For simplicity, let $G$ be a matrix Lie group (closed subgroup of $GL$) with associated Lie algebra $\mathfrak{g}$. Let Ad and ad denote the Adjoint and adjoint representation on $G,\mathfrak{g}$. We know that the Lie algebra of the kernel of Ad is isomorphic to the kernel of ad, but how is the image of Ad related to the image of ad? I suspect that the centers of Ad, ad must be involved, but I'm not sure of the details.
How is the Lie algebra of the image of the Adjoint representation related to the image of the adjoint representation
lie-algebraslie-groups
Best Answer
Suppose that we have the definition $\mathfrak{g} = Lie(G) = \{ X \in Mat(n) \mid e^{tX} \in G \ \forall t \in \mathbb{R} \}$. And recall that if $\varphi: G \to H$ is a homomorphims of Lie groups then $$ \varphi(e^{tX}) = e^{t d\varphi X}, \qquad X \in \mathfrak g.$$
For $ad_X \in ad (\mathfrak{g})$ we have
$$ e^{tad_X} = Ad_{e^{tX}} \in Ad(G), $$
for all $t$, since the derivative of $Ad$ is $ad$, so $ad(\mathfrak{g}) \subseteq Lie(Ad(G))$. On the other hand since $ad(\mathfrak{g}) \cong \mathfrak{g}/\mathfrak{z(g)}$ and $Ad(G) \cong G/N$ for some subgroup $Z(G) \subseteq N$ by dimensions $$ \dim (Ad(G)) = \dim G - \dim N \leq \dim G - \dim Z(G) = \dim \mathfrak g - \dim \mathfrak{z(g)} = \dim (ad (\mathfrak{g})) $$ we have that $\dim Lie(Ad(G)) \leq \dim (ad(\mathfrak{g}))$ and therefore the Lie algebra of $Ad(G)$ is $ad(\mathfrak{g})$.