[proofplan] We construct a vector field from a tangent vector at the identity by translating it to every point of the group. Smoothness follows from the smoothness of multiplication, viewed through a smooth curve representing the tangent vector. Left-invariance is then a direct consequence of associativity of multiplication. Finally, every left-invariant vector field is recovered uniquely from its value at the identity, so the construction is bijective and linear. [/proofplan]

[step:Translate a tangent vector at the identity to a smooth vector field]Fix $X\in\mathfrak g=T_eG$. Define a section $\widetilde X:G\to TG$ by $\widetilde X(g)=(dL_g)_e(X)$. For each $g\in G$, the differential $(dL_g)_e:T_eG\to T_gG$ has codomain $T_gG$, so $\widetilde X_g\in T_gG$ and $\widetilde X$ is a vector field as a set-theoretic section of $TG\to G$. It remains to verify smoothness. By the curve characterization of tangent vectors on a smooth manifold, choose a smooth curve $c:(-\varepsilon,\varepsilon)\to G$ for some $\varepsilon>0$ such that $c(0)=e$ and $c'(0)=X$. Define $F:G\times(-\varepsilon,\varepsilon)\to G$ by $F(g,t)=gc(t)$. The map $F$ is smooth because it is the composition of the smooth map $\operatorname{id}_G\times c:G\times(-\varepsilon,\varepsilon)\to G\times G$ with the smooth multiplication map of the Lie group. For every $g\in G$, \begin{align*} \frac{d}{dt}\Big|_{t=0}F(g,t)=\frac{d}{dt}\Big|_{t=0}gc(t)=(dL_g)_e(X)=\widetilde X_g. \end{align*} Let $g_0\in G$, and let $(U,\varphi)$ be a smooth coordinate chart with $g_0\in U$. Since $F(g_0,0)=g_0\in U$ and $F$ is continuous, there exist an open neighbourhood $V\subset G$ of $g_0$ and a number $\delta\in(0,\varepsilon)$ such that $F(V\times(-\delta,\delta))\subset U$. In the chart $(U,\varphi)$, the coordinate components of $\widetilde X$ on $V$ are the $t$-derivatives at $t=0$ of the smooth coordinate functions of $\varphi\circ F:V\times(-\delta,\delta)\to\mathbb R^n$, hence are smooth functions of $g\in V$. Since $g_0$ was arbitrary, $\widetilde X$ is a smooth vector field on $G$.[/step]

[guided]We begin with a tangent vector $X\in\mathfrak g=T_eG$. The only natural way to create a vector at an arbitrary point $g\in G$ from $X$ is to use left translation. For each $g\in G$, left translation is the smooth map $L_g:G\to G$, $h\mapsto gh$. Its differential at the identity is a [linear map](/page/Linear%20Map) $(dL_g)_e:T_eG\to T_gG$. Thus the formula $\widetilde X_g=(dL_g)_e(X)$ really does define an element of the correct tangent space $T_gG$ for every $g\in G$. The main point is smoothness in the variable $g$. By the curve characterization of tangent vectors on a smooth manifold, choose a smooth curve $c:(-\varepsilon,\varepsilon)\to G$ with $c(0)=e$ and $c'(0)=X$. Such a curve represents the tangent vector $X$ at $e$. Now define $F:G\times(-\varepsilon,\varepsilon)\to G$ by $F(g,t)=gc(t)$. This map is smooth because $G$ is a Lie group: multiplication $G\times G\to G$ is smooth, and $(g,t)\mapsto (g,c(t))$ is smooth. For fixed $g\in G$, the curve $t\mapsto F(g,t)=gc(t)$ is exactly $L_g\circ c$. Therefore its velocity at $t=0$ is \begin{align*} \frac{d}{dt}\Big|_{t=0}F(g,t)=\frac{d}{dt}\Big|_{t=0}(L_g\circ c)(t)=(dL_g)_e(c'(0))=(dL_g)_e(X)=\widetilde X_g. \end{align*} So $\widetilde X_g$ is obtained by differentiating a smooth family of curves with respect to the parameter $t$ at $t=0$. To see that this produces a smooth vector field, fix a point $g_0\in G$ and choose a smooth coordinate chart $(U,\varphi)$ with $g_0\in U$. We need the family $F(g,t)$ to remain inside this chart before we write coordinate functions. Since $F(g_0,0)=g_0$ and $F$ is continuous, there are an open neighbourhood $V\subset G$ of $g_0$ and a number $\delta\in(0,\varepsilon)$ such that $F(V\times(-\delta,\delta))\subset U$. Now the coordinate expression $\varphi\circ F:V\times(-\delta,\delta)\to\mathbb R^n$ is smooth. Its [partial derivative](/page/Partial%20Derivative) with respect to $t$ at $t=0$ exists and depends smoothly on $g\in V$. These partial derivatives are exactly the coordinate components of the tangent vector $\widetilde X_g$, because $\widetilde X_g$ is the velocity at $t=0$ of the curve $t\mapsto F(g,t)$. Hence the coordinate components of $\widetilde X$ are smooth on $V$. Since $g_0$ was arbitrary, $\widetilde X$ is smooth on all of $G$.[/guided]

