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Dominance depends on a positive system

Statement

Assume the Axiom of Choice. Dominance of weights is defined canonically, without choosing positive roots.

Facts & Assumptions

Given: The Axiom of Choice, the root system $Φ = {α, - α}$ of ${s l}_{2} (C)$ with respect to $h = C h$ , where $α (h) = 2$ (The special linear Lie algebra sl_2, Reduced crystallographic Euclidean root system), the two opposite positive systems $Φ_{1}^{+} = {α}$ and $Φ_{2}^{+} = {- α}$ (Positive systems and simple roots), the coroots $h_{α} = h$ and $h_{- α} = - h$ of Coroot of a Lie-algebra root, and the functional $λ \in h^{*}$ with $λ (h) = 1$ .

[A1]

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

[L1]

$λ$ is dominant integral with respect to a positive system with base ${β}$ exactly when $⟨ λ, β^{\lor} ⟩ = λ (h_{β}) \in Z_{\geq 0}$ (Integral, dominant, and strictly dominant weights).

[L2]

For the root $- α$ the coroot is $h_{- α} = - h_{α} = - h$ , since the coroot is the normalised Killing dual and the dual vector changes sign with the functional (Coroot of a Lie-algebra root).

Refutation

technique · direct

1.1

With respect to the positive system ${α}$ the functional $λ$ is dominant integral: its only simple root is $α$ and $⟨ λ, α^{\lor} ⟩ = λ (h) = 1 \in Z_{\geq 0}$ by [L1].

L1A1

1.2

With respect to the positive system ${- α}$ the functional $λ$ is not dominant: its simple root is $- α$ and $⟨ λ, (- α)^{\lor} ⟩ = λ (- h) = - 1 \notin Z_{\geq 0}$ by [L1] and [L2].

L1L2

2.1

The same functional is thus dominant for one choice of positive roots and non-dominant for the opposite choice, so there is no choice-free notion of dominance; the failed conclusion is that dominance could be decided without fixing positive roots.

step 1.1step 1.2∎

Depends on

Used by

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Dependency tree · two levels

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Sources

Anthony W. Knapp, Lie Groups Beyond an Introduction, 2nd ed. (standard reference, not scraped)