math 240b: differentiable manifolds and riemannian geometry · math 240b: differentiable...
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Math 240B: Differentiable Manifolds andRiemannian Geometry
Simon Rubinstein–Salzedo
Winter 2006
0.1 Introduction
These notes are based on a graduate course on differentiable manifolds and Rieman-nian geometry I took from Professor Doug Moore in the Winter of 2006. The textbookwas Riemannian Geometry by Manfredo Perdigao do Carmo. Many other books arealso mentioned in the notes. Since the professor handed out very good notes, I havemade very few changes to these notes.
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Chapter 1
January 10, 2006
Let Mn be a smooth manifold.
A Riemannian metric on M is a function 〈 , 〉 which assigns to each p ∈ M apositive-definite symmetric bilinear form
〈·, ·〉p : TpM × TpM → R
which varies smoothly with p ∈ M .
The last condition means that if (U, (x1, . . . , xn)) is a smooth coordinate system onM , then each of the functions gij : U → R defined by
gij(p) =
⟨∂
∂xi
∣∣∣∣p
,∂
∂xj
∣∣∣∣p
⟩or gij =
⟨∂
∂xi,
∂
∂xj
⟩is smooth. We sometimes write
〈 , 〉 |U=n∑
i,j=1
gij dxi ⊗ dxj.
A Riemannian manifold is a pair (M, 〈 , 〉), where M is a smooth manifold and〈 , 〉 is a Riemannian metric on M .
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1.1 Geometry of a Riemannian Manifold
One can use the Riemannian manifold to define lengths of vectors and lengths ofcurves. Thus if v ∈ TpM , the length of v is given by the formula
|v| =√〈v, v〉p.
If γ : [a, b] → M is a smooth curve, the length of γ is
L(γ) =
∫ b
a
√〈γ′(t), γ′(t)〉γ(t) dt.
If v, w ∈ TpM , we can define the angle θ between v and w by
cos θ =〈v, w〉p|v| |w|
.
We can also use Riemannian metrics to calculate areas and volumes.
Simplest Example. M = Rn with standard coordinates (x1, . . . , xn) and Rieman-nian metric 〈 , 〉 defined by⟨
n∑i=1
ai∂
∂xi
∣∣∣∣p
,n∑
j=1
bj ∂
∂xj
∣∣∣∣p
⟩=
n∑i=1
aibj
= (a1, . . . , an) · (b1, . . . , bn),
where · stands for the dot product. In this case,
gij = δij =
1 if i = j,
0 if i 6= j.
If γ : [a, b] → Rn is a smooth curve,
L(γ) =
∫ b
a
√γ′(t) · γ′(t) dt.
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An Important Class of Examples. Suppose Mn is a smooth manifold and F :Mn → RN is an imbedding. We then define the induced Riemannian metric on Mby
〈v, w〉p = F∗p(v) · F∗p(w), for v, w ∈ TpM.
We can identify TF (p)RN with RN . Then
F∗p
(∂
∂xi
∣∣∣∣p
)=
∂F
∂xi(p),
where on the right, F is regarded as a vector-valued function. Then⟨∂
∂xi,
∂
∂xj
⟩=
∂F
∂xi· ∂F
∂xj,
which is clearly smooth.
Note that if 〈 , 〉 is the Riemannian metric on M induced by F : Mn → RN andγ : [a, b] → M is a smooth curve, then
L(γ) =
∫ b
a
√〈γ′(t), γ′(t)〉γ(t) dt
=
∫ b
a
√F∗γ(t)(γ′(t)) · F∗γ(t)(γ′(t)) dt
=
∫ b
a
√(F γ)′(t) · (F γ)′(t) dt
= L(F γ).
Thus the length of a curve γ in M is the length of its image in RN .
In the case of Euclidean space Rn, with coordinates (x1, . . . , xn) and the Riemannianmetric
〈 , 〉 = dx1 ⊗ dx1 + · · ·+ dxn ⊗ dxn,
the curves of smallest length joining p and q are straight lines.
What are the curves of shortest length in a Riemannian manifold?
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This question can be formulated with the calculus of variations.
Let Ω(M ; p, q) = smooth paths γ : [0, 1] → M | γ(0) = p and γ(1) = q. DefineL : Ω(M ; p, q) → R by L(γ) =length of γ. The function L : Ω(M ; p, q) → R is difficultto deal with because
1. There is a troublesome square root in the integrand.
2. If γ is a reparametrization of γ, L(γ) = L(γ).
Fortunately, there is a function J : Ω(M ; p, q) → R, closely related to L, which doesnot share these problems. This function is defined by
J(γ) =1
2
∫ 1
0
〈γ′(t), γ′(t)〉 dt
and is called the action (or sometimes the energy).
Proposition. (L(γ))2 ≤ 2J(γ), with equality holding if and only if γ is of constantspeed; that is, 〈γ′(t), γ′(t)〉γ(t) is constant.
Proof. It follows from the Cauchy-Schwarz inequality that
(L(γ))2 =
[∫ 1
0
√〈γ′(t), γ′(t)〉 dt
]2
≤∫ 1
0
1 dt
∫ 1
0
〈γ′(t), γ′(t)〉 dt
= 2J(γ),
with equality holding if and only if the functions 1 and 〈γ′(t), γ′(t)〉 are linearly inde-pendent.
Proposition. If γ ∈ Ω minimizes L and has constant speed, it minimizes J .
Proof. If λ ∈ Ω,2J(γ) = [L(γ)]2 ≤ [L(λ)]2 ≤ 2L(γ).
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Proposition. If γ ∈ Ω minimizes J , it minimizes L and has constant speed.
Sketch of Proof. Suppose we can reparametrize γ so that it has constant speed.Let γ be such a reparametrization. Then
2J(γ) = [L(γ)]2 = [L(γ)]2 ≤ 2J(γ).
If γ minimizes J , we must have equality, and by the first proposition, γ has constantspeed. Suppose that λ ∈ Ω and L(λ) < L(γ). If λ has a unit speed reparametrization
λ, then
2J(λ) = (L(λ))2
= (L(λ))2
< (L(γ))2
= 2J(γ),
contradicting the fact that γ minimizes J .
There is only one problem with this sketch: it assumes we can reparametrize γ andλ to have constant speed. This can always be done, so long as γ′(t) and λ′(t) nevervanish.
If dim M ≥ 2, we can complete the sketch by approximating γ and λ by immersions(possible by an approximation theorem of Whitney, proven in Hirsch, DifferentialTopology, Theorem 2.12).
But there is also an easier argument we will be able to give later. The upshot is thatγ ∈ Ω minimizes J iff γ ∈ Ω minimizes L and has constant speed. This motivateslooking at the calculus of variations problem for
J : Ω(M ; p, q) → R.
We think of Ω(M ; p, q) as an “infinite-dimensional manifold.” Let γ ∈ Ω. A variationof γ is a map α : (−ε, ε) → Ω such that
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1. α(0) = γ, and
2. the map α : (−ε, ε)× [0, 1] → M defined by α(s, t) = α(s)(t) is smooth.
We can think of a variation of γ as a “smooth path” α : (−ε, ε) → Ω with α(0) = γ.
Definition. We say that γ ∈ Ω is a critical point for J if
d
ds(J(α(s)))
∣∣∣∣s=0
= 0
for every variation α of γ.
Note that if γ ∈ Ω minimizes J , it must be a critical point for J .
Definition. An element γ ∈ Ω is a (constant speed) geodesic if it is a criticalpoint for J .
Theorem. Suppose that F : Mn → RN is an imbedding and Mn is given the inducedRiemannian metric. Then γ ∈ Ω is a geodesic iff
[(F γ)′′(t)]T = 0,
where [·]T is the orthogonal projection from TF (γ(t))RN to Tγ(t)M ⊆ TF (γ(t))RN .
To simplify notation, we often identify p ∈ M with F (p) ∈ RN . In this case, thecondition that γ be a geodesic is just
[γ′′(t)]T = 0,
where [·]T is the orthogonal projection to the tangent space.
Proof. We consider a variation α : (−ε, ε) → Ω(M ; p, q) of γ. Then
α : (−ε, ε)× [0, 1] → M ⊆ RN
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is a smooth map. Dropping F , we can regard α as a smooth map
α : (−ε, ε)× [0, 1] → RN
whose image lies in the submanifold M . Note that
d
ds(J(α(s))) =
1
2
∫ 1
0
∂
∂s
(∂α
∂t(s, t) · ∂α
∂t(s, t)
)dt
=
∫ 1
0
∂2α
∂s∂t(s, t) · ∂α
∂t(s, t) dt
= −∫ 1
0
∂α
∂s(s, t) · ∂2α
∂t2(s, t) dt.
The last step follows by integration by parts, since
∂α
∂s(s, 0) ≡ 0 =
∂α
∂s(s, 1).
Thus
d
ds(J(α(s)))
∣∣∣∣s=0
= −∫ 1
0
∂α
∂s(0, t)
∂2α
∂t2(0, t) dt
= −∫ 1
0
V (t) · γ′′(t) dt,
where V (t) = ∂α∂s
(0, t) ∈ Tγ(t)(M), since α takes values in M ⊆ RN . We can constructa variation α(s) such that
V (t) =∂α
∂s(0, t)
is any smoothly varying family of tangent vectors to M such that V (0) = 0 = V (1).Thus
(γ′′(t))T = 0.
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Chapter 2
January 12, 2006
Let Mn ⊆ RN with the induced Riemannian metric.
Motivated by the theorem we proved last time, we say that a smooth map γ : (a, b) →Mn ⊆ RN is a constant speed geodesic if it satisfies the condition
(γ′′(t))T = 0 for all t ∈ (a, b).,
where (·)T is the orthogonal projection to the tangent space Tγ(t)(M) ⊆ Tγ(t)RN .Thus γ : (a, b) → Mn ⊆ RN is a constant speed geodesic if the tangential componentof its acceleration is zero.
Thus, for example, if (e1, e2, e3) is an orthonormal basis for R3, S2 is the unit spherein R3, then the great circle γ : R → S2 ⊆ R3,
γ(t) = cos t e1 + sin t e2
is a constant speed geodesic, since
γ′′(t) = − cos t e1 − sin t e2 = −γ(t),
which is perpendicular to Tγ(t)S2. Indeed, it has unit speed (γ′(t) · γ′(t) = 1) and is
a smooth closed geodesic, since γ(t + 2π) = γ(t) for all t ∈ R.
By a similar argument, we can show that if
M2 = (x, y, z) ∈ R3 : x2 + y2 = 1,
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the right circular cylinder, then the curve
γ : R → M, γ(t) = (cos(at), sin(at), bt)
is a constant speed geodesic for all choices of constants a and b. Thus the geodesicson the right circular cylinder are helices and degenerate cases of these (circles andlines).
Recall that any Riemannian manifold (Mn, 〈 , 〉) arises from some imbedding F :Mn → RN :
Nash Imbedding Theorem. (1956) If (Mn, 〈 , 〉) is any Riemannian manifold,there is an imbedding F : Mn → RN for some N such that 〈 , 〉 is the metric on Mn
induced by F .
This is a very famous theorem by the mathematician John Nash who received a NobelPrize in Economics. (See the movie A Beautiful Mind.) However, given (Mn, 〈 , 〉),it is usually quite difficult to construct F , and in any case, F is by no means unique!
Thus we would like an intrinsic definition of geodesics, one that does not requirean “isometric” imbedding F : Mn → RN .
Problem. If (M, 〈 , 〉) is a Riemannian manifold and γ : (a, b) → M , can we definethe acceleration of γ? (Geodesics would then, one would hope, be curves of “zeroacceleration.”)
In Euclidean space Rn with Euclidean coordinates (x1, . . . , xn) and Euclidean metric
〈 , 〉 = dx1 ⊗ dx1 + · · ·+ dxn ⊗ dxn,
we can set
acceleration of γ at t =n∑
i=1
d2(xi γ)
dt2(t)
∂
∂xi
∣∣∣∣γ(t)
=n∑
i=1
(d2xi
dt2∂
∂xi
)(t). (∗)
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However, under a change to curvilinear coordinates,
d2xi
dt2=∑ ∂xi
∂yj
d2yj
dt2+∑ ∂2xj
∂yj∂yk
dyj
dt
dyk
dt,
so (∗) does not hold in curvilinear coordinates. We need a more subtle definition ofacceleration, based upon the notion of connection.
Definition. Let X(Mn) be the set of smooth vector fields on M and F (Mn) the setof smooth real-valued functions on M . A connection (in the tangent bundle TM toM) is a map
∇ : X(Mn)× X(Mn) → X(Mn)
(we write ∇XY for ∇(X,Y )) such that
1. ∇fX+gY Z = f∇XZ + g∇Y Z,
2. ∇Z(fX + gY ) = (Zf)X + f∇ZX + (Zg)Y + g∇ZY
for X, Y, Z ∈ X(Mn), f, g ∈ F (Mn).
Lemma. A connection ∇ is local: if U is an open subset of M ,
X |U≡ 0 ⇒ (∇XY ) |U≡ 0 and (∇Y X ≡ 0)
for all Y ∈ X(Mn).
Proof. Let p ∈ U and choose f : M → R such that f ≡ 0 on a neighborhood of pand f ≡ 1 outside U . Then X |U= 0 implies that fX = X, and hence
(∇XY )(p) = (∇fXY )(p)
= f(p)(∇XY )(p)
= 0,
(∇Y X)(p) = (∇Y (fX))(p)
= (Y f)(p)X(p) + f(p)∇Y X(p)
= 0.
Since p is an arbitrary point of U , (∇XY ) |U≡ 0 and (∇Y X)(p) = 0.
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The lemma can be restated as
supp(∇XY ) ⊆ supp(X), supp(∇Y X) ⊆ supp(X).
It follows from the lemma that a connection ∇ restricts to a well-defined connectionon any coordinate neighborhood U for local coordinate systems (x1, . . . , xn).
If (U, (x1, . . . , xn)) is such a local coordinate chart, we can define smooth functionsΓk
ij : U → R by
∇ ∂
∂xi
∂
∂xj=
n∑k=1
Γkij
∂
∂xk.
If X =∑n
i=1 f i ∂∂xi and Y =
∑nj=1 gj ∂
∂xj , then
∇XY =n∑
i,j=1
∇f i ∂
∂xi
(gj ∂
∂xj
)
=n∑
i,j=1
(f igj∇ ∂
∂xi
∂
∂xj+ f i ∂gj
∂xi
∂
∂xj
)(∗)
=n∑
i=1
[n∑
j=1
f j ∂gi
∂xj+∑
Γijkf
jgk
]∂
∂xi.
Lemma. (∇XY )(p) depends only on X(p) and on Y on some curve tangent to X(p).
Proof. Immediate from (∗).
Thus if v ∈ TpM and X ∈ X(Mn), we can define ∇vX ∈ TpM . In particular, ifγ : (a, b) → M is any smooth curve, we can define the acceleration of γ (withrespect to ∇) to be ∇γ′(t)γ
′(t), also written as (∇γ′γ′)(t).
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In terms of local coordinates,
(∇γ′γ′)(t) =
n∑i=1
d2(xi γ)
dt2(t)
∂
∂xi
∣∣∣∣γ(t)
+n∑
i,j,k=1
Γkij(γ(t))
d(xi γ)
dt(t)
d(xi γ)
dt(t)
∂
∂xk
∣∣∣∣γ(t)
.
We have defined acceleration in terms of a connection, but we really want to defineacceleration in terms of a Riemannian metric. To do this we use the Levi-Civitaconnection.
Fundamental Theorem of Riemannian Geometry. If (M, 〈 , 〉) is a Riemannianmanifold, there is a unique connection ∇ on TM (called the Levi-Civita connection)such that
A. ∇ is symmetric: ∇XY −∇Y X = [X,Y ].
B. ∇ is metric: X〈Y, Z〉 = 〈∇XY, Z〉+ 〈Y,∇XZ〉.
We will prove this next time and show that for this connection, if γ : (a, b) → Mn,
(∇γ′γ′)(t) = 0 ⇐⇒ (γ′′(t))T = 0
whenever Mn ⊆ RN with the induced Riemannian metric.
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Chapter 3
January 17, 2006
Recall that a connection on the tangent bundle TM to a smooth manifold M is amap
∇ : X(Mn)× X(Mn) → X(Mn),
where X(Mn) is the set of smooth vector fields on Mn, which satisfies the followingaxioms (with ∇XY = ∇(X, Y )):
1. ∇X+Y Z = ∇XZ +∇Y Z,
2. ∇Z(X + Y ) = ∇ZX +∇ZY ,
3. ∇fXY = f∇XY ,
4. ∇X(fY ) = (Xf)Y + f∇XY ,
for X, Y, Z ∈ X(Mn) and f a smooth real-valued function on Mn.
Last time we saw that a connection is local; in other words,
supp(∇XY ) ⊆ supp(X) ∩ supp(Y ).
Thus we can restrict a connection∇ to the domain U of a coordinate system (x1, . . . , xn).With respect to these coordinates, we can define the components of the connectionas the functions
Γkij : U → R defined by ∇ ∂
∂xi
∂
∂xj=
n∑k=1
Γkij
∂
∂xk.
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Fundamental Theorem. If (M, 〈 , 〉) is a Riemannian manifold, there is a uniqueconnection ∇ on TM which satisfies the conditions
A. ∇XY −∇Y X = [X, Y ],
B. X〈Y, Z〉 = 〈∇XY, Z〉+ 〈Y,∇XZ〉
for X, Y, Z ∈ X(Mn).
We can write these conditions in local coordinates as
A′. Γkij = Γk
ji.
B′.∂gij
∂xk =∑k
`=1 gi`Γ`jk +
∑n`=1 gj`Γ
`ik.
For A′, note that[∂
∂xi,
∂
∂xj
]= 0 ⇒ ∇ ∂
∂xi
∂
∂xj−∇ ∂
∂xj
∂
∂xi= 0 ⇒
n∑k=1
(Γkij − Γk
ji)∂
∂xk= 0.
We leave the verification of B′ to the reader.
It follows from A′ and B′ that
∂gjk
∂xi=
n∑`=1
g`jΓ`ik +
n∑`=1
g`kΓ`ij
∂gik
∂xj=
n∑`=1
g`iΓ`jk +
n∑`=1
g`kΓ`ij
−∂gij
∂xk= −
n∑`=1
g`iΓ`jk −
n∑`=1
g`jΓ`ik,
and so
∂gjk
∂xi+
∂gik
∂xj− ∂gij
∂xk= 2
n∑`=1
g`kΓ`ij.
