(Co)tangent things
killingform
June 20, 2018
Contents
1 The tangent bundle 2
1.1 Motivation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2
1.2 The abstract tangent space . . . . . . . . . . . . . . . . . . . . . 4
1.3 Reconciling the motivation with the abstract . . . . . . . . . . . 6
1.4 The pushforward . . . . . . . . . . . . . . . . . . . . . . . . . . . 7
1.5 The tangent bundle and vector fields . . . . . . . . . . . . . . . . 8
2 The cotangent bundle 10
2.1 Linear algebra review . . . . . . . . . . . . . . . . . . . . . . . . 10
2.2 The cotangent bundle and 1-forms . . . . . . . . . . . . . . . . . 12
2.3 k-forms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15
2.4 In R
n
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16
2.5 Pullbacks, orientations, and the integration of forms . . . . . . . 17
2.6 Cohomology . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20
A Further Reading 23
Reminder
Recall that a (topological) manifold is a locally Euclidean, second-countable,
Hausdorff topological space,
M
. By locally Euclidean, we mean for each point
p
in M there exists an open neighbourhood U of p which is homeomorphic to an
open subset
V
of
R
n
. The pair (
U, ϕ
) of the open set and the homemorphism
1
ϕ: U V
is called a
chart
, the collection of charts on
M
being called an
atlas
.
We can discuss the “smooth structure” of
M
by giving it a maximal smooth
atlas (the transition maps between any two charts are smooth in the usual
sense of being infinitely differentiable, and this atlas is maximal with respect to
inclusion).
1
A homemorphism is a continuous bijection with a continuous inverse.
1
A map
F : M N
between manifolds is called smooth at
p M
if and only
if for any two charts (
U, ϕ
) about
p
(on
M
) and (
V, ψ
) about
F
(
p
) (on
N
) such
that F (U) V , we have that
ψ F ϕ
1
is a smooth map in the usual sense.
Recall also that if (
U, ϕ
) is a chart on
M
and
f C
(
M
) (the algebra of
smooth functions
M R
), and
r
1
, . . . , r
n
are the standard coordinates on
R
n
,
then
x
i
:=
r
i
ϕ
characterize the coordinates on the chart (
U, ϕ
). So if
p U
,
then we define
f
x
i
(p) :=
(f ϕ
1
)
r
i
(ϕ(p)),
the partial derivative of
f
at
p
with respect to
x
i
. Because of the compatibility
of charts, this will not depend on the choice of (U, ϕ).
The idea now is that we want to study manifolds by linearising them. This is
not a strange thing to anyone who has ever taken calculus as this is exactly what
is done in a first course in calculus, except only in the special case of
R
n
. That
is, we study
R
n
by looking at the linearisations of functions (i.e. derivatives) on
R
n
. In order to do this in the general context of manifolds, we will define and
study their tangent bundles.
The main purpose of this overview is to learn about the tangent bundle,
it’s dual notion the cotangent bundle, differential forms, their integration, and
cohomology. We end with calculating the de Rham cohomology which makes
use of the theory we develop.
1 The tangent bundle
1.1 Motivation
Consider a smooth curve γ in R
n
. We will write γ as a function
γ : (ε, ε) R
n
: t 7→ (γ
1
(t), γ
2
(t), . . . γ
n
(t)),
where we call the functions
γ
i
:
(
ε, ε
)
R
the component functions of
γ
. The
velocity vector of γ at time t = 0 is given by:
γ
0
(0) := (γ
0
1
(0), γ
0
2
(0), . . . , γ
0
n
(0)) R
n
, γ
0
i
(0) :=
i
dt
(0).
We can imagine the vector
γ
0
(0) to be an arrow tangent to the curve
γ
, starting
at γ(0) R
n
.
2
γ
γ
0
(0)
γ(0)
With this picture in mind, we write γ
0
(0) as a pair,
(γ(0), γ
0
(0)) R
n
× R
n
,
which is an example of what we will call a
tangent vector at p
=
γ
(0). The
set of all possible tangent vectors at a
p R
n
is called the
tangent space
of
R
n
at
p
and is denoted by
T
p
R
n
. It represents all the possible velocity vectors
of smooth curves which pass through
p
. If we “bundle” these tangent spaces for
each p R
n
together in a disjoint union, we get the tangent bundle of R
n
:
T R
n
=
G
pR
n
T
p
R
n
.
In the particular case for
R
n
, the tangent bundle is trivial,
T R
n
=
R
n
× R
n
, and
so it may not seem very interesting as of yet (but it will later!).
