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Let $G$ be a Lie group with Lie algebra $\mathfrak{g}$. Let $H$ be a Lie subgroup of $G$ and its Lie algebra $\mathfrak{h}$ is naturally a subalgebra of $\mathfrak{g}$. Let $G^\circ$ denote the identity component of $G$, $H^\circ$ denote the identity component of $H$. Assume now $\mathfrak{h}$ is an ideal of $\mathfrak{g}$, then we know $H^\circ$ is a normal subgroup of $G^\circ$. Is it true that $H^\circ$ is also a normal subgroup of $G$?

I think it is highly possible that this is not true, but I cannot find a counter example since one need to find a suitable unconnected Lie group here.

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  • $\begingroup$ What about $G=\Bbb C\rtimes \Bbb Z/4\Bbb Z$ wher ethe action of $\Bbb Z/4\Bbb Z$ on $\Bbb C$ is multiplicationj with $i$, and we let $H=\Bbb R\rtimes 1$? Then $H^0=H$ is nrmal in $G^0=\Bbb C$, but not in $G$. $\endgroup$
    – Hagen von Eitzen
    Nov 7, 2017 at 7:08
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    $\begingroup$ I think you want $\mathfrak h$ is an ideal in $\mathfrak g$, not vice-versa. $\endgroup$
    – Robert Lewis
    Nov 7, 2017 at 7:08
  • $\begingroup$ Thank you for reminding me of the typo Robert, I have corrected them. And thank you for your suggestion Hagen, I am trying to figure out the meaning of the operator $\rtimes$ since I am a beginner for Lie groups. $\endgroup$
    – Cohen Lu
    Nov 7, 2017 at 7:18
  • $\begingroup$ That operator is not characteristic in any way to Lie groups; it is the semidirect product, and is there already when you consider just groups. $\endgroup$
    – B. Pasternak
    Nov 7, 2017 at 7:33

1 Answer 1

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Suppose $G$ is a Lie group and $H$ is a normal subgroup of $G$. If $g$ is an element of the group, the map $x\in G\mapsto gxg^{-1}\in G$ restricts to a map $x\in H\mapsto gxg^{-1}\in H$. The latter is continuous and maps the identity element to the identity element, so it maps the connected component of $H$ of the identity element to itself.

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  • $\begingroup$ So this essentially boils down to the connected component being "topologically characteristic". $\endgroup$
    – Tobias Kildetoft
    Nov 7, 2017 at 7:46
  • $\begingroup$ Thanks Mariano. Here in the answer we suppose $H$ is a normal subgroup of $G$, but generally we cannot get this information only from $\mathfrak{h}$ is an ideal of $\mathfrak{g}$. $\endgroup$
    – Cohen Lu
    Nov 7, 2017 at 17:16

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