# Is total curvature of a closed space curve a multiple of $2\pi$?

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

## 1 Answer

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 $$6 \times \frac{\pi}{2} = 3\pi$$.

• 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? – Edward Z. Miao May 9 '20 at 13:07
• @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 $\theta$. – achille hui May 9 '20 at 13:15
• 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! – Edward Z. Miao May 11 '20 at 0:55
• @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. – achille hui May 11 '20 at 2:51
• Get it! That solves my puzzles, thanks a lot! – Edward Z. Miao May 11 '20 at 3:10