Observe that there are many distinct curves through a point
p
that yield
the same velocity vector. We have a natural representative (natural in the
sense that it is the simplest) smooth curve
γ
which passes through some point
p
= (
p
1
, p
2
, . . . , p
n
)
R
n
with velocity vector
v
p
= (
v
1
, v
2
, . . . , v
n
)
R
n
(we use
the subscript p to keep track of where our vector is based at):
γ(t) := (p
1
+ v
1
t, p
2
+ v
2
t, . . . , p
n
+ v
n
t),
where clearly γ
0
(0) = v
p
.
Associated to each such representative
γ
(equivalently, to each
v
p
) is the
directional derivative
v
p
: C
(R
n
) R at p, defined by:
v
p
(f) :=
d(f γ)
dt
(0) R, f C
(R
n
).
If we apply the chain rule to this, we get:
v
p
(f) =
n
X
i=1
f
x
i
(p)
i
dt
(0)
=
n
X
i=1
v
i
f
x
i
(p),
where
x
1
, x
2
, . . . , x
n
are the standard coordinate functions
2
in
R
n
. This coincides
with the sometimes-used definition of:
v
p
(f) = v
p
· (f)(p),
2
The superscript indices is a, albeit painful, convention.
3
where
(
f
)(
p
) is the gradient of
f
evaluated at
p
. Note then that we can just
consider
v
p
as an operator:
v
p
=
n
X
i=1
v
i
x
i
(p).
This establishes a correspondence
v
p
v
p
which allows us to identify the
concrete tangent vectors with a more abstract operator. This correspondence
does not depend on the actual particular representative
γ
. Some particular
properties of
v
p
include:
(i) If α, β R and f, g C
(R
n
) then:
v
p
(αf + βg) = α
v
p
(f) + β
v
p
(g).
(ii) If f, g C
(R
n
) then:
v
p
(f · g) =
v
p
(f) · g(p) + f (p) ·
v
p
(g).
In summary,
v
p
is an
R
-linear map on the vector space
C
(
R
) which satisfies
the Leibniz (product) rule.
1.2 The abstract tangent space
Now we generalize the correspondence
v
p
v
p
to arbitrary manifolds using
these properties of
v
p
. First we will define the tangent space of a manifold at
a point.
Let
M
be a smooth manifold and
p M
. We define a
tangent vector
to
M at p to be a map v
p
: C
(M) R such that:
(i) If α, β R and f, g C
(M) then:
v
p
(αf + βg) = αv
p
(f) + βv
p
(g).
(ii) If f, g C
(M) then:
v
p
(f · g) = v
p
(f) · g(p) + f (p) · v
p
(g).
The set of all tangent vectors is called the
tangent space
of
M
at
p
, and is
denoted by
T
p
M
. It not difficult to show that
T
p
M
is an
R
-vector space
3
and so
T
p
M
is a linearised version of
M
at the point
p
. As with the case of
R
n
, taking
the disjoint union of all the tangent spaces gives us what is called the
tangent
bundle of M:
T M :=
G
pM
T
p
M.
3
Indeed there are many equivalent ways to define
T
p
M
, but this way is the easiest to see
the vector space structure.
4
Proposition 1.1.
Let
M
be an smooth
n
-manifold and
p M
. If (
U, ϕ
) =
(U, x
1
, . . . , x
n
) is a chart about p, then
x
i
(p)
1 5 i 5 n
is a basis for T
p
M.
Proof. First a little lemma:
Let
V R
n
be a convex neighbourhood of 0 and
f : U R
a
smooth function with f (0) = 0. Note that:
f(x) =
Z
1
0
d
dt
f(tx) dt,
and so by the chain rule,
f(x) =
Z
1
0
X
i
f (tx)
x
i
x
i
dt =
X
i
Z
1
0
f (tx)
x
i
dt
x
i
,
so
f
(
x
) =
P
a
i
(
x
)
x
i
for some smooth
a
i
. Moreover, note that
after differentiating this and evaluating at 0,
a
i
(0) = (
f /∂x
i
)(0).
