[proofplan] We prove first that differentiating conjugation gives a representation: the identity $C_{g_1g_2}=C_{g_1}\circ C_{g_2}$ differentiates at $e$ to give $\operatorname{Ad}_{g_1g_2}=\operatorname{Ad}_{g_1}\circ\operatorname{Ad}_{g_2}$. This also shows that each $\operatorname{Ad}_g$ is invertible, with inverse $\operatorname{Ad}_{g^{-1}}$. Finally, smoothness is checked in local coordinates by viewing $(g,h)\mapsto ghg^{-1}$ as a smooth family of maps fixing $e$ and observing that its derivative in the $h$-variable depends smoothly on the parameter $g$. [/proofplan]

[step:Define conjugation as a smooth family fixing the identity] For each $g\in G$, the map \begin{align*} C_g:G\to G,\qquad h\mapsto ghg^{-1}, \end{align*} is smooth, since it is the composition of the smooth maps $h\mapsto gh$, $k\mapsto kg^{-1}$, and the inversion map used to form the fixed element $g^{-1}$. Moreover, \begin{align*} C_g(e)=geg^{-1}=e. \end{align*} Therefore the differential at $e$ is a [linear map](/page/Linear%20Map) \begin{align*} d(C_g)_e:T_eG\to T_eG. \end{align*} Since $\mathfrak g=T_eG$, this is exactly a linear map \begin{align*} \operatorname{Ad}_g:\mathfrak g\to\mathfrak g. \end{align*} [/step]

[step:Differentiate the conjugation identity to get the homomorphism law]Let $g_1,g_2\in G$. For every $h\in G$, \begin{align*} C_{g_1g_2}(h)=(g_1g_2)h(g_1g_2)^{-1}=g_1(g_2hg_2^{-1})g_1^{-1}=C_{g_1}(C_{g_2}(h)). \end{align*} Thus \begin{align*} C_{g_1g_2}=C_{g_1}\circ C_{g_2}. \end{align*} Both maps $C_{g_2}$ and $C_{g_1}$ send $e$ to $e$, so the [chain rule for differentials](/theorems/3907) at $e$ gives \begin{align*} d(C_{g_1g_2})_e=d(C_{g_1})_e\circ d(C_{g_2})_e. \end{align*} By the definition of $\operatorname{Ad}$, this is \begin{align*} \operatorname{Ad}_{g_1g_2}=\operatorname{Ad}_{g_1}\circ\operatorname{Ad}_{g_2}. \end{align*}[/step]

[guided]We want to prove that $g\mapsto \operatorname{Ad}_g$ respects multiplication. The input is the convention \begin{align*} C_g(h)=ghg^{-1}. \end{align*} With this convention, for $g_1,g_2,h\in G$ we compute \begin{align*} C_{g_1g_2}(h)=(g_1g_2)h(g_1g_2)^{-1}. \end{align*} Using the group identity $(g_1g_2)^{-1}=g_2^{-1}g_1^{-1}$, this becomes \begin{align*} C_{g_1g_2}(h)=g_1g_2hg_2^{-1}g_1^{-1}. \end{align*} The middle part is $C_{g_2}(h)=g_2hg_2^{-1}$, so \begin{align*} C_{g_1g_2}(h)=g_1C_{g_2}(h)g_1^{-1}=C_{g_1}(C_{g_2}(h)). \end{align*} Since this holds for every $h\in G$, we have the identity of smooth maps \begin{align*} C_{g_1g_2}=C_{g_1}\circ C_{g_2}. \end{align*} Now we differentiate this identity at the identity element $e$. The chain rule applies because $C_{g_1}$ and $C_{g_2}$ are smooth. Also $C_{g_2}(e)=e$, so the middle base point in the chain rule is still $e$. Hence \begin{align*} d(C_{g_1g_2})_e=d(C_{g_1})_{C_{g_2}(e)}\circ d(C_{g_2})_e=d(C_{g_1})_e\circ d(C_{g_2})_e. \end{align*} By definition, $d(C_g)_e=\operatorname{Ad}_g$ for each $g\in G$. Therefore \begin{align*} \operatorname{Ad}_{g_1g_2}=\operatorname{Ad}_{g_1}\circ\operatorname{Ad}_{g_2}. \end{align*} This proves the multiplication law for the adjoint map.[/guided]

