Riemannian Geometry - ex 3 - solution sketch
(1) (Geodesics on surfaces of revolution - Do Carmo Chapter 3 ex 1) Denote
by (u, v) the cartesian coordinates of R2 . Show that the function φ :
U ⊂ R2 −→ R3 given by φ(u, v) = (f (v) cos u, f (v) sin u, g(v)), where
U = {(u, v) ∈ R2 | u0 < u < u1 ; v0 < v < v1 }, and f, g are differentiable
functions with f 0 (v)2 + g 0 (v)2 6= 0 and f (v) 6= 0, is an immersion. The
image φ(U ) is the surface generated by the rotation of the curve (f (v), g(v))
around the z-axis. It is called the surface of revolution S. The image by
φ of the curves u = constant and v = constant are called meridians and
parallels, respectively, of S.
(a) Show that the induced metric in the coordinates (u, v) is given by
g11 = f 2 , g12 = 0, g22 = (f 0 )2 + (g 0 )2 .
solution: In this coordinate system
∂
∂
= (−f sin u, f cos u, 0),
= (f 0 cos u, f 0 sin u, g 0 )
∂u
∂v
and the metric coefficients are obtained by taking the regular inner
products of these metrics. For the next calculation we record the
derivative of the metric coefficients depend only on v therefore
∂g11
∂g22
∂g11
∂g22
=
= 0,
= 2f f 0 ,
= 2(f 0 f 00 + g 0 g 00 )
∂u
∂u
∂v
∂v
(b) Show that the local equations for a geodesic γ are
d2 u
dt2
2
d v
dt2
−
0
ff
(f 0 )2 +(g 0 )2
2f f 0 du dv
f 2 dt dt
f 0 f 00 +g 0 g 00
du 2
+ (f
0 )2 +(g 0 )2
dt
+
=0
dv 2
dt
=0
Solution:
We first calculate the Chirstoffel symbols. Note that
since the coordinate system is orthogonal (in every tangent space) the
matrix gk,m is diagonal and so its inverse is easy to calculate. Thus
we have
1X ∂
∂
∂
m
Γi,j =
gj,k +
gk,i −
gi,j g k,m
2
∂xi
∂xj
∂xk
k
∂
∂
∂
1
gj,m +
gm,i −
gi,j (gm,m )−1
=
2 ∂xi
∂xj
∂xm
1 ∂g11
f0
Γ112 = Γ121 =
(g11 )−1 = f f 0 f −2 =
2 ∂v
f
0
∂g11
ff
1
−
(g22 )−1 = − 0 2
Γ211 =
2
∂v
(f ) + (g 0 )2
1 ∂g22
f 0 f 00 + g 0 g 00
Γ222 =
(g2 2)−1 = 0 2
2 ∂v
(f ) + (g 0 )2
Where I wrote only the non-zero coefficients. Now substitute into the
geodesic equations:
d2 xk X k dxi dxj
+
Γij
=0
dt2
dt dt
i,j
1
2
to obtain the two desired equations.
(c) Obtain the following geometric meaning of the equations above: the
second equation is, except for meridians and parallels, equivalent to
2
the fact that the “energy” |γ 0 (t)| of a geodesic is constant along γ;
the first equation signifies that if β(t) is the oriented angle β(t) < π,
of γ with a a parallel P intersecting γ at γ(t), then
r cos β = const,
where r is the radius of the parallel P . The equation above is called
Clairaut’s relation.
