The Lumenary · First-year graduate

Riemannian geometry

Name: ______________________

Date: ______________________

You will learn to

Define Riemannian metrics, the Levi-Civita connection, parallel transport and geodesics, prove that the Levi-Civita connection exists and is unique, and compute Christoffel symbols and geodesics on spheres, hyperbolic space and surfaces of revolution
Define the curvature tensor and sectional, Ricci and scalar curvature, prove the symmetries of the curvature tensor, and interpret curvature through Jacobi fields and the spreading of geodesics
Prove the global Gauss–Bonnet theorem for closed oriented surfaces
State the Hopf–Rinow theorem and prove the Bonnet–Myers theorem from the second variation formula

1.
The metric $g = dt^2 + t^4\, d\theta^2$ (for $t > 0$ ) has the form $dt^2 + f(t)^2 d\theta^2$ . What is its Gaussian curvature at $t = 1$ ?
2.
Let $M$ be a closed orientable surface of genus $3$ with any Riemannian metric. Compute $\dfrac{1}{\pi}\displaystyle\int_M K\, dA$ .
3.
Which of these curves on the unit sphere $S^2$ , parametrised with constant speed, are geodesics? (Choose all that apply.)
- A.The equator
- B.A meridian from pole to pole
- C.The circle of latitude $45^\circ$ N
- D.The great circle through $(1, 0, 0)$ and $(0, 1, 1)/\sqrt2$
4.
The paraboloid $z = x^2 + y^2$ , with the metric induced from $\R^3$ , is complete and has Gaussian curvature $K > 0$ at every point, yet it is not compact. Which hypothesis of the Bonnet–Myers theorem fails?
- A.There is no constant $k > 0$ with $K \ge k$ everywhere
- B.Completeness
- C.The theorem needs dimension at least $3$
- D.On surfaces Ricci curvature is not defined
5.
In the proof of the fundamental theorem, the Koszul formula expresses $2\langle\nabla_X Y, Z\rangle$ in terms of the metric and brackets. Which property is needed to conclude that $\nabla_X Y$ itself is determined?
- A.The metric is non-degenerate: a vector $v$ with $\langle v, Z\rangle = 0$ for all $Z$ is $0$
- B.The metric is positive definite
- C.The manifold is complete
- D.The curvature tensor is tensorial
6.
On the unit sphere, let $\gamma$ be the unit-speed great-circle arc of length $3\pi/2$ from $p$ to $q$ . Which statement is true?
- A. $\gamma$ is a geodesic but not minimising: it passes the point conjugate to $p$ at length $\pi$
- B. $\gamma$ is not a geodesic, because geodesics minimise length
- C. $\gamma$ is the shortest curve from $p$ to $q$
- D. $\gamma$ minimises length among nearby curves, though not globally
7.
A geodesic triangle in the hyperbolic plane (curvature $-1$ ) has angles $\pi/3$ , $\pi/4$ and $\pi/6$ . What is its area?
- A. $\pi/4$
- B. $-\pi/4$
- C. $3\pi/4$
- D.It depends on the side lengths, which are not given
8.
In semicircle_geodesic, the curve is $x = \sinh t/\cosh t$ , $y = 1/\cosh t$ , with $x' = 1/\cosh^2 t$ and $y' = -\sinh t/\cosh^2 t$ . Which conjunct states that it has unit speed for the hyperbolic metric $(dx^2 + dy^2)/y^2$ ?
- A.((1 / cosh t ^ 2) ^ 2 + (-sinh t / cosh t ^ 2) ^ 2) / (1 / cosh t) ^ 2 = 1
- B.(sinh t / cosh t) ^ 2 + (1 / cosh t) ^ 2 = 1
- C.(1 / cosh t ^ 2) ^ 2 + (-sinh t / cosh t ^ 2) ^ 2 = 1
- D.-2 * sinh t / cosh t ^ 3 - 2 / (1 / cosh t) * (1 / cosh t ^ 2) * (-sinh t / cosh t ^ 2) = 0
9.
Write excess(a, b, c) for three unit vectors $a, b, c \in \R^3$ (tuples) spanning a geodesic triangle on the unit sphere. Return $\alpha + \beta + \gamma - \pi$ , where the angle at $a$ is the angle between the tangent vectors $b - \langle a, b\rangle a$ and $c - \langle a, c\rangle a$ (the initial directions of the great-circle arcs from $a$ ), and similarly at $b$ and $c$ . By local Gauss–Bonnet the excess is the area.
Computational — Practice

