Differential Geometry by Andrews PDF
Differential Geometry by Andrews PDF
Differential Geometry by Andrews PDF
Introduction
his work making rigorous the theory of Riemann surfaces), but this did not
really come into common use until much later, after a series of papers by Has-
sler Whitney around 1936. However the idea has its roots much earlier. Gauss
had a major interest in differential geometry, and published many papers on
the subject. His most renowned work in the area was Disquisitiones generales
circa superficies curva (1828). This paper contained extensive discussion on
geodesics and what are now called ‘Gaussian coordinates’ and ‘Gauss curva-
ture’, which he called the ‘measure of curvature’. The paper also includes the
famous theorema egregium:
“If a curved surface can be developed (i.e. mapped isometrically) upon
another surface, the measure of curvature at every point remains un-
changed after the development.”
This result led Gauss to a fundamental insight:
“These theorems lead to the consideration of curved surfaces from a
new point of view ... where the very nature of the curved surface is
given
by means of the expression of any linear element in the form
Edp2 + 2F dpdq + Gdq 2 .”
In other words, he saw that it is posible to consider the geometry of a
surface as defined by a metric (here he is locally parametrizing the surface
with coordinates p and q), without reference to the way the surface lies in
space.
In 1854 Riemann worked with locally defined metrics (now of course
known as Riemannian metrics) in any number of dimensions, and defined
the object we now call the Riemann curvature. These computations were lo-
cal, and Riemann only gave a rather imprecise notion of how a curved space
can be defined globally.
The definition of a manifold encapsulates the idea that there are no pre-
ferred coordinates, and therefore that geometric computations must be in-
variant under coordinate change. Since this is built automatically into our
framework, we never have to spend much time checking that things are geo-
metrically well-defined, or invariant under changes of coordinates. In contrast,
in the works of Riemann, Ricci and Levi-Civita these computations take some
considerable effort.
The abstract notion of a manifold, without reference to any ‘background’
Euclidean spaces, also arises naturally from several directions. One of these
is of course general relativity, where we do not want the unnecessary baggage
associated with postulating a larger space in which the physical spacetime
should lie. Another comes from the work of Riemann in complex analysis,
in what is now called the theory of Riemann surfaces. Consider an ana-
lytic function f on a region of the complex plane. This can be defined in a
neighbourhood of a point z0 by a convergent power series. This power series
converges in some disk of radius r0 about z0 . If we now move to some other
point z1 in this disk, we can look at the power series for f about z1 , and
this in general converges on a different region, and can be used to extend
f beyond the original disk. Using analytic extensions in this way, we can
move around the complex plane as long as we avoid poles and singularities
of f . However, it may happen that the value of the function that we obtain
depends on the path we took to get there, as in the case f (z) = z 1/2 . In this
4 Lecture 1. Introduction
case we can think of the function as defined not on the plane, but instead on
an abstract surface which projects onto the plane (if f (z) = z 1/2 , this surface
covers C\{0} twice). Another place where it is natural to work with abstract
manifolds is in the theory of Lie groups, which are groups with a manifold
structure.
Ua
M
Ub
j jb
a
n
R Vb
n
Va R
1
If you are not familiar with these topological notions, it is sufficient to consider
metric spaces, or even subsets of Euclidean spaces. If you want to know more, see
Appendix A.
2
I will use the terms ‘differentiable’ and ‘smooth’ interchangeably, and both will
mean ‘infinitely many times differentiable’.
1.1 Differentiable Manifolds 5
Example 1.1.1 Consider the set M = R (with the usual topology). This can be
made a manifold is many different ways: The obvious way is to take the atlas
A = {Id : R → R}. Another way is to take the atlas B = {x → x3 : R → R}. More
generally, any homeomorphism of R to itself (or to an open subset of itself) can be
used to define an atlas. The atlases A and B are incompatible because the union is
{x → x, x → x3 }, which is not an atlas because the first map composed with the
inverse of the second is the map x → x1/3 , which is not smooth. This example can
be extended to show there are infinitely many different differentiable structures on
the real line. This seems ridiculously complicated, but it will turn out that these
differentiable structures are all equivalent in a sense to be defined later.
Example 1.1.3: An atlas for S n . The sphere S n is the set {x ∈ Rn+1 : |x| =
1}. Since this is a closed subset of Euclidean space, the topological require-
ments are satisfied. Define maps ϕ+ : S n \{N } → Rn and ϕ− : S n \{S} → Rn
as follows, where N is the “north pole“ (0, . . . , 0, 1) and S the “south pole”
(0, . . . , 0, −1): Writing x ∈ Rn+1 as (y, z) where y ∈ Rn and z ∈ R, we take
(2w,|w|2 −1)
ϕ+ (y, z) = 1−z y y
and ϕ− (y, z) = 1+z . Then we have ϕ−1
+ (w) = |w|2 +1 and
(2w,1−|w|2 )
ϕ−1
− (w) = |w|2 +1 . It follows that ϕ± is a homeomorphism, since clearly
6 Lecture 1. Introduction
ϕ−1 n
± is continuous, and ϕ± is the restriction to S of a continuous map de-
fined on all of R n+1
. Also ϕ+ ◦ ϕ− (w) = |w|2 and ϕ− ◦ ϕ−1
−1 w w
+ (w) = |w|2 for
all w ∈ Rn \{0}. Since these maps are differentiable, the two charts ϕ± form
a differentiable atlas, and so define a differentiable manifold.
n+
R N
(y,z)
j(y,z)
- R
n
j(y,z)
+
n
S
The maps ϕ± in the last example are the stereographic projections from
the north and south poles. Next we consider an example which does not come
from a subset of Rn+1 :
Example 1.1.4: The real projective spaces. The n-dimensional real projective
space RP n is the set of lines through the origin in Rn+1 (an observer at the
origin sees anything along such a line as being at the same position in the
field of view – thus real projective space captures the geometry of perspective
drawing). Equivalently, RP n = (Rn+1 \{0})/ ∼ where x ∼ y ⇐⇒ x = λy for
some λ ∈ R\{0}. A point in RP n is denoted [x1 , x2 , . . . , xn+1 ], meaning the
equivalence class under ∼ of the point (x1 , . . . , xn+1 ) ∈ Rn+1 . We place a
topology on RP n by taking the open sets to be images of open sets in Rn+1
under the projection onto RP n . This topology is Hausdorff: Take any two non-
zero points x and y in Rn+1 not lying on the same line. Choose an open set U
about x which is disjoint from the line through y and the origin. Then choose
an open set V about y which is disjoint from the set {λz : z ∈ U, λ ∈ R}.
Then U/ ∼ and V / ∼ are disjoint open sets in RP n with [x] ∈ U/ ∼ and [y] ∈
V / ∼. Define subsets Vi of RP n for i = 1, . . . , n + 1 by Vi = {[x1 , . . . , xn+1 ] :
xi = 0}. Note that Vi is well-defined. Define maps ϕi : Vi → R
n
for i =
x1 x2 xi−1 xi+1 xn+1
1, . . . , n + 1 by ϕi ([x1 , . . . , xn+1 ]) = xi , xi , . . . , xi , xi , . . . , xi . This
has the inverse ϕ−1 i (x1 , . . . , xn ) = [x1 , x2 , . . . , xi−1 , 1, xi , . . . , xn ]. The maps
ϕ and their inverses are continuous; the open sets Vi cover RP n ; and (for
i < j)
x1 x2 xi−1 xi+1 xj−1 1 xj xn
ϕi ◦ ϕ−1
j (x1 , . . . , xn ) = , , . . . , , , . . . , , , , . . . ,
xi xi xi xi xi xi xi xi
1.1 Differentiable Manifolds 7
N
M
W
F
U
x
h
j
V Z
Remark. Although the definition requires that f ◦ ϕ−1 be smooth for every
chart, it is enough to show that this holds for at least one chart around each
point: If ϕ : U → V is a chart with f ◦ϕ−1 smooth, and η : W → Z is another
10 Lecture 2. Smooth functions and maps
Exercise 2.1.2 Show that the map x → [x] from Rn+1 to RP n is smooth.
Example 2.1.1 The group GL(n, R). On GL(n, R) we have a natural family
of maps: Fix some M ∈ GL(n, R), and define ρM : GL(n, R) → GL(n, R) by
2
ρM (A) = MA. The trivial chart ι given by inclusion of GL(n, R) in Rn is the
map which takes a matrix A to its components (A11 , . . . , A1n , . . . , Ann ). To
check that ρM is smooth, we need to check that ι ◦ ρM ◦ ι−1 is smooth. But
this is just the map which takes the components of A to the components of
MA, which is
n
n
n
ι−1 ◦ ρM ◦ ι (A11 , A12 , . . . , Ann ) = M1j Aj1 , M1j Aj2 , . . . , Mnj Ajn
j=1 j=1 j=1
Remark. The definitions above reflect the fact that the rank of the derivative
of a smooth map between manifolds is well-defined: If we change coordinates
from ϕ to ϕ̃ on M and from η to η̃ on N , then the chain rule gives:
Dϕ̃(x) η̃ ◦ F ◦ ϕ̃−1 = Dη(x) η̃ ◦ η −1 Dϕ(x) η ◦ F ◦ ϕ−1 Dϕ̃(x) ϕ ◦ ϕ̃−1 .
The first and third matrices on the right are non-singular, so the rank of the
matrix on the left is the same as the rank of the second matrix on the right. I
will not dwell on this, because it is a simple consequence of the construction
of tangent spaces for manifolds, which we will work towards in the next few
lectures.
Example 2.2.2 We will show that the projection [.] from Rn+1 \{0} to RP n
is a submersion. To see this, fix x = (x1 , . . . , xn+1 ) in Rn+1 \{0}, and assume
that xn+1 = 0. On Rn+1 \{0} we take the trivial chart ι, and onRP n we take
ϕn+1 : Vn+1 → Rn . Then ϕn+1 ◦ [.] ◦ ι−1 (x) = xn+1 x1
, . . . , xxn+1
n
, and
1 0 . . . 0 − xn+1
x1
v1
0 1 . . . 0 − xx2 v2
1
Dx ϕn+1 ◦ [.] ◦ ι−1 v = ..
n+1
. .
.. .. . . . ... .
xn+1 ..
.
0 0 . . . 1 − xxn+1n vn+1
Exercise 2.2.4 Let M and N be differentiable manifolds. Show that for any
x ∈ M the map ix : N → M × N given by y
→ (x, y) is an embedding. Show
that the projection pi : M × N → M given by π(x, y) = x is a submersion.
Here we can take η(x) = x to be the trivial chart for Rn+1 , and ϕ to be
one of the stereographic projections defined in Example 1.1.3, say ϕ− . Then
(2w,1−|w|2 )
η ◦ ι ◦ ϕ−1 (w) = 1+|w|2 . Differentiating, we find
∂(η ◦ ι ◦ (ϕ−1 )i ) 2 2wi wj
= δij −
∂wj 1 + |w|2 1 + |w|2
and
∂(η ◦ ι ◦ (ϕ−1 )n+1 ) 4wj
= − .
∂wj (1 + |w|2 )2
In particular we find for any non-zero vector v ∈ Rn
4
Dw (η ◦ ι ◦ ϕ−1 )(v)2 = |v|2 > 0.
(1 + |w|2 )2
Equator
More generally, consider a circle which moves from south to north pole on
S 2 . At each ‘time’ in this sweepout we have a corresponding torus in space,
which starts as a thin torus around the unit circle, grows until it agrees with
the Clifford torus as our curve passes the equator, and continues growing,
becoming larger and larger in size with a smaller and smaller ‘tube’ down
the middle around the z axis as we approach the north pole. This can be
obtained by rotating the following picture about the z axis:
Next we should think about how the circles line up on these tori. The
crucial thing to keep in mind is that this must be continuous as we move over
S 2 . In particular, as we approach the south pole the circles have to approach
the unit circle, so they have to wind around the torus ‘the long way’. On the
other hand, as we approach the north pole the circle must start to lie ‘along’
the z axis. These two things seem to contradict each other, until we realise
that we can wind our curves around the tori ‘both ways’:
One of the important questions in differential geometry is the study (or classi-
fication) of manifolds up to diffeomorphism. That is, we introduce an equivalence
relation on the space of k-dimensional differentiable manifolds by taking M ∼ N if
and only if there exists a diffeomorphism from M to N , and study the equivalence
classes. Among the results:
• There are only two equivalence classes of connected one-dimensional mani-
folds, namely those represented by S 1 and by R. In particular the different
differentiable structures on R introduced in Example 1.1.1 are diffeomorphi-
cally equivalent (the map x → x1/3 is a diffeomorphism between the two);
• The equivalence class of a compact connected two-dimensional manifold is
determined by its genus (an integer representing the number of ‘holes’ in the
surface) and its orientability (which will be defined later).
A more subtle question is the following: Given a topological manifold M , are all
differentiable structures on M related by diffeomorphism? If not, how many non-
diffeomorphic differentiable structures are there on a given manifold? The answer
is that in general there can be more than one differentiable structure on a mani-
fold: It is known that the spheres S n generally have more than one differentiable
structure if n ≥ 7, and in particular John Milnor proved in 1956 that there are 28
2.2 Further classification of maps 15
In this lecture we will look at some of the most important examples of man-
ifolds, namely those which arise as subsets of Euclidean space.
Example 3.1.2 Consider the following examples of curves in the plane, which
illustrate the conditions in the definition of submanifold above: The home-
omorphism property can be violated if the function ξ is not injective, for
example if ξ(t) = (t2 , t3 − t) for t ∈ (−2, 2):
18 Lecture 3. Submanifolds
Often the definition we have given is not the most convenient way to check
whether a given subset of Euclidean space is a submanfold. There are several
alternative characterisations of submanifolds, which are given by the following
result:
Our main tool in the proof will be the Inverse function theorem:
Exercise 3.2.1 Consider the subset of R4 given by the image of the map
ϕ : R → R4 defined by
√ √
ϕ(t) = (cos t, sin t, cos( 2t), sin( 2t)).
Is this a submanifold of R4 ?
Example 3.2.1 The groups SL(n, R) and O(n). Each of these groups is con-
tained as a submanifold in GL(n, R). This implies that SL(n, R) × SL(n, R)
and O(n) × O(n) are submanifolds of GL(n, R) × GL(n, R). Therefore the
restriction of the multiplication map ρ on GL(n, R) × GL(n, R) to either
of these submanifolds is smooth, and has image contained in SL(n, R) or
O(n) respectively. Hence by the result of Exercise 3.2.2, the multiplication
on SL(n, R) and O(n) are smooth maps.
3.3 Orientability 21
3.3 Orientability
Exercise 3.3.2 Show that every connected manifold has either zero or two
orientations.
Example 3.3.2 The Möbius strip and the Klein bottle. The Möbius strip is the
topological quotient of R × R by the equivalence relation ∼ which identifies
(s, t) with (s+1, −t) for every s and t in R. M can be given an atlas as follows:
We take a chart ϕ1 : (0, 1) × R/ ∼→ (0, 1) × R to be the inverse of the map
which takes (s, t) to [(s, t)], and ϕ2 : (−1/2, 1/2) × R/ ∼→ (−1/2, 1/2) × R
similarly. Then
(s, t) if s ∈ (0, 1/2)
ϕ1 ◦ ϕ−1 (s, t) =
2 (s + 1, t) if s ∈ (−1/2, 0)
The Möbius strip is not orientable. I will not prove this rigorously yet.
Heuristically, the idea is that if we take an oriented pair of vectors at some
point (s, 0), and ‘slide’ them around the Möbius strip to (s + 1, 0), then if
there were an oriented atlas it would have to be the case that the vectors
obtained in this way were still oriented with respect to the original pair. But
this is not the case.
The Klein bottle is another non-orientable surface, given by making the
following identifications on the square:
The second definition expresses even more explicitly the idea of a ‘velocity
vector’ in the manifold: We define Mx to be the space of smooth paths in
M through x (i.e. smooth maps γ : I → M with γ(0) = x) modulo the
equivalence relation which identifies any two curves if they agree to first
order (as measured in some chart): γ ∼ σ ⇔ (ϕ ◦ γ) (0) = (ϕ ◦ σ) (0) for some
chart ϕ : U → V with x ∈ U . The equivalence does not depend on the choice
of chart: If we change to a chart η, then we have
(η ◦ γ) (0) = (η ◦ ϕ−1 ) ◦ (ϕ ◦ γ) (0) = Dϕ(x) η ◦ ϕ−1 (ϕ ◦ γ) (0)
Proposition 4.1.1 There are natural isomorphisms between the three spaces
Mx , Tx M , and Dx M .
