[proofplan] We prove the identity by testing both tangent vectors on an arbitrary smooth germ at the identity. The curve $t\mapsto \operatorname{Ad}_{\exp(tX)}Y$ is computed by differentiating the conjugation curve $s\mapsto \exp(tX)\exp(sY)\exp(-tX)$ at $s=0$. The Lie bracket with the left-invariant convention is computed from the flows of $X^L$ and $Y^L$, and the two resulting mixed derivatives agree with the positive sign. [/proofplan]

[step:Represent tangent vectors as derivations on smooth functions] Let $X,Y\in\mathfrak g$. Let $\mathfrak X(G)=\Gamma(TG)$ denote the real [vector space](/page/Vector%20Space) of smooth vector fields on $G$. For each $g\in G$, let $L_g:G\to G$ denote the left translation map $L_g(h)=gh$. Let $X^L,Y^L\in\mathfrak X(G)$ denote the unique left-invariant vector fields satisfying $X^L_e=X$, $Y^L_e=Y$, $X^L_g=d(L_g)_e(X)$, and $Y^L_g=d(L_g)_e(Y)$ for every $g\in G$. Let $f:G\to\mathbb R$ be a smooth function defined on an open neighbourhood of $e$. We regard a tangent vector $v\in T_eG$ as the derivation \begin{align*} v(f)=df_e(v). \end{align*} It is enough to prove that, for every such $f$, \begin{align*} \operatorname{ad}_X(Y)(f)=[X^L,Y^L]_e(f), \end{align*} because tangent vectors at $e$ are determined by their values on smooth germs at $e$. [/step]

[step:Compute the infinitesimal adjoint action as a mixed derivative]Define the one-parameter curve \begin{align*} a:\mathbb R\to G \end{align*} by $a(t)=\exp(tX)$. For each $t\in\mathbb R$, the curve \begin{align*} \gamma_t:\mathbb R\to G \end{align*} defined by \begin{align*} \gamma_t(s)=C_{a(t)}(\exp(sY))=a(t)\exp(sY)a(t)^{-1} \end{align*} satisfies $\gamma_t(0)=e$. Therefore, by the definition of pushforward, \begin{align*} \operatorname{Ad}_{a(t)}Y = d(C_{a(t)})_e(Y) = \left.\frac{d}{ds}\right|_{s=0}a(t)\exp(sY)a(t)^{-1}. \end{align*} Applying this tangent vector to $f$ gives \begin{align*} (\operatorname{Ad}_{a(t)}Y)(f) = \left.\frac{d}{ds}\right|_{s=0} f(a(t)\exp(sY)a(t)^{-1}). \end{align*} Since the map $(t,s)\mapsto f(a(t)\exp(sY)a(t)^{-1})$ is smooth near $(0,0)$, differentiating with respect to $t$ at $0$ gives \begin{align*} \operatorname{ad}_X(Y)(f) = \left.\frac{d}{dt}\right|_{t=0} \left.\frac{d}{ds}\right|_{s=0} f(a(t)\exp(sY)a(t)^{-1}). \end{align*}[/step]

[guided]The adjoint action is defined as the differential of conjugation at the identity, so to understand $\operatorname{Ad}_{a(t)}Y$ we must choose a curve through $e$ with initial velocity $Y$. The exponential curve \begin{align*} s\mapsto \exp(sY) \end{align*} has initial velocity $Y$ at $s=0$. Conjugating this curve by $a(t)=\exp(tX)$ gives the curve \begin{align*} \gamma_t:\mathbb R\to G \end{align*} defined by \begin{align*} \gamma_t(s)=a(t)\exp(sY)a(t)^{-1}. \end{align*} Because $\gamma_t(0)=a(t)ea(t)^{-1}=e$, its initial velocity lies in $T_eG$. By the definition of the differential, \begin{align*} \left.\frac{d}{ds}\right|_{s=0}\gamma_t(s) = d(C_{a(t)})_e(Y) = \operatorname{Ad}_{a(t)}Y. \end{align*} Testing this tangent vector against the smooth function $f$ gives \begin{align*} (\operatorname{Ad}_{a(t)}Y)(f) = \left.\frac{d}{ds}\right|_{s=0} f(a(t)\exp(sY)a(t)^{-1}). \end{align*} Now the infinitesimal adjoint action is the derivative in $t$ of this curve in $T_eG$. Since multiplication, inversion, the exponential map, and $f$ are smooth, the two-variable function \begin{align*} (t,s)\mapsto f(a(t)\exp(sY)a(t)^{-1}) \end{align*} is smooth near $(0,0)$. Hence the derivative defining $\operatorname{ad}_X(Y)$ is the mixed derivative \begin{align*} \operatorname{ad}_X(Y)(f) = \left.\frac{d}{dt}\right|_{t=0} \left.\frac{d}{ds}\right|_{s=0} f(a(t)\exp(sY)a(t)^{-1}). \end{align*} This formula is the place where the sign is fixed: the conjugating element $a(t)$ appears on the left and its inverse appears on the right.[/guided]

