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I started reading "Elementary Differential Geometry" by Andrew Pressley. I've been confused about Proposition 1.3.6, which reads:

A parametrized curve has a unit-speed reparametrization if and only if it is regular.

I've been looking over the converse portion of the proof. We're given a regular curve $\gamma: (\alpha, \beta) \rightarrow \mathbb{R}^n$. The proof uses arc-length s to set up a reparametrization map $s^{-1} : (\tilde{\alpha}, \tilde{\beta}) \rightarrow (\alpha, \beta)$

After setting up the reparametrization map, however, the proof follows:

We take $\phi = s^{-1}$ and let $\tilde{\gamma}$ be the corresponding reparametrization of $\gamma$, so that $\tilde{\gamma}(s) = \gamma(t)$.

What I do not understand is how we're able to automatically assume the existence of reparametrization. I'm confused how we can ensure that the reparametrization would map to the same image of the original curve. The proof seems to imply that a reparametrization exists if we're given a reparametrization mapping.

Another part of the proof that I'm confused about is establishing the arc-length as the unique unit-speed parameter on a regular curve. By the end of the proof, arc-length is confirmed to be the unit-speed parameter. However, I don't really understand its uniqueness as a unit-speed parameter.

phillip1618
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1 Answers1

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Note $\gamma$ is a regular curve if its derivative does not vanish.

So $s^{'}(t)=\left| \gamma^{'}(t)\right|>0$ and $s(t)$ is strictly increasing.

This implies that a reparametrization exists. Using Chain rule, $\left|\tilde{\gamma}^{'}(s)\right| = \frac{\left|\gamma^{'}(t)\right|}{\left|s^{'}(t)\right|}=1$.

For uniqueness part, it is unique in the following sense: since the reparametrization map is bijective, if there exist two unit-speed reparametrization map, just take composite of one and inverse of another, then the composite is also bijective.

Tim
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