LECTURE 17: THE INDEX FORM
1. Proof of The Jacobi theorem
We want to prove
Theorem 1.1 (Jacobi). Let γ : [a, b] → M be a geodesic from p = γ(a) to q = γ(b).
(1) If there is no conjugate points of p along γ, then there exists ε > 0 so
that for any piecewise smooth curve γ̄ : [a, b] → M from p to q satisfying
dist(γ(t), γ(t)) < ε, we have L(γ̄) ≥ L(γ), with equality holds if and only if
γ̄ is a reparametrization of γ.
(2) If there exists t̄ ∈ (a, b) so that q̄ = γ(t̄) is a conjugate point of p, then there
is a proper variation of γ so that L(γs ) < L(γ) for 0 < |s| < ε.
Before we prove the theroem, we will first prove
Lemma 1.2. Suppose Ep = E ∩Tp M contains a line segment [0, l]Xp . Let ϕ : [a, b] →
Ep be a piecewise smooth curve with ϕ(a) = 0, ϕ(b) = lXp . Consider the geodesic
γ(t) := expp (tXp ),
0≤t≤l
γ(t) := expp (ϕ(t))
a ≤ t ≤ b.
and the curve
Then
L(γ) ≥ L(γ).
Moreover, if expp is nonsingular along [0, l]Xp , then
L(γ) > L(γ)
unless γ is a reparametrization of γ.
Proof. WLOG, we may assume ϕ(t) 6= 0 for all t ∈ (a, b). Write
ϕ(t) = r(t)e(t),
where r(t) = |ϕ(t)| and e(t) ∈ Sp M has unit length. Then
ϕ̇(t) = ṙ(t)e(t) + r(t)ė(t),
and e(t) ⊥ ė(t). According to Gauss lemma,
h(d expp )ϕ(t) e(t), (d expp )ϕ(t) e(t)i = he(t), e(t)i = 1,
h(d expp )ϕ(t) e(t), (d expp )ϕ(t) ė(t)i = he(t), ė(t)i = 0.
It follows that
|γ̇(t)| = |(d expp )ϕ(t) ϕ̇(t)| ≥ |(d expp )ϕ(t) (ṙ(t)e(t))| = |ṙ(t)|.
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LECTURE 17: THE INDEX FORM
Therefore,
Z
b
|ṙ(t)|dt ≥ |r(b) − r(a)| = ||ϕ(b)| − |ϕ(a)|| = l|Xp | = L(γ).
L(γ) ≥
a
If expp is nonsingular along [0, l]Xp , then for ε small enough expp is nonsingular
in the ε-neighborhood of [0, l]Xp . Now suppose the equality holds, then we have
(d expp )ϕ(t) ė(t) = 0,
thus ė(t) = 0 for all t. It follows that e(t) is a constant unit vector, i.e.
ϕ(t) = r(t)e
for some e ∈ Sp M . Obviously e = Xp /|Xp | is the direction vector of Xp . Moreover,
r(t)
ṙ cannot change sign. So γ(t) = expp ( |X
Xp ) is a reparametrization of γ.
p|
Proof of the Jacobi theorem.
(1) Denote l = b − a. Then by the assumption, expp is regular at sγ̇(a) for all
0 ≤ s ≤ l. As a consequence, one can find a subdivision
a = t0 < t1 < · · · < tk < tk+1 = b
and open neighborhoods Vi , 1 ≤ i ≤ k, of the line segment [ti , ti+1 ]γ̇(a) in Tp M so
that expp is a diffeomorphism on each Vi . We denote Ui = expp (Vi ). According to
our assumption on γ, for ε small enough, γ([ti , ti+1 ]) ⊂ Ui . Now we define
ti−1 ≤ t ≤ ti .
ϕ(t) = (expp V )−1 (γ̄(t)),
i
Then ϕ(t) is a piecewise smooth curve in Tp M connecting 0 to lγ̇(a) so that
expp (ϕ(t)) = γ(t).
So the conclusion follows from the lemma above.
(2) Let X be a nonzero Jacobi field along γ with X(a) = 0, X(t̄) = 0. Note that
∇γ̇(t̄) X 6= 0, otherwise X will be identically zero. Now let Z be an arbitrary smooth
vector field along γ with
Z(a) = 0, Z(b) = 0, Z(t̄) = −∇γ̇(t̄) X.
We will denote the restriction of Z to intervals [a, t̄ ] and [ t̄, b] by Z1 and Z2
respectively. For any η > 0 we put
1
Yη = X(t) + ηZ1 (t), for a ≤ t ≤ t̄,
Yη (t) :=
Yη2 = ηZ2 (t),
for t̄ ≤ t ≤ b.
