 Solution

The map $\alpha :\mathbb{R}\to \alpha (\mathbb{R})$ is a local arc-length coordinate and, when the image is closed, a covering parametrization. The restriction of $\varphi$ to the one-dimensional submanifold $\alpha (\mathbb{R})$ therefore lifts to a map $h:\mathbb{R}\to \mathbb{R}$ satisfyingBoth curves have unit speed and $\varphi$ is an isometry, so $|h^{\prime }(s)|=1$. Continuity makes the sign constant on the connected domain, hence $h^{\prime }(s)=\pm 1$ throughout andfor some constant ${\sigma }_0$.

A proper Euclidean motion of Euclidean three-space preserves both curvature and torsion. Reversal of an arc-length parameter also preserves them: in the determinant formula for torsion, both the first and third derivatives change sign. Comparing the two expressions for the transformed curve givesand, wherever $\kappa (s)>0$,

If the sign is plus and ${\sigma }_0=0$, then $\varphi$ fixes every point of the curve. A nonidentity proper Euclidean motion with any fixed points is a rotation, whose fixed set is its rotation axis. Thus $\alpha (\mathbb{R})$ lies on that axis and is a straight line.