Smooth Vector Bundles and Sections

Surjectivity is part of the definition of a smooth fibre bundle, so $\pi$ is surjective. Fix $e\in E$ with $p=\pi (e)$, and choose a local trivialization $\mathrm{\Phi }:{\pi }^{-1}(U)\to U\times {\mathbb{R}}^r$ around $e$. In this chart, $\pi$ becomes the product projection ${\mathrm{pr}}_1:U\times {\mathbb{R}}^r\to U$.

In product coordinates the differential of ${\mathrm{pr}}_1$ is the coordinate projection $T_{p}U\oplus T_v{\mathbb{R}}^r\to T_{p}U$, which is surjective. Therefore $d{\pi }_e$ is surjective, and since $e$ was arbitrary, $\pi$ is a submersion by [L2].

On $U_{\alpha }$, the map ${\mathrm{\Phi }}_{\alpha }\circ {\mathrm{\Phi }}_{\alpha }^{-1}$ is the identity, so [L1] gives $(p,g_{\alpha \alpha }(p)v)=(p,v)$ for every $v\in {\mathbb{R}}^r$. Hence $g_{\alpha \alpha }(p)=I_r$.

On a triple overlap, ${\mathrm{\Phi }}_{\gamma }\circ {\mathrm{\Phi }}_{\alpha }^{-1}=({\mathrm{\Phi }}_{\gamma }\circ {\mathrm{\Phi }}_{\beta }^{-1})\circ ({\mathrm{\Phi }}_{\beta }\circ {\mathrm{\Phi }}_{\alpha }^{-1})$. Applying [L1] to $(p,v)$ gives $(p,g_{\gamma \alpha }(p)v)=(p,g_{\gamma \beta }(p)g_{\beta \alpha }(p)v)$ for every $v$, so $g_{\gamma \alpha }(p)=g_{\gamma \beta }(p)g_{\beta \alpha }(p)$.

Let $X={\coprod }_{\alpha \in A}(U_{\alpha }\times {\mathbb{R}}^r)$ and declare $(p,v{)}_{\alpha }\sim (q,w{)}_{\beta }$ when $p=q\in U_{\alpha }\cap U_{\beta }$ and $w=g_{\beta \alpha }(p)v$. By [L1], this relation is reflexive, symmetric, and transitive, so it is an equivalence relation on $X$.

Let $E=X\mathrm{/}\sim$ and write $q:X\to E$ for the quotient map. For each $\alpha$, define ${\psi }_{\alpha }:q(U_{\alpha }\times {\mathbb{R}}^r)\to U_{\alpha }\times {\mathbb{R}}^r$ by ${\psi }_{\alpha }([p,v{]}_{\alpha })=(p,v)$. The cocycle relation shows that every class meeting $U_{\alpha }\times {\mathbb{R}}^r$ has a unique representative there, so ${\psi }_{\alpha }$ is bijective. Its domain is open because $q^{-1}(q(U_{\alpha }\times {\mathbb{R}}^r))$ is the union over $\beta$ of the open sets $\{(p,g_{\alpha \beta }(p)w{)}_{\beta }:p\in U_{\alpha }\cap U_{\beta },w\in {\mathbb{R}}^r\}$, whence [F1] makes ${\psi }_{\alpha }$ a homeomorphism onto an open subset.

On overlaps, ${\psi }_{\beta }\circ {\psi }_{\alpha }^{-1}(p,v)=(p,g_{\beta \alpha }(p)v)$, so the chart changes are smooth with smooth inverses. The projection $\pi ([p,v{]}_{\alpha })=p$ is well defined and in chart $\alpha$ is the product projection $U_{\alpha }\times {\mathbb{R}}^r\to U_{\alpha }$, while the fibre maps are linear by construction. Therefore the ${\psi }_{\alpha }$ form a smooth rank-$r$ vector-bundle atlas on $E\to M$.

Because the cover is countable and each chart image has a countable base, these bundle charts give $E$ a countable base. Hausdorffness is local in the charts when two classes are distinct over one base point, and over different base points it comes from Hausdorffness of $M$. Hence $E$ is a smooth manifold, and $\pi :E\to M$ is the required smooth vector bundle.