Answer key — Riemannian geometry

1.
Here $f(t) = t^2$ , so $f'' = 2$ and $K = -f''/f = -2/t^2$ , which is $-2$ at $t = 1$ . (Derivation: with $e_1 = \partial_t$ , $e_2 = f^{-1}\partial_\theta$ , the connection form is $\omega = f'\,d\theta$ and $d\omega = f''\,dt\wedge d\theta = -K f\, dt\wedge d\theta$ .)
2.
$\chi(M) = 2 - 2 \cdot 3 = -4$ , so $\int_M K\, dA = 2\pi \cdot (-4) = -8\pi$ and the answer is $-8$ . The metric can be changed freely; the total curvature cannot.
3.
Geodesics of the unit sphere are the constant-speed great circles: the equator, the meridians and every tilted great circle. A parallel at latitude $\lambda \ne 0$ lies in a plane not through the centre; its acceleration has a component tangent to the sphere (towards the nearer pole), so $D_t\dot\gamma \ne 0$ . Clairaut's picture says the same: $f(t) = \sin t$ has $f' = 0$ only on the equator.
4.
For $z = x^2 + y^2$ , $K = 4/(1 + 4(x^2 + y^2))^2$ , which is positive but tends to $0$ . With no positive lower bound $k$ , the integral $\frac L2(\pi^2/L^2 - k)$ cannot be made negative for any fixed $L$ , and long geodesics can be minimising — as the meridians of the paraboloid are.
5.
The Koszul formula determines the covector $Z \mapsto \langle\nabla_X Y, Z\rangle$ . Because $g$ is non-degenerate, $v \mapsto \langle v, \cdot\rangle$ is injective, so the vector $\nabla_X Y$ is determined. Only non-degeneracy is used, which is why a pseudo-Riemannian (e.g. Lorentzian) metric also has a unique Levi-Civita connection.
6.
Every great-circle arc is a geodesic. The complementary arc has length $\pi/2 < 3\pi/2$ , so $\gamma$ is not minimising. Moreover $\gamma$ passes the antipode of $p$ , conjugate to $p$ at $t = \pi$ , and by Jacobi's theorem a geodesic containing an interior conjugate point is not even locally minimising: rotating the part beyond the antipode slightly off the great circle shortens it.
7.
With $K \equiv -1$ , $\int_T K\, dA = -\operatorname{area}(T) = \pi/3 + \pi/4 + \pi/6 - \pi = -\pi/4$ , so the area is $\pi/4$ . Every hyperbolic triangle has area less than $\pi$ ; the CurvedTriangles widget lets you check this numerically.
8.
The squared hyperbolic speed is $g(\dot\gamma, \dot\gamma) = (\dot x^2 + \dot y^2)/y^2$ , which is conjunct (a): $(1/\cosh^4 t + \sinh^2 t/\cosh^4 t)\cosh^2 t = (1 + \sinh^2 t)/\cosh^2 t = 1$ . Conjunct (b) places the curve on the unit circle, (d) is a geodesic equation, and (c) is false.

import math

def dot(u, v):
    return sum(x * y for x, y in zip(u, v))

def angle(u, v):
    c = dot(u, v) / math.sqrt(dot(u, u) * dot(v, v))
    return math.acos(max(-1.0, min(1.0, c)))

def excess(a, b, c):
    def tangent(p, q):
        k = dot(p, q)
        return tuple(qi - k * pi for pi, qi in zip(p, q))
    total = 0.0
    for p, q, r in ((a, b, c), (b, c, a), (c, a, b)):
        total += angle(tangent(p, q), tangent(p, r))
    return total - math.pi

The octant triangle has three right angles, excess $\pi/2$ — one eighth of the sphere's area $4\pi$ . For a tiny triangle the excess is almost exactly its area $h^2/2$ , because the sphere is nearly flat at small scales. These are checks of the formula area $=$ excess on examples; the formula itself is the local Gauss–Bonnet theorem.