Then
f (t, s) = expγ(t) sYη (t)
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3
is a proper variation of the geodesic γ whose variation field is Yη . Recall: in lecture
12 we showed that for a variation γs of a geodesic γ with variation field V , one has
the following second variation formula
Z b
d2 E(γs )
(hV (t), R(γ̇, V )γ̇(t)i − h∇γ̇ V, ∇γ̇ V i) dt
(0) = −
ds2
a
∂f
∂f
− h∇V (a) , γ̇(a)i + h∇V (b) , γ̇(b)i.
∂s
∂s
For simplicity we will denote
Z b
(h∇γ̇ X, ∇γ̇ Xi − R(γ̇, X, γ̇, X)) dt.
I(X, X) :=
a
Then since f is a proper variation, the second variation formula simplifies to
d2 E(γs )
(0) = I(Yη , Yη ) = I(Yη1 , Yη1 ) + I(Yη2 , Yη2 ).
2
ds
Since X is a Jacobi field, using integration by parts one gets (see (*) below)
I(Yη1 , Yη1 ) = −η|∇γ̇(t̄) X|2 + η 2 I(Z1 , Z1 ).
It follows that
d2 E(γs )
(0) = I(Yη , Yη ) = −η|∇γ̇(t̄) X|2 + η 2 I(Z1 , Z1 ) + η 2 I(Z2 , Z2 ) < 0
2
ds
for η small enough. This proves the theorem.
2. The Index Form
Let γ : [a, b] → M be a geodesic. We can polarize I(X, X) that appeared in the
proof above and define the following symmetric bilinear form
Definition 2.1. The index form of geodesic γ is
Z b
I(X, Y ) =
(h∇γ̇ X, ∇γ̇ Y i − R(γ̇, X, γ̇, Y )) dt.
a
where X, Y are two piecewise smooth vector fields along γ.
Note that for a proper variation of a γ with variation field V ,
d2 E(γs )
(0) = I(V, V ).
ds2
Using a simple integration by parts argument, one easily gets
Z b
k
X
b
I(X, Y ) = −
h∇γ̇ ∇γ̇ X+R(γ̇, X)γ̇, Y idt+h∇γ̇ X, Y i|a − h∇γ̇(t+j ) X−∇γ̇(t−j ) X, Y i,
a
where tj ’s are those points where X is not smooth.
j=1
(∗)
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LECTURE 17: THE INDEX FORM
Now let V be the set of all piecewise smooth vector fields along γ, and let
V 0 = {X ∈ V | X(a) = 0, X(b) = 0.}
The following theorem relates the index form with Jacobi fields along γ:
Theorem 2.2. Let X ∈ V. Then X is a Jacobi field along γ if and only if for any
Y ∈ V 0 , I(X, Y ) = 0.
Proof. According to (∗), if Y ∈ V 0 and X is a Jacobi field (which has to be smooth),
then I(X, Y ) = 0.
Conversely suppose X ∈ V satisfies I(X, Y ) = 0 for all Y ∈ V 0 . We let
a = t0 < t1 < · · · < tk < tk+1 = b
be a subdivision of [a, b] so that X is smooth on each [tj , tj+1 ]. We take a smooth function f : [a, b] → R with f (ti ) = 0 for all i and f (t) > 0 for all
t 6∈ {t0 , t1 , · · · , tk+1 }. We define
Y = f (t)(∇γ̇ ∇γ̇ X + R(γ̇, X)γ̇).
Then Y ∈ V 0 and so
Z
b
0 = I(X, Y ) = −
f (t)|∇γ̇ ∇γ̇ X + R(γ̇, X)γ̇|2 dt.
a
It follows that X is a Jacobi field on each (tj , tj+1 ). Next let’s choose Z ∈ V 0 with
Z(ti ) = ∇γ̇(t+i ) X − ∇γ̇(t−i ) X.
Then
0 = I(X, Z) = −
k
X
|∇γ̇(t+i ) X − ∇γ̇(t−i ) X|2 .
i=1
1
It follows that X is of class C at each ti . By uniqueness, X is smooth. So V is a
Jacobi field.
Next let’s relate the index form with conjugate points.
Theorem 2.3. Let γ : [a, b] → M be a geodesic from p = γ(a) to q = γ(b).
(1) p has no conjugate point along γ ⇐⇒ the index form I is positive definite
on V 0 .
(2) q is the first conjugate point of p along γ ⇐⇒ the index form I is positive
semidefinite but not positive definite on V 0 .
(3) There is a point q̄ = γ(t̄)(a < t < b) conjugate to p = γ(a) along γ ⇐⇒
I(X, X) < 0 for some X ∈ V 0 .
