Past exam of the mathematics course of the University of Cambridge / 2004 / iii / Paper 14 / 1

A smooth vector field on a smooth manifold $M$ is a smooth section of a vector bundle $X:M\to TM$ of the tangent bundle projection $\pi$, so $\pi \circ X={\mathrm{id}}_M$. In a coordinate chart it has the form $X={\sum }_i{X}^i{\partial }_i$ with smooth component functions $X^i$.

Suppose a smooth vector bundle isomorphism $\Phi :M\times {\mathbb{R}}^n\to TM$ covering the identity is given. For the standard basis $e_1,\dots ,e_n$ of ${\mathbb{R}}^n$, put $X_i(p)=\Phi (p,e_i)$. These are smooth vector fields, and the linear isomorphism on each fibre makes $X_1(p),\dots ,X_n(p)$ a basis of $T_{p}M$.

Conversely, given such a global frame of a vector bundle, defineIt is smooth, covers the identity and is a linear isomorphism on each fibre. Its inverse is smooth as well: in any coordinate chart, the component columns of the $X_i$ form an invertible smooth matrix $C(p)$, and the fibre coordinates of the inverse are $C(p{)}^{-1}v$. matrix inversion is smooth on the invertible matrices. Thus the tangent bundle is trivial exactly when a global frame exists, the defining property of a parallelizable manifold.

For a Lie group $G$, write $L_h(g)=hg$. A left-invariant vector field is a field satisfyingSuch a field is determined by $v=X(e)\in T_{e}G$, since $X(g)=(d{L}_g{)}_{e}v$. Conversely, this formula defines a left-invariant field, by the chain rule and $L_h{L}_g=L_{hg}$.

It also proves smoothness even if smoothness was not initially assumed. Choose a smooth curve $c(t)$ with $c(0)=e$, $c^{\prime }(0)=v$. ThenMultiplication is a smooth map on a Lie group, so differentiating its coordinate functions in the second variable yields a smooth function of $g$. This gives smooth components for $X$ in every local chart.

Choose a basis $v_1,\dots ,v_m$ of $T_{e}G$, where $m=\dim G$, and left translate it. Each $L_g$ is a diffeomorphism, with inverse $L_{g^{-1}}$, so $(d{L}_g{)}_e$ is a linear isomorphism and the translated fields are a basis at every $g$. The explicit parallelization of a Lie group by left translations isIts inverse sends $w\in T_{g}G$ to $(g,(d{L}_{g^{-1}}{)}_{g}w)$. Identifying $T_{e}G$ with ${\mathbb{R}}^m$ gives the required product bundle. The differential facts used are linearity, the chain rule, $d(\mathrm{id})=\mathrm{id}$ and smooth dependence of the differential of a smooth map on its base point.