Past exam of the mathematics course of the University of Cambridge / 2019 / ii / Paper 2 / 25H (Differential Geometry)

Put $T={\alpha }^{\prime }(s)$. Since $s$ is arc length, $|T|=1$, so $T^{\prime }\perp T$. The curvature of a space curve isWhere $k(s)\ne 0$, define the principal normal and binormal byand define the torsion of a space curve byThus torsion requires nonzero curvature, so that $N$ and $B$ are defined.

Differentiating the orthonormality relations for the frame $(T,N,B)$ shows that its derivative matrix is skew-symmetric. Since $T^{\prime }=kN$, differentiating $B=T\times N$ and using the definition of $\tau$ gives the Frenet-Serret formulas

If $\tau =0$ everywhere, the third of the Frenet-Serret formulas gives $B^{\prime }=0$, so $B$ is a constant vector. MoreoverThus $\alpha (s)\cdot B$ is constant, and the image of $\alpha$ lies in the plane perpendicular to $B$.

The fundamental theorem of regular space curves states that smooth functions $k>0$ and $\tau$ on an interval determine a smooth arc-length-parametrized curve in ${\mathbb{R}}^3$, unique up to a proper Euclidean motion. Consequently two such curves $\alpha$ and $\tilde{\alpha }$ are related parameterwise by a proper Euclidean motion if and only iffor every $s$.

False. Vanishing torsion only controls each connected region on which the curvature is nonzero. A smooth curve can pass through a flat point or straight segment and emerge in a different plane.

For an explicit local construction on $(-1,1)$, put $\eta (s)=e^{-1/s^2}$ for $s\ne 0$ and $\eta (0)=0$, and define a unit tangent fieldThis field is smooth because $\eta$ is flat at zero. Its integral $\alpha (s)={\int }_0^{s}T(u)du$ is parametrized by arc length, has nonzero curvature for $s\ne 0$, and has zero torsion on both sides. Its negative half lies nontrivially in the $xy$-plane and its positive half lies nontrivially in the $xz$-plane, so the whole curve is not planar.

False. The curve constructed in part (i) has curvature nonzero, and hence torsion defined and equal to zero, at every $s\ne 0$. Torsion is undefined only at $s=0$, yet the curve is not contained in a plane.

False. Use the same flat-point construction for both curves and make their tangent fields identical for $s<0$. For $s>0$, let one curve bend in the $xz$-plane and let the other bend in the plane spanned by $(1,0,0)$ and $(0,\cos \theta ,\sin \theta )$, where this second plane is neither the $xz$-plane nor its reflection fixing the $xy$-plane. The two curves have the same curvature everywhere and zero torsion away from $s_0=0$.

Any rigid motion relating them parameterwise would be fixed by their common nonstraight arc for $s<0$; its orthogonal part can only fix the $xy$-plane pointwise and possibly reflect its normal. It therefore cannot carry the chosen positive-side plane to the other one. The curves need not be related by a rigid motion.