= 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 by
$$
\boxed{[X,Y]f=X(Yf)-Y(Xf).}
$$
Expanding 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,
$$
[X,Y]=\sum_i\left(\sum_jX^j\partial_jY^i-Y^j\partial_jX^i\right)\partial_i.
$$
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 $(dL_g)_hX_h=X_{gh}$ for every $g,h$. Evaluation $X\mapsto X_e$ is linear and injective, since $X_g=(dL_g)_eX_e$. Conversely any $B\in T_eG$ gives a smooth <left-invariant vector field> $X_B(g)=(dL_g)_eB$. Smoothness follows from smooth multiplication and its differential. Thus
$$
\boxed{\{\text{left-invariant vector fields on }G\}\cong T_eG.}
$$
For any <diffeomorphism> $F$, the intrinsic bracket identity on functions gives $F_*[X,Y]=[F_*X,F_*Y]$. 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 $\mathfrak g=T_eG$.
For the <special orthogonal group>, differentiation of $Q(t)^TQ(t)=I$ 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 $Q(t)^TQ(t)=I$ and lies in the determinant-one component, because $Q(0)=I$ and its determinant varies continuously in $\{\pm1\}$. Its initial derivative is $B$, proving
$$
\boxed{\mathfrak{so}(n)=\{B\in M_n(\mathbb R):B^T=-B\}.}
$$
For 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 $HB$. Hence
$$
[X_{B_1},X_{B_2}](Q)=DX_{B_2}(Q)[QB_1]-DX_{B_1}(Q)[QB_2]=Q(B_1B_2-B_2B_1).
$$
Restriction to the embedded <special orthogonal group> preserves this bracket because the fields are tangent. Evaluating at $I$ gives \b[$[B_1,B_2]=B_1B_2-B_2B_1$], 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.
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