Math 6397 Riemannian Geometry II
The Jacobi Field
1
Covariant Derivatives along Curves
Lemma 4.9 (see P. 57 on the textbook) Let 5 be a linear connection on M .
For each curve γ : I → M , 5 determines a unique operator
Dt : T (γ) → T (γ)
where T (γ) is th eset of smooth vector fields along γ, such that it is linear over
R, it satisfies the product rule, and if V is extendible, then Dt V (t) = 5γ 0 (t) Ṽ ,
where Ṽ is an extension of V . Dt is called the Covariant Derivatives along
the Curve γ.
The motivation of Dt comes from the following: if f (t) is a function
defined on γ, then, we can express f (t) = f˜ ◦ γ(t), where f˜ is some function
in the neighborhood of γ (assuming that f is extendible). Then f 0 (t) =
df (γ 0 (t)) = 5γ 0 (t) f = Dt f . Similarily, if X is a vector field along the curve
γ, if we write
X = X i ∂i
then
dX i
∂i + X i Dt ∂i .
dt
Thus we see that Dt is a natural differential operator along the curve γ.
Dt X =
2
Jacobi field
• Introduction: The Jacobi fields arise naturally from the variation of the geodesic γ through neighborhouring geodesics. Let
now γ : I → M be a geodesic. A variation of γ is a differential map
Γ(s, t) : (−, ) × I → M with Γ(0, t) = γ(t) for all t ∈ I. Γ is said to
be a geodesic variation if for every s ∈ (−, ), the curve F (·, s) is
geodesic.
1
Write T (s, t) = ∂t Γ(s, t), S(s, t) = ∂s Γ(s, t). The geodesic equation
tells us that
Dt T = 0.
We can take the covariant derivative of this equation to get
Ds Dt T = 0.
The variation vector field of the variation Γ is defined as
V = S(0, t).
Using the fact that Ds Dt T = 0 and two lemmas (Lemma 6.3 (symmetry lemma): Ds ∂t Γ = Dt ∂s Γ and Lemma 10.1: Ds Dt V − Dt Ds V =
R(S, T )V for any vector field V along γ), we can prove that
Theorem 1 Let γ : [a, b] → M be a geodesic on an m-dimensional
Riemmanian manifold (M, g). Then the variation vector field V of its
geodesic variation satisfies the Jacobi equation
Dt2 V + R(V, γ 0 )γ 0 = 0.
• Definition of a Jacobi field along a geodesic
Let J = J(t) be a vector field along γ. Denote by J 0 (t) = Dt J and
J 00 (t) = Dt2 J.
Definition. Let γ : [a, b] → M be a geodesic on an m-dimensional
Riemmanian manifold (M, g). A vector field J = J(t) along γ is called
Jacobi field if it satisfies the following Jacobi equation:
J 00 (t) + R(J(t), γ 0 (t))γ 0 (t) = 0,
where J 00 (t) = Dt2 J.
Remarks:
2
1. It is obvious that if J is a Jacobi field along γ, then J ⊥ := J− <
J, T > T is also a Jacobi field along γ where T = γ 0 .
2. One of the important result about Jacobi field is the following:
d2
< J, γ 0 >≡ 0,
dt2
so so < J(t), γ 0 (t) > is a linear function of t, i.e. < J(t), γ 0 (t) >=
λt + µ for some constants λ and µ. The proof is as follows: Using
the compatibility with the metric and the fact that 5 ∂ γ 0 (t) = 0, we
∂t
compute
Dt2 < J, γ 0 > = < J 00 (t), γ 0 >
= − < R(J, γ 0 )γ 0 , γ 0 >
= −R(J, γ 0 , γ 0 , γ 0 ) = 0.
Next, we want to find a local expression the the Jacobi equation. Let
γ be a geodesic on M . We express the Jacobi field equation in terms
of the local coordinate. Choose an orthonormal basis {ei |p } for Tp M ,
and extend it to a parallel orthonormal frame along all γ. Then
e0i (t) = 5 ∂ ei (t) = 0.
∂t
Let J(t) be a Jacobi field along γ. Write
J(t) =
m
X
J i (t)ei (t).
i=1
Then
0
J (t) =
m
X
i0
J (t)ei (t),
00
J (t) =
i=1
0
Write γ (t) =
Pm
k=1
m
X
00
J i (t)ei (t).
i=1
0k
γ ek , then Thus
J 00 (t) + R(J(t), γ 0 (t))γ 0 (t) =
X
00
{J i (t) +
X
j
i
3
k
l
i
Rljk
(γ(t))J j γ 0 γ 0 }ei (t).
Hence the Jacobi equation is reduced to
00
J i (t) =
X
k
l
i
Rljk
(γ(t))J j γ 0 γ 0 ,
1 ≤ i ≤ m.
j
Theorem 2. Let γ : [a, b] → M be a geodesic on an m-dimensional
Riemmanian manifold (M, g). Then for every v, w ∈ Tγ (a) M , there
exists a unique Jacobi field J(t) along γ satisfying
J(a) = v, J 0 (a) = w.
There are always two trivial Jacobi fields along any geodesic γ, which
can be written down immediately: Because Dt γ 0 (t) = 0 and R(γ 0 , γ 0 )γ 0 =
0 by antisymmetry of R, the vector field J0 (t) = γ 0 (t) satisfies the Jacobi equation with J0 (0) = γ 0 (0), J00 (0) = 0. Similarly, J1 (t) = tγ 0 (t) is
a Jacobi field satisfying J1 (0) = 0, J10 (0) = γ 0 (0). It is easy to see that
J0 is the variation field of the variation Γ(s, t) = γ(t + s), and J1 is the
variation field of the variation F (t, u) = γ(teu ). Therefore, these two
Jacobi fields just reflect the possible reparametrizations of γ, and don’t
tell us anything about the behavior of geodesics other than γ itself.
To distinguish these two trivial cases from more informative ones, we
make the following definitions.
Definition If J is orthogonal to γ everywhere, then J is called a normal Jacobi field.
Theorem 3. Let γ : I → M be a geodesic on an m-dimensional
Riemmanian manifold (M, g), and let a ∈ I.