Now suppose that (
U, ϕ
) = (
U, x
1
, . . . , x
n
)
4
is a chart around
p M
such that
ϕ
(
p
) = 0. Pick an
f C
(
M
) defined on
U
such that
f
(
p
) = 0 (if it doesn’t,
set
f
0
=
f f
(
p
) and continue). By the above lemma, there is a neighbourhood
V ϕ
(
U
) such that the function
g
:=
f ϕ
1
has the form
g
=
P
g
i
r
i
for
smooth g
i
such that g
i
(0) = (g/∂r
i
)(0). For any q U
f(q) = g(ϕ(q)) =
X
g
i
(ϕ(q))x
i
(q),
and so if v
p
T
p
M, we have:
v
p
(f) = v
p
X
(g
i
ϕ)x
i
=
X
g
i
(ϕ(p)) · v
p
(x
i
)
=
X
v
p
(x
i
) ·
f
x
i
(0),
as
x
i
(
p
) = 0 and
g
i
(
h
(
p
)) =
g
i
(0) = (
(
f ϕ
1
)
/∂r
i
)(0). Hence the set of
partials is a spanning set. Note that (
(
x
j
)
/∂x
i
)(0) =
δ
ij
, and hence the set is
linearly independent.
Hence we have that any v
p
T
p
M looks like v
p
=
P
i
v
i
x
i
(p) for v
i
R.
4
We reserve the coordinates r
i
for those on R
n
5
Example 1.2
(Tangent space of
S
1
)
.
Consider
M
=
S
1
, the unit circle. Ge-
ometrically, it’s clear that any velocity vector of a smooth curve at
p S
1
is
a tangent vector to
S
1
at
p
. The tangent space
T
p
S
1
at
p
is then all possible
magnitudes of this tangent vector—that is, T
p
S
1
=
R.
With respect to the abstract formulation of the tangent space, if (
U, ϕ
) =
(U, θ) is a chart on S
1
around p, then
T
p
S
1
= {λ(/∂θ)(p) | λ R}
=
R,
just as our geometric intuition suggests!
If we imagine
T
p
S
1
as the real line
R
lying tangent to
S
1
at
p
, then the
tangent bundle,
T S
1
can then be imagined as a cylinder (
S
1
× R
) by rotating
these lines around.
1.3 Reconciling the motivation with the abstract
Recall that to motivate tangent vectors, we considered smooth curves in
R
n
and then showed a correspondence between velocity vectors
γ
0
(0) =
v
p
of these
curves and directional derivatives
v
p
. This correspondence doesn’t fall apart
in the case of manifolds, but the curve notion just ends up being unwieldy for
the algebraic approach to manifolds that we will be taking.
Consider a smooth curve
γ :
(
ε, ε
)
M
on
M
such that
γ
(0) =
p
. We
identify the notations of γ
0
(0) = v
p
and
v
p
and define for f C
(M):
γ
0
(0)f :=
d(f γ)
dt
(0).
It’s not difficult to show that
γ
0
(0)
T
p
M
. If we consider a chart (
U, ϕ
) =
(
U, x
1
, . . . , x
n
) around
p M
, we can show that the local-coordinates represen-
tation of γ
0
(0) is
γ
0
(0) =
n
X
i=1
γ
0
i
(0)
x
i
(0),
where γ
i
:= x
i
γ, and so in terms of a matrix:
γ
0
(0) =
γ
0
1
(0)
γ
0
2
(0)
.
.
.
γ
0
n
(0)
.
6
Note this is just the usual (as in Section 1.1) “velocity vector”
α
0
(0) of the curve
α := ϕ γ, so it’s rather fitting that we use the same notation.
As it turns out, for any
p M
and
v
p
T
p
M
, we can find a curve
γ
such that
γ
(0) =
p
and
γ
0
(0) =
v
p
. It ends up being an equivalent method to defining the
tangent space, but despite its computational perks it lacks the algebraic perks
we want to come easy (such the vector space structure on
T
p
M
). Nonetheless,
we will sometimes now abuse what we have found by sometimes viewing tangent
vectors as both velocity vectors and operators (since they are for all intents and
purposes one the same).
1.4 The pushforward
Consider a smooth map
F : M N
between manifolds. A smooth curve
γ
on
M
with
γ
(0) =
p
induces another smooth curve
F γ
on
N
, and furthermore
a tangent vector
v
p
:=
γ
0
(0)
T
p
M
induces an analogous tangent vector
w
F (p)
:= (F γ)
0
(0) T
F (p)
N.
γ
γ
0
(0)
p
M
F γ
(F γ)
0
(0)
F (p)
N
F
We can treat this correspondence
v
p
7→ w
F (p)
more formally as a function
F
,p
: T
p
M T
F (p)
N where we see for g C
(N):
F
,p
(v
p
)g = F
,p
(γ
0
(0))g
:= w
F (p)
(g)
= (F γ)
0
(0)(g)
=
d(g F γ)
dt
(0)
= γ
0
(0)(g F )
= v
p
(g F ),
which gives us the more general definition:
F
,p
(v
p
)g := v
p
(g F ),
called the
pushforward
of
F
at
p
. When the context is clear, we might choose
to write F
to ease the notation. One can check that:
7
(i) F
is a linear map.
