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Let $G$ be a simply connected lie group, must all lie subgroups of $G $ be embedded submanifolds?

Amr
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1 Answers1

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First of all, one needs to be clear about the notion of a Lie subgroup: The use of this notion in the literature is inconsistent. From your post, it is clear that what you mean is:

Question. Suppose that $G, H$ are Lie groups, such that $G$ is simply connected, $f: H\to G$ is an injective morphism of Lie groups (i.e. a smooth monomorphism). Is $f$ an embedding?

The answer is easily negative. There are simpler examples with $H$ disconnected but I suspect that connectivity of $H$ was an unstated assumption.

Consider for instance $G=SU(2)\times SU(2)$ (it is simply connected). The group $G$ contains an (embedded) subgroup $T$ isomorphic to $S^1\times S^1$. The latter contains a dense Lie subgroup $H$ isomorphic to ${\mathbb R}$ (a line with irrational slope). Thus, $G$ contains a non-embedded (connected) Lie subgroup $H$.

Moishe Kohan
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  • Thank you, yes you interpreted my question correctly. Now I have a follow up question: Let L be a finite dimensional lie algebra over the real numbers and K a subalgebra. Then there exists a unique (up to iso) simply connected lie group G with L as a lie algebra and a unique lie subgroup H of G having K as a subalgebra. What algebraic condition on (L,K) determines if H will be embedded or not ? – Amr Dec 07 '19 at 07:11
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    @Amr: This should be posted as a separate question. Also, it is probably better suited for Mathoverflow (unlike the original question which is just right for the MSE). – Moishe Kohan Dec 07 '19 at 11:34