Solution: As mentioned above the second equation is just the fact
2 2 dv 2
that speed |γ 0 (t)|2 = du
f + dt ((f 0 )2 +(g 0 )2 ) is constant. Indeed
dt
1 d
|γ 0 (t)|2
2 dt
2
2
2
dv
dv
dv
du
du d2 u
0 00
0 00
0 2
0 2 dv d v
0
+ (f f + g g )
+ f2
+
((f
)
+
(g
)
)
= ff
dt
dt
dt
dt
dt dt2
dt dt2
2
2
2
dv
dv
dv
du
dv
dv d2 v
du
+ (f 0 f 00 + g 0 g 00 )
− 2f f 0
+ ((f 0 )2 + (g 0 )2 )
= ff0
dt
dt
dt
dt
dt
dt
dt dt2
(
)
2
2
dv
d2 v
ff0
du
f 0 f 00 + g 0 g 00 dv
=
((f 0 )2 + (g 0 )2 )
−
+
dt
dt2
(f 0 )2 + (g 0 )2 dt
(f 0 )2 + (g 0 )2 dt
Where in we substituted the first equation to obtain the second equality above.
The second equation is more interesting. To see that the meaning
is as written above note that r = f = f (v(t)). Now since γ 0 (t) =
∂
dv ∂
( du
dt ∂u + dt ∂v ) and we have
r cos β = f
du ∂
dt ∂u
∂
∂
+ dv
1 du 2
dt ∂v
∂u
·
= 0
f
|γ 0 (t)|
f
|γ (t)| dt
As |γ 0 (t)| is constant the first equation is obtained by equating the
derivative of this expression to zero.
(d) Use Clairaut’s relation to show that a geodesic of the paraoloid
(f (v) = v, g(v) = v 2 , 0 < v < ∞, − < u < 2π + ),
which is not a meridian, intersects itself an infinite number of times.
Solution: Since the surface of revolution is obviously complete, by
the Hopf-Rienow theorem, every geodesic can be continued indefinitely.
Let a : R −→ M be a geodesic which is not a meridian, assume that in
the given coordinates a = (u(t), v(t)). And let β be the angle defined
above. We note that if β = π/2, or β = −π/2 at even one point
then the geodesic will be a Meridian by the uniqueness of geodesics.
Thus by continuity of β the geodesic a is either turning all the time to
the right, or all the time to the left around the surface of revolution.
Without loss of generality we may assume the first case holds, namely
that −π/2 < β(t) < π/2 ∀t ∈ R. We claim that if β(t2 ) > 0 for
some t2 ∈ R then β is increasing on [t2 , ∞) and in particular β(t) >
0, ∀t > t0 . Indeed by continuity v(t) > v(t0 ) at least for small right
neighborhood of t0 . But by the geometry of the surface this means
3
that r(t) > r(t0 ) in this neighborhood so that by the Clairaut equation
cos β(t) < cos β(t0 ) and β(t) > β(t0 ). A similar argument implies that
if β(t1 ) < 0 then beta is decreasing on the interval (−∞, t1 ). If we have
t1 , t2 with β(t1 ) < 0, β(t2 ) > 0 this implies that the curve intersects
itself infinitely many times as it turns around the surface.
Without loss of generality assume that there exists a point t2 with
β(t2 ) > 0. If β is everywhere positive, and hence v(t) everywhere
increasing so that when t −→ −∞ the curve α becomes asymptotic
to some parallel P . By the Clairaut equation r(t) > r(t) cos(t) =
r(t2 ) cos(β(t2 )) > 0 is bounded away from zero so the parallel P is
some parallel of positive hight (rather than the degenerate parallel
which is the point at the origin). But it is impossible for such a curve
to be a geodesic as P itself is not a geodesic.
(2) (Lie Groups with a bi-invariant metric - Do Carmo Chapter 3, ex 3). Let
G be a Lie group, G its Lie algebra (identified with the collection of left
invariant vector fields) and let X ∈ G. The trajectories of X determine a
mapping φ : (−, ) −→ G with φ(0) = e, φ0 (t) = X(φ(t)).
(a) Prove that φ(t) is defined for all t ∈ R and that φ : (R, +) −→ G is
a group homomorphism. Such a φ is called a 1-parameter subgroup of
G.
