Fenchel's theorem

In differential geometry, Fenchel's theorem is an inequality on the total absolute curvature of a closed smooth space curve, stating that it is always at least $2\pi$. Equivalently, the average curvature is at least $2\pi /L$, where $L$ is the length of the curve. The only curves of this type whose total absolute curvature equals $2\pi$ and whose average curvature equals $2\pi /L$ are the plane convex curves. The theorem is named after Werner Fenchel, who published it in 1929.

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 Template:Math.

Proof

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

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

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

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

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

References

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