How is the Lie algebra of the image of the Adjoint representation related to the image of the adjoint representation?

Question

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.

Answer 1

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 $ad(\mathfrak{g})$.