TM x a
Mx
c b
DM
x
r(x)
x(x)
x
-1 0 1 x
For this to make sense we need to check that there is some smooth function
f˜ on M which agrees with f on a neighbourhood of x, and that the definition
does not depend on which such function we choose.
Without loss of generality we suppose f is defined on U as above, and
define ρ̃ as above. Then we define
f (y)ρ̃(y) for y ∈ U ;
f˜(y) = (4.8)
0 for y ∈ M \U .
Lemma 4.1.3 Suppose f and g are two smooth functions on M which agree
on a neighbourhood of x. Then v(f ) = v(g).
Proof. Without loss of generality, assume that f and g agree on an open set
U containing x, and construct a ‘bump’ function ρ̃ in U as above. Then we
observe that ρ̃(f − g) is identically zero on M , and that v(0) = v(0.0) =
0.v(0) = 0. Therefore
This shows that the definition of v(f ) makes sense, and so our definition of
χ(v) makes sense. However we still need to check that χ(v) does not depend
on the choice of a chart ϕ. Suppose we instead use another chart η. Then we
have in a small region about x,
n
η i (y) = η i (x) + Gij (y)(ϕj (y) − ϕj (x)) (4.9)
j=1
Now apply v to Eq. (4.9). Note that the first term is a constant.
Proof.
v(1) = v(1.1) = 1.v(1) = 1.v(1) = 2v(1) =⇒ v(1) = 0 =⇒ v(c) = cv(1) = 0.
v applied to η i gives
n
v(η i ) = Gij (x)v(ϕj ) + (ϕj (x) − ϕj (x))v(Gij )
j=1
n
∂ i
−1
= j
η ◦ ϕ v(ϕj ).
j=1
∂z ϕ(x)
Therefore we have
n n
∂ i
−1
v(η i )ei = j
η ◦ ϕ v(ϕj )ei
i=1 i,j=1
∂z ϕ(x)
n
−1
= Dϕ(x) η ◦ ϕ i
v(ϕ )ei
i=1
n i
n i
and so [(ϕ, i=1 v(ϕ )ei )] = [(η, i=1 v(η )ei )] and χ is independent of the
choice of chart.
In order to prove the proposition, it is enough to show that the three
triple compositions χ ◦ β ◦ α, β ◦ α ◦ χ, and α ◦ χ ◦ β are just the identity map
on each of the three spaces.
We have
χ ◦ β ◦ α([(ϕ, u)]) = χ ◦ β [t
→ ϕ−1 (ϕ(x) + tu)]
= χ f
→ Dϕ(x) f ◦ ϕ−1 (u)
n
i
−1
= ϕ, Dϕ(x) ϕ ◦ ϕ (u)ei
i=1
= [(ϕ, u)].
Similarly we have
d
α ◦ χ ◦ β([σ]) = α ◦ χ f
→ (f ◦ γ)
dt t=0
n
=α ϕ, (ϕ ◦ γ) (0)
i=1
= t
→ ϕ−1 ϕ(x) + t (ϕ ◦ γ) (0) ,
4.2 The differential of a map 29
We need to show that this is the same as v. To show this, we note (using the
Taylor expansion for f ◦ ϕ−1 ) that
n
f (y) = f (x) + Gi (y) ϕi (y) − ϕi (x)
i=1
for y in a sufficiently
small neighbourhood of x, where Gi is a smooth function
with Gi (x) = ∂z∂ i f ◦ ϕ−1 |ϕ(x) . Applying v to this expression, we find
n n
∂
−1
v(f ) = Gi (x)v(ϕi ) = i
f ◦ ϕ v(ϕi ) (4.10)
i=1 i=1
∂z ϕ(x)
N f
M
g F(g)
x F(x)
F R
j h
u w
Dx F (v) : f
→ Dϕ(x) f ◦ F ◦ ϕ−1 (u)
= Dη(F (x)) f ◦ η −1 ◦ Dϕ(x) η ◦ F ◦ ϕ−1 (u)
which is the vector corresponding to η, Dϕ(x) η ◦ F ◦ ϕ−1 (u) .
Alternatively, if we think of a vector v as the tangent vector of a curve
γ, then we have v : f
→ (f ◦ γ) (0), and so Dx F (v) : f
→ (f ◦ F ◦ γ) (0),
which is the tangent vector of the curve F ◦ γ. In other words,
Dx (G ◦ F ) = DF (x) G ◦ Dx F.
These two theorems follow directly from the corresponding theorems for
smooth maps between Euclidean spaces.
Thus the first n coordinates describe the point p, and the last n give the
components of the vector v with respect to the basis of coordinate tangent
vectors {∂i }ni=1 , since by Eq. (4.10),
32 Lecture 4. Tangent vectors
n
∂
−1
n
v(f ) = i
f ◦ϕ i
v(ϕ ) = v(ϕi )∂i (f ) (4.12)
i=1
∂z ϕ(x)
i=1
n
for any smooth f , and hence v = i=1 v(ϕi )∂i . For convenience we will often
write the coordinates on T M as (x1 , . . . xn , ẋ1 , . . . , ẋn )
To check that these charts give T M a manifold structure, we need to
compute the transition maps. Suppose we have two charts ϕ : U → V
and η : W → Z, overlapping non-trivially. Then η̃ ◦ ϕ̃−1 first takes a 2n-
n
tuple (x1 , . . . , xn , ẋ1 , . . . , ẋn ) to the element ϕ−1 (x1 , . . . , xn ), i=1 ẋi ∂i
(ϕ)
of T M , then maps this to R2n by η̃. Here we add the superscript (ϕ) to
distinguish the coordinate tangent vectors coming from the chart ϕ from
those given by the chart η. The first n coordinates of the result are then just
η ◦ ϕ−1 (x1 , . . . , xn ). To compute the last n coordinates, we need to ∂i in
(ϕ)
(η)
terms of the coordinate tangent vectors ∂j : We have
∂i f = Dϕ(x) f ◦ ϕ−1 (ei )
(ϕ)
= Dϕ(x) (f ◦ η −1 ) ◦ (η ◦ ϕ−1 ) (ei )
= Dη(x) f ◦ η −1 ◦ Dϕ(x) η ◦ ϕ−1 (ei )
n
= Dη(x) f ◦ η −1 Dϕ(x) η ◦ ϕ−1 i ej
j
j=1
n
j
= Dϕ(x) η ◦ ϕ−1 i Dη(x) f ◦ η −1 (ej )
j=1
n
j (η)
= Dϕ(x) η ◦ ϕ−1 i ∂j f
j=1
and
η̃ ◦ ϕ̃−1 (x, ẋ) = η ◦ ϕ−1 (x), Dϕ(x) η ◦ ϕ−1 (ẋ) .
Since η ◦ ϕ−1 is smooth by assumption, so is its matrix of derivatives D(η ◦
ϕ−1 ), so η̃ ◦ ϕ̃−1 is a smooth map on R2n .
4.5 Vector fields 33
TM
x
TM
x M
Vpi = Vp (ϕi )
(V (f ))(x) = Vx (f )
satisfies the two conditions in the proposition and takes a smooth function
to a smooth function.
5.1 Examples
Recall that a Lie Group is a group with the structure of a smooth manifold
such that the composition from M ×M → M and the inversion from M → M
are smooth maps.
Example 5.1.1 The general linear, special linear, and orthogonal groups. The
general linear group GL(n) (or GL(n, R)) is the set of non-singular n × n
matrices with real components. The is a Lie group under the usual multipli-
cation of matrices. We showed in Example 2.1.1 that the multiplication on
GL(n) is a smooth map from GL(n) × GL(n) → GL(n). To show that GL(n)
is a Lie group, we nned to show that the inversion is also a smooth map.
2
GL(n) is an open subset of M n
Rn , so is covered by a single chart.
With respect to this chart, the inversion is given by the formula
−1 j 1 τ (1) τ (2) τ (n−1) i τ (n)
M i
= sgnσ sgnτ Mσ(1) Mσ(2) . . . Mσ(n−1) δσ(n) δj
n! det M σ,τ
where δij = 1 if i = j and 0 otherwise, and the sum is over pairs of permuta-
tions σ and τ of the set {1, . . . , n}. Each component of M −1 is therefore given
by sums of productss of components of M , divided by the non-zero smooth
function det M :
1 τ (1) τ (2) τ (n−1) τ (n)
det M = sgnσ sgnτ Mσ(1) Mσ(2) . . . Mσ(n−1) Mσ(n)
n! σ,τ
j
and so M −1 i is a smooth function of the components of M , and GL(n) is
a Lie group.
From Example 3.2.1 we know that the multiplications on SL(n) and O(n)
are also smooth; these are both submanifolds of GL(n), so it follows that the
36 Lecture 5. Lie Groups
restriction of the inversion to each of these is smooth, so SL(n) and O(n) are
Lie groups.
Example 5.1.2 The group O(2). Recall that the orthogonal group O(n) (or
O(n, R) is the group of n × n matrices with real components satisfying
M T M = In . In particular O(2) is the group of orthogonal 2 × 2 matrices.
a b
M= is in O(2) if the rows (or columns) of M form an orthonor-
c d
mal basis:
a2 + b2 = 1;
c2 + d2 = 1;
ac + bd = 0.
In particular M is determined by its first row [a b] and its determinant δ =
det M = ±1:
a b
M= .
−δb δa
This gives a natural map ϕ : O(2) → S 1 × {−1, 1} given by
a b
ϕ = ((a, b), ad − bc) ∈ S 1 × {−1, 1}.
c d
Then ϕ is smooth and has the smooth inverse
−1 a b
ϕ ((a, b), δ) = .
−δb δa
ρ(k, z) = eikα z
(x, x) · (x, x) = x2 + x2
Remark. We have seen two examples of spheres which are also Lie groups:
S 1 is a Lie group, and so is S 3 (one could also say that S 0 = {1, −1} is a 0-
dimensional Lie group). This raises the question: Are all spheres Lie groups?
If not which ones are? The following proposition, which also introduces some
important concepts in the theory of Lie groups, will be used to show that the
two-dimensional sphere S 2 can’t be made into a Lie group:
Proposition 5.1.1 Let G be a Lie group. Then there exist n smooth vec-
tor fields E1 , . . . , En on G each of which is everywhere non-zero, such that
E1 , . . . , En form a basis for Tx M for every x ∈ M .
Proof. Consider the left and right translation maps from G to itself given by
lg (g ) = gg and rg (g ) = g g for all g ∈ G. These are smooth maps, and
(lg )−1 = lg−1 and (rg )−1 = rg−1 are smooth, so lg and rg are diffeomorphisms
from G to G. The derivative of a diffeomorphism is a linear isomorphism, and
in particular this gives us isomorphisms De lg , De rg : Te G → Tg G.
Choose a basis {e1 , . . . , en } for Te G. Then define Ei (g) = De lg (ei ) for
each i and each g ∈ G. The vector fields E1 , . . . , En have the properties
claimed in the proposition.
to the dimension of G (in fact {E1 , . . . , En } are a basis). Similarly one can
define right-invariant vector fields.
Now return to the 2-sphere S 2 . We have the following result from algebraic
topology:
Proof (Sketch):
Suppose V is a non-vanishing continuous vector field on S 2 . Then in the
chart given by stereographic projection from the north pole, we have in the
neighbourhood of the origin (possibly after rotating the sphere about the
polar axis)3
(Dϕ+ (V )) (x) = (1, 0) + o(|x|) as x → 0.
Now look at this vector field through the chart ϕ− : The transition map is
(ϕ− ◦ϕ−1 2 2 2 2
+ )(x, y) = (x/(x +y ), y/(x +y )). Therefore Ṽ = Dϕ− (V ) is given
by
−1
Ṽ (w, z) = D(w,z) (ϕ+ ◦ ϕ−1
− ) ((1, 0) + o(|x|))
2
z −w 2
−2wz 1 + o(|(w, z)|−1 )
=
−2wz w2 − z 2 o(|(w, z)|−1 )
2
z − w2
= + o(|(w, z)|) as |(z, w)| → ∞.
−2wz
There can be no zeroes of the vector field Ṽ on R2 , so we can divide Ṽ by its
length at each point to obtain a continuous vector field V which has length
1 everywhere, and such that
z2 −w2
V (w, z) = w22wz
+z 2 + o(1) as |(w, z)| → ∞
− w2 +z2
For each r ∈ R+ , consider the restriction of V to the circle of radius r
about the origin. Since V has length 1 everywhere, this defines for each r a
continuous map fr from the circle S 1 to itself, given by
fr (z) = V (rz)
fr (cos θ, sin θ) = (sin2 θ−cos2 θ, −2 cos θ sin θ)+o(1) = −(cos 2θ, sin 2θ)+o(1)
(x)
3
We use the notation f (x) = o(g(x)) as x → 0 to mean limx→0 fg(x) = 0. Another
f (x)
useful notation is f (x) = O(g(x)) as x → 0 to mean lim supx→0 g(x) < ∞.
40 Lecture 5. Lie Groups
as r → ∞. Thus the limit as r → ∞ of the map fr is just the map that sends
θ to π−2θ. In other words, fr (θ) winds twice around the circle backwards as θ
traverses the circle forwards once: fr has winding number −2 for r sufficiently
large.
But now consider what happens as r → 0: f0 (θ) is just the constant vector
V (0, 0) ∈ S 1 , so limr→0 fr (θ) = θ0 , a constant. But this means that for r
sufficiently small the function fr (θ) does not traverse the circle at all as θ
moves around S 1 , and fr has winding number zero for r small. Now we use
the result from algebraic topology that the winding number of a map from
S 1 to itself is a homotopy invariant. The family of maps {fr } is a continuous
deformation from a map of winding number −2 to a map of winding number
0, which is impossible. Therefore there is no such vector field.
S2
j
-
j
+
It is useful to see how to find the left-invariant vector fields for some examples:
Example 5.2.1 More on Rn . Let us consider the Lie group Rn under addition.
Then for any x ∈ Rn we have lx (y) = x + y, and so D0 lx (v) = v. Taking the
standard basis {e1 , . . . , en } for T0 Rn
Rn , we find that the left-invariant
vector fields are given by
(Ei )x = ei .
That is, the left-invariant vector fields are just the constant vector fields.
5.2 Examples of left-invariant vector fields 41
DI lM (A) = M A
for any A ∈ TI G. Therefore the left-invariant vector fields have the form
AM = M A
In particular, we have
TI SL(n) = {A ∈ Mn : tr(A) = 0} .
TI O(n) = {A ∈ Mn : B T = −B},
Example 5.2.3 The group C\{0}. The multiplicative group of non-zero com-
plex numbers is an open set in C
R2 , so we need not worry about charts.
We have
lz (w) = zw,
and hence D1 lz (v) = zv for all v ∈ R2 and z ∈ C\{0}. The left invariant
vector fields are linear combinations of the two vector fields (E1 )z = z and
(E2 )z = iz.
42 Lecture 5. Lie Groups
for w ∈ C fixed, and this has the solution z(t) = etw . This gives the one-
parameter subgroups of C\{0}: If w = 1 we get the positive real axis, with
Γ (t) = et ; If w = i we get the unit circle, with Γ (t) = eit , and for other
complex numbers we get logarithmic spirals.
This corresponds to the vector field (E1 )(β,α) = (α, 0). Similarly we get a left
invariant vector field (E2 )(β,α) = (0, α) corresponding to the vector (0, 1).
44 Lecture 5. Lie Groups
The one-parameter subgroups are lines through (0, 1), with Γ (t) =
(ab−1 (ebt − 1), ebt ).
Ṁ = M A (5.2)
with the initial condition M (0) = I. This is exactly the problem which arises
in the process of solving a general linear system of ordinary equations: Recall
that to solve the system v̇ = Av with any initial condition v(0) = v0 , one first
solves the system (4.1) to obtain a matrix M (t). Then the required solution
is v(t) = M (t)v0 . The solution of (4.1) is denoted by eAt , and is given by the
convergent power series
∞
At 1 n n
e = t A .
n=0
n!
