PropositionStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedverified 2026-09-26 (gpt-6-sol)

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Simple-reflection embeddings of Verma modules

Statement

If $⟨ λ + ρ, α_{i}^{\lor} ⟩ \in Z_{> 0}$ , there is an embedding $M (s_{i} \cdot λ) ↪ M (λ)$ .

Facts & Assumptions

Given: The positive integer $m = ⟨ λ + ρ, α_{i}^{\lor} ⟩$ , the singular vector of The simple-root singular vector in a Verma module, and the universal property The universal property of Verma modules.

[F1]

The PBW model identifies $M (μ)$ with $U (n^{-}) v_{μ}$ as a vector space for every $μ$ (The PBW model of a Verma module).

[F2]

The PBW filtration has $gr U (n^{-}) ≅ S (n^{-})$ (PBW gives an ordered monomial basis for the enveloping algebra).

Proof

technique · direct

1.1givenconstruct

The singular vector $f_{i}^{m} v_{λ}$ has weight $s_{i} \cdot λ$ , so the universal property gives a homomorphism $M (s_{i} \cdot λ) \to M (λ)$ taking its highest vector to it.

2.1F1F2step 1.1algebra∎

By [F1], every source vector is uniquely $u v_{s_{i} \cdot λ}$ with $u \in U (n^{-})$ . Its image is $u f_{i}^{m} v_{λ}$ , which vanishes only if $u f_{i}^{m} = 0$ in $U (n^{-})$ , again by [F1]. If $u \neq 0$ , its highest nonzero PBW symbol and the symbol of $f_{i}^{m}$ have nonzero product in the polynomial domain $S (n^{-})$ of [F2]. Therefore $u f_{i}^{m} \neq 0$ , and the map is injective.

Depends on

Used by

A singular integral A2 central-character summand Example
The A2 regular integral-dominant Verma embedding poset Example
The BGG resolution for sl2 Example
The regular integral sl2 block Example

Dependency tree · two levels

13 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

Yiannis Sakellaridis, Verma Modules and the Category O, Proposition 3.2 (standard reference, not scraped)