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In the text "O'neill-Elementary differential geometry (2nd edition)", there are following problems.

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As we know, any unit speed-reparametrized curve can be viewed as an arc-length reparametrizated curve. Since the arc-length reparametrized curve must achieve the positive orientation, the problem 5 can be easily proved. But, I have a question for the problem 7. $\alpha$, $\bar{\alpha}$ are unit-speed curves. Then, why we have to admit the "minus sign"(opposite orientation) in the problem 7? I want to know the exact difference of two problems.

Chris kim
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2 Answers2

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$$\alpha(s)=\bar{\alpha}(h(s))$$

Differentiate w.r.t. $s$

$$\alpha'(s)=\bar{\alpha}'(h(s))h'(s)$$

Take absolute value

$$|\alpha'(s)|=|\bar{\alpha}'(h)||h'(s)|$$

$|h'(s)|=1$

Here $h(s)$ should be continuous, otherwise $\alpha$ won't satisfy as a regular curve

$h'(s)=1$ or $-1$

$h(s)=\pm s +s_0$ for some constant $s_0$

Arclength parametrization only means $|\alpha'(s)|=1$ at each point on the curve, it doesn't tell us anyone about the orientation. Reversing the direction would still give us the arclength parametrization.

Chris kim
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lEm
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  • If given reparametrization $\beta(u)=\alpha(h(u))$ is unit-speed, then $ds/du=1>0$. I think this shows the positive orientation. Is it false? – Chris kim Apr 02 '16 at 12:49
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    @Chriskim Not necessarily, because remember $\frac {ds}{du} = \frac{|d\beta |}{du}$ , it represents the magnitude of the speed, which doesn't show that it must have a positive orientation. As I have show, if we have $h'(u)=-1$, it still satisfy the definition of arclength reparametrization, but the direction would be reversed – lEm Apr 02 '16 at 13:10
  • Ah, I understand! $\beta_1$ and $\beta$ can have the relation in #5#. But, the "derivation" should have two possibilities! – Chris kim Apr 02 '16 at 13:11
  • @Chriskim yes exactly – lEm Apr 02 '16 at 13:12
  • Furthermore, $ds/du=d\tilde{s}/du $ as you said($\tilde{s}$ is a reparametrization) . Thank you! :) – Chris kim Apr 02 '16 at 13:13
  • / In my opinion, the problem #5 may be better if it has the delicate expression $\beta_{1}(s)=\beta_2(\pm s+s_0)$. I think the author ignore the choice of orientation about the reparameter "h" because we use the positively orientated reparametrization for the curve in many cases :) Am I right? – Chris kim Apr 03 '16 at 12:53
  • @Chriskim yes, you are correct – lEm Apr 03 '16 at 14:19
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(1) $s_0$ is a length of arc between $\beta_1(t),\ \beta_2(t)$ for any $t$

(2) direction going and direction escaping plane are changed by parametrization $\pm s$. It depends on position seeing the curve. But $N$ indicate the bending direction which is independent on position seeing the curve.

HK Lee
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  • If given reparametrization $β(u)=α(h(u))$ is unit-speed, then $ds/du=1>0$ I think this shows the positive orientation. Is it false? Also, this fact is illustrated in the text "O'neill". – Chris kim Apr 02 '16 at 13:06
  • Oh, previous question was wrong. Your answer also helped me. Thank you! :) – Chris kim Apr 02 '16 at 13:18