Alphabeta Math
False statementConstruction: Literature-sourcedVerification: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-22
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  • Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
  • AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
  • AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

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Every element of a complex semisimple Lie algebra is semisimple

Statement

Every element of a complex semisimple Lie algebra is semisimple.

Facts & Assumptions

Given: An element x of a Lie algebra is called semisimple when its adjoint operator ad⁡x is a semisimple endomorphism as defined in Semisimple endomorphisms as endomorphisms diagonalisable over an algebraic closure, and nilpotent endomorphisms; the same convention underlies Regular element and rank. In sl2(C)=Ch⊕Ce⊕Cf of The special linear Lie algebra sl_2 the brackets are [h,e]=2e, [h,f]=−2f, [e,f]=h.

[L1]

The Killing form of sl2(C) is nondegenerate, and a finite-dimensional characteristic-zero Lie algebra is semisimple exactly when its Killing form is nondegenerate (Killing form of sl_2, Cartan's semisimplicity criterion).

Refutation

technique · explicit witness
1.1

The element e=(0100) is nonzero, and [L1] shows that sl2(C) is a complex semisimple Lie algebra.

givenL1
1.2

Its adjoint operator is nilpotent and nonzero: on the basis (h,e,f) one has ad⁡e(h)=−2e, ad⁡e(e)=0 and ad⁡e(f)=h, so ad⁡e3=0 while ad⁡e≠0.

givenalgebra
2.1

A nonzero nilpotent endomorphism is not semisimple: over C its only eigenvalue is 0, so if it were diagonalisable it would be the zero operator; hence ad⁡e is not semisimple and the element e is not semisimple. This refutes the statement that every element is semisimple.

givenstep 1.1step 1.2algebra∎

Depends on

Used by

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Dependency tree · two levels

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Sources