Suppose $G$ is a compact connected Lie group and let $\{X_i\}$ be a basis for its Lie algebra $\mathfrak g$. We know that the exponential $\exp:\mathfrak g \to G$ is surjective but when is it the case that $G$ is generated by $\{\exp(tX_i) : t\in \mathbb R\}$?
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The map $\mathbb{R}^n \to G$ sending $(t_1, \dots, t_n)$ to $\mathrm{exp}(t_1 X_1) \dots \mathrm{exp}(t_n X_n)$ has nonsingular Jacobian at $0$, so its image contains a neighborhood of the origin. By a standard argument, a neighborhood of the origin in a connected topological group generates the full group.
Akhil Mathew
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Can you explain why for the compact connected Lie group the exponential map has to be surjective? (though there can be cases where the exponential map has a finite injectivity radius) – Student Feb 17 '11 at 08:14
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@Anirbit: One way to see this is to use the Hopf-Rinow theorem in Riemannian geometry (since the geodesics with respect to a bi-invariant metric are precisely $\exp(tX)$ for $X$ in the Lie algebra). Sorry for the very delayed response! – Akhil Mathew Feb 25 '12 at 16:00
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@Anirbit Akhil's argument is great but if you do not wish to explicitly use Riemannian geometry, then there is an alternative proof if you assume Cartan's theorem on the maximal tori in a compact connected Lie group (Cartan's theorem can be proved using the Lefschetz fixed point theorem in algebraic topology). Indeed, each element of $G$ is an element of some maximal torus $T\subseteq G$ which means that we need only prove the surjectivity of the exponential map $\text{exp}:\mathfrak{t}\to T$ where $T$ is a torus. Of course, this is very easy to do. – Amitesh Datta Jun 27 '12 at 02:22
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hi so if $G$ is not compact but just connected $exp$ may not be surjective but is it still true that it is generated by a basis of $g$? thanks – l4teLearner Sep 06 '22 at 08:52
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@AmiteshDatta forgot the mention – l4teLearner Sep 06 '22 at 11:28
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@l4teLearner I would like to know whether you have found the answer to this question. It seems to me that this might be true for any semisimple real Lie group via the Iwasawa decomposition, which decomposes any element into a product of an element from a compact group, abelian group and nilpotent group generated by elements in root subgroups. Now, for the first two, the proof above about tori and Cartan's theorem works, while for the nilpotent case I'm still not sure. A general proof based on differential topology/geometry would be nice. – Another User Dec 17 '23 at 19:17
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@AnotherUser I don't have a proof but this seems to be true, see https://mathoverflow.net/questions/416754/lie-groups-generated-by-finitely-many-lie-algebra-elements or https://math.stackexchange.com/questions/4027262/does-a-lie-algebra-generate-the-entire-identity-component . maybe if we relax the constraint on the maximum k a sketch or a proof can be found here? : https://math.stackexchange.com/questions/3076360/connected-linear-lie-groups-g-generated-by-exp-mathfrakg let me know! – l4teLearner Dec 18 '23 at 07:56
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That seems to answer however a different question, which is that any Lie group element, you can write it as a product of one (or more) exponentials of a single Lie algebra element each. Here I'm asking whether, for a fixed Lie algebra basis/sequence, you can find suitable coefficients such that you can write any element as a product of exponentials of each basis element (with fixed order of appearance in the product). I'm guessing the two are related, but I can't figure out how. – Another User Dec 18 '23 at 16:20
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@AnotherUser you are right, it's a different thing. I don't have the answer, this probably deserves a question on its own. maybe for commutative group the two are related/equivalent given that $e^A+e^B=e^{AB}$? but I don't know in general. – l4teLearner Dec 20 '23 at 07:46
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looking at your first comment you had already found this result, probably... – l4teLearner Dec 20 '23 at 07:59