Let $G\leq GL(n,\mathbb{R})$ be a connected linear Lie group and let $N$ be a normal subgroup of $G$ such that their Lie algebras are isomorphic: $$\mathfrak{g}\cong\mathfrak{n}$$ Does it follow that $N$ is connected? I feel like it should be, although I don't have much intuition for Lie groups.
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Presumably you want $G$ itself connected? For instance, $\mathbb{Z}$ is a subgroup of $\mathrm{GL}_2(\mathbb{R})$ with trivial Lie algebra, and every subgroup is normal. – Stephen Mar 23 '13 at 14:39
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Yes, $G$ itself is connected and $\mathfrak{g}$ is nontrivial. – Jimmy R Mar 23 '13 at 14:40
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$\mathfrak{g}$ being nontrivial seems unnecessary in that case. – Stephen Mar 23 '13 at 14:41
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No it's not, I edited my question and added that $G$ should be connected. – Jimmy R Mar 23 '13 at 14:42
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Well, if $G$ is connected and $\mathfrak{g}$ is trivial, then $G$ is trivial and there is nothing to prove. So there is no need to add that $\mathfrak{g}$ is trivial (and you haven't, in the body of your question). – Stephen Mar 23 '13 at 14:43
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Yes, I agree with your argument, that's why I didn't add that $\mathfrak{g}$ is nontrivial. – Jimmy R Mar 23 '13 at 14:45
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It does not define $N$ uniquely, do you think it might be true only for some normal subgroups with this property? – Jimmy R Mar 23 '13 at 14:52
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Is $N$ a normal subgroup of $G$ or of $GL(n\mathbb{R})$? – Jason DeVito - on hiatus Mar 23 '13 at 14:53
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julien, I understand where the confusion comes from, I edited the question – Jimmy R Mar 23 '13 at 14:58
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Jason, $N$ is a normal subgroup of $G$. – Jimmy R Mar 23 '13 at 14:59
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Edit: As pointed out by Mirjam below, there is a straighforward proof, which does not need to assume that $N$ be closed in $G$. The exponential map is a local homeomorphism $0\longmapsto 1$ between the Lie algebras and the groups. If the former are equal, it follows that $N$ is open in $G$, hence $G\setminus N=\bigsqcup_i g_iN$ (where $g_i$'s are a system of representative of $G/N$) is closed in $G$. So $N$ is open/closed and nonempty in $G$. By connectedness, $N=G$.
Quotient argument: If you assume $N$ to be a closed normal subgroup in $G$, then $G/N$ is a Lie group with Lie algebra naturally isomorphic to the quotient of the Lie algebra of $G$ by the one of $N$. See here. Now if the Lie algebras of $N$ and $G$ are the same, the Lie algebra of $G/N$ is trivial. And if $G$ is connected, so is $G/N$. We deduce from this that $G/N$ is trivial, i.e. $N=G$.
Non-connected case: This is no longer true if $G$ is disconnected. For instance, consider $$ G=\mathbb{R}\times\mathbb{Z}_2 \qquad N=\mathbb{R}\times\{0\}.$$
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Take $G\times \Gamma' \leq G\times \Gamma$ where $\Gamma$ is a finite Group. – Alexander Thumm Mar 23 '13 at 15:28
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@AlexanderThumm Is your comment addressed to me or to Jimmy R? Is there a problem with my argument? – Julien Mar 23 '13 at 15:32
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It is addressed to Jimmy R. A general principle in Lie theory is, that the Lie algebra can only measure local properties. To 'integrate' these properties we will always need some connectedness assumption. In our case the property is simply equality, thus '$N$ and $G$ are equal locally' plus '$G$ is connected' impies '$N$ and $G$ are equal'. – Alexander Thumm Mar 23 '13 at 15:38
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@AlexanderThumm Nicely stated, thanks. If you want Jimmy R to be notified, I think you need to add @ Jimmy R. – Julien Mar 23 '13 at 15:42
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@Julien thank you for your excellent answer, and I'm reading all comments here anyway. – Jimmy R Mar 23 '13 at 15:46
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1Why do you make the detour via quotients? The exponential map $\mathfrak{g} \to G$ sends an open neighborhood of zero in $\mathfrak{n=g}$ to an open neighborhood of $1$ in $N$, so $N$ is open in $G$, hence it is also closed and therefore all of $G$. – Mirjam Mar 23 '13 at 17:34
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@Mirjam You're right, that's the best argument. For some reason, I went in a more algebraic direction. – Julien Mar 23 '13 at 17:45
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Note that proving the quotient $G/N$ to be a Lie group is much more cumbersome (and far less algebraic) than the argument @Mirjam suggests. Also, her argument does not need to assume that $N$ be closed in $G$. – Martin Mar 23 '13 at 21:04
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@Martin Ok, ok, fine... Far less algebraic, really? Why not way far less? The argument I gave works, I think. If you don't like it, you should write another answer instead. – Julien Mar 23 '13 at 21:33
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@Martin I did not suggest to prove the structure of the quotient, by the way. Just to use it. And if you read my comment above yours, I already said that Mirjam had the best argument. What else do you want? – Julien Mar 23 '13 at 21:36
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Yes, of course, your argument works and I don't not like it nor do I want anything more. I only agreed with her and tried to point out a subtlety that might be hidden. In no way did I mean to offend you, sorry about the unfortunate wording. – Martin Mar 23 '13 at 21:55
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