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If (gmk) = (gmk)−1, we conclude that
Γmij =
1
2
n∑`=1
gmk
(∂gjk
∂xi+
∂gik
∂xj− ∂gij
∂xk
). (∗)
This establishes uniqueness of the connection satisfying A and B.
For existence, use (∗) to define Γkij and hence∇ on a coordinate chart (U, (x1, . . . , xn)).
Check that ∇ satisfies A and B.
Local uniqueness implies that the locally defined ∇’s fit together to give a globallydefined connection ∇ on TM which satisfies A and B.
The connection ∇ whose existence is guaranteed by the preceding theorem is calledthe Levi-Civita connection.
If X =∑n
i=1 f i ∂∂xi and Y =
∑ni=1 gi ∂
∂xi , then
∇XY =n∑
i,j=1
f i ∂gj
∂xi
∂
∂xi+
n∑i,j,k=1
f igjΓkij
∂
∂xk
=n∑
k=1
[X(gk) +
n∑i,j=1
f igjΓkij
]∂
∂xk.
This formula shows that (∇XY )(p) depends only on X(p) and on Y along somecurve tangent to X(p). Thus if γ : (a, b) → M is a smooth curve, we can define∇γ′γ
′(t) ∈ Tγ(t)M .
Definition. A smooth curve γ : (a, b) → M is called a (constant speed) geodesicif ∇γ′γ
′ = 0.
16
We need to check that this agrees with the definition previously given when Mn ⊆ RN
has the induced metric. To do this, we first note that the Levi-Civita connection forEuclidean space Rn with Euclidean coordinates (x1, . . . , xn) and Riemannian metric
n∑i,j=1
gij dxi ⊗ dxj, gij = δij =
1 if i = j,
0 if i 6= j.
is given by Γkij = 0. (This follows from (∗).) Thus if X and Y =
∑ni=1 gi ∂
∂xi are vectorfields on Rn, the Levi-Civita connection for Euclidean space is
DXY =n∑
i=1
X(gi)∂
∂xi.
Condition B can be stated as
X(Y · Z) = (DXY ) · Z + Y · (DXZ).
Suppose now that Mn ⊆ RN is an imbedded submanifold with the induced metric: ifX, Y ∈ X(Mn),
〈X,Y 〉(p) = X(p) · Y (p), for p ∈ Mn,
whenever X and Y are extensions of X and Y to vector fields on RN .
We claim that the Levi-Civita connection ∇ on Mn is given by
(∇XY )(p) = ((DX Y (p)))T for X,Y ∈ X(Mn), p ∈ Mn, (∗∗)
where (·)T is the orthogonal projection into the tangent space. One verifies that(∗∗) defines a connection on TM which satisfies A′ and B′. Thus (∗∗) defines theLevi-Civita connection on TM . In particular,
(∇γ′γ′)(t) = (Dγ′γ
′(t))T = (γ′′(t))T ,
and a geodesic γ : (a, b) → Mn ⊆ RN is simply a curve which has tangential acceler-ation zero.
17
Chapter 4
January 19, 2006
Suppose that (M, 〈 , 〉) is a Riemannian manifold with
〈 , 〉 =n∑
i,j=1
gij dxi ⊗ dxj.
Last time we saw that there is a unique connection ∇, the so-called Levi-Civitaconnection, on TM such that
A. ∇XY −∇Y X = [X, Y ],
B. X〈Y, Z〉 = 〈∇XY, Z〉+ 〈Y,∇XZ〉, for X, Y, Z ∈ X(M).
In terms of local coordinates,
∇ ∂
∂xi
∂
∂xj=
n∑k=1
Γkij
∂
∂xk,
where
Γkij =
1
2
n∑`=1
gk`
(∂gi`
∂xj+
∂gj`
∂xi− ∂gij
∂x`
).
In the special case where Mn ⊆ RN and 〈 , 〉 is the metric induced by the Euclideandot product on RN ,
∇XY = (DXY )T ,
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where (·)T is the orthogonal projection to TM and D is the usual directional derivativeon RN . In terms of Euclidean coordinates (u1, . . . , uN) on RN ,
DX
(N∑
I=1
f I ∂
∂uI
)=
N∑I=1
(Xf I)∂
∂uI.
The Levi-Civita connection is used to make the following key definition:
Definition. A smooth curve γ : (a, b) → M is a (constant speed) geodesic if∇γ′γ
′(t) ≡ 0.
In terms of local coordinates (x1, . . . , xn) on M ,
γ′(t) =n∑
i=1
d(xi γ)
dt(t)
∂
∂xi
∣∣∣∣γ(t)
,
(∇γ′γ′)(t) =
n∑i=1
γ′(t)
(∑ d(xi γ)
dt(t)
∂
∂xi
∣∣∣∣γ(t)
)
+n∑
i,j,k=1
Γijk(γ(t))
d(xj γ)
dt(t)
d(xk γ)
dt(t)
∂
∂xi
∣∣∣∣γ(t)
=n∑
i=1
[d(xi γ)
dt2(t) +
n∑j,k=1
Γijk(γ(t))
d(xi γ)
dt(t)
d(xk γ)
dt(t)
]∂
∂xi
∣∣∣∣γ(t)
.
We write xi(t) for (xγ)(t). Then γ : (a, b) → M is a geodesic if and only if
d2xi
dt2(t) +
n∑j,k=1
Γijk(γ(t))
dxj
dt(t)
dxk
dt(t) = 0. (∗)
If we let xi(t) = dxi
dt(t), we can rewrite (∗) as
dxi
dt= xi,
dxi
dt= −
∑nj,k=1 Γi
jkxjxk.
(∗∗)
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Suppose p ∈ M , v ∈ TpM . It follows from the fundamental existence and uniquenesstheorem for ordinary differential equations that for some ε > 0, there is a uniquesolution xi(t), −ε < t < ε to (∗) such that
xi(0) = p,dxi
dt(0) = xi(0) = dxi |p (v).
Thus there is a unique geodesic γv : (−ε, ε) → M such that
γv(0) = p, γ′v(0) = v,
for any p ∈ M , v ∈ TpM .
Definition. Let (M, 〈 , 〉) be a Riemannian manifold. A diffeomorphism F : M → Mis an isometry of F ∗〈 , 〉 = 〈 , 〉, or in other words, if for every p ∈ M ,
v ∈ TpM ⇒ |Fp∗(v)| = |v|.
Isometries preserves lengths of curves and take geodesics to geodesics.
It is usually quite difficult to find explicit solutions to (∗). The explicit solutionsusually occur only when M has lots of isometries.
Examples.
Mn = Rn with the Euclidean coordinates (x1, . . . , xn) and the Riemannian metric
〈 , 〉 =n∑
i=1
dxi ⊗ dxi, gij = δij =
1 if i = j,
0 if i 6= j.
In this case, Γkij = 0, and (∗) becomes d2xi
dt2= 0, which has xi(t) = ait + bi as its
solutions, where ai and bi are constants. Thus the geodesics in Euclidean space are
20
just the constant speed straight lines.
Define F : R2 → R2 by(x1 Fx2 F
)=
(cos θ − sin θsin θ cos θ
)(x1
x2
)+
(c1
c2
),
where θ, c1, and c2 are constants. Then F is an isometry. One can construct anisometry which takes any straight line into any other straight line.
Suppose next that M2 = S2 = (x, y, z) ∈ R3 : x2 + y2 + z2 = 1, and give S2 theRiemannian metric it inherits from R3. We have seen that if (e1, e2, e3) is a positivelyoriented orthonormal basis, then the constant speed great circle,
γ : R → S2, γ(t) = cos(at)e1 + sin(at)e2
is a geodesic in S2.
Let SO(3) = A ∈ GL(3, R) : AT = A, det A = 1. If A ∈ SO(3), define
FA : S2 → S2 by FA
xyz
= A
xyz
.
Then FA is an isometry. As we saw in 240A, SO(3) is a three-dimensional manifold.Thus we have a three-dimensional family of isometries of S2 which take any greatcircle to any other great circle.
Suppose next that H2 = (x, y) ∈ R2 : y > 0, the so-called Poincare upperhalf-plane, with the Riemannian metric
〈 , 〉 =1
y2(dx2 + dy2).
We can calculate the geodesics by brute force. Let x1 = x, x2 = y. Then a straight-forward calculation gives
Γ111 = Γ1
22 = Γ121 = Γ1
12 = 0
Γ222 = Γ1
21 = Γ112 = −1
y
Γ211 = y.
21
Hence the equations for geodesics ared2xdt2− 2
ydxdt
dydt
= 0d2ydt2− 1
y
(dydt
)2+ 1
y
(dxdt
)2= 0.
It dxdt
= 0 at one point, it follows from the first equation that dxdt≡ 0 and from the
second thatd2y
dt2− 1
y
(dy
dt
)2
= 0.
Reduce the order, setting z = dydt
, sodydt
= zdzdt
= 1yz2
⇒ dz
dy=
z
y.
Hence dzz
= dyy
, log |z| = log |y|+ C, z = (±ec)y = by, dydt
= by, y = aebt. Thus
x = 0, y = aebt
is a geodesic for all a, b.
If dxdt6= 0 and y is a function of x,
d2y
dt2=
d
dt
(dy
dx
dx
dt
)=
d
dt
(dy
dx
)dx
dt+
dy
dx
d2x
dt2
=d2y
dx2
(dx
dt
)2
+dy
dx
d2x
dt2,
and hence
1
y
(dy
dt
)2
− 1
y
(dx
dt
)2
=d2y
dx2
(dx
dt
)2
+dy
dx
(d2x
dt2
)=
d2y
dx2
(dx
dt
)2
+2
y
dy
dx
dx
dt
dy
dt.
22
Dividing by(
dxdt
)2allows us to eliminate t:
1
y
(dy
dx
)2
− 1
y=
d2y
dx2+
2
y
(dy
dx
)2
,
yd2y
dx2+
(dy
dx
)2
= −1,
d
dx
(ydy
dx
)= −1,
ydy
dx= x + 2a,
y2 = −x2 + ax + b.
Thus we finally obtain the equation
y2 + x2 + ax + b = 0.
These are the equations of circles with centers on the x-axis.
Conclusion. Geodesics in the Poincare upper half-plane are suitably parametrizedvertical lines and semicircle centered on the x-axis.
Of course, we still need to find the constant speed parametrizations of the semicircles.These could be determined by exploiting the three-dimensional group of isometrieswhich operates on H2. Let
SL(2, R) =
(a bc d
): ad− bc = 1
,
the so-called special linear group. If A =
(a bc d
)∈ SL(2, R), define FA : H2 → H2
by
x F1 + iy Fa =az + b
cz + d, z = x + iy.
It is an exercise to show that each FA is an isometry. Moreover, given any semicircle,there is an A ∈ SL(2, R) such that FA takes a vertical line to that semicircle.
23
The three Riemannian manifolds
R2with the Euclidean metric,
S2with the metric induced by the inclusion S2 ⊆ R3,
H2with the Poincare half-plane metric
have three-dimensional groups of isometries. Along with RP 2, they are the only Rie-mannian manifolds with three-dimensional groups of isometries.
Theorem. (Hilbert, 1901) There is no imbedding F : H2 → R3 which makes thePoincare half-plane metric the induced metric on H2.
24
Chapter 5
January 24, 2006
Let F0 : R2 → R3 be the parametrized surface defined by
F0(t, s) = (t, s, 0).
The Riemannian metric that F0 induces on R2 is
〈 , 〉 = dt⊗ dt + ds⊗ ds,
which is, of course, just the Euclidean metric.
But there are many other imbeddings from R2 into R3 which induce exactly the samemetric. Consider, for example, the catenary z = cosh y. We can parametrize thecatenary by arc length; one can check that
y =√
s1 + 1, z = log(s +√
s2 + 1)
is such a parametrization. We can then construct a cylinder over the catenary F1 :R2 → R3,
F1(t, s) = (t,√
s2 + 1, log(s +√
s2 + 1)).
A direct, if slightly tedious, calculation shows that the induced metric on R2 is onceagain
〈 , 〉 = ds⊗ ds = dt⊗ dt.
We say that the parametrized surfaces F0 and F1 have the same intrinsic geometrybecause the Riemannian metrics induced on R2 are the same, but different extrinsic
25
geometry because the shapes of the images F0(R2) and F1(R2) are different.
How can we recognize when the intrinsic geometry of a surface is not that of Euclideanspace?
More generally, we say that a Riemannian manifold (M, 〈 , 〉) is locally isometricto Euclidean space (or locally Euclidean) if given any p ∈ M , there is an openneighborhood U of p such that (U, 〈 , 〉 |U) is isometric to an open subset of Euclideanspace.
How do we recognize when a given Riemannian manifold (M, 〈 , 〉) is locally Eu-clidean?
The invariant that answers this question is the Riemann-Christoffel curvature tensor.
Recall that associated to a Riemannian metric
〈 , 〉 =n∑
i,j=1
gij dxi ⊗ dxj
is a Levi-Civita connection ∇,
∇ ∂
∂xi
∂
∂xj=
n∑k=1
Γkij, Γk
ij =1
2
n∑`=1
gk`
(∂gi`
∂xj+
∂gj`
∂xi− ∂gij
∂x`
).
Definition. The Riemann-Christoffel curvature tensor is the map R : X(M)×X(M)× X(M) → X(M) defined by
R(X, Y )Z = ∇X∇Y Z −∇Y∇XZ −∇[X,Y ]Z for X,Y, Z ∈ X(M).
26
Just like ∇, R is a local operator and it is clearly linear over the field R of real num-bers. However, it is also linear over the ring F (M) = smooth real-valued functionsf : M → R.
Proposition.
(a) R(fX, Y )Z = fR(X, Y )Z,
(b) R(X, fY )Z = fR(X, Y )Z,
(c) R(X, Y )fZ = fR(X, Y )Z.
We prove only (c); the other proofs are quite similar.
R(X, Y )fZ = ∇X∇Y (fZ)−∇Y∇X(fZ)−∇[X,Y ](fZ)
= ∇X((Y f)Z + f∇Y Z)−∇Y ((Xf)Z + f∇XZ)− ([X, Y ](f)Z + f∇[X,Y ]Z)
= XY (f) + (Y f)∇XZ + (Xf)∇Y Z + f∇X∇Y Z
− Y X(f)− (Xf)∇Y Z − (Y f)∇XZ − f∇Y∇XZ
− [X, Y ](f)Z − f∇[X,Y ]Z
= f(∇X∇Y Z −∇Y∇XZ −∇[X,Y ]Z)
= fR(X, Y )Z,
as desired.
The Proposition justifies defining the components R`kij of R by
R
(∂
∂xj,
∂
∂xj
)∂
∂xk=
n∑`=1
R`kij
∂
∂x`.
Proposition.
R`kij =
∂
∂xi(Γ`
kj)−∂
∂xj(Γ`
ki) +n∑
m=1
(ΓmjkΓ
`im − Γm
ikΓ`jm).
27
Proof.
n∑`=1
R`kij
∂
∂x`= R
(∂
∂xi,
∂
∂xj
)∂
∂xk
= ∇ ∂
∂xi
(∇ ∂
∂xj
∂
∂xk
)−∇ ∂
∂xj
(∇ ∂
∂xi
∂
∂xk
)= ∇ ∂
∂xi
(m∑
`=1
Γ`jk
∂
∂x`
)−∇ ∂
∂xj
(m∑
`=1
Γ`ik
∂
∂x`
)
=n∑
`=1
[∂
∂xi(Γ`
jk)−∂
∂xj(Γ`
ik) +∑
(Γ`imΓm
jk − Γ`jmΓm
ik)
]∂
∂x`,
as desired.
If M is Euclidean space,
gij = δij =
1 if i = j,
0 if i 6= j⇒ Γ`
ij = 0 ⇒ R`kij = 0.
Thus any locally Euclidean Riemannian manifold must have R = 0.
Proposition. The curvature tensor R satisfies the following symmetries:
1. R(X, Y )Z = −R(Y, X)Z.
2. R(X, Y )Z = R(Y, Z)X + R(Z,X)Y = 0.
3. 〈R(X, Y )Z,W 〉 = −〈R(X, Y )W, Z〉.
4. 〈R(X, Y )Z,W 〉 = 〈R(Z,W )X, Y 〉.
Note. It suffices to prove these symmetries for coordinate fields X = ∂∂xi , Y = ∂
∂xj ,Z = ∂
∂xk , and W = ∂∂x` .
28
Proof of 1.
R
(∂
∂xi,
∂
∂xj
)∂
∂xk= ∇ ∂
∂xi∇ ∂
∂xj
∂
∂xk−∇ ∂
∂xj∇ ∂
∂xi
∂
∂xk
= −R
(∂
∂xj,
∂
∂xi
)∂
∂xk.
Proof of 2.
R
(∂
∂xi,
∂
∂xj
)∂
∂xk+ R
(∂
∂xj,
∂
∂xk
)∂
∂xi+ R
(∂
∂xk,
∂
∂xi
)∂
∂xj
= ∇ ∂
∂xi∇ ∂
∂xj
∂
∂xk−∇ ∂
∂xj∇ ∂
∂xi
∂
∂xk
+∇ ∂
∂xj∇ ∂
∂xk
∂
∂xi−∇ ∂
∂xk∇ ∂
∂xj
∂
∂xi
+∇ ∂
∂xk∇ ∂
∂xi
∂
∂xj−∇ ∂
∂xi∇ ∂
∂xk
∂
∂xj= 0,
where we have used the fact that ∇ ∂
∂xi
∂∂xj = ∇ ∂
∂xj
∂∂xi .
Proof of 3. It suffices to show that⟨R
(∂
∂xi,
∂
∂xj
)∂
∂xk,
∂
∂xk
⟩= 0.
But
0 =
(∂2
∂xi∂xj− ∂2
∂xj∂xi
)⟨∂
∂xk,
∂
∂xk
⟩= 2
∂
∂xi
⟨∇ ∂
∂xj
∂
∂xk,
∂
∂xk
⟩− 2
∂
∂xj
⟨∇ ∂
∂xi
∂
∂xk,
∂
∂xk
⟩= 2
⟨∇ ∂
∂xi∇ ∂
∂xj
∂
∂xk,
∂
∂xk
⟩+ 2
⟨∇ ∂
∂xj
∂
∂xk,∇ ∂
∂xi
∂
∂xk
⟩− 2
⟨∇ ∂
∂xj∇ ∂
∂xi
∂
∂xk,
∂
∂xk
⟩− 2
⟨∇ ∂
∂xi
∂
∂xk,∇ ∂
∂xj
∂
∂xk
⟩= 2
⟨R
(∂
∂xi,
∂
∂xj
)∂
∂xk,
∂
∂xk
⟩.