(ii) For smooth maps M
F
N
G
P between manifolds and p M :
(G F )
,p
= G
,F (p)
F
,p
.
(iii) If id : M M is the identity, then id
: T
p
M T
p
M is also the identity.
(iv) If F is a diffeomorphism, then F
is a linear isomorphism.
So in this process, a smooth
F : M N
between curvy manifolds is turned into
a linear
F
: T
p
M T
F (p)
N
between linear approximations of manifolds at a
point. The result is a “linearisation” of F .
Picking suitable charts (
U, x
1
, . . . , x
m
) around
p
and (
V, y
1
, . . . , y
n
) around
F
(
p
) such that
F
(
U
)
V
, we can express the image of the basis of
T
p
M
under
F
in terms the basis of
T
F (p)
N
. In particular, we know that for suitable
a
j
R
,
F
x
i
(p)
=
n
X
j=1
a
j
y
j
(F (p)),
and application of both sides to the coordinate function y
k
yields:
(y
k
F )
x
i
(p) = a
k
.
Hence, setting
F
k
:=
y
k
F
we get that the matrix representation of
F
in these
bases is
[F
] =
F
i
x
j
,
the Jacobian matrix of
F
! So
F
is indeed the manifold-equivalent
5
to the
derivative!
1.5 The tangent bundle and vector fields
Let M be a manifold. Define the tangent bundle of M to be:
T M :=
G
pM
T
p
M,
which is associated with the projection
π : T M M : v
p
7→ p
. By gluing the
pushforwards
F
,p
for each
p M
, we get a global pushforward,
F
: T M T N
.
The following theorem we leave unproved.
Theorem 1.3.
If
M
is a smooth
n
-manifold, then
T M
can be given the structure
of a smooth 2n-manifold.
5
In general, this is how we generalize things from R
n
to manifolds.
8
It turns out that the smooth structure that
T M
inherits is directly from
M
itself so that the projection
π
is a smooth map, and so that for any smooth
F : M N , the following diagram commutes:
T M T N
M N
F
π π
F
Moreover, this
F
is a smooth map with respect to the smooth structures of
T M
and T N.
The thing that is not obvious from our definition is that with respect to the
smooth structure, the tangent bundle
T M
of an
n
-manifold
M
is not necessarily
diffeomorphic to
M × R
n
. It is true that the only two tangent bundles we have
defined—for
R
n
and
S
1
—have been diffeomorphic to this “trivial” bundle, but
in general it does not hold. If it did, one might guess that the topic of tangent
bundles would be really bland and boring!
The idea behind the tangent bundle is that of the more general “vector
bundles”. A vector bundle of rank n is a triple (E, B, π) where:
(i) E and B are topological spaces.
(ii) π : E B is a continuous surjective map.
(iii)
For each
x B
,
π
1
(
x
) has the structure of an
n
-dimensional vector space.
(iv)
For each
x B
there exists an open neighbourhood
U
of
x
such that
π
1
(U) is homeomorphic to U × R
n
.
The final condition states that every point
x B
has a neighbourhood which is
“trivial” with respect to the vector bundle—that when you look at
π
1
(
U
) for a
certain
U
, it looks like a “bundle” of copies of
R
n
, one for each
y U
. If a global
trivializing neighbourhood U = B exists, then the bundle is called trivial.
In our case, we are considering the smooth vector bundles (
T M, M, π : T M
M
) where we additionally require smoothness instead of just continuity. An
example of where
T M 6
=
M × R
n
is the case for
M
=
S
2
(i.e. non-trivial). This
is not obvious, but relates to the (non)existence of a non-vanishing vector field
and the “hairy ball” theorem
6
.
Intuitively, a vector field is a map which takes a point in
M
and spits out a
tangent vector. Hence, a “field of vectors” on
M
. Formally, a
vector field
is
defined to be a smooth map
v : M T M
such that
π v
=
id
M
(that is, it is a
smooth “section” of π).
Example 1.4. Consider M = R
2
, defining v : R
2
T R
2
by
v(x, y) := v
(x,y)
= (y, x),
6
One cannot comb a hairy ball without making a cowlick!