(a) A Jacobi field J along γ is normal if and only if
J(a) ⊥ γ 0 (a),
J 0 (a) ⊥ γ 0 (a).
(b) Any Jacobi field orthogonal to γ 0 at two points is normal.
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Proof. As we showed earlier, f (t) :=< J(t), γ 0 (t) > is a linear function
of t. Note that f (a) =< J(a), γ 0 (a) > and f 0 (a) =< J 0 (a), γ 0 (a) >.
Thus J(a) and J 0 (a) are orthogonal to γ 0 (a) if and only if f and its
first order derivative vanish at a, which happens if and only if f ≡ 0.
Similarly, if J is orthogonal to γ 0 at two points, then f vanishes at two
points, so f ≡ 0. This proves the theorem.
• Geodesic variations, Gauss lemma, normal coordinates, and
conjugate points.
We first recall the definition and some important statement regading
the differential of the exponent map. From chapter 4 in the textbook
(John M. Lee) we have that any initial point p ∈ M and any initial
tangent vector v ∈ Tp M determine a unique maximal geodesic γ v with
γ v (0) = p and γ 0v (0) = v. It has the important property that γ cv (t) =
γ v ct, whenever it is defined. By taking t = 1 and using c = t, we get
γ tv (1) = γ v t, whenever it is defined. Let
E := {V ∈ T M ; γ v is defined on an interval containing [0, 1]}
and then we define the exponential map exp : E → M by
exp(v) = γ v (1).
Then we have, using the property which we just mentioned, exp(tv) =
γ v (t). (this is one of the most important property of exp.
Another importnat result is as follows: (see Lemma 5.10 on the textbook P. 76)
Under the natural identification T0 (Tp M ) ∼
= Tp M , we have (expp )∗,0 :
∼
T0 (Tp M ) = Tp M → Tp M is the identity map.
Recall the proof of the above result: We write (expp )∗ to denote (expp )∗,0 .
To compute (expp )∗ (v) for v ∈ Tp M , we just need to choose a curve
τ in Tp M starting at 0 whose initial tangent vector is v and compute
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the inital tangent vector of the component of the curve expp ◦τ . An
obvious such curve is τ (t) = tv. Thus
(expp )∗ (v) =
d
d
d
|t=0 expp ◦τ (t) = |t=0 expp (tv) = |t=0 γv (t) = γv0 (0) = v.
dt
dt
dt
Hence the result is proved.
We now give a new proof of Gauss lemma using Jacobi field.
Lemma (Gauss). Let p ∈ M and v ∈ Tp M such that expp v is defined.
Let w ∈ Tp M ∼
= Tv (Tp M ). Then
< (expp )∗v (v), (expp )∗v (w) >=< v, w > .
New Proof. Let v ∈ Tp (M ) and consider the geodesic γ(t) := expp (tv),
then γ(0) = p and γ 0 (0) = v. Identifying Tv (Tp M ) with Tp M as usual,
we can compute
(exp)∗,v (w) =
d
|s=0 expp (v + sw).
ds
To compute this, we define a geodesic variation of γ, as before, by
Γ(s, t) = expp t(v +sw). (we shift w to sw and connects v and sw to get
a straightline in T (M ) an then use the exponential map to project down
to M , see the picture on P. 184 in the textbook). Then, its variation
field JW (t) = ∂s Γ(s, t)|s=0 is a Jacobi field and JW (1) = exp∗,v W.
We now compute J(t) and J 0 (t). Hence
J 0 (0) = Dt J|t=0 = (Dt ∂s Γ(s, t))|s=0,t=0 = Ds ∂t Γ(s, t))|s=0,t=0 ,
note that the purpose to switch the order of differentiation is that we
want to use the result aboue (so we need to differentiate Γ in t).
Due to the fact that (expp )∗ (see the result above) is the identity at the
origin of Tγ (s) M , we have that, by definition and the result above,
∂t Γ(s, t)|t=0 = (expp )∗,0 (v + sw) = v + sw.
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Hence
J 0 (0) = J 0 (1) = Ds (v + sw)|s=0 = w
Therefore,
< v, w >=< γ 0 (0), J 0 (0) > .
Using the property for the Jacobi fields we proved earlier,
d2
< J(t), γ 0 (t) >= 0,
dt2
so < J(t), γ 0 (t) > is a linear function, say < J(t), γ 0 (t) >= at + b, so,
< J 0 (t), γ 0 (t) >= a which is a constant (using γ is a geodesic) and
b =< J(0), γ 0 (0) >= 0. Therefore,
< v, w > = < γ 0 (0), J 0 (0) >=< J 0 (1), γ 0 (1) >= a =< J(1), γ 0 (1) >
= (expγ (0) )∗,v (w), (expp )∗,v v > .
This proves Gauss’ lemma.
In the above new proof of Gauss’ lemma, we used Γ(s, t) = expp (t(v +
sw)) as the variation of the curve γ v (t) = expp (tv). Its variation field
J(t) = ∂s Γ(s, t)|s=0 is a Jacobi field along the curve γv which satisfies
J(0) = 0, J(1) = exp∗,v (w) and J 0 (0) = w. Using the normal coordinates, so γ v (t) = (v1 t, . . . , vn t), we have the following simple expression
of J:
Theorem Let p ∈ M and let (xi ) be a normal coordinate on a neighborhood of p, and let γ be a radical geodesic starting at p. For any
W = W i ∂i ∈ Tp M , the Jacobi field J along γ with J(0) = 0 and
J 0 (0) = W is given by
J(t) = tW j ∂i .
The geodesic variation Γ(s, t) = expp (t(v + sw)) as the variation of the
curve γ v (t) = expp (tv) and its variation field J(t) = ∂s Γ(s, t)|s=0 is also
used in the study of conjugate points (see the section about ”conjugate
points”).
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• Every Jacobi field arises from the variation of some geodesic
variation of γ.
We now prove our statement as the following theorem.
Theorem Let γ : [a, b] → M be a geodesic on a complete Riemmanian
manifold (M, g). Let J = J(t) be a Jacobi field along γ. Then J must
be a variation vector field of some geodesic variation of γ.