[step:Compute the left-invariant bracket by differentiating flows] For a left-invariant vector field $Z^L$ with $Z\in\mathfrak g$, its flow is the smooth map \begin{align*} \Phi^Z:\mathbb R\times G\to G \end{align*} given by \begin{align*} \Phi^Z_r(g)=g\exp(rZ). \end{align*} Indeed, $\Phi^Z_0(g)=g$, and the velocity of $r\mapsto g\exp(rZ)$ at time $r$ is $d(L_{g\exp(rZ)})_e(Z)=Z^L_{g\exp(rZ)}$. Using the derivation definition of the vector-field bracket, we have \begin{align*} [X^L,Y^L]_e(f)=X^L_e(Y^L f)-Y^L_e(X^L f). \end{align*} The first term is \begin{align*} X^L_e(Y^L f) = \left.\frac{d}{dt}\right|_{t=0} (Y^L f)(\exp(tX)). \end{align*} Since the flow of $Y^L$ through $\exp(tX)$ is $s\mapsto \exp(tX)\exp(sY)$, \begin{align*} (Y^L f)(\exp(tX)) = \left.\frac{d}{ds}\right|_{s=0} f(\exp(tX)\exp(sY)). \end{align*} Thus \begin{align*} X^L_e(Y^L f) = \left.\frac{d}{dt}\right|_{t=0} \left.\frac{d}{ds}\right|_{s=0} f(\exp(tX)\exp(sY)). \end{align*} Similarly, \begin{align*} Y^L_e(X^L f) = \left.\frac{d}{ds}\right|_{s=0} \left.\frac{d}{dt}\right|_{t=0} f(\exp(sY)\exp(tX)). \end{align*} Therefore \begin{align*} [X^L,Y^L]_e(f) = \left.\frac{d}{dt}\right|_{t=0} \left.\frac{d}{ds}\right|_{s=0} f(\exp(tX)\exp(sY)) - \left.\frac{d}{ds}\right|_{s=0} \left.\frac{d}{dt}\right|_{t=0} f(\exp(sY)\exp(tX)). \end{align*} [/step]

[step:Match the two mixed derivatives and obtain the identity]Define \begin{align*} F:\mathbb R^2\to\mathbb R \end{align*} by \begin{align*} F(t,s)=f(\exp(tX)\exp(sY)\exp(-tX)). \end{align*} For each fixed $t$, the $s$-derivative at $s=0$ is exactly \begin{align*} \left.\frac{d}{ds}\right|_{s=0}F(t,s) = (\operatorname{Ad}_{\exp(tX)}Y)(f), \end{align*} so \begin{align*} \operatorname{ad}_X(Y)(f) = \left.\frac{d}{dt}\right|_{t=0} \left.\frac{d}{ds}\right|_{s=0}F(t,s). \end{align*} To compare this with the bracket, fix $s$ near $0$ and apply the product rule to the smooth map \begin{align*} t\mapsto \exp(tX)\exp(sY)\exp(-tX) \end{align*} at $t=0$. More explicitly, write the multiplication map as $m:G\times G\to G$, $m(g,h)=gh$, and define $p_s:G\times G\to G$ by $p_s(g,h)=g\exp(sY)h$. Then the curve above is $p_s(\exp(tX),\exp(-tX))$. Since $\frac{d}{dt}|_{t=0}\exp(tX)=X\in T_eG$ and $\frac{d}{dt}|_{t=0}\exp(-tX)=-X\in T_eG$, the chain rule gives its velocity at $t=0$ as \begin{align*} d(p_s)_{(e,e)}(X,-X). \end{align*} By linearity of the differential in the product tangent space $T_{(e,e)}(G\times G)\cong T_eG\oplus T_eG$, this is the sum of the contribution from $(X,0)$ and the contribution from $(0,-X)$. After applying $df_{\exp(sY)}$, the first contribution is represented by $t\mapsto f(\exp(tX)\exp(sY))$, and the second contribution is the negative of the contribution represented by $t\mapsto f(\exp(sY)\exp(tX))$. Therefore \begin{align*} \left.\frac{d}{dt}\right|_{t=0}F(t,s) = \left.\frac{d}{dt}\right|_{t=0}f(\exp(tX)\exp(sY)) - \left.\frac{d}{dt}\right|_{t=0}f(\exp(sY)\exp(tX)). \end{align*} The functions of $(t,s)$ appearing here are smooth near $(0,0)$, because multiplication, inversion, the exponential map, and $f$ are smooth. Hence equality of mixed partial derivatives gives \begin{align*} \left.\frac{d}{ds}\right|_{s=0}\left.\frac{d}{dt}\right|_{t=0}f(\exp(tX)\exp(sY)) = \left.\frac{d}{dt}\right|_{t=0}\left.\frac{d}{ds}\right|_{s=0}f(\exp(tX)\exp(sY)). \end{align*} Differentiating the displayed identity for $\partial_tF(0,s)$ with respect to $s$ at $s=0$ and using this equality of mixed partial derivatives gives \begin{align*} \left.\frac{d}{dt}\right|_{t=0} \left.\frac{d}{ds}\right|_{s=0} f(\exp(tX)\exp(sY)\exp(-tX)) = [X^L,Y^L]_e(f). \end{align*} Combining this with the computation of the infinitesimal adjoint action yields \begin{align*} \operatorname{ad}_X(Y)(f)=[X^L,Y^L]_e(f). \end{align*} Since this holds for every smooth germ $f$ at $e$, the tangent vectors are equal: \begin{align*} \operatorname{ad}_X(Y)=[X^L,Y^L]_e=[X,Y]. \end{align*} This proves the theorem.[/step]

