Alphabeta Math
False statementConstruction: Literature-sourcedVerification: AI-adaptedPipeline-generatedaudited 2026-09-22
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If alpha and beta are roots then alpha plus beta is always a root

Statement

Assume AC (The Axiom of Choice). If α and β are roots of a complex semisimple Lie algebra relative to a Cartan subalgebra, then α+β is again a root.

Facts & Assumptions

Given: AC; roots are nonzero functionals with gα≠0 (Root and root space), and [gα,gβ]⊆gα+β with gγ=0 for γ neither a root nor 0 (Brackets of root spaces). The set of indices k with β+kα∈Φ∪{0} is a nonempty interval {−p,…,q} (The root-string property). For every root α, the opposite −α is a root (Opposite root spaces pair nondegenerately), and the only scalar multiples of α that are roots are ±α (The only scalar multiples of a root that are roots are plus or minus the root).

Refutation

technique · explicit witness
1.1givenalgebra

Let α be any root and put β=−α, which is a root because the opposite root space is nonzero. Then α+β=0, and 0 is not a root by definition, since a root is required to be nonzero.

2.1givenstep 1.1algebra

A second, nontrivial failure occurs with β=α: then α+β=2α, and 2α is not a root because the only scalar multiples of the root α that are roots are ±α.

3.1givenstep 1.1step 2.1algebra∎

Neither failure contradicts the bracket inclusion of the given facts, which only asserts [gα,gβ]⊆gα+β and therefore says that the bracket vanishes when α+β is not a root or 0. Hence the claim that α+β is always a root is false.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

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Sources