MA3D9 Example Sheet 5
1. Let v and w be tangent vector fields along a curve γ : I → S. Prove that
d
< v(t), w(t) >=< ∇γ v, w > + < v, ∇γ w > .
dt
Solution:
d
dt
< v(t), w(t) > =< dv(t)
, w(t) > + < v(t), dw(t)
>
dt
dt
dv
=< ∇γ v + ( dt · N)N, w > + < v, ∇γ w + ( dw
· N)N >
dt
=< ∇γ v, w > + < v, ∇γ w >
since N is orthogonal to the tangent plane Tγ S.
2. Let γ(s) be a curve parametrized by arclength s, with nonzero curvature.
parametrized surface
Consider the
σ(s, v) = γ(s) + vb(s), s ∈ I, − < v < , > 0,
where b is the binormal vector of γ. Prove that if is small, σ(I × (−, )) = S is a regular
surface over which γ(I) = σ(I × 0) is a geodesic.
Solution:
At σ(I × 0), σs = t(s) − vτ n(s), σv = b(s). So σv × σs = n(s) + vτ t(s) 6= 0. Since it is
continuous, when is small, σv × σs is still nonzero at σ(I × (−, )) = S. So S is a regular
surface.
When v = 0, the normal vector of surface N = n. So γ 00 = κn, which is perpendicular to the
tangent plane. So γ(I) = σ(I × 0) is a geodesic.
3. Let T be a torus of revolution which we shall assume to be parametrized by
σ(u, v) = ((r cos u + a) cos v, (r cos u + a) sin v, r sin u).
Prove that
a. If a geodesic is tangent to the parallel u = π2 , then it is entirely contained in the region of
T given by − π2 ≤ u ≤ π2 .
b. A geodesic that intersects the parallel u = 0 under an angle θ (0 < θ < π2 ) also intersects
a−r
the parallel u = π if cos θ < a+r
.
Hint: use Clairaut’s relation.
Solution:
a.) Clairaut’s relation says that for a geodesic, let R be the radius of the parallel of intersections
and θ be the angle with the parallel, R cos θ is fixed, which is a in this case (notice θ = 0). So
for R cos θ = a to hold in general, R ≥ a. So cos u ≥ 0, which is − π2 ≤ u ≤ π2 .
b.) In this case R cos θ = (a + r) cos θ. To intersect u = π, a − r ≥ (a − r) cos θπ = (a + r) cos θ.
But if the equality holds, the geodesic must tangent to u = π2 . Notice u = π2 is also a geodesic
since it is a critical point of f (u) = r cos u + a. It contradicts to the fact that there is unique
geodesic passing through a given point and tangent to a given direction. Hence cos θ < a−r
.
a+r
4. Surface of Liouville are those surfaces for which it is possible to obtain a system of local
coordinates σ(u, v) such that the coefficients of the first fundamental form are given as
E = G = U (u) + V (v), F = 0.
Prove that
a. The geodesic of a surface of Liouville may be obtained by
Z
Z
du
dv
√
=± √
+ c1 ,
U −c
V +c
where c and c1 are constants that depend on the initial conditions.
b. If θ, 0 ≤ θ ≤ π2 , is the angle which a geodesic makes with the curve v = const, then
U sin2 θ − V cos2 θ = const.
(this is the analogue of Clairaut’s relation.)
Solution:
a.) We only need to check that the curves given are geodesics since c1 and c gives the initial
conditions for point and tangent direction respectively. To check they are geodesics, we play
the following trick: introduce new coordinates
√
√
U − c du ± V + c dv
du
dv
dv 0 = √
∓√
U −c
V +c
du0 =
The first fundamental form becomes du02 + (U − c)(V + c)dv 02 . In this fundamental form,
v 0 = const (the so called u0 curves) are geodesics by geodesic equations. These are exactly the
curves in our statement.
b.) Suppose our geodesic is unit-speed, so (U + V )(u02 + v 02 ) = 1. So
v 02 =
1
V +c
=
du 2
(U + V )2
(U + V )(1 + ( dv ) )
2
u02 =
U −c
1
=
dv 2
(U + V )2
(U + V )(1 + ( du ) )
We calculate that cos2 θ = (U + V )v 02 and sin2 θ = (U + V )u02 . So
U sin2 θ − V cos2 θ =
U (V + c) V (U − c)
−
= c.
U +V
U +V
3