Lemma 1. Let p ∈ M , let (xi ) be normal coordinates on a nbhd U of p, and let
γ be a radial geodesic starting at p. For any W = W i ∂i ∈ Tp M , the Jacobi field J
along γ such that J(0) = 0 and Dt J(0) = W is given in normal coordinates by
J(t) = tW i ∂i .
(1)
• Suppose γ is a unit speed geodesic in M . The normal Jacobi fields along γ
vanishing at t = 0 are precisely the vector fields J(t) = u(t)E(t), where E is any
parallel normal vector field along γ, and then Dt J(0) = u0 (0)E(0).
• Let (xi ) be Riemannian normal coordinates on a normal nbhd U of p,
let | · |g be the Euclidean norm in these coordinates,
and let τ be the radial distance function.
– For any g ∈ U \ {p} and V ∈ Tp M , write V = V > + V ⊥ , where V > is tangent
∂
to the sphere {r = constant} through q and V ⊥ is a multiple of ∂r
.
>
⊥
— By the Gauss lemma, the decomposition V = V + V is orthogonal, so
|V |2g = |V > |2g + |V ⊥ |2g .
Since
∂
∂r
is a unit vector in both the g and g norms, it is immediate that
|V ⊥ |g = |V ⊥ |g .
Thus we need only compute |V > |g .
Set X = V > , and let γ denote the unit speed radial geodesic from p to q.
— By Lemma 1, X is the value of a Jacobi field J along γ that vanishes at p, namely
X = J(r), where r = d(p, q) and
t
(2)
J(t) = X i ∂i .
r
Because J is orthogonal to γ 0 at p and q, it is normal.
— Now J can also be written in the form J(t) = u(t)E(t) as in Lemma 1. In this
representation,
Dt J(0) = u0 (0)E(0).
— Therefore, since E ia parallel and thus of constant length,
(3)
|X|2g = |J(r)|2g = |u(r)|2 |E(r)|2g = |u(r)|2 |E(0)|2g = |u(r)|2 |u0 (0)|2 |Dt J(0)|2g .
— Observe that
1 i X ∂i p
r
by (2). Since g agrees with g at p, we have
1 i
1
|Dt J(0)|g = X ∂i |g = |X|g .
r
r
g
Dt J(0) =
Inserting this into (6) and using (5) for u(r), we obtain
g(V, V ) = |V ⊥ |2g +
(u(r))2 (u0 (0))2 > 2
|V |g .
r2
— If u0 (0) = 1, we have
(4)
g(V, V ) = |V ⊥ |2g +
|J(r)|2g >
(u(r))2 >
⊥ 2
|V
|
=
|V
|
+
|V |g .
g
g
r2
r2
Typeset by AMS-TEX
1
2
Geodesic Spherical Coordinates
• Given p ∈ M , and a coordinate system ξ : O → ∂B1 (0p ), a coordinate system
H on M is determined by
H(t, u) = exp(tξ(u)).
The domain of the map H consists of the collection {(t, u)} in (0, +∞) × O for
which tξ(u) is in the domain of the exponential map E.
• Given ξ ∈ ∂B1 (0p ) we let
Pt;ξ : Tp M → Tγξ (t) M
denote the parallel translate along γξ , and we write
∂α ξ = ξ∗
∂t H = H∗
∂
,
∂t
∂
∂uα
∂ α H = H∗
∂
,
∂uα
∂
∂
∂
α = 1, · · · , n − 1, where ∂t
and ∂u
1 , · · · , ∂un−1 are natural coordinate vector
fields on (0, +∞) and O, respectively. Then
(∂t H)(t; ξ) = ∂t H and
exp tξ
(∂α H)(t; ξ) = ∂α H exp tξ
= γξ0 (t),
= Yα (t; ξ),
where Yα (t; ξ) is the Jacobi field along γξ determined by the initial conditions
Yα (0; ξ) = 0,
(Dt Yα )(0; ξ) = ∂α ξ.
— We have
|∂t H| = 1;
(4)
and since ∂α ξ ⊥ ξ, we have
(5)
h∂t H, ∂α Hi = 0.
(This is the content of the Gauss lemma).
— So the full knowledge of the Riemannian metric along γξ requires the study of
h∂α H, ∂β Hi(exp tξ) = hYα (t; ξ), Yβ (t; ξ)i
3
If M has constant sectional curvarure κ along γξ , then
Yα (t; ξ) = Sκ (t)Pt;ξ (∂α ξ);
hence
h∂α H, ∂β Hi(exp tξ) = S2κ (t)h∂α ξ, ∂β ξi,
where
Sκ (t) =

t,


√
√1 sin κt,
κ
√
√1 sinh −κt,
−κ


κ = 0;
κ > 0;
κ < 0.
