I am wondering whether the following claim is true:

Let $G$ be a Lie group, $\mathfrak{g}$ its Lie algebra and $V$ some vector subspace of $\mathfrak{g}$.

Claim: $V$ is an ideal of $\mathfrak{g}$ if and only if $V$ is invariant under $\mathrm{Ad} \colon G \to \mathrm{GL}(\mathfrak{g})$, i.e. for all $g \in G$

$$ \mathrm{Ad}_g(V) \subseteq V $$

It is clear to me, that $V$ has to be invariant under $\mathrm{ad} \colon \mathfrak{g} \to \mathrm{End}(\mathfrak{g})$, since $\mathrm{ad}_X(Y)= [X,Y]$, but I struggle to proof this for $\mathrm{Ad}$. It may be that the claim only holds for semisimple Lie algebras, but I am not sure about that.

Any help or references are greatly appreciated!

  • $\begingroup$ I love this question because it feels so obviously true but indeed the proof is not obvious. Can we agree that, as you write, $V$ being an ideal is equivalent to $V$ being ad-invariant and that being ad-invariant follows quickly from being Ad-invariant, but that the converse (does being $ad$-invariant imply being Ad-invariant?) is the one that is not clear? In other words: can I interpret your question as 'Let $V$ be an ideal of $\mathfrak{g}$ and $g \in G$. Do we necessarily have that $Ad_g(V) \subset V$? $\endgroup$ – Vincent Jun 13 '15 at 14:19
  • $\begingroup$ The answer to the last question is of course yes if $g$ is the exponential of some Lie algebra element. This holds for all $g$ if $G$ is compact and also, at the opposite end of the spectrum, if $\mathfrak{g}$ is nilpotent, but for non-compact semi-simple groups we already have $g$ not of this form. I will think about it! $\endgroup$ – Vincent Jun 13 '15 at 14:21
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    $\begingroup$ @Vincent, if $G$ is connected, there are no problems since in this case $\mathrm{Ad}_G$ is generated by $\mathrm{exp}(\mathrm{ad}_{\mathfrak{g}})$. The only case to study is the case of a non-connected Lie group. $\endgroup$ – gniourf_gniourf Jun 13 '15 at 14:27
  • $\begingroup$ Do you mean by "quickly" using $\mathrm{ad}_X(Y) = \frac{d}{d t} \mathrm{Ad}(e^{tX})(Y) \big|_{t=0}$? $\endgroup$ – JBantje Jun 13 '15 at 14:36
  • $\begingroup$ Jbante: yes. @gniourf, ah yes, I didn't think of that. But it seems unlikely that the disconnectedness is going to be a problem. Generally we have that $G$ is the direct product of its identity component and a discrete group that is in its center and hence doesn't do much for Ad, right? $\endgroup$ – Vincent Jun 13 '15 at 14:38

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