29
Proof of 4.
〈R(X, Y )W, Z) = −〈R(Y,X)W, Z〉= 〈R(X, W )Y, Z〉+ 〈R(W, Y )X, Z〉= −〈R(X, Y )Z,W 〉= 〈R(Y, Z)X, W 〉+ 〈R(Z,X)Y,W 〉,
so
2〈R(X, Y )W, Z〉 = 〈R(X, W )Y, Z〉+〈R(W, Y )X, Z〉+〈R(Y, Z)X, W 〉+〈R(Z,X)Y,W 〉.
Interchanging (X, Y ) and (W, Z) gives
2〈R(W, Z)X,Y 〉 = 〈R(W, X)Z, Y 〉+〈R(X, Z)W, Y 〉+〈R(Z, Y )W, X〉+〈R(Y,W )Z,X〉.
The terms on the right sides are equal in pairs, so
2〈R(X, Y )W, Z〉 = 2〈R(W, Z)X, Y 〉.
The first of the above propositions shows that R restricts to a multilinear map
R : TpM × TpM × TpM → TpM
for each p ∈ M . We can define
R : TpM × TpM × TpM × TpM → R
by R(x, y, z, w) = 〈R(x, y)w, z〉. Then the last of the above propositions implies thatR satisfies the symmetries
R(x, y, z, w) = −R(y, x, z, w) = −R(x, y, w, z) = R(z, w, x, y)
andR(x, y, z, w) + R(y, z, x, w) + R(z, x, y, w) = 0.
Reference: Milnor, Morse Theory, §9.
30
Chapter 6
January 26, 2006
A pseudo-Riemannian metric 〈 , 〉 on M is a smoothly varying family p 7→ 〈 , 〉pof nondegenerate symmetric bilinear forms on TpM :
〈 , 〉p : TpM × TpM → R.
The metric is Riemannian if each 〈 , 〉p is positive definite. In local coordinates,
〈 , 〉 =n∑
i,j=1
gij dxi ⊗ dxj.
Fundamental Theorem. Associated to each pseudo-Riemannian metric is a Levi-Civita connection ∇, a connection which satisfies
A. ∇XY −∇Y X = [X, Y ],
1. X〈Y, Z〉 = 〈∇XY, Z〉+ 〈Y,∇XZ〉.
In local coordinate,∇ ∂
∂xi
∂∂xj =
∑nk=1 Γk
ij∂
∂xk , where Γkij = 1
2
∑n`=1 gk`
(∂gi`
∂xj +∂gj`
∂xi − ∂gij
∂x`
).
The connection in turn defines the Riemann-Christoffel curvature tensor R:
R(X, Y )Z = ∇X∇Y Z = ∇Y∇XZ −∇[X,Y ]Z.
31
Examples. First, suppose MN = RN with coordinates (x1, . . . , xN) and metric
〈 , 〉 = dx1 ⊗ dx1 + · · ·+ dxN ⊗ dxN .
Then Γkij = 0, and the Levi-Civita connection D is defined by
DX
(N∑
I=1
f I ∂
∂xI
)=
N∑I=1
X(f I)∂
∂xI.
Next, suppose that Mn ⊆ RN , a submanifold with the induced metric. Vector fieldsX, Y on Mn can be extended to vector fields X, Y on RN . Note that if p ∈ Mn,(DXY )(p) depends only on X(p) and on Y along some curve tangent to X(p). Thus(DXY )(p) depends only on the values of X and Y along Mn itself.
If p ∈ Mn and v ∈ TpRN , we can write v = vT + v⊥, where vT is tangent to M andv⊥ is perpendicular to M . IF X : Mn → RN is a smooth vector-valued function onMn, XT is a smooth vector field on M , while X⊥ is a smooth “section” of the normalbundle
NM =⊔NpM : p ∈ M,
NpM being the orthogonal complement to TpM in TpRN ∼= RN .
We can define ∇ : X(Mn)× X(Mn) → X(Mn) by
(∇XY )(p) = (DXY (p))⊥,
and check that it satisfies the axioms for a connection on TM . Moreover, ∇ satisfiesaxioms A and B. It suffices to check A when X = ∂
∂xi and Y = ∂∂xj , where (x1, . . . , xn)
are curvilinear coordinates on M which extend to curvilinear coordinates (x1, . . . , xN)on RN , and then
∇ ∂
∂xi
∂
∂xj−∇ ∂
∂xj
∂
∂xi=
(D ∂
∂xi
∂
∂xj−D ∂
∂xj
∂
∂xi
)T
= 0.
32
Similarly, axiom B holds: if X, Y, Z are tangent to Mn,
X〈Y, Z〉 = X(Y · Z)
= (DXY ) · Z + Y · (DXZ)
= 〈∇XY, Z〉+ 〈Y,∇XZ〉.
Thus ∇ is the Levi-Civita connection on M .
Let Γ(NM) = smooth maps X : M → RN : X(p) is perpendicular to TpM forp ∈ M = smooth sections of the normal bundle NM.
Definition. The second fundamental form of Mn in RN is the map
α : X(M)× X(M) → Γ(NM), α(X, Y )(p) = (DXY (p))⊥.
Proposition. α is bilinear over functions and symmetric:
(a) α(fX, Y ) = fα(X,Y ) = α(X, fY ),
1. α(X, Y ) = α(Y, X).
Proof of (a).
α(fX, Y ) = (DfXY )⊥
= f(DXY )⊥
= fα(X, Y ).
α(X, fY ) = (DX(fY ))⊥
= ((Xf)Y + fDXY )⊥
= f(DXY )⊥
= fα(X, Y ).
33
Proof of (b). We can take X = ∂∂xi and Y = ∂
∂xj . Then
α(X, Y )− α(Y,X) =
(D ∂
∂xi
∂
∂xj−D ∂
∂xj
∂
∂xi
)⊥
= 0,
as desired.
If γ : (a, b) → Mn ⊆ RN is a unit-speed curve, the acceleration γ′′(s) is also calledthe curvature of γ at s.
κ(s) = curvature of unit speed γ at s = γ′′(s).
We set
κg(s) = (γ′′(s))T
= (∇γ′γ′)(s)
= geodesic curvature of γ at s,
κn(s) = (γ′′(s))⊥
= α(γ′(s), γ′(s))
= normal curvature of γ at s.
The Proposition implies that α restricts to a symmetric bilinear map
α : TpM × TpM → NpM
with the following interpretation: If v ∈ TpM has unit length, α(v, v) is the normalcurvature of a unit-speed curve γ : (−ε, ε) → M with γ(0) = p and γ′(0) = v.
Theorem. Suppose Mn ⊆ RN with the induced Riemannian metric. Then
〈R(x, y)w, z〉 = α(x, z) · α(y, w)− α(x, w) · α(y, z)
for x, y, z, w ∈ TpM .
34
This is known as the Gauß equation. It provides an often quite effective means ofcalculating the curvature of Mn.
Proof. Since Euclidean space has zero curvature, DXDY W−DY DXW−D[X,Y ]W = 0for X, Y, W ∈ X(M), and hence if Z ∈ X(M),
0 = (DXDY W −DY DXW −D[X,Y ]W ) · Z= X(DY X · Z)− (DY W ) · (DXZ)
− Y (DXW · Z) + (DXW ·DY Z)− 〈∇[X,Y ]W, Z〉= X〈∇Y W, Z〉 − 〈∇Y W,∇XZ〉 − α(Y,W ) · α(X, Z)
− Y 〈∇XW, Z〉+ 〈∇XW,∇Y Z〉+ α(X, W ) · α(Y, Z)− 〈∇[X,Y ]W, Z〉= 〈∇X∇Y W, Z〉 − α(Y, W ) · α(X, Z)
〈∇Y∇XW, Z〉+ α(X, W ) · α(Y, Z)− 〈∇[X,Y ]W, Z〉= 〈R(X,Y )W, Z〉 − α(Y,W ) · α(X,Z) + α(X, W ) · α(Y, Z),
as desired.
Suppose, for example, that
Mn = Sn(a) =
(x1, . . . , xn+1) ∈ Rn+1 :
n+1∑I=1
(xI)2 = a2
,
the sphere of radius a. If γ : (−ε, ε) → Sn(a) is a unit speed great circle
γ(s) = a cos(s
a
)e1 + a sin
(s
a
)e2,
where e1 and e2 are unit-length vectors perpendicular to each other, then
γ′′(s) = −1
acos(s
a
)e1 −
1
asin(s
a
)= −1
aN(γ(s)),
where N(p) is the outward-pointing unit normal to Sn(a) at p. Then
α(γ′(s), γ′(s)) = −1
aN(γ(s)),
and
α(v, v) = −1
aN(p), when v ∈ TpM, 〈v, v〉 = 1.
35
By polarization,
α(x, y) = −1
a〈x, y〉N(p) for x, y ∈ TpM.
It therefore follows from the Gauß equation that
〈R(x, y)w, z〉 =1
a2[〈x, z〉〈y, w〉 − 〈x, w〉〈y, z〉]
for x, y, z, w ∈ TpM .
36
Chapter 7
January 31, 2006
There is one case in which the curvature of a Riemannian manifold is easy to calcu-late:
Theorem. Suppose Mn is a submanifold of Rn and is given the induced Riemannianmetric 〈 , 〉. Then the curvature R of M is given by the formula
〈R(x, y)w, z〉 = α(x, z) · α(y, w)− α(x, w) · α(y, z)
for x, y, z, w ∈ TpM , the dot on the right denoting the usual dot product in RN .
Proof. If D is the Levi-Civita connection on RN ,
DXDY W −DY DXW −D[X,Y ]W = 0 for X, Y, W ∈ X(Mn).
Hence if Z ∈ X(M),
0 = (DXDY W −DY DXW −D[X,Y ]W ) · Z= X(DY W · Z)− (DY W ) · (DXZ)− Y (DXW · Z) + (DXW ) · (DY Z)− (D[X,Y ]W ) · Z= X〈∇Y W, Z〉 − 〈∇Y W,∇XZ〉 − α(Y,W ) · α(X, Z)
− Y 〈∇XW, Z〉+ 〈∇XW,∇Y Z〉+ α(X, W ) · α(Y, Z)
− 〈D[X,Y ]W, Z〉= 〈∇X∇Y W, Z〉 − α(Y,W ) · α(X, Z)−〉∇Y∇XW, Z〉
+ α(X, W ) · α(Y, Z)− 〈∇[X,Y ]W, Z〉= 〈R(X, Y )W, Z〉 − α(X, Z) · α(Y, W ) + α(X,W ) · α(Y, Z),
37
as desired.
As we saw last time, this enables us to calculate the curvature of
Sn(a) = (x1, . . . , xn+1) ∈ Rn+1 : (x1)2 + · · ·+ (xn+1)2 = a2,
the result being
〈R(x, y)w, z〉 =1
a2[〈x, z〉〈y, w〉 − 〈x, w〉〈y, z〉]
for x, y, z, w ∈ TpSn(a).
The above Theorem can be extended to submanifolds of Minkowski space-time,the arena for special relativity. Minkowski space-time is Rn+1 with coordinates(t, x1, . . . , xn) and pseudo-Riemannian metric
〈 , 〉 = −dt⊗ dt + dx1 ⊗ dx1 + · · ·+ dxn ⊗ dxn
=(dt dx1 · · · dxn
)J
dtdx1
...dxn
,
where
J =
−1 0 · · · 00 1 · · · 0...
......
0 0 · 1
.
Let O(1, n) = A ∈ GL(n + 1, R) : AT JA = J, a subgroup of the general lineargroup called the Lorentz group. If
tx1
...xn
= A
tx1
...xn
,
then A ∈ O(1, n) if and only if
〈 , 〉 = −dt⊗ dt + dx1 ⊗ dx1 + · · ·+ dxn ⊗ dxn.
38
Elements of O(1, n) are called Lorentz transformations.
We will sometimes identify Rn+1 with its tangent space and denote the metric 〈 , 〉on Minkowski space-time by a dot: X · Y = 〈X, Y 〉.
The Levi-Civita connection D of Minkowski space-time is defined by
DX
(f
∂
∂t+
n∑i=1
f i ∂
∂xi
)= (Xf)
∂
∂t+
n∑i=1
X(f i)∂
∂xi.
Of course, the curvature of Minkowski space-time is zero.
Analogous to Sn(a) is the submanifold
Hn(a) = (t, x1, . . . , xn) : t > 0 and t2 − (x1)2 − · · · − (xn)2 = a2= p = (t, x1, . . . , xn) : t > 0 and p · p = −a2.
The submanifold Hn(a) inherits an induced metric from Rn+1, and we claim that itis positive definite. Indeed, in terms of the coordinates (x1, . . . , xn), we can write
t =√
a2 + (x1)2 + · · ·+ (xn)2,
and hence
dt =
∑ni=1 xi dxi√
a2 + (x1)2 + · · ·+ (xn)2,
and hence the induced metric is
〈 , 〉 = −∑n
i,j=1 xixj dxi ⊗ dxj
a2 + (x1)2 + · · ·+ (xn)2+ dx1 ⊗ dx1 + · · ·+ dxn ⊗ dxn.
Thus
gij = δij −xixj
a2 + (x1)2 + · · ·+ (xn)2,
and one readily checks that the matrix (gij) is positive definite.
If p ∈ Hn(a), and γ : (−ε, ε) → Hn(a) is a smooth curve with γ(0) = p, thenγ(t) · γ(t) ≡ −a2 implies that γ′(0) · p = 0, so p is perpendicular to Hn(a).
39
Lemma. If p ∈ Hn(a), there is an element A ∈ O(1, n) such that
p = A
a0...0
.
Proof. Suppose
p =
tx1
...xn
=
a cosh α
a sinh α u1...
a sinh α un
,
where u1
...un
is a unit-length vector in Rn. Let B be a rotation matrix of Rn such thatu1
...un
= B
10...0
.
Then1 0 · · · 00... B0
cosh α sinh α 0 · · · 0sinh α cosh α 0 · · · 0
0 0 1 · · · 0...
......
...0 0 0 · · · 1
a
0...
0
=
a cosh α
a sinh α u1...
a sinh α un
,
so we can take
A =
1 0 · · · 00... B0
cosh α sinh α 0 · · · 0sinh α cosh α 0 · · · 0
0 0 1 · · · 0...
......
...0 0 0 · · · 1
,
40
as desired.
Lemma. If p ∈ H and v ∈ TpHn(a), there is an element A ∈ O(1, n) such that
p = A
10· · ·0
, v = A
0c0...0
for some c ∈ R.
Proof. Exercise.
We can now calculate the geodesics of Hn(a). Define γ : R → Hn(a) by
γ(s) =(a cosh
(s
a
), a sinh
(s
a
), 0, . . . , 0
). (∗)
Then γ′′(s) = 1a
(cosh
(sa
), sinh
(sa
), 0, . . . , 0
)is a scalar multiple of γ(s); hence (γ′′(s))T =
0. Thus γ is a geodesic. The other geodesics of γ are just the curves
γ(s) = Aγ(s),
where A ∈ O(1, n) and A(Hn(a)) ⊆ Hn(a). These are just parametrizations of inter-sections of Hn(a) with planes passing through the origin.
The Levi-Civita connection ∇ on Hn(a) is defined by
∇XY = (DXY )T for X, Y ∈ X(Hn(a)),
while the second fundamental form α is defined by
α(X, Y ) = (DXY )⊥ for X, Y ∈ X(Hn(a)).
If γ is defined by (∗),
γ′(s) =(sinh
(s
a
), cosh
(s
a
), 0, . . . , 0
)41
and
α(γ′(s), γ′(s)) = (γ′′(s))⊥ =1
a2γ(s).
Thus if x ∈ TpHn(a) and 〈x, x〉 = 1,
α(x, x) =1
aN(p),
where N is the future-pointing normal vector at p with N(p) · N(p) = −1. Bypolarization,
α(x, y) =1
a〈x, y〉N(p) for x, y ∈ TpH
n(a).
Since N(p) · N(p) = −1, it follows from the theorem presented at the beginning ofthe lecture that
〈R(x, y)w, z〉 = − 1
a2[〈x, z〉〈w, y〉 − 〈x, w〉〈y, z〉].
Definition. If σ ⊆ TpM is a two-dimensional subspace, the sectional curvature ofσ is
K(σ) =〈R(x, y)y, x〉
|x|2|y|2 − 〈x, y〉2,
whenever x, y is a basis for σ.
En has K(σ) ≡ 0 for all 2-planes. Sn(a) has K(σ) ≡ 1a2 for all 2-planes. Hn(a) has
K(σ) ≡ − 1a2 for all 2-planes. These are the spaces of constant curvature, the
Euclidean and non-Euclidean geometries in n dimensions.
Proposition. Let
R,S : TpM × TpM × TpM × TpM → R
be two quadrilinear functionals satisfying the curvature symmetries. If
R(x, y, x, y) = S(x, y, x, y) for all x, y ∈ TpM,
then R = S.
42
Thus the sectional curvatures determine the full curvature tensor.
Proof. Let T = R− S. Then T satisfies the curvature symmetries, and
T (x, y, x, y) = 0 for all x, y ∈ TpM.
Hence
0 = T (x, y + z, x, y + z)
= T (x, y, x, y) + T (x, y, x, z) + T (x, z, x, y) + T (x, z, x, z)
= 2T (x, y, x, z),
and
0 = T (x + z, y, x + z, w) = T (x, y, z, w) + T (z, y, x, w),
0 = T (x + w, y, z, x + w) = T (x, y, z, w) + T (w, y, z, w).
Thus
0 = 2T (x, y, z, w) + T (z, y, x, w) + T (w, y, z, x)
= 2T (x, y, z, w)− T (y, z, x, w)− T (z, x, y, w)
= 3T (x, y, z, w),
as desired.
43
Chapter 8
February 2, 2006
Let (M, 〈 , 〉) be a pseudo-Riemannian manifold. A map ϕ : M → M is said to bean isometry if ϕ∗〈 , 〉 = 〈 , 〉.
