Past exam of the mathematics course of the University of Cambridge / 2023 / ii / Paper 2 / 26G (Differential Geometry)

Let $\gamma$ be a $C^3$ regular curve parametrized by arc length, so $|{\gamma }^{\prime }(s)|=1$. Define the Frenet frameand define the torsion byThus curvature requires two derivatives, while $N,B,$ and torsion require $\kappa >0$; the displayed classical torsion requires three derivatives. At a point where $\kappa =0$, $N,B,$ and $\tau$ are not defined by this construction.

Since $(T,N,B)$ is an orthonormal frame, differentiating its inner products shows that its derivative matrix is skew-symmetric. The definition gives $T^{\prime }=\kappa N$. Next,so $N^{\prime }=-\kappa T+cB$. As $B^{\prime }=T\times N^{\prime }$ and $T\times B=-N$, we have $B^{\prime }=-cN$; hence $c=-B^{\prime }\cdot N=\tau$. The Frenet-Serret formulas are therefore

If $\kappa \equiv 0$, then $T^{\prime }=0$, so the unit tangent is constant andThus both curves are straight line segments. Equal arc length makes their parameter intervals equally long, and a rotation in $SO(3)$ followed by a translation maps one to the other. Hence they are related by a proper Euclidean motion of Euclidean three-space.

The conclusion fails for $\kappa \equiv 1$, because curvature alone does not determine a space curve. A unit circle has curvature one and torsion zero. The circular helix with $a=b=1/2$,is unit speed and has curvature one but torsion one. Proper Euclidean motions preserve torsion, so equal-length pieces of these two curves cannot be related by one.

Write a proper Euclidean motion of Euclidean three-space as $E(x)=Ax+b$, where $A\in SO(3)$. ThenBecause $A$ preserves norms, dot products, and cross products, the formulasshow directly that curvature and torsion are unchanged by $E$.

For the shifted curve ${\gamma }_{s_0}(s)=\gamma (s-s_0)$, every derivative at $s$ equals the corresponding derivative of $\gamma$ at $s-s_0$. ConsequentlyBoth are therefore pointwise Euclidean invariants in the stated sense.

No. For sufficiently smooth curves,is unchanged by proper Euclidean motions and transforms correctly under translations of the arc-length parameter, so it is a pointwise Euclidean invariant of a curve. It is not determined by the two numbers $\kappa (s)$ and $\tau (s)$.

For a concrete comparison near $s=0$, use the plane curve reconstructed from curvature construction withThe resulting unit-speed planar curves both have $\kappa (0)=1$ and $\tau (0)=0$, but their curvature derivatives at zero are respectively $0$ and $1$. Thus no single function $f$ can satisfy $Q(\gamma ,s)=f(\kappa (s),\tau (s))$ for all curves.

Indeed, the covariance conditions alone even permit nonlocal examples such as $Q(\gamma ,s)={\kappa }_{\gamma }(s+1)$. This is the distinction recorded by pointwise Euclidean invariants need not depend only on current curvature and torsion.