On the $\alpha$-th trivializing piece define $H_{\alpha }(p,v)=(p,h_{\alpha }(p)v)$. If $(p,v{)}_{\alpha }\sim (p,g_{\beta \alpha }(p)v{)}_{\beta }$, then the gauge relation gives $H_{\beta }(p,g_{\beta \alpha }(p)v)=(p,g_{\beta \alpha }^{\mathrm{\prime }}(p)h_{\alpha }(p)v)=H_{\alpha }(p,v)$, so the local maps descend to a well-defined bundle map $H:E\to E^{\mathrm{\prime }}$.

In the quotient charts of [L1], the descended map is $(p,v)\mapsto (p,h_{\alpha }(p)v)$, hence smooth and fibrewise linear. Replacing $h_{\alpha }$ by $h_{\alpha }^{-1}$ gives the inverse construction, so $H$ is a smooth bundle isomorphism.

In a local trivialization $\mathrm{\Phi }:{\pi }^{-1}(U)\to U\times {\mathbb{R}}^r$, the zero section is represented by $p\mapsto (p,0)$. This map is smooth, injective, and its image is the slice $U\times \{0\}$.

The coordinate slice $U\times \{0\}$ is an embedded submanifold of the product, so $p\mapsto (p,0)$ is an immersion and a homeomorphism onto its image. Transporting this property through the bundle charts proves that $0_M$ is a smooth embedding by [L2].

Fix $e\in E$ with base point $p=\pi (e)$. Choose a chart of $M$ sending a neighborhood of $p$ diffeomorphically onto an open subset of ${\mathbb{R}}^n$, and combine it with a vector bundle chart from [L1]. This identifies a neighborhood of $e$ in $E$ with an open subset of ${\mathbb{R}}^n\times {\mathbb{R}}^r$.

Since ${\mathbb{R}}^n\times {\mathbb{R}}^r\cong {\mathbb{R}}^{n+r}$, every point of $E$ has a chart of dimension $n+r$. Therefore the total space $E$ is an $(n+r)$-manifold.

If $(s_1,\dots ,s_r)$ is a local frame on $U$, then [L1] gives for each $p\in U$ and $v\in E_p$ unique scalars $a_1,\dots ,a_r$ with $v={\sum }_i{a}_i{s}_i(p)$. Around any $p\in U$, choose an existing vector bundle chart $\mathrm{\Psi }:E{\mathrm{\mid }}_W\to W\times {\mathbb{R}}^r$ with $W\subseteq U$. Writing $\mathrm{\Psi }(s_i(x))=(x,a_i(x))$, the column vectors $a_i(x)$ form a smooth matrix $A(x)\in GL(r,\mathbb{R})$ because the $s_i(x)$ are a basis of $E_x$. Thus the local coordinate map is ${\mathrm{\Phi }}_W(v)=(x,A(x{)}^{-1}u)$ when $\mathrm{\Psi }(v)=(x,u)$, so ${\mathrm{\Phi }}_W$ and its inverse $(x,\lambda )\mapsto {\mathrm{\Psi }}^{-1}(x,A(x)\lambda )$ are smooth. By uniqueness of the coordinates from [L1], these local formulas agree on overlaps and patch to a vector bundle chart $\mathrm{\Phi }:E{\mathrm{\mid }}_U\to U\times {\mathbb{R}}^r$.

Conversely, if $\mathrm{\Phi }:E{\mathrm{\mid }}_U\to U\times {\mathbb{R}}^r$ is a vector bundle chart, let $e_1,\dots ,e_r$ be the standard basis of ${\mathbb{R}}^r$ and set $s_i(p)={\mathrm{\Phi }}^{-1}(p,e_i)$. Then each $s_i$ is a smooth local section, and the vectors $s_i(p)$ form a basis of $E_p$ because $\mathrm{\Phi }{\mathrm{\mid }}_{E_p}$ is a linear isomorphism.

Applying the second construction to the chart from step 1.1 recovers the original frame because $\mathrm{\Phi }(s_i(p))=(p,e_i)$ by construction. Applying the first construction to the sections from step 1.2 recovers the original chart because the resulting coordinates are exactly the fibre coordinates already read by $\mathrm{\Phi }$. Hence local frames and local trivializations are inverse constructions.

If $E$ is trivial, then the global bundle chart $E\cong M\times {\mathbb{R}}^r$ exists. Applying [L1] on $U=M$ to that chart produces a global frame.

If $E$ has a global frame, then applying [L1] on $U=M$ to that frame gives a global trivialization $E\cong M\times {\mathbb{R}}^r$. Therefore $E$ is trivial.

Steps 1.1 and 1.2 prove both directions of the biconditional.

