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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 $d\pi: TTX \to TX$?

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  • $\begingroup$ 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 $\endgroup$ – funktor Aug 15 '16 at 17:02
  • $\begingroup$ 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. $\endgroup$ – JDoe Aug 16 '16 at 8:17
  • $\begingroup$ @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 $X = F = \mathbb{R}^n$? $\endgroup$ – JDoe Aug 19 '16 at 15:30
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    $\begingroup$ 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. en.wikipedia.org/wiki/Hairy_ball_theorem $\endgroup$ – funktor Aug 31 '16 at 7:01
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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 $T(X)$

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  • $\begingroup$ Thank you for your help. The proof is very nice. Now I just have to include the local submersion theorem into my thesis. $\endgroup$ – JDoe Aug 18 '16 at 19:08
  • $\begingroup$ Note that you can directly show that $\nu$ is a submersion by computing its Jacobian matrix. $\endgroup$ – Sou Dec 28 '17 at 17:35
  • $\begingroup$ @KelvinLois could you provide the details of this method please ? $\endgroup$ – Intuition Nov 28 '18 at 13:21
  • $\begingroup$ @KelvinLois could you provide the details of this method please ? $\endgroup$ – Secretly Nov 28 '18 at 13:52

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