Alphabeta Math
CorollaryStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedaudited 2026-09-22
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Opposite root spaces pair nondegenerately

Statement

Assume the Axiom of Choice. Let α be a root of the finite-dimensional complex semisimple Lie algebra g with respect to a Cartan subalgebra h (Root and root space). Then −α is a root, and the Killing form restricts to a nondegenerate pairing gα×g−α→C; in particular g−α≠0 and B is nonzero on gα×g−α.

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 orthogonality and root-space decomposition used in [L1] and [L2].

[L1]

B(gγ,gδ)=0 whenever γ+δ≠0, and B∣h is nondegenerate (Orthogonality of root spaces and nondegeneracy on the Cartan subalgebra).

[L2]

g=h⊕⨁γ∈Φgγ is a direct sum (Root-space decomposition), and B is nondegenerate on the semisimple algebra g (Cartan's semisimplicity criterion).

Proof

technique · direct
1.1A1L1L2algebra

Let 0≠x∈gα. By nondegeneracy of B there is y∈g with B(x,y)≠0; write y=y0+∑γyγ according to [L2]. By [L1] all summands vanish in the pairing with x except possibly y−α, whose weight space would make −α a root; hence B(x,y−α)≠0, so −α is a root and g−α≠0.

2.1L1L2step 1.1algebra∎

The restriction of B to gα⊕g−α is nondegenerate: if x∈gα pairs to zero with all of g−α, then it pairs to zero with every weight space and with h by [L1], hence with g, so x=0; the same argument applies to g−α. Since gα≠0 by definition of a root, this nondegenerate pairing is nonzero.

Depends on

Used by

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Sources