False statementConstruction: Literature-sourcedVerification: AI-adaptedPipeline-generatedaudited 2026-09-22

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All integer multiples of a root are roots

Statement

Assume AC (The Axiom of Choice). If $α$ is a root of a complex semisimple Lie algebra, then every integer multiple $k α$ with $k \in Z$ is again a root.

Facts & Assumptions

Given: AC; for a root $α$ the only scalar multiples of $α$ that are roots are $\pm α$ , so in particular $2 α$ is not a root (The only scalar multiples of a root that are roots are plus or minus the root); the root-string property describes the roots of the form $β + k α$ (The root-string property). The root spaces are the eigenspaces of Root and root space, and ${s l}_{2} (C) = C h \oplus C e \oplus C f$ is the Lie algebra of The special linear Lie algebra sl_2.

Refutation

technique · explicit witness

1.1

Take $g = {s l}_{2} (C)$ with Cartan subalgebra $C h$ and the root $α$ determined by $α (h) = 2$ . Then the root spaces are $g_{α} = C e$ and $g_{- α} = C f$ , and there are no other roots.

givenalgebra

2.1

The integer multiple $2 α$ is not a root: $g_{2 α}$ would be the eigenspace of ${ad}_{h}$ with eigenvalue $4$ , whereas the eigenvalues of ${ad}_{h}$ on ${s l}_{2} (C)$ are $2, 0, - 2$ ; alternatively $2 α$ is a scalar multiple of the root $α$ other than $\pm α$ .

givenstep 1.1algebra

3.1

Likewise $k α$ is not a root for every integer $k$ with $∣ k ∣ \geq 2$ , while $0 = 0 \cdot α$ is not a root either because roots are nonzero by definition. Hence not all integer multiples of a root are roots, and the statement is false.

givenstep 1.1step 2.1algebra∎

Depends on

Used by

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Sources

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