HOMEWORK 4 SOLUTIONS
I 1a: On a Riemannian manifold, show that the values hR(X, Y )X, Y i completely
determine R.
F
hR(X, Y + Z)X, Y + Zi
= hR(X, Y )X, Y i + hR(X, Z)X, Zi + hR(X, Y )X, Zi + hR(X, Z)X, Y i
= hR(X, Y )X, Y i + hR(X, Z)X, Zi + 2hR(X, Y )X, Zi
therefore R(X, Y )X is completely determined by the given type of expression.
R(X + Z, Y )(X + Z) = R(X, Y )X + R(Z, Y )Z + R(Z, Y )X + R(X, Y )Z,
so terms of the form R(Z, Y )X +R(X, Y )Z are determined. Thus from the given
expressions we also can compute
(R(Z, Y )X + R(X, Y )Z) − (R(Y, X)Z + R(Z, X)Y )
= R(Z, Y )X + R(X, Y )Z + R(X, Y )Z + R(X, Z)Y
= 2R(X, Y )Z + R(Z, Y )X + R(X, Z)Y
= 3R(X, Y )Z + R(Z, Y )X + R(X, Z)Y + R(Y, X)Z
= 3R(X, Y )Z.
3
B b: On R+ = {(x, y, z); z > 0}, define a metric so that ∂x , ∂y and ∂z are all
orthogonal, ||∂x ||2 = z12 , ||∂y ||2 = 1, and ||∂z ||2 = z12 . Calculate all Γkij .
F Clearly ∇∂y ∂x = ∇∂y ∂y = ∇∂y ∂z = 0. Using compatibility with the metric, we
see that 2h∇∂x ∂z , ∂z i = 2h∇∂x ∂x , ∂x i = 0. Also 2h∇∂z ∂x , ∂x i = 2h∇∂z ∂z , ∂z i = −2
z3 .
Therefore ∇∂x ∂z = ∇∂z ∂x = −1
∂
.
Finally
we
know
h∇
∂
,
∂
i
+
h∂
,
∇
∂
i
∂x x z
x
∂x z =
z x
1
−1
h∇∂z ∂x , ∂z i + h∂x , ∇∂z ∂z i = 0. Therefore ∇∂z ∂z = z ∂z and ∇∂x ∂x = z ∂z . This
defines all Γkij .
B c: Show that II = 0 on the circles {x2 + z 2 = r2 , y = 0}. Using this, write down
explicitly all geodesics. Show that this metric is complete.
F First, parametrize the semi-circles by ωr (t) = (x, y, z) = (r cos t, 0, r sin t). Then
ω̇r (t) = −r sin t∂x + r cos t∂z , and
Dω̇r
ω̈r (t) =
(t) = ∇ω̇r (t) (−r sin t∂x + r cos t∂z )
dt
= −r cos t∂x − r sin t∂z − r sin t∇−r sin t∂x +r cos t∂z ∂x + r cos t∇−r sin t∂x +r cos t∂z ∂z
1
1
1
1
= −r cos t∂x − r sin t∂z + r2 (sin2 t ∂z + sin t cos t ∂x + cos t sin t ∂x − cos2 t ∂z )
z
z
z
z
cos2 t
= −r cos t∂x − r sin t∂z + r(sin t∂z + cos t∂x + cos t∂x −
∂z )
sin t
cos2 t
= r(cos t∂x −
∂z ) = r cot t(sin t∂x − cos t∂z ).
sin t
This is parallel to ω̈r and therefore II = 0. Next we construct geodesics. Let
Z t
Z t
Z t
1
1
1
t
s(t) =
||ω̈r ||dt =
dt =
dt = log(tan ).
π
π z
π r sin t
r
2
2
2
2
1
2
HOMEWORK 4 SOLUTIONS
Since by definition ṡ = ||ω̇r ||, we see that γr = ωr ◦ s−1 satisfies ||γ̇r || = 1. Since
γr has constant speed and also has image with II = 0, we conclude that γr is a
geodesic.
