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TheoremStatement: AI-adaptedProof: AI-adaptedPipeline-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-14
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Left-invariant vector fields are complete

Statement

Assume ACω. Every left-invariant smooth vector field on a finite-dimensional real Lie group is complete. Equivalently, its maximal flow is defined on all of R×G. The countable-choice assumption is used exactly through the supplied invariant-field and smooth-tangent-bundle framework.

Facts & Assumptions

Given: ACω, a finite-dimensional real Lie group G with identity e, and a left-invariant smooth vector field X on G.

[F1]

ACω is countable choice. The Axiom of Countable Choice (ACω).

[F2]

Left invariance means d(La)q(Xq)=Xaq for all a,qG. Left- and right-invariant vector fields.

[F3]

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.

[F4]

An integral curve c satisfies c(t)=Xc(t). Integral curves of a vector field.

[F5]

Differentials obey the chain rule. The chain rule for differentials of smooth maps.

[F6]

Completeness means that every maximal integral curve has domain all of R. Complete vector fields.

[F7]

A vector field is complete if and only if its maximal flow domain is all of R×G. A vector field is complete if and only if its flow is global.

Proof

technique · direct
1.1

Let c:IG be the maximal integral curve of X with c(0)=e, supplied by [F3]. Since I is open and contains 0, fix δ>0 with (δ,δ)I.

F3choose
2.1

For sI, define ηs(t)=c(s)c(ts) on (sδ,s+δ). By [F4], [F5], and left invariance [F2], ηs(t)=d(Lc(s))c(ts)c(ts)=d(Lc(s))c(ts)Xc(ts)=Xηs(t). Also ηs(s)=c(s)c(0)=c(s). After shifting the parameter by s, uniqueness in [F3] shows that ηs and c agree wherever their domains overlap near s, and hence on their whole interval overlap by the same local uniqueness argument.

F2F3F4F5step 1.1
3.1

Suppose the right endpoint b=supI were finite. Choose sI with bδ/2<s<b. Then s+δ>b, while step 2.1 makes c and ηs agree on the nonempty overlap. Splicing them therefore gives an integral curve through e on the strictly larger interval I(sδ,s+δ), contradicting maximality in [F3]. The identical argument at the left endpoint, using s with a<s<a+δ/2 if a=infI were finite, excludes a finite left endpoint. Thus I=R.

F3step 1.1step 2.1constructcontradiction
4.1

For an arbitrary pG, define cp(t)=pc(t) on all of R. The calculation of step 2.1 with c(s) replaced by the fixed element p proves that cp is an integral curve of X, and cp(0)=p. Its domain is already all of R, so maximal uniqueness [F3] and [F6] show that X is complete.

F2F3F4F5F6step 2.1step 3.1
5.1

By [F7], completeness is equivalent to the maximal flow domain being all of R×G, which proves the final formulation in the statement.

F7step 4.1
6.1

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 ACω is inherited through [F2] and the smooth tangent-field framework; choosing one δ and one s 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.

F1F2F3F4F5F6F7step 1.1step 2.1step 3.1step 4.1step 5.1

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