Alphabeta Math
ExampleConstruction: AI-adaptedVerification: AI-adaptedverified 2026-09-23 (gpt-6-sol)
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Degree-two and degree-three Harish-Chandra generators for sl3

Example

Assume the Axiom of Choice (The Axiom of Choice). For sl3, the Weyl group is S3 permuting t1,t2,t3. On the Cartan subalgebra, the invariant polynomials

s2=t12+t22+t32,s3=t13+t23+t33

generate S(h)W, so their inverse Harish-Chandra images give independent degree-two and degree-three generators of Z(U(sl3)).

Facts & Assumptions

Given: The Axiom of Choice and the Cartan subalgebra of diagonal traceless matrices diag⁡(t1,t2,t3) in sl3, with t1+t2+t3=0.

Verification

technique · direct
1.1givenalgebra

For diagonal h=diag⁡(t1,t2,t3), the root vectors satisfy [h,Eij]=(ti−tj)Eij. The corresponding root reflections from Root reflections and the Weyl group action interchange ti and tj, so W=S3 on the trace-zero Cartan. The symmetric polynomials in t1,t2,t3 are generated freely by the elementary symmetric functions e1,e2,e3. The trace-zero relation sets e1=0. Every invariant class modulo this relation has an invariant representative, obtained by averaging any representative over the finite group S3. If an invariant polynomial is a multiple of e1 in the full polynomial ring, averaging its quotient makes it a multiple of e1 already in the invariant ring. Thus the quotient invariant ring is exactly C[e2,e3]. Newton's identities give s2=−2e2 and s3=3e3 there. Hence s2,s3 are algebraically independent generators.

2.1step 1.1given∎

Under Choice, The center of the enveloping algebra is polynomial on rank-many generators says the center of U(sl3) is polynomial on two generators of definite PBW degrees. The filtered isomorphism Harish-Chandra isomorphism for the center and step 1.1 show that the preimages of s2 and s3 are algebraically independent generators of respective PBW degrees two and three.

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