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Consider the unit sphere $S^1$ of $\mathbb{R}^2$. This is a 1-dimensional manifold. And an orientation $\sigma$ of $S^1$ is given by the orientated atlas $\left\{\phi_1,\phi_2\right\}$ with the maps $$ \phi_1\colon (-\pi,\pi)\to S^1\setminus (-1,0), t\mapsto (\cos t,\sin t)\\ \phi_2\colon (0,2\pi)\to S^1\setminus (1,0), t\mapsto (\cos t,\sin t) $$ Moreover an orientation of $S^1$ is given by the external normal field $$ \nu\colon S^1\to\mathbb{R}^2. $$ Show that the orientation given by the external normal field is identical with the orientation $\sigma$ above.

Do I have to show that the external normal field $\nu$ is positive orientated related to $\sigma$? Or something different?

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I think you want to show that, if you pick a point $p\in S^{1}\subset\mathbb{R}^{2}$, then $\nu(p),v(p)$ is positively oriented in $T_{p}\mathbb{R}^2$, where $v(p)$ is the oriented basis for $T_{p}S^{1}$ you get from the atlas $\{\phi_1,\phi_2\}$. Is this what you meant by "the external normal field $\nu$ is positive oriented related to $\sigma$"?

Munchlax
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  • I guess you mean exactly what I meant: Does $S^1$ have an orientation $\sigma$, then the external normal field $\nu$ is called positive orientated related to $\sigma$" if for any $a\in S^1$ the following condition is fullfilled: Let $(w_1)$ be a positive orientated basis of $T_a S^1$, then $(\nu(a),w_1)$ is a positive orientated basis of $\mathbb{R}^2$, i.e. $det(\nu(a),w_1) > 0$. --- And now I wanted to use the positive orientated map $\phi_1$ and the positive orientated basis $\left(\frac{\partial\phi_1(c)}{\partial t}\right)$, $a:=\phi_1(c), c\in (-\pi,\pi)$ to show that. –  Jul 27 '13 at 17:38
  • Is this what I have to show? –  Jul 27 '13 at 18:13
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    I think we agree on what we're saying, so you have my vote. – Munchlax Jul 27 '13 at 21:35
  • I think it's not that difficult: Consider any $\phi_1(c)=:a\in S^1$, $c\in (-\pi,\pi)$. $\phi_1$ is a positive orientated map and $\left(\frac{\partial\phi_1(c)}{\partial t}\right)$ is a positve orientated basis of $T_aS^1$. Because of $\frac{\partial\phi_1(c)}{\partial t}=(-\sin c,\cos c)$ it is $det\left(\nu(a),\frac{\partial\phi_1(c)}{\partial t}\right)=det(a,(-\sin c,\cos c))=det((\cos c,\sin c),(-\sin c,\cos c))=1 > 0$ and that's what I wanted to show. --- That's it? –  Jul 28 '13 at 10:55
  • I think that's it. – Munchlax Jul 30 '13 at 01:57