[proofplan] We identify the tangent space at the identity of the product Lie group $G\times H$ with the direct sum of the tangent spaces at the identities of $G$ and $H$. This gives the underlying vector-space isomorphism $\operatorname{Lie}(G\times H)\cong\mathfrak g\oplus\mathfrak h$. To prove that it preserves brackets, we write the left-invariant vector field on $G\times H$ determined by $(X,Y)$ and observe that it is the product of the left-invariant vector fields determined by $X$ on $G$ and by $Y$ on $H$. The Lie bracket of product vector fields is computed componentwise, which gives the desired formula. [/proofplan]

[step:Identify the tangent space at the product identity] Let $e_G\in G$ and $e_H\in H$ denote the identity elements, and let \begin{align*} K:=G\times H \end{align*} denote the product Lie group with identity \begin{align*} e_K:=(e_G,e_H). \end{align*} Let \begin{align*} \operatorname{pr}_G:K\to G \end{align*} and \begin{align*} \operatorname{pr}_H:K\to H \end{align*} be the smooth coordinate projection maps. Define a [linear map](/page/Linear%20Map) \begin{align*} \Phi:T_{e_K}K\to T_{e_G}G\oplus T_{e_H}H \end{align*} by \begin{align*} \Phi(Z):=((d\operatorname{pr}_G)_{e_K}(Z),(d\operatorname{pr}_H)_{e_K}(Z)). \end{align*} By the definition of the product smooth manifold structure, tangent vectors at $(e_G,e_H)$ are pairs of tangent vectors at $e_G$ and $e_H$, and the map $\Phi$ is the corresponding linear isomorphism. Since \begin{align*} \operatorname{Lie}(G)=T_{e_G}G \end{align*} and \begin{align*} \operatorname{Lie}(H)=T_{e_H}H, \end{align*} the map $\Phi$ gives a vector-space isomorphism \begin{align*} \operatorname{Lie}(K)=T_{e_K}K\cong \mathfrak g\oplus\mathfrak h. \end{align*} [/step]

[step:Write the left-invariant vector field determined by a product tangent vector]Let $\mathfrak X(M)$ denote the space of smooth vector fields on a smooth manifold $M$. Let $X\in\mathfrak g$ and $Y\in\mathfrak h$. Let $\widetilde X\in\mathfrak X(G)$ be the left-invariant vector field on $G$ determined by $X$, and let $\widetilde Y\in\mathfrak X(H)$ be the left-invariant vector field on $H$ determined by $Y$. Thus, for each $g\in G$ and $h\in H$, \begin{align*} \widetilde X_g=(dL^G_g)_{e_G}(X) \end{align*} and \begin{align*} \widetilde Y_h=(dL^H_h)_{e_H}(Y), \end{align*} where $L^G_g:G\to G$ is the left translation map $a\mapsto ga$, and $L^H_h:H\to H$ is the left translation map $b\mapsto hb$. Let $\widetilde{(X,Y)}\in\mathfrak X(K)$ be the left-invariant vector field on $K$ determined by $(X,Y)\in T_{e_K}K$. For $(g,h)\in K$, the left translation map \begin{align*} L^K_{(g,h)}:K\to K \end{align*} is given by \begin{align*} L^K_{(g,h)}(a,b)=(ga,hb). \end{align*} Therefore its differential at $e_K$ acts componentwise: \begin{align*} (dL^K_{(g,h)})_{e_K}(X,Y)=((dL^G_g)_{e_G}(X),(dL^H_h)_{e_H}(Y)). \end{align*} Hence \begin{align*} \widetilde{(X,Y)}_{(g,h)}=(\widetilde X_g,\widetilde Y_h). \end{align*}[/step]

[guided]The purpose of this step is to make the product identification compatible with the definition of the Lie bracket. The [Lie algebra](/page/Lie%20Algebra) bracket of a Lie group is defined by taking the bracket of the corresponding left-invariant vector fields and evaluating at the identity. So we must understand exactly which left-invariant vector field on $G\times H$ corresponds to a pair $(X,Y)$. Let $X\in\mathfrak g=T_{e_G}G$ and $Y\in\mathfrak h=T_{e_H}H$. The left-invariant vector field on $G$ determined by $X$ is the smooth vector field $\widetilde X\in\mathfrak X(G)$ satisfying \begin{align*} \widetilde X_g=(dL^G_g)_{e_G}(X) \end{align*} for every $g\in G$, where $L^G_g:G\to G$ is the smooth map $a\mapsto ga$. Likewise, the left-invariant vector field on $H$ determined by $Y$ is the smooth vector field $\widetilde Y\in\mathfrak X(H)$ satisfying \begin{align*} \widetilde Y_h=(dL^H_h)_{e_H}(Y) \end{align*} for every $h\in H$, where $L^H_h:H\to H$ is the smooth map $b\mapsto hb$. Now consider the product Lie group $K=G\times H$. Its left translation by $(g,h)\in K$ is the smooth map $L^K_{(g,h)}:K\to K$ defined by \begin{align*} L^K_{(g,h)}(a,b)=(ga,hb). \end{align*} This formula shows that $L^K_{(g,h)}$ is exactly the product map $L^G_g\times L^H_h$. Therefore its differential at the identity $e_K=(e_G,e_H)$ sends a pair of tangent vectors to the pair of differentials: \begin{align*} (dL^K_{(g,h)})_{e_K}(X,Y)=((dL^G_g)_{e_G}(X),(dL^H_h)_{e_H}(Y)). \end{align*} By the definitions of $\widetilde X$ and $\widetilde Y$, the right-hand side is \begin{align*} (\widetilde X_g,\widetilde Y_h). \end{align*} Thus the left-invariant vector field on $K$ determined by $(X,Y)$ is the product vector field \begin{align*} \widetilde{(X,Y)}_{(g,h)}=(\widetilde X_g,\widetilde Y_h). \end{align*} This is the key compatibility statement: the product tangent vector produces the product of the corresponding left-invariant vector fields.[/guided]