Proof. WLOG, we assume that a = 0. The key is to construct a
geodesic variation Γ of γ such that its variation field V satisfies V (0) =
J(0) and V 0 (0) = J 0 (0). In the proof of Gauss lemma, we used the
following natural variation of γ: Γ(s, t) = expp t(γ 0 (0) + sw), then we
see that V 0 (0) = w. By letting w = J 0 (0). Then V 0 (0) = J 0 (0).
The problem is that, for Γ(s, t) = expp t(γ 0 (0) + sw), we always have
Γ(s, 0) = p (the initial point is always fixed, so V 0 (0) = 0. To get
V 0 (0) = J(0), we need to let the initial point moving. So we replace
the initial point with a curve σ = σ(s), s ∈ (−, ) such that σ(0) =
γ(0), σ 0 (0) = J(0). After that, take a parallel translation of w := J(0)
and γ 0 (0) along the curve σ to get two vector fields W (s) and T (s),
s ∈ (−, ) which are parallel along σ. Now we define that
Γ(s, t) = expσ (s) t(T (s) + sW (s))
then, for every u ∈ (−, ), the curve Γ(·, s) is geodesic, and Γ(0, t) =
γ(t), so Γ is a geodesic variation of the geodesic γ. Let V (t) =
∂s Γ(s, t)|s=0 be the variation vector field. Then V is a Jacobi field.
We verify that V (0) = J(0), V 0 (0) = J 0 (0). To see this,
V (0) = ∂s Γ(s, 0)|s=0 .
Using Γ(s, 0) = expσ (s) (0) = σ(s), we have V (0) = σ 0 (0) = J(0). To
see V 0 (0) = J 0 (0), we compute
V 0 (0) = Dt V |t=0 = (Dt ∂s Γ(s, t))|s=0,t=0 = Ds ∂t Γ(s, t))|s=0,t=0 .
Due to the fact that (expσ (s) )∗ (see the result above) is the identity at
the origin of Tσ (s) M , we have that
∂t Γ(s, t))|t=0 = (expσ (s) )∗ ((T (s) + sW (s)) = T (s) + sW (s).
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Hence
V 0 (0) = Ds (T (s) + sW (s))|s=0 = W (0)
since Ds T = 0, Ds W = 0 due to the fact that T and W are parallel
along σ. Hence V 0 (0) = J 0 (0). By the uniqueness result, we have
J(t) = V (t).
Hence J(t) is a variation vector field of the geodesic variation F of γ.
This finishes the proof.
As another application, we prove
Corollary 3.3 Let p ∈ M and v ∈ Tp M be contained in the domain of
definition of expp , and c(t) = expp tv. Let the piecewise sommth curve γ :
[0, 1] → Tp M be likewise contained in the domain of definition of expp , and
assume that γ(0) = 0, γ(1) = v. Then
kvk = L(expp tv|t∈[0,1] ) ≤ L(expp ◦γ),
and the equality holds if and only if γ is different from the curve tv, t ∈ [0, 1]
only by a reparameterization.
Proof. We shall show that any piecewise sommth curve γ : [0, 1] → Tp M
with γ(0) = 0 satisfies
L(expp ◦γ) ≥ kγ(1)k.
Write γ(t) = r(t)φ(t), r(t) ∈ R and φ(t) ∈ Tp M , with kφ(t)k ≡ 1 (polar
coordinates in Tp M ). We have
γ 0 (t) = r0 (t)φ(t) + r(t)φ0 (t)
and < φ, φ0 >≡ 0. Thus, by Corollary 3.2, also
< exp∗,γ (t) φ(t), exp∗,γ (t) φ0 (t) >= 0,
kexp∗,γ (t) φ(t)k = kφ(t)k = 1,
and it follows that
kexp∗,γ (t) γ 0 (t)k ≥ |r0 (t)|.
Hence
L(expp ◦γ) =
Z 1
0
kexp∗,γ (t) γ 0 (t)kdt ≥
9
Z 1
0
|r0 (t)|dt ≥ r(1) − r(0) = kγ(1)k.
This proves the statement.
Remark: Corollary 4.3.3 by no means implies that the geodesic c(t) =
exp +ptv is the shortest connection between its end points. It only is shorter
than any other curve that is the exponential image of a curve with the same
initial and end points as the ray tv, 0 ≤ t ≤ 1.
3
Jacobi Fields of the Manifolds with Constant Sectional Curvature
Let (M, g) be a Riemannian manifold with constant sectional curvature C.
Recall that the sectional curvature at p ∈ M with respect to Xp , Yp ∈ Tp M
ie defined by
r(Xp , Yp , Yp , Xp )
K(Xp , Yp )) =
.
|Xp |2 |Yp |2 − < Xp , Yp >2
We can show that if (M, g) be a Riemannian manifold with constant sectional,
then R(Z, W, X, Y ) =< R(X, Y )Z, W >= −c(< X, Z >< Y, W > − <
X, W >< Y, Z >), or equivalently, R(X, Y )Z = −c(< X, Z > Y − < Y, Z >
X) (Here is the outline of the proof: We first prove the following result: Let
V be a m-dimensional vector space and let f : V × V × V × V → be multilinear map (i.e. f is a (0, 4)-tensor). f is called curvature-type tensor if it
satisfies, for every u, v, z, w ∈ V ,
(i) Antisymmetric: f (u, v, w, z) = −f (v, u, w, z) = −f (u, v, z, w);
(ii) First Bianchi identity: f (u, v, w, z) + f (w, v, z, u) + f (z, v, u, w) = 0,
(iii) Symmetric: f (u, v, w, z) = f (w, z, u, v).
Now let f, g : V × V × V × V → be two curvature-type (0, 4)-tensors.
Assume that, for every u, v ∈ V , f (u, v, u, v) = g(u, v, u, v). Prove that
f ≡ g.
Hint: Let F (u, v, w, z) = f (u, v, w, z) − g(u, v, w, z). The goal is to show
that F ≡ 0. Step 1: From Note that F (u, v, u, v) = 0 implies that F (u, v +
z, u, v +z) = 0. From here to show that F (u, v, u, z) = 0 for every u, v, z ∈ V .