By [L1], the chosen frame gives a bundle chart $\mathrm{\Phi }:E{\mathrm{\mid }}_U\to U\times {\mathbb{R}}^r$ in which $s_i(p)$ corresponds to the $i$-th standard basis vector. Therefore $\mathrm{\Phi }(\sigma (p))=(p,f_1(p),\dots ,f_r(p))$ exactly when $\sigma (p)={\sum }_i{f}_i(p)s_i(p)$.

In this chart, $\sigma$ is smooth exactly when the coordinate map $p\mapsto (f_1(p),\dots ,f_r(p))$ is smooth. Equivalently, each component $f_i$ is smooth, and the criterion is local on $U$ by [L2].

Let $\sigma ,\tau$ be smooth sections and let $f\in C^{\mathrm{\infty }}(M)$. On a local frame, write $\sigma =\sum a_i{s}_i$ and $\tau =\sum b_i{s}_i$ with smooth components $a_i,b_i$. Then $\sigma +\tau =\sum (a_i+b_i)s_i$ and $f\sigma =\sum (f{a}_i)s_i$.

By [L2], the component functions $a_i+b_i$ and $f{a}_i$ are smooth, so [L1] shows that $\sigma +\tau$ and $f\sigma$ are again smooth sections. The module axioms hold fibrewise because each fibre is a vector space.

Choose a local frame $(s_1,\dots ,s_r)$ on an open set $U$ containing $p$. Write $v={\sum }_i{a}_i{s}_i(p)$ and define a local section $\tau ={\sum }_i{a}_i{s}_i$ on $U$. Then $\tau (p)=v$.

Choose a smooth bump function $\chi :M\to \mathbb{R}$ with $\chi (p)=1$ and $\mathrm{supp}(\chi )\subseteq U$. On $U$ define $\sigma =\chi \tau$, which is smooth by [L3]. Because $\mathrm{supp}(\chi )\subseteq U$, there is an open neighborhood $V$ of $M\setminus U$ on which $\chi =0$; define $\sigma =0$ on $V$, which is smooth by [L4]. On $U\cap V$ the two formulas agree, so they paste to a smooth global section. Its support is contained in $\mathrm{supp}(\chi )$, hence compact, and $\sigma (p)=\chi (p)\tau (p)=v$.

Fix a local frame $(s_1,\dots ,s_r)$ on an open set $U$. Write ${\sigma }_i{\mathrm{\mid }}_U={\sum }_j{a}_{ij}{s}_j$ with smooth coefficient functions $a_{ij}$. Then on $U$ the formal sum has components ${\sum }_i{f}_i{a}_{ij}$. Because the supports of $f_i{\sigma }_i$ are locally finite, only finitely many terms are nonzero near each point.

Each component ${\sum }_i{f}_i{a}_{ij}$ is therefore a locally finite sum of smooth scalar functions, so it is smooth by [L2]. Applying [L1] again shows that ${\sum }_i{f}_i{\sigma }_i$ is a smooth section on $U$, and hence on all of $M$.

By [L1], the chosen frames identify $E{\mathrm{\mid }}_U$ with $U\times {\mathbb{R}}^r$ and $F{\mathrm{\mid }}_V$ with $V\times {\mathbb{R}}^s$. In these trivializations, fibrewise linearity forces $\mathrm{\Phi }$ to have the form $\mathrm{\Phi }(p,v)=(f(p),A(p)v)$ for a unique matrix $A(p)=(a_{ji}(p))$.

The local representative $(p,v)\mapsto (f(p),A(p)v)$ is smooth exactly when its matrix entries $a_{ji}$ are smooth on $U$. Equivalently, the images $\mathrm{\Phi }(e_i)={\sum }_j{a}_{ji}{u}_j\circ f$ have smooth local components, which is the criterion in [L2].

If the common fibre rank is $0$, then every fibre is the zero vector space, so $\mathrm{\Phi }$ is already the unique smooth bundle map between zero bundles over $f$ and hence a bundle isomorphism. Otherwise choose local frames so that on one trivializing neighborhood, $\mathrm{\Phi }(p,v)=(f(p),A(p)v)$. Fibrewise bijectivity means that each matrix $A(p)$ is invertible, so [L2] gives $\det A(p)\mathrm{\ne }0$ for every $p$.