[step:Compute the bracket of product left-invariant vector fields componentwise] Let $X_1,X_2\in\mathfrak g$ and $Y_1,Y_2\in\mathfrak h$. Let $\widetilde X_1,\widetilde X_2\in\mathfrak X(G)$ and $\widetilde Y_1,\widetilde Y_2\in\mathfrak X(H)$ be their corresponding left-invariant vector fields. Define vector fields $A,B\in\mathfrak X(K)$ by \begin{align*} A_{(g,h)}=(\widetilde X_{1,g},\widetilde Y_{1,h}) \end{align*} and \begin{align*} B_{(g,h)}=(\widetilde X_{2,g},\widetilde Y_{2,h}) \end{align*} for every $(g,h)\in K$. By the previous step, $A=\widetilde{(X_1,Y_1)}$ and $B=\widetilde{(X_2,Y_2)}$. Choose a product coordinate chart around $(g,h)$ with coordinates $(x_1,\dots,x_m,y_1,\dots,y_n)$, where $(x_1,\dots,x_m)$ are coordinates from a chart on $G$ and $(y_1,\dots,y_n)$ are coordinates from a chart on $H$. In these coordinates, there are smooth coefficient functions $a_i,b_i$ on the $G$-chart and $\alpha_j,\beta_j$ on the $H$-chart such that \begin{align*} A = \sum_{i=1}^{m} a_i(x)\,\partial_{x_i} + \sum_{j=1}^{n} \alpha_j(y)\,\partial_{y_j} \end{align*} and \begin{align*} B = \sum_{i=1}^{m} b_i(x)\,\partial_{x_i} + \sum_{j=1}^{n} \beta_j(y)\,\partial_{y_j}. \end{align*} Using the coordinate formula for the Lie bracket of vector fields, we obtain \begin{align*} [A,B] = \sum_{i=1}^{m}\bigl(A(b_i)-B(a_i)\bigr)\,\partial_{x_i} + \sum_{j=1}^{n}\bigl(A(\beta_j)-B(\alpha_j)\bigr)\,\partial_{y_j}. \end{align*} Because $a_i$ and $b_i$ depend only on the $G$-coordinates and $\alpha_j$ and $\beta_j$ depend only on the $H$-coordinates, the mixed derivatives vanish. Therefore the $G$-component of $[A,B]$ is $[\widetilde X_1,\widetilde X_2]$ and the $H$-component is $[\widetilde Y_1,\widetilde Y_2]$, so \begin{align*} [A,B]_{(g,h)}=([\widetilde X_1,\widetilde X_2]_g,[\widetilde Y_1,\widetilde Y_2]_h). \end{align*} Evaluating at $(e_G,e_H)$ gives \begin{align*} [A,B]_{e_K}=([\widetilde X_1,\widetilde X_2]_{e_G},[\widetilde Y_1,\widetilde Y_2]_{e_H}). \end{align*} [/step]

[step:Conclude that the product identification preserves the Lie bracket] By definition of the Lie algebra bracket on a Lie group in terms of left-invariant vector fields, \begin{align*} [(X_1,Y_1),(X_2,Y_2)]_{\operatorname{Lie}(K)}=[\widetilde{(X_1,Y_1)},\widetilde{(X_2,Y_2)}]_{e_K}. \end{align*} Using the componentwise bracket computation from the previous step, this becomes \begin{align*} [(X_1,Y_1),(X_2,Y_2)]_{\operatorname{Lie}(K)}=([\widetilde X_1,\widetilde X_2]_{e_G},[\widetilde Y_1,\widetilde Y_2]_{e_H}). \end{align*} Again by the definition of the Lie algebra brackets on $G$ and $H$, \begin{align*} [\widetilde X_1,\widetilde X_2]_{e_G}=[X_1,X_2]_{\mathfrak g} \end{align*} and \begin{align*} [\widetilde Y_1,\widetilde Y_2]_{e_H}=[Y_1,Y_2]_{\mathfrak h}. \end{align*} Therefore \begin{align*} [(X_1,Y_1),(X_2,Y_2)]_{\operatorname{Lie}(G\times H)}=([X_1,X_2]_{\mathfrak g},[Y_1,Y_2]_{\mathfrak h}). \end{align*} Thus the natural vector-space isomorphism \begin{align*} \operatorname{Lie}(G\times H)\cong\mathfrak g\oplus\mathfrak h \end{align*} is an isomorphism of Lie algebras, where $\mathfrak g\oplus\mathfrak h$ has the componentwise bracket. [/step]