PropositionStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedjudge pass (gpt-6.1-sol)

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Annihilators of simple highest-weight modules are primitive

Statement

Let $g$ be a finite-dimensional complex semisimple Lie algebra, let $λ \in h^{*}$ , let $M (λ)$ be the Verma module of highest weight $λ$ , and let $L (λ)$ be its unique simple quotient. Then $I (λ) : = {Ann}_{U (g)} (L (λ))$ is a primitive ideal of $U (g)$ ; moreover ${Ann}_{U (g)} M (λ) \subseteq I (λ)$ , with equality whenever $M (λ)$ is simple (in which case $L (λ) ≅ M (λ)$ ).

Facts & Assumptions

Given: A finite-dimensional complex semisimple Lie algebra $g$ , a weight $λ \in h^{*}$ , the Verma module $M (λ)$ , and its unique simple quotient $L (λ)$ .

[F1]

$M (λ)$ has a unique maximal submodule $J (λ)$ , and $L (λ) = M (λ) / J (λ)$ is simple; it is the unique simple quotient (A Verma module has a unique simple quotient, Verma modules).

[F2]

A two-sided ideal is primitive when it is the annihilator of some simple module (Primitive ideals of an enveloping algebra).

[F3]

The annihilator of a module is a two-sided ideal, annihilators grow when passing to quotients, and the annihilator of a quotient $M / N$ contains the annihilator of $M$ (The annihilator of a module over an enveloping algebra).

Proof

technique · direct

1.1F1F2given

By [F1], $L (λ)$ is a simple $U (g)$ -module, so by [F2] its annihilator $I (λ) = {Ann}_{U (g)} (L (λ))$ is a primitive ideal.

1.2F1F3algebra

The natural surjection $M (λ) ↠ L (λ)$ has kernel $J (λ)$ ; if $u$ annihilates every element of $M (λ)$ , then it annihilates the image of every element in $L (λ)$ , so $u \in {Ann}_{U (g)} L (λ)$ . Hence ${Ann}_{U (g)} M (λ) \subseteq I (λ)$ by [F3].

2.1step 1.2F1algebra∎

If $M (λ)$ is simple, then its unique maximal submodule $J (λ)$ is a proper submodule by [F1] and therefore must be $0$ , since a simple module has no nonzero proper submodule; hence $L (λ) = M (λ) / J (λ) ≅ M (λ)$ . The two modules then have the same annihilator, and by step 1.2 the inclusion is an equality.

Depends on

Used by

An intersection of two primitive ideals need not be primitive Counterexample
The central character does not determine the primitive ideal Counterexample
Primitive ideals of U(sl2) at a generic central character Example
Highest weights can have the same primitive ideal Remark

Dependency tree · two levels

12 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

P. Etingof, Representations of Lie Groups (18.757, MIT OCW 2023 full notes) (standard reference, not scraped)
D. Barbasch, Cells in Weyl groups and primitive ideals (AIM workshop notes, 2006) (standard reference, not scraped)