7

I am wondering how to see that the natural projection from $\pi:TX \to X$ is a submersion. To see that $\pi$ is surjective I can identify $TX$ as product space $X \times F$ where $F$ is a vector space with the same dimension as $X$ and so for all $(x,v) \in TX \simeq X \times F$, $\pi(x,v) = x$ is surjective. Correct? But I don't now how to proof that $d\pi$ is surjective. Do I need some concept of the double tangent space Or can I avoid the map $d\pi: TTX \to TX$?

JDoe
  • 724
  • You can't really identify $T(X)$ as $X\times F$. Locally, yes. Globally, not always. But luckily, locally is good enough in this case.

    EDIT: I'm hesitant to give out the entire solution straight away because something tells me that this is a homework problem

     –  Aug 15 '16 at 17:02
  • I am actually try to understand this for my bachelor thesis. I am slightly passing this topic and don't know how far I have to go into the theory. That's why I asked if I need the concept of the double tangent space. So your help is really appreciated. – JDoe Aug 16 '16 at 08:17
  • @funktor I have another question regarding your first comment. If $X$ is a $C^k$ manifold ($k \in \mathbb{N}$) isn't it correct that the Tangent fiber bundle $(T(X), \pi_{TX}, X)$ is globally trivial and therefore I can identify it with $X \times F$, $F$ is a vector space or is it just possible for $X = F = \mathbb{R}^n$? – JDoe Aug 19 '16 at 15:30
  • 1
    Your tangent bundle may not be always trivial. As an example, take $S^2$, the sphere. In this case the tangent bundle $T(S^2)$ is not diffeomorphic to $S^2\times \mathbb R^2$ globally. The fact that its tangent bundle is not trivial is what the the Hairy Ball theorem is I believe. https://en.wikipedia.org/wiki/Hairy_ball_theorem –  Aug 31 '16 at 07:01

2 Answers2

9

Let $\pi:T(X)\to X$, be the projection map in question.

Let $U\subset T(X)$ be a neighborhood of some $(x',v')\in T(X)$ and $V\subset X$ a neighborhood of $\pi(x',v')=x'\in X$. We know that there exists a chart $\psi:V\to \psi(V)\subset \mathbb R^k$.

Let $\phi:U\to \phi(U)\in\mathbb R^{2k}$ be a local coordinate chart for $T(X)$ where for any $(x,v)\in U$, $\phi((x,v))=(x_1,\ldots,x_k,\ldots, x_{2k})$ where $(x_1,\ldots,x_k)=\psi(x)$ for some $x\in V$. In that case, notice that the following diagram commutes $\newcommand{\ra}[1]{\kern-1.5ex\xrightarrow{\ \ #1\ \ }\phantom{}\kern-1.5ex} \newcommand{\ras}[1]{\kern-1.5ex\xrightarrow{\ \ \smash{#1}\ \ }\phantom{}\kern-1.5ex} \newcommand{\da}[1]{\bigg\downarrow\raise.5ex\rlap{\scriptstyle#1}}$ \begin{array}{c} U & \ra{\pi|_U} & V \\ \da{\phi} & & \da{\psi} \\ \phi(U) & \ras{\nu} & \psi(V) \\ \end{array} where $\nu:\phi(U)\to\psi(V)$ is the canonical submersion i.e, $$\nu(x_1,\ldots,x_k,\ldots, x_{2k})=(x_1,\ldots,x_k)=\psi(x)$$ Clearly, $\psi\circ \pi|_U\circ \phi^{-1}=\nu$. Thus by the local submersion theorem, $\pi$ is a submersion at $(x',v')$. Since our proof was independent of the choice of $(x',v')$, we conclude that $\pi$ is a submersion everywhere in $T(X)$

Community
  • 1
1

I want to provide a easier alternative proof here.

Suppose dim$M=n.$

Let $(U,(x^i))$ be a chart on $M$ and $p\in U$. Note that $(x^i)$ provides a smooth local coordinates on the open subset $U\subseteq M$. And then we get the corresponding natural coordinates $(x^i,v^i)$ on $\pi ^{-1}(U)\subseteq TM$. The coordinate representation of $\pi$ is $\hat{\pi}(x,v)=x.$ Note that the top-left $n\times n$ submatrix in the Jacobian Matrix of $\hat{\pi}$ is $I_n$, the $n\times n$ identity matrix. So the rank of the Jacobian Matrix of $\hat{\pi}$ is $n$ because its rank is bounded by the minimum of $\{ \text{dim}M, \text{dim}TM\}$.

The corresponding matrix of the linear map $d\pi$ w.r.t these two bases is exactly the Jacobian Matrix of $\hat{\pi}$. By the discussion above and the fact that $\dim T_p M=\dim M=n$, it is easy to see that for every $\alpha \in\pi^{-1}(U)$ with $\pi(\alpha)=p$, we have that $d\pi[T_\alpha(TM)]=T_p M$. And since $p$ is arbitrary, we conclude that the map $d\pi$ is surjective. Hence $\pi$ is a smooth submersion.

$\tag*{$\blacksquare$}$

Sam Wong
  • 2,481