Definition. A vector field Z on Mn is a Killing field if
〈∇XZ, Y 〉+ 〈X,∇Y Z〉 = 0 for all X, Y ∈ X(Mn),
or equivalently〈∇XZ,X〉 = 0 for all X ∈ X(Mn).
Note that 〈∇XZ, Y 〉(p) depends only on X(p) and Y (p). Thus to show that Z isKilling it suffices to show that
〈∇XZ, Y 〉(p) + 〈X,∇Y Z〉(p) = 0
for all vector fields X and Y such that 〈X, X〉, 〈Y, Y 〉, and 〈X,Y 〉 are constant.
Proposition. Let Z be a smooth vector field on a pseudo-Riemannian manifold(M, 〈 , 〉) with one-parameter group ϕt : t ∈ R. Then if each ϕt is an isometry, Zis a Killing field.
Remark. The converse is also true, but we will not prove it.
44
Proof. We can assume that 〈X, Y 〉 is constant. Then 〈ϕt∗(X), ϕt∗〉 is constant, and
〈X(p), Y (p)〉 = 〈ϕt∗p(X(p)), ϕt∗p(Y (p))〉= 〈(ϕt∗X)(ϕt(p)), (ϕt∗Y )(ϕt(p))〉= 〈(ϕt∗X)(p), (ϕt∗Y )(p)〉,
so
0 =d
dt〈(ϕt∗X)(p), (ϕt∗Y )(p)〉
∣∣∣∣t=0
=
⟨d
dt(ϕt∗X(p))
∣∣∣∣t=0
Y (p)
⟩+
⟨X(p),
d
dt(ϕt∗Y )(p)
∣∣∣∣t=0
⟩= −〈[Z,X], Y 〉(p)− 〈X, [Z, Y ]〉(p),
and we conclude that〈[Z,X], Y 〉+ 〈X, [Z, Y ]〉 = 0. (1)
But Z〈X, Y 〉 = 0, so〈∇ZX, Y 〉+ 〈X,∇ZY 〉 = 0. (2)
Subtracting (1) from (2) yields
〈∇XZ, Y 〉+ 〈X,∇Y Z〉 = 0.
Proposition. Suppose that γ : (a, b) → M is a geodesic in the pseudo-Riemannianmanifold (M, 〈 , 〉). Then
(a) 〈γ′, γ′〉 is constant (geodesics have constant speed).
(b) If Z is a Killing field, 〈Z, γ′〉 is constant.
Proof.
(a)d
dt〈γ′, γ′〉 = 2〈∇γ′γ
′, γ′〉 = 0;
45
(b)d
dt〈Z, γ′〉 = 〈∇γ′Z, γ′〉+ 〈Z,∇γ′γ
′〉 = 0.
This proposition can be used to study qualitative behavior of geodesics on surfacesof revolution, a surface M2 ⊆ R3 invariant under the rotation
ϕθ(x, y, z) = (cos θ x− sin θ y, sin θ x + cos θ y, z).
If γ : (a, b) → M2 ⊆ R3 is the path of a particle of mass m and Z has one-parametergroup ϕθ : θ ∈ R,
1
2m〈γ′, γ′〉 = kinetic energy, m〈γ′, Z〉 = angular momentum.
Thus the above proposition states that kinetic energy and angular momentum areconstant.
Thus suppose that F : (a, b)× R → R3 is an immersion of the form
F (u, θ) = (r(u) cos θ, r(u) sin θ, z(u)),
where r > 0, and that (r′(u))2 + (z′(u))2 = 1. For example, if (a, b) = R andr(u) = 2 + cos u, z(u) = sin u, F describes a torus of revolution. The Riemannianmetric on M is
〈 , 〉 = F ∗(dx)⊗ F ∗(dx) + F ∗(dy)⊗ F ∗(dy) + F ∗(dz)⊗ F ∗(dz)
= (d(r(u) cos θ))2 + (d(r(u) sin θ))2 + d(z(u))2
= (r′(u) cos θ du− r(u) sin θ dθ)2 + (r′(u) sin θ du + r(u) cos θ dθ)2 + (z′(u))2 (du)2
= (r′(u)2 + z′(u)2) du⊗ du + (r(u))2 dθ ⊗ dθ
= du⊗ du + (r(u))2 dθ ⊗ dθ.
Z = ∂∂θ
is a Killing field on M2 ⊆ R3 in this case.
If γ(t) = (u(t), θ(t)) is a geodesic on M , 12〈γ′, γ′〉 = E,
⟨γ′, ∂
∂θ
⟩= L, where E and L
are constants. Hence 12
(dudt
)2+ 1
2(r(u))2
(dθdt
)2= E,
(r(u))2 dθdt
= L.
46
Eliminating dθdt
yields
1
2
(du
dt
)2
+L2
2(r(u))2= E. (∗)
This is first order equation for u in terms of t, and the solution can be reduced to anintegration.
8.1 Qualitative Behavior of Solutions to (∗)We call VL(u) = L2
2(r(u))2the “effective potential energy.” We can then interpret (∗) as
kinetic energy + potential energy = constant.
For example, in the case of the torus,
VL(u) =L2
2(2 + cos u)2.
If L2
18< E < L2
2, the motion will be confined to one of the potential wells.
One can also calculate the curvature of surfaces of revolution quite easily.
Definition. Suppose that M2 ⊆ R3 is a surface. The Gaussian curvature of M2
is the function K : M2 → R defined by
K(p) = sectional curvature of TpM.
Thus if M2 is the image of F : (a, b)× R → R3 defined by
F (u, θ) = (r(u) cos θ, r(u) sin θ, z(u)),
47
where (r′(u))2 + (z′(u))2 = 1, a direct calculation gives
∂F
∂u= (r′(u) cos θ, r′(u) sin θ, z′(u)),
∂F
∂θ= (−r(u) sin θ, r(u) cos θ, 0),
N(u, θ) = unit normal
=∂F∂u× ∂F
∂θ∣∣∂F∂u× ∂F
∂θ
∣∣= (−z′(u) cos θ,−z′(u) sin θ, r′(u)).
Thus
α
(∂
∂u,
∂
∂u
)·N = D ∂
∂u
∂
∂u·N
=∂2F
∂u2·N
= −r′′(u)z′(u) + r′(u)z′′(u),
α
(∂
∂u,
∂
∂θ
)·N =
∂2F
∂u∂θ·N
= 0,
α
(∂
∂θ,
∂
∂θ
)·N =
∂2F
∂θ2·N
= r(u)z′(u).
Hence the Gaussian curvature of M2 is given by the formula
K(p) =α(
∂∂u
, ∂∂u
)· α(
∂∂θ
, ∂∂θ
)− α
(∂∂u
, ∂∂θ
)· α(
∂∂u
, ∂∂θ
)⟨∂∂u
, ∂∂u
⟩ ⟨∂∂θ
, ∂∂θ
⟩−⟨
∂∂u
, ∂∂θ
⟩2=
z′(u)
r(u)(r′(u)z′′(u)− z′(u)r′′(u)).
If r(u) = 2 + cos θ and z(u) = sin θ, we find that
K(u, θ) =cos u
2 + cos u.
48
Chapter 9
February 7, 2006
Definition. A Lie group is a group G which is also a manifold such that the maps
µ : G×G defined by µ(σ, τ) = σ · τ
andν : G → G defined by ν(σ) = σ−1
are smooth.
Example. GL(n, R) = n × n real matrices A : det A 6= 0 with µ(A, B) = A · B(matrix multiplication), ν(A) = A−1 (matrix inverse).
If G is a Lie group and σ ∈ G, define diffeomorphisms Lσ, Rσ : G → G by Lσ(τ) = σ·τand Rσ(τ) = τ · σ. We call Lσ and Rσ left and right translation by σ.
From Math 240A, Lecture 12, we know that a diffeomorphism F : M → M inducesa map F∗ : X(M) → X(M),
F∗(X)(p) = F∗F−1(p)(X(F−1(p))),
and that F∗([X,Y ]) = [F ∗(X), F ∗(Y )].
If G is a Lie group, a vector field X ∈ X(G) is said to be left invariant if (Lσ)∗(X) =X for all σ ∈ G.
49
Let g = X ∈ X(G) : (Lσ)∗(X) = X for all σ ∈ G. g is called the Lie algebra ofG. If X, Y ∈ g, then
(Lσ)∗([X, Y ]) = [(Lσ)∗(X), (Lσ)∗(Y )] = [X, Y ] for σ ∈ G,
so [X, Y ] ∈ g.
We claim that g is finite-dimensional and that in fact g ∼= TeG, where e is the identityelement of G. Indeed, we can define linear maps
α : g → TeG, β : TeG → g
by α(X) = X(e), β(x)(σ) = (Lσ)∗e(x). One easily checks that α β = id, β α = id,so α and β are real vector space isomorphisms.
Definition. A Lie group homomorphism is a homomorphism F : G → H be-tween Lie groups which is smooth.
If F : G → H is a Lie group homomorphism, F induces a linear map F∗ : TeG → TeHand hence a linear map F∗ : g → h, where g and h are the Lie algebras of G and H.Note that
F∗(X)(F (σ)) = (LF (σ))∗(F∗(X)(e)) = F∗σ(Lσ∗(X(e))) = F∗(X(σ)),
because LF (σ) F = F Lσ, and hence X is F -related to F∗(X) and F∗([X, Y ]) =[F∗(X), F∗(Y )].
More generally, a Lie algebra is a real vector space g together with an operation[ , ] : g× g → g such that
A. [X,Y ] = −[Y, X],
B. [[X, Y ], Z] + [[Y, Z], X] + [[Z,X], Y ] = 0.
50
A Lie algebra homomorphism between Lie algebras g and h is a linear mapf : g → h such that
f([X, Y ]) = [f(X), f(Y )].
We have defined a “functor” which assigns to each Lie group G its Lie algebra g andto each Lie group homomorphism F : G → H the corresponding Lie algebra homo-morphism F∗ : g → h. This is called the Lie group–Lie algebra correspondence.
Let G be a Lie group. A one-parameter subgroup of G is a Lie group homomor-phism θ : R → G. If θ : R → G is a one-parameter subgroup of G, then
ϕt = Rθ(t) : t ∈ R
is a one-parameter group of diffeomorphisms of G. It therefore generates a vectorfield X. Moreover, Lσ Rθ(t) = Rθ(t) Lσ, so (Lσ)∗(X) = X for σ ∈ G. Thus aone-parameter subgroup θ : R → G generates a corresponding left invariant vectorfield Xθ ∈ g. Conversely,
Proposition. If X is a left invariant vector field on a Lie group G, then X is com-plete, and the integral curve θX : R → G such that θX(0) = e is a one-parametersubgroup of G.
Proof. Let θX : (a, b) → G be the maximal integral curve for X such that θX(0) = e.Assume b < ∞. If σ ∈ G,
(σ · θX)′(t) = (Lσ θX)′(t) = Lσ∗(θ′X(t)) = (Lσ)∗(X(θX(t))) = Xσ·θX(t).
Hence σ · θX = Lσ θX : (a, b) → G is an integral curve for X such that σ · θX(0) = σ.In particular, if t0 ∈ (a, b), t0 > 0,
θ : (a + t0, b + t0) → G defined by θ(t) = θX(t0)θX(t− t0)
is an integral curve for X satisfying θ(t0) = θX(t0). By uniqueness of integral curves,
θX = θ on (a + t0, b). Hence θX can be extended to (a, b + t0), contradicting maxi-mality of b. Hence b = ∞. Similarly, a = −∞, and X is complete.
51
The one-parameter group of diffeomorphisms corresponding to X is now seen to beϕt = RθX(t) : t ∈ R, and
ϕt ϕs = ϕt+s ⇒ RθX(t) RθX(s) = RθX(t+s) ⇒ θX(s) · θX(t) = θXt + s,
so θX : R → G is a one-parameter subgroup of G.
Definition. Suppose that G is a Lie group. A biinvariant pseudo-Riemannianmetric on G is a Riemannian metric 〈 , 〉 such that Lσ and Rσ are isometries for allσ ∈ G.
For example, we have seen in 240A that
O(n) = A ∈ GL(n, R) : AT A = I
is a submanifold of GL(n, R) ⊆ Rn2. It inherits a Riemannian metric 〈 , 〉 which we
claim is biinvariant. Indeed, if we define
xij : O(n) → R by xi
j
a11 · · · a1
n...
...an
1 · · · ann
= aij,
then
〈 , 〉 =n∑
i,j=1
dxij ⊗ dxi
j.
Moreover,
xij(AB) =
n∑k=1
aikx
kj (B) ⇒ xi
j La =∑
aikx
kj
⇒ L∗A(xi
j) =∑
aikx
kj ⇒ L∗
A(dxij) =
∑ai
k dxkj .
52
Hence
L∗A〈 , 〉 =
n∑i,j=1
L∗A(dxi
j)⊗ L∗A(dxi
j)
=n∑
i,j,k,`=1
aika
i` dxk
j ⊗ dx`j
=n∑
j,k=1
dxkj ⊗ dxk
j
= 〈 , 〉,
so LA is an isometry. Similarly, RA is an isometry.
Theorem. Let G be a Lie group with a biinvariant pseudo-Riemannian metric 〈 , 〉.Then
(a) geodesics passing through the origin are just the one-parameter subgroup of G.
(b) the Levi-Civita connection on G is defined by
∇XY =1
2[X, Y ] for X, Y ∈ g.
(c) the curvature tensor R is given by
〈R(X, Y )W, Z〉 = 〈[X, Y ], [Z,W ]〉 for X,Y, Z, W ∈ g.
Proof. Let X be a left-invariant vector field with corresponding one-parameter groupθX . Since each RθX(t) is an isometry, X is a Killing field. Hence
〈∇Y X,Z〉+ 〈Y,∇ZX〉 = 0 for Y, Z ∈ g. (∗)
But since 〈X, X〉 is constant, 0 = Y 〈X, X〉 = 〈2∇Y X, X〉. Hence it follows from (∗)that 〈∇XX, Y 〉 = 0, which implies ∇XX = 0. But then
0 = ∇X+Y (X + Y ) = ∇XY +∇Y X,
53
and
∇XY −∇Y X = [X, Y ] ⇒ ∇XY =1
2[X, Y ].
This gives (b), and ∇XX = 0 ⇒ ∇θ′X(t)θ′X(t) = 0, so θX : R → G is a geodesic,
yielding (a). Finally
R(X, Y )W = ∇X∇Y W −∇Y∇XW −∇[X,Y ]W
=1
4[X, [Y, W ]]− 1
4[Y, [X, W ]]− 1
2[[X, Y ], W ]
=1
4[X, [Y, W ]] +
1
4[Y, [W, X]]− 1
2[[X, Y ], W ]
= −1
4[W, [X,Y ]]− 1
2[[X, Y ], W ]
= −1
4[[X, Y ], W ].
If X, Y, Z ∈ g,
0 = 2X〈Y, Z〉 = 2〈∇XY, Z〉+ 2〈Y,∇XZ〉 = 〈[X, Y ], Z〉+ 〈Y, [X, Z]〉,
and hence
〈R(X, Y )W, Z〉 = −1
4〈[[X,Y ], W ], Z〉
=1
4〈[W, [X,Y ]], Z〉
= −1
4〈[X, Y ]〉, [W, Z]〉
=1
4〈[X, Y ], [Z,W ]〉,
finishing (c).
54
Chapter 10
February 9, 2006
If A is an n× n real matrix, let
eA = I + A +1
2!A2 +
1
3!A3 + · · ·+ 1
n!An + · · · .
It is not difficult to see that this infinite series converges.
Let θA(t) = etA. Then θA(t) is a real analytic function of t, and
d
dt(etA) = AetA. (∗)
Indeed,
d
dt(etA) =
d
dt
(I + tA +
t2
2!A2 +
t3
3!A3 + · · ·+ tk
k!Ak + · · ·
)= A + tA2 +
t2
2!A3 + · · ·+ tk−1
(k − 1)!Ak + · · ·
= A
(I + tA +
t2
2!A2 + · · ·+ tk−1
(k − 1)!Ak−1 + · · ·
)= AetA.
Thus if b ∈ Rn,
x(t) = etA · b is a solution todx
dt= Ax.
The matrix exponential θA(t) = etA thus yields the solutions to systems of homoge-neous linear first order ordinary differential equations with constant coefficients.
55
Proposition. etAesA = e(t+s)A.
Proof. X1(t) = etA · esA and X2(t) = e(t+s)A are both solutions to the initial valueproblem
d(X(t))
dt= A ·X(t), X(0) = esA,
so they must be equal by uniqueness of solutions to systems of ordinary differentialequations.
In particular, etA · e−tA = I, so etA ∈ GL(n, R) for all t.
Conclusion. If A is an n× n real matrix, θA(t) = etA is a one-parameter subgroupof GL(n, R).
Let xij : GL(n, R) → R by
xij
a11 · · · a1
n...
...an
1 · ann
= aij.
If A =
a11 · · · a1
n...
...an
1 · · · ann
and B =
b11 · · · b1
n...
...bn1 · · · bn
n
, then
xij(B · θA(t)) =
n∑k=1
xik(B)xk
j (θA(t)),
d
dt(xi
j(B · θA(t)))
∣∣∣∣t=0
=n∑
k=1
xik(B)xk
j (A)
=n∑
k=1
xik(B)ak
j .
Hence the tangent vector to the curve
t 7→ BθA(t) = LB(θA(t)) at t = 0
56
is
X |B=n∑
i,j,k=1
xik(B)ak
j
∂
∂xij
∣∣∣∣B
.
In other words, if A = (aij) is an n×n real matrix defining the one-parameter subgroup
θA(t), the corresponding left invariant vector field is
XA =n∑
i,j,k=1
xika
kj
∂
∂xij
.
If
XB =n∑
i,j,k=1
xikb
kj
∂
∂xij
,
then a direct calculation yields
[XA, XB] =∑i,j,k
∑λ,µ,ν
xika
kj
∂
∂xij
(xλ
νbνµ
∂
∂xµλ
)−∑λ,µ,ν
∑i,j,k
xλνb
νµ
∂
∂xµλ
(xi
kakj
∂
∂xij
)=∑i,j,k,`
(xi
kakj b
j`
∂
∂x`i
− xikb
kj a
j`
∂
∂x`i
)=∑i,j,k
xik(x
kj (AB)− xk
j (BA))∂
∂xij
= X[A,B],
where [A, B] = AB−BA. Hence we see that the Lie algebra of GL(n, R) is gl(n, R) =n × n real matrices with bracket [A, B] = AB − BA. Most important Lie groupsare subgroups of GL(n, R), and their Lie algebras are subalgebras of gl(n, R).