[step:Use associativity to prove left-invariance]Let $a,g\in G$. Since left translations satisfy \begin{align*} L_a\circ L_g=L_{ag}, \end{align*} the chain rule gives \begin{align*} (dL_a)_g(\widetilde X_g)=(dL_a)_g((dL_g)_eX)=(d(L_a\circ L_g))_e(X)=(dL_{ag})_e(X)=\widetilde X_{ag}. \end{align*} Thus $\widetilde X$ is left-invariant, so $\Phi(X)=\widetilde X$ belongs to $\mathfrak X_L(G)$.[/step]

[guided]Fix $a,g\in G$. To prove left-invariance, we must verify the defining identity for the vector field $\widetilde X$: \begin{align*} (dL_a)_g(\widetilde X_g)=\widetilde X_{ag}. \end{align*} The reason this should hold is that translating first by $g$ and then by $a$ is the same group operation as translating once by $ag$. Indeed, for every $h\in G$, \begin{align*} (L_a\circ L_g)(h)=L_a(gh)=a(gh)=(ag)h=L_{ag}(h), \end{align*} where the equality $a(gh)=(ag)h$ is associativity in the group $G$. Therefore \begin{align*} L_a\circ L_g=L_{ag}. \end{align*} Applying the chain rule to this equality at the point $e\in G$ gives \begin{align*} (d(L_a\circ L_g))_e=(dL_a)_{L_g(e)}\circ(dL_g)_e. \end{align*} Since $L_g(e)=ge=g$, this becomes \begin{align*} (d(L_a\circ L_g))_e=(dL_a)_g\circ(dL_g)_e. \end{align*} Evaluating both sides on $X\in T_eG$ and using the definition $\widetilde X_g=(dL_g)_e(X)$, we obtain \begin{align*} (dL_a)_g(\widetilde X_g)=(dL_a)_g((dL_g)_e(X))=(d(L_a\circ L_g))_e(X). \end{align*} Because $L_a\circ L_g=L_{ag}$, the right-hand side is \begin{align*} (d(L_a\circ L_g))_e(X)=(dL_{ag})_e(X)=\widetilde X_{ag}. \end{align*} This is exactly the left-invariance condition. Hence $\widetilde X\in\mathfrak X_L(G)$.[/guided]

[step:Recover a left-invariant vector field from its value at the identity]Let $Y\in\mathfrak X_L(G)$ be a left-invariant smooth vector field. Define $X:=Y_e\in T_eG=\mathfrak g$. Using left-invariance with $a=g$ and with the identity element as the second group element, we obtain \begin{align*} Y_g=Y_{ge}=(dL_g)_e(Y_e)=(dL_g)_e(X)=\widetilde X_g. \end{align*} Since this holds for every $g\in G$, we have $Y=\widetilde X=\Phi(X)$. Therefore every left-invariant smooth vector field lies in the image of $\Phi$.[/step]

