= Solution
For the embedded <Lie group>, the <tangent space> $T_eG$ consists of velocities $\gamma'(0)$ of smooth curves in $G$ satisfying $\gamma(0)=e$. The ambient coordinates identify these velocities with vectors in $\mathbb R^n$, but the tangent space itself is intrinsic. Let $c_g(a)=gag^{-1}$. Since this conjugation map fixes $e$, its derivative defines the <Adjoint representation of a Lie group>:
$$
\boxed{\operatorname{Ad}_g=(dc_g)_e:T_eG\to T_eG.}
$$
The inverse is $\operatorname{Ad}_{g^{-1}}$, and $c_{gh}=c_g\circ c_h$ gives $\operatorname{Ad}_{gh}=\operatorname{Ad}_g\operatorname{Ad}_h$. Thus this is a smooth group homomorphism into the linear automorphisms of $T_eG$.
Differentiate that representation at the identity. The tangent space of $GL(T_eG)$ at its identity is $\operatorname{End}(T_eG)$, so $\operatorname{ad}=(d\operatorname{Ad})_e$ is a linear map from $T_eG$ into those endomorphisms. The <Lie bracket> is
$$
\boxed{[X,Y]=\operatorname{ad}(X)Y=\left.\frac d{ds}\right|_{s=0}\operatorname{Ad}_{\gamma(s)}Y,\quad\gamma(0)=e,\quad\gamma'(0)=X.}
$$
Equivalently, if $\eta(0)=e$ and $\eta'(0)=Y$, this is the mixed derivative at $(0,0)$ of $\gamma(s)\eta(t)\gamma(s)^{-1}$ in ambient coordinates. For a matrix <Lie group>, conjugation is ordinary matrix conjugation and this formula reduces to $[X,Y]=XY-YX$.
The checks required are that the derivative is independent of the representing curves, that it is bilinear and alternating, and that it satisfies the <Jacobi identity>. One must also check that each $\operatorname{Ad}_g$ preserves this bracket. Equivalently identify $X$ with its left-invariant vector field $X^L(g)=(dL_g)_eX$; the bracket above agrees with the <Lie bracket of vector fields> at $e$. These checks make $T_eG$ a <Lie algebra> and upgrade the adjoint maps to its Lie algebra automorphisms.
Back to article page
