Fenchel's theorem

The Fenchel theorem is enhanced by the Fáry–Milnor theorem, which says that if a closed smooth simple space curve is nontrivially knotted, then the total absolute curvature is greater than 4π.

Proof

Given a closed smooth curve ${\displaystyle \alpha :[0,L]\to {\mathbb{R}}^3}$ with unit speed, the velocity ${\displaystyle \gamma =\dot{\alpha }:[0,L]\to {\mathbb{S}}^2}$ is also a closed smooth curve (called tangent indicatrix). The total absolute curvature is its length ${\displaystyle l(\gamma )}$.

The curve ${\displaystyle \gamma }$ does not lie in an open hemisphere. If so, then there is ${\displaystyle v\in {\mathbb{S}}^2}$ such that ${\displaystyle \gamma \cdot v>0}$, so ${\displaystyle 0=(\alpha (1)-\alpha (0))\cdot v={\int }_0^L\gamma (t)\cdot v\mathrm{d}t>0}$, a contradiction. This also shows that if ${\displaystyle \gamma }$ lies in a closed hemisphere, then ${\displaystyle \gamma \cdot v\equiv 0}$, so ${\displaystyle \alpha }$ is a plane curve.

Consider a point ${\displaystyle \gamma (T)}$ such that curves ${\displaystyle \gamma ([0,T])}$ and ${\displaystyle \gamma ([T,L])}$ have the same length. By rotating the sphere, we may assume ${\displaystyle \gamma (0)}$ and ${\displaystyle \gamma (T)}$ are symmetric about the axis through the poles. By the previous paragraph, at least one of the two curves ${\displaystyle \gamma ([0,T])}$ and ${\displaystyle \gamma ([T,L])}$ intersects with the equator at some point ${\displaystyle p}$. We denote this curve by ${\displaystyle {\gamma }_0}$. Then ${\displaystyle l(\gamma )=2l({\gamma }_0)}$.

We reflect ${\displaystyle {\gamma }_0}$ across the plane through ${\displaystyle \gamma (0)}$, ${\displaystyle \gamma (T)}$, and the north pole, forming a closed curve ${\displaystyle {\gamma }_1}$ containing antipodal points ${\displaystyle \pm p}$, with length ${\displaystyle l({\gamma }_1)=2l({\gamma }_0)}$. A curve connecting ${\displaystyle \pm p}$ has length at least ${\displaystyle \pi }$, which is the length of the great semicircle between ${\displaystyle \pm p}$. So ${\displaystyle l({\gamma }_1)\ge 2\pi }$, and if equality holds then ${\displaystyle {\gamma }_0}$ does not cross the equator.

Therefore, ${\displaystyle l(\gamma )=2l({\gamma }_0)=l({\gamma }_1)\ge 2\pi }$, and if equality holds then ${\displaystyle \gamma }$ lies in a closed hemisphere, and thus ${\displaystyle \alpha }$ is a plane curve.