Past exam of the mathematics course of the University of Cambridge / 2015 / iii / Paper 17 / 2 / Solution

A vector field is a smooth section $X:M\to TM$ of the tangent bundle. It differentiates smooth functions by $Xf(p)=X_p(f)$. Define the Lie bracket of vector fields intrinsically byExpanding this expression on a product $fh$ shows that the two mixed terms cancel, giving $[X,Y](fh)=f[X,Y]h+h[X,Y]f$. It is therefore a derivation, hence a vector field. In local coordinates,Its coefficients are smooth. The intrinsic definition depends on no chart, which proves coordinate independence of this formula and defines the bracket on the whole manifold.

For a Lie group $G$, let $L_g(h)=gh$ be Left translation on a Lie group. A left-invariant vector field satisfies $(d{L}_g{)}_h{X}_h=X_{gh}$ for every $g,h$. Evaluation $X\mapsto X_e$ is linear and injective, since $X_g=(d{L}_g{)}_e{X}_e$. Conversely any $B\in T_{e}G$ gives a smooth left-invariant vector field $X_B(g)=(d{L}_g{)}_{e}B$. Smoothness follows from smooth multiplication and its differential. ThusFor any diffeomorphism $F$, the intrinsic bracket identity on functions gives $F_{\ast }[X,Y]=[F_{\ast }X,F_{\ast }Y]$. Applying this naturality of the Lie bracket to $F=L_g$ proves that the bracket of two left-invariant vector fields remains left invariant. Evaluation at $e$ therefore defines the Lie algebra bracket on $\mathfrak{g}=T_{e}G$.

For the special orthogonal group, differentiation of $Q(t{)}^{T}Q(t)=I$ at $Q(0)=I$ gives $B^T+B=0$, so its tangent space is contained in the skew-symmetric matrices. Conversely, for any such $B$, the matrix exponential $Q(t)=e^{tB}$ satisfies $Q(t{)}^{T}Q(t)=I$ and lies in the determinant-one component, because $Q(0)=I$ and its determinant varies continuously in $\{\pm 1\}$. Its initial derivative is $B$, provingFor matrices, $X_B(Q)=QB$. Extend these fields to the open matrix group $\mathrm{GL}(n,\mathbb{R})$, where the differential of $Q\mapsto QB$ in direction $H$ is $HB$. HenceRestriction to the embedded special orthogonal group preserves this bracket because the fields are tangent. Evaluating at $I$ gives $[B_1,B_2]=B_1{B}_2-B_2{B}_1$, the required bracket on the Special orthogonal Lie algebra. The commutator is again skew symmetric. This calculation is differentiating left-invariant matrix fields; reversing the order of the two differentials would give the wrong sign.