Hopefully the title says it all in terms of the question I'm asking. I feel that the answer is no, but I'm not sure why; I don't really know many deep theorems about the structure of lie groups.
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4A Lie subgroup is compact if and only if it's a closed subgroup. – Jan 28 '16 at 16:59
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Depends on what you mean by "subgroup". If your definition of a Lie subgroup requires that the subgroup is closed or embedded (which is equivalent), then this is not possible. If not, then consider the example of one-parameter subgroup whose image is (after the obvious identifications) a straight line with an irrational slope inside the torus $S^1 \times S^1$.
levap
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I had a feeling there was some subtlety with the topology here, probably because I read (and forgot) about the irrational line in the torus before. Thanks. – Brian Klatt Jan 28 '16 at 17:05
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1Are there authors who take "Lie subgroup" to mean embedded, rather than immersed? What's an example? – Jan 28 '16 at 18:28
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@MikeMiller I'm not sure - it's definitely not a good idea from a categorical point of view. Maybe I shouldn't have written my answer suggesting that it is a reasonable thing to do. – levap Jan 28 '16 at 22:20
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I'm not going to worry about what the definition of "Lie subgroup" is, but a compact Lie group can certainly have a subgroup which is a non-compact Lie group. Consider the torus $$\Bbb T^2=\{(e^{it},e^{is}):t,s\in \Bbb R\}.$$ Suppose $\alpha\in\Bbb R$ is irrational and consider the subgroup $$\{(e^{it},e^{i\alpha t}):t\in\Bbb R\}.$$
David C. Ullrich
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