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

1.1

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

L1L2

2.1

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.

step 1.1

3.1

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

step 2.1∎

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
Simple-root integrability bounds the dominant cyclic module 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)