It is a consequence of our construction that the matrix exponential of a matrix
A ∈ Ti G is in G – in particular, if A is traceless then eAt ∈ SL(n), and if A
is antisymmetric then eAt ∈ SO(n).
ẇ = wx.
d
First note that dt w̄ = x̄w̄, and so
d
|w|2 = x̄w̄w + w̄wx = (x + x̄)|w|2 = 2x0 |w|2 .
dt
w(t)
Therefore |w(t)| = ex0 t . Now consider u(t) = |w(t)| :
d wx w w
u= − x0 |w| = (x − x0 ) = u(x − x0 ).
dt |w| |w|2 |w|
Therefore u(t) = exp(x − x0 ), the exponential of the ‘vector’ part of x. So
now assume x0 = 0. Then we can write
1
w1 (t) = (w̄i − iw)
2
and
d 1 1
w (t) = (−iwx + x̄w̄i) .
dt 2
Differentiating again with respect to t, we get
d2 1 1
2
w (t) = −iwx2 + x̄2 w̄i .
dt 2
But now x has no real part, so x̄ = −x, and x2 = x̄2 = −|x|2 , giving
d2 1
w (t) = −|x|2 w1 (t),
dt2
1
and so (since w1 (0) = 0 and (w1 ) (0) = x1 ), w1 (t) = |x|
x
sin(|x|t). The other
components are similar, so we have (in the case |x| = 1)
Exercise 5.3.1 Show that the following map is a group homomorphism and
a diffeomorphism from the group S 3 to the group SU (2) of unitary 2 × 2
matrices with determinant 1:
0
x + ix3 x1 + ix2
ϕ((x0 , x1 , x2 , x3 )) = .
−x1 + ix2 x0 − ix3
46 Lecture 5. Lie Groups
Proposition 5.3.2 CP 1
S 2 .
for (x1 , x2 , x3 ) ∈ S 2 .
Exercise 5.3.2 Show that the following map from SU (2) to SO(3) is a group
homomorphism, and a local diffeomorphism, but not a diffeomorphism:
1
−b2 − c2 b1 − c1
a1 + ia2 b1 + ib2 2 (a1 + d1 )
Ψ = b2 + c2 1
2 (a1 + d1 ) d2 − a2 .
c1 + ic2 d1 + id2
c1 − b1 a2 − d2 1
2 (a1 + d1 )
Notes on Chapter 5: 47
Consider the composition of this with the map ϕ in Exercise 5.3.1: Deduce
that the group of rotations in space, SO(3), is diffeomorphic to RP 3 .
S
TS =R
1
S =CP
R =H
Fig. 5.5: A rotation of R3 is associated with a point in S 3 by following
the great circle through (1, 0, 0, 0) in the direction of the axis of rotation
v ∈ S 2 ⊂ T S 3 for a distance α equal to half the angle of rotation. This is
also naturally associated to the matrix U ∈ SU (2) which sends z to eiα z
for any z ∈ C2 with [z] = v ∈ CP 1
S 2 ⊂ T S 3 .
Notes on Chapter 5:
Lie groups are named after the Norwegian mathematician Sophus Lie (1842–
1899). He defined ‘infinitesimal transformations’ (what are now called Lie
algebras) first, as part of his work in describing symmetries of partial dif-
ferential equations, and then went on to develop his theory of ‘continuous
transformation groups’.
The classification of compact simple Lie groups was accomplished by Elie
Cartan in 1900 (building on earlier work by Killing).
48 Lecture 5. Lie Groups
Our aim is to prove the basic existence, uniqueness and regularity results for
ordinary differential equations on a manifold.
Ψt ◦ Ψs = Ψs+t
Vx = x2 ∂x .
y (t) = y(t)2 ;
y(0) = x.
x
This gives Ψ (x, t) = y(t) = 1−tx , on the time interval (1/x, ∞) if x < 0, or
(−∞, 1/x) if x > 0 (or (−∞, ∞) in x = 0). Thus the flow of the vector field
cannot be defined on R × (δ, δ) for any δ > 0, and cannot be defined on U × R
for any open subset U of R. Here the problem seems to arise because the
vector field V is not bounded.
Example 6.1.2: Problems with M . Next consider the example M = (0, 1) with
the vector field Vx = ∂x : Now we have the flow
Ψ (x, t) = x + t
which is defined on the region {(x, t) : t ∈ (−x, 1 − x)}. Again, this flow
cannot be defined on M × (−δ, δ) for any δ > 0, or on U × R for any U ⊂ M .
Here the problem seems to arise because the domain manifold has ‘edges’.
To begin the proof of Proposition 5.1.1, we will first reduce the problem to
an ordinary differential equation on Rn , by looking at the flow of the vector
field through a chart:
6.1 Ordinary differential equations on a manifold. 51
−1 1
n
V ϕ n
(z , . . . , z ) = Ṽ k (z 1 , . . . , z n )∂k .
k=1
The curves constructed in this proposition are called the integral curves
of the vector field F i ∂i .
for all k ≥ 0 and t ∈ (−δ, δ). This is true for k = 0, since γ (1) − γ (0) =
t
0 F (z)ds ≤ |t|M0 . We proceed by induction: Suppose the inequality holds
for k − 1. Then
6.2 Initial value problems. 53
# # # " t #
# (k+1) # # #
#γ (t) − γ (k) (t)# = #
# F (γ (k)
(s)) − F (γ (k−1)
(s)) ds #
#
0
" t # #
# #
≤ M1 #γ (k) (s) − γ (k−1) (s)# ds
0
"
t M M k−1 |s|k
0 1
≤ M1 ds
0 k!
M0 M1k |t|k+1
≤ .
(k + 1)!
This implies that the sequence {γ (k) } is a Cauchy sequence in the com-
plete space of continuous maps with respect to uniform convergence, and so
converges to a continuous limit γ. The continuity of F and the dominated
convergence theorem then imply that
" t
γ(t) = z + F (γ(s)) ds (6.5)
0
for all t ∈ (−δ, δ), so that γ is differentiable and satisfies the equation
γ (t) = F (γ(t))
for each t, and the initial condition γ(0) = z. Smoothness follows, since
γ ∈ C (k) implies γ ∈ C k+1 by the identity (6.5) and the smoothness of F .
This establishes the existence of a solution.
To prove uniqueness, suppose γ and σ are two solutions of the initial value
problem. Then
#" t #
# #
γ(t) − σ(t) = ## F (γ(s)) − F (σ(s)) ds #
#
0
" t (6.6)
≤ M1 γ(s) − σ(s) ds
0
Let C = sup γ − σ. Then an induction similar to that above shows that
CM1k |t|k
γ(t) − σ(t) ≤
k!
for any k. Taking k → ∞ gives γ ≡ σ.
54 Lecture 6. Differential Equations
The result of the previous section produced the unique flow Ψ of the vector
field V , but did not address the smoothness of Ψ except in the t direction.
Since smoothness is measured by reference to charts, it is enough to show
smoothness of the map u from Proposition 6.1.2:
Proposition 6.3.1 The function u which takes a pair (z, t) to the solution
γ(t) of the initial value problem (6.3) is smooth on the open set
S = {(z, t) ∈ Z × R : |t| < d(z, ∂Z)/M0 } .
Proof. I will show that u is the C k limit of a family of smooth functions for
any k: Define u(0) (z, t) = z, and successively approximate using
" t
(k+1)
u (z, t) = z + F u(k) (z, s) ds. (6.7)
0
Clearly u(k) maps S to Z for each k, u(k) is smooth for each k, and by (6.4)
we have
M0 M1k |t|k+1
u(k+1) (z, t) − u(k) (z, t) ≤ .
(k + 1)!
Differentiating (6.7) with t fixed gives
# # # #
" t #
#
" t # #
# (k+1) # # # # (k) #
#Dz ut # = #I + DF ◦ Dz us ds# ≤ 1 + M1
(k)
#Dz us # ds
0 0
This completes the proof of Proposition 6.3.1, since Exercise 6.3.2 shows
that u = lim u(k) is in C j for every j.
Next we show that the flow Ψ we have constructed satisfies the local group
property
Ψt ◦ Ψs = Ψt+s
whenever both sides of this equation make sense.
This is very easy: Ψt ◦ Ψs and Ψt+s both satisfy the same differential
equation
∂t Ψ = V ◦ Ψ,
and have the same initial condition at t = 0. Hence by the uniqueness part
of Proposition 6.2.1, they are the same.
Proposition 6.1.1 gives the existence of the flow of a vector field locally, and
we have seen that there are examples which show that one cannot in general
expect better than this. However there are some very important situations
where we can do better:
Proof. Cover suppV by sets of the form ϕ−1 α (Brα /2 (0)) where ϕα : Wα → Zα
is a chart for M and Brα (0) ⊂ Zα . By compactness there is a finite subcover
{ϕ−1 N
i (Bri /2 (0))}i=1 . Let r = inf i=1,...,N ri /2 > 0.
Proposition 6.1.1 gives a local flow Ψi on each region ϕ−1 i (Bri /2 (0)) ×
(−δ, δ), where δ = inf i=1,...,N sup r Ṽ > 0. The uniqueness of solutions
Zi i
implies that these local flows agree on the overlaps of these sets, so they
combine to give a local flow on the set M × (−δ, δ) (by taking Ψ to be the
identity map away from suppV ). The local group property implies that these
maps are diffeomorphisms,
m since Ψt−1 = Ψ−t for |t| < δ. Finally, for any t ∈ R,
define Ψt = Ψt/m , where m is sufficiently large to ensure that |t|/m < δ.
Then Ψ satisfies the required differential equation and is defined on M × R.
Exercise 6.6.1 Show that the exponential map on a Lie group G is a smooth
map defined on all of Te G [Hint: First prove existence in a neighbourhood of
the origin. Then use the one-parameter subgroup property to extend to the
whole of Te G].
Fig. 6.3: Images of a standard grid at a sequence of times under the flow of the
vector field from Fig. 6.2
In this lecture we will introduce the Lie bracket of two vector fields, and
interpret it in several ways.
To make sense of this definition we need to check that this does indeed
define a derivation. Linearity is clear, but we need to verify the Leibniz rule:
[X, Y ](f g) = X(Y (f g)) − Y (X(f g))
= X(f Y (g) + Y (f )g) − Y (f X(g) + X(f )g)
= f (XY g) + (Xf )(Y g) + (XY f )g + (Y f )(Xg)
− (Y f )(Xg) − f (Y Xg) − (Y Xf )g − (Xf )(Y g)
= f (XY g − Y Xg) + (XY f − Y Xf )g
= f [X, Y ]g + ([X, Y ]f )g.
∂2 ∂2
[∂i , ∂j ]f = ∂i ∂j f − ∂j ∂i f = f ◦ ϕ−1 − f ◦ ϕ−1 = 0.
∂i ∂j ∂j ∂i
Therefore we have
n
[X, Y ] = X i ∂i Y j ∂j − Y j ∂j X i ∂i = X i ∂i Y i − Y i ∂i X j ∂j . (7.1)
i,j=1
The Lie bracket measures the extent to which the derivatives in directions X
and Y do not commute. The following proposition makes this more precise:
60 Lecture 7. Lie brackets and integrability
Proposition 7.1.1 Let X, Y ∈ X (M), and let Ψ and be the local flow of X
in some region containing the point x ∈ M . Then
d
−1
[X, Y ]x = (Dx Ψt ) YΨt (x) .
dt t=0
The idea is this: The flow Ψt moves us from x in the direction of the
vector field X. We look at the vector field Y in this direction, and use the
map Dx Ψt : Tx M → TΨt (x) M , which is non-singular, to bring Y back to Tx M .
This gives us a family of vectors in the same vector space, so we can compare
them by differentiating. In particular, this gives us the nice interpretation:
The Lie bracket [X, Y ] vanishes if and only if Y is invariant under the flow
of X. For this reason, the Lie bracket is also often called the Lie derivative,
and denoted by LX Y .
Proof (Method 1): I will first give a proof which is quite illustrative, but which
has the drawback that we need to consider two separate cases. Fix x ∈ M .
Case 1: Suppose Xx = 0. I will construct a special chart about x as follows:
First take any chart ϕ : U → V about x, and by composition with a linear
map assume that Xx = ∂n (x). Let
Σ = ϕ−1 {(w1 , . . . , wn ) ∈ V : wn = ϕn (x)} .
with respect to the natural bases {∂1 , . . . , ∂n−1 , ∂t } for T(x,0) (Σ × R) and
{∂1 , . . . , ∂n−1 , ∂n = Xx } for Tx M . In particular, D(x,0) Ψ̃ is non-singular, and
so by the inverse function theorem there is a neighbourhood of (x, 0) in Σ ×R
on which Ψ̃ is a diffeomorphism. We take ϕ̃ to be the chart obtained by first
taking the inverse of Ψ̃ to obtain a point in Σ × R, and then applying ϕ to
the point in Σ to obtain a point in Rn−1 × R. The special feature of this
chart is that Xy = ∂n on the entire chart.
Now we can compute in this chart: The flow of X can be written down
immediately: ϕ̃ ◦ Ψt ◦ ϕ̃−1 (x1 , . . . , xn ) = (x1 , . . . , xn−1 , xn + t). If we write
Y = Y k ∂k , then we have
k
−1
(Dx Ψt ) YΨt (x) = Y k (x1 , . . . , xn + t),
and so
7.1 The Lie bracket. 61
d k ∂Y k ∂Y k
−1
(Dx Ψt ) YΨt (x) = n
= Xj = [X, Y ]k
dt t=0 ∂x ∂xj
by (7.1), since ∂i X j = 0 for all j.
Case 2: Now suppose Xx = 0. Then we have Ψt (x) = x for all t, and the local
group property implies that the linear maps {Dx Ψt } form a one-parameter
group in the Lie group GL(Tx M )
GL(n). Hence Dx Ψt = etA for some A,
and we get
−1
(Dx Ψt ) YΨt (x) = etA Yx ,
and
d −1
(Dx Ψt ) YΨt (x) = −AYx .
dt t=0
It remains to compute A: Working in local coordinates,
A(∂j ) = ∂t ∂j Ψt = ∂j ∂t Ψt = ∂j X.
Therefore we get
d
−1
(Dx Ψt ) YΨt (x) = −Y j ∂j X k ∂k = [X, Y ]
dt t=0
Proof (Method 2): Here is another proof which is somewhat more direct but
perhaps less illuminating: Choose any chart ϕ for M about x. In this chart
we can write uniquely X = X j ∂j and Y = Y k ∂k , and here the coefficients
X j and Y k are smooth functions on a neighbourhood of x. Then
d j
j
j
DΨt (x) Ψ−t YΨt (x) = ∂t ∂k Ψ−t Yxk + ∂k Ψ−t ∂t Y k (Ψt x)
dt t=0
t=0 t=0
j k j i k
= ∂k ∂t Ψ−t Yx + δ X ∂ Y
k x i x
t=0
= −Yxk ∂k Xxj + Xxi ∂i Yxj
= [X, Y ]jx .
We could now deduce the following result on the naturality of the Lie
bracket directly from Proposition 7.6.2. However I will give a direct proof:
[W, Z]y f = Wy Zf − Zy W f
= Dx F (Xx )Zf − Dx F (Yx )W f
= Xx ((Zf ) ◦ F ) − Yx ((W f ) ◦ F )
= Xx (Y (f ◦ F )) − Yx (X(f ◦ F ))
= [X, Y ]x (f ◦ F )
= Dx F ([X, Y ]x )f.
Here we used the fact that for every p ∈ M ,
The Lie bracket gives the space of smooth vector fields X (M ) on a manifold
M the structure of a Lie algebra:
The first two properties are immediate. The third (known as the Jacobi
identity) can be verified directly:
Differentiate this with respect to t to get another proof of the Jacobi identity.
Lie algebras play an important role in the theory of Lie groups: Consider
the space g of left-invariant vector fields on a Lie group G. We have already
seen that this is a finite-dimensional vector space isomorphic to the tangent
space at the identity Te G by the natural construction
v ∈ Te G → V ∈ g : Vg = De lg (v).
We will show that g is a Lie algebra. It is sufficient to show that the vector
subspace g of X (M ) is closed under the Lie bracket operation:
Proposition 7.2.2
[g, g] ⊆ g.
