Worksheet 2 Differential Geometry II, SS 2008

Worksheet 2
Differential Geometry II, SS 2008
K. Polthier and M. Wardetzky
Due May 6
This worksheet deals with the law of cosines in the Euclidean plane, E 2 , the unit 2-sphere,
S 2 , and the hyperbolic plane, H 2 . This worksheet is optional since there is no class on May
1 st . You can work on it to gain extra points.
Exercise 1 (6 points)
Let (M, g) be a complete 2-dimensional Riemannian manifold. Consider a point C ∈ M and
a disk D(0, ɛ) around the origin in TCM where the exponential map exp C is injective. In the
following, we work in U = exp C(D(0, ɛ)). Assume that the Riemannian metric is rotationally
symmetric around C, i.e., in polar coordinates (r, θ) it takes the form
g = dr 2 + ϕ 2 (r)dθ 2 .
Consider a triangle T = ABC in U whose edges are geodesics of length a, b, c, respectively.
Consider moving the point A(t) along the geodesic from A to B at unit speed such that
A(0) = A and A(c) = B. Let T (t) be the triangle with vertices A(t), B, C, geodesic edges of
lengths a, b(t), c(t), and interior angles α(t), β, γ(t), respectively. Show that
γ ′ sin α(t)
(t) = −
ϕ(b(t)) ,
where ′ denotes the derivative with respect to t. Use this fact to show that
′
κ dvol =
T (t)
sin α(t)
ϕ(b(t)) · (ϕr(b(t)) − 1) ,
where dvol is the volume form on M, ϕr denotes the partial derivative of ϕ with respect to
r, and κ is the Gauss curvature. Hint: express the integral in polar coordinates and use the
Jacobi equation for ϕ.
Exercise 2 (2 points)
With the above notations, show that
α ′ (t) =
sin α(t)
ϕ(b(t)) · ϕr(b(t)) .
Hint: use Exercise 1 and the Gauss-Bonnet theorem for geodesic triangles.

Exercise 3 (3 points)
With the above notations, let Φ(r) = r
0 ϕ(s) ds, and let f(t) = Φ(b(t)). Show that
f ′ (t) = −ϕ(b(t)) · cos α(t) , and
Hint: use Exercise 2.
f ′′ (t) = ϕr(b(t)) .
Exercise 4 (6 points)
Apply the above to M = E 2 , M = S 2 , and M = H 2 , to show that
f ′′ (t) = 1 , f ′′ (t) + f(t) = 1 , and f ′′ (t) − f(t) = 1 ,
respectively. Solve these ODEs using the initial conditions f(0) and f ′ (0). Plug t = c into the
solutions to show the law of cosines:
M = E 2 : a 2 = b 2 + c 2 − 2bc cos α ,
M = S 2 : cos a = cos b cos c + sin b sin c cos α ,
M = H 2 : cosh a = cosh b cosh c − sinh b sinh c cos α .