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Consider a Lie subalgebra $\mathfrak{g} \subset \mathfrak{su}(n)$ such that $G = \exp\mathfrak{g} \subset \mathrm{SU}(n)$ is Lie. Is $G$ a compact Lie group?

I'm thinking one can use Cartan's closed subgroup theorem in the opposite direction: since $G$ is a Lie subgroup, it is closed in $\mathrm{SU}(n)$. Then we use the fact that closed $\Leftrightarrow$ compact for Lie subgroups. However, I am not 100% sure if I can apply the theorem in this way.

Related questions:

Is every Lie subgroup of $SU(2)$ closed? (My question seems to be a generalization to arbitrary $n$, but the proof method in the answer seems too explicit to generalize.)

Can a Compact Lie Group have a Non-Compact Lie Subgroup? (I am not sure whether my above definition of Lie subgroup coincides with that of the counterexamples.)

Banach space fan
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    But how do you know that $G$ is in fact a Lie group? In general, a subgroup of a Lie group need not itself be a Lie group (e.g. line of irrational slope in a flat torus). – Nate Eldredge Apr 04 '20 at 23:55
  • If $\mathfrak{g}$ is a Lie algebra, is it not always true that $\exp\mathfrak{g}$ is a Lie group? [Of course, this is a particular class of all possible subgroups (some of which may indeed not be Lie), but it's the class I am particularly interested in for this question. Perhaps I should reword the title to be more explicit?] – Banach space fan Apr 05 '20 at 01:24
  • No - my example is of this form (a line in $S^1 \times S^1$ is generated by a one-dimensional subalgebra of the Lie algebra $\mathbb{R}^2$). – Nate Eldredge Apr 05 '20 at 01:27
  • I see. I suppose I misunderstood Lie's third theorem. If we explicitly required $G$ to be Lie, then would my argument hold? – Banach space fan Apr 05 '20 at 01:34
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    It depends on your definition of a Lie subgroup. Incidentally, what you wrote makes no sense: "... such that $G = \exp\mathfrak{g} \subset \mathrm{SU}(n)$ is Lie." – Moishe Kohan Apr 05 '20 at 02:51
  • Sorry, I'm still a beginner to the subject. Why does the statement make no sense? All I mean to say is that I have a particular Lie algebra $\mathfrak{g}$ which generates a Lie group via $\exp$. – Banach space fan Apr 05 '20 at 04:17
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    @Banachspacefan: Lie was a mathematician who dies over 100 years ago. It makes no sense to say that "$G$ is ... Lie." One can say, for instance, that $G$ is (isomorphic to) a Lie group, or that $G$ a Lie subgroup or whatever else you had in mind. Also, if you have a Lie group $G$ and its subgroup $H$, the sentence "$H$ is a Lie group" is rather ambiguous: It might mean that $H$ with its subspace topology is isomorphic to a Lie group or that $H$ is isomorphic to a Lie group as an abstract group, etc. – Moishe Kohan Apr 05 '20 at 18:52

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