Step 2: From step 1, we get F (u + v, w, u + v, z) = 0. From here, try to
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get (by expanding it) F (u, w, v, z) = F (v, w, z, u), for every u, v, z, w ∈ V .
Similarily, one can get F (u, w, v, z) = F (v, w, z, u) = F (z, w, u, v). Step 3:
Use the First Bianchi identity and the previous identity to get
3F (u, v, w, z) = F (u, v, w, z) + f (w, v, z, u) + f (z, v, u, w) = 0.
This finishes the proof. Then we prove the foolwoign statement: We say
that M is wondering at p ∈ M if the sectional curvature Kp (E) at p is constant(i.e. it is independent of E). Hence R(Xp , Yp , Xp , Yp ) = −Kp G(Xp , Yp , Xp , Yp )
for every Xp , Yp ∈ Tp (M ). Prove that if M is wondering at p ∈ M , then for
any Xp , Yp , Zp , Wp ∈ Tp (M ),
R(Xp , Yp , Zp , Wp ) = −Kp G(Xp , Yp , Zp , Wp ).
Hint: Use #2 by verifying that both R and G are curvature-type (0, 4)tensors. )
Let γ be a geodesic with unit-speed in M . Substitute this into the Jacobi
equation, we find a normal Jacobi field J satisfies
0 = J 00 − C(< J, γ 0 > γ 0 − < γ 0 , γ 0 > J) = J 00 + CJ.
Choosing a parallel normal vector field E along γ and setting J(t) = u(t)E(t),
and assuming that J(0) = 0, then the Jacobi equation becomes that
u00 (t) + cu(t) = 0
has fundamental
solutions with u(0) = 0 as follows: if c > 0, then
√ u(t) =
√
1
√
√1 sin
ct, if c = 0, then u(t) = t, if c < 0, then u(t) = −c sinh −ct. So
c
we proved the following statement: Let (M, g) be a Riemannian manifold
with constant sectional curvature C. Let γ be a geodesic with unit-speed in
M . Then the normal Jacobi field along along γ with J(0) = 0 are precisely
the vector fields J(t) = u(t)E(t), where E is any parallel normal
√ vector field
1
along γ, and u(t) is as follows: if c > 0, then u(t) = √c sin ct, if c = 0,
√
then u(t) = t, if c < 0, then u(t) = √1−c sinh −ct.
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4
Conjugate points
Let (M, g) be a complete m-dimensional Riemannian manifold. According
to the Hopf-Rinow theorem, for every p ∈ M , the exponential map is defined
on Tp M everywhere. Because, (expp )∗0 the derivative of expp at the zero
vector 0, is the identity map on Tp (M ) = T0 (Tp M ), so it is non-singular in
a neighborhood of 0. But, in general, sometimes, it may be (expp )∗ may be
singular at some points other than zero. Let us look at the case of unit-sphere
in Rn+1 .
Let
M = S n = {x ∈ Rn+1 |
n+1
X
(xα )2 = 1}.
α=1
Then (M, g) is a space with constant curvature (the sectional curvature is 1).
We know that for every p ∈ S n , all geodesic curves starting from p converge
to q = −p, and the length of these geodesics are all equal to π. Hence, on
S n−1 (π) ⊂ Tp (M ), the image under the map expp is the single point set {q}.
In particular, for every smooth curve σ(t) on S n−1 (π),
expp (σ(t)) = q,
hence
(expp )∗σ (t) (σ 0 (t)) = 0.
This means that the map expp is degenerate on S n−1 (π) everywhere.
Motivated from above example, we introduce the following definition.
Definition. Let (M, g) be a complete m-dimensional Riemannian manifold.
Let p ∈ M, v ∈ Tp M . If the exponential map expp is degenerate at v, i.e.
there exists w ∈ Tp M = Tv (Tp M ), such that
(expp )∗v (w) = 0,
then we call q = expp (v) ∈ M is the conjugate point of p along the curve
γ(t) = expp (tv).
Hence, by the inverse function theorem,
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Theorem. Suppose p ∈ M , v ∈ Tp M , and q = expp v. Then expp is local
diffeomorphism in a neighborhood of v if and only if q is not conjugate to p
along the geodesic γ(t) = expp (tv), t ∈ [0, 1].
The following theorem uses Jacobi field to describe the conjugate points.
Theorem. Let γ : [0, b] → M be a geodesic on a complete Riemannian
manifold (M, g). p = γ(0), q = γ(b). Then q is a conjugate point of p
along the curve γ if and only if there exists a nonzero Jacobi field J = J(t),
satisfying
J(0) = J(b) = 0.
Proof. We assume that b = 1. Identifying Tv (Tp M ) with Tp M as usual, we
can compute
d
(exp)∗,v (w) = |s=0 expp (v + sw).
dt
To compute this, we define a geodesic variation of γ, as before, by
Γ(s, t) = expp t(v + sw).
Then, its variation field JW (t) = ∂s Γ(s, t)|s=0 is a Jacobi field and JW (1) =
exp∗,v W. Therefore, exp∗,v fails to be an isomorphism when exp∗,v W = 0
for some W , which occurs precisely when there is a nonzero Jacobi field JW ,
satisfying JW (0) = JW (1) = 0. Conversely, assume there exists a nonzero
Jacobi field J = J(t), satisfying J(0) = J(b) = 0. We let w = J 0 (0),
and consider Γ(s, t) = expp t(v + sw). Then, its variation field JW (t) =
∂s Γ(s, t)|s=0 is a Jacobi field with Jw0 (0) = w. By the uniqueness, we have
J = Jw . Hence JW (1) = exp∗,v w = J(1) = 0. So q = γ(1) is a conjugate
point of p. This proves the theorem.
Corollary. Let p, q ∈ M . If q = γ(b) is the conjugate point of p = γ(0)
along the geodesic γ(t), t ∈ [0, b], then p is the conjugate point of q along the
geodesic γ̃(t) = γ(b − t).
Theorem. Let γ : [0, b] → M be a geodesic on a complete Riemannian
manifold (M, g). p = γ(0), q = γ(b). If q is not a conjugate point of p along
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the curve γ, then for every v ∈ Tp M, w ∈ Tq M , there exists a unique Jacobi
field with J(0) = v, J(b) = w.