This is what we encapsulated as
ds2 = dt2 + S2κ (t)|dξ|2 .
• For the general situation, we proceed as follows:
Given p ∈ M , ξ ∈ ∂B1 (0p ), let ξ ⊥ denote the orthogonal complement of Rξ ∈
Tp M . For each t > 0, let
R(t) =R(γξ0 (t), ·)γξ0 (t),
−1
R(t) =Pt;ξ
◦ R(t) ◦ Pt;ξ .
— Since R(t) maps Rξ to 0, one only considers R(t) as genuinely acting on ξ ⊥ .
• We let A(t; ξ) to be the solution of the matrix (more precisely: linear transformation) ODE on ξ ⊥ :
A00 + R(t)A = 0,
satisfying the initial conditions
A(0; ξ) = 0,
A0 (0; ξ) = I.
Then, for each η ∈ ξ ⊥ , the vector field Y (t) along γξ , given by
Y (t) = Pt;ξ A(t; ξ)η
is the Jacobi field along γξ in J ⊥ = {Y ∈ J : hY, γ 0 i = 0 on [a, b]}, determined
by the initial conditions
Y (0) = 0,
(Dt Y )(0) = η.
— We therefore have
(∂α H)(t; ξ) = Yα (t; ξ) = τt;ξ A(t; ξ)∂α ξ,
which implies
(6)
h∂α H, ∂β Hi(exp tξ) = hA(t; ξ)∂α ξ, A(t; ξ)∂β ξi.
We write
ds2 = dt2 + |A(t; ξ)dξ|2 .
For M with constant sectional curvature κ, we have
A(t; ξ) = Sκ (t)I.
4
Riemannian measure
• Start with the formula for change of variables of integral calculus:
Let D, Ω be domains in Rn , n ≥ 1, and let
ϕ:D→Ω
be a C 1 diffeomorphism, and let Jϕ (x) denote the Jacobian matrix associated to
ϕ at x. Then, for any L1 function f on Ω, we have
Z
Z
(7)
(f ◦ ϕ)| det Jϕ |dV =
f dV.
D
Ω
• Now let M be a Riemannian manifold, and let x : U → Rn be a chart on M .
Then, for each p ∈ U , we let Gx (p) denote the matrix given by
∂ ∂ x
x
Gx (p) = (gij
,
(p)), gij
(p) =
,
∂xi p ∂xj p
and we set
g x = det Gx > 0.
Question: What if we are given a different chart y : U → Rn on the same
set U in M ?
Then we relate the formulae as follows: Set J to be the Jacobian matrix
Jrj =
then we have
∂(y ◦ x−1 )r
;
∂xj
X ∂
∂
=
Jrj
∂xj
∂y r
r
which implies
Gx = J T Gy J,
where J T denotes the transpose of J, which implies
√
gx =
√
g y | det J|.
Thus we have the local densities
(8)
√
g x dx1 · · · dxn =
√
g y dy 1 · · · dy n ,
by which we mean that the integral
Z
√
(f g x ) ◦ x−1 dx1 · · · dxn
I(f ; U ) =
x(U )
depends only on f and U —not on the particular choice of chart x.
5
• We now turn the local Riemannian measure to a global Riemannian measure
on all of M .
(i) First, pick an atlas
{xα : Uα → Rn : α ∈ A},
and subordinate partition of unity {φα : α ∈ A}.
(ii) Then define the global Riemannian metric dV by
X √
dV :=
φα g xα dx1α · · · dxnα ,
α∈A
or, eqivalently,
Z
f dV =
M
X
I(φα ; f ; Uα ).
α∈A
— One easily check that the measure is well-defined; that is, it is independent of
both the particular choices of atlas and subordinate partition of unity.
— One easily checks that a function f is measurable w.r.t. dV iff f ◦ x−1 is measurable on x(U ) for any chart x : U → Rn .
In what follws, we work with this measure.
Definition. For any measurable B in M , we let V (B) denote the measure of B
and refer to V (B) as the volume of B.
If Γ is an (n − 1)-dimensional submfd of M , then we usually denote its Riemannian
measure by dA;
and for any measurable Λ in Γ, we denote its measure by A(Λ), and refer to A(Λ)
as the area of Λ.
The Effective Calculation of Integrals
• If the manifold M is diffeomorphic to Rn , then one has, possibly, a convenient
way to literally calculate an integral, by referring the calculation to one coordinate system.
