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I am learning Elementary Differential Geometry by O’Neill and having a hard time with this exercise.

Suppose that $\beta_1$ and $\beta_2$ are unit-speed reparametrizations of the same curve $\alpha$.

Show that there is a number $s_0$ such that $\beta_2(s)=\beta_1(s+s_0)$ for all $s$.

My questions are

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1 The two curves may not be defined on the same interval and each of them may traverse part of the curve $\alpha$. What is the meaning of for all $s$?

2 Why must the two beta curves have the same orientation?

3 How to prove the statement rigorously and completely?

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Let’s recall some facts and definitions: Let $\alpha$ be defined on some interval $I$ and define for $t_0\in I$

$$\lambda_{t_{0}}(t)=\int_{t_0}^t\|\alpha'(x)\|\,dx.$$

Any unit-speed-reparametrization (usp) $\phi_{t_{0}}$ is then given by

$$\phi_{t_{0}}=\lambda_{t_{0}}^{-1}.$$

That’s for the facts.

Now let $i\in\{1,2\}$ and choose some usp $\phi_{t_{i}}$ such that $\beta_i=

\alpha\mathop{\circ}\phi_{t_i}$. Let $t_0\in I$. Then there exists $s_{i}$ such that $t_0=\phi_{t_i}(s_i)$. So that gives

$$s_i=\phi_{t_i}^{-1}(t_0)=\lambda_i(t_0)=\int_{t_i}^{t_0}\|\alpha'(x)\|dx\,$$

and that means that $s_1$ and $s_2$ only differ by constant, namely

$$

s_0=\int_{t_1}^{t_2}\|\alpha'(x)|\,dx,

$$

i.$\,$ e., $|s_0|=$ length of $c$ between $t_0$ and $t_1$. Extremely intuitive. It then follows that $\beta_1(s)=\beta_2(s+s_0)$.

Ad (2): By definition of ups: it preserves orientation.

Michael

Yet another approach: let $\beta_2(s_2)=\beta_1(s_1)$. Then there exists a differentiable map $\gamma$ such that $s_2=\gamma(s_1)$ (why?). Thus

$$\beta_2\bigl(\gamma(s_1)\bigr)=\beta_1(s_1)\Rightarrow\beta_2’\bigl(\gamma(s_2)\bigr)\gamma'(s_2)=\beta_1′(s_1).$$

Taking lengths on both sides gives $|\gamma'(s_1)|=1$, thus $\gamma(s_1)=\pm s_1+s_0$. Since an usp preserves orientation, we must have $\gamma(s_1)=s_1+s_0$.

Michael

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