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Left-invariant vector fields are complete
Statement
Assume . Every left-invariant smooth vector field on a finite-dimensional real Lie group is complete. Equivalently, its maximal flow is defined on all of . The countable-choice assumption is used exactly through the supplied invariant-field and smooth-tangent-bundle framework.
Facts & Assumptions
Given: , a finite-dimensional real Lie group with identity , and a left-invariant smooth vector field on .
is countable choice. The Axiom of Countable Choice ().
Left invariance means for all . Left- and right-invariant vector fields.
Through every point there is a unique maximal integral curve on an open interval containing zero. Through each point there is a unique maximal integral curve.
An integral curve satisfies . Integral curves of a vector field.
Differentials obey the chain rule. The chain rule for differentials of smooth maps.
Completeness means that every maximal integral curve has domain all of . Complete vector fields.
A vector field is complete if and only if its maximal flow domain is all of . A vector field is complete if and only if its flow is global.
Proof
Let be the maximal integral curve of with , supplied by [F3]. Since is open and contains , fix with .
For , define on . By [F4], [F5], and left invariance [F2], Also . After shifting the parameter by , uniqueness in [F3] shows that and agree wherever their domains overlap near , and hence on their whole interval overlap by the same local uniqueness argument.
Suppose the right endpoint were finite. Choose with . Then , while step 2.1 makes and agree on the nonempty overlap. Splicing them therefore gives an integral curve through on the strictly larger interval , contradicting maximality in [F3]. The identical argument at the left endpoint, using with if were finite, excludes a finite left endpoint. Thus .
For an arbitrary , define on all of . The calculation of step 2.1 with replaced by the fixed element proves that is an integral curve of , and . Its domain is already all of , so maximal uniqueness [F3] and [F6] show that is complete.
By [F7], completeness is equivalent to the maximal flow domain being all of , which proves the final formulation in the statement.
A Lie group is nonempty. In dimension zero every smooth vector field is zero and its integral curves are constant; in dimension one the extension proof above is unchanged. Lie groups are boundaryless, so no boundary or finite-time endpoint exception remains, and no metric or nondegeneracy enters. The stated is inherited through [F2] and the smooth tangent-field framework; choosing one and one inside a single nonempty interval uses no family choice, and the endpoint argument adds no choice. The theorem is a direct assertion plus the supplied equivalence in [F7]; both directions of that cited equivalence are available.
Depends on
- The Axiom of Countable Choice ($\mathrm{AC}_\omega$)
- Left- and right-invariant vector fields
- Complete vector fields
- Integral curves of a vector field
- Through each point there is a unique maximal integral curve
- The chain rule for differentials of smooth maps
- A vector field is complete if and only if its flow is global
Used by
Dependency tree · two levels
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Sources
- Robert L. Bryant, An Introduction to Lie Groups and Symplectic Geometry (standard reference, not scraped)
- Alexander Kirillov Jr., An Introduction to Lie Groups and Lie Algebras (standard reference, not scraped)