[Tu Differential Geometry] 2. Curves

 This article is one of Manifold, Differential Geometry, Fibre Bundle.

In common usage a parametrized curve in a manifold $M$ is a smooth map $c:[a,b]\rightarrow M$, or it is the set of point s in $M$ that is the image of this map. The image of a parametrized curve is a gemetric curve.

2.1 Regular Curves

Definition 2.1. A parametrized curve $c:[a,b]\rightarrow M$ is regular if its velocity $c'(t)$ is never zero for all $t$ in the domain $[a,b]$. In other words, a regular curve in $M$ is an immersion: $[a,b]\rightarrow M$.

It $t=t(u)$ is a diffeomorphism of one closed interval with another, then $\beta(u):=c(t(u))$ is a reparametrizeation of the curve $c(t)$. The same geometric curve can have many different parametrizations.

2.2 Arc Length Parametrization

We define the speed of a curve $c:[a,b]\rightarrow M$ in a Riemannian manifold $M$ to be the magnitude $\lVert c'(t)\rVert$ of its velocity $c'(t)$, and the arc length of the curve to be $$l=\int_a^b \lVert c'(u)\rVert du.$$ For each $t\in [a,b]$, let $s(t)$ be the arc length of the curve $c$ restriced to $[a,t]$: $$s(t)=\int_a^t \lVert c'(u)\rVert du.$$ The function $s:[a,b]\rightarrow [0,l]$ is the arc length function of the curve $c$. By the fundamental theorem of calculus, the derivative of $s$ with respect to $t$ is $s'(t)=\lVert c'(t)\rVert$, the speed of $c$.

Proposition 2.3 (Existence of Arc Length). The arc length function $s:[a,b]\rightarrow [0,l]$ of a regular curve $c:[a,b]\rightarrow M$ has a $C^\infty$ inverse.

Proof. Because $c(t)$ is regular, $s'(t)=\lVert c'(t)\rVert$ is never zero. Then $s'(t)>0$ for all $t$. This implies that $s(t)$ is a monotonically increasing function, and so has an inverse $t(s)$. By the inverse function theorem, $t$ is a $C^\infty$ function of $s$.

Thus, given a regular curve $c(t)$, we can write $t$ as a $C^\infty$ function of the arc length $s$ to get the arc length parametrization $\gamma(s)=c(t(s))$.

Proposition 2.4 (Condition of Arc Length). A curve is parametrized by arc length if and only if is it has unit speed and its parameter starts at $0$.

Proof. As noted above, the spped of a curve $c:[a,b]\rightarrow M$ can be computed as the rate of change of the arc length $s$ with respect to $t\in [a,b]$: $$\lVert c'(t)\rVert =\frac{ds}{dt}.$$ Let $\gamma(s)$ be the arc length reparametrization of $c$. Since $s(a)=0$, the parameter $s$ starts at $0$. By the chain rule, the velocity of $\gamma$ is $\gamma'(s)=c'(t(s))t'(s).$$Hence, the speed of$\gamma$is$$\lVert \gamma'(s)\rVert=\lVert c'(t(s))\rVert \left\vert t'(s)\right\vert=\frac{ds}{dt}\left\vert \frac{dt}{ds}\right\vert =\left\vert \frac{ds}{dt}\frac{dt}{ds}\right\vert=1.$$

Conversely, if a curve $c(t)$ has unit speed, then its arc length is $$s(t)=\int_a^t \lVert c'(u)\rVert du=\int_a^t 1du=t-a.$$ If $a=0$, then $s=t$. So a unit-speed curve starting curve starting at $t=0$ is parametrized by arc length.

2.3 Signed Curvature of a Plane Curve

Let plane curve $\gamma:[0,l]\rightarrow \mathbb{R}^2$ will be parametrized by the arc length $s$. Then the velocity vector $T(s)=\gamma'(s)$ has unit length and is tangent to the curve at the point $p=\gamma(s)$. A resonable measure of curvature at $p$ is the magnitude of the derivative $$T'(s)=\frac{dT}{ds}(s)=\gamma''(s),$$ since the faster $T$ changes, the more the curve bends. However, in order to distinguish the directions in which the curve can bend, we will define a curvature with a sign.

There are two unit vectors in the plane perpendicular to $T(s)$ at $p$. We can choose either one to be $\mathbf{n}(s)$, but usually $\mathbf{n}(s)$ is chosen so that the pair $(T(s),\mathbf{n}(s))$ is oriented positively in the plane, i.e., counterclockwise.

Denote by $\left\langle\ ,\ \right\rangle$ the Euclidean inner product on $\mathbb{R}^2$. Since $T$ has unit length, $$\left\langle T,T\right\rangle=1.$$ Using the product rule to differentiate this equation with respect to $s$ gives $$\left\langle T',T\right\rangle+\left\langle T,T'\right\rangle=0,$$ or $$2\left\langle T',T\right\rangle = 0.$$ Thus, $T'$ is perpendicular to $T$ and so it must be a multiple of $\mathbf{n}$. The scalar $\kappa$ such that $$T'=\kappa \mathbf{n}$$ is called the signed curvarue, or simply the curvature, of the plane curve at $p=\gamma(s)$. We can also write $$\kappa=\left\langle T',\mathbf{n}\right\rangle = \left\langle \gamma'',\mathbf{n}\right\rangle.$$ The sign of the curvature depends on the choice of $\mathbf{n}$; it indicates whether the curve is bending towards $\mathbf{n}$ or away from $\mathbf{n}$ (Figure 2.2).

2.4 Orientation and Curvature

A curve whose endpoints are fixed has two possible arc length parametrizations, depending on how the curve is oriented. If the arc length of the curve is $l$, then the two parametrizations $\gamma(s)$, $\tilde{\gamma}(s)$ are related by $$\tilde{\gamma}(s)=\gamma(l-s).$$ Differentiating with respect to the arc length $s$ gives $$\tilde{T}(s):=\tilde{\gamma}(s)=-\gamma'(l-s)=-T(l-s)\quad \mbox{and}\quad \tilde{\gamma}''(s)=\gamma''(l-s).$$

Rotating the tangent vector $\tilde{T}(s)$ by $90^\circ$ amounts to multiplying $\tilde{T}(s)$ on the left by rotation matrix $\mbox{rot}(\pi/2)$. Thus, $$\tilde{\mathbf{n}}(s)=\mbox{rot}\left( \frac{\pi}{2}\right) \tilde{T}(s)=-\mbox{rot}\left(\frac{\pi}{2}\right) T(l-s)=-\mathbf{n}(l-s).$$ It follows that the signed curvature of $\tilde{\gamma}$ at $\tilde{\gamma}(s)=\gamma(l-s)$ is $$\tilde{\kappa}(s)=\left\langle \tilde{\gamma}''(s),\tilde{\mathbf{n}}(s)\right\rangle=\left\langle \gamma''(l-s),-\mathbf{n}(l-s)\right\rangle=-\kappa (l-s).$$ In summary, reversing the orientation of a plan curve reverses the sign of its signeed curvature at any point.