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The Lie bracket of left-invariant fields is left invariant
Statement
Assume . If and are left-invariant smooth vector fields on a Lie group , then their Lie bracket is left invariant.
The countable-choice assumption is used exactly through the supplied invariant-field and smooth translation-trivialization results.
Facts & Assumptions
Given: , a Lie group , and left-invariant smooth vector fields on .
is countable choice. The Axiom of Countable Choice ().
Left invariance means that every left translation carries the field to itself pointwise. Left- and right-invariant vector fields.
Every left translation is a diffeomorphism. Translations are diffeomorphisms and their differentials trivialize the tangent bundle.
Pushforward by a diffeomorphism preserves the Lie bracket. Diffeomorphism pushforward preserves Lie brackets.
Proof
Fix . By [F3], is a diffeomorphism. The pointwise invariance identities in [F2] say exactly that and .
Naturality [F4] and step 1.1 give . Since was arbitrary, [F2] says that is left invariant.
A Lie group is nonempty. In dimension zero all vector fields and brackets vanish, while dimension one requires no change. No metric or nondegeneracy condition occurs, and the group is boundaryless by convention. The stated is inherited through [F2] and [F3]; fixing one arbitrary group element and applying bracket naturality makes no family selection. The proposition is a one-way closure statement, not a biconditional.
Depends on
Used by
Dependency tree · two levels
17 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
- Anthony W. Knapp, Lie Groups Beyond an Introduction, 2nd ed. (standard reference, not scraped)
- Alexander Kirillov Jr., An Introduction to Lie Groups and Lie Algebras (standard reference, not scraped)