The significance of the Lie bracket, for our purposes, comes from its role in
indicating when a family of vector fields can be simultaneously integrated
to give a map: Suppose we have k vector fields E1 , . . . , Ek on M , and we
want to try to find a map F from Rk to M for which ∂i F = Ei for i =
1, . . . , k. The naive idea is that we try to construct such a map as follows:
Pick some point x in M , and set F (0) = x. Then we can arrange ∂k F = Ek ,
by setting F (tek ) = ΨEk (x, t) for each t. The next step should be to follow
the integral curves of the vector field Ek−1 from F (tek ) for time s to get the
point F (tek + sek−1 ):
In fact, there is an easy way to tell this could not have worked: If there
was such a map, then we would have
But in the example we have [E1 , E2 ] = ∂2 = 0. It turns out that this is the
only obstruction to constructing such a map:
Proof. This is clearly true for t = 0, since ΨX,0 (y) = y for all y. Therefore it
suffices to show that
∂t ΨX,t ◦ ΨY,s = ∂t ΨY,s ◦ ΨX,t .
The left-hand side is equal to the vector field X by assumption, while the
right is equal to DΨY,s (X). We have
" s
d
DΨY,s (X) = DΨY,0 (X) + DΨY,r (X) dr
0 dr
" s
d
=X+ DΨY,r DΨY,r (X) dr,
0 dr r =0
where I used the local group property of the flow of Y to get DΨY,r+r =
DΨY,r ◦ DΨY,r . By Proposition 6.1.1, we have
DΨY,r (X) = −[Y, X] = 0,
r =0
Remark. One can make this definition more natural (i.e. with less explicit reference
to charts) as follows: Given a manifold M , there is a natural manifold Gk (T M )
constructed from M , called the Grassmannian bundle (of k-planes) over M
Gk (T M ) = {(x, E) : x ∈ M, E a k−dimensional subspace of Tx M } .
Note that the k-dimensional subspaces of Tx M are in one-to-one correspondence
with the space of equivalence classes of rank (n−k) linear maps from Tx M to Rn−k ,
under the equivalence relation
M1 ∼ M2 ⇐⇒ M1 = LM2 , L ∈ GL(n − k) :
We denote the equivalence class of M by [M ]. Given a k-dimensional subspace E,
we choose any n−k independent linear functions f1 , . . . , fn−k on Tx M which vanish
66 Lecture 7. Lie brackets and integrability
for i = 1, . . . , k and j = 1, . . . , n − k, where the sums are over all pairs of permuta-
tions of n − k objects.
Note that these charts cover Gk (T M ), because even if [F ] ∈ Gk (Tx M ) does not
have its first n − k columns non-singular with respect to the coordinate tangent
basis of ϕ, we can choose a new chart for which this is true, of the form A ◦ ϕ for
some A ∈ GL(n).
Then we can define a k-dimensional distribution to be a smooth map D from
M to Gk (T M ) such that Dx ∈ Gk (Tx M ) for all x ∈ M .
Now we compute:
n
n
[Ei , Ej ] = ∂i + api (y)∂p , ∂j + aqj (y)∂q
p=k+1 q=k+1
n
n
= ∂i apj − ∂j api + aqi ∂q apj − aqj ∂q api ∂p .
p=k+1 q=k+1
Proof. Since h is closed under Lie brackets, the distribution D defined by the
vectors in h is involutive: If we take a basis {E1 , . . . , Ek } for h, then any
X, Y ∈ X (M ) can be written in the form X = X i Ei , Y = Y j Ej for some
smooth functions X 1 , . . . , X k and Y 1 , . . . , Y k . Then we have
[X, Y ] = [X i Ei , Y j Ej ] = (X i Ei (Y j ) − Y i Ei (X j ))Ej + X i Y j [Ei , Ej ]
which is in D. By Frobenius’ Theorem, there is a submanifold Σ passing
through e ∈ G with tangent space D.
To show that Σ is a subgroup, we show that xy ∈ Σ for all x and y in Σ:
Write y = exp(sY ) for Y ∈ De . Then
d d
(x exp(sY )) = (x exp(sY ) exp(rY )) = D0 lx exp(sY ) Y ∈ D
ds dr r=0
for each s. Thus the curve s
→ x exp(sY ) starts in Σ and is tangent to
D = T Σ, and so stays in Σ. Therefore xy ∈ Σ, and Σ is a subgroup.
Lecture 8. Connections
In other words, we just differentiate the coefficient functions. The key point
is that we think of the vectors {e1 , . . . , en } as being constant.
When we are dealing with vector fields on a manifold, this no longer makes
sense: There are no natural vector fields which we can take to be ‘constant’.
Example 8.1.1 A failed attempt. One might think that a chart would supply
natural vector fields that we can think of as being constant – every vector
field can be written as a linear combination of the coordinate tangent vectors
{∂1 , . . . , ∂n }. Unfortunately, defining derivatives in terms of these does not
really make sense: The difficulty is that changing to a different chart would
give a different answer for the derivative of a vector field.
To see this, suppose we have two charts, ϕ and η, and that a vector field X
n (ϕ) n (ϕ)
is given by i=1 X i ∂i , and a vector v = i=1 v i ∂i . Then we can define
(ϕ)
a derivative ∇v X by
n
(ϕ) (ϕ)
∇(ϕ)
v X = v j ∂j Xi ∂i .
i=1
(ϕ) n (η)
Now work in the chart η: We have ∂i = Λji ∂j , where Λ is the
j=1
matrix D η ◦ ϕ−1 . Therefore we have
72 Lecture 8. Connections
n
(η) (η)
∇(ϕ)
v X = v j Λkj ∂k X i Λli ∂l .
i=1
So these two notions of derivative don’t agree if Λ is not constant (i.e. if the
second derivatives of the transition map do not vanish).
Since there does not seem to be any particularly natural way to define deriva-
tives of vector fields on a manifold, we have to introduce extra structure on
the manifold to do this. The extra structure required is a connection on M .
Let us check explicitly that this is a connection: The linearity (with con-
stant coefficients) in each argument is immediate. The Leibniz rule also holds:
N
∇v (f X) = ∇v (f X α )eα
α=1
N
α
=π v(f X )eα
α=1
N
α α
=π (v(f )X + f v(X )) eα
α=1
N
N
= v(f )π X α eα + fπ v(X α )eα
α=1 α=1
= v(f )X + f ∇v X
since πX = X.
This makes sense because (De lγ(t) )−1 Xγ(t) ∈ Te G for all t, and we know how
to differentiate vectors in a vector space.
Note that we could just as well have considered right-invariant vector
fields to define the connection. This gives the right-invariant connection on
G, which is generally different from the left-invariant connection if G is a
non-commutative Lie group.
∇∂i ∂j = Γij k ∂k .
The main point of this proposition is that the derivative of a vector field
V in the direction of a vector v can be computed if one only knows the values
of V along some curve with tangent vector v.
The covariant derivative along γ is defined by
dV i
∇t V i (t)∂i = ∂i + γ̇ j V i Γji k (γ(t))∂k
dt
where γ̇ = γ̇ k ∂k . The first two statements of the proposition follow imme-
diately, and it remains to check that our definition is consistent with the
third.
Let Ṽ be a smooth vector field in M . Then the induced vector field along
a curve γ is given by V (t) = Ṽγ(t) , or in terms of the coordinate tangent
basis, V (t) = V i (t)∂i , where Ṽx = Ṽ i (x)∂i , and V i (t) = Ṽ i (γ(t)).
Then
∇γ̇ Ṽ = ∇γ̇ Ṽ i ∂i
= (Dγ(∂t )) (Ṽ i )∂i + γ̇ k Ṽ i Γki l ∂l
= ∂t Ṽ i ◦ γ ∂i + γ̇ k V i Γki l (γ(t))∂i
= ∇t V.
For example, the constant vector fields in Rn are parallel with respect to
the standard connection, and the left-invariant vector fields on a Lie group
are parallel with respect to the left-invariant connection.
dV i
0= + Γkl i γ̇ k V l . (8.5.1)
dt
This is a linear system of n first order ordinary differential equations for
the n functions V 1 . . . , V n along I, and so (since the coefficient functions
i
γ̇ k Γkl are bounded and piecewise continuous) there exist unique solutions
with arbitrary initial values V 1 (0), . . . , V n (0). We define Pt by Pt (V i (0)∂i ) =
V i (t)∂i , where V i (t) satisfy the system (8.5.1). We leave it to the reader to
check that this defines a linear isomorphism for each t.
Proof. Choose a basis {e1 , . . . , en } for Tγ(0) M , and define Ei (t) = Pt (ei ) ∈
Tγ(t) M . Then we can write V (t) = V i (t)Ei (t) = V i (t)Pt (ei ) = Pt (V i (t)ei ),
or in other words
Pt−1 (V ) = V i (t)ei .
Therefore we have
d −1 d i
Pt V (t) = V ei ,
dt dt
and
d −1 d i
Pt Pt V (t) = V (t) Ei (t).
dt dt
On the other hand we have
d i d i
∇t V = ∇t V i (t)Ei (t) = V (t) Ei (t) + V i (t)∇t Ei = V (t) Ei (t)
dt dt
since Ei is parallel.
Example 8.5.4 Parallel transport on the sphere. Let M = S 2 , and take the
submanifold connection given by Example 8.2.2. Then a vector field along a
curve in S 2 is parallel if and only if its rate of change (as a vector in R3 ) is
always normal to the surface of S 2 .
Consider the path γ on S 2 which starts at the north pole, follows a line of
longitude to the equator, follows the equator for some distance (say, a quarter
of the way around) and then follows another line of longitude back to the
north pole. Note that each of the three segments of this curve is a geodesic.
We compute the vector field given by parallel transport along γ of a vector
which is orthogonal to the initial velocity vector at the north pole.
On the first segment, the parallel transport keeps the vector constant as
a vector in R3 (this must be the parallel transport since it remains tangent
to S 2 , and has zero rate of change, so certainly the tangential component of
its rate of change is zero).
On the segment around the equator, we start with a vector tangent to the
equator, and parallel transport will give us the tangent vector to the equator
of the same length as we move around the equator: Then we have V = L dx dt
for some constant L (the length), and so
d2 x
V̇ = L = −Lx
dt2
since x(t) = (cos t, sin t, 0). In particular, V̇ is normal to S 2 .
On the final segment, the situation is the same as the first segment: We
can take V to be constant as a vector in R3 .
So parallel transport around the entire loop has the effect of rotating the
vector through π/2.
78 Lecture 8. Connections
Dv V = v i ∂i + v(V i )∂˙i .
∇v X = π(p,Xp ) (Dv X) .
8.6 The connection as a projection. 79
TM
p
T(p,v)(TM) TM
(p,v)
TM
p
(p,X)
T(p,X)(TM) Dv X TM
∇X
v
TM
p
v
We will show that every smooth manifold can be equipped with a connection
– in fact, there are many connections on any manifold, and no preferred or
canonical one (later, when we introduce Riemannian metrics, we will have a
way of producing a canonical connection).
We assume that our manifolds are paracompact – that is, every covering
of M by open sets has a refinement which is locally finite. In particular, we
have an atlas A for M which is locally finite (i.e. each point of M lies in only
finite many of the coordinate charts of the atlas).
Proof. By refining if necessary, assume that the coordinate charts have images
in Rn which are bounded open sets which have smooth boundary. Consider
one of these charts, ϕα : Uα → Vα . We can choose a number > 0 such that
the distance function d(y) = d(y, ∂Vα ) to the boundary of Vα is smooth at
points where d(y) < ε. Then define
ψα = Fε ◦ d ◦ ϕα ,
ψα (x)
ρα (x) = .
α ψα (x)
This makes sense because the sum is actually finite at each point of M . The
result is clearly linear (over R) in both v and X, and
∇v (f X) = ∇(α)
v (f ρα X)
α
= v(f )ρα X + f ∇(α)
v (ρα X)
α
= v(f )X + f ∇v X
where I used the fact that α ρα = 1 to get the first term in the last equality.
Suppose ∇(1) and ∇(2) are two connections on a smooth manifold M . Then
we have the following important result:
(1) k (2) k
∇(2)
v X − ∇v X = v(X )∂j + v X Γji ∂k − v(X )∂v − v X Γji
(1) j j i j j i
∂k
(2) k (1) k
= v j X i Γji − Γji ∂k .
It follows that the difference of two connections defines a map for each
x ∈ M from Tx M × Tx M to Tx M , which varies smoothly in x and is linear
in each argument. We also have a kind of converse to this:
x, X → ∇v X + A(v, X)
is also a connection on M .
The proof is a simple calculation to check that the Leibniz rule still holds.
It is easy to show that there are many such suitable maps A (for example, one
can write in a local coordinate chart A(ui ∂i , v j ∂j ) = ui v j Aij k ∂k where the
coefficient functions Aij k are smooth and have support inside the coordinate
chart. Then sums and linear combinations of such things are also suitable.
Thus there are many different connections on a manifold.
8.11 Geodesics and the exponential map 83
∇X Y − ∇Y X − [X, Y ],
This map is called the torsion T (X, Y ) of ∇ applied to X and Y . This can be
written in terms of its components Tij k , defined by ∇∂i ∂j − ∇∂j ∂i = Tij k ∂k .
γ̈ i + Γkl i γ̇ k γ̇ l = 0
Then our local calculations above show that the geodesic equation is given
exactly by the flow for time 1 of the vector field G. More precisely, it is given
by the projection onto M of this flow.
It follows immediately that the exponential map is smooth, and that the
map
d
(x, v)
→ expx (v), expx (tv)
dt t=1
Therefore D0 expx is just the identity map (here we are identifying the tan-
gent space to Tx M at 0 with Tx M itself, in the usual way). In particular, this
derivative is non-degenerate, so expx is a local diffeomorphism.
Roughly speaking, this says that points which are sufficiently close to
each other can be joined by a unique “short” geodesic (we cannot expect to
drop the shortness condition – consider for example the case where M is a
cylinder or a sphere, when there are generally many geodesics joining any
pair of points.
˜ : T M → M × M defined by
Proof. Consider the map exp
exp(p,
˜ v) = (p, expp (v)).
Let us compute the derivative of this map at a point (p, 0): Breaking up the
2n × 2n derivative matrix into n × n blocks, the top left corresponds to the
derivative of the map (p, v)
→ p with respect to p, which is just the identity
map, and the bottom left block is the derivative of the same with respect
to v, which is zero. The top right block might be quite complicated, but as
86 Lecture 8. Connections
Y3
p3
X3 Y2
p2
q(t) X2
Y
Y1
p X
p1 X1
Lecture 9. Riemannian metrics
9.1 Definition
There is one more crucial ingredient which we need to introduce for dealing
with manifolds: Lengths and angles. Given a smooth manifold, since we know
what it means for a curve in the manifold to be smooth, and we have a well-
defined notion of the tangent vector to a curve, all we need in order to have
a notion of distance on the manifold is a way of defining the speed of a curve
— that is, the length of its tangent vector.
Every smooth manifold carries a Riemannian metric (in fact, many of them).
We will prove this using an argument very similar to that used in showing
the existence of connections.
Choose an atlas {ϕα : Uα → Vα }, and a subordinate partition of unity
{ρα }. On each of the regions Vα in Rn , choose a Riemannian metric g (α) (as
in example 9.1.2). Then define
g(u, v) = ρα g (α) (Dϕα (u), Dϕα (v)) .
α
Thus γ1 #γ2 is a curve with length L[γ1 ] + L[γ2 ]. Given points p, q, and r in
M , for any ε > 0 we can choose curves γ1 joining p to q and γ2 joining q to
r, such that L[γ1 ] < d(p, q) + ε and L[γ2 ], d(q, r) + ε. Therefore
g q g
r
p g# g
9.4 Submanifolds
Similarly, one can define right-invariant metrics; in general these are not the
same.
Example 9.5.1 A metric on hyperbolic space. Recall that the hyperbolic plane
H 2
is upper
half-plane, identified with the group of matrices of the form
y x
for y > 0. If we choose an inner product at the identity (0, 1) such
0 1
that (0, 1) and (1, 0) are orthonormal, then the corresponding left-invariant
Riemannian metric on H2 is the one for which the left-invariant vector fields
E1 = (y, 0) and E2 = (0, y) are orthonormal. Thus in terms of the basis of
coordinate tangent vectors e1 = (1, 0) and e2 = (0, 1), the metric has the
form gij = y −2 δij .
u, ve = De lg (u), De lg (v)g = (De rg )−1 De lg (u), (De rg )−1 De lg (v)e .