Thus θA(t) = etA takes values in O(n) iff etA · etAT= I. Differentiating and setting
t = 0 yields the relation
d
dt(etA · etAT
)
∣∣∣∣t=0
= A + AT = 0.
Thus the Lie algebra of O(n) is
o(n) = A ∈ gl(n, R) : A + AT = 0.
57
Proposition. Suppose (M, 〈 , 〉) is a pseudo-Riemannian manifold, p ∈ Mn. Thenthere is an open neighborhood V of 0 in TpM such that if v ∈ TpM , the uniquegeodesic γv which satisfies the initial conditions
γv(0) = p, γ′v(0) = v
is defined for t ∈ [0, 1].
Proof. According to the theory of ordinary differential equations applied to
d2xi
dt2+
n∑j,k=1
Γijk
dxj
dt
dxk
dt= 0,
there is a neighborhood W of 0 in TpM and ε > 0 such that the geodesic γw is definedon [0, ε] for all w ∈ W . Let V = εW . If v ∈ V , v = εw for some w ∈ W and
d
dtγw(εt)
∣∣∣∣t=0
= εγ′w(0) = εw = v,
so γv(t) = γw(εt). γV is defined on [0, 1].
Define expp : V → M by expp(v) = γv(1). expp is called the exponential map.
Note that if v ∈ V , t 7→ expp(tv) is a geodesic (because expp(tv) = γtV (1) = γv(t)),and hence expp takes line segments through the origin in TpM to geodesic segmentsin M . For example, suppose M is a Lie subgroup G of GL(n, R) with a biinvariantmetric. (For example, G = O(n).) Let p = I, the identity matrix and v = A ∈ g ⊆gl(n, R). Then the geodesic γA satisfying the initial conditions
γA(0) = I, γ′A(0) = A
is just the one-parameter group θA(t) corresponding to A:
γA(t) = θA(t) = etA = I + tA +t2
2!A2 + · · · .
Thus expI(A) = eA in this case.
58
Proposition. There is an open neighborhood U of 0 ∈ TpM which expp maps dif-feomorphically onto an open neighborhood U of p in M .
Proof. By the inverse function theorem, it suffices to show that
(expp)∗0 : T0(TpM) → TpM
is an isomorphism. We identify T0(TpM) with TpM . If v ∈ TpM , define λv : R → TpMby λv(t) = tv. Then
expp λv(t) = expp(tv) = γv(t),
so(expp)∗0(v) = (exp p)∗0(λ
′v(0)) = γ′
v(0) = v,
and (expp)∗0 is indeed an isomorphism.
59
Chapter 11
February 14, 2006
Let (M, 〈 , 〉) be a Riemannian manifold, expressed in terms of local coordinates(x1, . . . , xn) as
〈 , 〉 =n∑
i,j=1
gij dxi ⊗ dxj.
Idea. Find coordinates near a point p which make the gij’s as simple as possible.
To do this we use the exponential map
expp : TpM → M
described last time.
We can always choose local coordinates (y1, . . . , yn) on a neighborhood of p such that
yi(p) = 0 and
⟨∂
∂yi
∣∣∣∣p
,∂
∂yj
∣∣∣∣p
⟩= δij.
These allow us to define coordinates
x1, . . . , xn : TpM → R by xi
(∑aj ∂
∂yj
∣∣∣∣p
)= ai.
60
(x1, . . . , xn) is a global coordinate system on TpM .
Suppose that expp maps the open neighborhood U of 0 in TpM diffeomorphicallyonly the open neighborhood U of p in M . Define coordinates (x1, . . . , xn) on U byxi = xi exp−1
p . The coordinates (x1, . . . , xn) are called normal coordinates at p.
Let r : U → R be defined by r =√
(x1)2 + · · ·+ (xn)2, and define a radial vectorfield R on U − p by
R =n∑
i=1
xi
r
∂
∂xi.
Gauß Lemma. dr = 〈R, ·〉 on U − p; thus the radius field R is perpendicular tothe level sets r =constant.
Definition. If f is any real-valued function on a Riemannian manifold (M, 〈 , 〉),the gradient of f is the vector field grad f defined by
df = 〈grad f, ·〉.
Thus the Gauß Lemma states that R = grad r.
To prove the Gauß Lemma, we make use of the rotation vector field
Eij = xi ∂
∂xj− xj ∂
∂xi.
Lemma 1 [Eij, R] = 0.
Proof. The radial vector field R is invariant under rotation. Thus if ϕt : t ∈ R isthe one-parameter group corresponding to Eij,
[Eij, R] = − d
dt(ϕt∗(R))
∣∣∣∣t=0
= 0.
61
Lemma 2 If ∇ is the Levi-Civita connection on M , ∇RR = 0.
Proof. If (a1, . . . , an) are constants with∑n
i=1(ai)2 = 1, then the curve γ defined by
xi(γ(t)) = ait
is an integral curve for R as one verifies by direct calculation. On the other hand,
γ(t) = expp
(∑ait
∂
∂xi
∣∣∣∣p
),
so γ is a geodesic. We conclude that the integral curves to R are geodesics, so∇RR = 0.
Lemma 3 〈R,R〉 ≡ 1 on U − p.
Proof. Suppose γ is an integral curve for R as above. Then ddt〈γ′(t), γ′(t)〉 =
2〈∇γ′(t)γ′(t), γ′(t)〉 = 0, so γ′(t) has constant length. But 〈γ′(0), γ′(0)〉 =
∑(ai)2 = 1,
so 〈γ′(t), γ′(t)〉 ≡ 1. It follows that 〈R,R〉 ≡ 1.
Lemma 4 〈R,Eij〉 = 0.
Proof. We calculate the derivative in the radial direction:
R〈R,Eij〉 = 〈∇RR,Eij〉+ 〈R,∇REij〉= 〈R,∇Eij
R〉
=1
2Eij〈R,R〉
= 0.
Then 〈R,Eij〉 is constant along the geodesic rays emanating from p. Let ‖X‖ =√〈X, X〉. Then
|〈R, Eij〉| ≤ ‖R‖ ‖Eij‖ ≤ ‖Eij‖ → 0
62
as r → 0. It follows that 〈R,Eij〉 ≡ 0.
Proof of Gauß Lemma. First note that R(r) = 1, Eij(r) = 0, by direct calculation.Suppose that X is a smooth vector field on part of U − p of the form
X = f∂
∂r+
n∑i,j=1
fijEij,
where f and fij are smooth. Then
dr(X) = X(r) = fR(r) +∑
fijEij(r) = f,
〈R,X〉 =⟨R, fR +
∑fijEij
⟩= f〈R,R〉 = f.
Thus dr = R.
Theorem. Suppose that (M, 〈 , 〉) is a Riemannian manifold and that U is an openball of radius ε centered at 0 in TpM such that expp maps diffeomorphically onto an
open neighborhood U of p in M . Suppose that v ∈ U and that
γ : [0, 1] → M by γ(t) = expp(tv).
Let q = expp(v). If λ : [0, 1] → Mn is any smooth curve such that λ(0) = p andλ(1) = q, then
(i) L(λ) ≥ L(γ), with equality holding only if λ is a reparametrization of γ.
(ii) J(λ) ≥ J(γ) with equality holding only if λ = γ.
Proof. We use normal coordinates (x1, . . . , xn) defined on U . Note that L(γ) =lengthof γ = r(q). Suppose that λ : [0, 1] → M is a smooth curve with λ(0) = p andλ(1) = q. We need to show that L(λ) ≥ L(γ).
63
Case I. λ does not leave U . Then
L(λ) =
∫ 1
0
√〈λ′(t), λ′(t)〉 dt
=
∫ 1
0
‖λ′(t)‖ dt
≥∫ 1
0
〈λ′(t), R(λ(t))〉 dt
=
∫ 1
0
dr(λ′(t)) dt
= r(λ(1))− r(λ(0))
= r(q)
= L(γ).
Moreover, equality holds if and only if λ′ is a nonnegative multiple of Rλ, i.e. if andonly if λ is a reparametrization of γ.
Case II. λ leaves U at some time t0 < 1. Then
L(λ) =
∫ 1
0
√〈λ′(t), λ′(t)〉 dt
≥∫ t0
0
√〈λ′(t), λ′(t) dt
≥∫ t0
0
〈λ′(t), R(λ(t))〉 dt
=
∫ t0
0
dr(λ′(t)) dt
= (r γ)(t0)
= ε
> L(γ).
This finishes the proof for L. The proof for J is quite similar.
Recall that if (M, 〈 , 〉) is a connected Riemannian manifold, we can define a mapd : M ×M → R by
d(p, q) = infL(γ) : γ : [0, 1] → M is a smooth map with γ(0) = p, γ(1) = q.
64
The only difficult part in showing that d is a metric is the implication
d(p, q) = 0 ⇒ p = q.
But if q lies in a normal coordinate system centered at p, this implication follows fromthe preceding theorem, while if q does not lie in such a coordinate system, the proofof the theorem shows that d(p, q) > ε for some ε > 0.
65
Chapter 12
February 16, 2006
We can give a small extension of the results established in Lecture 10. Namely, onan open neighborhood U of the zero-section in TM , we can define
exp : U → M ×M by exp(v) = (p, expp(v)),
for v ∈ TpM .
Proposition. The exponential map makes a small neighborhood V of the zero-section in TM (with V ⊆ U) diffeomorphically onto a neighborhood of the diagonal∆ ⊆ M ×M .
The proof is a straightforward application of the inverse function theorem.
It follows from this Proposition and the results of Lecture 11 that any point p has aneighborhood U with the following property: if q1, q2 ∈ U , q1 and q2 can be joined bya unique length-minimizing geodesic (which depends smoothly on (q1, q2) ∈ M ×M).
Recall that a Riemannian manifold (M, 〈 , 〉) can be made into a metric space (M, d)by setting
d(p, q) = inflength of γ | γ : [a, b] → M is a smooth curve with γ(a) = p, γ(b) = q.
The proof of the theorem given last time shows that if γ : [a, b] → M is a smoothcurve with γ(a) = p and γ(b) = q and length of γ is d(p, q), then γ is a reparametriza-tion of a constant speed geodesic.
66
Definition. A curve γ : [a, b] → M is a minimal geodesic if it is a constant speedgeodesic and its length is d(p, q).
Question. Given p, q ∈ M , does there exist a minimal geodesic from p to q?
Not Always. If M = R2 − (0, 0) with the standard Euclidean metric, p = (−1, 0)and q = (1, 0), there is no minimal geodesic from p to q.
Definition. A Riemannian manifold (M, 〈 , 〉) is geodesically complete if geodesicsin M can be extended indefinitely.
Examples of geodesically complete Riemannian manifolds include En (Rn with theEuclidean metric), Sn(a), and Hn(a).
Theorem 1 If (M, 〈 , 〉) is geodesically complete, then any two points p, q ∈ Mn canbe connected by a minimal geodesic.
Theorem 2 (Hopf and Rinow) Let (Mn, 〈 , 〉) be a connected Riemannian manifold.The following conditions are equivalent:
(a) (Mn, 〈 , 〉) is geodesically complete.
(b) expp : TpM → M is globally defined for some p ∈ M .
(c) (M, d) is complete as a metric space.
Proof of Theorem 1 We will in fact prove the following assertion (also needed inthe proof of Theorem 2): If p is a point in a Riemannian manifold (Mn, 〈 , 〉) suchthat expp : TpM → M is globally defined, then p can be joined to any point q ∈ Mby a minimal geodesic.
67
A candidate for the minimal geodesic: Let Bε be a closed ball of radius ε centeredat 0 ∈ TpM lying in an open set which is mapped diffeomorphically by expp onto anopen neighborhood of p.
Let Sε = ∂Bε, S = expp(Sε). S is a compact subset of M , so there is a point m ∈ Sof minimal distance from q. m = expp(εv) where v is a unit-length element of TpM .If a = d(p, q), define γ : [0, a] → M by γ(t) = expp(tv). γ is a candidate for theminimal geodesic from p to q.
It suffices to show that γ(a) = q. Thus it suffices to establish
d(γ(t), q) = a− t (∗)
for all t ∈ [0, a].
Note first that d(γ(t), q) ≥ a−t (because if d(γ(t), q) < a−t then d(p, q) ≤ d(p, γ(t))+d(γ(t), q) < t+a−t = a). Next note that if (∗) holds to t0 ∈ [0, a] it holds for t ∈ [0, t0],because if t ∈ [0, t0],
d(γ(t), q) ≤ d(γ(t), γ(t0)) + d(γ(t0), q) ≤ (t0 − t) + (a− t0) ≤ a− t.
Let t0 = supt ∈ [0, a] | (∗) holds for t. By continuity, (∗) also holds for t0. We willshow
(1) t0 ≥ ε,
(2) 0 < t0 < a leads to a contradiction.
To prove (1), note that
d(p, q) = infd(p, r) + d(r, q) : r ∈ S= infε + d(r, q) : q ∈ S= ε + d(m, q).
68
Hence a = ε + d(m, q), a− ε = d(m, q) = d(γ(ε), q), and (∗) holds for t = ε.
To prove (2), construct a sphere S about γ(t0) as we did for p, and let m be a point inS of minimal distance from q. Then d(γ(t0), q) = infd(γ(t0), r) + d(r, q) | r ∈ S =ε + d(m, q), so a− t0 = ε + d(m, q), so d(m, q) = a− (t0 + ε).
Note that d(p, m) ≥ t0 + ε, because otherwise d(p, q) < t0 + ε + a − (t0 + ε) = a.The broken path from p to γ(t0) to m has length t0 + ε. If this broken path had acorner, one could use the Proposition at the beginning of the lecture to construct ashorter curve from p to m, contradicting d(p, m) = t0 + ε. Hence m must lie on γ,m = γ(t0 + ε), and t0 + ε must satisfy (∗), contradicting maximality of (∗). Hencet0 = a, d(γ(a), q) = 0 and γ(a) = q.
Proof of Theorem 2. (a)⇒(b): Trivial.
(b)⇒(c): We assume that expp is globally defined for some p ∈ M . If A is a closedbounded subset of M , let K = supd(p, q) : q ∈ A. It follows from the proof ofTheorem 1 that
A ⊆ exppv ∈ TpM : ‖v‖ ≤ K.
A is a closed bounded set of a compact set and hence A must be compact.
If p1, p2, . . . , pi, . . . is a Cauchy sequence in Mn, its closure is contained in a closedbounded subset of M , hence a compact subset of M by the above argument. Thusp1, p2, . . . , pi, . . . must possess a convergent subsequence. This shows that (M, d) iscomplete as a metric space.
(c)⇒(a): This follows from the theory of ordinary differential equations, as we willsee next time.
69
Chapter 13
February 21, 2006
To complete the proof of the Hopf-Rinow theorem, we need to show that if (M, d) isa complete metric space, then geodesics on M can be extended indefinitely.
The equation for geodesics
d2xi
dt2+
n∑j,k=1
Γijk
dxj
dt
dxk
dt= 0
can be regarded as a first-order system in the variables xi and xi:dxi
dt= xi,
dxi
dt= −
∑nj,k=1 Γi
jkxjxk.
This first-order system corresponds to the vector field
X =n∑
i=1
xi ∂
∂xi−
n∑i,j,k=1
Γijkx
jxk ∂
∂xi
on the tangent bundle TM . Since geodesics do not depend upon the choice of coor-dinates on M , neither can X. X is called the geodesic spray.
Let T 1M = v ∈ TM : ‖v‖ = 1. T 1M is a codimension one submanifold of M calledthe unit tangent bundle. The geodesic spray is tangent to T 1M because geodesicshave constant speed. Thus we can regard X as a vector field on T 1M .
70
Let ϕt : t ∈ R be the local one-parameter group of diffeomorphisms correspondingto X. It follows from the theory of ordinary differential equations (see 240A, Lecture10) that given any v0 ∈ T 1M , there is an open neighborhood U of v0) in T 1M andan ε > 0 such that
ϕt(v) is defined for v ∈ V and t ∈ (−ε, ε).
Indeed, since X is never zero, it follows from a Lemma in Lecture 11 that there arecoordinates (y1, . . . , y2n−1) on a neighborhood U of v0 such that
X =∂
∂y1on U.
From this we see that U is decomposed into a disjoint union of integral curves for X.
Suppose now that γ : [0, a) → M is a unit-speed geodesic which cannot be extendedto a, a < ∞. Let t0, t1, . . . , ti, . . . be a sequence of real numbers with ti → a. Thenγ(t0), γ(t1), . . . , γ(ti), . . . is a Cauchy sequence in M (since d(γ(ti), γ(tj)) ≤ |ti − tj|),and a subsequence must converge to a limit point p0 ∈ M .
After possibly passing to a subsequence, we can assume that γ′(ti) → v0 ∈ T 1M .The curve γ′ : [0, a) → T 1M enters a coordinate neighborhood (U, (y1, . . . , y2n−1)) asdescribed above with X = ∂
∂yi , and γ′(ti) must lie on the orbit through v0. Hence γ
can be extended so that γ′(a) = v0, a contradiction.
Suppose that γ : [a, b] → M is a smooth curve. A smooth vector field along γ is asmooth function X : [a, b] → TM such that X(t) ∈ Tγ(t)M .
Alternatively, we can define γ∗TM by
γ∗TM = (t, v) ∈ [a, b]× TM : v ∈ Tγ(t)M.
We can regard γ∗TM as the total space of a vector bundle over [a, b], with projectionπ : γ∗TM → [a, b] defined by π(t, v) = t. Then a vector field X along γ is a sectionof γ∗TM ; that is, a smooth map X : [a, b] → γ∗TM such that π X = id.
71
We say that a vector field X along γ is parallel if
∇γ′X ≡ 0.
In terms of local coordinates (x1, . . . , xn), we can write
X =∑
f i ∂
∂xi, γ′ =
∑ dxi
dt
∂
∂xi,
and
∇γ′X =n∑
i=1
[df i
dt+
n∑j,k=1
Γijkf
j dxk
dt
]∂
∂xi.