[step:Use the identity element to prove that each adjoint map is invertible] The conjugation map associated to the identity element is \begin{align*} C_e:G\to G,\qquad h\mapsto ehe^{-1}=h, \end{align*} so $C_e=\operatorname{id}_G$. Therefore \begin{align*} \operatorname{Ad}_e=d(\operatorname{id}_G)_e=\operatorname{id}_{\mathfrak g}. \end{align*} Applying the homomorphism law to $g$ and $g^{-1}$ gives \begin{align*} \operatorname{Ad}_g\circ \operatorname{Ad}_{g^{-1}}=\operatorname{Ad}_{gg^{-1}}=\operatorname{Ad}_e=\operatorname{id}_{\mathfrak g}, \end{align*} and \begin{align*} \operatorname{Ad}_{g^{-1}}\circ \operatorname{Ad}_g=\operatorname{Ad}_{g^{-1}g}=\operatorname{Ad}_e=\operatorname{id}_{\mathfrak g}. \end{align*} Thus $\operatorname{Ad}_g$ is invertible and \begin{align*} (\operatorname{Ad}_g)^{-1}=\operatorname{Ad}_{g^{-1}}. \end{align*} Hence $\operatorname{Ad}_g\in GL(\mathfrak g)$ for every $g\in G$. [/step]

[step:Check smoothness in local coordinates] Define the smooth map \begin{align*} F:G\times G\to G,\qquad (g,h)\mapsto ghg^{-1}. \end{align*} We show that \begin{align*} G\to \operatorname{End}(\mathfrak g),\qquad g\mapsto d(F(g,\cdot))_e, \end{align*} is smooth. Let $g_0\in G$. Choose a chart $(U,\varphi)$ of $G$ around $e$ with $\varphi(e)=0$, and choose a chart $(P,\psi)$ of $G$ around $g_0$. After replacing $P$ by a smaller open neighbourhood of $g_0$ and $U$ by a smaller open neighbourhood of $e$, assume \begin{align*} F(P\times U)\subset U. \end{align*} Define the coordinate representative \begin{align*} \widetilde F:\psi(P)\times \varphi(U)\to \varphi(U),\qquad (a,y)\mapsto \varphi(F(\psi^{-1}(a),\varphi^{-1}(y))). \end{align*} This map is smooth. Since $F(g,e)=e$ for all $g\in G$, we have \begin{align*} \widetilde F(a,0)=0 \end{align*} for all $a\in \psi(P)$. Under the linear identification $\mathfrak g=T_eG\cong \mathbb R^n$ induced by $d\varphi_e$, the linear map $\operatorname{Ad}_{\psi^{-1}(a)}$ is represented by the Jacobian matrix of the partial map $y\mapsto \widetilde F(a,y)$ at $0$. Its entries are \begin{align*} \frac{\partial \widetilde F_i}{\partial y_j}(a,0), \end{align*} where $\widetilde F_i$ denotes the $i$-th coordinate function of $\widetilde F$. These entries are smooth functions of $a$, because $\widetilde F$ is smooth. Therefore $g\mapsto \operatorname{Ad}_g$ is smooth as a map from $G$ to $\operatorname{End}(\mathfrak g)$ near $g_0$. Since $g_0$ was arbitrary, it is smooth on all of $G$. The space $GL(\mathfrak g)$ is the open subset of $\operatorname{End}(\mathfrak g)$ consisting of invertible linear maps. Since the preceding step proved that $\operatorname{Ad}_g\in GL(\mathfrak g)$ for every $g\in G$, the same map is smooth as a map \begin{align*} \operatorname{Ad}:G\to GL(\mathfrak g). \end{align*} [/step]

[step:Conclude that the adjoint map is a smooth Lie group homomorphism] The homomorphism law \begin{align*} \operatorname{Ad}_{g_1g_2}=\operatorname{Ad}_{g_1}\circ\operatorname{Ad}_{g_2} \end{align*} holds for all $g_1,g_2\in G$, and the identity element satisfies \begin{align*} \operatorname{Ad}_e=\operatorname{id}_{\mathfrak g}. \end{align*} Each $\operatorname{Ad}_g$ lies in $GL(\mathfrak g)$, and the map $g\mapsto \operatorname{Ad}_g$ is smooth. Hence \begin{align*} \operatorname{Ad}:G\to GL(\mathfrak g) \end{align*} is a smooth [group homomorphism](/page/Group%20Homomorphism). [/step]