3Geodesics

III Riemannian Geometry 3.2 Jacobi fields
Fix a Riemannian manifold
M
. Let’s imagine that we have a “manifold” of all
smooth curves on
M
. Then this “manifold” has a “tangent space”. Morally,
given a curve
γ
, a “tangent vector” at
γ
in the space of curve should correspond
to providing a tangent vector (in M) at each point along γ:
Since we are interested in the geodesics only, we consider the “submanifold” of
geodesics curves. What are the corresponding “tangent vectors” living in this
“submanifold”?
In rather more concrete terms, suppose
f
s
(
t
) =
f
(
t, s
) is a family of geodesics
in
M
indexed by
s
(
ε, ε
). What do we know about
f
s
s=0
, a vector field
along f
0
?
We begin by considering such families that fix the starting point
f
(0
, s
), and
then derive some properties of
f
s
in these special cases. We will then define a
Jacobi field to be any vector field along a curve that satisfies these properties.
We will then prove that these are exactly the variations of geodesics.
Suppose
f
(
t, s
) is a family of geodesics such that
f
(0
, s
) =
p
for all
s
. Then
in geodesics local coordinates, it must look like this:
For a fixed p, such a family is uniquely determined by a function
a(s) : (ε, ε) T
p
M
such that
f(t, s) = exp
p
(ta(s)).
The initial conditions of this variation can be given by a(0) = a and
˙a(0) = w T
a
(T
p
M)
=
T
p
M.
We would like to know the “variation field” of
γ
(
t
) =
f
(
t,
0) =
γ
p
(
t, a
) this
induces. In other words, we want to find
f
s
(
t,
0). This is not hard. It is just
given by
(d exp
p
)
ta
0
(tw) =
f
s
(t, 0),
As before, to prove something about
f
, we want to make good use of the
properties of
. Locally, we extend the vectors
f
s
and
f
t
to vector fields
t
and
s
. Then in this set up, we have
˙γ =
f
t
=
t
.
Note that in
f
t
, we are differentiating
f
with respect to
t
, whereas the
t
on
the far right is just a formal expressions.
By the geodesic equation, we have
0 =
dt
˙γ =
t
t
.
Therefore, using the definition of the curvature tensor R, we obtain
0 =
s
t
t
=
t
s
t
R(
s
,
t
)
t
=
t
s
t
+ R(
t
,
s
)
t
We let this act on the function f . So we get
0 =
dt
ds
f
t
+ R(
t
,
s
)
f
t
.
We write
J(t) =
f
s
(t, 0),
which is a vector field along the geodesic γ. Using the fact that
ds
f
t
=
dt
f
s
,
we find that J must satisfy the ordinary differential equation
2
dt
2
J + R( ˙γ, J) ˙γ = 0.
This is a linear second-order ordinary differential equation.
Definition
(Jacobi field)
.
Let
γ
: [0
, L
]
M
be a geodesic. A Jacobi field is a
vector field J along γ that is a solution of the Jacobi equation on [0, L]
2
dt
2
J + R( ˙γ, J) ˙γ = 0. ()
We now embark on a rather technical journey to prove results about Jacobi
fields. Observe that ˙γ(t) and t ˙γ(t) both satisfy this equation, rather trivially.
Theorem.
Let
γ
: [0
, L
]
N
be a geodesic in a Riemannian manifold (
M, g
).
Then
(i) For any u, v T
γ(0)
M, there is a unique Jacobi field J along Γ with
J(0) = u,
J
dt
(0) = v.
If
J(0) = 0,
J
dt
(0) = k ˙γ(0),
then
J
(
t
) =
kt ˙γ
(
t
). Moreover, if both
J
(0)
,
J
dt
(0) are orthogonal to
˙γ
(0),
then J(t) is perpendicular to ˙γ(t) for all [0, L].
In particular, the vector space of all Jacobi fields along
γ
have dimension
2n, where n = dim M.
The subspace of those Jacobi fields pointwise perpendicular to
˙γ
(
t
) has
dimensional 2(n 1).
(ii) J
(
t
) is independent of the parametrization of
˙γ
(
t
). Explicitly, if
˜γ
(
t
) =
˜γ(λt), then
˜
J with the same initial conditions as J is given by
˜
J(˜γ(t)) = J(γ(λt)).
This is the kind of theorem whose statement is longer than the proof.
Proof.
(i)
Pick an orthonormal basis
e
1
, ··· , e
n
of
T
p
M
, where
p
=
γ
(0). Then
parallel transports
{X
i
(
t
)
}
via the Levi-Civita connection preserves the
inner product.
We take e
1
to be parallel to ˙γ(0). By definition, we have
X
i
(0) = e
i
,
X
i
dt
= 0.
Now we can write
J =
n
X
i=1
y
i
X
i
.
Then taking g(X
i
, ·) of () , we find that
¨y
i
+
n
X
j=2
R( ˙γ, X
j
, ˙γ, X
i
)y
j
= 0.
