Adjoint representation of a Lie group

Remarks

From $C_{gh} = C_g \circ C_h$ and the chain rule, $\operatorname{Ad}(gh) = \operatorname{Ad}(g)\operatorname{Ad}(h)$, so $\operatorname{Ad}$ is a representation; it is also smooth.

Example: for $G = GL_n(\R)$ and $\mathfrak{g} = M_n(\R)$, the map $C_g$ is the restriction of the linear map $X \mapsto g X g^{-1}$ of $M_n(\R)$. So $\operatorname{Ad}(g)X = g X g^{-1}$.

Non-example: the map $g \mapsto (dL_g)_e$, with $L_g(h) = gh$, does not define a representation on $\mathfrak{g}$. For $g \ne e$, $L_g$ does not fix $e$, so $(dL_g)_e$ sends $T_eG$ to $T_gG$, not to $\mathfrak{g}$.

The differential of $\operatorname{Ad}$ at $e$ is the adjoint representation $\operatorname{ad}$ of the Lie algebra $\mathfrak{g}$, given by $\operatorname{ad}(X)(Y) = [X, Y]$. The two are related by $\operatorname{Ad}(\exp X) = e^{\operatorname{ad} X}$, and $g \exp(X) g^{-1} = \exp(\operatorname{Ad}(g)X)$.