6. Smooth Vector Bundles

6.1Vector Bundles

定义 6.1.1. A smooth vector bundle of rank $k$ is a pair of smooth manifolds $E$, $B$ together with a submersion $\pi :E\to B$, s.t. the following hold:

(1)	for every $x\in B$, the fibre ${\pi }^{-1}(x)$ has the structure of a $k$-dimensional $\mathbb{R}$-vector space.
(2)	$B$ has an open cover $\{U_i\mid i\in I\}$ and diffeomorphisms ${\psi }_i:{\pi }^{-1}(U_i)\to U_i\times {\mathbb{R}}^k$ which restrict to linear isomorphisms on every ${\pi }^{-1}(x)$, $x\in U_i$ and satisfy ${\pi }_1\circ {\psi }_i=\pi$.

where ${\pi }^{-1}(x)=E_x$ is the fibre over $x\in B$.$$\dim E=\dim B+\dim E_x=\dim B+k$$

$U_i\cap U_j=\varnothing$.

$U_i\cap U_j\cap U_k\ne \varnothing$ $\Rightarrow$ ${\gamma }_{ji}\circ {\gamma }_{il}={\gamma }_{jl}$. Setting $j=l$ gives ${\gamma }_{ji}={\gamma }_{ij}^{-1}$. ${\gamma }_{ii}=\mathrm{Id}$, $\forall i\in I$. From the open covering of $B$ by the $U_i$ and the transition maps ${\gamma }_{ij}$, one can reconstruct the vector bundle $\pi :E\to B$.

定义 6.1.2. Let $\pi :E\to B$, ${\pi }^{\prime }:E^{\prime }\to B$ be smooth vector bundles over the same base $B$. An isomorphism of vector bundles is a diffeomorphism $f:E\to E^{\prime }$ which is a linear isomorphism on every fibre and satisfy ${\pi }^{\prime }\circ f=\pi$, i.e.

例 6.1.3.  

(1)	Product bundles $E=B\times {\mathbb{R}}^k$, $\pi ={\pi }_1$.

定义 6.1.4. A vector bundle is trivial if it is isomorphic to a product bundle.

(2)

Let $B=M$ be any smooth manifold, $E=TM$ is a vector bundle of rank$=\dim M$.

(3)

Let $B=S^1$ and take $U_1\times \mathbb{R}$, $U_2\times \mathbb{R}$. Then $U_1\cap U_2=V_1\bigsqcup V_2$. $$\begin{array}{rl}{\gamma }_{ij}:& U_i\cap U_j\to G{L}_k(\mathbb{R})\subset {\mathbb{R}}^{k^2}\text{smooth}\\ {\gamma }_{12}:& U_1\cap U_2=\begin{array}{rl}[t]V_1\bigsqcup V_2& \to G{L}_1(\mathbb{R})={\mathbb{R}}^{\ast }(\mathbb{R}\text{without origin})\\ x& \mapsto \{\begin{array}{ll}1& \text{for}x\in V_1\\ -1& \text{for}x\in V_2\end{array}\end{array}\end{array}$$Construct $E$ from this structure cocycle. Then $E$ is the Möbius strip. Rank $1$ vector bundles over $S^1$: $S^1\times \mathbb{R}$, $T{S}^1$, $E=M$. Then $S^1\times \mathbb{R}$ is isomorphic to $T{S}^1$, but $T{S}^1$ is not isomorphic to $E=M$.

定义 6.1.5. Let $\pi :E\to B$ be a vector bundle. A section of $E$ is a smooth map $s:B\to E$, s.t. $\pi \circ s={\mathrm{Id}}_B$.

引理 6.1.6. A vector bundle $E\stackrel{\pi }{\to }B$ of rank $k$ is trivial if and only if it admits $k$ sections $s_1,\dots ,s_k\in \Gamma (E)$ which are pointwise linearly independent, where $\Gamma (E)=\{s:B\to E\mid \pi \circ s={\mathrm{Id}}_B\}$ is a $\mathbb{R}$-vector space and a ${\mathcal{C}}^{\infty }(B)$-module.

证明. First, assume $E$ is trivial, and $f:E\to B\times {\mathbb{R}}^k$ is an isomorphism. Define $s_i(x):=f^{-1}(x,e_i)\in \Gamma (E)$, where $e_1,\dots ,e_k$ is any basis of ${\mathbb{R}}^k$. Then $s_1,\dots ,s_k$ are pointwise linearly independent.