9
keeping in mind that we always implicitly have
v
p
as a pair, (
p, v
p
). Here,
v
assigns to the point
p
= (
x, y
) a vector
v
p
= (
y, x
) (which we draw emanating
from the point (x, y) on the plane). Pictured below is a portion of R
2
with the
vector field v.
1 0.5 0 0.5 1
1
0.5
0
0.5
1
On
R
2
\ {
(0
,
0)
}
the vector
v
(x,y)
= (
y, x
) is non-zero, but at (0
,
0) however
v
is (0
,
0) and we say
v
vanishes at
p
= (0
,
0). Abstractly, without appealing
to the diffeomorphism
T R
2
=
R
2
× R
2
, this vector field can be written as
v = y
x
+ x
y
.
This last way of writing
v
hints at an alternative definition for a vector field.
One can define a vector field to be an
R
-linear map
v : C
(
M
)
C
(
M
) which
satisfies the Leibniz rule,
v(f · g) = v(f) · g + f · v(g).
In this way for a
p M
,
v
p
:=
v
(
p
) is “evaluated” in the codomain at
p
so that
it is a map v
p
: C
(M) R, a tangent vector!
2 The cotangent bundle
2.1 Linear algebra review
The usual concept of linear maps can be generalized to the concept of bilinear
maps. If
V, W,
and
X
are vector spaces, a map
T : V × W X
is
bilinear
if it
is linear with respect to each component. That is, if for each
v V
and
w W
,
we have that T (v, ·) and T (·, w) are linear in the usual sense
7
.
The tensor product of two vector spaces
V
,
W
is a pair (
V W,
) such
that:
7
This can again be extended to a concept of multilinearity
10
(a) V W is a vector space.
(b) : V × W V W
is a bilinear map such that for any bilinear map
f : V × W X
(for some vector space
X
), there exists a unique linear map
˜
f : V W X such that f =
˜
f.
That is, the tensor product
V W
“factors” any bilinear map through a
unique linear map, as shown in the following commutative diagram:
V × W X
V W
f
˜
f
The idea is that the tensor product gives us a “free” vector space where bilinear
maps V × W correspond to linear maps.
Theorem 2.1. The tensor product of V and W exists and is unique.
Most notably, if
{v
1
, . . . , v
n
}
is a basis for
V
and
{w
1
, . . . , w
m
}
is a basis for
W , then
{v
i
v
j
| 1 5 i 5 n, 1 5 j 5 m}
is a basis for V × W . Hence dim V W = dim V · dim W .
Proposition 2.2. For finite-dimensional real vector spaces U, V, W :
(i) V W
=
W V .
(ii) U (V W )
=
(U V ) W .
If {V
α
}
αI
is a family of vector spaces, then the direct sum,
M
αI
V
α
,
is the vector space consisting of tuples (
a
α
)
αI
with
a
α
V
α
such that all but
finitely many of the
a
α
are zero. The vector space operations are the component-
wise operations. With this notation, the
tensor algebra T
(
V
) is defined to be
the algebra:
T (V ) :=
M
n=0
V
n
= V
0
V
1
V
2
. . . ,
where
V
0
:=
R
and
V
n
=
V . . . V
| {z }
n times
, having the obvious multiplication
(
v, w
)
7→ v w
. This is the free algebra over
V
. Elements of the tensor algebra
look like
k
X
i=0
a
i
v
i,1
v
i,2
. . . v
i,n
i
,
for various v
i,j
V and a
i
R.
11
What we would now like to do is get the “alternating algebra” where we
mimic the property of the cross product that
v × w
=
w × v
. To this end, we
take the quotient
8
of
T
(
V
) by the two-sided ideal
9
I
generated by the relations
v w + w v for all v, w V :
I := span {u
1
. . . u
`
(w
1
w
2
+ w
2
w
1
) v
1
. . . v
k
| u
i
, v
j
, w
1
, w
2
V } .
That is, we end up “identifying”
v w
+
w v
with 0 which is the same as
saying that
v w
=
w v
. The resulting graded algebra Λ
V
:=
T
(
V
)
/I
is the
called the exterior algebra over V with the natural “exterior” product,
v w := v w + I.
The n
th
exterior power of V is the n
th
grade subspace of the exterior algebra,
Λ
n
(V ) := span{v
1
. . . v
n
| v
i
V }.
Because the multiplication
on the exterior algebra makes it into an algebra,
we will take for granted that it is a bilinear map.