[guided]Let $Y\in\mathfrak X_L(G)$ be a left-invariant smooth vector field. The construction $X\mapsto\widetilde X$ starts from a tangent vector at the identity, so the only possible tangent vector that could recover $Y$ is its value at the identity. Define $X:=Y_e\in T_eG=\mathfrak g$. We now use the left-invariance of $Y$. By definition, for all $a,h\in G$, \begin{align*} (dL_a)_h(Y_h)=Y_{ah}. \end{align*} Take $a=g$ and $h=e$. Since $ge=g$, this gives \begin{align*} (dL_g)_e(Y_e)=Y_g. \end{align*} Substituting $X=Y_e$, we obtain \begin{align*} Y_g=(dL_g)_e(X)=\widetilde X_g. \end{align*} This equality holds for every $g\in G$, so the two sections $Y:G\to TG$ and $\widetilde X:G\to TG$ agree pointwise. Therefore $Y=\widetilde X=\Phi(X)$, and every left-invariant smooth vector field is in the image of $\Phi$.[/guided]

[step:Check uniqueness and linearity]If $\Phi(X)=0$, then evaluating at $e$ gives \begin{align*} 0=\widetilde X_e=(dL_e)_e(X). \end{align*} Let $\operatorname{id}_G:G\to G$ denote the identity map on $G$. Since $L_e=\operatorname{id}_G$, its differential at $e$ is the identity map on $T_eG$, so $X=0$. Hence $\Phi$ is injective. For $X_1,X_2\in\mathfrak g$ and $\lambda,\mu\in\mathbb R$, the linearity of each differential $(dL_g)_e:T_eG\to T_gG$ gives, for every $g\in G$, \begin{align*} \Phi(\lambda X_1+\mu X_2)_g=(dL_g)_e(\lambda X_1+\mu X_2)=\lambda(dL_g)_e(X_1)+\mu(dL_g)_e(X_2)=\lambda\Phi(X_1)_g+\mu\Phi(X_2)_g. \end{align*} Thus $\Phi$ is linear. Together with the surjectivity proved above, $\Phi:\mathfrak g\to\mathfrak X_L(G)$ is an isomorphism of real vector spaces.[/step]

[guided]We first prove injectivity. Suppose $\Phi(X)=0$ for some $X\in\mathfrak g$. This means that the vector field $\widetilde X$ is the zero section of $TG\to G$. Evaluating at the identity element $e\in G$ gives \begin{align*} 0=\widetilde X_e=(dL_e)_e(X). \end{align*} Let $\operatorname{id}_G:G\to G$ denote the identity map on $G$. Since $L_e(g)=eg=g$ for every $g\in G$, we have $L_e=\operatorname{id}_G$. The differential of the identity map at $e$ is the identity linear map on $T_eG$, so \begin{align*} (dL_e)_e=\operatorname{id}_{T_eG}. \end{align*} Therefore $0=(dL_e)_e(X)=X$, and $\Phi$ is injective. We next prove linearity. Let $X_1,X_2\in\mathfrak g$ and let $\lambda,\mu\in\mathbb R$. For each $g\in G$, the differential \begin{align*} (dL_g)_e:T_eG\to T_gG \end{align*} is a linear map between real vector spaces. Hence \begin{align*} \Phi(\lambda X_1+\mu X_2)_g=(dL_g)_e(\lambda X_1+\mu X_2)=\lambda(dL_g)_e(X_1)+\mu(dL_g)_e(X_2)=\lambda\Phi(X_1)_g+\mu\Phi(X_2)_g. \end{align*} Using the definition of $\Phi$, this is \begin{align*} \Phi(\lambda X_1+\mu X_2)_g=\lambda\Phi(X_1)_g+\mu\Phi(X_2)_g. \end{align*} Since the equality holds at every $g\in G$, we have \begin{align*} \Phi(\lambda X_1+\mu X_2)=\lambda\Phi(X_1)+\mu\Phi(X_2). \end{align*} Thus $\Phi$ is linear. The previous step proved that $\Phi$ is surjective, and the first paragraph of this step proved that $\Phi$ is injective. Therefore $\Phi:\mathfrak g\to\mathfrak X_L(G)$ is an isomorphism of real vector spaces.[/guided]