The Lumenary · Advanced undergraduate

Differential geometry of curves and surfaces

Name: ______________________

Date: ______________________

You will learn to

Compute arc length, curvature and torsion of space curves using the Frenet–Serret frame
Compute the first and second fundamental forms and the principal, Gaussian and mean curvatures of standard surfaces
State and explain Gauss's Theorema Egregium, and deduce that no map of a region of the sphere to the plane preserves distances
State the Gauss–Bonnet theorem, prove its spherical case (Girard's theorem), and check it on spheres and tori
Distinguish the curvature of a curve from the Gaussian curvature of a surface, and intrinsic from extrinsic quantities

1.
A running track has semicircular ends of radius $36.5$ m. Along the straight parts the curvature is $0$ . What is the curvature on the bends?
- A. $1/36.5 \approx 0.027$ per metre
- B. $36.5$ per metre
- C. $1/36.5^2$ per square metre
- D. $\pi$ , because the direction turns through half a circle
2.
Which of these surfaces have Gaussian curvature $K = 0$ at every point? (Choose all that apply.)
- A.A plane
- B.A circular cylinder
- C.A cone (without its apex)
- D.A sphere
- E.The saddle $z = xy$
3.
For the cylinder of radius $2$ , $X(\theta, z) = (2\cos\theta, 2\sin\theta, z)$ , one finds $E = 4$ , $F = 0$ , $G = 1$ , $L = -2$ , $M = 0$ , $N = 0$ . What are $K$ and $H$ ?
- A. $K = 0$ and $H = -\tfrac14$
- B. $K = \tfrac14$ and $H = -\tfrac12$
- C. $K = 0$ and $H = 0$
- D. $K = -\tfrac12$ and $H = -\tfrac14$
4.
The helix $\gamma(s) = (3\cos(s/5), 3\sin(s/5), 4s/5)$ has unit speed (check: $3^2 + 4^2 = 5^2$ ), curvature $\kappa = a/c^2$ and torsion $\tau = b/c^2$ with $a = 3$ , $b = 4$ , $c = 5$ . Compute $\dfrac{1}{\kappa^2 + \tau^2}$ .
5.
A triangle on a sphere of radius $R$ is bounded by three great-circle arcs and has angles $100^\circ$ , $80^\circ$ and $90^\circ$ . By Girard's theorem its area is $R^2$ times its angle excess (in radians). What is the angle excess, in degrees?
6.
A closed orientable surface of genus $3$ (a pretzel with three holes) sits in $\mathbb{R}^3$ . Its total curvature $\iint K\, dA$ equals $m\pi$ for which integer $m$ ?
7.
Why can a flat sheet of paper be rolled into a cylinder without stretching or tearing, but not wrapped smoothly around a ball?
- A.Bending without stretching preserves $E, F, G$ , hence (by the Theorema Egregium) $K$ ; the plane and the cylinder have $K = 0$ , the sphere $K > 0$ .
- B.Because the plane and the cylinder have the same mean curvature.
- C.Because the sphere is compact and the plane is not.
- D.Because the sphere has Euler characteristic $2$ .
8.
The Lean theorem sphere_gauss_curvature has the hypothesis hφ : cos φ ≠ 0. Why is it needed?
- A.At the poles ( $\cos\varphi = 0$ ) the latitude–longitude chart is not regular: $EG - F^2 = R^4\cos^2\varphi = 0$ , and the formula divides by it.
- B.Because the sphere is flat at the poles.
- C.Because Lean cannot divide by zero, so the statement would not typecheck.
- D.Because $R$ might be negative.
9.
Write signed_curvature(dx, dy, ddx, ddy) returning the signed curvature $\kappa = (x'y'' - y'x'')/(x'^2 + y'^2)^{3/2}$ of a plane curve at a point, given the values of $x', y', x'', y''$ there. Then use it to find the curvature of the ellipse $(3\cos t, \sin t)$ at $t = 0$ and $t = \pi/2$ .
Computational — Practice
10.
Why must every torus in $\mathbb{R}^3$ , however it is shaped, have points of negative Gaussian curvature?
- A.Gauss–Bonnet gives $\iint K\, dA = 2\pi\chi = 0$ , and a compact surface in $\mathbb{R}^3$ has a point with $K > 0$ (the point farthest from the origin), so $K \lt 0$ somewhere else.
- B.Because the hole in the middle is a saddle.
- C.Because the torus has negative Euler characteristic.
- D.It need not: the flat torus has $K = 0$ everywhere.