In the positive-rank case, [L3] makes the entries of $A^{-1}$ smooth on the same neighborhood. Thus the local inverse is $(q,w)\mapsto (f^{-1}(q),A(f^{-1}(q){)}^{-1}w)$, which is smooth because $f^{-1}$ is smooth. These local inverses agree on overlaps, so $\mathrm{\Phi }$ is a smooth bundle isomorphism. Together with the rank-$0$ branch of step 1.1, this proves the proposition.

If $k=0$, then every fibre map is zero, so $\ker \mathrm{\Phi }=E$ and $\mathrm{im}\mathrm{\Phi }$ is the zero subbundle of $F$. Assume now that $k\ge 1$, fix $p\in M$, and write $\mathrm{\Phi }$ in local frames near $p$ as a smooth matrix $A(x)$. Because $\mathrm{rank}A(p)=k$, after reordering coordinates some $k\times k$ minor $B(x)$ is nonzero at $p$, hence nonzero on a smaller neighborhood.

Writing source coordinates as $(u,w)$ for that split, the kernel equation becomes $B(x)u+C(x)w=0$, so [L2] makes $B(x{)}^{-1}$ smooth and gives $u=-B(x{)}^{-1}C(x)w$. Therefore the kernel fibres are spanned by smooth local sections depending on the free variables $w$.

The same chosen $k$ columns of $A(x)$ remain linearly independent nearby, so they form a smooth local frame of the image bundle. Thus in the positive-rank case both the kernel and the image are locally spanned by part of a frame, and step 1.1 already handled $k=0$. Therefore $\ker \mathrm{\Phi }$ and $\mathrm{im}\mathrm{\Phi }$ are smooth vector subbundles.

Let ${\mathrm{\Phi }}_{\alpha }:E{\mathrm{\mid }}_{U_{\alpha }}\to U_{\alpha }\times {\mathbb{R}}^r$ be a vector bundle chart. For $q\in f^{-1}(U_{\alpha })$ and $(q,e)\in f^{\ast }E$, define ${\tilde{\mathrm{\Phi }}}_{\alpha }(q,e)=(q,v)$ when ${\mathrm{\Phi }}_{\alpha }(e)=(f(q),v)$. This is a bijection $(f^{\ast }E){\mathrm{\mid }}_{f^{-1}(U_{\alpha })}\to f^{-1}(U_{\alpha })\times {\mathbb{R}}^r$.

On overlaps, ${\tilde{\mathrm{\Phi }}}_{\beta }\circ {\tilde{\mathrm{\Phi }}}_{\alpha }^{-1}(q,v)=(q,g_{\beta \alpha }(f(q))v)$. These chart changes are smooth and fibrewise linear because the original transition functions are smooth. Therefore the pulled-back charts define a smooth rank-$r$ vector bundle over $N$.

For the identity map, define $I:{\mathrm{id}}_M^{\ast }E\to E$ by $I(p,e)=e$. This is well defined because $p=\pi (e)$ in the identity pullback, and its inverse is $e\mapsto (\pi (e),e)$.

An element of $g^{\ast }(f^{\ast }E)$ is a pair $(x,(g(x),e))$ with $f(g(x))=\pi (e)$. Send it to $(x,e)\in (f\circ g{)}^{\ast }E$. The inverse is $(x,e)\mapsto (x,(g(x),e))$. In the pulled-back bundle charts of [L1], both maps are the identity on the fibre coordinate, so they are smooth vector bundle isomorphisms.

On a common trivializing neighborhood $U$, use [L1] to identify $E{\mathrm{\mid }}_U\oplus F{\mathrm{\mid }}_U$ with $U\times ({\mathbb{R}}^r\oplus {\mathbb{R}}^s)\cong U\times {\mathbb{R}}^{r+s}$. This gives a local trivialization of the Whitney sum.

If the transition matrices for $E$ and $F$ are $g_{\beta \alpha }$ and $h_{\beta \alpha }$, then the transition matrix for $E\oplus F$ is the block diagonal matrix $\mathrm{diag}(g_{\beta \alpha },h_{\beta \alpha })$, which is smooth on overlaps. Therefore these local trivializations define a smooth rank-$(r+s)$ bundle.

If $\lambda \in E_p^{\ast }$ has row-coordinate vector $\mathrm{\ell }$ in one dual basis, then after changing the primal basis by $g_{\beta \alpha }(p)$, the same functional has coordinate vector $\mathrm{\ell }(g_{\beta \alpha }(p){)}^{-1}$. By [L2], the dual transition matrix is therefore $(g_{\beta \alpha }(p{)}^{-1}{)}^T$.