= 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 $\phi$ to the one-dimensional submanifold $\alpha(\mathbb R)$ therefore lifts to a map $h:\mathbb R\to\mathbb R$ satisfying
$$
\phi(\alpha(s))=\alpha(h(s)).
$$
Both curves have unit speed and $\phi$ is an isometry, so $|h'(s)|=1$. Continuity makes the sign constant on the connected domain, hence $h'(s)=\pm1$ throughout and
$$
\boxed{\phi(\alpha(s))=\alpha(\pm s+\sigma_0)}
$$
for 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 gives
$$
\boxed{\kappa(\pm s+\sigma_0)=\kappa(s)},
$$
and, wherever $\kappa(s)>0$,
$$
\boxed{\tau(\pm s+\sigma_0)=\tau(s)}.
$$
If the sign is plus and $\sigma_0=0$, then $\phi$ 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.
Back to article page