Proof. Recall that from Corollary 3.2, we know that
J(t) = t(expγ (0) )∗tγ 0 (0) (J 0 (0))
is a Jacobi field along γ with J(0) = 0. So to construct Jacobi field with
J(0) = v, J(b) = w, we construct J1 with J1 (0) = 0, J1 (b) = w and J2 (0) =
v, J2 (b) = 0 respectively. To do so, we need pass the condition “J1 (b) = w to
J10 (b) = w̃”.
From the assumption, since γ(t) = expp (tγ 0 (0)) so q = γ(b) = expp (bγ 0 (0)),
the map (expp )∗bγ 0 (0) is non-singular, so it is a linear isomorphism. So there
exists w̃ ∈ Tbγ 0 (0) (Tp M ), such that
w = (expp )∗bγ 0 (0) (w̃).
For every t ∈ [0, b], let
t
J1 (t) = (expp )∗tγ 0 (0) (w̃),
b
Then J1 is Jacobi field along the curve γ with
J1 (0) = 0, J1 (b) = (expp )∗bγ 0 (0) (w̃) = w.
On the other hand, since p is not a conjugate point of q, there exists a Jacobi
field J2 along the curve γ with J2 (0) = v, J2 (b) = 0. Let
J(t) = J1 (t) + J2 (t).
Then J is Jacobi field along the curve γ with J(0) = v, J(b) = w.
We now prove the uniqueness. If J˜ 6= J. Then J¯ = J˜ − J is a Jacobi
¯
¯
field along the curve γ with J(0)
= J(b)
= 0. This contradicts with the
assumption that q is a conjugate point of p. This finishes the proof.
14
5
The first and second variation formulas
Let γ : [a, b] → M be a differentiable curve with kγ 0 (t)k = 1 (unit-speed).
The lenght of γ is
l = L(γ) =
Z b
kγ 0 (t)kdt = |γ 0 (0)|(b − a).
a
A variation of γ is a differential map Γ(s, t) : (−, ) × [a, b] → M with
Γ(0, t) = γ(t) for all t ∈ [a, b]. Write γ s := Γ(s, ·), and its arc-length is
L(s) = L(γ s ) =
Z b
a
kγ 0s (t)kdt.
Let
T (s, t) = ∂t Γ(s, t), S(s, t) = ∂s Γ(s, t)
and let V (t) = S(0, t) be the variation vector field of the variation of Γ. Then
the first variation formula is, assuming that Γ is smooth on [ai−1 , ai ],
Z ai
d
< Dt S, T >
L(γ s )|[ai−1 ,ai ] =
dt.
ds
ai−1 < T, T >1/2
In particular, it shows that a geodesic is the critical point of the arc-length
functional. The geodesic is normal (namely, the parameter t is its arc-length)
if |γ 0 (t)| = 1, i.e. l = b − a.
Using the the first variation formula, we can show that every minimizing
curve is geodeisc when it it given by a unit-speed parametrization.
Assume Γ is smooth on [ai−1 , ai ]. Then
d2
L(γ s )|[ai−1 ,ai ]
ds2
!
Z ai
< Ds Dt S, T > < Dt S, Ds T > 1 < Dt S, T > 2 < Ds T, T )
=
+
−
dt
< T, T >1/2
< T, T >1/2
2
< T, T >3/2
ai−1
!
Z ai
< Dt Ds S + R(S, T )S, T > < Dt S, Dt S > < Dt S, T >2 < Ds T, T )
+
−
dt.
=
< T, T >1/2
< T, T >1/2
< T, T >3/2
ai−1
15
Now restricting to s = 0, where |T | = 1:
Z ai
d2
|
L(γ
)|
=
(< Ds Dt S, T > −R(S, T, T, S)
s=0
s [ai−1 ,ai ]
ds2
ai−1
+ |Dt S|2 − < Dt S, T >2 )dt|s=0 .
Because Dt T = 0 when s = 0, the first term above can be integrated as
follows:
Z ai
ai−1
< Ds Dt S, T > dt =
Z ai
ai−1
∂
i
< Ds S, T > dt =< Ds S, T > |t=a
t=ai−1 .
∂t
Notice that S(s, t) = 0 for all s at the end points t = a0 and t = ak = b,
so Ds S = 0 there. Moreover, along the boundaries {t = ai } of the smooth
regions, Ds S = Ds (∂s Γ) depends only on the value of Γ when t = ai , and
it is smooth up to the line {t = ai } from both sides; therefore Ds S is continuous for all (s, t). Thus when we inerst the above identity, the boundary
contributions from the first term all cancel, and we get
Z b
d2
|
L(γ
)
=
(|Dt S|2 − < Dt S, T >2 −R(S, T, T, S))dt|s=0
s=0
s
ds2
a
=
Z b
a
(|Dt V |2 − < Dt V, γ 0 >2 −R(V, γ 0 , γ 0 , V ))dt.
Let V ⊥ be the normal component of V , which is orthogonal to γ 0 , i.e.
U ⊥ = U − < U, γ 0 > γ 0 .
Then
0
V 0 = V ⊥ + < V 0, γ 0 > γ 0.
So
0
0
0
< V 0 , V 0 > = < V ⊥ , V ⊥ > + < V ⊥ , γ 0 >< V 0 , γ 0 >
+ < V 0 , γ 0 >2 < γ 0 , γ 0 >
0
0
0
= < V ⊥ , V ⊥ > + < V ⊥ , γ 0 >2 ,
where we used
0
< V ⊥ , γ 0 >=
∂
< U ⊥ , γ 0 > − < U ⊥ , Dt γ 0 >= 0.
∂t
16
Hence, we have that the second variation formula is
Z b
d2
|
L(γ
)
=
(|Dt V ⊥ |2 − R(V ⊥ , γ 0 , γ 0 , V ⊥ ))dt.
s=0
s
ds2
a
It should come as no surprise that the second variation depends only on
the normal component of U ; intuitively, the tangential components of U
contributes only to a reparametrization of γ, and the length is independent
of parametrization. For this reason, we general apply the second variation
formula only to variations whose variation fields are proper (fix-end points
variation) and normal.
