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 sl_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 sl_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

1.1

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.

A1L1L2L4

1.2

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 $h = {span}_{C} {h_{α_{1}}, \dots, h_{α_{N}}}$ .

A1L4

2.1

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.

A1L2L3step 1.1

3.1

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 $j$ .

A1L3step 2.1

4.1

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.

A1step 3.1

5.1

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

step 4.1∎

Depends on

Used by

Every finite-dimensional irreducible module has a highest-weight vector Lemma

Dependency tree · two levels

49 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

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)