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Let $S\subset R^3$ be a regular surface with $p\in S$. Let $(U,\phi)$ be a local chart such that $\phi(q) = p$ where $q\in U\subset R^2$ (Basically, $(U,\phi)$ is a local chart when $\phi$ is a local parameterization of $S$ that has $p$ in its image set).

We know that the tangent plane at $p$ $T_p(S)$ is defined as

$T_p(S) = \{\alpha'(0) \mid \alpha : (-\epsilon, \epsilon) \to S \text{ is smooth and } \alpha(0) = p\}$

We know that the tangent plane is spanned by $T_p(S) = \text{span} \{\phi_x(q), \phi_y(q)\}$

We define the normal unit vector to $T_p(S)$ with respect to local chart $(U, \phi)$ by

$$N_\phi(p) = \frac{\phi_x(q)\times \phi_y(q)}{\|\phi_x(q)\times \phi_y(q)\|} $$

Now, note that for another local chart $(V,\gamma)$ such that $\gamma(r) = p$,

$$N_\gamma(p) = \frac{\gamma_x(r)\times \gamma_y(r)}{\|\gamma_x(r)\times \gamma_y(r)\|}$$

is also unit normal to $T_p(S)$ at $p$.

Hence, $$N_\phi(p) = \pm N_\gamma(p)$$

The transition map between $\phi $ and $\gamma $ which is given by $\gamma^{-1}\circ \phi : \phi^{-1}(W)\to \gamma^{-1}(W)$ where $p\in W = \phi(U)\cap \gamma(V)$ is a diffeomorphism.

I want to prove that if $\text{det } [D(\gamma^{-1}\circ \phi )(q)] >0 $ then we must have $N_\phi(p) = N_\gamma(p)$

How can I show this precisely?

Note that $D(\gamma^{-1}\circ \phi )(q)$ is a $2\times 2 $ matrix and doesn't have $0$ determinant because $\gamma^{-1}\circ \phi $ is diffeomorphism.

Bernard
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chesslad
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1 Answers1

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Take $\gamma^{-1}\circ\phi$ as the change of variables $$(x,y)^{\top}\mapsto(v,w)^{\top},$$ then $D(\gamma^{-1)}\circ\phi)$ is a $2\times 2$ matrix which allows to express $$\phi_x=P\gamma_v+Q\gamma_w,$$ $$\phi_y=S\gamma_v+T\gamma_w,$$ since both sets $\{\phi_x,\phi_y\}$ and $\{\gamma_v,\gamma_w\}$ are basis for the tangent space. Now you have $$\phi_x\times\phi_y=(P\gamma_v+Q\gamma_w)\times (S\gamma_v+T\gamma_w)$$ $$\qquad=(PT-QS)\ \gamma_v\times\gamma_w,$$ which gives you control over the sign of vector $N$ in the two coordinates.

janmarqz
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