Second, suppose $s_1,\dots ,s_k$ are linearly independent sections. Define$$\begin{array}{rl}g:B\times {\mathbb{R}}^k& \to E_k\\ (x,({\lambda }_1,\dots ,{\lambda }_k))& \mapsto \sum _{i=1}^k{\lambda }_i{s}_i(x)\end{array}$$This is a smooth map and satisfies $\pi \circ g\to {\mathrm{Id}}_B$.

Moreover, $g$ is a linear isomorphism $x\times {\mathbb{R}}^k\to E_x$, $\forall x\in B$. $f:=g^{-1}$ is a global trivialization of $E$.

$\square$

推论 6.1.7. A rank $1$ vector bundle is trivial if and only if it has a nowhere zero section, i.e. $\exists s\in \Gamma (E)$, s.t. $s(x)\ne 0$, $\forall x\in B$.

注 6.1.8. The zero $0\in \Gamma (E)$ is the section $\begin{array}{rl}0:B& \to E\\ x& \mapsto 0\in E_x\end{array}$. This is called the zero-section.

Let $S^1\subset {\mathbb{R}}^2$ be the unit circle as the following figure shown.

Then $T{S}^1\subset T{\mathbb{R}}^2={\mathbb{R}}^2\times {\mathbb{R}}^2$ and we have$$\begin{array}{rl}{T}_p{S}^1& =\mathbb{R}\cdot (-y,x)\\ T{S}^1& =\{(x,y,s,t)\in {\mathbb{R}}^4\mid x^2+y^2=1,s=-\lambda y,t=\lambda x,\text{for some}\lambda \in \mathbb{R}\}\end{array}$$with the mapwhere $s(x,y)=(x,y,-y,x)$.

引理 6.1.9. The Möbius strip $M$ is not a trivial vector bundle.

证明. Suppose $M$ were trivial. Then let $s:S^1\to M$ be a nowhere zero-section.

$s$ is smooth hence it is continuous. The intermediate value theorem says it has a zero. This leads to a contradiction.

$\square$

6.2Metric

定义 6.2.1. A metric on a vector bundle $\pi :E\to B$ is a fibrewise positive definite scalar product on $E_x$ which depends smoothly on $x\in B$.

Smoothness can be checked/defined in one of two ways:

(1)

With local trivialization: Let $\begin{array}{rl}\psi :{\pi }^{-1}(U)& \to U\times {\mathbb{R}}^k\\ \subset E_y& \stackrel{\cong }{\to }\{y\}\times {\mathbb{R}}^k\end{array}$ be a local trivialization with $x\in U$. A metric $⟨,⟩$ on $E$ induces a scalar product $⟨,{⟩}_x$ on $E_x$, which we think of as a scalar product $g_x$ on ${\mathbb{R}}^k$, via the isomorphism $E_x\underset{\psi }{\cong }{\mathbb{R}}^k$. $g_y$, as $y$ varies in $U$, gives a family of positive definite scalar products on ${\mathbb{R}}^k$, dependency on $y$. $$\begin{array}{rl}g:U& \to V=\text{symmetric bilinear forms on}{\mathbb{R}}^k.\\ y& \mapsto g_y\end{array}$$Smoothness of $⟨,⟩$ means that in every local trivialization, $g$ is a smooth map.

(2)

Smoothness of $⟨,⟩$ means that for any two $s_1,s_2\in \Gamma (E)$, $⟨s_1,s_2⟩\in {\mathcal{C}}^{\infty }(B)$. $$\begin{array}{rl}⟨s_1,s_2⟩:B& \to \mathbb{R}\\ x& \mapsto ⟨s_1(x),s_2(x){⟩}_x\end{array}$$

命题 6.2.2. Every vector bundle admits a metric.

证明. Let $\{U_i\mid i\in I\}$ be a covering of $B$ by trivializing open sets for $E$, ${\psi }_i:{\pi }^{-1}(U_i)\to U_i\times {\mathbb{R}}^k$.

For $y\in U_i$, let $⟨,{⟩}_{i,y}$ be the scalar product on $E_y$ obtained from the standard scalar product on ${\mathbb{R}}^k$ via the isomorphism $\psi :E_y\to \{y\}\times {\mathbb{R}}^k$.