[guided]We now compare the conjugation calculation with the bracket calculation. Define $F:\mathbb R^2\to\mathbb R$ by \begin{align*} F(t,s)=f(\exp(tX)\exp(sY)\exp(-tX)). \end{align*} For fixed $s$, the curve in the argument of $f$ has three factors. The middle factor $\exp(sY)$ is constant in $t$, so we isolate the two moving outer factors. Define $p_s:G\times G\to G$ by $p_s(g,h)=g\exp(sY)h$. Then \begin{align*} F(t,s)=f(p_s(\exp(tX),\exp(-tX))). \end{align*} The velocities of the two factor curves at $t=0$ are $X\in T_eG$ and $-X\in T_eG$. Hence the chain rule gives the velocity of $t\mapsto p_s(\exp(tX),\exp(-tX))$ at $t=0$ as \begin{align*} d(p_s)_{(e,e)}(X,-X). \end{align*} Using the vector-space identification $T_{(e,e)}(G\times G)\cong T_eG\oplus T_eG$ and the linearity of $d(p_s)_{(e,e)}$, this tangent vector is the sum of the contribution from $(X,0)$ and the contribution from $(0,-X)$. After applying $df_{\exp(sY)}$, the first contribution is represented by \begin{align*} \left.\frac{d}{dt}\right|_{t=0}f(\exp(tX)\exp(sY)), \end{align*} and the second contribution is \begin{align*} -\left.\frac{d}{dt}\right|_{t=0}f(\exp(sY)\exp(tX)). \end{align*} Therefore, for every $s$ near $0$, \begin{align*} \left.\frac{d}{dt}\right|_{t=0}F(t,s) = \left.\frac{d}{dt}\right|_{t=0}f(\exp(tX)\exp(sY)) - \left.\frac{d}{dt}\right|_{t=0}f(\exp(sY)\exp(tX)). \end{align*} Now differentiate this identity with respect to $s$ at $s=0$. Because multiplication, inversion, the exponential map, and $f$ are smooth, all two-variable functions involved are smooth near $(0,0)$. Thus the mixed partial derivatives commute. In particular, \begin{align*} \left.\frac{d}{ds}\right|_{s=0}\left.\frac{d}{dt}\right|_{t=0}f(\exp(tX)\exp(sY)) = \left.\frac{d}{dt}\right|_{t=0}\left.\frac{d}{ds}\right|_{s=0}f(\exp(tX)\exp(sY)). \end{align*} The bracket computation from the previous step states that \begin{align*} [X^L,Y^L]_e(f) = \left.\frac{d}{dt}\right|_{t=0} \left.\frac{d}{ds}\right|_{s=0} f(\exp(tX)\exp(sY)) - \left.\frac{d}{ds}\right|_{s=0} \left.\frac{d}{dt}\right|_{t=0} f(\exp(sY)\exp(tX)). \end{align*} The equality of mixed partial derivatives just established converts the derivative in $s$ of the right-hand side of the identity for $\partial_tF(0,s)$ into this displayed bracket expression. Consequently, \begin{align*} \left.\frac{d}{dt}\right|_{t=0} \left.\frac{d}{ds}\right|_{s=0} f(\exp(tX)\exp(sY)\exp(-tX)) = [X^L,Y^L]_e(f). \end{align*} The left-hand side is the value of $\operatorname{ad}_X(Y)$ on $f$, so the two tangent vectors have the same action on the arbitrary smooth germ $f$ at $e$.[/guided]