2.2 The cotangent bundle and 1-forms
Recall that if
V
is a vector space over
R
, then the
dual space
of
V
is the vector
space
V
:= {f : V R | f is linear}.
Given that
V
is finite dimensional with a basis
{v
1
, v
2
, . . . , v
n
}
, there exists a
corresponding dual basis {f
1
, f
2
, . . . , f
n
} of V
such that
f
i
(v
j
) = δ
ij
.
Because T
p
M is a vector space, we can take its dual,
T
p
M := {α
p
: T
p
M R | α
p
is linear},
which we call its
cotangent space
at
p M
. This is also a vector space and
we can form the cotangent bundle of M :
T
M :=
G
pM
T
p
M,
which has its associated bundle projection π : T
M M : α
p
7→ p.
8
Recall that if
W
is a linear subspace of a vector space
V
, then the
quotient space V /W
is the vector space
V/W := {v + W | v V },
where
v
+
W
=
{v
+
w | w W }
is the coset of
v
. This quotient space is a vector space with
the operations α(v + W ) := αv + W and (u + W ) + (v + W ) := (u + v) + W .
9
A
two-sided ideal
of the tensor algebra
T
(
V
) is a subspace
U T
(
V
) such that for each
x T (V ) and u U, we have that u x, x u U.
12
With the tangent bundle, we found that the sections of the bundle projection
π
had the nice description of being vector fields as we know them. In the case of
the cotangent bundle, the geometric interpretation of sections of the projection
is not so obvious but the sections are certainly no less important!
A smooth section
α: M T
M
of the cotangent bundle is called a differential
1-form
on
M
. It is our first example of a differential form. We can easily create
examples of 1-forms from functions
f C
(
M
) by using the
differential
”,
d
,
defined by:
df(p)v
p
:= (df)
p
v
p
:= v
p
(f) R, v
p
T
p
M.
It turns out that the differential is just the pushforward of
f
in disguise. This is
more apparent if we write:
f
,p
(v
p
) = (df)
p
(v
p
)
d
dt
(f(p)).
This gives an easy identification of the differential and pushforward of f.
Fixing a point
p M
and considering a chart (
U, ϕ
) = (
U, x
1
, . . . , x
n
) around
p, we might wonder a basis for T
p
M. By direct calculation, one sees that:
(dx
i
)
p
x
j
(p)
= δ
ij
,
and so the set
{(dx
i
)
p
| 1 5 i 5 n}
forms the dual basis to the standard basis on
T
p
M
. As a result, a 1-form
α
can
be written locally in a chart (U, x
1
, . . . , x
n
) as
α =
n
X
i=1
a
i
dx
i
, a
i
C
(M).
Example 2.3.
Consider
M
=
R
3
and the 1-form
α
=
dz y dx
. We cannot
readily give a geometric meaning to the dual space, however we can identify
α
with its kernel (a sub-bundle of
T R
3
) which we can certainly draw a picture for!
Recall that the kernel of α is
ker α := {v
p
T R
3
| α
p
(v
p
) = 0}.
In particular,
ker α
p
is a 2-dimensional subspace of
T
p
R
3
and hence a plane.
So
ker α
p
is a plane and hence
ker α
is a
plane field
. To visualize this plane
field, we consider an arbitrary vector
v
p
=
a
x
(
p
) +
b
y
(
p
) +
c
z
(
p
) where
p = (p
1
, p
2
, p
3
) R
3
:
α
p
(v
p
) = 0 (dz(p) p
2
dx(p))(v
p
) = 0
dz(p)(v
p
) p
2
dx(p)(v
p
) = 0
c p
2
a = 0,
13
and so it’s not hard to see that
ker α
p
= span
y
(p),
x
(p) + p
2
z
(p)
.
Along the
x, z
-plane where
p
2
= 0,
ker α
p
can be identified as the
x, y
-plane. If
we look at the lines in the
x, y
-plane
p
3
= 0 which are parallel to the
y
-axis,
ker α
p
rotates (if you are viewing from the origin) counter-clockwise as
p
2
to get close and closer to the y, z-plane. Below is a picture of the plane field
10
.
So what is the proper geometric interpretation of 1-forms? There is no
definitive way, but the popular one (especially with physicists) is as follows. To
each point
p M
, a 1-form
α
gives a linear form
α
p
:=
α
(
p
)
: T
p
M R
. For
any
x R
we have that
α
1
p
(
x
) (the level-set of
α
p
at
x
, if
x
= 0 then this is
ker α
p
) is an affine subspace of
T
p
M
. Moreover, all level-sets can be imagined to
be “parallel” in a way.
