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TheoremStatement: Literature-sourcedProof: AI-adaptedverified 2026-09-23 (gpt-6-sol)
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The enveloping algebra is free over its center

Statement

Assume the Axiom of Choice. For a finite-dimensional complex semisimple Lie algebra g, the enveloping algebra U(g) is a free left, hence also right, module over its center Z(U(g)).

Facts & Assumptions

Given: The Axiom of Choice, a finite-dimensional complex semisimple Lie algebra g, and the PBW filtration on U(g).

[A1]

The Axiom of Choice is The Axiom of Choice and is inherited through [F1].

[F1]

Under [A1], Kostant's harmonic decomposition gives a graded subspace H⊆S(g) for which multiplication is an isomorphism H⊗S(g)g→∼S(g) (Kostant harmonic decomposition of the symmetric algebra).

Proof

technique · direct
1.1givenalgebra

The PBW theorem PBW gives an ordered monomial basis for the enveloping algebra identifies gr⁡U(g) with S(g). Its symmetrization map sym⁡(x1⋯xm)=1m!∑σ∈Smxσ(1)⋯xσ(m) is a filtration-preserving vector-space isomorphism whose associated graded map is the identity. Because the adjoint action is a derivation on both sides, sym⁡ is g-equivariant. It therefore restricts to a filtered vector-space isomorphism S(g)g→∼U(g)g=Z(U(g)), where the last equality holds because g generates U(g). Consequently gr⁡Z(U(g))=S(g)g.

2.1F1step 1.1construct

Fix a finite ordered basis of g and order its monomials degree by degree. In each finite-dimensional space Hd⊆Sd(g), row reduction in this fixed monomial order specifies a basis; taking the union over d≥0 gives a homogeneous basis (hj) of H without a simultaneous choice. Define its PBW lifts by aj=sym⁡(hj). The PBW filtration is nonnegative and exhaustive, as is its restriction FnZ=Z∩FnU to the center. The symmetrization in step 1.1 places aj in filtration degree dj=deg⁡hj and gives it leading symbol hj.

3.1A1F1step 1.1step 2.1∎

By [A1, F1] and step 1.1, the symbols (hj) are a homogeneous free basis of gr⁡U(g)=S(g) over gr⁡Z(U(g))=S(g)g. The filtrations and specified lifts in step 2.1 meet the hypotheses of Filtered freeness lifts from associated graded algebras. Its finite-degree induction lifts the graded expansion of each element of U(g), while its highest-degree argument proves independence; it follows that (aj) is a left and right basis over the center.

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