Alphabeta Math
ExampleConstruction: Literature-sourcedVerification: AI-adaptedPipeline-generatedaudited 2026-09-22
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Exterior powers and fundamental weights of sl_n

Example

Assume the Axiom of Choice. Let g=sln(C) with its diagonal Cartan h, coordinate functionals εi and upper-triangular positive system as in Standard and dual representations of sl_n. For every k with 1kn1 the exterior power Λk(Cn) is an irreducible module of highest weight ωk=ε1++εk (Fundamental weights).

Facts & Assumptions

Given: The Axiom of Choice, such g,h, the standard module V=Cn with basis e1,,en, and W=Λk(V) with basis the wedges eI=ei1eik for increasing index sets I={i1<<ik}{1,,n}.

[A1]

The Axiom of Choice is assumed; it enters through the root-space and highest-weight theory used below (The Axiom of Choice).

[L1]

The weights of the standard module are ε1,,εn, ε1=ω1, and ε1++εk=ωk for 1kn1, because the simple coroots are hαj=EjjEj+1,j+1 and (ε1++εk)(hαj)=δkj (Standard and dual representations of sl_n, Fundamental weights).

[L2]

The wedge eI is a weight vector of weight iIεi, these weights are pairwise distinct for distinct index sets I, and EabeI=l:il=bei1eaeik, the sum being zero when the replacement produces a repeated index (Weight and weight space, Root systems of the classical complex Lie algebras).

[L3]

For a finite set of pairwise distinct weights and one of them, an element of U(h) acts as the projection onto the corresponding weight component, since U(h) is the polynomial algebra on h (Poincaré–Birkhoff–Witt theorem).

[L4]

A nonzero submodule of an irreducible module is the whole module (Irreducible, completely reducible, and faithful representations, Highest-weight vectors and modules).

Verification

technique · direct
1.1

The vector e{1,,k}=e1ek has weight ε1++εk=ωk by [L1] and [L2], and it is killed by every positive root vector Eij with i<j: if jk then i<jk gives i{1,,k} and the replacement repeats an index, while if j>k then ej is not a factor at all; either way Eije{1,,k}=0 by [L2].

L1L2A1
2.1

From any basis wedge eIe{1,,k} one reaches e{1,,k} by positive root vectors: let j be the smallest positive integer not in I, so jk, and choose iI with i>j, which exists since I has k elements; then EjieI=±eI with I=I{i}{j} a nonvanishing basis wedge whose index sum is strictly smaller, and repeating finitely many times reaches {1,,k}.

L2step 1.1
2.2

From e{1,,k} one reaches every basis wedge by negative root vectors: for a{1,,k} and b>k the operator Eba replaces the factor ea by eb without repeated indices (as b{1,,k}), giving ±eI with I=I0{a}{b}; successive replacements of this kind produce every increasing index set.

L2step 1.1
3.1

W is irreducible: if 0UW is a submodule, then by [L3] some basis wedge eI lies in U, so by step 2.1 the highest vector e{1,,k} lies in U, and by step 2.2 every basis wedge lies in U; hence U=W by [L4].

L3L4step 2.1step 2.2
4.1

By step 1.1 the vector e{1,,k} is a highest weight vector of weight ωk, and by step 3.1 the module is irreducible; hence Λk(Cn) has highest weight ωk, as asserted.

step 1.1step 3.1

Depends on

Used by

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