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Prove that:

a regular surface $S\subset \mathbb{R}^3$ is an orientable manifold if and only if there exists a differentiable mapping of $N:S\rightarrow \mathbb{R}^3$ with $N(p)\perp T_p(S)$ and $|N(p)|=1$, for all $p\in S$.

If part:

I guess I have to first let $X_a,X_b$ be parametrization of $S$, and $<dX_a,N>=<dX_b,N>=0$. Differentiability of $N$ imply (not sure) $N\circ X_a$ and $N\circ X_b$ are also differentiable, which their differentials are linear function. I then have no idea how to show that $\det(d(X_b^{-1}X_a))>0$.

Only if part: Zero idea.

Please give me some insight!!

JSCB
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  • If you have an orthogonal basis $e_1(x),\ e_2(x)$ in tangent space $T_xS$, then $N_x = e_1\times e_2$. Now we need to find a local basis which diffrentiable in $x$. – TZakrevskiy Dec 03 '14 at 10:48
  • Hint: if ${ a_1, a_2 }$ and ${ b_1, b_2 }$ are two bases of the same orientation for some 2-dimensional subspace of $\mathbb{R}^3$, then $a_1 \wedge a_2$ and $b_1 \wedge b_2$ point in the same direction. – user18063 Dec 03 '14 at 10:56

2 Answers2

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Hint : I guess your definition is the following: A regular surface $S\subset \mathbb R^3$ is orientable if one can find a family $(X_a)_{a\in \alpha}$ so that $\det(dX^{-1}_bX_a)) >0$.

So if $S$ is orientable, then you are given that family of charts $X_a$'s. For each chart, one can define locally the normal vector

$$N_a = \frac{\partial_u X_a\times \partial_vX_a}{|\partial_u X_a\times \partial_vX_a|}$$

So is it true that you can extends this map to the whole $S$? Or, in another chart $X_b$, do you have $N_a = N_b$? If you can show that, then $N: S\to \mathbb R^3$ is the mapping you want.

On the other hand, if such a $N$ is given, then you can consider all charts $X_a$ of $S$ so that

$$N = \frac{\partial_u X_a\times \partial_vX_a}{|\partial_u X_a\times \partial_vX_a|}. $$

Will this families of charts $X_a$'s satisfy $\det(dX^{-1}_bX_a)) >0$?

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Let $\{(U_{\alpha},x_{\alpha})\}$ be an atlas of a regular surface $S\subset\mathbb{R}^3$. We will construct a differentiable mapping $N:S\to\mathbb{R}^3$ as its "unit normal vector bundle". Arbitrarily choose $$ p=(y^1,y^2,y^3)\in x_{\alpha}(U_{\alpha})\subset\mathbb{R}^3 $$ and write $$ x_{\alpha}^{-1}(p)=(x_{\alpha}^1,x_{\alpha}^2)\in U_{\alpha}\subset\mathbb{R}^2. $$ Then a normal vector of $S$ at $p$ can be expressed as $$ m=\left( \frac{\partial(y^2,y^3)}{\partial(x_{\alpha}^1,x_{\alpha}^2)}, \frac{\partial(y^3,y^1)}{\partial(x_{\alpha}^1,x_{\alpha}^2)}, \frac{\partial(y^1,y^2)}{\partial(x_{\alpha}^1,x_{\alpha}^2)} \right), $$ which follows that $n:=m/|m|$ is a unit normal vector of $S$ at $p$. Define $N(p)=n$. Clearly $N:S\to\mathbb{R}^3$ is differentiable in each $U_{\alpha}$. We only need to testify that it is well-defined. Let $q\in x_{\alpha}(U_{\alpha})\cap x_{\beta}(U_{\beta})$. Through direct computations, we have $$ \left( \frac{\partial(y^2,y^3)}{\partial(x_{\beta}^1,x_{\beta}^2)}, \frac{\partial(y^3,y^1)}{\partial(x_{\beta}^1,x_{\beta}^2)}, \frac{\partial(y^1,y^2)}{\partial(x_{\beta}^1,x_{\beta}^2)} \right)= \det(x_{\alpha}^{-1}x_{\beta}) \left( \frac{\partial(y^2,y^3)}{\partial(x_{\alpha}^1,x_{\alpha}^2)}, \frac{\partial(y^3,y^1)}{\partial(x_{\alpha}^1,x_{\alpha}^2)}, \frac{\partial(y^1,y^2)}{\partial(x_{\alpha}^1,x_{\alpha}^2)} \right). $$ Hence $N:S\to\mathbb{R}^3$ is globally differentiable if and only if $S$ is orientable.

User1999
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