We define a symmetric bilinear form I, called the index form, on the space
of proper normal vector fields along γ by
I(V, W ) =
Z b
a
{< Dt V, Dt W > + < R(γ 0 , V )γ 0 , W >}dt.
You should think of I(V, W ) as sort of “Hessian” or second order derivative
of the length functional. Then, for every proper variation of γ,
L00 (0) = I(V ⊥ , V ⊥ ).
Because every proper normal vector field along γ is the variation field of
some proper variation, it can be rephrased on terms of the index form in the
following way
Corollary. If F is a proper variation of a geodesic γ whose variation field
is a proper normal variation field V , the second variation of L(γ) is I(V, V ),
i.e.L00 (0) = I(V, V ). In particular, if γ is minimizing, then I(V, V ) ≥ 0 for
any proper normal vector field along γ.
The next Proposition gives another expression for I, which makes the role
of the Jacobi equation more evident.
Proposition 5.1. For any pair of proper normal vector fields V, W along a
geodesic segment γ,
I(V, W ) = −
Z b
a
< Dt2 V + R(V, γ 0 )γ 0 , W > dt −
k
X
i=1
17
< 4i Dt V, W (ai ) >,
where {ai } are the points where V is not smooth, and 4i V 0 is the jump on
V 0 at t = ai .
Proof. On any interval [ai−1 , ai ] where V and W are smooth,
d
< Dt V, W >=< Dt2 V, W > + < Dt V, Dt W > .
dt
Thus, by the fundamental theorem of calculus,
Z ai
ai−1
< Dt V, Dt W > dt = −
Z ai
ai−1
< Dt2 V, W > + < Dt V, W > |aaii−1 .
Summing over i, and noting that W is continuous at t = ai , and W (a) =
W (b) = 0, we get the identity. This finishes the proof.
6
Geodesics do not minimize past conjugate
points
In this section, we use the second variation to prove another extremely important fact about conjugate points: No geodesic is minimizing past its first
conjugate point.
Theorem. If γ is a geodesic segment from p to q that has an interior
conjugate point to p, then there exists a proper normal vector field X along
γ such that I(X, X) < 0. In particular, γ is not minimizing.
Proof. Suppose γ : [0, b] → M is a unit-speed parametrization of γ and γ(a)
is the first conjugate to γ(0) for some 0 < a < b. This means that there is
a non-trivial normal Jacobi field J along γ|[0,a] that vanishes at t = 0 and
t = a. Define a vector field V along γ by V (t) = J(t) for t ∈ [0, a], V (t) = 0
for t ∈ [a, b]. and X(t) = 0 for t ∈ [a + , b]. This is a proper, normal,
piecewise smooth vector field along γ.
Let W (t) be a smooth proper normal vector field along γ such that W (b)
is equal to the jump 4V 0 at t = a. Such a vector field is easily constructed
in a local coordinates and extended to all γ by a bump function. Note that
4V 0 = −J 0 (a) is not zero, because otherwise J would be a Jacobi field
satisfying J(a) = J 0 (a) = 0, and thus would be identically zero.
18
For small positive , let X = V + W . Then
I(X , X ) = I(V, V ) + 2I(V, W ) + 2 I(W, W ).
Since V satisfies the Jacobi equation on each subinterval [0, a] and [a, b], and
V (a) = 0, by the Proposition above,
I(V, V ) = −
Z a
< V 00 + R(V, γ 0 )γ 0 , V > dt −
0
Z b
< V 00 + R(V, γ 0 )γ 0 , V > dt
a
− < 4V 0 , V (a) >= − < 4V 0 , V (a) > .
Similarly,
I(V, W ) = − < 4V 0 , W (b) >= −|W (b)|2 .
Thus
I(X , X ) = −2|W (b)|2 + 2 I(W, W ).
If we choose small enough, this is strictly negative. This proves the theorem.
In contrast to the above theorem, we prove
Theorem. Let γ : [a, b] → M be geodesic. It there does not exist a point
conjugate to γ(a) along γ, then there exists > 0 with the property that
for any piecewise smooth curve g : [a, b] → M with g(a) = γ(a), g(b) =
γ(b), d(g(t), γ(t)) < for all t ∈ [a, b], we have that L(g) ≥ L(c) with equality
if and only if g is a reparametrization of c.
Proof. We want to apply Corollary 3.2.3. WLOG, we assume that a =
0, b = 1 and v := γ(0). Since there are no points conjugate to γ(a), the
exponential map expp is of maximal rank along any radical curve tv, 0 ≤ t ≤
1. Thus, by inverse function theorem, for each t, expp is diffeomorphism in a
suitable neighborhood of tv. We cover {tv, 0 ≤ t ≤ 1} by finitely many such
neighborhoods {Ωi , 1 ≤ i ≤ k} and let Ui := expp Ωi .
Let us assume that tv ∈ Ωi for ti−1 ≤ t ≤ ti . If > 0 is sufficiently small,
we have for any curve g : [0, 1] → M satisfying the assumption d(g(t), γ(t)) <
,
g([ti−1 , ti ]) ⊂ Ui .
Let c(t) = (expp |Ωi )−1 (g(t)) for ti−1 ≤ t ≤ ti , then c : [0, 1] → Tp M with
expp ◦ c = g, c(0) = 0, c(1) = v. Thus Corollary 3.2.3 implies our statement.
19
7
Curvature and topology
By means of the second variation of geodesics we can prove the following
Theorem (Myers and Bonnet). Let (M, g) be a complete Riemannian
manifold. If
(i) (Myers) r(M ) ≥ (n − 1)k for some constant k > 0, or
√(ii) (Bonnet) KM ≥ k > 0, then the diameter of M satisfying d(M ) ≤
π/ k, so that M is actually compact.