However, as soon as one cannot cover the manifold with one “naturally” chosen chart, one would then be forced to literally pick an atlas and subordinate
partition of unity.
This would not go well at all.
• The simplest overarching approach is to use the geometry of the Riemannian
manifold to indicate a judicious choice of a set of measure 0 to delete,
which will thereby leave an open set that is the domain of a chart on M .
— The quickest example that comes to mind is the stereographic projection of the
sphere Sn to Rn , in which the domain of the chart covers all of Sn minus the
pole of the projection.
So any integral on the sphere may be referred to this chart.
• Before proceeding, we note that (8) implies that the notion of a set of measure 0 depends only on the differentiable structure of the mfd.
— It makes no difference whether we are referring to a local measure on M induced
by Lebesgue measure on the image of a chart on M , or whether we are referring
to Riemannian measure.
6
• Continung, we work, in our setting, with spherical coordinates as follows.
For convenience, we assume that M is complete.
For any p ∈ M , introduce normal coordinates about p, which describe (locally)
a differentiable map of (0, ∞) × ∂B1 (0p ) into M \ {p}, given by
(t, ξ) 7→ exp tξ.
(i) This map may fail to be the inverse map of a chart on M \ {p}
since the map may fail to be a diffemorphism.
(ii) Also, since exp(∂B1 (0p )) is not diffeomorphic to a subset of Rn−1 ,
one cannot use ξ, literally, as an (n − 1)-dimensional coordinate.
– The second difficulty is simply addressed by picking a chart on exp(∂B1 (0p ).
It need never be explicit, since the final formulation never require it.
– The first difficulty must be dealt with by restricting the geodesic spherical
coordinates to Dp \ {p}.
• Thus, a chart on M \ Cut(p) = D(p) is given by
−1
: Dp \ {p} → Dp \ {p};
exp Dp \{p}
and the Riemannian measure is given on Dp by
√
dV (exp ξ) = g(t; ξ)dt dµp (ξ),
√
for some function g on Dp , where dµp (ξ) denote the Riemannian measure on
∂B1 (0p ) induced by the Eucliden Lebesgue measure on Tp M .
(i) The set {p} has measure 0; so we never have to explicitly include it in, or exclude
it from, our discussion of integrals.
(ii) More significantly, C(p) has measure 0. Indeed, the function c(ξ) is continuous
on all of SM = {ξ ∈ T M : |ξ| = 1}, so its restriction to ∂B1 (0p ) is continuous.
• Thus, the tangential cut locus of p is the image of the continuous map
ξ 7→ c(ξ)ξ
from ∂B1 (0p ) to Tp M , and therefore has the Lebesgue measure 0.
The image of the tangential cut locus of p under the differentiable exponential
map is C(p), the cut locus of p in M . Therefore
Proposition 1. For any p ∈ M , the cut locus C(p) of p is a set of measure 0.
Thus, for any p ∈ M , and integrable function f on M , we have
Z
Z
√
f dV =
f (exp tξ) g(t; ξ)dt dµp (ξ)
M
=
Z
=
Z
Dp
dµp (ξ)
∂B1 (0p )
Z
∞
c(ξ)
√
f (exp tξ) g(t; ξ)dt
0
√
f (exp tξ) g(t; ξ)dµp (ξ)
dt
0
−1
Z
t−1 ∂B
t (0p )∩Dp
where t ∂Bt (0p ) ∩ Dp is the subset of ∂B1 (0p ) obtained by dividing each of the
elements of ∂Bt (0p ) ∩ Dp by t.
7
Theorem 2. We have
√
g(t; ξ) = det A(t; ξ),
where A(t; ξ) is the solution of the system of ODEs on ξ ⊥ :
A00 + R(t)A = 0,
satisfying the initial conditions A(0; ξ) = 0, A0 (0; ξ) = I.
Proof. Let ψ be a chart on ∂B1 (0p ), ξ = ψ −1 , and let x be a chart on Dp \ {p}
given by
(exp )−1 −1
Dp
x= ψ◦
,
|(exp
)
|
.
Dp
|(exp Dp )−1 |
Then, what was called ∂t H is here equal to ∂/∂xn , and what was referred to as
∂α H is here equal to ∂/∂xα , α = 1, · · · , n − 1.
— Let G be the matrix of the Riemannian metric on M associated to the chart x,
b be the matrix of the Riemannian metric on ∂B1 (0p ) associated to the
and let G
chart u.
