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