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 (\mod 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: $$ \int_{0}^L \frac{1}{\kappa(s)} ds \equiv 0 (\mod 2\pi) $$


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? $\endgroup$ – Edward Z. Miao May 9 '20 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 $\theta$. $\endgroup$ – achille hui May 9 '20 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! $\endgroup$ – Edward Z. Miao May 11 '20 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. $\endgroup$ – achille hui May 11 '20 at 2:51
  • $\begingroup$ Get it! That solves my puzzles, thanks a lot! $\endgroup$ – Edward Z. Miao May 11 '20 at 3:10

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.