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So I just finished my differential geometry exam few days ago, and there is this one bonus problem that bugs me, its a part of a series of questions.

BONUS PROBLEM: We give $\Lambda$ is a regular surface, where $(0,0,0) \notin \Lambda.$ Note the following definition:

i) For each $k$ in $\Lambda,$ we define $N_k\Lambda$ as the vector space with dimension $1$ spanned by the normal vector$(\ne 0)$ at point $k$ on $\Lambda.$

ii) $\zeta=\{(k,n_k)\in{k}\times N_k\Lambda: k\in \Lambda\}$, $\Psi = \{(k,ak)\in\Lambda \times \mathbb{R}^3: a\in \mathbb{R}\}$

If we give a well defined function (smooth) $\overline{n}:\Lambda \rightarrow \mathbb{R}^3$ $(\overline{n}$ = unit normal vector at $k$ in $\Lambda)$, then $\Psi$ is diffeomorphic to $\zeta.$ True or False?

I answered true, and got it right, but it was a just a guess. I now want to know why as the answer sheet just says T. Instructor does not seem very responsive, so I ask here. Can anyone show me why they are diffeomorphic?

Aurora Borealis
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1 Answers1

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In order to provide a rigorous argument, we should start by describing the smooth structures of $\zeta$ and $\Psi$. We can use $\overline{n}$ to construct local parametrizations of $\zeta$, as follows. Let $$\varphi:U\subset\mathbb{R}^2\to\Lambda$$ be a local parametrization. Then $$\overline{\varphi}:U\times\mathbb{R}\to\zeta,\quad(p,t)\mapsto(\varphi(p),t\overline{n}(\varphi(p))),$$is a parametrization of $\zeta$. Similarly, $$\psi:U\times\mathbb{R}\to\Psi,\quad(p,t)\mapsto(\varphi(p),t\varphi(p))$$is a parametrization of $\Psi$. With respect to these parametrizations, the map $$(k,ak)\mapsto(k,a\overline{n}(k))$$ looks like $(p,a)\mapsto(p,a)$, so it is clearly a diffeomorphism.

Amitai Yuval
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  • Thank you now let me try to understand your argument. I am confused about how we can deduce its a diffeomorphism, as I dont see why the inverse is smooth? I am sorry I am still a beginner so I can only see things by criteria mentione

    Also my professor replied and he said hint: consider non parallel vectors $s$ and $t$ in $\mathbb{R}^3,$ I have no idea why this is a hint.

     – Aurora Borealis Apr 14 '18 at 05:59