When is $\mathrm{exp}: \mathfrak g \rightarrow G$ surjective?

Asked Apr 03 '24 at 21:36

Active Apr 03 '24 at 21:36

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Let $G$ be a connected Lie group. What other conditions on $G$ are necessary and sufficient for the exponential map $$\mathrm{exp}: \mathfrak g \rightarrow G$$ from the Lie algebra $\mathfrak g$ of $G$ to be surjective?

Is it sufficient that $G$ be compact or nilpotent? What about solvable?

asked Apr 03 '24 at 21:36

Rodrigo

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Cf. https://math.stackexchange.com/questions/924557/if-a-connected-lie-group-is-divisible-is-its-exponential-map-surjective – Travis Willse Apr 03 '24 at 22:04
Compactness is sufficient, and so is nilpotence. I.I.r.c. connected and solvable is enough to guarantee that the image $\operatorname{im} \exp$ is dense in $G$. An equivalent condition is divisibility (plus connectedness); see https://math.stackexchange.com/questions/924557/if-a-connected-lie-group-is-divisible-is-its-exponential-map-surjective – Travis Willse Apr 03 '24 at 22:08
@TravisWillse, what is I.I.r.c.? – Rodrigo Apr 03 '24 at 22:27
"If I recall correctly" – Travis Willse Apr 03 '24 at 22:52
See also this duplicate. – Dietrich Burde Apr 04 '24 at 13:09

When is $\mathrm{exp}: \mathfrak g \rightarrow G$ surjective?

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