Past exam of the mathematics course of the University of Cambridge / 2003 / iii / Paper 1 / 3 / i / Solution

For the embedded Lie group, the tangent space $T_{e}G$ consists of velocities ${\gamma }^{\prime }(0)$ 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 $c_g(a)=ga{g}^{-1}$. Since this conjugation map fixes $e$, its derivative defines the Adjoint representation of a Lie group:The inverse is ${\mathrm{Ad}}_{g^{-1}}$, and $c_{gh}=c_g\circ c_h$ gives ${\mathrm{Ad}}_{gh}={\mathrm{Ad}}_g{\mathrm{Ad}}_h$. Thus this is a smooth group homomorphism into the linear automorphisms of $T_{e}G$.

Differentiate that representation at the identity. The tangent space of $GL(T_{e}G)$ at its identity is $\mathrm{End}(T_{e}G)$, so $\mathrm{ad}=(d\mathrm{Ad}{)}_e$ is a linear map from $T_{e}G$ into those endomorphisms. The Lie bracket isEquivalently, if $\eta (0)=e$ and ${\eta }^{\prime }(0)=Y$, 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 ${\mathrm{Ad}}_g$ preserves this bracket. Equivalently identify $X$ with its left-invariant vector field $X^L(g)=(d{L}_g{)}_{e}X$; the bracket above agrees with the Lie bracket of vector fields at $e$. These checks make $T_{e}G$ a Lie algebra and upgrade the adjoint maps to its Lie algebra automorphisms.