Proposition. If γ : [a, b] → M is a smooth curve, t0 ∈ [a, b], v ∈ Tγ(t0)M , then thereis a unique vector field X along γ such that
∇γ′X = 0 and γ(t0) = v. (∗)
Proof. Suppose v =∑
ai ∂∂xi . Then (∗) is equivalent to the linear initial value
problemdf i
dt+∑
Γijk
dxj
dtfk = 0, f i(t0) = ai,
which has a unique solution on [a, b] by the theory of ordinary differential equations.
If γ : [a, b] → M is a smooth curve, we can define a vector space isomorphism
τγ : Tγ(a)M → Tγ(b)M
by τγ(v) = X(b), where X is the unique parallel vector field along γ such thatX(a) = v. Similarly, we can define τγ if γ is only piecewise smooth. We call τγ
parallel translation along γ.
If X and Y are parallel vector fields along γ, then
γ′〈X, Y 〉 = 〈∇γ′X, Y 〉+ 〈X,∇γ′Y 〉 = 0,
72
so 〈X, Y 〉 is constant along γ. Thus τγ : Tγ(a)M → Tγ(b)M is an isometry of innerproduct spaces.
Parallel translation around a geodesic triangle on S2 (with its metric of constantcurvature one) rotates through an angle ϕ. (It can be shown that the angle is equalto the area of the triangle.) By contrast, parallel translation around any closed pathin Euclidean space is the identity map.
73
Chapter 14
February 23, 2006
Linear ordinary differential equations are easier to solve than nonlinear ordinary dif-ferential equations. Given a solution to a nonlinear ordinary differential equation,one often studies nearby solutions by means of “linearization.” The linearization ofthe nonlinear geodesic equation is the Jacobi equation.
Suppose that Map([a, b], M) denotes the space of smooth maps γ : [a, b] → M andthat
Geod([a, b], M) = γ ∈ Map([a, b], M) | γ is a geodesic.
By completing with respect to a suitable Banach space norm, we can make Map([a, b], M)into an infinite-dimensional smooth manifold.
If we are lucky, Geod([a, b], M) will be a smooth manifold.
The tangent space of Map([a, b], M) is the space of sections of γ∗TM , and we ask:What is the tangent space to Geod([a, b], M)? To answer this question, we study“deformations through geodesics.”Let α : (−ε, ε) → M be a smooth map. then for each s ∈ (−ε, ε), we have a smoothcurve α(s) : [a, b] → M defined by α(s)(t). We use the notation ∂α
∂s(s0, t0) = (tangent
vector to s 7→ α(s, t0) at s = s0),∂α∂t
(s0, t0) = (tangent vector to t 7→ α(s0, t) att = t0). Although ∂α
∂sand ∂α
∂tare only defined along the image of α, their integral
curves commute on[
∂α∂s
, ∂α∂t
]along the image of α. We can let
α∗TM = (s, t, v) ∈ (−ε, ε)× [a, b]× TM | v ∈ Tα(s,t)M,
74
which is the total space of a vector bundle over (−ε, ε) × [a, b]. If we let Γ(α∗TM)denote the space of sections of α∗TM , then
∂α
∂s,∂α
∂t∈ Γ(α∗TM).
We are interested in the case where γ(t) = α(0, t) is a geodesic. We let
X(t) =∂α
∂s(0, t) = deformation vector field.
Note that X is a vector field along γ, so X ∈ Γ(γ∗TM).
Proposition. If each α(s) is a geodesic, then X will satisfy the Jacobi equation
∇γ′∇γ′X + R(X, γ′)γ′ = 0.
Proof. α(s) is a geodesic for every s implies that ∇ ∂α∂t
∂α∂t≡ 0, so ∇ ∂α
∂s∇ ∂α
∂t
∂α∂t
= 0, so
∇ ∂∂t∇ ∂
∂s
∂α∂t
+ R(
∂α∂s
, ∂α∂t
)∂α∂t
= 0. Thus ∇ ∂α∂t∇ ∂α
∂t
∂α∂s
+ R(
∂α∂s
, ∂α∂t
)∂α∂t
= 0. Now evaluate
along γ.
The proof suggests a broader definition of connection. Let (E, π, M) be a smoothvector bundle over M , Γ(E) the space of smooth sections of E.
Definition. A connection in the vector bundle E is a map∇ : X(M)×Γ(E) → Γ(E)(with ∇Xσ denoting ∇(X, σ), for X ∈ X(M), σ ∈ Γ(E)) such that
1. ∇fX+Y σ = f∇Xσ +∇Y σ,
2. ∇X(fσ + τ) = f∇Xσ + (Xf)σ +∇Xτ
for X, Y ∈ X(M), σ, τ ∈ Γ(E), f ∈ F (M).
Given a deformation α such as above, we can define a connection ∇ on α∗TM by
∇ ∂∂s
σ = ∇ ∂α∂s
σ, ∇ ∂∂t
σ = ∇ ∂α∂t
σ.
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This is called the pullback connection.
Solutions to the Jacobi equation are called Jacobi fields.
Suppose γ : [a, b] → M is a geodesic parametrized by arclength and that (e1, . . . , en)is an orthonormal basis for Tγ(a)M with e1 = γ′(a). We can parallel transport thesevectors to vector fields (E1, E2, . . . , En) along γ so that E ′
1 = γ. We can define thecomponent functions Ri
jk`(t) by
R(Ek, E`)Ej =∑
Rijk`Ei.
If X =∑n
i=1 f iEi, then Jacobi’s equation becomes
d2f i
dt2(s) +
n∑j=1
Ri1j1(t)f
j(t) = 0,
a first order linear system. This system has a linear space of solutions of dimension2n. The Jacobi fields which vanish at a given point form a linear subspace of dimen-sion n.
Example. If Mn has “constant sectional curvature”; that is,
R(X, Y )W = k[〈Y,W 〉X − 〈X, W 〉Y ],
then X =∑n
i=1 f iEi will be a Jacobi field iff
∇γ′∇γ′X + R(X, γ′)γ′ = k[〈X, γ′〉γ′ − 〈γ′, γ′〉X],
or equivalently, d2f1
dt2= 0,
d2f i
dt2= −kf for 2 ≤ i ≤ n.
The solutions are
f 1(t) = a1 + b1t,
f i(t) =
ai cos(
√kt) + bi sin(
√kt) if k > 0,
ai + bit if k = 0,
ai cosh(√−kt) + bi sinh(
√−kt) if k < 0.
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Here (a1, b1, . . . , an, bn) are constants of integration to be determined by the initialconditions.
If a = 0 and X(a) = 0, the solutions are
f 1(t) = b1t,
f i(t) =
bi sin(
√kt) if k > 0,
bit if k = 0,
bi sinh(√−kt) if k < 0.
Definition. Suppose that γ : [a, b] → M is a geodesic in a Riemannian manifoldwith γ(a) = p and γ(b) = q. We say that p and q are conjugate along γ if there is anonzero Jacobi field X along γ which vanishes at p and q.
Example. Antipodal points on Sn(1) are conjugate along the geodesics which jointhem, while Hn(1) and En do not have conjugate points. Great circles which startat he north pole p focus at the south pole q.
Suppose that M is geodesically complete, p ∈ M , v ∈ TpM . We have a geodesicγv : [0, 1] → M defined by γv(t) = expp(tv). We say that v belongs to the conjugatelocus in TpM if γv(0) and γv(1) are conjugate along γv.
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Chapter 15
February 28, 2006
Suppose that (M, 〈 , 〉) is a complete Riemannian manifold, p ∈ M . If v ∈ TpM , welet γv : R → M by γv(t) = expp(tv). We say that v belongs to the conjugate locusin TpM if γv(0) and γv(1) are conjugate along γv.
Proposition. A vector v ∈ TpM belongs to the conjugate locus iff expp∗ is singularat v.
Proof. ⇐: If expp∗ is singular at v, there is a nonzero vector w ∈ Tv(TpM) such thatexpp∗v(w) = 0. Define α : (−ε, ε)× [0, 1] → M by
α(s, t) = expp(t(v + sw)).
If X(t) = ∂α∂t
(0, t), X is a Jacobi field along γv which satisfies the condition X(0) = 0.Moreover,
X(1) = expp∗v(w) = 0.
Hence X is a nonzero Jacobi field which vanishes at γv(0) and γv(1), so γv(0) andγv(1) are conjugate along γv.
⇒: Conversely, if X is a nonzero Jacobi field along γv with X(0) = 0, then (∇γ′X)(0) =w for some nonzero w ∈ TpM , and
X(t) =∂α
∂s(0, t),
where α : (−ε, ε)× [0, 1] → M is the deformation defined by
α(s, t) = expp(t(v + sw)).
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Then expp∗v(w) = ∂α∂s
(0, 1) = X(1) = 0, so expp∗ is singular at v.
Recall that if F : M1 → M2 is a smooth map, a point q ∈ M2 is a regular value forF if
p ∈ F−1(q) ⇒ dFp is onto.
Otherwise it is a critical value. According to Sard’s Theorem (240A, Lecture 9),the set of critical values has measure zero. It therefore follows from the precedingProposition that
Corollary. If p ∈ M , the set of q ∈ M which are conjugate to p along some geodesichas measure zero.
Suppose that (Mn, 〈 , 〉) is a complete Riemannian manifold. A point p ∈ M is apole if the conjugate locus in TpM is empty.
Proposition. If (M, 〈 , 〉) is a complete Riemannian manifold whose curvature tensorsatisfies
〈R(x, y)y, x〉 ≤ 0 for all x, y ∈ TqM, q ∈ M,
then any point p ∈ M is a pole.
Proof. Let p ∈ M , v ∈ TpM , γ(t) = expp(tv). We need to show that p = γ(0) andq = expp(v) = γ(1) are not conjugate along γ.
Suppose, on the contrary, that X is a nonzero Jacobi field along γ which vanishes atγ(0) and γ(1). Then
∇γ′∇γ′X + R(X, γ′)γ′ = 0,
〈∇γ′∇γ′X,X〉 = −〈R(X, γ′)γ′, X〉 ≥ 0.
Write X =∑n
i=1 f iEi, where (E1, E2, . . . , En) is a parallel orthonormal from fieldalong γ. Then the above inequality becomes
n∑i=1
f i d2f i
dt2≥ 0.
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Since X(0) = X(1) = 0, f i(0) = f i(1) = 1, and we can integrate by parts
0 ≤∫ 1
0
∑f i d
2f i
dt2dt = −
∫ 1
0
∑(df i
dt
)2
dt le0.
We conclude that df i
dt≡ 0, f i ≡ 0, and X ≡ 0.
Theorem. (Hadamard-Cartan) If (Mn, 〈 , 〉) is a complete Riemannian manifoldwhose curvature R satisfies the inequality
〈R(x, y)y, x〉 ≤ 0 for all x, y ∈ TqM, q ∈ M, (∗)
then for any p ∈ M , expp : TpM → M is a smooth covering.
Recall that a smooth map π : M → M between connected manifolds is a smoothcovering, and each p ∈ M lies in an open set U which is evenly covered by π; that is,π−1(U) is a disjoint union of open sets, each of which is mapped diffeomorphically byπ onto U .
Corollary. If (M, 〈 , 〉) is a complete simply connected Riemannian manifold whosesectional curvatures satisfy (∗), then Mn is diffeomorphic to Rn.
This follows from the fact that if π : M → M is a smooth covering of connectedmanifolds and M is simply connected, then π : M → M is a diffeomorphism.
The Hadamard-Cartan Theorem follows from
Proposition. If (Mn, 〈 , 〉) is a complete Riemannian manifold and p ∈ M is a pole,then expp : TpM → M is a smooth covering.
We will prove this next time.
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Chapter 16
March 2, 2006
One of the most studied problems in Riemannian geometry is the relationship be-tween curvature and topology. To discuss this problem requires an understanding ofsome of the basic topological invariants: the cohomology groups Hk(M ; R) that wesaw in 240A and the homotopy groups πn(M).
16.1 The Fundamental Group
Most of you have seen this invariant described in other courses, so our presentationwill be brief.
Let X be a metric space, x0, x1 ∈ X. We let
Ω(X, x0, x1) = continuous paths γ : [0, 1] → M | γ(0) = x0, γ(1) = x1.
Suppose γ, λ ∈ Ω(X, x0, x1). We say γ and λ are homotopic and write γ ' λ ifthere is a continuous map α : [0, 1]× [0, 1] → X such that α(s, 0) = x0, α(s, 1) = x1
for all s ∈ [0, 1], α(0, t) = γ(t), α(1, t) = λ(t) for all t ∈ [0, 1].
It is not hard to check that ' is an equivalence relation: γ ' γ, γ ' λ ⇒ λ ' γ,γ ' λ and λ ' µ ⇒ γ ' µ. Let π(X, x0, x1) denote the set of equivalence classes, [γ]the equivalence class of γ.
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If [γ] ∈ π(X, x0, x1) and [λ] ∈ π(X, x1, x2), we define
[γ] · [λ] ∈ π(X, x0, x2)
to be the equivalence class of γ · λ, where
γ · λ =
γ(2t) for t ∈
[0, 1
2
],
λ(2t− 1) for t ∈[
12, 1].
One can check that this product is well-defined on equivalence classes: γ ' γ′ andλ ' λ′ implies γ · λ ' γ′ · λ′.
The fundamental groupoid of X is the “category” whose objects are the points ofX and whose morphisms from x0 to x1 are the elements of π(X, x0, x1). The funda-mental group of X at x0 is π(X, x0) = π(X, x0, x0).
To check that the fundamental groupoid is really a category, we need to verify asso-ciativity:
([γ] · [λ]) · [µ] = [γ] · ([λ] · [µ]) (∗)
and existence of an identity [εx0 ] ∈ π(X, x0), for each x0 ∈ X, such that
[εx0 ] · [γ] = [γ], [λ] · [εx0 ] = [λ], (∗∗)
when the products are defined. To establish (∗), we need to show that
(γ · λ) · µ ' γ · (λ · µ),
while to establish (∗∗), we need to verify that
εx0 · γ ' γ, λ · εx0 ' λ.
Explicit homotopies are given in Spanier, Algebraic Topology, Chapter 1, and manyother places.
To make π(X, x0) into a group, we need to define an inverse to [γ] ∈ π(X, x0). Thisis the equivalence class of γ−1 ∈ Ω(X, x0, x0), where
γ−1(t) = γ(1− t).
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One verifies that γ · γ−1 ' εx0 and γ−1 · γ ' εx0 , so [γ] · [γ−1] = [εx0 ] = [γ−1] · [γ].
A continuous map F : X → Y induces a covariant functor F∗ from the fundamentalgroupoid of X to the fundamental groupoid of Y by
F∗([γ]) = [F γ] ∈ π(Y, F (x0), F (x1)) for [γ] ∈ π(X, x0, x1).
This of course defines a homomorphism of fundamental groups
F∗ : π(X, x0) → π(Y, F (x0)).
We have a covariant functor from pointed topological spaces to groups such that
(X, x0) 7→ π(X, x0),
(F : (X, x0) → (Y, F (x0))) 7→ (F∗ : π(X, x0) → π(Y, F (x0))).
If X is arcwise connected and x0, x1 ∈ X and [γ] ∈ π(X, x0, x1), we can define anisomorphism
h[γ] : π(X, x0) → π(X, x1) by h[γ][λ] = [γ] · [λ] · [γ−1].
The fundamental group π1(X, x0) is actually the first of a series of homotopy groupsπn(X, x0), n = 1, 2, 3 . . . To define the other homotopy groups, one makes
Ω(X, x0) = Ω(X, x0, x0)
into a metric space by setting
d(γ, λ) = supd(γ(t), λ(t)) : t ∈ [0, 1].
One then defines πn(X, x0) to be πn−1(Ω(X, x0), εx0). This gives a series of groupswhich are usually very difficult to calculate. Thus it is still not known how to calcu-late πn(Sm, x0) for all values of n and m, for example.
Definition. A metric space X is simply connected if it is arcwise connected andπ1(X, x0) ∼= e, the trivial group.
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Let X and X be metric spaces. A covering from X to X is a surjective continuousmap π : X → X such that any p ∈ X lies in an open neighborhood U which is evenlycovered; even covered means that π−1(U) is a disjoint union of open sets each ofwhich is mapped homeomorphically by π onto U .
Example. Define π : R → S1 = z ∈ C : |z| = 1 by π(t) = e2πit.
The first two key facts about coverings are:
Proposition 1 (Unique path lifting) If π : X → X is a covering of arcwise connectedspaces, γ : [0, 1] → X is a continuous path and p ∈ π−1(γ(0)), then there exists a
unique path γ : [0, 1] → X such that γ(0) = p and π γ = γ.
To prove this, we cover the image of γ with a finite collection of evenly covered opensets U1, U2, . . . , Un. We then lift γ inductively over U1, U2, . . . using the defining prop-erty of evenly covered sets.
Proposition 2 (Unique homotopy lifting) If π : X → X is a covering of arcwise
connected spaces, α : [0, 1]× [0, 1] → X is a continuous map, and σ : [0, 1] → X is acontinuous path such that α(s, 0) = π σ(s) for s ∈ [0, 1], then there is a unique lift
α : [0, 1]× [0, 1] → X such that π α = α and α(s, 0) = σ(s) for s ∈ [0, 1].
The proof is quite similar to that of the previous proposition.
Proposition 3 Suppose π : X → X is a covering of arcwise connected spaces. If Xis simply connected, π is a homeomorphism.
Proof. We need only show that π is one-to-one. If x, y ∈ X and π(x) = π(y), then
since X is arcwise connected, there is a smooth path γ : [0, 1] → X with σ(0) = x,σ(1) = y. Let γ = π γ. If x = γ(0) = γ(1), γ ∈ Ω(X, x0). Since X is simply
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connected, there is a homotopy α : [0, 1]× [0, 1] → X such that
α(s, 0) = γ(s), α(s, 1) = x, α(0, t) = α(1, t) = x.
By unique homotopy lifting, there is an α : [0, 1] × [0, 1] → X such that π α = α,α(s, 0) = γ(s). The path t 7→ α(0, t) projects to a constant and hence must be con-stant by unique path lifting. Similarly, t 7→ α(1, t) and s 7→ α(s, 1) are constant. Itfollows that x = α(0, 0) = α(1, 0) = y.
If M and M are smooth manifolds, a smooth covering from M to M is just a smoothmap which is also a covering. If π : M → M is a smooth covering and M is simplyconnected, it follows from Proposition 3 that π is a diffeomorphism, a fact we citedin the previous lecture.