Exercise 9.6.1 Show that the adjoint action of S 3 on its Lie algebra R3 gives
a homomorphism ρ : S 3 → SO(3) (compare Exercise 5.3.2 and the remark
following it). Deduce that the only bi-invariant metric on S 3 is the standard
one.
is the right circular cone |x|2 = t2 . Directions which point from the origin
above the upper surface or below the lower surface of this cone are timelike,
while those that point between the two surfaces are spacelike. In particular,
a hypersurface given by the graph of a function u over Rn × {0} is spacelike
if the gradient of u is everywhere less than 1 in magnitude.
timelike
direction
n+1
R spacelike
direction
null
cone
Exercise 9.7.3 Show that the induced Riemannian metric on the unit ball
in Rn by the stereographic projection described above is
4
gij = δij .
(1 − |x|2 )2
1−|x|2 −y 2
Exercise 9.7.4 Show that the map (x, y)
→ 2x
|x|2 +(y−1)2 , |x|2 +(y−1)2 for
(x, y) ∈ Rn−1
× R
R diffeomorphically maps the open unit ball to the
n
upper half space, and induces from the metric on the unit ball in exercise
9.7.3 the following metric on the upper half-space:
1
gij = δij .
y2
9.7 Semi-Riemannian metrics 93
In the case n = 2, show that this is the same as the left-invariant metric on
H2 from example 9.5.1.
z
Hn
x(z)
(0,-1)
Lecture 10. The Levi-Civita connection
Here on the left hand side we are applying the vector v (as a derivation) to
the smooth function x
→ gx (Xx , Yx ). This is something which is well-defined
without reference to any connection. The right-hand side does depend on the
connection.
Dv (X · Y ) = Dv X · Y + X · Dv Y.
Exercise 10.1.2 Show that any connection for which the lengths and an-
gles between parallel vector fields are constant must be compatible with the
metric.
10.2 Submanifolds
∇X Y − ∇Y X = [X, Y ];
∇Y Z − ∇Z Y = [Y, Z];
∇Z X − ∇X Z = [Z, X],
and the compatibility conditions
g(∇X Y, Z) + g(Y, ∇X Z) = Xg(Y, Z);
g(∇Y Z, X) + g(Z, ∇Y X) = Y g(Z, X);
g(∇Z X, Y ) + g(X, ∇Z Y ) = Zg(X, Y ).
Take the sum of the first two of the latter equations, and subtract the third.
Then apply the symmetry conditions, yielding:
2g(∇X Y, Z) = Xg(Y, Z) + Y g(Z, X) − Zg(X, Y )
+ g(Z, [X, Y ]) + g(Y, [Z, X]) + g(X, [Z, Y ]).
This determines the inner product of ∇x Y with any vector field Z purely in
terms of the metric, and so implicitly determines ∇x Y . This completes the
proof of uniqueness. To prove existence, it is only necessary to check that the
formula above does indeed define a connection with the desired properties.
Denote the right-hand side of the formula by C(X, Y, Z). C is C ∞ in X
and in Z:
C(f X, Y, hZ) = f X (hg(Y, Z)) + Y (f hg(Z, X)) − hZ (f g(X, Y ))
+ hg(Z, [f X, Y ]) + g(Y, [hZ, f X]) + f g(X, [hZ, Y ])
= f hXg(Y, Z) + f hY g(Z, X) − f hZg(X, Y )
+ f (Xh)g(Y, Z) + f Y (h)g(Z, X)
+ hY (f )g(Z, X) − hZ(f )g(X, Y )
+ hf g(Z, [X, Y ]) + hf g(Y, [Z, X]) + f hg(X, [Z, Y ])
− hY (f )g(Z, X) − f X(h)g(Y, Z)
+ hZ(f )g(Y, X) − f Y (h)g(X, Z)
= f hC(X, Y, Z)
It follows that the map x
→ C(X, Y, Z)x depends only on Y and the values
Xx and Zx of the vector fields X and Z at x In the second slot we have
instead:
C(X, f Y, Z) = f C(X, Y, Z) + X(f )g(Y, Z) − Z(f )g(X, Y )
+ X(f )g(Z, Y ) + Z(f )g(X, Y )
= f C(X, Y, Z) + 2X(f )g(Y, Z).
where ∇(h) is the Levi-Civita connection of the metric h on N . Show that this
is well-defined (i.e. makes sense even though DF (X) is only defined on the
image F (M ) ⊂ N ) and defines a connection on M , which is the Levi-Civita
connection on M .
Proof. Recall that the derivative of the exponential map at the origin is just
the identity map. Therefore
gij (0) = g(∂i , ∂j ) = g(D0 expx (ei ), D0 expx (ej )) = δij
since the {ei } were chosen to be orthonormal.
We also have
∇ai ∂i aj ∂j = 0
at the origin for any constants a1 , . . . , an , since ai ∂i is the tangent vector to
the geodesic through 0 in that direction. By symmetry we have at the origin
0 = ∇∂i +∂j (∂i + ∂j )
= ∇∂i ∂i + ∇∂j ∂j + ∇∂i ∂j + ∇∂j ∂i
= 2∇∂i ∂j .
for every i and j. Therefore all the connection coefficients vanish at the origin.
Remark. By the same proof, the connection coefficients at the origin vanish
with respect to exponential coordinates for a connection (not necessarily a
Levi-Civita connection) if and only if the connection is symmetric. This gives
another interpretation of the torsion of a connection.
Lecture 11. Geodesics and completeness
Proposition 11.1.1 In geodesic polar coordinates the metric takes the form
g(∂r , ∂r ) = 1;
g(∂r , u) = 0,
In other words, the image under the exponential map of the unit radial
vector in Tx M is always a unit vector, and the image of a vector tangent
to a sphere about the origin is always orthogonal to the image of the radial
vector. Think of a polar ’grid’ of radial lines and spheres in Tx M , mapped
onto M by the exponential map Tx M . Then this map says that the images
of the spheres are everywhere orthogonal to the images of the radial lines
(which are of course geodesics).
Proposition 11.1.1 is often called the Gauss Lemma.
Proof. The first part of the claim follows from the fact that ∂r is the tangent
vector to a geodesic, so ∇r ∂r = 0. Hence by compatibility,
∂r g(∂r , ∂r ) = 2g(∂r , ∇r ∂r ) = 0,
Tx M
expx
Note that we cannot expect that geodesics minimise length on long inter-
vals – consider the example of the sphere S 2 : Geodesics are great circles, and
these achieve the distance between their endpoints on intervals of length no
greater then π, but not on longer intervals.
2
S
Minimising
segments A non-minimising
segment
g(t )
p q
Now let σ be any other curve joining the points p and q. Suppose first that
σ remains in the set Br (p) = expp (Br (0)). Then we can write σ(t) = r(t)ω(t)
where r > 0 and |ω(t)| = 1. The squared length of the tangent vector σ is
then
|σ (t)|2 = (r )2 g(∂r , ∂r ) + 2rr g(∂r , ω ) + r2 g(ω , ω )
= (r )2 + r2 |ω |2
≥ (r )2 ,
with equality if and only if ω vanishes. Therefore we have
104 Lecture 11. Geodesics and completeness
"
L[σ] ≥ |r |dt ≥ 2ε,
Proof. First observe that γ achieves the distance between any pair of its
points: If there were some subinterval on which this were not true, then
replacing γ by a shorter path on that subinterval would also yield a shorter
path from γ(0) to γ(1).
For any t ∈ (0, 1), there is a sufficiently small neighbourhood J of t in [0, 1]
such that γ J is contained in a diffeomorphic image of the exponential map
from one of its endpoints, and so by the Gauss Lemma γ J is a reparametrised
geodesic.
We know that points which are sufficiently close to each other can be con-
nected by a unique ‘short’ geodesic. This result can be stated somewhat
more cleany in the special case of the Levi-Civita connection than it can in
the general case:
Proof. The idea is exactly the same as the proof of Proposition 8.11.5: The
map exp
˜ is a local diffeomorphism from a neighbourhood of (p, 0) in T M to a
neighbourhood of (p, p) in M × M . Choose η sufficiently small to ensure that
11.3 Convex neighbourhoods 105
Proposition 11.3.2 For any p ∈ M there exists a constant ε > 0 such that
for every pair of points q and r in Bε (p), there exists a unique geodesic γqr
for which γ(0) = q, γ(1) = r, and d(p, γ(t)) ≤ max{d(p, q), d(p, r)} for all
t ∈ [0, 1]. Furthermore L[γqr ] = d(q, r).
d2 d 2 k 2
2
d(p, γ(t))2 = 2 x
dt dt
k
d k k
=2 x ẋ
dt
k
2 d2 k
=2 ẋk +2 xk x
dt2
k k
k 2
=2 ẋ −2 xk Γij k ẋi ẋj
k k
1 2
≥ 2 ẋk
2
ẋk −
2
k [
≥ 0,
since we are within the ball of radius 2η about p. Therefore d(p, γ(t))2 is a
convex function along γqr , so the maximum value is attained at the endpoints.
106 Lecture 11. Geodesics and completeness
Proof. (2) =⇒ (3) trivially. We prove (1) =⇒ (2), and then (3) =⇒ (∗p ) and
((3) and (∗p )) =⇒ (1), where (∗p ) is the statement that every point q ∈ M
can be connected to p by a length-minimising geodesic.
Suppose (M, d) is a complete metric space. If (2) does not hold, then there
is p ∈ M , v ∈ Tp M with |v| = 1, and T < ∞ such that γ(t) = expp (tv) exists
for t < T but not for t = T .
Be (y)
But then γ can be extended beyond y by taking t > T in this formula. This
is a contradiction, so condition (2) must hold.
Next we prove (3) =⇒ (∗p ). Let p ∈ M be such that (3) holds, and
q ∈ M any other point. Choose ε > 0 such that expp is a diffeomorphism
on a set containing the closure of Bε (0). If q ∈ Bε (p) then we are done, so
assume not. Then Sε (p) = expp (Sε (0)) is the image of a compact set under a
continuous map, and so the function d(., q) attains its minimum on this set.
In other words, there exists r = expp (εv) such that d(r, q) = d(Sε (p), q) =
inf{d(r , q) : r ∈ Sε (p)}. Then we must have
p g
r
q
S (p)
e
d(p, q) = t + d(γ(t), q)
= s + (t − s) + d(γ(t), q)
≥ s + d(γ(s), q)
≥ d(p, q),
so equality must hold throughout, and s ∈ A.
Finally, we will prove that A is open in [0, d(p, q)]. This will imply A =
[0, d(p, q)], so in particular d(γ(d(p, q)), q) = 0, and γ(d(p, q)) = q.
Suppose T ∈ A, and write p = γ(T ). Choose δ > 0 such that expp is a
diffeomorphism on Bδ (0). If q ∈ Bδ (p ), then write q = expp (d(p , q)w), and
define
108 Lecture 11. Geodesics and completeness
γ(t), for t ≤ T ;
σ(t) =
expp ((t − T )w) for t ≥ T .
Then σ is a curve joining p to q, and by assumption
Therefore σ is a minimising curve, hence a geodesic, so σ(t) = expp (t) for all
t ∈ [0, d(p, q)] and we are done.
Otherwise, q ∈ / Bδ (p ), and we can choose r ∈ Sδ (p ) such that d(r , q) =
d(Bδ (p ), q). Then as before, we have
and so by assumption
Let σ be the unit speed curve given by following γ from p to p , then following
the radial geodesic from p to r . Then L[σ] = d(p, p ) + δ. Therefore,
S (p')
d
g p'
p
r'
q
It follows that L[σ] = d(p, r ), since if there were any shorter path σ from
p to r , we would have
We have already seen several examples of the idea we are about to introduce,
namely linear (or multilinear) operators acting on vectors on M .
For example, the metric is a bilinear operator which takes two vectors
to give a real number, i.e. gx : Tx M × Tx M → R for each x is defined by
u, v
→ gx (u, v).
The difference between two connections ∇(1) and ∇(2) is a bilinear op-
erator which takes two vectors and gives a vector, i.e. a bilinear operator
Sx : Tx M × Tx M → Tx M for each x ∈ M . Similarly, the torsion of a connec-
tion has this form.
Tx : Tx M × . . . × Tx M × Tx∗ M × . . . × Tx∗ M → R.
. /0 1 . /0 1
k times l times
Note that a covector is just a tensor of type (1, 0), and a vector is a tensor
of type (0, 1), since a vector v acts linearly on a covector ω by v(ω) := ω(v).
Multilinearity means that
T ci1 vi1 , . . . , cik vik , aj1 ω j1 , . . . , ajl ω jl
i1 ik j1 jl .
i1 ik j1 jl
= c . . . c aj1 . . . ajl T (vi1 , . . . , vik , ω , . . . , ω )
i1 ,...,ik ,j1 ,...,jl
110 Lecture 12. Tensors
dxi : Tx M → R
by
dxi (∂j ) = δji .
The notation here comes from the fact that dxi is the differential of the
smooth function xi , in other words
df (v) = v(f ).
Definition 12.2.1 Let T and S be two tensors at x of types (k, l) and (p, q)
respectively. Then the tensor product T ⊗ S is the tensor at x of type
(k + p, l + q) defined by
Proof. The vector space structure of the space of (k, l)-tensors is clear.
Note that a tensor T is completely determined by its action on the basis
vectors and covectors:
T (v1i ∂i , . . . , vkj ∂j , ωa1 dxa , . . . , ωbl dxb )
= v1i . . . vkj . . . ωa1 . . . ωbl T (∂i , . . . , ∂j , dxa , . . . , dxb ),
by multilinearity.
Then observe that this allows T to be written as a linear combination of
the proposed basis elements: Define
Then in particular the value of this linear combination on any set of vectors
∂a1 , . . . , ∂ak and covectors dxb1 , . . . , dxjl must give zero. But this value is
exactly the coefficient:
Aa1 ...ak b1 ...bl = 0.
This establishes the linear independence, and show that we do indeed have a
basis.
12.3 Contractions
Example 12.4.1 If T is a tensor of type (k, 1), then it takes k vectors and
gives another one, which we denote by T (v1 , . . . , vk ). The index-lowering map
12.6 Tensor bundles 113
should produce from this a map which takes (k + 1) vectors and gives a
number, which we denote by T (v1 , . . . , vk+1 ). Following the procedure above,
we see that this isomorphism is just given by
In analogy with the definition of the tangent bundle, we can define (k, l)-
tensor bundles:
2k 2l 3 2k 2l
TM ⊗ T ∗M = Tx M ⊗ Tx∗ M.
x∈M
If we have two such charts ϕ and η, the transition maps between them
are given by η ◦ ϕ−1 on the first n components, and by the map
114 Lecture 12. Tensors
Ti1 ...ik j1 ...jl → Λai11 . . . Λaikk (Λ−1 )jb11 . . . (Λ−1 )jbll Ta1 ...ak b1 ...bl
Often the tensors which we work with will arise in the following way: We are
given an operator T which takes k vector fields and l 1-forms, and gives a
smooth function. Suppose that the operator is also multilinear (over R), so
that multiplying any of the vector fields or 1-forms by a constant just changes
the result by the same factor, and taking one of the vector fields or 1-forms
to be given by a sum of two such, gives the sum of the results applied to the
individual vector fields or 1-forms.
T (f1 X1 , . . . , fk Xk , g1 ω 1 , . . . , gl ω l ) = f1 . . . fk g1 . . . gl T (X1 , . . . Xk , ω 1 , . . . , ω l )
j1 ...jl
(Tx )i1 ...ik = T (∂i1 , . . . , ∂ik , dxj1 , . . . , dxjl ) x .
Example 12.7.2 The Lie bracket of two vector fields takes two vector fields
and gives another vector field. This is not a tensor, because
in general. Similarly, the connection ∇ takes two vector fields and gives a
vector field, but is not a tensor, because
∇X (f Y ) = f ∇X Y + X(f )Y = f ∇X Y
in general.
We can also extend the connection (defined to give derivatives of vector fields)
to give a connection on each tensor bundle – i.e. to allow the definition of
derivatives for any tensor field on M .
We want the following properties for the connection: ∇ should take a
tensor field T of type (k, l), and give a tensor field ∇T of type (k + 1, l), such
that
(1). The Leibniz rule holds for tensor products: If S and T are tensor fields
of type (k, l) and type (p, q) respectively, then
As with the case of vector fields, the connection allows us to define derivatives
of tensors aloong curves, and parallel transport of tensors along curves (by
solving the first order linear ODE system corresponding to ∇t T = 0. As in
the case of vector fields, parallel transport preserves inner products between
tensors.