Proof. Suppose the contrary. There there are points p, q ∈ M and (HopfRinow theorem)
a minimal unit-speed curve γ connecting p and q with length
√
l > π/ k. Fix a normalized minimal geodesic γ : [0, l] → M (with unitspeed) and let {ei } be an orthonormal basis of parallel fields along γ such
that e1 = γ 0 = T . Let Wi (t) = sin(πt/l)ei (t) be vector fields along γ. Then
2
Wi0 (t) = πl cos(πt/l)ei (t), Wi00 (t) = − πl2 sin(πt/l)ei (t). Hence
I(Wi , Wi ) = −
=
Z l
Z l
0
< 52T Wi − R(T, Wi )T, Wi > dt
(sin(πt/l))2 {
0
π2
+ R(T, ei )T, ei >}dt
l2
from which it follows that
n
X
i=2
I(Wi , Wi ) =
Z l
(sin(πt/l))2 {(n − 1)
0
π2
− r(T, T )}dt.
l2
√
So if r(M ) ≥ (n − 1)k > 0 and l > π/ k, then the sum therefore at
least one summand must be negative. But this is impossible since some
I(Wi , Wi ) < 0 implies that the second variation of arclength is in some
direction negative, so that γ cannot be minimal. This proves the theorem.
Remark: (a) In case (ii) (i.e. the Bonnet’s condition), we can just take
W (t) = sin(πt/l)E(t), where E(t) is any parallel normal unit vector field
along γ.
(b) The sphere S n (r) := {x ∈ Rn+1 | |x| = r} of radius r has (sectional)
curvature r12 , hence Ricci curvature n−1
and diameter πr. The theorem above
r2
20
means that if If M has Ricci curvature not less than the one of S n (r), then
the diameter of M is at most one of S n (r).
Theorem (Cartan-Hadama). Let (M, g) be a complete Riemnannian
manifold with nonpositive sectional curvature. Then M has no conjugate
points.
Proof. Let γ : [0, l] → M be a normalized geodesic, and X be a vector field,
normal to γ, then
I(X, X) =
Z l
0
{| 5T X|2 − |X|2 KM (X ∧ T )}ds ≥ 0.
Then by the corollary above, M has no conjugate points. This proves the
theorem.
8
Some Comparison Theorems
We’ll prove, in this section, that an upper bound on sectional curvature
produces a lower bound on Jacobi fields. In particular, we recall that we
three model space S n , Rn and H n of curvature 1, 0, -1. Let γ(t) be a geodesic
with |γ 0 (t)| = 1, w ∈ Tγ (0) M where M ∈ {S n , Rn , H n }. Then the Jacobi
field J(t) along γ with J(0) = 0, J 0 (0) = w is given by (sin t)W, tW, (sinh t)W
respectively, where W is the parallel vector field along γ with W (0) = w. So
it is natural to compare with these spaces.
Our starting point is the following very classical comparison theorem for
ordinary differential equations.
Theorem (Sturm Comparison Theorem). Suppose u and v are differentiable real-valued functions on [0, t], twice differentiable on (0, T ), and u > 0
on (0, T ). Suppose further that u and v satisfy
u”(t)+a(a)u(t) = 0, v”(t)+a(t)v(t) ≥ 0, u(0) = v(0) = 0, u0 (0) = v 0 (0) > 0
for some function a : [0, t] → R. Then v(t) ≥ u(t) on [0, T ].
Proof. Consider the function f (t) = v(t)/u(t) defined on (0, T ). It follows
from l’Hopital’s rule that limt→0 f (t) = v 0 (0)/u0 (0) = 1. Since f is differentiable on (0, T ), if we could show that f 0 ≥ 0 there it would follow from
21
elementary calculus that f ≥ 1 and therefore v ≥ u on (0, T ), and by continuity also on [0, T ]. Differentiating
d v
v 0 u − vu0
=
.
dt u
u2
Thus to show that f 0 ≥ 0 it would suffice to show that v 0 u − vu0 ≥ 0.
Since v 0 (0)u(0) − v(0)u0 (0) = 0, we need only show this expression has no
nonnegative derivative. Differentiating again and substituting the ODE for
u,
d 0
(v u − vu0 ) = v 00 u + v 0 u − v 0 u0 − vu00 = v 00 u + avu ≥ 0.
dt
This proves the theorem.
Theorem (Jacobi Field Comparison Theorem). Suppose that (M, g)
is a Riemannian manifold with all sectional curvature bounded above by a
constant C. If γ is a unit speed geodesic in M , and J is any normal Jacobi
field along γ such that J(0) = 0, then
(1) If C = 0, |J(t)| ≥ t|J 0 (0)| for t ≥ 0;
(2) If C =
1
,
R2
|J(t)| ≥ R(sin Rt )|J 0 (0)| for 0 ≤ t ≤ πR;
If C = − R12 , |J(t)| ≥ R(sinh Rt )|J 0 (0)| for t ≥ 0;
Proof. The function |J(t)| is smooth wherever J(t) 6= 0. Using the Jacobi
equation, we compute
d2
d < J 0, J >
|J|
=
dt2
dt < J, J >1/2
< J 00 , J >
< J 0, J 0 >
< J 0, J >
=
+
−
< J, J >1/2 < J, J >1/2 < J, J >3/2
< R(J, γ 0 )γ 0 , J > |J 0 |2 | < J 0 , J >2
=
+
−
.
|J|
|J|
|J|3
Using the Schwarz inequality, < J 0 , J >2 ≤ |J 0 |2 |J|2 , so the sum of the last
two terms above is nonnegative. Thus
R(J, γ 0 , γ 0 , J)
d2
|J|
≥
−
.
dt2
|J|
22
Since < J, γ 0 >= 0 and |γ 0 | = 1, R(J, γ 0 , γ 0 , J)/|J|2 is the sectional curvature
of the plane spanned by J and γ 0 . Therefore our assumption on the sectional
curvature of M guarantees that R(J, γ 0 , γ 0 , J) ≤ C|J|2 , so
d2
|J| ≥ −C|J|
dt2
wherever |J| > 0.
We wish to use the Sturm comparison theorem to compare |J| with the
solution u to u00 + Cu = 0, where u(t) = t if C = 0, u(t) = R sin Rt if
C = 1/R2 , and u(t) = R sinh Rt if C = −1/R2 . To do so, we need to arrange
d|J|/dt = 1 at t = 0, because u0 (0) = 1. Multiplying J by a positive constant,
we may assume, WLOG, that |J 0 (0)| = 1. Also J can be written near t = 0
as J(t) = tW (t) where W is a smooth vector field (the proof is omitted).
