PropositionStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-22

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

Root vectors shift weights

Statement

Let $g$ be a finite-dimensional complex semisimple Lie algebra with Cartan subalgebra $h$ , let $V$ be a representation of $g$ , let $α$ be a root and let $g_{α}$ be its root space (Root and root space). If $x \in g_{α}$ and $v \in V_{μ}$ for some $μ \in h^{*}$ , then $x \cdot v \in V_{μ + α}$ (Weight and weight space); in particular $x \cdot v = 0$ is allowed.

Facts & Assumptions

Given: Such $g, h, V$ , a root $α$ , $x \in g_{α}$ , $μ \in h^{*}$ and $v \in V_{μ}$ .

[L1]

The action of $g$ on $V$ is a Lie algebra homomorphism into the endomorphisms of $V$ with commutator bracket (Representations of Lie algebras): for all $X, Y$ and $w$ , $X \cdot (Y \cdot w) - Y \cdot (X \cdot w) = [X, Y] \cdot w .$

[L2]

$g_{α} = {x : [H, x] = α (H) x for all H \in h}$ (Root and root space), so $H \cdot v = μ (H) v$ and $[H, x] = α (H) x$ for $H \in h$ .

Proof

technique · direct

1.1L1L2

Let $H \in h$ ; using [L1] with $X = H$ , $Y = x$ , $w = v$ and [L2], $H \cdot (x \cdot v) = [H, x] \cdot v + x \cdot (H \cdot v) = α (H) x \cdot v + μ (H) x \cdot v = (α + μ) (H) (x \cdot v)$ .

2.1step 1.1

Equation 1.1 says precisely that $x \cdot v$ is annihilated by $ρ (H) - (μ + α) (H) {id}_{V}$ for every $H \in h$ , that is, $x \cdot v \in V_{μ + α}$ (Weight and weight space); this includes the possibility $x \cdot v = 0$ , which lies in every subspace.

3.1step 2.1∎

Steps 1.1 and 2.1 prove the stated containment, including the zero case.

Depends on

Used by

Not every weight vector is highest False statement
Every finite-dimensional irreducible module has a highest-weight vector Lemma
Highest weight modules lie below the top weight Lemma
Minuscule weights have exactly the Weyl orbit as their weights Lemma
Positive root strings sum the Freudenthal correction Lemma
Simple-root integrability bounds the dominant cyclic module Lemma
The Casimir comparison on a weight space Lemma
Unique simple quotient of the dominant cyclic module Lemma
The highest-weight space is one-dimensional Proposition
Top summand in a tensor product Proposition

Dependency tree · two levels

10 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

Alexander Kirillov Jr., An Introduction to Lie Groups and Lie Algebras (standard reference, not scraped)