Now for constants a, b, c, d, m, r let (x(t), y(t), z(t)) = (γrx (at + b) + c, mt +
d, γrz (at + b)), and notice that this defines a geodesic for any choice of constants. In
the xz-plane, we can choose a circle centered on the x-axis passing through a given
point and tangent to a given vector at that point, as long as the vector is not parallel to ∂z . This shows that we can choose constants a geodesic matching any initial
conditions, as long as the initial vector is not in the span of ∂z and ∂y . For these
initial conditions we can instead have geodesics (x(t), y(t), z(t)) = (c, mt + d, beat ).
Since all geodesics are defined for all time we see that the metric is complete.
B d: Calculate the curvature (from a it suffices to write 3 functions R3+ → R). For
the geodesic (x, y, z) = (0, 0, et ), calculate all Jacobi vector fields.
F We calculate R(V, W, )U , where each of V, W , or U may be ∂x , ∂y , or ∂z , using
linearity this determines R. Notice that if any of them are ∂y then the curvature
is zero. Using the formulas from (b) we compute
1
−1
1
1
−1
∂x ) − ∇∂z ( ∂z ) =
∇∂x ∂x + 2 ∂z − ∇∂z ∂z
R(∂x , ∂z )∂x = ∇∂x (
z
z
z
z
z
1
= 2 ∂z .
z
Using the symmetries this completely determines R. Letting ω be the geodesic,
we see that ω̇ = et ∂z , so the Jacobi equations become J¨ + e2t R(J, ∂z )∂z = 0.
Writing J in components J = J x ∂x + J y ∂y + J z ∂z , we have R(J, ∂z )∂z = −J x z12 ∂x
and
J˙ = J˙x ∂x + J˙y ∂y + J˙z ∂z + et (J x ∇∂z ∂x + J y ∇∂z ∂y + J z ∇∂z ∂z )
et
et
et
= J˙x ∂x + J˙y ∂y + J˙z ∂z − (J x ∂x + J z ∂z ) = (J˙x − J x )∂x + J˙y ∂y + (J˙z − J z )∂z .
z
z
z
Iterating this again we get
et
e2t
et
e2t
J¨ = (J¨x − 2 J˙x + 2 J x )∂x + J¨y ∂y + (J¨z − 2 J˙z + 2 J z )∂z .
z
z
z
z
t
Noting that z = e , our Jacobi equations become
(J¨x − 2J˙x + J x ) − J x = J¨y = J¨z − 2J˙z + J z = 0.
The solution space is then J ∈ span{∂y , t∂y , ∂x , e2t ∂x , et ∂z , tet ∂z }. In terms of
families of geodesics, we can describe these vector fields as (in order): ∂d and ∂m
(Euclidean motions in the y direction), ∂c (translation in the x direction, notice
the original metric doesn’t depend on x), e2t ∂x is deformation of a vertical line into
a circle of large radius, and finally et ∂z and tet ∂z are ∂b and ∂a : translation and
acceleration along the geodesic.
B e: Let C ⊆ {y = 0} be any properly embedded curve. Show that C × R =
{(x, y, z); (x, 0, z) ∈ C} is isometric to R2 (not just locally!). Show that {y = 0} is
not even locally isometric to R2 .
F Note that our metric on R3 is isometric to {y = 0} × R. If C is a properly
embedded curve, then a parametrization by arclength defines an isometry to the
flat line R. This completes the first claim. For the second, simply notice that the
curvature is non-zero everywhere.
I 2a: Let M be a complete Riemannian manifold, and let U
R 1: M → R be a smooth
function. Let A : Ω(p, q) → R be defined by A(ω) = 0 ||ω̇(t)||2 − U (ω(t))dt.
HOMEWORK 4 SOLUTIONS
3
Calculate dA : T Ω → T R. Then show the critical points of A are smooth, and they
satisfy the equation ω̈ = −∇U .
