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 $g_{α} \neq 0$ (Root and root space), and $[g_{α}, g_{β}] \subseteq g_{α + β}$ with $g_{γ} = 0$ for $γ$ neither a root nor $0$ (Brackets of root spaces). The set of indices $k$ with $β + k α \in Φ \cup {0}$ is a nonempty interval ${- p, \dots, q}$ (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 $\pm α$ (The only scalar multiples of a root that are roots are plus or minus the root).

Refutation

technique · explicit witness

1.1

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.

givenalgebra

2.1

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

givenstep 1.1algebra

3.1

Neither failure contradicts the bracket inclusion of the given facts, which only asserts $[g_{α}, g_{β}] \subseteq g_{α + β}$ 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.

givenstep 1.1step 2.1algebra∎

Depends on

Used by

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Dependency tree · two levels

21 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

Anthony W. Knapp, Lie Groups Beyond an Introduction, 2nd ed., Chapter II (standard reference, not scraped)