Therefore
d
|J(t)| − |J(0)|
|J||t=0 = lim
t→0
dt
t
t|W (t)
= lim
= |W (0)| = |J 0 (0)| = 1.
t→0
t
Now the Sturm comparison theorem implies that |J| ≥ u, provided |J| is
nonzero (to ensure that it is smooth). The fact that d|J|/dt = 1 at t = 0
means that |J| > 0 on some interval (0, ), and |J| cannot attain its first zero
before u does without contradicting the estimate |J| ≥ u. Thus |J| ≥ u as
long as u ≥ 0, which proves the theorem.
9
The Hessian and Laplace Comparison Theorem
The Comparison theorem is one of the most important tools in the study
of analysis on manifolds. Basically it is through the connections between
the Jacobi field and the curvature of the manifold, as well as its curvature
properties, to give more general properties of the manifold itself. On the
other hand, from the Jacobi equation point view, it is also the application
of the differential equations to the geometry. For example, the proof of the
Bonnet theorem is the application of the Sturm-Liouville theory.
23
The purpose of the section is to study the curvature and harmonic functions, so we first derive the Hessian and Laplace Comparison Theorem, and
then give some of its consequences.
Let M be an n-dimensional complete Riemannian manifold with the metP
ric ds2 = gij dxi dxj . The Laplace-Bertrami operator is defined as
1 ∂ √ ij ∂
4= √
Gg
∂xj
G ∂xi
!
,
where G = det(gij ), (g ij = (gij )−1 .
There is a very natural function on M which is the distance function on
M with a point fixed, i.e. fix a point O ∈ M , define, fo rx ∈ M ,
ρ(x) = dist(o, x).
Obviously, ρ is not only continuous, but it also satisfies the Lipschitz condition. From the measure theory, it is differentiable everywhere.
For O ∈ M , consider the exponential map expo . From the Hopf-Rinow
theorem, expo is defined on To (M ), i.e. expo : To (M ) → M . For every
X ∈ To (M ), let γ(t) = exp(tX), then γ : R → M is a smooth geodesic
from X, when |t| is small, γ is the unique minimal geodesic connection γ(t)
and O, and d expo |tX : TtX (To M ) → Texpo (tX) M is a diffeomorphism. When
|t| increases, there’ll be two possibilities: (i) γ(t) is no longer the minimal
geodesic connecting γ(t) and O, (ii) d expo |tX is no longer a diffeomorphism,
in this case, we say γ(t) is a conjugate point to O.
Let
t0 = sup{t| γ is the unique minimal geodesic connection γ(t) and O}.
Then t0 ∈ R ∪ {+∞}. If t0 < +∞, we call γ(t0 ) is a cut point along γ,
relative to O. The set of all cut points along γ, relative to O is called a cut
locus, and is denoted by cut(O). From the definition, if x ∈ Cut(O), then
either (i) x is the first conjugate point to O along γ, or (ii) there are at least
two minimal geodesics with the same arc-length connecting x and O.
Obviously, for X ∈ To M with |X| = 1, there is at most one cut point
along expo (tX) (t > 0), hence cut(O) is the image of the a closed set of
24
S n−1 under the map expo , so its n-dimensional measure is 0. Further, let
µ(X) = dist(O, γ(t0 )), where X ∈ S n−1 ⊂ To M . Define
E = {tX | 0 ≤ t < µ(X), X ∈ S n−1 ⊂ To M },
then expo : E → expo (E) is a diffeomorphism, hence it induces a normal
coordinate frame. Obviously, it is the largest possible normal coordinate
frame with o as the origin. Furthermore,
M = expo (E) ∪ cut(O).
From the definition, we can see that expo (E) is a star domain with origin
O, and ρ is smooth on expo (E). As the boundary of expo (E), there may
have X1 6= X2 ∈ cut(O), but expo (µ(X1 )X1 ) = expo (µ(X2 )X2 ). On the
points such that ρ is smooth, we have, since the geodesic has arc-length as
parameter,
| 5 ρ| = 1,
i.e.
g ij ρi ρj = 1, where ρi is the covariant derivative of ρ along the ∂i
direction.
P
Also, recall that
Ric : Tp M × Tp M → R
defined by Ric(X, Y ) =
< R(ei , X)Y, ei > where {ei } is an orthonormal
basis for Tp M . When e = en , then
P
Ric(e, e) =
n−1
X
K(ei , e),
i=1
where K is the sectional curvature.
Let f ∈ C 2 (M ), we now define the Hessian of f . For X, Y ∈ Tx M , extend
X, Y to vector fields X̃, Ỹ , we define
H(f )(X, Y ) = (X̃ Ỹ ((f )(x) − (5X̃ Ỹ f )(x).
Remark: H(f ) = 52 f = 5df which is a (2,0)-tensor field. Also, 4(f ) =
tr(52 f ) = tr(H(f )).
25
We now calculate H(ρ) (which is also H(r). Fix p ∈ M , for any point
x which is in the cut locus, let σ be the minimal geodesic connecting p and
x with σ(0) = p, σ(r) = x. Take X ∈ Tx M, < X.∂/∂r > (x) = 0. Since x
is not a conjugate point, we can extend X to a Jacobi field X̃ along σ with
X̃(σ(0)) = 0, X̃(σ(r)) = X. Since every Jacobi field X̃ is the variation field
of some variation of σ, by the symmetry lemma, we have
"
#
∂
X̃,
= 0 (0 ≤ t ≤ r).
∂t
Hence
H(r)(X, X) = X̃ X̃r − (5X̃ X̃r)
+ *
+
*
∂
∂
− 5X̃ X̃,
= X X̃,
∂r
∂r
*
+
E
D
∂
=
X̃, 5X̃
= X̃, 5 ∂ X̃ ,
∂r
∂r
"
#
∂
where in the last equality, we used X̃,
= 0 (0 ≤ t ≤ r).
∂t
26