Let ${\rho }_i$ be a partition of unity subordinate to the covering of $B$ by the $U_i$. Define $⟨,⟩:=\sum _i{\rho }_i\cdot ⟨,{⟩}_i$. This is a metric! It satisfies $\sum {\rho }_i\equiv 1$.

$\square$

注 6.2.3. This proof uses positive-definiteness.

6.3Constructions with Vector Bundles

(1)	Subbundles If $\pi :E\to B$ is a vector bundle of rank $k$, then a subbundle of rank $l⩽k$ is a submanifold $F\subset E$ such that $\pi {|}_F:F\to B$ is a vector bundle of rank $l$. For every $x\in B$, $F\cap E_x=F_x$ is a $l$-dimensional subspace of $E_x\cong {\mathbb{R}}^k$. Let $\pi :E\to B$, ${\pi }^{\prime }:E^{\prime }\to B$ be vector bundles and $f:E\to E^{\prime }$ a smooth map with ${\pi }^{\prime }\circ f=\pi$ and $f{|}_{E_x}$ is linear for all $x\in B$. If $\mathrm{rank}(f{|}_{E_x})$ is a constant function of $x\in B$, then $\mathrm{im}(f)\subset E^{\prime }$ is a subbundle of $\mathrm{rank}=\mathrm{rank}(f)$ and $\ker (f)\subset E$ is a subbundle of $\mathrm{rank}=\mathrm{rank}E-\mathrm{rank}f$.
(2)	Quotient bundles If $E$ is a vector bundle, $F\subset E$ a subbundle, the $\bigcup _{x\in B}(E_x/F_x)$ is a vector bundle over $B$, called the quotient bundle.
(3)	If $E$ has a metric $⟨,⟩$, $F^{\perp }=\{v\in E_x\mid x\in B,⟨v,w{⟩}_x=0,\forall w\in F_x\}$ is a subbundle, and $F^{\perp }\cong E/F$.
(4)	Whitney sums $E\stackrel{\pi }{\to }B$, $E^{\prime }\stackrel{{\pi }^{\prime }}{\to }B$ are vector bundles. $E\oplus E^{\prime }\to B$ is the vector bundle with $(E\oplus E^{\prime }{)}_x=E_x\oplus E_x^{\prime }$, for all $x\in B$. Let $\{U_i\mid i\in I\}$ be an open cover of $B$ which is simultaneously trivialization for $E$ and for $E^{\prime }$. Let ${\gamma }_{ij}:U_i\cap U_j\to G{L}_k(\mathbb{R})$, ${\gamma }_{ij}^{\prime }:U_i\cap U_j\to G{L}_{k^{\prime }}(\mathbb{R})$ be the corresponding cocycles of transition maps. Then $E\oplus E^{\prime }$ is the vector bundle of $\mathrm{rank}k+k^{\prime }$ defined by $$\begin{array}{rl}{U}_i\cap U_j& \to G{L}_{k+k^{\prime }}(\mathbb{R})\\ x& \mapsto \left(\begin{array}{cc}{\gamma }_{ij}(x)& 0\\ 0& {\gamma }_{ij}^{\prime }(x)\end{array}\right)\end{array}$$
(5)	Dual bundles If $\pi :E\to B$ is a vector bundle of $\mathrm{rank}k$, the dual bundle $E^{\ast }\stackrel{\pi }{\to }B$ is the $\mathrm{rank}k$ vector bundle given by ${\gamma }_{ij}(x)\in G{L}_k(\mathbb{R})=\mathrm{Hom}({\mathbb{R}}^k,{\mathbb{R}}^k)$. $$\begin{array}{rl}\lambda :\mathbb{R}& \to {\mathbb{R}}^k\\ {\lambda }^{\ast }:({\mathbb{R}}^k{)}^{\ast }& \mapsto ({\mathbb{R}}^k{)}^{\ast }\text{defined by}{\lambda }^{\ast }(\phi )(x)=\phi (\lambda (x))\\ & \mathrm{Hom}(({\mathbb{R}}^k{)}^{\ast },({\mathbb{R}}^k{)}^{\ast })\ni G{L}_k(\mathbb{R})\ni {\gamma }_{ij}^{\ast }(x)\end{array}$$If $F\subset E$ is a subbundle, then $$F\oplus (E/F)\cong F\oplus F^{\perp }\cong E$$

Let $G\subset G{L}_k(\mathbb{R})$ be a subgroup.