The idea is that when we zoom in at
p M
on to
α
p
, what we see is a little
collection of these parallel hyperplanes, stacked like pancakes. Compare this to
when you zoom in on a vector field v at p, and we get an arrow v
p
.
Now handed
v
p
, how does one visualize
α
p
(
v
p
)? Imagine that these level-sets
have their values associated to them and that the vector
v
p
is based at the
0-hyperplane. The vector
v
p
then pierces the stack of pancakes, and we can
“measure”
v
p
using the number of pancakes pierced as a ruler (pictured below
left, α
p
(u
p
) = 2, α
p
(v
p
) = 0, α
p
(w
p
) = 3).
w
p
u
p
v
p
x
y
x
y
10
For those who care: we just realized the standard “contact manifold” structure of R
3
!
14
In
R
2
with the usual coordinates
x
and
y
we get that the differential form
dx
near some point
p M
is imagined as a collection of vertical lines. This way,
y
(
p
) has the obvious property that (
dx
)
p
y
(
p
) = 0 as the vector is parallel to
the vertical lines and hence doesn’t pierce any (pictured above right). That is, a
1-form can be seen as a coordinate-free way of measuring the intuitive lengths of
tangent vectors of a vector field.
2.3 k-forms
We now consider the
k
th
exterior power of
T
M
, Λ
k
T
M
which has a natural
vector bundle structure. A
k-form
is a smooth section of Λ
k
T
M
. There
is an isomorphism Λ
k
T
M
k
T M
)
and so each
k
-form
ω
evaluates at
some
p M
to
ω
p
:=
ω
(
p
)
:
Λ
k
T
p
M R
which can be seen as an alternating
multilinear map. The set of all smooth
k
-forms on
M
we will denote by
k
(
M
).
The set
0
(M) of 0-forms is identified with C
(M).
Given a
k
-form
α
and an
-form
β
, we can take the exterior product
α β
k+`
of the forms (within the exterior algebra) where we characterize the the
anticommutativity (alternating) property of the exterior algebra by:
α β = (1)
k`
β α,
If k = = 1, at a point p M, the exterior product becomes
(α β)
p
(v
p
, w
p
) :=
1
2
(α
p
(v
p
)β
p
(w
p
) α
p
(w
p
)β
p
(v
p
)) , v
p
, w
p
T
p
M,
when viewed as an alternating map (α β)
p
: T
p
M × T
p
M R.
We can extend our geometric interpretation of 1-forms for
k
= 2
,
3. Imagine
that
α, β, ω
are 1-forms. We know how to imagine them locally at a point
p
as a collection of level sets. When taking
α β
, we imagine this as taking the
intersection of our level sets. That is, if we had a stack of planes to begin with,
now we intersect two stack of planes to get a bundle of lines around
p
. Our
vectors that (
α β
)
p
take as arguments look like
v
p
w
p
which we can look at as
infinitesimal parallelograms (see: cross product as the area of a parallelogram),
so evaluated, we count the number of lines in our bundle that this parallelogram
passes through to calculate something like an area.
Extending this once more, we get that (
α β ω
)
p
appears as an intersection
of 3 stacks of planes, which makes a lattice of points. So when we evaluate this
3-form at a
u
p
v
p
w
p
(something akin to a tiny parallelepiped), we measure
the density or volume of a parallelepiped.
15
2.4 In R
n
Now we turn our attention to
k
(R
n
). In particular, let’s let n = 3:
0
(R
3
) = C
(R
3
),
1
(R
3
) = {a dx + b dy + c dz | a, b, c C
(R
3
)},
2
(R
3
) = {a dx dy + b dx dz + c dy dz | a, b, c C
(R
3
)},
3
(R
3
) = {a dx dy dz | a C
(R
3
)}.
Recall that we have the differential, a map
d:
0
(
R
3
)
1
(
R
3
). In
R
3
, this
looks like the gradient:
df =
f
x
dx +
f
y
dy +
f
z
dz (f) =
f
x
,
f
y
,
f
z
.
We generalize this (for arbitrary n) to a map d:
k
(R
n
)
k+1
(R
n
):
d(f
i
1
...i
k
dx
i
1
. . . dx
i
k
) := df
i
1
...i
k
dx
i
1
. . . dx
i
k
,
called the exterior derivative.
Proposition 2.4. (i) d is R-linear.
(ii) If ω
k
(R
n
), µ
`
(R
n
), then d(ω µ) = µ + (1)
k
ω .