Solution: The existence and uniqueness theorem for solutions of
ODE’s guarantees the local existence of the follow, namely the map φ :
(−, ) −→ G is defined such that φ(0) = e and φ0 (t) = X(φ(t)), ∀t ∈
(−, ). But since everything is left invariant we will have that for any
g ∈ G the map ψ(t) := Lg ◦ φ(t) : (−, ) −→ G given by t 7→ gφ(t)
satisfies ψ(t) = g, ψ 0 (t) = X(ψ(t)). It follows directly that we can
extend φ to the whole real line - as the set of points to which we
can extend φ is both open and closed. Finally to see that this is
a one parameter subgroup consider ψ1 (t) = φ(s + t) and ψ2 (t) =
φ(s)φ(t) as functions of t with s = s0 fixed. Both are flows along
the X vector space with initial position φ(s). Namely they satisfy
ψi (0) = φ(s), ψ 0 (t) = X(ψ(t)). For ψ1 this is true by definition, and
for ψ2 by the left invariance. Thus but the uniqueness of the solution
to this ODE both are identical.
Important note: The map G −→ G given by X 7→ φX (1) where
φ is the flow defined above is sometimes denoted by exp : G −→
G. This exponential map, which you probably saw in the smooth
manifold course, is defined using the Lie group structure without using
a connection or a Riemannian metric. it is a different object than the
Riemannian exponential map which is defined using geodesics. As we
show below when G admits a bi-invariant Riemannian metric the two
exponential maps are one and the same. But this is no longer true for
more general Riemannian metrics on Lie groups - for example it might
fail for left invariant Reimannian metrics!
(b) Assume that G Riemannian metric. This Reimannian metric induces
in a natural way an inner product on G denoted also by h·, ·i. Show
that if the Reimannian metric is bi-invariant (with respect to both left
translation Lv (x) = vx and right translation Rv (x) = xv on the Lie
4
group) then
h[U, X], V i = −hU, [V, X]i,
∀ U, V, X ∈ G.
Hint:
Prove that hU, V i = hdRφ(t) U, dRφ(t) V i. Now prove that
[Y, X] = limt−→0 1t (dRφ(t) Y −Y ). Differentiate the first equation along
the φ-flow using the second.
Solution: This is explained in some detail in Do-Carmo Chapter 1
Sec 2 (page 41).
(c) Prove that if G has a bi-invariant Reimannian metric then the geodesics
of G that start from e are 1-parameter subgroups of G.
Hint: Use the relation
2hX, ∇Z Y i =
+
ZhX, Y i + Y hX, Zi − XhY, Zi
hZ, [X, Y ]i + hY, [X, Z]i − hX, [Y, Z]i.
and that the metric is left invariant to prove that hX, ∇Y Y i = hY, [X, Y ]i,
where X, Y, Z are left invariant vector fields. It follows that ∇Y Y = 0,
for all Y ∈ G. Thus 1-parameter subgroups are geodesics. By uniqueness, geodesics are 1-parameter subgroups.
Solution: Recall that the relation above was the relation that describes the Levi-Civita connection in terms of the Reimannian metric
(in a coordinate free way). Plugging in Y = Z we get
1
{Y hX, Y i + Y hX, Y i − XhY, Y i
hX, ∇Y Y i =
2
+ hY, [X, Y ]i + hY, [X, Y ]i − hX, [Y, Y ]i} = 0
Where all the summand in the first line vanish because X, Y are left
invariant so that the functions hX, Y i, hY, Y i are constant on the Lie
group. And the second line vanishes due to the relation established in
the previous item. Now since X is arbitrary, and the inner product is
by definition non-degenerated, we deduce that ∇Y Y = 0 which means
that the Y flow is equal to the geodesic flow as Y is parallel along the
flow lines.
(3) (Nice coverings - Do Carmo Chapter 3, ex 4). Prove that any differentiable
manifold admits an open covering {Uα } with the property that ∩r1 Uαi is
contractible (whenever it is not empty) for any r and any r-open subsets
from the covering Uα1 , . . . , Uαr .
Solution: We saw that any differentiable manifold admits a Riemannian
metric. Let us fix such a metric. Now let us cover the manifold by stongly
convex balls - we proved the existence of such balls in class. The nice
property of the intersections follows directly from convexity.