In the last few lectures we have seen how a connection can be used to dif-
ferentiate tensors, and how the introduction of a Riemannian metric gives a
canonical choice of connection. Before exploring the properties of Rieman-
nian spaces more thoroughly, we will first look at a special class of tensors
for which there is a notion of differentiation that makes sense even without
a connection or a metric. These are called differential forms, and they play
an extremely important role in differential geometry.
We will first look a little more at the linear algebra of tensors at a point.
We will consider a natural subspace of the space of k-tensors, namely the
alternating tensors.
Example 13.1.1 The geometric meaning of this definition is probably not clear
at this stage. An illustrative example is the following: Choose an orthonormal
120 Lecture 13. Differential forms
Note that this is zero if the k-tuple is not distinct, and that if we change the
order of the k-tuple then the result merely changes sign (depending whether
the k-tuple is rearranged by an even or an odd permutation).
The factor k! is included in our definition for the following reason: If we
apply the alternating k-tensor dxi1 ∧ . . . ∧ dxik (where i1 , . . . , ik are distinct)
to the k vectors ∂i1 , . . . , ∂ik , the result is 1. If we apply it to the same k
vectors in a different order (say, rearranged by some permutation σ), then
the result is just the sign of σ. Any other k vectors yield zero.
These ‘elementary alternating k-tensors’ have a geometric interpretation
similar to that in Example 13.1.1: The value of dxI (v1 , . . . , vk ) is the determi-
nant of the matrix with (m, n) coefficient dxim (vn ), and this gives the signed
k-dimensional volume of the projection of the parallelopiped generated by
v1 , . . . , vk onto the subspace generated by ∂i1 , . . . , ∂ik . This relationship be-
tween alternating forms and volumes will be central in the next lecture when
we define integration of differential forms and prove Stokes’ theorem.
Proposition 13.1.1
(1). dxi1 ∧. . .∧dxik = σ∈Sk sgnσ dxiσ(1)⊗. . .⊗dxiσ(k) = k!A(dxi1⊗. . .⊗dxik ).
(2). For each k, {dxi1 ∧ . . . ∧ dxik : 1 ≤ i1 < . . . < ik ≤ n} is a basis for
Λk Tx∗ M . In particular the space of alternating k-tensors at x has dimension
( nk ) = k!(n−k)!
n!
.
Proof. (1) is immediate, since the value on any k-tuple of coordinate basis
vectors agrees. To prove (2), we note that by Proposition 12.2.2, any alter-
nating tensor can be written as a linear combination of the basis elements
dxi1 ⊗ . . . ⊗ dxik . Invariance under A shows that this is the same as a linear
combination of k-forms of the form A(dxi1 ⊗ . . . ⊗ dxik ), and these are all of
the form given. It remains to show the supposed basis is linearly independent,
but this is also immediate since if I = (i1 , . . . , ik ) then dxI (∂i1 , . . . , ∂ik ) = 1,
but dxJ (∂i1 , . . . , ∂ik ) = 0 for any increasing k-tuple J = I.
for some coefficients Ti1 ...ik . Some caution is required here, because T can
also be written in the form
1
T = Ti ...i dxi1 ∧ . . . ∧ dxik
k! i ,...,i 1 k
1 k
where the coefficients are the same as above for increasing k-tuples, and to
be given from these by antisymmetry in other cases. Thus the coefficients in
this expression differ by a factor k! from those given in the expression after
Proposition 12.2.2.
This may not seem the obvious definition, because of the factor on the
right. This is chosen to make our notation consistent with that in our def-
inition of the basis elements: Take an increasing k-tuple i1 , . . . , ik and an
increasing l-tuple j1 , . . . , jl , and assume for simplicity that ik < j1 . Then
we can form the alternating tensors dxi1 ∧ . . . ∧ dxik , dxj1 ∧ . . . ∧ dxjl and
dxj1 ∧ . . . ∧ dxik ∧ dxj1 ∧ . . . ∧ dxjl , and we would like to know that the third
of these is the wedge product of the first two.
Proposition 13.2.1
The wedge product is characterized by the following properties:
(i). Associativity: f ∧ (g ∧ h) = (f ∧ g) ∧ h;
(ii). Homogeneity: (cf ) ∧ g = c(f ∧ g) = f ∧ (cg);
(iii). Distributivity: If f and g are in Λk Tx∗ M then
(f + g) ∧ h = (f ∧ h) + (g ∧ h);
g ∧ f = (−1)kl f ∧ g;
122 Lecture 13. Differential forms
The homogeneity and distributivity properties of the wedge product are im-
mediate from the definition. From this we can deduce the following expression
for the wedge product in local coordinates: For S = Si1 ...ik dxi1 ∧. . .∧dxik and
T = Tj1 ...jl dxj1 ∧ . . . ∧ dxjl (summing over increasing k-tuples and l-tuples
respectively)
1
S∧T = Si1 ...ik Tj1 ...jl dxi1 ∧ . . . ∧ dxik ∧ dxj1 ∧ . . . ∧ dxjl .
k!l! i
1 ,...,ik ,j1 ,...,jl
(−1)k
= gi ...i fj ...j dxj1 ∧ dxi1 ∧. . .∧dxik ∧ dxj2 ∧ . . . ∧ dxjl
k!l!i ,...,i ,j ,...,j 1 k 1 l
1 k 1 l
...
(−1)kl
= gi1 ...ik fj1 ...jl dxj1 ∧ . . . ∧ dxjl ∧ dxi1 ∧ . . . ∧ dxik
k!l! i1 ,...,ik ,j1 ,...,jl
= (−1) f ∧ g.kl
13.4 The exterior derivative 123
(iii). d2 = 0.
One can easily check that this formula defines an operator which satisfies
the required conditions. In particular we can compute d2 to check that it
vanishes:
d2 ωi1 ...ik dxi1 ∧ . . . ∧ dxik
= d ∂i ωi1 ...ik dxi ∧ dxi1 ∧ . . . ∧ dxik
= ∂j ∂i ωi1 ...ik dxj ∧ dxi ∧ dxi1 ∧ . . . ∧ dxik
1 ∂ 2 ωi1 ...ik ∂ 2 ωi1 ...ik
= − dxj ∧ dxi ∧ dxi1 ∧ . . . ∧ dxik
2 ∂xj ∂xi ∂xi ∂xj
= 0.
To extend this definition to all of M we need to check that it does not
depend on the choice of coordinate chart. Let η be any other chart, with
components y 1 , . . . , y n . On the common domain of η and ϕ, we have xi =
F i (y), where F = ϕ ◦ η −1 , and
∂F i j
dxi = dy .
∂y j
This implies that
∂F i1 ∂F ik j1
dxi1 ∧ . . . ∧ dxik = . . . dy ∧ . . . ∧ dy jk .
i1 ,...,ik ,j1 ,...,jk
∂y j1 ∂y jk
Now we can check that if we define the operator d in the y coordinates, then
d(dxi1 ∧ . . . ∧ dxik ) = 0:
i1
i1 ∂F ∂F ik j1
d dx ∧ . . . ∧ dx ik
= d ... dy ∧ . . . ∧ dy jk
∂y j1 ∂y jk
I,J
k
∂ 2 F im 5 ∂F ip j0
= jm ∂y j0 jp
dy ∧ . . . ∧ dy jk
m=1
∂y ∂y
I,J,j0 p =m
1
k 2
∂ F ∂ 2 F im
im
= − j0 jm
2 ∂y jm ∂y j0 ∂y ∂y
I,J,j0 m=1
5 ∂F ip
× dy j0 ∧ . . . ∧ dy jk
∂y jp
p =m
=0
It follows (by linearity and distributivity) that the differential operators de-
fined in the two charts agree.
Now we will prove a remarkable result which really makes the theory of
differential forms work:
F∗ (ω ∧ η) = F∗ (ω) ∧ F∗ η
and
d(F∗ ω) = F∗ (dω).
126 Lecture 13. Differential forms
Proof. The proof of the first statement is immediate from the definition. The
second statement is proved by an argument identical to that used to prove
that the definition of the exterior derivative does not depend on the chart in
Proposition 13.4.1, except that the map F may be a smooth map between
Euclidean spaces of different dimension.
Proof. Suppose there exists such an n-form ω. Let A be the set of charts
ϕ : U → V for M for which ω(∂1 , . . . , ∂n ) > 0. Then A is an atlas for M ,
since any chart for M is either in A or has its composition with a reflection
in A — in particular charts in A cover M . Furthermore A is an oriented
atlas: For any pair of charts ϕ and η in A with non-trivial common domain
of definition in M , we have
j (ϕ)
∂i = Di η ◦ ϕ−1 i ∂j ,
(η)
where ωi1 ...il = 0 whenever i1 , . . . , il ≤ k. This implies that ∂m ωi1 ...il = 0 for
i1 , . . . , il ≤ k and arbitrary m. Applying the exterior derivative, we find
dω(X, Y ) = X i Y j (∂i ωj − ∂j ωi )
= X i ∂i (Y j ωj ) − Y j ∂j (X i ωi ) − X i (∂i Yj )ωj + Y j (∂j Xi )ωi
= Xω(Y ) − Y ω(X) − ω([X, Y ])
= 0 at x,
k
dω(X0 , . . . , Xk ) = (−1)i Xi ω(X0 , . . . , X̂i , . . . , Xk )
i=0
+ (−1)i+j ω([Xi , Xj ], X0 , . . . , X̂i , . . . , X̂j , . . . , Xk ).
0≤i<j≤k
Here the open subsets of Rn+ are the sets of the form W ∩ Rn+ where
W ⊆ Rn is open.
Now we can give some further meaning to differential forms by defining what
is meant by integration of differential forms on oriented manifolds. A key
point to keep in mind here is that none of our definitions depend on us having
a metric on the manifold, so we do not in general have any notion of volume
of surface area or length. Nevertheless the structure of differential forms is
exactly what is required to produce a well-defined notion of integration.
Let M n be a compact, oriented differentiable manifold with boundary,
and let ω ∈ Ω n (M ). Then we define the integral of ω over M , denoted M ω,
as follows: Let {ρα : α ∈ I} be a partition of unity subordinate to an
oriented boundary atlas for M , so that for each α there exists an oriented
chart (either a regular chart or a boundary chart) ϕα : Uα → Vα for M , such
that suppρα ⊂ Uα .
Define
" "
ω= (ϕ−1 1 n
α )∗ (ρα ω) (e1 , . . . , en ) dx . . . dx .
M α∈I Vα
To put this into words: We write ω as a sum α ρα ω of forms which are
supported in charts. Each of these can be integrated over the chart by inte-
grating the smooth function obtained by plugging in the coordinate tangent
vectors for that chart. Then we add the resulting numbers together to get
the integral of ω.
We need to check that this does not depend on the choice of partition
of unity. Suppose {φβ : β ∈ J } is any other partition of unity for M , with
corresponding oriented coordinate charts ηβ : Wβ → Zβ . Then we have
132 Lecture 14. Stokes’ Theorem
"
(ϕ−1 1
α )∗ (ρα ω) (e1 , . . . , en ) dx . . . dx
n
α Vα
"
= (ϕ−1 1 n
α )∗ (ρα φβ ω) (e1 , . . . , en ) dx . . . dx .
α,β ϕα (Uα ∩Wβ )
Fix α and β. Then by definition of the pull-back, we have for any form σ
−1
(ϕα )∗ σ (e1 , . . . , en ) = (ηβ−1 )∗ σ (D(ηβ ◦ ϕ−1 −1
α )(e1 ), . . . , D(ηβ ◦ ϕα )(en )).
Proof. We can assume that ω is non-zero. Consider the map from GL(n) to
R defined by
˜ : L
→ ω(Le1 , . . . , Len ) .
det
ω(e1 , . . . , en )
The denominator is non-zero by assumption.
The multilinearity and antisymmetry of ω imply det ˜ is linear in each row,
is unchanged by adding one row to another, and has the value 1 if L = I.
These are the axioms that define the determinant, so det ˜ = det.
It follows that
−1
−1
(ϕα )∗ σ (e1 , . . . , en ) = det(D(ηβ ◦ ϕ−1
α )) (η β )∗ σ (e1 , . . . , en ).
We also know, since the charts are oriented, that the determinant on the
right-hand side is positive. The change of variables formula therefore gives
"
−1
(ϕα )∗ (ρα φβ ω) (e1 , . . . , en ) dx1 . . . dxn
ϕα (Uα ∩Wβ )
"
−1
= | det(D(ηβ ◦ ϕ−1 1
α ))| (ηβ )∗ σ (e1 , . . . , en ) dx . . . dx
n
ϕα (Uα ∩Wβ )
"
= (ηβ−1 )∗ σ (e1 , . . . , en ) dx1 . . . dxn .
ηβ (Uα ∩Wβ )
Consequently
"
(ϕ−1 1
α )∗ (ρα ω) (e1 , . . . , en ) dx . . . dx
n
α Vα
"
= (ηβ−1 )∗ (φβ ω) (e1 , . . . , en ) dx1 . . . dxn ,
β Zβ
where the integral on the right-hand side is taken using the induced orientation
on ∂Ω, integrating the restriction of ω to ∂M (i.e. the pull-back of ω by the
inclusion map).
α Vα
"
= d (ϕ−1 1 n
α )∗ (ρα ω) (e1 , . . . , en ) dx . . . dx .
α Vα
For each α there are two possibilities: The chart ϕα is either a regular chart
or a boundary chart.
In the first case, Vα is an open set in Rn . Write ω in components in the
chart ϕα :
n
ω= ˆ j ∧ . . . ∧ dxn .
ωj dx1 ∧ dx
j=1
n "
" " 0
∂(ρα ωj ) 1 ∂(ρα ωn ) n 1
dx . . . dxn = dx dx . . . dxn−1
j=1 Vα ∂xj Rn−1 ×{0} −∞ ∂xn
"
= ρα ωn dx1 . . . dxn−1
Rn−1 ×{0}
"
= ρα ω.
∂M
Turning the result of Stokes’ theorem around, we can interpret the exterior
derivative in the following way: Let ω be a k-form in a manifold M . Fix
linearly independent vectors v1 , . . . , vk+1 in Tx M , and choose any chart ϕ
about x. Write vj = vjl ∂l in this chart. For r small we can define smooth
maps xr from the k-dimensional sphere S k into M , by
xr (z i ei ) = ϕ−1 (ϕ(x) + rz i vik ek ).
Using Stokes’ theorem, we can deduce
"
1
dω(v1 , . . . , vk+1 ) = lim (xr )∗ ω
r→0 r k+1 |B k+1 | Sk
where |B k+1 | is the volume of the unit ball in Rk+1 . In this sense the exterior
derivative measures the ‘boundary integral per unit volume’ of a form (where
‘volume’ is measured in comparison to that of the parallelepiped generated
by v1 , . . . , vk+1 , not using any notion of measure or metric on the manifold).
This is easy to understand in the case of a 0-form (i.e. a function). Then
the ‘boundary integral’ becomes ‘difference in values at the endpoints’, while
‘per unit volume’ means ‘per unit time along a curve with velocity v1 ’. So this
just recaptures the usual notion of the directional derivative of a function in
terms of difference quotients.
ωij = εijk V k
This is just the divergence of the vector field V times the volume form.
Stokes’ theorem then says that the integral of dω over a region U (i.e. the
integral of the divergence of V over U ) is equal to the integral of the 2-form
ω over ∂U . The latter is equal to the integral of V, n over ∂U , where n
is the outward-pointing unit normal vector. This is just the classical Gauss
theorem (or divergence theorem).
Lecture 15. de Rham cohomology
In this lecture we will show how differential forms can be used to define topo-
logical invariants of manifolds. This is closely related to other constructions
in algebraic topology such as simplicial homology and cohomology, singular
homology and cohomology, and Čech cohomology.
(Here we really mean the integral over Σ of the form obtained by pulling
back ω under the inclusion map).
Now suppose we have two such submanifolds, Σ0 and Σ1 , which are
(smoothly) homotopic. That is, we have a smooth map F : Σ × [0, 1] → M
with F |Σ×{i} an immersion describing Σi for i = 0, 1. Then d(F∗ ω) is a
(k + 1)-form on the (k + 1)-dimensional oriented manifold with boundary
Σ × [0, 1], and Stokes’ theorem gives
" " "
d(F∗ ω) = ω− ω.
