Let $G$ be a Lie group and $H\subset G$ be a connected Lie subgroup, with Lie algebras $\mathfrak{g}$ and $\mathfrak{h}$, respectively. In this case we know that $$ Lie(N(H)) = \mathfrak{n} (\mathfrak{h}) $$ Where $N(H)=\{ g\in G : gHg^{-1} \subseteq H \} $ is the normalizer of $H$, $Lie(N(H))$ is the Lie algebra of such a closed subgroup and $\mathfrak{n} (\mathfrak{h}) = \{ X\in \mathfrak{g} : ad(X)\mathfrak{h} \subset \mathfrak{h} \}$. With $ad(X)Y=[X,Y]$ the Lie bracket of the corresponding vector fields.

However, I'm looking for an example on the case where $H$ is a non-connected subgroup and the above inequality does not hold. Any help would be greatly appreciated. At the moment I don't know the theory of solvable nor nilpotent Lie groups, so any example avoiding those is preferred.

An attempt

I want a non abelian group as I want at least one of the sets to be nontrivial.

I find computing normalizers rather hard, so if we have a discrete subgroup (which will be non-connected), we have that the Lie algebra $\mathfrak{h}$ is trivial and then $\mathfrak{n} (\mathfrak{h}) = \mathfrak{g}$. Now I want an element that is not in the Lie algebra of the normalizer. But I'm stuck at finding explicitly what is the Lie algebra of the Normalizer, even on a simple matrix group.

Thanks in advance!


1 Answer 1


What about $C_2 \times C_2 \simeq H = \{ \pmatrix{a&0\\0&d} : a,d \in \{\pm 1\}\} \subset G = GL_2(\mathbb R)$?

  • 1
    $\begingroup$ Thank you, sounds good for me. I found that the normalizer $N(H)$ is the set of diagonal matrices with $a,d\neq 0$, and then $Lie (N(H))$ is the set of diagonal matrices; additionaly, as the subgroup is discrete $\mathfrak{n}(\mathfrak{h})=\mathfrak{g}$. Thus, it suffices to pick any non diagonal matrix $A\in \mathfrak{g}$ but $A\not\in Lie(N(H))$. Is that so? $\endgroup$ Commented Jun 10, 2023 at 13:53

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