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PropositionStatement: Literature-sourcedProof: AI-adaptedverified 2026-09-23 (gpt-6-sol)
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The Harish-Chandra map on the center is injective

Statement

Assume the Axiom of Choice. For a finite-dimensional complex semisimple Lie algebra g and any Cartan subalgebra h in the nilpotent self-normalizing sense, if a central element z∈Z(U(g)) has zero Harish-Chandra projection relative to a chosen positive root system, then z=0. Equivalently, the Harish-Chandra map on the center is injective.

Facts & Assumptions

Given: The Axiom of Choice, a finite-dimensional complex semisimple Lie algebra g, any Cartan subalgebra h in the nilpotent self-normalizing sense, chosen positive roots, and a central element z∈Z(U(g)) with pr⁡(z)=0.

[A1]

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

[F1]

Under [A1], the shifted Harish-Chandra map for these data is an algebra isomorphism (Harish-Chandra isomorphism for the center).

Proof

technique · direct
1.1A1F1given

By [A1, F1], the shifted Harish-Chandra map HC⁡ρ(w)(λ)=pr⁡(w)(λ−ρ) is an algebra isomorphism from Z(U(g)) to S(h)W. Therefore it is injective.

2.1step 1.1∎

If pr⁡(z)=0, then HC⁡ρ(z)=0 as a polynomial function on h∗. Injectivity from step 1.1 then forces z=0.

Depends on

Used by

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Dependency tree · two levels

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