— Then, equation (6) translates to our language here as
X
gαβ =
A∗αγ gbγδ Aδβ , α, β, γ, δ = 1, · · · , n − 1,
γ,δ
and (4) and (5) translate to
gnn = 1, gαn = gnα = 0, α = 1, · · · , n − 1.
We conclude that
√
which implies the claim.
g=
p
gb det A,
Notation. Given x ∈ M , we let V (x; r) denote the volume of Br (x) = B(x; r);
that is
Z
V (x; r) =
dV.
Br (x)
Notation. For each x ∈ M , r > 0, define Dx (r) to be the subset of ∂B1 (0x )
consisting of those elements ξ for which rξ ∈ Dx , i.e.
rDx (r) = ∂Br (0x ) ∩ Dx .
— We have
V (x; r) =
=
Z Z
det A(t; xi)dtdµx (ξ)
Dx ∩Br (x)
Z
r
dt
0
Z
det A(t; ξ)dµx (ξ).
Dx (t)
8
Volume Comparison Theorem
• Let (M, g) be a complete Riemannian manifold,
and Bp (r) be a ball which does not meet Cut(p).
— Instead of working with A, we work with B =: A∗ A, which is self-adjoint.
— Take a geodesic γξ = exp(tξ) from p, and an orthonormal basis let {e1 , · · · , en−1 }
be an orthonormal basis of ξ ⊥ consisting of eigenvectors of B(r).
– Let Eα , 1 ≤ α ≤ n − 1, be parallel vector fields along γξ such that Eα (0) = eα .
– Let ρ(ξ) be the (possibly infinite) distance to the cut-locus in the direction ξ.
Suppose that
0 ≤ r ≤ ρ(ξ).
For such an r, there exists a unique Jacobi field Yαr such that
Yαr (0) = 0 and Yαr (r) = Eα (r).
(9)
Indeed, in normal coordinates around p, this Jacobi field is given by
t
1
Yαr (t) = eα = Yα (t)
r
r
where Yα (t) = teα is the Jacobi field satisfying the initial condition
Yα (0) = 0 and Dt Y α (0) = eα = Eα (0).
We have, in geodesic spherical coordinates,
Yαr (t) =
det A(t; ξ) =
1
tn−1
1
A(t; ξ)
Yα (t) =
Eα (t).
A(r; ξ)
A(r; ξ)
det(Y1 , · · · , Yn−1 ) =
rn−1
r
det(Y1r (t), · · · , Yn−1
(t))
tn−1
Lemma 3.99. Let Ajk : Rm → R ∈ C 1 , j, k = 1, · · · , n. Then, on the open set
for which det(Aij ) 6= 0, one has, setting A = (Ajk ),
∂A
∂
ln det A = tr ` A−1 , for j = 1, · · · , m.
∂x`
∂x
Lemma 3.102. Denoting by I the index form, we have
n−1
X
(A(t; ξ))0
n−1
=
.
I(Yαr , Yαr ) −
A(t; ξ)
r
α=1
r
Proof. Instead of working with Ar = det(Y1r (t), · · · , Yn−1
(t)), we work with
B r =: Ar ∗ Ar ,
9
which is self-adjoint. We have
(det Ar )0
1 (det B r )0
=
,
r
det A
2 det B r
and
(det A(t; ξ))0
(det Ar )0
1 (det B r )0
n−1
n−1
=
=
.
−
−
det A(t; ξ)
det Ar
t
2 det B r
t
Then, by Lemma 3.99 and (9),
n−1
n−1
X hY r 0 , Y r i
X
1 (det B r )0
1
α
α
r0 r
trB
(τ
)
=
=
B
(τ
)
=
hYαr 0 , Yαr i.
r
r
r
2 det B
2
hY
,
Y
i
α
α
α=1
α=1
∴ I(Yαr , Yαr ) =
Z
r
hYαr 0 , Yαr 0 i − R(Yαr , γξ0 , Yαr , γξ0 ))ds
Z r
r
=−
hYαr 00 , Yαr i − R(Yαr , γξ0 , Yαr , γξ0 ))ds + hYαr 0 , Yαr i 0
0
r0
=hYα (r), Yαr (r)i.
0
Lemma 3.103. If γ : [a, b] → M is a minimizing geodesic, Y is a Jacobi field and
X is a vector field along γξ such that
Y (a) = X(a) and Y (b) = X(b),
then
I(X, X) ≥ I(Y, Y ).
Proof. Since X(a) − Y (a) = X(b) − Y (b) and γ is minimizing, we have
I(X − Y, X − Y ) ≥ 0.
On the other hand, we have
b
b
0
I(Y, Y ) = hY , Y i and I(X, Y ) = hY , Xi .