定义 6.3.1. A $\mathbf{G}$-structure on a rank $k$ vector bundle $E\stackrel{\pi }{\to }B$ is a system of local trivializations whose transition maps take values in $G$.

注 6.3.2. A $G$-structure is sometimes called a $\mathbf{G}$-reduction.

(1)	$G=\{e\}$. In this case, a $G$-structure is a global trivialization.
(2)	$G=G{L}_k^{+}(\mathbb{R})$ orientation-preserving isomorphism ${\mathbb{R}}^k\stackrel{\cong }{\to }{\mathbb{R}}^k$. In this case, a $G$-structure on $E$ is an orientation for $E$, i.e. a consistent choice of orientation for all $E_x$ varying smoothly $x\in B$.

The Möbius strip as a vector bundle over $S^1$ does not admit an orientation.

引理 6.3.3. A rank $1$ vector bundle is trivial if and only if it is orientable.

证明. If $E$ is trivial, then it is orientable. Conversely, suppose $E$ is orientable and of rank $1$. Then $E$ has a $G$-structure for $G{L}_1^{+}(\mathbb{R})={\mathbb{R}}^{>0}$.
$${\gamma }_{ij}:U_i\cap U_j\to {\mathbb{R}}^{>0}$$Without loss of generality, all $U_i\cap U_j$ are either $\varnothing$ or diffeomorphic to bundles. Then we can define the ${\gamma }_{ij}$ smoothly to be $\equiv 1$. Then $E$ is trivial since it has a $G$-structure for the trivial group.

$\square$

(3)	$G=O(k)$. In this case, a $G$-structure is a choice of metric $⟨,⟩$ on $E$. Every $E$ admits such a $G$-structure.
(4)	$G=SO(k)=G{L}_k^{+}(\mathbb{R})\cap O(k)$.

6.4Pullback Bundles

Suppose $f:M\to N$ is a smooth map, and $\pi :E\to N$ is a smooth vector bundle over $N$.

定义 6.4.1. $f^{\ast }E:=\{(x,v)\in M\times E\mid f(x)=\pi (v)\}$ is the pullback bundle of $E$ under $f$.

That is the following diagram commute:

And we have$${\pi }_1^{-1}(x)={\pi }^{-1}(f(x))=E_{f(x)}$$

If $E$ is a vector bundle of rank $k$ over $N$, then $f^{\ast }E$ is a vector bundle of rank $k$ over $M$.

6.5Bundles Homomorphisms

定义 6.5.1. If ${\pi }_E:E\to M$, ${\pi }_F:F\to N$ are smooth vector bundles, then a homomorphism of vector bundles is a smooth map $h:E\to F$, which restricts to every $E_x\subset E$ as a linear map into a fibre of $F$.

$f(x)={\pi }_F\circ h(v)$ for any $v\in E_x$. This is well-defined and smooth.

commute

例 6.5.2. ${\pi }_2:f^{\ast }E\to E$ is a homomorphism of vector bundle.

例 6.5.3. If $f:M\to N$ is any smooth map, then $Df:TM\to TN$ is a homomorphism of vector bundle.

Let $h:E\to F$ be a homomorphism of vector bundles covering $f:M\to N$.

Define $\overline{h}:E\to f^{\ast }F$ by $\overline{h}(v)=({\pi }_E(v),h(v))\in M\times F$. Then$${\pi }_1(\overline{h}(v))={\pi }_1({\pi }_E(v),h(v))={\pi }_E(v)$$

Let $N=\mathbb{R}$, then

where $\begin{array}{rl}Df:TM& \to M\times \mathbb{R}\\ v& \mapsto (\pi (v),\underset{\begin{array}{c}\text{linear form in}\\ \text{tangent space}\end{array}}{\underbrace{(D_{\pi (v)}f)(v)}})\end{array}$ and $T\mathbb{R}=\mathbb{R}\times \mathbb{R}$, the first $\mathbb{R}$ represents manifold and the second $\mathbb{R}$ represents vector space.

If $f:M\to \mathbb{R}$ is smooth, then its derivative $Df$ is a section in $\mathrm{Hom}(TM,M\times \mathbb{R})=T^{\ast }M=(TM{)}^{\ast }$. Three different interpretation of derivative of smooth function:$$Df:TM\to T\mathbb{R}Df:TM\to M\times \mathbb{R}\mathrm{d}f\in \Gamma (T^{\ast }M)$$