3. Tangent Spaces and Tangent Bundle

Let $M$ be a smooth manifold and $\mathcal{A}=\{(U_i,{\phi }_i)\mid i\in I\}$ a differentiable atlas. All the ${\phi }_i$ take values in ${\mathbb{R}}^m$, $n=\dim M$. Consider triples $(x,i,v)\in M\times I\times {\mathbb{R}}^n$ with $x\in U_i$. On the set of such triples define the relation $(x,i,v)\sim (y,j,w)$ by $x=y$ and $D_{{\phi }_i(x)}\underset{f_{ji}}{\underbrace{({\phi }_j\circ {\phi }_i^{-1})}}(v)=w$. Then$$(D_{{\phi }_j(y)}{f}_{ij})(w)=v.$$

Claim 3.0.1. This is an equivalence relation.

$(x,i,v)\sim (y,j,w)\sim (z,k,t)$

$x=y=z$$$\underset{D_{{\phi }_i(x)}{f}_{ki}}{\underbrace{D_{{\phi }_j(x)}{f}_{kj}\circ D_{{\phi }_i(x)}{f}_{ji}}}(v)=(D_{{\phi }_j(x)}{f}_{kj})w=t$$Let $TM$ be the set of equivalence classes, and$$\begin{array}{rl}\pi :TM& \to M\\ [x,i,v]& \mapsto x\end{array}$$

If $A\subset M$, then ${\pi }^{-1}(A)=T_{A}M$.

If $A=\{x\}$, then ${\pi }^{-1}(x)=T_{x}M$, the tangent space to $M$ at $x$.

If $A\subset M$ is open, then $A$ is itself a manifold, and $TA=T_{A}M$.

For every chart $(U_i,{\phi }_i)$, we have a bijective map$$\begin{array}{rl}T{\phi }_i:T{U}_i& \to {\phi }_i(U_i)\times {\mathbb{R}}^n\subset {\mathbb{R}}^n\times {\mathbb{R}}^n\\ [x,i,v]& \mapsto ({\phi }_i(x),v)\end{array}$$We give each $T{U}_i$ the unique topology which makes $T{\phi }_i$ into a homeomorphism. This is well-defined. On $TM$, we define topology by requiring each $T{U}_i$ to be open, and itself have the topology defined via $T{\phi }_i$.

We consider ${\mathcal{A}}^{\prime }=\{(T{U}_i,T{\phi }_i)\mid i\in I\}$ as an atlas for $TM$. This has ${\mathcal{C}}^{\infty }$ transition maps, and so values $TM$ into a ${\mathcal{C}}^{\infty }$ manifold.

With respect to this differentiable structure on $TM$, the projection $\pi :TM\to M$ is a differentiable map.

引理 3.0.2. For every $x\in M$, the tangent space $T_{x}M$ has a well-defined structure as a $\mathbb{R}$-vector space of $\dim n$.

证明. Suppose $x\in U_i$, then $T{\phi }_i{|}_{T_{x}M}:T_{x}M\to \{{\phi }_i(x)\}\times {\mathbb{R}}^n$ is bijective. Define the vector space structure on $T_{x}M$ to be the unique one that makes $T{\phi }_i{|}_{T_{x}M}$ a linear isomorphism. If $x\in U_j$, then $f_{ji}:{\phi }_i(U_i\cap U_j)\to {\phi }_j(U_i\cap U_j)$ is a diffeomorphism. The derivative is linear$$D_{{\phi }_i(x)}{f}_{ji}:{\mathbb{R}}^n\to {\mathbb{R}}^n.$$This is an isomorphism of the vector space. This shows that the vector space structure on $T_{x}M$ defined using $(U_j,{\phi }_j)$ instead of $(U_i,{\phi }_i)$ is isomorphic to the one gotten from $U_i$.

$\square$

For every $x\in M$, ${\pi }^{-1}(x)=T_{x}M$ is a vector space.

Suppose $f:M\to N$ is a differentiable map between differentiable manifolds.

Define $\begin{array}{rl}Df:TM& \to TN\\ [x,i,v]& \mapsto [f(x),i,D_{{\phi }_i(x)}({\psi }_{i^{\prime }}\circ f\circ {\phi }_i^{-1})(v)]\end{array}$

Suppose $(U_j,{\phi }_j)$ is another chart for $M$ with $x\in U_j$.$$\begin{array}{c}[x,i,v]=[x,,D_{{\phi }_i(x)}{f}_{ji}(v)]\mapsto [f(x),i^{\prime },D_{{\phi }_j(x)}({\psi }_{i^{\prime }}\circ f\circ {\phi }_j^{-1})D_{{\phi }_i(x)}{f}_{ji}(v)]\\ (\psi \circ f\circ {\phi }_j^{-1})\circ f_{ji}=(\psi \circ f\circ {\phi }_j^{-1})\circ ({\phi }_j\circ {\phi }_i^{-1})=\psi \circ f\circ {\phi }_i^{-1}\\ D_{{\phi }_j(x)}(\psi \circ f\circ {\phi }_j^{-1})\circ D_{{\phi }_i(x)}{f}_{ji}=D_{{\phi }_i(x)}(\psi \circ f\circ {\phi }_i^{-1})\end{array}$$In the same way, one checks that $Df$ does not depend on the chart used for $N$.
$\begin{array}{rl}Df{|}_{T_{x}M}:T_{x}M& \to T_{f(x)}N\subset TN\text{is a linear map between tangent spaces.}\\ [x,i,v]& \mapsto [f(x),i^{\prime },D_{{\phi }_i(x)}({\psi }_{i^{\prime }}\circ f\circ {\phi }_i^{-1})(v)]\end{array}$

定义 3.0.3. $D_{x}f:=Df{|}_{T_{x}M}$ is the derivative of $f$ at $x\in M$.