Past exam of the mathematics course of the University of Cambridge / 2018 / ii / Paper 2 / 25I (Differential Geometry)

Fix $t_0\in [a,b]$ and defineBecause $\gamma$ is a regular curve, $s^{\prime }(t)=\Vert {\gamma }^{\prime }(t)\Vert >0$. The inverse function theorem gives a smooth inverse $t=t(s)$ onto the corresponding interval. For $\tilde{\gamma }(s)=\gamma (t(s))$,Thus the arc-length parametrization exists.

For a unit-speed curve with nonzero curvature of a space curve $\kappa$, define the Frenet frameThe torsion of a space curve iswith this choice of convention.

For the requested example, take the unit-speed circleand the unit-speed circular helixBoth have the constant nonzero curvature $\kappa =1$, but their torsions are respectively $0$ and $1$. A proper Euclidean motion of Euclidean three-space preserves torsion, so no such rigid motion relates the two curves.

Choose unequal positive numbers $a,b$. The planar ellipseis a simple closed curve and is not a circle. The nonidentity proper rigid motionis rotation through $\pi$ about the $z$-axis, and $R\gamma (t)=\gamma (t+\pi )$. It therefore preserves the curve as a set.

Let $M_t(x)=R(t)x+T(t)$ be the given family. Composing throughout with $M_0^{-1}$ lets us assume $M_0$ is the identity; the family remains regular and preserves $\gamma$. Differentiating at $t=0$ produces a nonzero infinitesimal proper Euclidean motionFor every $x\in \gamma$, the path $t\mapsto M_t(x)$ lies in $\gamma$, so $V(x)$ is tangent to $\gamma$. The flow of this complete vector field consequently preserves $\gamma$ for all times.

The orbit of a one-parameter proper Euclidean motion is a line if $a=0$, and otherwise a helix about an axis parallel to $a$. Since $\gamma$ is compact, it cannot contain an entire line or a helix with nonzero axial pitch. Thus the flow is a pure rotation about an axis. The curve cannot lie entirely in the fixed axis, because no subset of a line is a simple closed curve. Choose a point off the axis. Its orbit is a circle contained in $\gamma$; as a one-dimensional orbit it is open in the connected one-dimensional manifold $\gamma$, and it is closed by compactness. It must therefore equal the whole curve. Hence the continuous rigid-motion symmetry of a simple closed space curve gives