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.

  • $\begingroup$ 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. $\endgroup$ – Stephen Mar 23 '13 at 14:39
  • $\begingroup$ Yes, $G$ itself is connected and $\mathfrak{g}$ is nontrivial. $\endgroup$ – Jimmy R Mar 23 '13 at 14:40
  • $\begingroup$ $\mathfrak{g}$ being nontrivial seems unnecessary in that case. $\endgroup$ – Stephen Mar 23 '13 at 14:41
  • $\begingroup$ No it's not, I edited my question and added that $G$ should be connected. $\endgroup$ – Jimmy R Mar 23 '13 at 14:42
  • $\begingroup$ 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). $\endgroup$ – Stephen Mar 23 '13 at 14:43

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 $$ G=\mathbb{R}\times\mathbb{Z}_2 \qquad N=\mathbb{R}\times\{0\}.$$

  • $\begingroup$ What happens if we drop connectedness of $G$? $\endgroup$ – Jimmy R Mar 23 '13 at 15:18
  • $\begingroup$ Take $G\times \Gamma' \leq G\times \Gamma$ where $\Gamma$ is a finite Group. $\endgroup$ – Alexander Thumm Mar 23 '13 at 15:28
  • $\begingroup$ @AlexanderThumm Is your comment addressed to me or to Jimmy R? Is there a problem with my argument? $\endgroup$ – Julien Mar 23 '13 at 15:32
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    $\begingroup$ Why 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 $G$. $\endgroup$ – Mirjam Mar 23 '13 at 17:34
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    $\begingroup$ @Martin No offense, no worries. I'll edit my answer. $\endgroup$ – Julien Mar 23 '13 at 22:17

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