisable if and only if its members commute pairwise).

[L4]

The roots of $(g, h)$ form a reduced crystallographic Euclidean root system on $E = {span}_{R} Φ$ whose simple coroots span $h$ over $C$ ; in particular the coroots $h_{α}$ with $α$ ranging over the finitely many roots span $h$ (The roots form a reduced crystallographic Euclidean root system).

Proof

technique · direct

1.1A1L1L2L4

If $V = 0$ , every weight space is zero and the asserted direct sum is the empty direct sum, so the conclusion is immediate. If the root set $Φ$ is empty, [L4] gives $h = 0$ , and then $V = V_{0}$ is already the required decomposition. Assume henceforth that $V \neq 0$ and $Φ \neq \emptyset$ . Fix a root $α$ and the ${s l}_{2}$ -triple $(e_{α}, f_{α}, h_{α})$ of [L1]; restricting the representation to this three-dimensional subalgebra makes $V$ a nonzero finite-dimensional ${s l}_{2}$ -module, so by [L2] the operator $ρ (h_{α})$ is diagonalisable.

1.2A1L4

The coroots span $h$ over $C$ by [L4] and the root set $Φ$ is finite, so there are roots $α_{1}, \dots, α_{N}$ with $h = {span}_{C} {h_{α_{1}}, \dots, h_{α_{N}}}$ .

2.1A1L2L3step 1.1

Every $H \in h$ is a linear combination $H = \sum_{i = 1}^{N} c_{i} h_{α_{i}}$ ; the operators $ρ (h_{α_{1}}), \dots, ρ (h_{α_{N}})$ are diagonalisable by step 1.1 and commute pairwise because $h$ is abelian and $ρ$ preserves brackets, so by [L3] they are simultaneously diagonalisable, and in a common eigenbasis every $ρ (H)$ is diagonal, hence diagonalisable.

3.1A1L3step 2.1

The family ${ρ (H) : H \in h}$ consists of pairwise commuting diagonalisable endomorphisms by step 2.1, so [L3] provides a basis $v_{1}, \dots, v_{n}$ of $V$ and functionals $μ_{1}, \dots, μ_{n} \in h^{*}$ with $ρ (H) v_{j} = μ_{j} (H) v_{j}$ for all $H \in h$ and all $j$ .

4.1A1step 3.1

A basis vector $v_{j}$ spans a nonzero weight space $V_{μ_{j}}$ (Weight and weight space), while a vector $v \in V_{μ}$ has $ρ (H) v = μ (H) v$ for all $H$ and is therefore a linear combination of the basis vectors $v_{j}$ with $μ_{j} = μ$ ; hence each $V_{μ}$ is the span of those $v_{j}$ with $μ_{j} = μ$ , distinct weights have disjoint sets of basis vectors, and $V = ⨁_{μ} V_{μ}$ , with only finitely many nonzero summands.

5.1step 4.1∎

The stated direct-sum decomposition and finiteness of the list of weights are proved.

Depends on

Used by

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Sources

Alexander Kirillov Jr., An Introduction to Lie Groups and Lie Algebras (standard reference, not scraped)
Anthony W. Knapp, Lie Groups Beyond an Introduction, 2nd ed. (standard reference, not scraped)