We now complete the proof of the Hadamard-Cartan Theorem. This will follow from:
Proposition. If (Mn, 〈 , 〉) is a complete Riemannian manifold and p ∈ M is a pole,then expp is a smooth covering.
Proof of Proposition. Since expp is nonsingular at each v ∈ TpM , we can define aRiemannian metric 〈〈 , 〉〉 on TpM by
〈〈x, y〉〉 = 〈expp∗v(x), expp∗v(y)〉 for x, y ∈ Tv(TpM).
Locally, expp is an isometry from (TpM, 〈〈 , 〉〉) to (M, 〈 , 〉), and it takes lines throughthe origin in TpM to geodesics in M . Thus lines through the origin must be geodesicsin the Riemannian manifold (TpM, 〈〈 , 〉〉). It follows from the Hopf-Rinow Theoremthat (TpM, 〈〈 , 〉〉) is complete. Thus the Proposition will be a consequence of
Lemma. If π : M → M is a local isometry of connected Riemannian manifolds withM complete, then π is a smooth covering.
Proof. One easily checks that π is surjective. Let q ∈ M . We need to show thatq possesses an open neighborhood U which is evenly covered by π. There existsε > 0 such that expq maps the open ball of radius 2ε diffeomorphically onto r ∈
85
M | d(r, q) < ε. Let qα : α ∈ A = π−1(q), U = r ∈ M | d(r, q) < ε,Uα = r ∈ M | d(r, qα) < ε. We have a commutative diagram
TqαMπ∗
isometry//
exp qα
TqM
expq
M π
// M
(Note that exp qα is defined on Bε = v ∈ TqαM | ‖v‖ < ε by completeness of M .)
Since expqαmaps Bε surjectively onto Uα and expq π∗ is a diffeomorphism from Bε
onto U , π maps Uα diffeomorphically onto U .
If r ∈ Uα ∩ Uβ, we would have geodesics γα, γβ of length less than ε from qα and qβ
to r. These would project to geodesics γα and γβ from q to r = π(r) of length lessthan ε. By uniqueness of geodesics emanating from q in a normal coordinate chart,we would have γα = γβ. Since π is a local isometry, γα and γβ would satisfy the same
initial conditions at r. thus γα = γβ and qα = qβ. We have proven that Uα ∩ Uβ 6= ∅,so α = β.
If r ∈ π−1(U), r = π(r) ∈ U , and there exists a geodesic γ from r to q of length lessthan ε. γ lifts to a geodesic from r to qα of length less than ε for some qα ∈ π−1(q).
Hence r ∈ Uα for some α ∈ A, and π−1(U) =⋃Uα : α ∈ A.
We have shown that each q ∈ M lies in an open neighborhood U such that π−1(U)is a disjoint union of open sets, each of which is mapped diffeomorphically by π ontoU . This shows that π is a smooth covering.
86
Chapter 17
March 7, 2006
17.1 Manifolds of Positive Ricci Curvature
Suppose (M, 〈 , 〉) is a Riemannian manifold with curvature tensor R. The Riccicurvature
Ric : TpM × TpM → R
is defined by
Ric(x, y) = trace of the linear map (v 7→ R(v, x)y).
Thus if (e1, . . . , en) is an orthonormal basis for TpM ,
Ric(e1, e1) =n∑
i=1
〈R(ei, e1)e1, ei〉 =n∑
i=2
K(σi),
where σi is the two-plane spanned by e1, ei. Hence if the sectional curvatures of Msatisfy the inequality
K(σ) ≥ 1
a2,
then the Ricci curvature will satisfy
Ric
(x
‖x‖,
x
‖x‖
)≥ n− 1
a2or Ric(x, x) ≥ n− 1
a2〈x, x〉
for x ∈ TpM , x 6= 0.
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The scalar curvature s : M → R is defined by
s(p) =n∑
i=1
Ric(ei, ei),
whenever (e1, . . . , en) is an orthonormal basis for TpM .
Myers’s Theorem. Let (Mn, 〈 , 〉) be a connected complete Riemannian manifoldwhose Ricci curvature satisfies
Ric(x, x) ≥ n− 1
a2〈x, x〉 for x ∈ TpM, (∗)
where a is a nonzero real number. Then Mn is compact and has finite fundamentalgroup. Moreover, d(p, q) ≤ πa for all p, q ∈ M .
To prove this theorem, we need to take the “second derivative” of the action function
J : Ω(M ; p, q) → R.
HereΩ(M ; p, q) = smooth paths γ : [0, 1] → M | γ(0) = p, γ(1) = q
and
J(γ) =1
2
∫ 1
0
〈γ′(t), γ′(t)〉 dt.
We think of J as a function on an “infinite-dimensional manifold” Ω(M ; p, q) andapply the second-derivative test for local minima. To carry this out, it is convenientto have a candidate for the tangent space to Ω(M ; p, q) at γ, and we take
TγΩ(M ; p, q) = smooth sections X of γ∗TM | X(0) = 0 = X(1)= smooth vector fields X along γ | X(0) = 0 = X(1).
The second derivative should then be a symmetric bilinear map
d2E(γ) : TγΩ(M ; p, q)× TγΩ(M ; p, q) → R.
If α : (−ε, ε) → Ω(M ; p, q) is a curve such that α(0) = γ and the function
α : (−ε, ε)× [0, 1] → M defined by α(s, t) = α(s)(t)
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is smooth, we say that α or α is a deformation of γ with deformation fieldX(t) = ∂α
∂s(0, t). The second derivative is defined by requiring
d2E(f)(X, X) =d2
ds2J(α(s))
∣∣∣∣s=0
whenever α is a deformation with deformation field X.
So we need to calculate d2
ds2 J(α(s))∣∣∣ s = 0, and we find that
d2
ds2J(α(s))
∣∣∣∣s=0
=d2
ds2
[1
2
∫ 1
0
⟨∂α
∂t,∂α
∂t
⟩dt
]∣∣∣∣s=0
=
∫ 1
0
∂
∂s
⟨∇ ∂α
∂s
∂α
∂t,∂α
∂t
⟩dt
∣∣∣∣s=0
=
∫ 1
0
[⟨∇ ∂α
∂s∇ ∂α
∂s
∂α
∂t,∂α
∂t
⟩+
⟨∇ ∂α
∂s
∂α
∂t,∇ ∂α
∂s
∂α
∂t
⟩]dt
∣∣∣∣s=0
=
∫ 1
0
[⟨∇ ∂α
∂s∇ ∂α
∂t
∂α
∂s,∂α
∂t
⟩+
⟨∇ ∂α
∂t
∂α
∂s,∇ ∂α
∂t
∂α
∂s
⟩]dt
∣∣∣∣s=0
=
∫ 1
0
[⟨R
(∂α
∂s,∂α
∂t
)∂α
∂s,∂α
∂t
⟩+
⟨∇ ∂α
∂t∇ ∂α
∂s
∂α
∂s,∂α
∂t
⟩+
⟨∇ ∂α
∂t
∂α
∂s,∇ ∂α
∂t
∂α
∂s
⟩]dt
∣∣∣∣s=0
=
∫ 1
0
[⟨R
(∂α
∂s,∂α
∂t
)∂α
∂s,∂α
∂t
⟩+
∂
∂t
⟨∇ ∂α
∂s
∂α
∂s,∂α
∂t
⟩−⟨∇ ∂α
∂s
∂α
∂s,∇ ∂α
∂t
∂α
∂t
⟩+
⟨∇ ∂α
∂t
∂α
∂s,∇ ∂α
∂t
∂α
∂s
⟩]dt
∣∣∣∣s=0
.
The second term integrates to zero because(∇ ∂α
∂s
∂α
∂s
)(0, 0) = 0 =
(∇ ∂α
∂s
∂α
∂s
)(0, 1).
The third term vanishes because γ is a geodesic. We are left with
d2
ds2J(α(s))
∣∣∣∣s=0
=
∫ 1
0
[〈R(X, γ′)X, γ′〉+ 〈∇γ′X,∇γ′X〉] dt,
and by “polarization,” we conclude that
d2E(γ)(X,Y ) =
∫ 1
0
[〈∇γ′X,∇γ′Y 〉 − 〈R(X, γ′)γ′, Y 〉] dt.
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We can integrate by parts to obtain
(d2J)(γ)(X, Y ) = −∫ 1
0
〈∇γ′∇γ′X + R(X, γ′)γ′, Y 〉 dt. (∗∗)
Proof of Myers’s Theorem. Suppose that p, q ∈ M and d(p, q) > πa. Let γ :[0, 1] → M be a geodesic which minimizes J . Since |γ′| is constant, |γ′| = d(p, q).Choose parallel orthonormal vector fields (E1, . . . , En) along γ such that E1 = 1
d(p,q)γ′.
Let Xi(t) = (sin πt)Ei(t) for i = 2, . . . , n. Then Xi(0) = 0 = Xi(1), so Xi ∈TγΩ(M ; p, q). Then
∇γ′Xi = −π cos πt Ei,
∇γ′∇γ′Xi = −π2 sin πt Ei,
〈∇γ′∇γ′Xi, Xi〉 = −π2 sin2 πt.
On the other hand,
〈R(Xi, γ′)γ′, Xi〉 = d(p, q)2 sin2 πt〈R(Ei, E1)E1, Ei〉.
Substituting into (∗∗) yields
d2J(γ)(Xi, Xi) =
∫ 1
0
[π2 sin πt− d(p, q)2 sin2 πt R(Ei, E1, E1, Ei)] dt
=
∫ 1
0
sin2 πt[π2 − d(p, q)2R(Ei, E1, E1, Ei)] dt.
Hence
n∑i=2
d2J(γ)(Xi, Xi) =
∫ 1
0
sin2 πt[(n− 1)π2 − d(p, q)2 Ric(E1, E1)] dt.
But
d(p, q) > πa ⇒ d(p, q)2 > π2a2
Ric(E1, E1) ≥ n−1a2
⇒ d(p, q)2 Ric(E1, E1) > (n− 1)π2.
Hencen∑
i=2
d2J(γ)(Xi, Xi) < 0 and d2J(γ)(Xi, Xi) < 0
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for some i, 2 ≤ i ≤ n. This contradicts minimality of the geodesic. Thus it must bethe case that d(p, q) ≤ πa.
It follows that M is compact because it lies in the image under the exponential mapof the ball of radius πa.
To prove that π1(M) is finite, we use the fact that any connected smooth manifold
M has a universal cover M : a simply connected manifold such that π : M → M isa smooth covering. Give M the metric π∗〈 , 〉, where 〈 , 〉 is the Riemannian metric
on M . Then M is also complete and has Ricci curvature satisfying (∗). Hence M isalso compact, and if p ∈ M , π−1(p) must be a finite set. But the points of π−1(p) arein one-to-one correspondence with the elements of π1(M, p). Thus π1(M) is finite.
91
Chapter 18
March 9, 2006
In the proof of Myers’s Theorem, we used the following theorem from topology:
Theorem. Let M be a connected smooth manifold. Then there is a smooth manifoldM which is simply connected and a smooth covering projection π : M → M . M isuniquely determined up to diffeomorphism.
We call M the universal cover of M .
If p ∈ M , p ∈ M , and π(p) = p, we can define a map h : π1(M, p) → π−1(p) byh([γ]) = γ(1), where γ : [0, 1] → M is the unique map such that π γ = γ andγ(0) = p.
Proposition 2 from Lecture 16 shows that h is well-defined, i.e. γ(1) does not depend
on the choice of γ ∈ [γ]. We claim that h is a bijection. h is onto because M is path
connected. h is injective because M is simply connected; hence if q ∈ π−1(p) and γ, λ
are two curves in M with γ(0) = λ(0) = p and γ(1) = λ(1) = q, then γ ' λ, and
hence π γ ' π λ.
Example. M = S1 = z ∈ C | |z| = 1 has universal cover M = R with projectionπ defined by π(t) = e2πit. (This is actually a group homomorphism.) In this case,h : π1(S
1, 1) → Z, and one can check that h is a group isomorphism.
92
This example shows that universal covers can sometimes be used to calculate funda-mental groups.
Using them, one can show, for example, that π1(S1 × S1) ∼= Z ⊕ Z. For n ≥ 2, Sn
is simply connected, and R × Sn is the universal cover of S1 × Sn. It follows thatπ1(S
1 × Sn) ∼= Z for n ≥ 2. Hence S1 × S2 does not have a metric of positive Riccicurvature by Myers’s Theorem. It therefore does not have a metric of positive sec-tional curvature.
Open Problem. Does S2 × S2 have a metric with positive sectional curvature?
Can we find further relationships between curvature and the fundamental group? Theanswer is “yes,” and a famous example is
Synge’s Theorem. Let (M, 〈 , 〉) be a compact Riemannian manifold with positivesectional curvatures.
1. If M is even-dimensional and oriented, it must be simply connected.
2. If M is odd-dimensional, it must be orientable.
Recall that RP n is Sn with antipodal points identified. There is a natural projectionπ : Sn → RP n which exhibits Sn as the universal cover of RP n. Since π is two-to-one,π1(RP n) ∼= Z2. Moreover, there is a Riemannian metric 〈 , 〉 on RP n such that π∗〈 , 〉is the standard Riemannian metric on Sn of constant curvature one.
If n is odd, RP n is orientable, in agreement with the second statement of Synge’stheorem. If n is even, RP n is not orientable, hence not a counterexample to the firststatement of Synge’s theorem.
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The proof of Synge’s theorem is based upon the notion of “free homotopy class.”Let Map(S1, M) = continuous maps γ : S1 → M, where S1 = [0, 1] with thepoints 0 and 1 identified. We say that two elements γ, λ ∈ Map(S1, M) are freelyhomotopic, and write γ ' λ, if there is a continuous map
α : [0, 1]× S1 → M such that
α(0, t) = γ(t),
α(1, t) = λ(t).
Free homotopy is just homotopy without regard to base points. It is easily verifiedthat ' is an equivalence relation. We let [S1, M ] be the set of equivalence classes andwrite [γ] for the equivalence class of γ ∈ Map(S1, M).
It is not too difficult to show that [S1, M ] is in one-to-one correspondence with the setof conjugacy classes in π1(M), when M is a connected manifold. Thus a connectedmanifold is simply connected iff [S1, M ] consists of a single point.
If M is connected, [S1, M ] can be regarded as the set of components of Map(S1, M).One of the components contains all of the constant paths
M = γ : S1 → M | γ(t) ≡ p for all t ∈ [0, 1].
Elements of the component of Map(S1, M) which contains M are said to be nullhomotopic.
We will establish the following theorem next time:
Lemma. If (M, 〈 , 〉) is a compact Riemannian manifold and λ : S1 → M is not freelyhomotopic to a constant path, then there is a smooth closed geodesic γ : S1 → Mwhich is freely homotopic to λ and minimizes the action integral J among all smoothcurves freely homotopic to γ.
Here the action integral J is defined by
J =1
2
∫ 1
0
〈γ′(t), γ′(t)〉 dt.
94
Let us assume the Lemma for now and prove Synge’s Theorem with its help. Theproof uses the “second derivative” of J in a manner quite similar to that employedin Myers’s Theorem.
We give a slightly new meaning to the symbol Map(S1, M). We need to regard itas the space of smooth maps γ : S1 → M . We want to think of Map(S1, M) as an“infinite-dimensional smooth manifold” with tangent space
Tγ Map(S1, M) = smooth sections X of γ∗TM.
A deformation of γ is a map α : (−ε, ε) → Map(S1, M) such that α(0) = γ whichis smooth in the following sense: if we define
α : (−ε, ε)× S1 → M by α(s, t) = α(s)(t),
then α is smooth.
To each such deformation is associated a deformation vector field
t 7→ X(t) =∂α
∂s(0, t),
which is an element of Tγ Map(S1, M). Conversely, given X ∈ Tγ Map(S1, M), wecan construct a deformation α : (−ε, ε) → Map(S1, M) which has deformation vectorfield X.
As in the proof of Myers’s Theorem, we can calculate
d2
ds2J(α(s))
∣∣∣∣s=0
= d2J(γ)(X, X).
he steps are virtually the same, except that we do not have to worry about endpoints.The result is
d2J(γ)(X, X) =
∫ 1
0
[〈∇γ′X,∇γ′X〉 − 〈R(X, γ′)γ′, X〉] dt. (∗)
95
If γ : S1 → M minimizes in its free homotopy class, we must have d2J(γ)(X, X) ≥ 0for all X ∈ Tγ Map(S1, M).
Let us now turn to part 1 of Synge’s Theorem. If M is not simply connected, theLemma says there is a smooth closed geodesic γ : S1 → M which minimizes in its freehomotopy class. Let p = γ(0). Parallel translation around γ gives an isomorphismτγ : TpM → TpM .
τγ is defined by τγ(x) = X(1), where X(t) is the unique section of γ∗TM → [0, 1] suchthat ∇γ′X ≡ 0 and X(0) = x. Note that |τγ(x)| = |x|; that is, τγ is an orthogonaltransformation of TpM . Since M is oriented, τγ preserves orientation.
We can choose an orthonormal basis (e1, . . . , en) for TpM such that e1 = 1L(γ)
γ′. Withrespect to such a basis, τγ is represented by an element of
SO(n) = A ∈ GL(n, R) : AT A = I and det A = 1.
A canonical form theorem for elements of SO(n) states that after conjugation, we canarrange that τγ is represented by a matrix of the form
A =
±1±1
. . .
±1cos θ1 − sin θ1
sin θ1 cos θ1
. . .
cos θk − sin θk
sin θk cos θk
.
Since the determinant of A is one, the number of (−1)’s is even. Since dim M iseven the number of (+1)’s is also even. There must be a +1 that corresponds toe1 = 1
L(γ)γ′, and hence there must be at least one more +1 corresponding to e2. Thus
there is a unit-length vector field E2 along γ such that ∇γ′E2 ≡ 0, E2(1) = E2(0),
96
and E2 ⊥ E1. Then
d2J(γ)(E2, E2) = −∫ 1
0
〈R(E2, γ′)γ′, E2〉 dt,
which is less than 0 since sectional curvatures are assumed positive. This yields acontradiction. Hence M must be simply connected.
The proof of part 2 is similar; one uses a smooth closed geodesic γ : S1 → M suchthat parallel translation around γ is orientation-reversing.