Σ×[0,1] Σ1 Σ1
In particular,
if dω = 0, then d(F∗ ω) = F∗ (dω) = 0, and we deduce that
Σ1
ω = Σ0
ω.
This says that k-forms with exterior derivative zero give a well-defined
functional on homotopy classes of compact oriented k-dimensional submani-
folds of M .
We know some examples of k-forms with exterior derivative zero, namely
those of the form ω = dη for some (k − 1)-form η. But Stokes’ theorem then
gives that Σ ω = Σ dη = 0, so in these cases the functional we defined on
homotopy classes of submanifolds is trivial.
138 Lecture 15. de Rham cohomology
Z k (M )
H k (M ) = .
dΩ k−1 (M )
Since the exterior derivative and Stokes’ theorem do not depend in any
way on the presence of a Riemannian metric on M , the cohomology groups
of M depend only on the differentiable structure on M . It turns out that
they in fact depend only on the topological structure of M , and not on
the differentiable structure at all — any two homeomorphic manifolds have
the same cohomology groups5 . The groups H k (M ) are therefore topological
invariants, which can be used to distinguish manifolds from each other: If two
manifolds have different cohomology groups, they cannot be homeomorphic
(let alone diffeomorphic).
The k-the cohomology group H k (M ) is a real vector space. The dimension
of this vector space is called the kth Betti number of M , and denoted bk (M ).
5
The de Rham theorem states that the de Rham cohomology groups are isomorphic
to the singular or Čech cohomology groups with real coefficients, and these are
defined in purely topological terms. It is also a consequence of this theorem that
the cohomology groups are finite dimensional.
15.4 The group H 1 (M ) 139
Since this is true for all closed loops γ, Proposition 15.4.1 applies to show
[ω] = 0 in H 1 (M ), and so H 1 (M ) = 0.
and therefore
" 1 " 1 " 1
∂ω0
f1∗ ω − f0∗ ω = ω0 (1) − ω0 (0) = dt = dM ω1 dt = dM ω1 dt.
0 ∂t 0 0
k
(dη)i1 ...ik = (−1)p ∂ip ηi0 ...iˆp ...ik
p=0
k " 1
= (−1)p tk (ωty )ip i0 ...iˆp ...ik dt
p=0 0
k " 1
+ (−1)p y j tk+1 ((∂ip ω)ty )ji0 ...iˆp ...ik dt.
p=0 0
We rewrite the last term using the fact that dω = 0: This means
0 = (dω)ji0 ...ik
k
= ∂j ωi0 ...ik − (−1)p ∂ip ωji0 ...iˆp ...ik .
p=0
142 Lecture 15. de Rham cohomology
Remark. Now that we have seen the explicit proof of the Poincaré Lemma,
I remark that there is a very simple proof using the homotopy invariance
result of Proposition 15.4.1: First, the cohomology of R0 is trivial to compute.
Second, there is a smooth homotopy equivalence f : B → R0 = {0} defined
by f (x) = 0: If we take g : R0 → B to be given by g(0) = 0, then we
have f ◦ g equal to the identity on R0 , and g ◦ f (x) = 0 on B. The latter is
homotopic to the identity under the homotopy F : B × [0, 1] → B given by
F (x, t) = (1 − t)x. Corollary 15.5.2 applies.
In the next few sections we will see some example of these long exact
sequences in cohomology and their applications.
144 Lecture 15. de Rham cohomology
forms via the inclusion map i. This map is surjective. The kernel consists
of those forms ω on M which have i∗ ω = 0. This is again a chain complex,
since i∗ (dω) = di∗ ω = 0 if i∗ ω = 0. This gives us a short exact sequence
relating the cohomologies of M , Σ, and the chain complex Ω0k (M, Σ) = {ω ∈
Ω ∗ (M ) : i∗ ω = 0}.
We will now show that the cohomology of the latter is isomorphic to the
compactly supported cohomology of M \Σ.
To see this, we note that Ωck (M \Σ) ⊂ Ω0k (M, Σ). We denote by C k the
quotient space Ω0k (M, Σ)/Ωck (M \Σ). We define an operator d : C k → C k+1
by d[ω] = [dω]. If η ∈ Ωck (M \Σ), then d(ω+η) = dω+dη ∈ dω+Ωck+1 (M \Σ),
so this operator is well defined and satisfies d2 = 0. Therefore the complex C
is a cochain complex, and we have a short exact sequence of chain complexes
We will prove that H k (C) = 0 for all k, and the long exact sequence above
then implies that Hck (M \Σ)
H0k (M, Σ).
Suppose ω ∈ Ω0k (M, Σ) satisfies d[ω] = 0, that is,
dω = η
for some η ∈ Ωck+1 (M \Σ). We want to show that [ω] = d[σ] for some [σ] ∈
C k−1 , which means we want to show that ω − dσ ∈ Ωck (M \Σ).
Since Σ is a smooth compact submanifold, the nearest-point projection
p (defined using any Riemannian metric on M ) is a smooth map from a
neighbourhood T of Σ in M to Σ, and is homotopic to the identity map on
T . We can assume that du = 0 on T since du ∈ Ωck+1 (M \Σ). Therefore by
the homotopy invariance we have
ω − p∗ ω = dv
0 → R → R2 → R2 → H 1 (S 1 ) → 0
0 → R → R2 → R → H 1 (S 2 ) → 0 → R → H 2 (M ) → 0.
Proposition 15.11.1
0, k<n
Hck (Rn ) =
R, k = n.
Proof. For k = 0 the result is immediate because constants are not compactly
supported in Rn .
1
n = 1 the result is also immediate: If ω = ω1 dx , then
For k = 1 and
ω = df implies R ω1 = 0, and conversely.
We will use the long exact sequence from section 15.9 together with the
results on cohomology groups of spheres from section 15.10: The sphere S n
contains an equatorial S n−1 as a submanifold, and the complement S n \S n−1
is diffeomorphic to two copies of Rn . Therefore the long exact sequence be-
comes (for n > 1)
15.12 The group H n (M ) 147
0→R→R
→Hc1 (Rn )2 → 0 → 0
...
→Hcn−1 (Rn )2 → 0 → R
→Hcn (Rn )2 → R → 0
It follows that Hck (Rn ) = 0 for k = 1, . . . , n − 1 and Hcn (Rn ) = R.
and hence
ω = d(v + vα )
α
148 Lecture 15. de Rham cohomology
and ω is exact.
The subspace X is defined by a finite collection of equations cjk xk = 0,
j = 1, . . . , K. Therefore an n-form ω on M is exact if and only if
"
(cjk ρk )ω = 0
M
∗
Tx M , hence Λn T(x,[ω]) M̃
Λn Tx∗ M , and so [ω] ∈ P Λn Tx∗ M gives a global
section of P Λn T ∗ M̃ .
In the case where M is not orientable, there is a natural involution i of
M̃ induced by the map ω
→ −ω of Λn T ∗ M , and this is orientation-reversing.
Since π ◦ i = π, a differential n-form ω̃ on M̃ arises from pull-back by π of a
differential form ω on M if and only if i∗ ω̃ = ω̃. But then we have
" " "
ω̃ = − i∗ ω̃ = − ω̃
M̃ M̃ M̃
since i is orientation-reversing, and hence M̃ ω̃ = 0. It follows from the case
we have already considered that ω̃ = dη for some η ∈ Ω n−1 (M̃ ). Then let
η̃ = (η + i∗ η)/2. Then we have i∗ η̃ = η̃, so η̃ = π ∗ η for some η ∈ Ω n−1 (M ),
and dη̃ = (dη + di∗ η)/2 = (ω̃ + i∗ ω̃)/2 = ω̃. It follows that dη = ω, so ω is
exact. Therefore H n (M ) = 0, as claimed.
15.13 Cohomology of surfaces 149
In this section we will use the results we have developed above about coho-
mology groups to compute the cohomology of compact surfaces.
We already know the cohomology groups of S 2 . Next we will compute
the cohomology groups of the torus T2 = S 1 × S 1 . This can be written as
the union of open sets U and V , where U and V are each diffeomorphic to
S 1 ×R, and U ∩V is diffeomorphic to two copies of S 1 ×R. We therefore know
H 0 (T2 ) = H 0 (U ) = H 0 (V ) = R, H 0 (U ∩ V ) = R2 , H 1 (U ) = H 1 (V ) = R
and H 1 (U ∩ V ) = R2 , and H 2 (U ) = H 2 (V ) = H 2 (U ∩ V ) = 0, H 2 (T2 ) = R.
Thus we have the long exact sequence
0 → R → R2 → R2 → H 1 (T2 ) → R2 → R2 → R → 0 → 0
which implies that H 1 (T2 ) = R2 .
We will proceed by induction on the genus of the surface: A surface Mg+1
of genus g + 1 can be written as the union of sets A and B, where A is diffeo-
morphic to T2 \{p}, B is diffeomorphic to Mg \{q}, and A∩B is diffeomorphic
to S 1 × R. To use this we first need to find the cohomology groups of T2 \{p}
and Mg \{p}.
and each of the terms of the right is a tensor in X and Y . This leaves one
further calculation:
R(X, Y )(f Z) = ∇Y (f ∇X Z + X(f )Z) − ∇X (f ∇Y Z + Y (f )Z)
− [X, Y ](f )Z − f ∇[X,Y ] Z
= f ∇Y ∇X Z + Y (f )∇X Z + Y X(f )Z + X(f )∇Y Z
− f ∇X ∇Y Z − X(f )∇Y Z − XY (f )Z − Y (f )∇X Z
− [X, Y ](f )Z − f ∇[X,Y ] Z
= f ∇Y ∇X Z − ∇X ∇Y Z − ∇[X,Y ] Z
+ (Y X(f ) − XY (f ) − [X, Y ](f )) Z
= f R(X, Y )Z.
Remark. Note that this calculation does not use the compatibility of the
connection with the metric, only the symmetry of the connection. Thus any
152 Lecture 16. Curvature
Proof. The first symmetry is immediate from the definition of curvature. For
the second, work in a coordinate tangent basis:
Rikjl + Rkjil + Rjikl = g (∇k ∇i ∂j − ∇i ∇k ∂j , ∂l )
+ g (∇j ∇k ∂i − ∇k ∇j ∂i , ∂l )
+ g (∇i ∇j ∂k − ∇j ∇i ∂k , ∂l )
= g (∇k (∇i ∂j − ∇j ∂i ) , ∂l )
+ g (∇j (∇k ∂i − ∇i ∂k ) , ∂l )
+ g((∇i (∇j ∂k − ∇k ∂j ) , ∂l )
=0
by the symmetry of the connection.
The third symmetry is a consequence of the compatibility of the connec-
tion with the metric:
0 = ∂i ∂j gkl − ∂j ∂i gkl
= ∂i (g(∇j ∂k , ∂l ) + g(∂k , ∇j ∂l ))
− ∂j (g(∇i ∂k , ∂l ) + g(∂k , ∇i ∂l ))
= g(∇i ∇j ∂k , ∂l ) + g(∇i ∂k , ∇j ∂l ) + g(∇j ∂k , ∇i ∂l ) + g(∂k , ∇i ∇j ∂l )
− g(∇j ∇i ∂k , ∂l ) − g(∇j ∂k , ∇i ∂l ) − g(∇i ∂k , ∇j ∂l ) − g(∂k , ∇j ∇i ∂l )
= Rjikl + Rjilk .
16.2 Sectional curvatures 153
More generally, we see that (in any dimension), if {ei } are orthonormal,
then Rijkl depends only the (oriented) two-dimensional plane generated by
ei and ej , and the one generated by ek and el .
Proof. We will give an explicit expression for a component Rijkl of the curva-
ture tensor, in terms of sectional curvatures. We work with an orthonormal
basis {e1 , . . . , en } at a point of M .
For convenience we will refer to the oriented plane generated by e1 and
ej by the notation ei ∧ ej . We compute the sectional curvature of the plane
2 (ei + ek ) ∧ (ej + el ):
1
(ei + ek ) ∧ (ej + el ) 1
K = R(ei + ek , ej + el , ei + ek , ej + el )
2 4
1 1 1 1
= K(ei ∧ ej ) + K(ei ∧ el ) + K(ej ∧ ek ) + K(ek ∧ el )
4 4 4 4
1 1 1 1
+ Rijil + Rijkj + Rilkl + Rklkj
2 2 2 2
1 1
+ Rijkl + Rkjil .
2 2
Now add the same expression with ek and el replaced by −ek and −el :
(ei + ek ) ∧ (ej + el ) (ei − ek ) ∧ (ej − el )
Rijkl + Rkjil = K +K
2 2
1 1 1 1
− K(ei ∧ ej ) − K(ei ∧ el ) − K(ej ∧ ek ) − K(ek ∧ el ).
2 2 2 2
Finally, subtract the same expression with ei and ej interchanged: On the
left-hand side we get
Here the sum is over all i and j with i < j, and all k and l with k < l.
In particular, this curvature operator, since symmetric, can be diago-
nalised. It is important to note that the eigenvalues of the curvature operator
need not be sectional curvatures! The sectional curvatures are the values of
the curvature operator on simple 2-planes, but there is no reason why the
eigen-vectors of the curvature operator should be simple 2-planes. In partic-
ular, it is possible to have all the sectional curvatures positive (or negative)
at a point, while not having all of the eigenvalues of the curvature operator
positive (negative).
In the special case of three dimensions, however, every 2-plane is simple,
and so the eigenvalues of the curvature operator are sectional curvatures.
In this case we refer to the eigenvectors of the curvature operator as the
principal 2-planes, and the eigenvalues the principal sectional curvatures.
∇∂i ∂j = Γij k ∂k .
gij = f δij
for some function f . Then the inverse metric is also easy to compute:
g ij = f −1 δ ij .
An orthonormal basis is given by {yei }, which gives for any sectional curva-
ture
K = −1.
Thus hyperbolic space has constant negative curvature.
Hence the curvature operator has all eigenvalues equal to −1, and all of the
sectional curvatures are −1.
where on the left hand side the Lie bracket is to be interpreted as the com-
mutator of the linear transformations Ad(X) and Ad(Y ).
Now when we differentiate the Ad-invariance condition, we get:
Lemma 16.8.3
ad(X)Y = [X, Y ].
and
k
(ΨX,s ΨY,−t (x)) = −tY k (x) + sX k (x) − stY i (x)∂i X k (x) + O(st2 , s2 )
and finally,
k
(ΨY,t ΨX,s ΨY,−t (x)) = −tY k (x) + sX k (x) − stY i (x)∂i X k (x) + O(st2 , s2 )
+ tY k (x) + t(−tY i (x) + sX i (x))∂i Y k (x) + O(st2 , t2 )
= sX k (x) − st Y i (x)∂i X k (x) − X i (x)∂i Y k (x)
+ O(s2 , t2 ).
162 Lecture 16. Curvature
And by left-invariance,
d d tX
ΨX,t (h) = Xh = De lh Xe = he ,
dt dt
so that
ΨX,t (h) = hetX .
Therefore
etX esY e−tX = ΨX,−t etX esY = ΨX,−t ΨY,s ΨX,t (e).
Proof. By Proposition 16.8.1 we have cijk = cjki , and we also have the sym-
metry cijk = −cjik . Therefore
1 kl
∇Ei Ej = g (cijl + clij + clji ) Ek
2
1
= g kl (cijl + cijl + cjil ) Ek
2
1 kl
= g cijl Ek .
2
Therefore
Rikjl = g ∇k ∇i Ej − ∇i ∇k Ej − ∇[Ek ,Ei ] Ej , El
1 1 1
= ckpl cijp − cipl ckjp − ckip cpjl
4 4 2
Now we note that the Jacobi identity (cf. Lecture 6) gives
16.8 Bi-invariant metrics 163
Example 16.8.6 We will apply this to a simple example, namely the Lie group
S 3 : Here the metric for which i, j and k are orthonormal at the identity is
easily seen to be bi-invariant, and we have the structure coefficients
[i, j] = ij − ji = 2k;
and
∇u v = πT M (Du (DX(v))) .
