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Let $G$ be a Lie group with Lie algebra $\mathfrak{g}$. Let $Ad: G \rightarrow GL(\mathfrak{g})$ be the Adjoint representation. I want to prove that $Ad(G)$ is a Lie subgroup of $GL(\mathfrak{g})$.

Here is what I tried: From Cartan's Theorem, it suffices to show that $Ag(G)$ is closed in $GL(\mathfrak{g})$. So take $x \in Ad(G)$ and $(Ad(g_n))_n$ a sequence in $Ad(G)$ that converges to $x$, with $g_n \in G$. I can't continue from here. It may be possible that some additional other hypothesis are required.

Thanks!

Bruno Suzuki
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  • Don't you just need to show Ad is a Lie group homomorphism? – Tim kinsella Oct 27 '14 at 04:54
  • @Timkinsella: One should clarify what "Lie subgroup" is supposed to mean. Any path-connected subgroup of a Lie group can be endowed with a topology and smooth structure so as to make the inclusion a smooth homomorphism - however, unless the subgroup is closed in the ambient group, this topology will not be the subspace topology, as the dense subgroups of the torus show. – Hanno Nov 01 '14 at 10:19
  • @Hanno I thought the standard definition was "an injective immersion which is a group homomorphism." Then a lie group homomorphism always descends to a subgroup, right? – Tim kinsella Nov 02 '14 at 00:24
  • @Timkinsella: Yes, I agree, taking this as the definition of Lie subgroup, the statement reduces to noting that $G / \text{ker}(Ad)=G/\text{Z(G)}$ is a Lie group. – Hanno Nov 02 '14 at 09:12

2 Answers2

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Hilgert, Neeb: The Structure and Geometry of Lie Groups, Exercise 18.1.2

Pick $\theta\in{\mathbb R}$ irrational and define $G := {\mathbb C}^2\rtimes_\alpha{\mathbb R}$, where $\alpha: {\mathbb R}\to {\text{GL}}({\mathbb C}^2)$ is given by $$\alpha(t) := \begin{pmatrix} e^{it} & 0 \\ 0 & e^{i\theta t}\end{pmatrix}.$$ Then $\text{Ad}(G)$ is claimed not to be closed in $\text{GL}({\mathfrak g})$, though I haven't had the time to check.

Hanno
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User proposition 7.1 in San Martin book. http://www.ime.unicamp.br/~smartin/cursos/grupolie-2013/gruplie0.pdf

Alcides de carvalho jr
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