Then the claims of the theorem follow from the standard existence and
uniqueness of solutions of differential equations.
In particular, for the orthogonality part, we know that
J
(0) and
J
dt
(0)
being perpendicular to
˙γ
is equivalent to
y
1
(0) =
˙y
1
(0) = 0, and then
Jacobi’s equation gives
¨y
1
(t) = 0.
(ii) This follows from uniqueness.
Our discussion of Jacobi fields so far has been rather theoretical. Now that
we have an explicit equation for the Jacobi field, we can actually produce some
of them. We will look at the case where we have constant sectional curvature.
Example.
Suppose the sectional curvature is constantly
K R
, for
dim M
3.
We wlog |˙γ| = 1. We let J along γ be a Jacobi field, normal to ˙γ.
Then for any vector field T along γ, we have
hR( ˙γ, J) ˙γ, T i = K(g( ˙γ, ˙γ)g(J, T ) g( ˙γ, J)g( ˙γ, T )) = Kg(J, T ).
Since this is true for all T , we know
R( ˙γ, J) ˙γ = KJ.
Then the Jacobi equation becomes
2
dt
2
J + KJ = 0.
So we can immediately write down a collection of solutions
J(t) =
sin(t
K)
K
X
i
(t) K > 0
tX
i
(t) K = 0
sinh(t
K)
K
X
i
(t) K < 0
.
for i = 2, ··· , n, and this has initial conditions
J(0) = 0,
J
dt
(0) = e
i
.
Note that these Jacobi fields vanishes at 0.
We can now deliver our promise, proving that Jacobi fields are precisely the
variations of geodesics.
Proposition.
Let
γ
: [
a, b
]
M
be a geodesic, and
f
(
t, s
) a variation of
γ(t) = f(t, 0) such that f (t, s) = γ
s
(t) is a geodesic for all |s| small. Then
J(t) =
f
s
is a Jacobi field along ˙γ.
Conversely, every Jacobi field along
γ
can be obtained this way for an
appropriate function f.
Proof.
The first part is just the exact computation as we had at the beginning of
the section, but for the benefit of the reader, we will reproduce the proof again.
2
J
dt
=
t
t
f
s
=
t
s
f
t
=
s
t
f
t
R(
t
,
s
) ˙γ
s
.
We notice that the first term vanishes, because
t
f
t
= 0 by definition of geodesic.
So we find
2
J
dt
= R( ˙γ, J) ˙γ,
which is the Jacobi equation.
The converse requires a bit more work. We will write
J
0
(0) for the covariant
derivative of
J
along
γ
. Given a Jacobi field
J
along a geodesic
γ
(
t
) for
t
[0
, L
],
we let ˜γ be another geodesic such that
˜γ(0) = γ(0),
˙
˜γ(0) = J(0).
We take parallel vector fields X
0
, X
1
along ˜γ such that
X
0
(0) = ˙γ(0), X
1
(0) = J
0
(0).
We put X(s) = X
0
(s) + sX
1
(s). We put
f(t, s) = exp
˜γ(s)
(tX(s)).
In local coordinates, for each fixed s, we find
f(t, s) = ˜γ(s) + tX(s) + O(t
2
)
as t 0. Then we define
γ
s
(t) = f(t, s)
whenever this makes sense. This depends smoothly on
s
, and the previous
arguments say we get a Jacobi field
ˆ
J(t) =
f
s
(t, 0)
We now want to check that
ˆ
J
=
J
. Then we are done. To do so, we have to
check the initial conditions. We have
ˆ
J(0) =
f
s
(0, 0) =
d˜γ
ds
(0) = J(0),
and also
ˆ
J
0
(0) =
dt
f
s
(0, 0) =
ds
f
t
(0, 0) =
X
ds
(0) = X
1
(0) = J
0
(0).
So we have
ˆ
J = J.
Corollary. Every Jacobi field J along a geodesic γ with J(0) = 0 is given by
J(t) = (d exp
p
)
t ˙γ(0)
(tJ
0
(0))
for all t [0, L].
This is just a reiteration of the fact that if we pull back to the geodesic local
coordinates, then the variation must look like this:
But this corollary is stronger, in the sense that it holds even if we get out of the
geodesic local coordinates (i.e. when exp
p
no longer gives a chart).
Proof.
Write
˙γ
(0) =
a
, and
J
0
(0) =
w
. By above, we can construct the variation
by
f(t, s) = exp
p
(t(a + sw)).
Then
(d exp
p
)
t(a+sw)
(tw) =
f
s
(t, s),
which is just an application of the chain rule. Putting
s
= 0 gives the result.
It can be shown that in the situation of the corollary, if
a w
, and
|a|
=
|w| = 1, then
|J(t)| = t
1
3!
K(σ)t
3
+ o(t
3
)
as t 0, where σ is the plane spanned by a and w.