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

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Root spaces can have arbitrary dimension in a complex semisimple Lie algebra

Statement

Assume AC (The Axiom of Choice). The root spaces of a complex semisimple Lie algebra relative to a Cartan subalgebra can have arbitrary dimension, so no uniform bound on $\dim g_{α}$ holds.

Facts & Assumptions

Given: AC; root spaces are the eigenspaces $g_{α} = {x : [H, x] = α (H) x for all H \in h}$ of Root and root space, and for a root $α$ the theorem Root spaces of a complex semisimple Lie algebra are one-dimensional asserts $\dim g_{α} = 1$ . In ${s l}_{2} (C) = C h \oplus C e \oplus C f$ (The special linear Lie algebra sl_2) the Cartan subalgebra $C h$ has a root $α$ with $α (h) = 2$ , and the root triple of The root sl_2 triple realizes the roots $\pm α$ (Coroot of a Lie-algebra root).

Refutation

technique · explicit witness

1.1

Take $g = {s l}_{2} (C)$ with the Cartan subalgebra $h = C h$ . Its roots are the nonzero functionals $α$ with $g_{α} \neq 0$ ; since $[h, e] = 2 e$ and $[h, f] = - 2 f$ , the functional $α$ with $α (h) = 2$ is a root with $g_{α} = C e$ and $- α$ is a root with $g_{- α} = C f$ .

givenalgebra

2.1

Both root spaces are one-dimensional, so in this example the dimension is $1$ and not, say, $2$ .

givenstep 1.1

3.1

More generally, the cited theorem gives $\dim g_{α} = 1$ for every root of every finite-dimensional complex semisimple Lie algebra, so no root space has dimension $2$ or any other value different from $1$ . The statement that root spaces can have arbitrary dimension is therefore false.

step 2.1algebra∎

Depends on

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Sources

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