Abstract Nonsense

Crushing one theorem at a time

Reparamaterization, Regular Curves, and Unit Speed Curves (Pt. I)

Point of Post: In this post we discuss the notion of reparamaterizations (definitions and intuition), regular curves, unit curves, and the interrelation between the three.

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Up to this point we’ve figured out that intuitively the better way to discuss a ‘curve’ thought of as a subset of some Euclidean space is to instead imagine a smooth mapping \gamma:I\to\mathbb{R}^n whose image is equal to the ‘curve’ we are focusing on. That said, an obvious question that comes with the territory is–which map do we choose? In other words, multiple maps can paramaterize the same ‘curve’. For example, it’s not hard to see that

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\begin{aligned}&\gamma:\mathbb{R}\to\mathbb{R}^2:t\mapsto (\cos(t),\sin(t))\\ &\alpha:\mathbb{R}\to\mathbb{R}^2:t\mapsto (\cos(2t),\sin(2t))\\ &\beta:\mathbb{R}\to\mathbb{R}^2:t\mapsto (\sin(t),\cos(t))\end{aligned}

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all have the unit circle \mathbb{S}^1 as their image. So, if these smooth maps are our vehicle to study ‘curves’, then clearly whatever properties we care about should be the same for whichever smooth curve we pick to represent the ‘curve’. But, it’s clear that all not curves are created equal. For example, I think it’s pretty clear that the functions \mathbb{R}^+\to\mathbb{R}^2 given by t\mapsto (t,t^2) and t\mapsto (t^2,t^4) have the same image, but are fundamentally different. In particular, if one looks at the tangent vectors to each of these curves at t=0 one finds that the first has non-zero tangent vector but the second doesn’t. So? Why does this matter? Well, it shall soon be important to us that we are able to make statements such as “there is precisely two unit vectors orthogonal to \gamma'(t)” which is, as I’m sure you are well aware, is always true–except when \gamma'(t)=0. Thus, as we shall see it will be a decidedly geometrically important fact as to whether or not a curve has a zero tangent vector. In particular, two curves one of which has this property and one of which don’t are fundamentally different objects. So, our notion of equivalence between curves, should be discerning enough to detect this, so that for, example (t,t^2) and (t^2,t^4) are not equivalent but (perhaps) (\cos(t),\sin(t)) and (\sin(t),\cos(t)) are. Moreover, we shall see that curves that are ‘nice’ (in the sense that they never have a zero tangent vector) are equivalent (whatever this will eventually mean) to the nicest possible type of curve, namely a curve whose tangent vectors are always unit vectors. This type of curve is nice for both now-hidden and now-apparent reasons. For example, for such a curve it’s pretty easy to calculate arc length since \|\gamma'(t)\|=1.

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Reparamaterizations, Regular Curves, and Unit Curves

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Recall that a diffeomorphism is a bijective map \phi:I\to J, for some intervals I,J\subseteq\mathbb{R}, such that both \phi and \phi^{-1} are C^\infty (where at the endpoints of the intervals we mean one-sided differentiability). We say that a curve \gamma_1:I\to\mathbb{R}^n is a reparamaterization of \gamma_:J\to\mathbb{R}^n if there exists a diffeomorphism \phi:J\to I such that \gamma_1\circ\phi=\gamma_2. We denote this relation by saying that \gamma_1\simeq\gamma_2. It’s clear that \simeq is an equivalence relation on the set of all curves. To see an example of this note that if \gamma_1:\mathbb{R}\to\mathbb{R}^2:t\mapsto (\cos(t),\sin(t)) and \gamma_2:\mathbb{R}\to\mathbb{R}^2:t\mapsto (\sin(t),\cos(t)) then \gamma_1\simeq\gamma_2 via the diffeomorphism \displaystyle \phi:\mathbb{R}\to\mathbb{R}:t\mapsto\frac{\pi}{2}-t.

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For a curve \gamma:I\to\mathbb{R}^n we say that t\in I is a regular point if \gamma'(t)\ne 0 and a singular point otherwise. If t\in I is a regular point then \gamma(t) is known as a regular value and similarly if t\in I is a singular point then \gamma(t) is a singular value. We call a curve \gamma:I\to\mathbb{R}^n regular if every point of t\in I is a regular point. What we now claim is that regularity is reparamaterization invariant. To be more explicit:

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Theorem: Let \alpha:I\to\mathbb{R}^n and \beta:J\to\mathbb{R}^n be curves with \alpha\simeq\beta. Then, \alpha is regular if and only if \beta is regular.

Proof: It clearly suffices to prove that \alpha regular implies that \beta is regular. To do this we merely note that if \phi:I\to J is the reparamaterization map between \alpha and \beta then since \phi^{-1}(\phi(t))=t for every t\in I we have by the chain rule that (\phi^{-1})'(\phi(t))\phi'(t)=1 for all t\in I, and so in particular \phi'(t)\ne 0 for all t\in I. Note then, that by the chain rule one has that D_\beta(t)=D_{\alpha}(\phi(t))\circ D_\phi(t) (where D is the total derivative) but since \alpha is regular we know that D_\alpha(\phi(t)) is invertible and by the previous observation we also have that D_\phi(t) is invertible, and so D_\beta(t) is invertible. But, by our previous observation this is equivalent to \beta'(t). Since t is arbitrary the conclusion follows. \blacksquare

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The next important and interesting fact concerning regular curves is their relation to the arc length function. Namely:

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Theorem: Let \gamma:I\to\mathbb{R}^n be a regular curve. Then, the arc length function s_\gamma is a smooth function on I.

Proof: It clearly suffices to prove that \|\gamma'(t)\| is a smooth function on I since the rest follows from the fundamental theorem of calculus. That said, we know that \gamma' is differentiable on I and so it suffices by the chain rule to prove that the norm function \|\cdot\| is smooth on an open set containing the image of \gamma. That said, it’s a basic fact from analysis that \|\cdot\| is smooth on \mathbb{R}^n-\{0\} and by assumption \gamma(I)\subseteq\mathbb{R}-\{0\} and so the conclusion follows. \blacksquare

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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.



September 23, 2011 - Posted by | Differential Geometry | , , , , , , , ,


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