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The annihilator of the trivial sl(2)-module

Example

Assume the Axiom of Choice. Let g=sl2(C) with standard basis e,f,h and Casimir element Ω, so that Z(U(g))=C[Ω]. The annihilator of the trivial module C is the augmentation ideal Ann⁡U(g)C=U(g)g=(e,f,h), the two-sided ideal generated by g; it is a primitive ideal whose central character χ0 is the one with χ0(Ω)=0, and Ann⁡U(g)C∩Z(U(g))=(Ω) is a maximal ideal of Z(U(g)). The central ideal U(g)ker⁡χ0=(Ω)U(g) is strictly smaller than the annihilator: the highest vector of the simple Verma module M(−2) is an eigenvector of h with eigenvalue −2, so h∈Ann⁡C but h∉(Ω)U(g). Thus at the trivial central character the annihilator of the trivial module is not generated by the Casimir relation.

Facts & Assumptions

Given: The Axiom of Choice, g=sl2(C) with basis e,f,h, the trivial module C, and the ideal Ann⁡U(g)C of operators acting by 0 on C.

[F1]

The trivial module is the unital module on which e,f,h act by 0; the corresponding algebra homomorphism is the augmentation ε ⁣:U(g)→C with ε(g)=0, so its kernel is the two-sided ideal generated by g: after killing e,f,h, every nonempty tensor word is zero and the quotient is spanned by 1; ε(1)=1 makes the remaining scalar quotient exactly C (The annihilator of a module over an enveloping algebra, The special linear Lie algebra sl_2).

[F2]

Under AC the center is C[Ω], so ker⁡χ0=(Ω) and U(g)ker⁡χ0=(Ω)U(g); the trivial module is simple and nonzero, so its annihilator is primitive and the center acts on it through the central character χ0 with χ0(Ω)=0, the Casimir eigenvalue of the trivial module (The central reduction of U(sl2) is simple away from the finite-dimensional central characters, The center of the enveloping algebra is polynomial on rank-many generators, The Axiom of Choice, Primitive ideals of an enveloping algebra, A primitive ideal determines a central character, Central character of a Lie algebra module, The quadratic Casimir eigenvalue on a highest-weight module is (λ,λ+2ρ)).

[F3]

M(−2) is a simple Verma module by the Verma irreducibility criterion, since −2+1=−1∉Z>0; its highest vector v−2 satisfies hv−2=−2v−2, and its central character is χ−2, which equals χ0 because the normalized Casimir Ω=ef+fe+12h2 acts on both by 0, by the eigenvalue formula Ω↦λ(λ+2)/2 and −2⋅0/2=0 (Verma modules, The Verma irreducibility criterion from Shapovalov determinants, Highest-weight vectors and cyclic highest-weight modules, The central reduction of U(sl2) is simple away from the finite-dimensional central characters).

[F4]

U(g)ker⁡χλ⊆Ann⁡U(g)M(λ) for every λ (The Verma annihilator contains the central-character ideal).

Verification

technique · direct
1.1F1F2given

By [F1], Ann⁡U(g)C=ker⁡ε=U(g)g=(e,f,h); this is a proper two-sided ideal, and since C is simple and nonzero it is primitive by [F2].

1.2F2algebra

The trivial module has Casimir eigenvalue Ω↦0, so its central character is χ0 with χ0(Ω)=0; by [F2], Ann⁡U(g)C∩Z(U(g))=ker⁡χ0, which contains Ω and is a maximal ideal of Z(U(g)). The two-sided ideal generated by ker⁡χ0 is U(g)ker⁡χ0=(Ω)U(g), contained in the annihilator by [F2].

1.3F3F4algebra

By [F3], χ−2=χ0, so [F4] applied to M(−2) gives (Ω)U(g)⊆Ann⁡U(g)M(−2). The highest vector satisfies hv−2=−2v−2≠0, so h∉Ann⁡U(g)M(−2) and hence h∉(Ω)U(g).

2.1step 1.1step 1.3F1algebra∎

On the other hand h acts on the trivial module by 0, so h∈Ann⁡U(g)C by [F1]. Thus (Ω)U(g)⊊Ann⁡U(g)C: the central ideal does not exhaust the annihilator.

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