A particular case of this is an immersed hypersurface, which is the case
where M is of dimension N − 1. We will develop the theory of extrinsic
curvature first for the simpler case of hypersurfaces, and then extend this to
the more general case of immersed submanifolds.
and
n(z) = en+1 .
Therefore
∂2u
∂i ∂j X = en+1
∂xi ∂xj
and so at z,
∂2u
hij = − .
∂xi ∂xj
To put this another way, we have
1
u(y) = − hij (z)y i y j + O(y 3 )
2
as y → 0. This says that the second fundamental form gives the best approx-
imation of the hypersurface by a paraboloid defined over its tangent plane.
The normal bundle (and indeed the tangent bundle and the tensor bundles we
have already defined) is an example of a more general object called a vector
bundle. A vector bundle E of dimension k over M is defined by associating
to each x ∈ M a vector space Ex (often called the fibre at x), and taking
E = {(p, v) : p ∈ M, v ∈ Ep }. We require that E be a smooth manifold, and
that for each x ∈ M there is a neighbourhood U of x in M such that there
are k smooth sections φ1 , . . . , φk of E (i.e. smooth maps φi from M to E
such that π ◦ φi = id) such that φ1 (y), . . . , φk (y) form a basis for Ey for each
y ∈ U (it follows that the restriction of the bundle E to U is diffeomorphic
to U × Rk ).
We denote the space of smooth sections of E (i.e. smooth maps from M
to E which take each x ∈ M to the fibre Ex at x) by Γ (E).
A connection on a vector bundle E is a map which takes a vector u ∈ Tx M
and section φ ∈ Γ (E) and gives an element ∇u φ ∈ Ex , smoothly in the sense
that if U ∈ X (M ) and φ ∈ Γ (E) then ∇U φ ∈ Γ (E), which is linear in the
first argument and satisfies a Leibniz rule in the second:
168 Lecture 17. Extrinsic curvature of submanifolds
∇u (f φ) = f ∇u φ + u(f )φ
This is tensorial in all arguments — that is, the value of the resulting function
when evaluated at any point x ∈ M depends only on the values of X, Y , φ
and ψ at x. The proof of this is identical to the proof that the curvature of M
is a tensor (Lecture 16). This can be considered as an operator which takes
Λ2 Tx M to Λ2 Ex , since it is antisymmetric in the first two and the last two
arguments.
∇U (f V ) = πN M ((U f )V + f DU V )
= U (f )πN M V + f πN M DU V
= U (f )V + f ∇U V
so the Leibniz rule holds. This connection is compatible with the metric
induced on N M by the inner product on RN . By the argument above, this
defines a curvature tensor acting on Λ2 T M ⊗ Λ2 E, which we denote by R⊥
and call the normal curvature of M .
17.8 Second fundamental form of a submanifold 169
The second fundamental form is defined in an analogous way to that for the
hypersurface case: Given U, V ∈ X (M ) define
Then we find
1
f j (x1 , . . . , xn ) = − hkl (z), ej xk xl + O(x3 )
2
as x → 0. Thus the second fundamental form at z defines the best approxi-
mation to X(M ) as the graph of a quadratic function over DX(Tz M ).
Lecture 18. The Gauss and Codazzi equations
In this lecture we will prove the fundamental identities which hold for the
extrinsic curvature, including the Gauss identity which relates the extrinsic
curvature defined via the second fundamental form to the intrinsic curvature
defined using the Riemann tensor.
The definitions (from the last lecture) of the connection on the normal and
tangent bundles, and the second fundamental form h and the associated
operator W, can be combined into the following two useful identities: First,
for any pair of vector fields U and V on M ,
We will use the same method as above to deduce further important identities,
by applying U V − V U − [U, V ] to an arbitrary tangential vector field.
Let W be a smooth vector field on M . Then we have
0 = (U V − V U − [U, V ])W X
= U (V W X) − V (U W X) − [U, V ]W x
(18.1)
= U (−h(V, W ) + (∇V W )X) − V (−h(U, W ) + (∇U W )X)
− −h([U, V ], W ) + (∇[U,V ] W )X
(18.2)
= −DX(W(U, h(V, W ))) − (∇h(U, V, W ) + h(∇U V, W ) + h(V, ∇U W ))
+ DX(W(V, h(U, W ))) − (∇h(V, U, W ) + h(∇V U, W ) + h(U, ∇V W ))
+ U (∇V W )X − V (∇U W )X + h([U, V ], W ) − (∇[U,V ] W )X
(18.1)
= −DX(W(U, h(V, W ))) − (∇h(U, V, W ) + h(∇U V, W ) + h(V, ∇U W ))
+ DX(W(V, h(U, W ))) − (∇h(V, U, W ) + h(∇V U, W ) + h(U, ∇V W ))
− h(U, ∇V W ) + (∇U ∇V W )X + h(V, ∇U W ) − (∇V ∇U W )X
+ h([U, V ], W ) − (∇[U,V ] W )X
= DX (R(V, U )W − W(U, h(V, W )) + W(V, h(U, W )))
+ ∇h(U, V, W ) − ∇h(V, U, W ).
In deriving this we used the definition of curvature in the last step, and
used the symmetry of the connection to note that several of the terms cancel
out. The tangential and normal components of the resulting identity are the
following:
R(U, V )W = W(U, h(V, W )) − W(V, h(U, W )) (18.3)
and
∇h(U, V, W ) = ∇h(V, U, W ). (18.4)
Note that the tensor ∇h appearing here is the covariant derivative of the
tensor h, defined by
for any vector fields U , V and W . Here the ∇ appearing in the first term on
the right-hand side is the connection on the normal bundle, and the other
two terms involve the connection on T M .
18.3 The Ricci equations 173
Equation (18.3) is called the Gauss equation, and Equation (18.4) the
Codazzi equation. Note that the Gauss equation gives us a formula for the
intrinsic curvature of M in terms of the extrinsic curvature h.
It is sometimes convenient to write these identities in local coordinates:
Given a local chart with coordinate tangent vectors ∂1 , . . . , ∂n , choose also
a collection of smooth sections eα of the normal bundle, α = 1, . . . , N − n,
which are linearly independent at each point. Let g be the metric on T M and
g̃ the metric on N M , and write gij = g(∂i , ∂j ) and g̃αβ = g̃(eα , eβ ). Then we
can write
h(∂i , ∂j ) = hij α eα
and
W(∂i , eβ ) = Wiβ j ∂j .
The relation between h and W then tells us that Wiβ j = g jk g̃βα hik α .
The Gauss identity then becomes
Rijkl = hjk α hil β − hjk β hil α g̃αβ (18.5)
As before, this gives us two sets of identities, one from the tangential
component and one from the normal component. In fact the tangential com-
ponent is just the Codazzi identities again, but the normal component gives
a new identity, called the Ricci identity, which expresses the curvature of the
normal bundle in terms of the second fundamental form:
R⊥ (U, V )φ = h(V, W(U, φ)) − h(U, W(V, φ)). (18.7)
In local coordinates, with a local basis for the normal bundle as above, this
⊥
identity can be written as follows: If we write Rijαβ = g̃(R⊥ (∂i , ∂j )eα , eβ )
and hijα = g̃(h(∂i , ∂j ), eα ), then we have
⊥
Rijαβ = g kl (hikα hjlβ − hjkα hilβ ). (18.8)
18.4 Hypersurfaces
In particular, this basis diagonalizes the curvature operator, and the eigen-
values of the curvature operator are precisely λi λj for i < j. Note that all of
the eigenvectors of the curvature operator are simple planes in Λ2 T M . It fol-
lows that if M is any Riemannian manifold which has a non-simples 2-plane
as a eigenvector of the curvature operator at any point, then M cannot be
immersed (even locally) as a hypersurface in Rn+1 .
18.4 Hypersurfaces 175
Example 18.4.1 (Curvature of the unit sphere). Consider the unit sphere S n ,
which is a hypersurface of Euclidean space Rn+1 (so we take the immersion
X to be the inclusion). With a suitable choice of orientation, we find that
the Gauss map is the identity map on S n — that is, we have n(z) = X(z)
for all z ∈ S n . Differentiating this gives
DX(W(u)) = Du n = DX(u)
so that W is the identity map on T S n , and hij = gij . It follows that all of
the principal curvatures are equal to 1 at every point, and that all of the
sectional curvatures are equal to 1. Therefore the unit sphere has constant
sectional curvatures equal to 1.
We can now compute the second fundamental form for a very special
example, namely Hyperbolic space Hn , which is the set of unit future timelike
vectors in Minkowski space. In this case we have n(z) = z for every z ∈ Hn ,
since 0 = Du z, z = 2Du z, z = 2u, z. Differentiating, we find (exactly as
in the calculation for the sphere in Example 18.4.1 above) that hij = gij , so
that the principal curvatures are all equal to 1. The Gauss equation therefore
gives that the sectional curvatures are identically equal to −1.
More generally, a hypersurface with all principal curvatures of the same
sign (i.e. a convex hypersurface) in Minkowski space has negative curvature
operator, while a hypersurface with all principal curvatures of the same sign
in Euclidean space has positive curvature operator.
Lecture 19. The Cartan Moving Frame
Method
In this lecture we will explore the use of differential forms to compute con-
nection coefficients and curvatures for given Riemannian metrics.
We start with a technical Lemma (Cartan’s Lemma):
aijk − aikj = 0
aijk + ajik = 0.
Thus we have
Proof. We start by proving uniqueness: Suppose that ω̄ij is any other collec-
tion of one-forms satisfying the same conditions. Then let ηij = ω̄ij − ωij .
Then we have
and
ηij + ηji = 0.
By Lemma 19.1, we have ηij = 0, and therefore ωij = ω̄ij .
Now we prove existence. We set ωij = g(∇ek ei , ej )ωk , where ∇ is the
Levi-Civita connection of g. We use the identity
and
The fact that these two agree follows from the compatibility of the connection
with the metric, which gives
The 1-forms ωij are called the connection 1-forms corresponding to the
frame {ei }. Once these have been computed, we can compute the curvature
as follows:
Proposition 19.3 For any orthonormal frame {ei }, the curvature 2-form
1
Ωij = − Rijkl ωk ∧ ωl
2
can be computed from the connection 1-forms as follows:
Lecture 19. The Cartan Moving Frame Method 179
Proof. From the proof of the previous proposition, we have ωij (ek ) =
g(∇ek ei , ej ). Taking the exterior derivative, we find
Ω12 = −Kω1 ∧ ω2 .
Example 19.4 We will compute the curvatures of the Riemannian metric gij =
f 2 δij on a region of Rn , where f = f (xn ).
Here we have an obvious orthonormal frame given by ei = f −1 ∂i , and the
corresponding basis of 1-forms ωi = f dxi . Computing exterior derivatives,
we find
dωi = d(f dxi ) = f dxn ∧ dxi .
This gives the equations
180 Lecture 19. The Cartan Moving Frame Method
ωij ∧ ωj = f /f 2 ωn ∧ ωi
ηij ∧ ωj = 0.
dωij = 0
Ωij = (f )2 /f 2 ωi ∧ ωj
Rijij = −(f )2 /f 4
for 1 < i < j < n, while (except for symmetries) all other curvature compo-
nents are zero.
An important special case of this example is where f (x) = x−1 . Then we
find (f /f ) /f 2 = 1 and (f )2 /f 4 = 1, and therefore all sectional curvatures
of this metric are equal to −1.
Lecture 20. The Gauss-Bonnet Theorem
In this lecture we will prove two important global theorems about the geome-
try and topology of two-dimensional manifolds. These are the Gauss-Bonnet
theorem and the Poincaré-Hopf theorem.
Let us begin with a special case:
Suppose M is a compact oriented 2-dimensional manifold, and assume
that there exists a vector field V on M which is nowhere zero. Now let g
be any Riemannian metric on M . Then we can produce an orthonormal
frame globally on M by taking e1 = V /g(V, V )1/2 , and taking e2 to be the
unique unit vector orthogonal to e1 such that e1 and e2 are an oriented basis
at each point. The dual basis ω1 , ω2 is then also globally defined, and we
have a globally defined 1-form ω12 on M such that dω1 = ω12 ∧ ω2 and
dω2 = −ω12 ∧ ω1 . Then we have −K(g)ω1 ∧ ω2 = dω12 .
Stokes’ Theorem then applies to give
" "
− K(g)dV ol(g) = Ω12 = 0.
M M
So we have the remarkable conclusion that the integral of the Gauss curvature
over M is zero for every Riemannian metric on M .
.
The result we proved above is a special case of the famous Gauss-Bonnet
theorem. The general case is as follows:
Example 20.3 Here are some vector fields in the plane with nontrivial index
about 0:
2
y1
–2 –1 0 1 2
x
–1
–2
y1
–2 –1 0 1 2
x
–1
–2
y1
–2 –1 0 1 2
x
–1
–2
y1
–2 –1 0 1 2
x
–1
–2
y1
–2 –1 0 1 2
x
–1
–2
y1
–2 –1 0 1 2
x
–1
–2
We will prove that the sum of the indices of the vector field V at its
zeroes is independent of V , and the result can then be taken to define the
Euler characteristic. To see that this definition of the Euler characteristic
agrees with the one given above in terms of triangulations, suppose that we
are given a triangulation of M . Then choose a vector field as sketched below
with zeroes at the centre of each face, the middle of each edge, and at each
vertex. This vector field points outwards from the centre of each face, so has
index 1 there, and points inwards at each vertex, so also has index 1 there.
At the middle of each edge the vector field has index −1. Therefore the sum
of the indices equals F + V − E, agreeing with our previous definition of the
Euler characteristic.
184 Lecture 20. The Gauss-Bonnet Theorem
If we keep V fixed and vary g, we deduce that the left-hand side is independent
of g, and if we keep g fixed and vary V , we deduce that the right-hand side is
independent of V , so we can define the Euler characteristic and deduce both
results.
So let V and g be given, and let x1 , . . . , xN be the zeroes of V . About
each of these zeroes we can choose a small neighbourhood Ui of xi (say, given
by the image of a ball under some chart) such that V is nonvanishing on
M̃ = M \ ∪N i=1 Ui . Let γi be the boundary of Ui , parametrised anticlockwise
in some oriented chart.
Now M̃ is a compact oriented manifold with boundary, and V is nonva-
nishing on M̃ . Define e1 = V /g(V, V )1/2 , and take e2 to be the unit vector
orthogonal to e1 which is given by a positive right-angle rotation of e1 . Denote
the dual frame by ω1 and ω2 , and let ω12 be the corresponding connection
one-form. Then we have
Ω12 = dω12 ,
and Stokes’ theorem gives
" " "
− K(g) dV ol(g) = Ω12 = ω12 .
M̃ M̃ ∂ M̃
Lecture 20. The Gauss-Bonnet Theorem 185
(i)
Now on each of the regions Ui we can choose a non-vanishing frame ē1 ,
(i) (i)
ē2 .This gives a corresponding connection one-form ω̄12 , and we have for
each i " " " "
(i) (i)
− K(g) dV ol(g) = Ω12 = dω̄12 = ω̄12 .
Ui Ui Ui γi
(i)
To interpret the integral of ω̄12 − ω12 around γi , we need to consider
(i)
the relationship between the frames e1 and ē1 . Since these are both unit
vectors, they are related by a rotation. Therefore there exists a map from γi
to SO(1)
S 1 ⊂ R2 , say x
→ z = (α, β) ∈ S 1 , such that
(i)
ē1 = αe1 + βe2
(i)
ē2 = −βe1 + αe2 .
Then we have
(i)
ω̄1 = αω1 + βω2
(i)
ω̄2 = −βω1 + αω2 .
Since α2 + β 2 = 1, we have β −1 dα = −α−1 dβ = dθ. Applying the exterior
derivative we find
(i)
dω̄1 = dα ∧ ω1 + αdω1 + dβ ∧ ω2 + βdω2
= −(ω12 − β −1 dα) ∧ βω1 + (ω12 + α−1 dβ) ∧ αω2
(i)
= (ω12 + dθ) ∧ ω̄2 .
and similarly
(i) (i)
dω̄2 = −(ω12 + dθ) ∧ ω̄1 .
(i)
It follows that ω̄12 − ω12 = dθ, and so
" "
(i)
ω̄12 − ω12 = dθ
γi γi
We have proved
"
N
− K(g) dV ol(g) = −2π I(V, xi ),
M i=1
and the proofs of the Gauss Bonnet and Poincaré-Hopf theorems are com-
plete.