Answer key — Differential geometry of curves and surfaces

1.
On a circle of radius $r$ the unit tangent turns through $\pi$ over the arc length $\pi r$ of a semicircle, so $\kappa = \pi/(\pi r) = 1/r = 1/36.5 \approx 0.027\ \text{m}^{-1}$ . Equivalently, for the unit-speed circle $\gamma(s) = (r\cos(s/r), r\sin(s/r))$ , $|\gamma''| = 1/r$ (proved in Lean as circle_curvature).
2.
The plane, the cylinder and the cone have $K = 0$ : each has a direction in which it does not bend at all, and each can be made from flat paper. By the Theorema Egregium, surfaces made from paper without stretching must have $K = 0$ . The sphere has $K > 0$ and the saddle $K \lt 0$ .
3.
$K = \dfrac{(-2)(0) - 0}{4 \cdot 1 - 0} = 0$ and $H = \dfrac{4 \cdot 0 - 0 + 1 \cdot (-2)}{2 \cdot 4} = -\dfrac14$ . The principal curvatures are $\mp\tfrac12$ (around) and $0$ (along). $K$ is intrinsic and vanishes; $H$ is not intrinsic and is what distinguishes the cylinder from a plane.
4.
$\kappa = 3/25$ , $\tau = 4/25$ , so $\kappa^2 + \tau^2 = (9 + 16)/625 = 1/25$ and the answer is $25 = c^2$ . In general $\kappa^2 + \tau^2 = (a^2 + b^2)/c^4 = 1/c^2$ for every helix, a pleasant check of the Frenet computation (formalised in Lean as helix_frenet).
5.
The excess is $100 + 80 + 90 - 180 = 90^\circ = \pi/2$ , so the area is $\pi R^2/2$ — one eighth of the sphere's area $4\pi R^2$ . Girard's theorem is the constant-curvature case of the local Gauss–Bonnet theorem: angle excess $= \iint K\, dA$ .
6.
$\chi = 2 - 2 \cdot 3 = -4$ , so $\iint K\, dA = 2\pi \cdot (-4) = -8\pi$ , and $m = -8$ . However the pretzel is shaped, its curvature always adds up to $-8\pi$ ; in particular it must have regions of negative curvature (and, being compact in $\mathbb{R}^3$ , also some of positive curvature).
7.
Bending without stretching is a local isometry: it preserves lengths of curves, hence $E, F, G$ . By the Theorema Egregium it preserves $K$ . Paper has $K = 0$ , the cylinder has $K = 0$ , but every region of a sphere has $K = 1/R^2 \ne 0$ . Mean curvature changes freely under bending, so it is no obstruction.
8.
With $E = R^2\cos^2\varphi$ , $F = 0$ , $G = R^2$ , the denominator $EG - F^2 = R^4\cos^2\varphi$ vanishes at the poles, where $X_\theta = 0$ and the chart is singular. In Lean, division by zero returns $0$ , so without the hypothesis the claimed equality $K = 1/R^2$ would be false at the poles. Honest formalisation keeps track of exactly where coordinates are valid.
9.
```
def signed_curvature(dx, dy, ddx, ddy):
    return (dx * ddy - dy * ddx) / (dx ** 2 + dy ** 2) ** 1.5
```
For the ellipse $(3\cos t, \sin t)$ : at $t = 0$ (the end of the long axis) $\kappa = 3$ , a sharp bend with osculating radius $1/3$ ; at $t = \pi/2$ (the end of the short axis) $\kappa = 1/9$ , a gentle bend with radius $9$ . In general the ellipse $(a\cos t, b\sin t)$ has curvature $a/b^2$ and $b/a^2$ at these points.
10.
By Gauss–Bonnet the total curvature is $2\pi\chi(T^2) = 0$ . At the point $p$ farthest from the origin, the surface lies inside the sphere of radius $|p|$ and touches it, so $K(p) \ge 1/|p|^2 > 0$ . Since the integral is $0$ , $K$ must be negative somewhere. (The flat torus, an abstract surface with $K \equiv 0$ , therefore has no smooth realisation in $\mathbb{R}^3$ preserving its metric — although, by the Nash–Kuiper theorem, it has $C^1$ ones.)