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Connected Lie subgroups with the same Lie algebra are equal as immersed subgroups
Statement
Assume . Two connected immersed Lie subgroups of a Lie group with the same tangent Lie subalgebra have the same image and the same intrinsic immersed-subgroup structure: there is a unique Lie-group isomorphism between them commuting with their inclusions into .
Facts & Assumptions
Given: and connected immersed Lie subgroups and with the same tangent image .
A Lie subalgebra integrates to a connected immersed subgroup uniquely up to the unique isomorphism over . Lie subgroup–Lie subalgebra correspondence.
Proof
Apply [F1] to the common subalgebra . Its uniqueness clause supplies a Lie-group isomorphism satisfying .
The equality gives as subsets of . Because and are smooth, they identify the intrinsic manifold structures, not merely the underlying image. The zero-dimensional and full-dimensional cases are included, and the stated countable-choice assumption is exactly the one inherited from [F1].
Depends on
Used by
Nothing in the library uses this result yet.
Dependency tree · two levels
7 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
- John M. Lee, Introduction to Smooth Manifolds, 2nd ed. (standard reference, not scraped)