Fenchel's theorem

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 ${\displaystyle \textstyle 0=(\alpha (1)-\alpha (0))\cdot v=\int _{0}^{L}\gamma (t)\cdot v\,\mathrm {d} t>0}$, 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 ${\displaystyle l(\gamma _{1})\geq 2\pi }$, and if equality holds then ${\displaystyle \gamma _{0}}$ does not cross the equator.

Therefore, ${\displaystyle l(\gamma )=2l(\gamma _{0})=l(\gamma _{1})\geq 2\pi }$, and if equality holds then ${\displaystyle \gamma }$ lies in a closed hemisphere, and thus ${\displaystyle \alpha }$ is a plane curve.