For a regular closed space curve $x:[0,L] \rightarrow \mathbb{R}^3$ parametrized by arc-length $s$, we define total curvature $$ K = \int_0^L \kappa(s) ds $$ In particular, when $x$ is a plane closed curve, we know the total curvature is a multiple of 2$\pi$, namely, $K \equiv 0 (\mod 2\pi) $, to be more precise, $K = \int_0^L \kappa(s) ds = 2\pi i_r(x)$, where $i_r(x)$ stands for rotation index of curve $x$ or winding number of unit tangent vector.

I want to know: For a space regular closed curve, whether total curvature satisfying \begin{equation} K \equiv 0 \pmod{2\pi} \quad \quad \quad \quad(*) \end{equation}
I guess $(*)$ is not true in general, but it is awkward for me to give an example of "untrivial" closed space curve whose total curvature can be calculate explicitly. I have tried to compute closed curve lying on torus: $$ x(t) = \left([2+\cos 2t]\cos t,[2+\cos 2t]\sin t,\sin 2t \right), t\in [0,2\pi]$$ but it is still not easy to calculate $K$.

Can you help me with an easier counterexample or show $(*)$ is true?

More generally, I want to ask whether some variations holds: $$ \int_{0}^L \frac{1}{\kappa(s)} ds \equiv 0 \pmod{2\pi} $$


1 Answer 1


Let $D$ be the surface obtained by intersecting the unit cube $[0,1]^3$ with coordinates planes, i.e.

$$D = \{ (x,y,z) \in [0,1]^3 : xyz = 0 \}$$

Its boundary $\partial D$ consists of $6$ line segments of length $1$ joining at right angles. Smooth each corner by a planar curve, you get a regular curve with total curvature $6 \times \frac{\pi}{2} = 3\pi$.

  • $\begingroup$ Thank you, achille! Your answer is quite elegant and enlightening. I didn't realize I can use piecewise smooth curve and then "smooth the corner". But I am still not sure about why the total curvature remains invariant after smoothing the corner. Could you please elaborate that part? $\endgroup$ Commented May 9, 2020 at 13:07
  • $\begingroup$ @EdwardZ.Miao For planar curve, $\kappa(s) = \frac{d\theta}{ds}$ where $\theta$ is the angle of tangent vector wrt some reference axis. As long as you "rotate" the tangent vector in one direction (i.e keep the sign of $\frac{d\theta}{ds}$ the same), $\int |\kappa(s)| ds$ equals to the change of $\theta$. $\endgroup$ Commented May 9, 2020 at 13:15
  • $\begingroup$ I still don't understand about one thing: think of a regular plane n-polygon $C$ embedding in $\mathbb{R}^3$, if your conclusion is right, the total curvature of $C$ should be $(n-2)\pi$. However, since $C$ is actually a plane closed curve, thus total curvature is just $2\pi$ . Could you please explain this "contradiction"? Thanks! $\endgroup$ Commented May 11, 2020 at 0:55
  • 1
    $\begingroup$ @EdwardZ.Miao the contribution to total curvature from a corner (of a planer curve) is the change of $\theta$ at that corner. You need to sum over the absolute values of external angles, not the internal angles. $\endgroup$ Commented May 11, 2020 at 2:51
  • $\begingroup$ Get it! That solves my puzzles, thanks a lot! $\endgroup$ Commented May 11, 2020 at 3:10

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