0
a
a
∴ I(X − Y, X − Y ) = I(X, X) − 2I(x, Y ) + I(Y, Y ) = I(X, X) − I(Y, Y ).
10
Grunther-Bishop theorem 1. Assume that we have the geodesic γξ as described
above, with all sectional curvatures along γξ less than or equal to δ. Then
√
S0
(det A)0
≥ (n − 1) δ , on (0, π/ δ),
(10)
det A
Sδ
and
√
(11)
det A ≥ Sδ n−1 on (0, π/ δ].
• We have equality in (2) iff
A(t; ξ) = Sδ (t)I, R(t) = δI ∀t ∈ [0, t0 ].
Proof. Denoting by Y one of the Jacobi fields Yαr , we have
Z r
0
(hY 0 , Y 0 i − R(Y, γξ0 , Y, γξ0 ))ds
hY (r), Y (r)i =
0
Z r
≥
(|Y 0 |2 − δ|Y |2 )ds.
0
Write
A(t; ξ)
Eα (t).
A(r; ξ)
On the simply-connected manifold with constant mean curvature δ, take a geodesic
eα along γ
γ
eξ of length r, and define vector vector fields E
eξ in the same way as the
e
vectors Eα = Eα . Set
A(t; ξ) e
Ye =
Eα (t).
A(r; ξ)
Then
Z r
Z r
(|Ye 0 |2 − δ|Ye |2 )ds =
(|Y 0 |2 − δ|Y |2 )ds = I(Ye , Ye ).
Y (t) =
0
0
Applying Lemma 3.103 to the manifold Mδ with constant curvature δ, we obtain
eα , X
eα ),
I(Yeαr , Yeαr ) ≥ I(X
where
eα (t) = Sκ (t) E
eα (t)
X
Sκ (τ )
is the Jacobi field which takes at the ends of e
γξ the same value as Yeαr .
By Lemm 3.102, we have
n−1
n−1
X
((Sδ )n−1 )0
(det A(t; ξ))0
n − 1 X er er
n−1
=
≥
=
I(Yαr , Yαr ) −
I(Xα , Xα ) −
. det A(t; ξ)
r
r
(Sδ )n−1
α=1
α=1
Grunther-Bishop theorem 2. Assume that the sectional curvatures of M are
all less than or equal to δ. Then, ∀x ∈ M , we have
√
V (x; r) ≥ Vδ (r), ∀r ≤ min{inj x, π/ δ},
with equality for some fixed r iff B(x; r) is isometric to the disk of radius r in the
constant space form Mδ .
11
Bishop theorem 1. Assume that we are given a real constant κ and the fixed
geodesic γξ , with the Ricci curvatures along γξ greater than or equal to (n − 1)κ;
i.e.
Ric(γξ0 (t), γξ0 (t)) = trR(t) ≥ (n − 1)κ, ∀t ∈ (0, con j].
Then
(det A)0
S0
≤ (n − 1) κ , on (0, conj ξ),
det A
Sκ
and
det A ≤ Sκ n−1 on (0, conj ξ),
and
• We have equality in (5) iff
A(t : ξ) = Sκ (t)I, R(t) = κI ∀t ∈ [0, t0 ].
√
Proof. Let r = π/ κ and
Xαr (t) =
Sκ (t)
Eα (t).
Sκ (r)
Lemma 3.103 gives
n−1
X
I(Yαr , Yαr ) ≤
α=1
n−1
X
I(Xαr , Xαr ).
α=1
We have
n−1
X
I(Xαr , Xαr )
=
Z 1
0
α=1
≤
Sκ (t)
Sκ (r)
2
((n − 1)κ − Ric(γξ0 , γξ0 ))ds +
n−1
X
hXαr (r), Xαr 0 (r)i
α=1
n−1
X
hXαr (r), Xαr 0 (r)i.
α=1
Using this and Lemma 3.102, we obtain

√
√
(n − 1) κ cot( κt) − n−1

t

0
(det A)
≤ 0,

det A
√
√

(n − 1) −κ coth( −κt) −
if κ > 0,
if κ = 0,
n−1
τ
if κ < 0
n−1 0
=
(Sκ (t))
)
.
(Sκ (t))n−1
Bishop theorem 2. Assume that the Ricci curvatures of M are all greater than
or equal to (n − 1)κ. Then, ∀x ∈ M and ∀r > 0, we have
V (x; r) ≤ Vκ (r),
with equality for some fixed r iff B(x; r) is isometric to the disk of radius r in the
constant space form Mκ .