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Weight-lambda vectors are singular at a maximal label

Statement

Assume the Axiom of Choice (The Axiom of Choice). In the setting of Truncation at a finite downward-closed ideal of a linkage class, let $Γ$ be a finite downward-closed ideal of a linkage class $C$ and let $λ \in Γ$ be maximal in $Γ$ .

If $ν$ is a weight of an object $X$ of $O_{Γ}$ with $ν \geq λ$ in the root order of Root order on weights, then $ν = λ$ ; equivalently, no object of $O_{Γ}$ has a weight $λ + β$ with $β \in Q^{+} ∖ {0}$ .
Consequently every vector of weight $λ$ in every $X \in O_{Γ}$ is annihilated by $n^{+}$ : for $x \in n^{+}$ of weight $α > 0$ and $v \in X_{λ}$ , the vector $x v$ has weight $λ + α$ and hence vanishes by (1). Thus $X_{λ}^{n^{+}} = X_{λ}$ for every $X \in O_{Γ}$ .
The weight functor $X \mapsto X_{λ}$ is exact on all $h$ -semisimple $g$ -modules, hence on $O_{Γ}$ .

Maximality of $λ$ is used only in (1). Incomparable maximal labels do not invalidate (1): its antecedent requires $ν \geq λ$ , and any composition label above such a $ν$ is then comparable to $λ$ and forced equal to it. What can fail is the stronger assertion that every weight of every object lies below one specified maximal label; a simple with an incomparable highest weight refutes that stronger assertion.

Facts & Assumptions

Given: The Axiom of Choice, a finite downward-closed ideal $Γ$ of a linkage class $C$ , a maximal element $λ \in Γ$ , and an object $X \in O_{Γ}$ .

[F1]

$O_{Γ}$ is the full subcategory of objects of $O$ all of whose simple composition factors are $L (μ)$ with $μ \in Γ$ ; membership depends only on the isomorphism class, and $Γ$ is a finite lower set for the root order (Truncation at a finite downward-closed ideal of a linkage class). Maximality of $λ$ means that $μ \in Γ$ and $λ \leq μ$ imply $μ = λ$ .

[F2]

The simple objects of $O$ are exactly the $L (μ)$ ; $L (μ)$ is the unique simple quotient of the Verma module $M (μ)$ , and the weights of $M (μ)$ are exactly $μ - Q^{+}$ with finite weight spaces (The simple objects of O, A Verma module has a unique simple quotient, Weights of a Verma module lie below lambda).

[F3]

For a short exact sequence $0 \to A \to X \to B \to 0$ of $h$ -semisimple modules, a functional is a weight of $X$ exactly when it is a weight of $A$ or of $B$ : the corresponding sequence of weight spaces is exact at each weight (Category O is abelian and extension closed among weight modules). Iterating along a composition series, every weight of $X$ is a weight of some composition factor (Composition series and composition factors of an object).

[F4]

The root order is transitive and antisymmetric (Root order on weights), and a root vector of weight $α$ maps $X_{ν}$ into $X_{ν + α}$ (Weight and weight space, The classical BGG category O).

Proof

technique · direct: reduce a top weight to a composition factor, force equality by maximality, and read off singularity and exactness

1.1F1F2F3F4given

Let $ν$ be a weight of $X \in O_{Γ}$ with $ν \geq λ$ . By [F3] the weight $ν$ occurs in some composition factor $L (μ)$ of $X$ , and $μ \in Γ$ because $X \in O_{Γ}$ . By [F2], $ν$ is then a weight of $M (μ)$ , so $ν \leq μ$ . From $λ \leq ν \leq μ$ and transitivity in [F4] we get $λ \leq μ$ with $μ \in Γ$ , so maximality of $λ$ gives $μ = λ$ ; then $λ \leq ν \leq λ$ and antisymmetry give $ν = λ$ . Hence no object of $O_{Γ}$ has a weight $λ + β$ with $β \in Q^{+} ∖ {0}$ .

1.2F4givenalgebra

The weight functor $X \mapsto X_{λ}$ is exact on $h$ -semisimple $g$ -modules: given a short exact sequence $0 \to X^{'} \overset{i}{\to} X \overset{p}{\to} X^{''} \to 0$ , injectivity of $i_{λ}$ is immediate, and if $x^{''} \in X_{λ}^{''}$ lifts to $x \in X$ , then writing $x = \sum_{ν} x_{ν}$ as a finite sum of weight vectors gives $x^{''} = p (x) = \sum_{ν} p (x_{ν})$ with $p (x_{ν})$ of weight $ν$ ; by the directness of the weight decomposition of $X^{''}$ all terms with $ν \neq λ$ vanish and $x^{''} = p (x_{λ})$ , so $p_{λ}$ is surjective.

2.1F4step 1.1step 1.2∎

By step 1.1 no object of $O_{Γ}$ has a weight strictly above $λ$ in the sense of $λ + β$ with $β \in Q^{+} ∖ {0}$ : if $v \in X_{λ}$ is nonzero and $x \in n^{+}$ has weight $α > 0$ , then $x v \in X_{λ + α}$ by [F4], and $λ + α > λ$ ; if $x v \neq 0$ it would be a weight vector of weight $λ + α$ , contradicting step 1.1. Hence $x v = 0$ for every $x \in n^{+}$ and $X_{λ}^{n^{+}} = X_{λ}$ . Together with the exactness of the weight functor in step 1.2 this proves all three assertions.

Depends on

Used by

A maximal-label Verma is projective in its truncation Lemma

Dependency tree · two levels

41 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 (18.757, Fall 2023), Proposition 16.4 and its proof (standard reference, not scraped)
Lin Chen, lecture notes (Spring 2024), Lecture 8, Section 4 (standard reference, not scraped)