R1
F This is false, instead we should have A(ω) = 0 12 ||ω̇(t)||2 − U (ω(t))dt. Letting
α be a variation of ω, we get
d
dA ◦ α
=
du
du
We note that
D
du α̇
Z
=
1
1
Z
1
hα̇, α̇i − U (α)dt =
2
0
D 0
dt α
0
D
α̇, α̇ − dU (α0 )dt.
du
and then integrate the first term by parts
!
Z 1
Z 1
X
D
0
0
α̇, α̇ dt = −
hα , ∆ω̇i +
hα , ω̈idt
du
0
0
t
just as in the case for geodesic energy. For the second term we just note that
dU (α0 ) = h∇U, α0 i by definition. Therefore
dA(W ) = −
X
t
Z
hW, ∆ω̇i −
1
hW, ω̈ + ∇U idt.
0
The only way that this could be zero for all W ∈ T Ωω is if ∆ω̇ = 0 and ω̈ +∇U =
0 everywhere, that is, ω is a smooth Newtonian path.
B b: Suppose that M is complete and U is bounded. Show that the pair (M, U )
is dynamically complete: every solution to Newton’s equation can be defined for
t ∈ R. Show that this is not necessarily true if U is unbounded.
F Let ω be a curve satisfying Newton’s equation. Then
Z b
Z b
1
1
2
2
||ω̇(b)|| − ||ω̇(a)|| =
hω̈(t), ω̇(t)idt = −
h(∇U )(ω(t)), ω̇(t)idt
2
2
a
a
Z b
=−
dU (ω̇)dt = U (ω(a)) − U (ω(b))
a
that is, the change in kinetic energy is equal to the change in potential energy.
Since U is assumed to be bounded (say |U | < C), this implies that ||ω̇|| is bounded.
More precisely,
any physical motion ω : [a, b] → M must have finite length `(ω) 6
p
(b − a) ||ω̇(a)||2 + 2C.
We need to show that, if ω[0, a) is defined for a < ∞ then it can be defined
on [0, a + ε) (and do the same argument for negative time). By the above length
estimate we see that ω(a − k1 ) is a Cauchy sequence and therefore converges to a
point ω(a) since M is complete. ω̇(a − k1 ) is also Cauchy, since ω̈ = −∇U is locally
bounded. Using these initial conditions we can define a physical motion at ω(a)
with initial velocity ω̇(a), and by uniqueness of solutions to the differential equation
we know that this must give an extension of ω.
For a counterexample we let x(t) = log(t) (with x ∈ M = R with the standard
metric). Then ẍ = − t12 = −e−2x . Therefore if we define U (x) = − 21 e−2x , the curve
x : (0, 1] → R covers infinite arclength in finite time.
B c: With the same assumptions, show that (M, U ) is dynamically complete in
another sense: any two points p, q ∈ M are connected by a Newtonian path.
F First we show that all nearby points are connected by a Newtonian path. For
q ∈ M, t ∈ R, let expU
q,t : T Mq → M be the map which evaluates a Newtonian
path at time t with the given initial conditions. If ε is a constant, notice that
D2
1
2
cU
U
U 1
dt2 ω(ct) = c ω̈, therefore expq,t (v) = expq,ct ( c v). Therefore we see that expq,ε ( ε v)
4
HOMEWORK 4 SOLUTIONS
limits to expq (v) as ε → 0, C ∞ uniformly on compact subsets. In particular this
shows that for sufficiently small ε, expU
q,ε is a diffeomorphism onto its image when
n K
restricted to B ( ε ) ⊆ T Mq , where K is a constant depending on U, M and q but
n K
n K
not ε. This shows that expU
q,ε (B ( ε )) ⊇ expq (B ( 2 )) for sufficiently small ε, in
U
n K
particular q ∈ expq,ε (B ( ε )). Therefore any point of arclength distance less than
K
2 is connected to q by a Newtonian path, which evolves for any given sufficiently
small time.
