# Abstract Nonsense

## Reparamaterization, Regular Curves, and Unit Speed Curves (Pt. II)

Point of Post: This is a continuation of this post.

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As stated in the motivation there is another naturally desirable type of curve, namely we call a curve $\gamma:I\to\mathbb{R}^n$  unit speed if $\|\gamma'(t)\|=1$ for every $t\in I$. Or, said differently, if $\gamma':I\to\mathbb{S}^{n-1}$. These curves can be thought of as carving it’s image $\gamma(I)$ into pieces and move along them at a given speed, moving so long each ‘step’. In other words, they are the continuous analogue of the following situation: measure the length of a hallway, divide the total length by $m\in\mathbb{N}$ and each step move $m^{-1}$ units.

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So, what exactly do regular curves have to do with unit speed curves? Well, it should be clear that every unit speed curve is regular, but in a sense the converse is true. Namely, every regular curve is, up to reparamaterization, unit speed. So, every regular curve is equivalent (in the sense defined above) to a unit speed curve. Why this should intuitively make sense is that if $\|\gamma'(t)\|\ne 0$ then we should be able to ‘normalize’ it to get a curve who is unit speed. The real proof is:

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Theorem: Let $\gamma:I\to\mathbb{R}^n$ be a smooth curve. Then, $\gamma$ is regular if and only if it posses a unit speed reparamaterization.

Proof: Assume that $\mu:J\to\mathbb{R}^n$ is a unit speed curve such that there exists a diffeomorphism $\phi:J\to I$ such that $\gamma\circ\phi=\mu$. We see then that $\phi'(t)\gamma'(t)=\mu'(t)$ and so, in particular, if $\gamma'(t)=0$ then $\mu'(t)=0$ which is impossible since $\|\mu'(t)\|=1$.

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So, assume now that $\gamma$ is regular. Fix some $x_0\in I$ and let $s_\gamma$ be with respect to $x_0$. Since $\gamma'(t)\ne 0$ for all $t\in I$ we know that $s_\gamma:I\to\mathbb{R}$ is a smooth function with the property that $s_\gamma'(t)=\|\gamma'(t)\|>0$ for all $t\in I$. We know then $s_\gamma$ is injective and a local diffeomorphism, and so from analysis we know that $s_\gamma$ is an open map and so $s_\gamma(I)=J$ for some open interval $J$. So, consider the reparamaterization $\mu:J\to\mathbb{R}^n$ given by $\mu(t)=\gamma(s_\gamma^{-1}(t))$. We claim that $\mu$ is unit speed. Indeed:

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\displaystyle \begin{aligned}\left\|\mu'(t)\right\| &=\left\|\gamma'(s_\gamma^{-1}(t))\left(s_\gamma^{-1}(t)\right)'\right\|\\ &=\left\|\gamma(s_\gamma^{-1}(t))\frac{1}{s_\gamma'(s_\gamma^{-1}(t))}\right\|\\ &=\left\|\gamma'(s_\gamma^{-1}(t))\frac{1}{\|\gamma'(s_\gamma^{-1}(t))\|}\right\|\\ &=1\end{aligned}

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from where the conclusion follows. $\blacksquare$

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Note that we used the arc length to paramaterize the curve. Thus, often times people will say that, given a curve in $\mathbb{R}^n$ (in the sense of a subset of $\mathbb{R}^n$) that it is ‘paramaterized by arc length’.

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We end this post by showing that choosing arc length as our reparamaterization map was necessary, in the sense that every map which reparamaterizes a given regular curve must be very close to the arc length curve. In particular:

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Theorem: Let $\gamma:I\to\mathbb{R}^n$ be a smooth regular curve. Then, if $J$ is an interval and $\phi:J\to I$ a diffeomorphism such that $\gamma\circ \phi$ is unit speed then for any $x_0\in I$ one has that $J=\pm s_\gamma(I)+c$ for some constant and $s_\gamma$ is with respect to $x_0$. Moreover, $\phi=\pm s_\gamma+c$.

Proof: We note that

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$|\phi'(t)|=|\phi'(t)|\left\|\gamma'(\phi(t))\right\|=\|\gamma'(\phi(t))\phi'(t)\|=\|\left(\gamma(\phi(t))\right)'\|=\|\gamma'(t)\|=s'_\gamma(t)$

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and so $\phi'(t)=\pm s_\gamma'(t)$ which obviously implies that $\phi(t)=\pm s'_\gamma(t)+c$. $\blacksquare$

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Remark: Note that this ‘makes ok’ a brushed over fact. Namely, when proving that every regular curve has a unit speed reparamaterization we fixed some point in the interval and measured arc length from that point. In particular, it didn’t matter what point we picked. Thus, the arc length function started from any point will reparamaterize a regular curve to unit speed. What makes this ‘ok’, in the sense that it’s consistent with the above is that fooling around with the starting point of the integral one can get that any two arc length functions (measuring from different points) differ  at most by a sign and a constant. In other words, not trying to insult the readers intelligence,

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$\displaystyle \int_{3}^x \|\gamma'(t)\|\; dt=\int_5^x \|\gamma'(t)\|\;dt+\int_3^5\|\gamma'(t)\|\; dt$

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Now, arc length is an intuitively geometric concept. By this, I mean that we are intuitively measuring the arc length of the image of the curves we are dealing with. In particular, if $\alpha\simeq\beta$ then $\alpha(I)=\beta(J)$ and so we would expect that arc length from point $\gamma(t_0)$ to $\gamma(t_1)$ (on the curve, think geometrically!) should be the same as the length from $\gamma(\phi(t_0))$ to $\gamma(\phi(t_1))$. And, in deed, this is the case:

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Theorem: Let $\alpha:I\to\mathbb{R}^n$ a smooth curve and $\beta:J\to\mathbb{R}^n$ a reparamaterization of $\alpha$ with reparamaterization map $\phi:I\to J$. Then, picking some $x_0,x_1\in I$ one has that the arc length along $\alpha$ from $x_0$ to $x_1$ is equal, up to a sign, to the arc length of $\beta$ along $\phi(x_0)$ to $\phi(x_1)$.

Proof: We merely note that

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$\displaystyle \int_{\phi(x_0)}^{\phi(x_1)}\|\beta'(t)\|\; dt=\int_{\phi(x_0)}^{\phi(x_1)}|\phi'(t)|\|\alpha'(\phi(t)\|\; dt=\pm \int_{x_0}^{x_1}\|\alpha'(t)\|\; dt$

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$\blacksquare$

The sign discrepancy can be thought of as saying that the ‘absolute length’ is invariant under reparamaterization, but recall that our definition of arc length ‘signs’ the ‘absolute length’ between two points according to whether you are measuring ‘backwards’ or forwards (i.e. if you are measuring from a point $\gamma(t)$ on a curve to a point $\gamma(t')$ on the curve with $t' so you are ‘going backwards’) and this relation of backwards to forwards is not preserved under diffeomorphism (i.e. think of curves $\gamma:\mathbb{R}\to\mathbb{R}^2$ (e.g. $\gamma(t)=(\cos(t),\sin(t))$ and decreasing diffeomorphisms $\phi:\mathbb{R}\to\mathbb{R}$ (e.g. $\phi(t)=1-t$)).

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References:

1. Carmo, Manfredo Perdigão Do. Differential Geometry of Curves and Surfaces. Upper Saddle River, NJ: Prentice-Hall, 1976. Print.

2.  Montiel, Sebastián, A. Ros, and Donald G. Babbitt. Curves and Surfaces. Providence, RI: American Mathematical Society, 2009. Print.