Alphabeta Math
PropositionStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedaudited 2026-09-22
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Dominant weights in fundamental coordinates

Statement

Assume the Axiom of Choice. Let g be a finite-dimensional complex semisimple Lie algebra with Cartan subalgebra h and a chosen base of simple roots, and let ω1,…,ωr be the fundamental weights (Fundamental weights). For λ∈h∗ the following are equivalent:

(i) λ is dominant integral (Integral, dominant, and strictly dominant weights);

(ii) λ=∑i=1rniωi with integers ni≥0.

Facts & Assumptions

Given: The Axiom of Choice, such g,h and a chosen base of simple roots α1,…,αr with fundamental weights ω1,…,ωr.

[A1]

The Axiom of Choice is assumed; it enters through the root-space theory supplying the real form and the coroot basis of [L1] (The Axiom of Choice).

[L1]

The simple coroots hα1,…,hαr form a basis of h; the simple roots form a basis of E=span⁡RΦ; and the fundamental weights are the dual basis to the simple coroots, (ωi,αj∨)=δij, and form a basis of the weight lattice P (The roots form a reduced crystallographic Euclidean root system, Simple roots form a signed integral basis, Fundamental weights, Root, coroot, weight, and coweight lattices).

[L2]

λ is integral when ⟨λ,αi∨⟩=λ(hαi)∈Z for all i, dominant integral when all these integers are nonnegative, and every integral functional lies in E; for λ∈E the expansion in the dual basis is λ=∑i⟨λ,αi∨⟩ωi (Integral, dominant, and strictly dominant weights).

Proof

technique · direct
1.1A1L2

Assume (i): then by [L2] λ∈E and ni:=⟨λ,αi∨⟩∈Z≥0 for every i, so the expansion λ=∑iniωi of [L2] exhibits λ in the form (ii).

1.2L1

Conversely assume (ii), say λ=∑iniωi with ni∈Z≥0; then λ∈P⊆E by [L1], and by the duality (ωi,αj∨)=δij of [L1] we get ⟨λ,αj∨⟩=∑iniδij=nj∈Z≥0 for each j.

2.1L2step 1.1step 1.2∎

By [L2] the pairings of step 1.2 are exactly the values that make λ dominant integral; hence (ii) implies (i), and steps 1.1 and 2.1 prove the equivalence.

Depends on

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Sources