Since U is bounded and Newton’s equation only depends on ∇U , we can assume
U > 0 everywhere. Let Ω6K (p, q) be the subset of path space (defined for t ∈ [0, 1])
with action A 6 K. If ω ∈ Ω6K (p, q) then `(ω)2 6 E(ω) 6 Aω, so all ω are
contained in a compact subset of M . Therefore there is a uniform δ > 0 so that all
points within distance δ are connected by a Newtonian path defined for t ∈ [0, ε].
2
1
< δK .
We are free to shrink ε so we choose ε = m
Just as in the geodesic case, we let B K be the set of piecewise Newtonian paths
which are non-smooth only at mi ∈ (0, 1), and have total action no greater than
K. We see that the evaluation map ev : B K → M ×(m−1) is an injection with open
image, and A|B K is a smooth map. Furthermore, the differential of A is exactly
given by the formula from part (a) (both of these statements follows from the fact
that solutions to a differential equation depend smoothly on the initial conditions,
and the fact that expU
q,ε has smooth inverse).
Any ω ∈ B K which is critical for A|B K is also critical for A: by virtue of
R1
being in the set B K we have that 0 hW, ω̈ + ∇U idt = 0, and therefore dA(W ) =
P
− i hW, ∆ω̇( mi )i. If ∆ω̇ 6= 0 then we can choose a vector W ∈ T B K which makes
dA(W ) 6= 0, since we can move the mi endpoint in the direction of ∆ω̇( mi ).
Therefore, any critical point of A|B K is a Newtonian path. But B K is a manifold,
and ∂B K = A−1 K. Thus as long as we choose K large enough so that B K is nonempty, A must have a minimum which occurs on the interior and therefore it is a
Newtonian path.
B d: Equip R2 with the metric which is given in polar coordinates by h∂r , ∂θ i =
0, ||∂θ ||2 = sinh2 (r), ||∂r ||2 = 1. (This is the hyperbolic plane, it is isometric to
the submanifold {y = 0} from problem 1.) Show that r is equal to the arclength
distance to the origin (in a sentence). Suppose we have a point mass at the origin,
inducing a potential energy given by U = − GM
r for some constant GM .
Show that a circular orbit (r, θ) = (constant, ωt) satisfies Newton’s equation,
where ω is the constant angular velocity. Show that
ω2 =
r2
GM
.
sinh(r) cosh(r)
This is in contrast to the Euclidean case, where ω 2 = GM
r 3 (Kepler’s third law,
known to Copernicus in the circular case). Notice that these two equations are
equivalent up to first order at r = 0.
F Let γ(t) = (r(t), θ(t)) = (ρ, ωt). Then γ̇ = ω∂θ , so γ̈ = ω 2 ∇∂θ ∂θ . Differentiating
the expressions for the metric we see 2h∇∂θ ∂θ , ∂θ i = 0, h∇∂θ ∂θ , ∂r i + h∂θ , ∇∂θ ∂r i =
0, and 2h∇∂r ∂θ , ∂θ i = 2 sinh(r) cosh(r). Therefore ∇∂θ ∂θ = − sinh(r) cosh(r)∂r .
Then γ̈ = −ω 2 sinh(r) cosh(r)∂r = −∇U = − GM
r 2 ∂r , which justifies the claim.
I 3a: Let C1 and C2 be two embedded closed curves in S 3 which are disjoint.
Suppose there are compact oriented surfaces Σ1 , Σ2 ⊆ B 4 so that ∂Σj = S 3 ∩ Σj =
HOMEWORK 4 SOLUTIONS
5
Cj , which are transverse to each other, and to S 3 . Define `k(C1 , C2 ) to be the
signed count of intersections Σ1 ∩ Σ2 . Show that this invariant, called the linking
number, only depends on C1 and C2 , not the surfaces.
e 1 ⊆ B 4 with ∂Σ1 = ∂ Σ
e 1 = C1 , and transverse
F Suppose we have two surfaces Σ1 , Σ
3
e 2 . Think of Σ
e 1 and Σ
e 2 as living inside a different
to S . Similarly with Σ2 and Σ
e4.
