Past exam of the mathematics course of the University of Cambridge / 2007 / iii / Paper 15 / 1

Regard $T_{p}M$ as the real vector space of derivations at a point $p$: a vector $v$ acts on germs of smooth functions by a linear map satisfying $v(fh)=f(p)v(h)+h(p)v(f)$. In a coordinate chart $x=(x^1,\dots ,x^n)$, the coordinate derivations $\partial /\partial x^i{|}_p$ form a basis, so $v={\sum }_i{v}^i\partial /\partial x^i{|}_p$, with $v^i=v(x^i)$. To see that these exhaust all derivations, the first-order Taylor theorem locally writes $f-f(p)={\sum }_i(x^i-x^i(p))h_i$ with $h_i(p)={\partial }_{i}f(p)$. The product rule gives $v(f)={\sum }_{i}v(x^i){\partial }_{i}f(p)$; applying this to the coordinate functions also proves independence. Equivalently one can use velocities of smooth curves through $p$.

Set $TM={\coprod }_{p\in M}{T}_{p}M$, with projection $\pi (p,v)=p$. Over a coordinate chart $x:U\to x(U)$ defineGive the disjoint union the topology generated by the inverse images of open sets under these bijections. For overlapping charts $y=f\circ x$, the chain rule gives the tangent coordinate transitionIt is a smooth bijection with smooth inverse $(w,u)\mapsto (f^{-1}(w),D{f}^{-1}(w)u)$. Thus the induced charts form a smooth atlas of dimension $2n$. The chain rule gives the cocycle on triple overlaps, and any additional smooth coordinate chart gives compatible induced charts, so the structure is independent of the starting atlas.

The total space is Hausdorff: different base points can be separated downstairs, and two distinct vectors over one point can be separated in a common product chart. A countable coordinate cover, together with countable bases for its products, gives second countability. Hence this atlas defines a smooth manifold. In every local trivialization, $\pi$ is projection onto the first factor and the fibre change $v\mapsto Df(z)v$ is an invertible linear map depending smoothly on $z$. Fibrewise addition and scalar multiplication are smooth in these product coordinates. Consequently $TM\to M$ is naturally a smooth real vector bundle of rank $n$.

Define the cotangent space $T_p^{\ast }M={\mathrm{Hom}}_{\mathbb{R}}(T_{p}M,\mathbb{R})$ and the cotangent bundle $T^{\ast }M={\coprod }_p{T}_p^{\ast }M$, with projection $(p,\alpha )\mapsto p$. A coordinate chart $x$ gives the dual basis $d{x}^1{|}_p,\dots ,d{x}^n{|}_p$. Write $\alpha ={\sum }_i{\xi }_{i}d{x}^i{|}_p$ and define the induced local trivialization byIf $y=f(x)$ and $\alpha ={\sum }_j{\eta }_{j}d{y}^j$, the chain rule gives $\xi =Df(x{)}^T\eta$. Therefore the cotangent coordinate transition isThe inverse Jacobian matrix is smooth wherever its determinant is nonzero, so this transition and its inverse are smooth. The cocycle follows either from the chain rule or by dualizing the tangent transitions.

Use these product charts to define the topology and smooth atlas. The Hausdorff and second-countability arguments are the same as for the tangent bundle: separate distinct fibres using the base, separate distinct covectors in one product chart, and use a countable coordinate cover. The fibre transitions are smooth linear isomorphisms, so the projection is a smooth vector bundle of rank $n$. Using every compatible base coordinate chart shows that this structure is natural and independent of chart choices.

Yes, the two real smooth vector bundles are isomorphic, although the choice is not canonical. A standard smooth manifold is Hausdorff and second countable, hence admits a smooth locally finite partition of unity. Choose Euclidean Riemannian metrics on coordinate neighborhoods and take their partition-of-unity sum. At each point at least one positive coefficient occurs, so the resulting smooth symmetric tensor $g$ is positive definite.

The resulting musical isomorphism isIt covers the identity of $M$ and is linear on every fibre. In coordinates it is $(x,v)\mapsto (x,G(x)v)$, where $G=(g_{ij})$ is a smooth positive-definite matrix. Its inverse uses the smooth inverse matrix $G^{-1}$, giving the map ${♯}_g$. Thus it is a smooth vector bundle isomorphism.

The Riemannian metric supplies the extra choice: no positive-dimensional real vector space has a linear identification with its dual invariant under all changes of basis. Indeed a scalar change $tI$ acts by $t$ on vectors and $t^{-1}$ on covectors, so an equivariant map $L$ would satisfy $tL(v)=t^{-1}L(v)$ for every $t>0$, forcing $L=0$. An isomorphism therefore exists without being a metric-free canonical one.