97
Chapter 19
March 14, 2006
To finish the proof of Synge’s Theorem, we need to establish
Theorem. Let (M, 〈 , 〉) be a compact Riemannian manifold. If λ : S1 → M isa smooth map which is not freely homotopic to a constant, then there is a smoothclosed geodesic γ : S1 → M freely homotopic to γ which minimizes the action integralJ among all smooth curves freely homotopic to λ.
Here is S1 is regarded as [0, 1] with endpoints identified, and
J(γ) =1
2
∫ 1
0
〈γ′(t), γ′(t)〉 dt.
Let Map(S1, M) = piecewise smooth paths γ : S1 → M. Informally, we think ofMap(S1, M) as a smooth manifold with tangent space
Tγ Map(S1, M) = piecewise smooth sections of γ∗TM.
How do we make this picture rigorous? One approach is through finite-dimensionalapproximations. More precisely, we let
Map(S1, M)c = γ ∈ Map(S1, M) | J(γ) ≤ c.
The Cauchy-Schwarz inequality will enable us to approximate this space by a finite-dimensional manifold.
98
Let N be a large integer, and for integers i such that 0 ≤ i ≤ N , let ti = iN
. LetBGN(S1, M) be the set of continuous maps γ : [0, 1] → M with γ(0) = γ(1) andγ |[ti−1,ti] is a smooth constant speed geodesic for 1 ≤ i ≤ N and
BGN(S1, M)c = γ ∈ BGN(S1, M) | J(γ) ≤ c.
We have a maph : BGN(S1, M)c → M × · · · ×M︸ ︷︷ ︸
N
defined by h(γ) = (γ(t1), γ(t2), . . . , γ(tN)).
Recall that the Cauchy-Schwarz inequality asserts that(∫ b
a
f(t)g(t) dt
)2
≤∫ b
a
(f(t))2 dt
∫ b
a
(g(t))2 dt,
with equality holding if and only if f and g are linearly dependent.
We set f(t) =√〈γ′(t), γ′(t)〉 and g(t) = 1. Then(∫ b
a
√〈γ′(t), γ′(t)〉 dt
)2
≤∫ b
a
〈γ′(t), γ′(t)〉 dt
∫ b
a
a dt,
which can be written(L(γ))2 ≤ 2(b− a)J(γ),
where
L(γ) =
∫ b
a
√〈γ′(t), γ′(t)〉 dt, J(γ) =
1
2
∫ b
a
〈γ′(t), γ′(t)〉 dt.
Thus if J(γ) ≤ c,L(γ) ≤
√b− a · c. (∗)
By taking b − a sufficiently small, we can arrange that L(γ) < δ where δ > 0 is anyprescribed number. Since M is compact, there exists δ > 0 such that if p, q ∈ M ,d(p, q) < δ, p and q are connected by a unique geodesic γp,q of length d(p, q) whichdepends smoothly on p and q. Indeed, we can choose δ > 0 so that if Ω(M ; p, q)2δ =
99
piecewise smooth γ : [0, 1] → M such that γ(0) = p, γ(1) = q, and L(γ) ≤ 2δ withthe topology defined by the metric
d(γ1, γ2) = supd(γ1(t), γ2(t)) : t ∈ [0, 1],
we ca define a continuous map
H : Ω(M ; p, q)× [0, 1] → Ω(M ; p, q)
such that H(γ, 0) = γ and H(γ, 1) = γp,q.
We can choose N so that√
1N· c < δ, so that it follows from (∗) that
L(γ |[ i−1
N, iN ]
)< δ.
If γ ∈ Map(S1, M)c, we let g(γ) = γ, where γ |[ i−1N
, iN ] is the unique geodesic from
γ(
i−1N
)to γ
(iN
). This defines a map g : Map(S1, M))c → BGN(S1, M)c. In fact,
with the metric space topology defined above, the homotopy H shows that g is a“homotopy equivalence.”
We call γ a broken geodesic and regard it as an approximation to γ.
Proof of the Theorem. Let
µ = infL(γ) : γ ∈ Map(S1, M) and γ is freely homotopic to λ.
Choose c > µ. Choose a sequence γi ∈ Map(S1, M)c such that J(γi) → µ. Thenγi = g(γi) is a broken geodesic with J(γi) → µ. h(γi) = (γi(t1), γi(t2), . . . , γi(tN)) ∈M × · · · ×M︸ ︷︷ ︸
N
. But ˜M × · · · ×MN is compact. Hence a subsequence of h(γi) : i =
1, 2, . . . will converge to a point (p1, . . . , pN) ∈ M × · · · ×M . Let γ be the brokengeodesic such that γ |[ti−1,ti] is a constant speed minimal geodesic from pi−1 to pi.Then J(γ) = µ.
γ is piecewise smooth with possible corners at ti = iN
. But γ cannot have cornersbecause existence of a corner would enable us to construct a piecewise smooth path
100
γ : S1 → M with J(γ) < µ by smoothing off the corner. Hence γ is a smooth closedgeodesic with J(γ) = µ.
A similar argument could be used to give an alternate proof of the following theoremfrom Lecture 12:
Theorem. Let (M, 〈 , 〉) be a complete Riemannian manifold, p, q ∈ M . Then thereexists a smooth geodesic
γ : [0, 1] → M
such that γ(0) = p, γ(1) = q, and L(γ) = d(p, q).
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Chapter 20
March 16, 2006
Let (M, 〈 , 〉) be a smooth Riemannian manifold with
〈 , 〉 =n∑
i,j=1
gij dxi ⊗ dxj =n∑
α,β=1
gαβ dyα ⊗ dyβ,
in two different coordinate systems (x1, . . . , xn) and (y1, . . . , yn). Then
gαβ =n∑
i,j=1
gij∂xi
∂yα
∂xj
∂yβ,
so
det(gαβ) = det(gij)
(det
(∂xi
∂yα
))2
,
so √det(gαβ) =
√det(gij)
∣∣∣∣det
(∂xi
∂yα
)∣∣∣∣ .It follows that if M is oriented and the coordinate systems (x1, . . . , xn) and (y1, . . . , yn)are positively oriented,√
det(gij) dx1 ∧ · · · ∧ dxn =√
det gαβ dy1 ∧ · · · ∧ dyn.
On a positively oriented Riemannian manifold (M, 〈 , 〉), we can thus define a volumeform Θ by
Θ =√
det(gij) dx1 ∧ · · · ∧ dxn,
102
whenever (x1, . . . , xn) are positively oriented coordinates on M and 〈 , 〉 =∑
gij dxi⊗dxj.
If K is a compact region within M , then
volume of K =
∫K
Θ.
We can apply this to the case of a compact Lie group G.
If G is any Lie group, any inner product
〈 , 〉 : TeG× TeG → R
extends to a unique left invariant Riemannian metric on G such that
〈Lσ∗(x), Lσ∗(y)〉σ = 〈x, y〉e for x, y ∈ TeG,
where Lσ is left translation, defined by Lσ(τ) = σ · τ .
Can we construct a metric which is also invariant under right translation, Rσ : G → G,Rσ(τ) = τ · σ? Note that Rσ(e) = e, so (Rσ)∗ : TeG → TeG. Thus given an innerproduct 〈 , 〉 on TeG, we can define
f : G → L2(TeG; R) by f(σ) = R∗σ〈 , 〉.
If G is compact, we can then set
〈〈 , 〉〉 =
∫G
f(σ) Θ∫G
1 Θ∈ L2(TeG; R).
This will be an inner product on TeG which extends to a biinvariant Riemannianmetric 〈〈 , 〉〉 on G.
The conclusion is that any compact Lie group has a biinvariant Riemannian metric.For example, if G = O(n),
o(n) = TIO(n) ∼= n× n matrices A | A + AT = 0,
103
and 〈〈A, B〉〉 =trace(AT B) defines a biinvariant metric on O(n).
It follows from Lecture 9 that if G is a compact Lie group with a biinvariant Rieman-nian metric 〈 , 〉, then the curvature of G is given by
〈R(X, Y )W, Z〉 =1
4〈[X, Y ], [Z,W ]〉
for X, Y, Z, W ∈ g, the Lie algebra of G. Thus curvature is given by a simple explicitformula in this case.
We can also determine curvatures of certain submanifolds of G. To do this, we needto generalize the Gauß equation from Lecture 7. Suppose (M, 〈 , 〉) is a Riemannianmanifold, M0 an imbedded submanifold. Let X(M0) be the set of smooth vectorfields on M0 and X(M0)
⊥ the set of smooth vector fields in M along M0 which areperpendicular to M . Let i : M0 → M be the inclusion. Then
(i∗TM)p = TpM0 ⊕NpM0,
where NpM0 = v ∈ TpM : v ⊥ TpM0. As p varies over M0, we get a “direct sum”of vector bundles i∗TM = TM0 ⊕NM0 and
X(M0) = Γ(TM0), X(M0)⊥ = Γ(NM0).
Let D : X(M) × X(M) → X(M) be the Levi-Civita connection on M . We can thendefine
∇ : X(M0)× X(M0) → X(M0) by ∇XY = (DXY )T ,
α : X(M0)× X(M0) → X(M0)⊥ by α(X,Y ) = (DXY )⊥,
where (·)T and (·)⊥ are orthogonal projections to tangent and normal space, respec-tively.
Lemma. ∇ is the Levi-Civita connection on M and α is symmetric and bilinear overfunctions.
104
The proof is exactly the same as before (when M was Euclidean space).
The symmetric bilinear form α : X(M0) × X(M0) → X(M0)⊥ is called the second
fundamental form of M0 in M . Note that α is defined pointwise and restricts to abilinear map
α : TpM0 × TpM0 → NpM0
for each p ∈ M0.
Theorem. Suppose (M, 〈 , 〉) is a Riemannian manifold, M0 is an imbedded sub-manifold with the induced metric (also denoted by 〈 , 〉). If R is the curvature of Mand R0 is the curvature of M0, then
〈R0(x, y)w, z〉 = 〈R(x, y)w, z〉+ 〈α(z, z), α(y, w)〉 − 〈α(x, w), α(y, z)〉, (∗)
for x, y, z, w ∈ TpM .
Proof. Same as before.
Definition. The submanifold M0 ⊆ M is said to be totally geodesic if α ≡ 0.
Note that if γ : (−ε, ε) → M0 ⊆ M , then
(Dγ′γ′)(t) = ((Dγ′γ
′)(t))T + ((Dγ′γ′)(t))⊥
= ∇γ′γ′(t) + α(γ′(t), γ′(t)).
Thus if M0 is totally geodesic, Dγ′γ′ = ∇γ′γ
′.
Indeed, we easily see that M0 ⊆ M is totally geodesic iff geodesics in M0 are alsogeodesics in M .
105
If M0 ⊆ M is totally geodesic, the Gauß equation (∗) becomes
〈R0(x, y)w, z〉 = 〈R(x, y)w, z〉 for x, y, z, w,∈ TpM0.
How do we construct totally geodesic submanifolds?
Theorem. Let (M, 〈 , 〉) be a compact Riemannian manifold, F : M → M anisometry, M0 a connected component of
Fix(F ) = p ∈ M | F (p) = p.
Then M0 is an imbedded totally geodesic submanifold of M .
Proof. Suppose p ∈ M0, and let
Bε = v ∈ TpM : ‖v‖ < ε.
Choose ε > 0 sufficiently small that expp maps Bε diffeomorphically onto an opensubset U of p in M . Since F is an isometry, it takes geodesics to geodesics. Hence ifv ∈ TpM , γ(t) = F (expp(tv)) is a geodesic in M and γ′(0) = F∗p(v). Thus
F (expp(tv)) = γ(t) = expp(tF∗p(v)),
and henceF expp = expp F∗p : Bε → U.
Let W = v ∈ TpM | F∗p(v) = v, a linear subspace of TpM . Moreover,
ϕ : exp−1p : U → TpM
maps U ∩ M0 bijectively onto Bε ∩ M0 if ε is sufficiently small. Thus (U,ϕ) is asubmanifold chart for M0 and M0 is indeed an imbedded submanifold of M . Byconstruction, geodesics through p in M0 are also geodesics through p in M . So M0 istotally geodesic.
We now return to the example of a compact Lie group with a biinvariant Riemannianmetric 〈 , 〉. Can we construct isometries of G which have interesting fixed point
106
sets?
Note that ν : G → G define by ν(σ) = σ−1 takes biinvariant metrics to biinvariantmetrics and
ν∗e = −id : TeG → TeG,
so ν must be an isometry. Similarly, if s : G → G is an automorphism of G such that(s∗)id is an isometry, s will be an isometry.
Suppose now that s is an automorphism such that
(i) s∗〈 , 〉 = 〈 , 〉 and
(ii) s2 = id.
Let K = σ ∈ G : s(σ) = σ, a compact subgroup of G. Of course, K is a totallygeodesic submanifold of G, but we can construct an even more interesting one asfollows:
Define an isometry F : G → G by F (σ) = s(σ−1) and let
M0 = σ ∈ G | F (σ) = σ = σ ∈ G | s(σ) = σ−1,
a totally geodesic submanifold. Define
β : G×M0 → M0 by β(σ, τ) = στs(σ−1).
(Note that s(στs(σ−1)) = s(σ)s(τ) · σ−1 = s(σ)τ−1σ−1 = (στs(σ−1))−1, so β(σ, τ) ∈M0.) The map τ 7→ β(σ, τ) is an isometry for each σ ∈ G. Note that β(σ, e) = e iffσs(σ−1) = e iff e ∈ K. This implies that β induces a map h : G/K → M0 by
h(σK) = σs(σ−1).
One can check that h is a bijection. Hence we havegiven a manifold structure to thequotient G/K and have a totally geodesic imbedding of G/K in G.
Example. Let G = SO(p + q) and define s : G → G by S(A) = JAJ , where
J =
(−Ip 00 Iq
).
107
Then K = A ∈ P (p) × O(q) : det A = 1 and G/K is the Grassmann manifold ofp-planes in Rp+q. Thus the Grassmann manifold is a totally geodesic submanifold ofSO(p + q).
Note that F∗I : TI SO(p + q) → TI SO(p + q) by F∗I(A) = JAJ and
TIM0 = A ∈ gl(p + q, R) | A + AT = 0, JAJ = −A;
thus the matrices in TIM0 are of the form(0 X
−XT 0
).
If x =
(0 X
−XT 0
)and y =
(0 Y
−Y T 0
)are elements of TIM0, we find therefore
that
〈R0(x, y)y, x〉 =1
4〈[x, y], [y, x]〉,
or
〈R0(x, y)y, x〉 =1
4trace
((XY T − Y XT )2 0
0 (XT Y − Y T X)2
).
This gives a very explicit formula for the curvature of the Grassmann manifold.
This construction gives a large class of manifolds for which we can actually calculatethe curvature quite explicitly.
108
Chapter 21
Problems
1. Let U = (r, θ) : r > 0, 0 < θ <√
2π. Define F : U → R3 by
F (r, θ) =1√2
r cos(√
2θ)
r sin(√
2θ)r
.
(a) Show that the metric induced on U is
〈 , 〉 = dr ⊗ dr + r2 dθ ⊗ dθ.
(b) Show that if we let x = r cos θ, y = r sin θ, the metric takes the form
〈 , 〉 = dx⊗ dx + dy ⊗ dy.
(c) Sketch the geodesics in U and their images under F in R3.
2. Let M2 be an orientable imbedded surface in Euclidean space R3 and N :M2 → R3 a choice of smoothly varying unit normals on Mn. Then the secondfundamental form α : TpM × TpM → NpM satisfies
α(x, y) = h(x, y)N(p), for x, y ∈ TpM,
where h : TpM × TpM → R is a real-valued symmetric bilinear form.
(a) Let (e1, e2) be an orthonormal basis for TpM . Use the Gauß equations toshow that
K(p) =
∣∣∣∣h(e1, e1) h(e1, e2)h(e2, e1) h(e2, e2)
∣∣∣∣ .109
(b) By citing an appropriate theorem from linear algebra, show that there isan orthonormal basis for TpM such that h(e1, e2) = 0.
(c) If (e1, e2) is such an orthonormal basis, h(e1, e1) and h(e2, e2) are calledprincipal curvatures of M2 at p, and are denoted by κ1(p) and κ2(p).The Gaussian curvature is then given by the formula K(p) = κ1(p)κ2(p).Suppose that F : R× S1 → R3 by
F (u, eiv) = (cosh u cos v, cosh u sin v, sinh u),
and let M2 be the image of F . Find an orthonormal basis (e1, e2) for TpMfor each p ∈ M , determine the principle curvatures κ1 and κ2, and find theGaussian curvature K : M2 → R. M2 is called the hyperboloid.
3. Suppose that (M, 〈 〉) is a connected Riemannian manifold which has the fol-lowing property: Given p, q ∈ Mn, there is an isometry F : M → M such thatF (p) = q. Show that (Mn, 〈 〉) is geodesically complete.
4. Consider the paraboloid of revolution z = x2+y2 in R3 which can be parametrizedby
F (u, v) = (u, v, u2 + v2)
or in terms of polar coordinates u = r cos θ, v = r sin θ,
F (r, θ) = (r cos θ, r sin θ, r2).
(a) Show that the curves θ = constant are geodesics.
(b) Show that ∂∂θ
is a Jacobi field.
(c) Show that the point (0, 0, 0) on the paraboloid is a pole.
5. Let F : (0,∞)× S1 → R3 by
F (u, eiv) =
(e−u cos v, e−u sin v,
∫ u
1
√1− e−2w dw
).
Let M2 be the image of F , a two-dimensional submanifold of R3.
(a) Sketch M2.
(b) Calculate the coefficients of the metric induced by F on (0,∞)× S1.
(c) Calculate the second fundamental form.
(d) Determine the Gaussian curvature κ.
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6. (a) Let (M, 〈 , 〉) be a Riemannian manifold, C a one-dimensional submani-fold which is the fixed-point set of an isometry. Show that C consists ofgeodesics.
(b) Suppose that M = (x, y) ∈ R2 | x2 + y2 < 1 and
〈 , 〉 =4
(1− x2 − y2)2(dx⊗ dx + dy ⊗ dy).
Find the geodesics which pass through the origin.
(c) Parametrize these geodesics by arclength.
(d) Show that (M, 〈 , 〉) is geodesically complete.
7. Let G be a compact Lie group with a biinvariant Riemannian metric 〈 , 〉.Suppose that the Lie algebra g of G satisfies the condition
X ∈ g : [X, Y ] = 0 for all Y ∈ g = 0.
Use Myers’s Theorem to show that G has finite fundamental group.
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