4-ball, which we label B
4
e 4 by a diffeomorphism identifying C1 and C2 . Since B
e4
We then glue ∂B to ∂ B
3
4
e
is glued to the opposite side of S , we must reverse the orientation of B to get
e4 ∼
a consistent orientation on B 4 ∪ B
= S 4 . Similarly, by reversing the orientation
e
e
e
of Σ1 and Σ2 , we can glue Σj to Σj along Cj to get a closed surface Mj ⊆ S 4 .
This follows from transversaility, which implies that after diffeomorphism, we can
identify Σ1 with C1 × [0, ε) ⊆ S 3 × [0, ε) near ∂B 4 .
e 1 , C2 ) be the linking number computed using the surfaces Σ
e j . We claim
Let `k(C
e
that `k(C1 , C2 ) − `k(C1 , C2 ) = [M1 ] · [M2 ]. Both sides of the equation count all
e 1 with Σ
e 2 . And on both sides the tilded
intersection points of Σ1 with Σ2 and Σ
intersections are counted with opposite sign: on the left the equation is just written
that way, on the right it comes from the fact that the orientations were all reversed
e 4 , making three reversals).
(including B
There is at least one point in S 4 which is disjoint from M1 and M2 , by deleting
this point we can think of Mj ⊆ R4 ∼
= S 4 \ {point}. But then [M1 ] · [M2 ] = 0, since
we showed that all manifolds have zero intersection product in a vector space.
B b: Let f : S 3 → S 2 be a smooth map, and let p, q ∈ S 2 be regular values. Show
that, if f is homotopic to a constant map, then `k(f −1 (p), f −1 (q)) = 0.
F Let F : S 3 × [0, 1] be the homotopy, so F0 = f and F1 is constant. By further
homotopy we can assume that p and q are regular values of F and F1 (parametric
transversality). Then F −1 (p) = Σp ⊆ S 3 × [0, 1] is a compact surface with ∂Σp =
Σp ∩ S 3 × {0, 1}. But F1 is a constant map which is transverse to p, so F1−1 (p) =
∅ = Σp ∩ S 3 × {1}. Therefore ∂Σp = Σp ∩ S 3 × {0} = f −1 (p). Everything similarly
holds for Σq = F −1 (q), which is disjoint from Σp .
Since S 3 × [0, 1] ⊆ B 4 , we can use Σp and Σq to calculate the linking number,
which is zero since they are disjoint.
B c: Let S 3 be the unit sphere in C2 , let S 2 = C ∪ {∞}, and define a map
f : S 3 → S 2 by f (z1 , z2 ) = zz21 . Show that f is not homotopic to a constant.
F First, df = z12 (z2 dz1 − z1 dz2 ), which is surjective everywhere (maybe not com2
pletely obvious for ∞ ∈ S 2 ). Let p = 0 and q = 1. Then f −1 (p) = {(z1 , z2 ) =
(0, eit )} and f −1 (q) = {(z1 , z2 ) = √12 (eit , eit )}. We define Dp2 in S 3 by Dp2 =
{(z1 , z2 ) = (1 − r2 , r2 eit )} with r ∈ [0, 1]. Then ∂Dp2 = f −1 (p), and Dp2 ∩ f −1 (q) =
{ √12 (1, 1)}.
Choose any surface Σq ⊆ B 4 with ∂Σq = f −1 (q) (it’s easy to show explicitly
that one exists). We can “push” Dp2 slightly into B 4 , so that it becomes transverse
to S 3 . To be rigorous, choose a collar neighborhood of S 3 where Σq ∩ S 3 × [0, ε) =
e 2 = {(z1 , z2 , r) = (1 − r2 , r2 eit , ε r)} ⊆ S 3 × [0, ε). Then
f −1 (q) × [0, ε), and let D
p
2
2
e p intersects Σq in exactly the same way Dp2 intersects f −1 (q), in one point.
D