3

A Lie group $H$ is called a Lie subgroup of a Lie Group $G$ if there is a map $i:H\to G$ which is (a) an injective immersion and (b) a group homomorphism.

My questios are: What happens if we replace (a) "injective immersion" by (a') "injective and differentiable"?

What happens if we go further and replace (a) "injective immersion" by (a'') "injective"?

Can anybody give examples where (a'') and (b) hold but not (a') and (b)? Or (a') and (b) but not (a) and (b)?

Klaas
  • 679

1 Answers1

2

For the first question, note that, because $i$ is a group homomorphism, $i$ fails to be an immersion if and only if the derivative of $i$ degenerates at every point of $H$. If this happens, $i$ cannot be injective, by the constant rank theorem, for instance (note that, in fact, the rank is constant). Therefore, {injective + smooth + group hom} seem to imply {immersion}.

For the second question: A continuous homomorphism of Lie groups is automatically smooth (see the section titled continuous homomorphisms in Warner's Foundations of Diffferential Geometry and Lie Groups). Therefore, given a map $H \rightarrow G$ between Lie groups, {injective + continuous + group hom} seem to imply {$H \rightarrow G$ is a Lie subgroup}, which is pretty remarkable.

Tim kinsella
  • 5,903
  • 1
    Your argument for the second question works only if $H$ is already known to be a Lie group. If $G=\mathbb R$ and $H=\mathbb Q$, for example, then $i\colon H\to G$ is a continuous injective homomorphism, but $H$ is not a Lie subgroup. – Jack Lee Jan 14 '15 at 04:34
  • 3
    @JackLee Yes, that was my assumption, but I should have made it explicit in the last line. Also, thanks for your wonderful textbooks. – Tim kinsella Jan 14 '15 at 05:29
  • You're welcome! – Jack Lee Jan 14 '15 at 05:43
  • Thanks for the answer! Can you say anything about the case where $i$ is not even assumed to be continuous? – Klaas Jan 14 '15 at 15:02
  • So you're asking what can be said about abstract subgroups of Lie groups which happen also to be group-isomorphic to a Lie group? Interesting question, but I have no idea. I guess I know next to nothing about the topological pathologies of abstract subgroups of Lie groups, beyond things like $\mathbb{Q}$ and the dense line on the torus. But even these cases are not so pathological in the sense that they're both images of Lie subgroups. You should post this question; I'd be interested in the answer. – Tim kinsella Jan 15 '15 at 16:19
  • But I suppose both of those examples show that you can't dispense with continuity in the definition of Lie subgroup, which is what you were asking, I think. – Tim kinsella Jan 15 '15 at 16:30
  • I do not quite understand your (counter)examples: $\mathbb Q$ is not even a Lie group, so the problem is not the continuity (besides, $\mathbb Q\hookrightarrow\mathbb R$ is continuous). And an irrational winding of a torus is continous and gives a Lie subgroup (not a regular one, i. e. not a submanifold). – Klaas Jan 16 '15 at 15:25
  • I have now posted the question here: http://math.stackexchange.com/q/1106817/74153 – Klaas Jan 16 '15 at 15:44
  • Yeah I'm not sure what I was thinking when I said those demonstrate that continuity can't be dispensed with. Just to clarify though, I meant Q with the discrete topology (which is a Lie group) being included into R. This is a Lie subgroup. – Tim kinsella Jan 16 '15 at 18:48