Alphabeta Math
TheoremStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedaudited 2026-09-22
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Root spaces of a complex semisimple Lie algebra are one-dimensional

Statement

Assume the Axiom of Choice. Let α be a root of a finite-dimensional complex semisimple Lie algebra g with respect to a Cartan subalgebra h (Root and root space). Then dim⁡gα=1.

Facts & Assumptions

Given: The Axiom of Choice, such g,h and a root α.

[A1]

The Axiom of Choice is The Axiom of Choice; it licenses the root triple, coroot, opposite-bracket, and root-decomposition facts in [L1]--[L3].

[L1]

There is a triple eα∈gα, fα∈g−α, hα=[eα,fα] with [hα,eα]=2eα and [hα,fα]=−2fα (The root sl2 triple, Coroot of a Lie-algebra root).

[L2]

[gγ,gδ]⊆gγ+δ with gη=0 for η neither a root nor 0, and g0=h (Brackets of root spaces, Root-space decomposition).

[L3]

[gα,g−α]=CHα⊆h for the corresponding dual vector (The bracket of opposite root spaces is the root line).

[L4]

A trace of a commutator of finite-dimensional endomorphisms vanishes, tr⁡(AB)=tr⁡(BA) (For A∈Mm×n(F) and B∈Mn×m(F), tr⁡(AB)=tr⁡(BA)).

Proof

technique · direct
1.1A1L1L2L3algebra

Put W=Ceα⊕Chα⊕⨁k<0gkα, a finite-dimensional subspace of g containing g−α. It is stable under ad⁡hα by [L2] together with [hα,eα]=2eα and [hα,hα]=0 from [L1]; it is stable under ad⁡eα because [eα,gkα]⊆g(k+1)α with (k+1)α either 0 or a negative multiple, [eα,g−α]⊆CHα by [L3], and [eα,hα]=−2eα; and it is stable under ad⁡fα because [fα,gkα]⊆g(k−1)α and [fα,hα]=2fα, [fα,eα]=−hα.

2.1L1L4step 1.1algebra

Since hα=[eα,fα], the restriction of ad⁡hα to the invariant subspace W is a commutator of the restrictions of ad⁡eα and ad⁡fα, so its trace vanishes by [L4].

3.1L1L2step 2.1algebra

On the other hand ad⁡hα acts on Ceα by the scalar 2, on Chα by 0, and on the eigenspace gkα, k<0, by the scalar kα(hα)=2k; hence 0=tr⁡(ad⁡hα∣W)=2−2∑j≥1jdim⁡g−jα, that is, ∑j≥1jdim⁡g−jα=1. As the summands are nonnegative integers, dim⁡g−α=1 and dim⁡g−jα=0 for j≥2.

4.1L1L2L3L4step 3.1algebra∎

The argument is symmetric in α and −α: the triple (fα,eα,−hα) satisfies the same relations with −α in place of α by [L1], and all the facts [L2]–[L4] are unchanged. Applying step 3.1 with −α therefore gives dim⁡gα=1 and dim⁡gjα=0 for j≥2, which proves the statement.

Depends on

Used by

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