LemmaStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-07

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Every Verma module contains a simple Verma submodule

Statement

Every Verma module contains a submodule isomorphic to a simple Verma module.

Facts & Assumptions

Given: Injectivity A nonzero homomorphism between Verma modules is injective, singular vectors Every nonzero Verma submodule contains a singular vector, Casimir scalars The quadratic Casimir eigenvalue on a highest-weight module is $(λ, λ + 2 ρ)$ , and the Verma weight cone Weights of a Verma module lie below lambda.

Proof

technique · contradiction

1.1

If no embedded Verma submodule were simple, repeatedly choose a nonzero proper submodule and then a singular vector in it; injectivity gives an infinite strictly descending chain of embedded Vermas $M (λ - β_{j}) \subset M (λ)$ .

givenassume-contra

2.1

Their Casimir scalars equal that of $M (λ)$ , so $2 (λ + ρ, β_{j}) = (β_{j}, β_{j})$ . The $β_{j}$ lie in the positive lattice cone and strictly increase in height, while this positive-definite quadratic equation has only finitely many lattice solutions. This contradiction yields a simple embedded Verma module.

step 1.1algebradischarge-contradiction∎

Depends on

Used by

Homomorphism spaces between Verma modules have dimension at most one Theorem

Dependency tree · two levels

15 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

Pavel Etingof, Representations of Lie Groups, Exercise 8.14(ii) (standard reference, not scraped)