Proof. (1) (=⇒) According to part (1) of Jacobi theorem, we know that if p = γ(0)
has no conjugate point along γ, then for any X ∈ V 0 , I(X, X) ≥ 0. (Otherwise one
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can construct a variatoin with L(γs ) < L(γ).) If I is not positive definite on V 0 ,
then I(Y, Y ) = 0 for some Y ∈ V 0 . It follows that for any Z ∈ V 0 and any λ ∈ R,
0 ≤ I(Y − λZ, Y − λZ) = −2λI(Y, Z) + λ2 I(Z, Z).
As a consequence, I(Y, Z) = 0 for any Z ∈ V 0 . In other words, Y is a Jacobi field.
Since q = γ(b) is not a conjugate point of p, and Y (a) = 0, Y (b) = 0, we must have
Y ≡ 0. So I is positive definition on V 0 .
(1) (⇐=) Conversely, suppose γ(t0 ) is conjugate to γ(a) along γ, and X a nonzero
Jacobi field along γ so that X(a) = X(t0 ) = 0. If we define Y to be the vector field
along γ which equals X on between γ(a) and γ(t0 ), and vanishes after γ(t0 ), then
obviously 0 6= Y ∈ V 0 and I(Y, Y ) = 0. So I is not positive definite.
(2) (=⇒) Fix any c ∈ (a, b). For any X ∈ V 0 , we may write X = X i (t)ei (t),
where {ei } are orthonormal and parallel along γ with e1 = γ̇. We let
b−a
(t − a))ei (t), a ≤ t ≤ c.
c−a
Then X c is a vector field along γ|[a,c] with X c (a) = 0, X c (c) = 0. Since p has no
conjugate point along γ|[a,c] , we get from (1) that I(X c , X c ) ≥ 0. It follows that
I(X, X) = limc→a I(X c , X c ) ≥ 0. So I is positively semidefinite on V 0 .
Obviously I is not positively definite on V 0 , since if X is a nonzero Jacobi field
along γ with X(a) = 0, X(b) = 0, then we have I(X, X) = 0.
(3) (=⇒) We have already seen this is the proof of Jacobi theorem part (2).
(3) (⇐=) This follows from (1) (=⇒) and (2) (=⇒).
(2) (⇐=) This follows from (1) (=⇒) and (3) (=⇒).
X c (t) = X i (a +
Corollary 2.4. Suppose p = γ(a) has no conjugate point along γ. If X is a Jacobi
field along γ, and Y ∈ V satisfies Y (a) = X(a), Y (b) = X(b), then I(X, X) ≤
I(Y, Y ), with equality holds if and only if X = Y .
Proof. Since X is a Jacobi field and Y has the same boundary condition as X, the
formula (*) implies
I(X, X) = I(X, Y ).
It follows that
0 ≤ I(X − Y, X − Y ) = I(X, X) − 2I(X, Y ) + I(Y, Y ) = −I(X, X) + I(Y, Y ),
with equality holds if and only if X − Y = 0.
Finally let’s state the Morse index theorem, which is a generalization of the
Jacobi theorem.
Let (M, g) be a Riemannian manifold, and γ : [a, b] → M be a geodesic.
Definition 2.5. We will call
ind(γ) = max dim{A ⊂ V 0 | I|A is negatively definite}
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LECTURE 17: THE INDEX FORM
the index of γ, and call
N (γ) = dim{X ∈ V 0 | I(X, Y ) = 0 for all Y ∈ V 0 }
the nullity of γ.
We have seen that X ∈ V 0 is in the nullspace of I if and only if X ∈ V 0 is a
Jacobi field along γ. This can happen only if γ(b) is a conjugate point of γ(a). It
follows
Lemma 2.6. The nullity N (γ) equals the multiplicity of γ(b) as a conjugate point
of γ(a). In particular, if γ(b) is not conjugate to γ(a), then N (γ) = 0.
In what follows we will denote by γ t the geodesic γ|[a,t] , so that the corresponding
index and nullity are ind(γ t ) and N (γ t ). Now we can state the main theorem:
Theorem 2.7 (Morse Index Theorem). For any geodesic γ : [a, b] → M , we have
X
N (γ t ) < ∞.
ind(γ) =
a<t<b
Proof. The will be a term paper for one of you.
As a consequence, we immediately get
Corollary 2.8. γ(a) has only finitely many conjugate points along γ. If we denote
these conjugate points (except possibly γ(b)) by γ(t1 ), · · · , γ(tk ) (a < t1 < · · · < tk <
b), then
k
X
ind(γ) =
n(tj ),
j=1
where n(tj ) is the multiplicity of γ(ti ).