# Abstract Nonsense

## The Fundamental Theorem of Plane Curve Curvature

Point of Post: In this post we prove the fundamental theorem of plane curve curvature and use it to prove the curvature characterization of  segments of circles (positive constant curvature). Moreover, we also show that the fundamental theorem implies the existence of an osculating circle.

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Motivation

We now prove one of the most fulfilling theorems in the differential geometry of $\mathbb{R}^2$, namely that a smooth unit speed curve is entirely determined up to an orientation preserving isometry of $\mathbb{R}^2$ by its curvature. This is intuitively obvious for if one draws a curve and decides that it will follow some set ‘curvature pattern’ which is unit speed, then there is only two decisions to make: where to start, and what direction to start towards. But, this is precisely the statement in the first sentence. In fact, we’ll prove more. Not only does the curvature function characterize, up to an isometry, a unit speed curve but in fact given any smooth map $f:I\to\mathbb{R}$ for some interval $I\subseteq\mathbb{R}$ one has that there exists a unit speed curve $\mu:(a,b)\to\mathbb{R}^2$ such that $\varkappa_{\mu}=f$. As a particular consequence of  this deep fact is the somewhat satisfying fact that the only curve with positive constant curvature is a segment of a curve. We will then talk about what shall serve to be a prime motivator for curvature in higher dimensions–the osculating circle. Intuitively, given a point on a regular curve there is a unique circle which ‘best approximates’ the curve near that point, this circle is precisely the circle tangent to the curve at that point with radius equal to the inverse of the curvature at that point.

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The Fundamental Theorem of Plane Curves

We begin by proving the fundamental theorem:

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Theorem(Fundamental Theorem of Plane Curves): Let $f:I\to\mathbb{R}$ be a smooth function for some interval $I\subseteq\mathbb{R}$. Then, there exists a unit speed curve $\mu:I\to\mathbb{R}^2$ such that $\varkappa_\mu=f$. Moreover, if $\nu:I\to\mathbb{R}^2$ is another unit speed curve with $\varkappa_\nu=f$ then there exists an orientation preserving isometry $T:\mathbb{R}^2\to\mathbb{R}^2$ such that $\nu=T\circ\mu$.

Proof: We begin by construction a function with $f$ as its signed curvature. Indeed, choose $x_0\in I$ and let

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$\displaystyle \varphi(x)=\int_{x_0}^x f(t)\; dt$

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Define then

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$\displaystyle \mu(s)=\left(\int_{x_0}^{s}\cos(\varphi(x))\; dx,\int_{x_0}^{s}\sin(\varphi(x))\; dx\right)$

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clearly then $\mu$ is smooth (since $f$ is) unit speed because $\mu'(s)=(\cos(\varphi(s)),\sin(\varphi(s))$ which is clearly a unit vector. But, recall that $\varphi(x)$ is ‘a’ turning angle for $\gamma$, and by uniqueness ‘the’ turning angle for $\gamma$ (the unique smooth turning angle that is). We thus have from previous discussion that $\varkappa_\gamma=\varphi'$ but, $\varphi'=f$ and so $\gamma$ is a unit speed curve with curvature equal to $f$.

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Now, assume that $\nu:I\to\mathbb{R}^2$ is another unit speed curve with $\varkappa_nu=f$. Let $\widehat{\varphi}$ be the smooth turning angle for $\nu$. We see then from $\widehat{\varphi}'=f$ that

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$\displaystyle \widehat{\varphi}(x)=\int_{x_0}^{x}f(t)\; dt+\widehat{\varphi}(x-0)$

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We see then from the fact that $\nu'(s)=(\cos(\widehat{\varphi}(s)),\sin(\widehat{\varphi}(s))$ that

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$\displaystyle \nu(s)=\left(\int_{x_0}^{s}\cos\left(\widehat{\varphi}(x)\right)\; dx,\int_{x_0}^{s}\cos\left(\widehat{\varphi}(x)\right)\; dx\right)+\nu(x_0)$

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But, we know that $\widehat{\varphi}(x)=\varphi(x)+c$ for some constant $c$ (in particular $c=\widehat{\varphi}(x_0)$, but this doesn’t matter). Thus, we see that

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\displaystyle \begin{aligned}\nu(s) &= \left(\int_{x_0}^{s}\cos\left(\varphi(x)+c\right)\; dx,\int_{x_0}^{s}\sin\left(\varphi(x)+c\right)\; dx\right)+c\\ &= \left(\cos(c)\int_{x_0}^{s}\cos(\varphi(x))\; dx-\sin(c)\int_{x_0}^{s}\sin(\varphi(x))\; dx,\sin(c)\int_{x_0}^{s}\cos(\varphi(x))\; dx+\cos(c)\int_{x_0}^{s}\sin(\varphi(x))\; dx\right)+\nu(x_0)\\ &= \begin{pmatrix}\cos(c) & \sin(c)\\ -\sin(c) & \cos(c)\end{pmatrix}\mu(s)+\nu(x_0)\end{aligned}

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and so $\nu=R_c\circ T_{\nu(x_0)}\circ\mu$ where $R_c$ is the counterclocwise rotation by the angle $c$ and $T_{\nu(x_0)}$ is translation by $\nu(x_0)$. Since $R_c\circ T_{\nu(x_0)}$ is an orientation preserving isometry the conclusion follows. $\blacksquare$

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From this we get the satisfying corollary we alluded to in the motivation:

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Theorem: Let $\gamma:I\to\mathbb{R}^2$ be a regular curve with constant positive unsigned curvature. Then, $\gamma(I)$ is part of a circle in $\mathbb{R}^2$.

Proof: The first step is to note that since $\kappa_\gamma$ is constant that $\varkappa_\gamma$ is constant. This is because if $\varkappa_\gamma$ was not constant then $\varkappa_\gamma(I)$ is a non-degenerate interval which is impossible since $|\varkappa_\gamma|$ is constant. So, choose a unit speed reparamaterization $\mu$ of $\gamma$. Since curvature is independent of reparamaterization we have that $\mu$ has constant positive curvature, say $c$. Since the circle $\displaystyle t\mapsto \left(\frac{1}{c}\cos\left(ct\right),\frac{1}{c}\cos(ct)\right)$ also has curvature equal to $c$ we may conclude from the previous theorem that there is an orientation preserving isometry carrying the circle onto $\mu$, and since circles are invariant under isometries we have that $\mu(I)=\gamma(I)$ is a subset of a circle. $\blacksquare$

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We end this topic by discussing the idea of the osculating circle. Roughly, for a regular curve $\gamma:I\to\mathbb{R}^2$ and a point $\gamma(t)$ with non-zero curvature the osculating circle at $\gamma(t)$ is the circle of radius $\displaystyle \frac{1}{\varkappa_\gamma(t)}$.

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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.