For a closed, regular space curve of length $\ell$, show that $\int_0^\ell\kappa(s)ds\geq 2\pi$, where $\kappa$ gives curvature

Question

Question:

How can one show that the following proposition is true (Only outline of the proof is needed)?

For every closed and regular space curve $c:[a,b] \to \mathbb R^3$ of total length $l$ one has the inequality $$\int_0^l \kappa (s) ds \ge 2 \pi,$$ with equality if and only if the curve is a convex, simple plane curve$.^1$

Definitions:

A regular parametrized curve is a continuously differentiable immersion $c: I \to \mathbb R^n$, defined on a real interval $I \subseteq \mathbb R$. This means that $\dot{c}=\frac{dc}{dt}\neq0$ holds everywhere. The length of the curve $c$ ($I=[a,b]$) is $$\int_a^b\left \Vert \frac {dc}{dt} \right \Vert dt \,.$$ The function $\kappa:=\left \Vert c'' \right \Vert$ is called the curvature of $c\,.^2$

A (regular) curve $c:[a,b] \to \mathbb R^n$ is called closed, if there is a (regular) curve $\tilde{c}|_{[a,b]}=c$ and $\tilde{c}(t+b-a)=\tilde{c}(t)$ for all $t \in \mathbb R $, where in particular $c(a)=c(b)$ and $c'(a)=c'(b)$. A closed curve $c$ is said to be simply closed, if $c|_{[a,b]}$ is injective$.^3$

A simply closed plane curve is called convex, if the image set of the boundary is a convex subset $C \subset \mathbb R^2 \, .^4$

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[1], [2], [3], [4] Wolfgang Kühnel, "Differential Geometry Curves-Surfaces-Manifolds", Second Edition, American Mathematical Society, 2006.

Answer 1

Given the parametrization c of your curve, let p be any point on the unit sphere. Consider the function $g(t)=p \cdot c(t)$. Since the interval $[a,b]$ is closed and bounded, and $g(t)$ is continuous, then it attains a maximum and minimum value within the interval. It is possible to compute this value by differentiating $g(t)$. Now \begin{equation} g'(t)=p \cdot c'(t) \end{equation} Now the points $x \in \mathbb{S}^2$ such that $x \cdot p =0 $ belongs to the great circle of the sphere. The great circle belong to a plane perpendicular to the line connecting the point p to the certer of the sphere. Since $c'$ and the unit tangent vector $T$ are collinear, while p is an arbitrary point on the sphere, we conclude that the tangent indicatrix intersects every great circle on $\mathbb{S}^2$. Since the length of the tangent indicatrix is the total curvature, we have now to prove that the length of the tangent indicatrix is larger of equal to $2 \pi$. To do that you can use the Horn lemma.

• Horn lemma Given a regular curve c on the unit sphere, if c has length less than 2π then c is contained in a hemisphere.

Answer 2

If $c$ is unit speed, then $T(t)=c'(t)$ is a spherical curve in $\mathbb{S}^2$, which is a closed curve. If ${\rm image}\ (T)$ is in some open hemisphere $H$ s.t. $\partial H$ is a great circle which is orthogonal to some $U\in \mathbb{S}^2$, then $c$ goes to the direction $U$ or $-U$. In any case, $c$ can not return to a point $c(0)$.

Hence ${\rm Image} (T)$ can not be in any open hemisphere, which means that ${\rm length}\ (T)\geq 2\pi$ (cf. Spherical lemma).