(iii) d() = d
2
ω = 0.
Consider an ω
1
(R
3
), ω = a dx + b dy + c dz. Then
=
b
x
a
y
dx dy +
a
z
c
x
dz dx +
c
y
b
z
dy dz,
which one recognizes immediately (under a certain identification of bivectors
x
i
x
j
and vectors) is the curl of ω.
Similarly, if µ
2
(R
3
), µ = c dx dy + a dy dz + b dz dx:
=
c
z
+
a
x
+
b
y
dx dy dz,
which is the divergence of µ!
Finally, from the properties of the exterior derivative,
d
, we have that
d
2
= 0
and so the usual theorem about grad/curl/divergence follow that taking any two
consecutive operators results in 0 for the following chain:
0
(R
3
)
grad
1
(R
3
)
curl
2
(R
3
)
div
3
(R
3
).
The definition for
d
induces (via charts) an exterior derivative on a general
smooth manifold
M
(with all the same properties), though we won’t go into the
details about this. The properties listed in Proposition 2.4 apply to the general
definition for the exterior derivative.
16
2.5 Pullbacks, orientations, and the integration of forms
We have seen that when given a smooth
F : M N
between manifolds, the
tangent space induces a linear map
F
,p
: T
p
M T
F (p)
N
called the pushforward
which “pushes” tangent vectors in
T
p
M
forward to tangent vectors in
T
F (p)
N
.
Because of duality, this gets mirrored in the dual space of
T
p
M
, the cotangent
space T
p
M.
In particular, given a smooth map
F : M N
between manifolds, we have
an induced map F
: T
F (p)
N T
p
M called the pullback of F , defined by:
F
(α
p
) := α
p
F
,p
, α
p
T
F (p)
N.
This is just the transpose of
F
,p
and clearly is linear. Some call this the
“co-differential” as it is dual to the “differential”.
Unique to the cotangent space is the ability to “pullback” 1-forms—it is
not true in general
11
that you can pushforward vector fields. The pullback of a
1-form α on M by F is given by:
(F
α)
p
(v
p
) := α
F (p)
(F
,p
(v
p
)), v
p
T
p
M.
This notion of pullback generalizes to
k
-forms, where we see for
ω
k
(
N
) that:
(F
ω)
p
(v
1
, . . . , v
k
) := ω
F (p)
(F
,p
(v
1
), . . . , F
,p
(v
k
)), v
1
, . . . , v
k
T
p
M.
Moreover, this generalized pullback has the following properties for differential
forms ω and η:
(i) F
is R-linear.
(ii) F
(ω η) = (F
ω) (F
η).
(iii) F
() = dF
(ω).
Before we discuss integration, we have a few restrictions. Firstly, we will only
be able to integrate over “oriented” manifolds. That is, things like integration
on a obius band (pictured below
12
) are out. Secondly, if M is an n-manifold,
we will only be able to integrate
n
-forms. Moreover, these
n
-forms will need to
have what is called compact support: zero outside of a compact set
13
.
11
Though, if F is a diffeomorphism, you can pushforward vector fields!
12
pgfplot code courtesy of https://tex.stackexchange.com/questions/118563/.
13
A subset of a topological space is said to be compact if any covering of the subset by open
subsets can be refined to a finite covering. This condition is important in order for the integral
to be bounded.
17
To formalize orientability of a manifold, first consider a finite dimensional vector
space
V
. If we consider all the ordered bases of
V
, then linear maps which take
the
i
th
basis element of one basis to the
i
th
of another can be characterized by
their determinant—it will either be positive or negative. Fixing a standard basis,
we can classify an order basis as either positively oriented (if the corresponding
determinant is positive) or negatively oriented (if the corresponding determinant
is negative). In this way we define an orientation of
V
as an additional structure.
Example 2.5.
Consider the vector space
R
3
with the standard ordered basis
(e
1
, e
2
, e
3
). Considering the alternative ordered basis
B = (e
1
+ 2e
2
, e
3
, e
1
e
2
),
the map T : R
3
R
3
which is defined by:
T (e
1
) = e
1
+ 2e
2
, T (e
2
) = e
3
, T (e
3
) = e
1
e
2
,
has the corresponding matrix
T =
1 0 1
2 0 1
0 1 0
,
which has determinant
det T
= 3
>
0 and so
B
is positively oriented with respect
to the standard basis.
Returning to manifolds, if
M
is a manifold, then an orientation on
M
consists of a choice of orientations on the tangent spaces
T
p
M
for each
p
which
are positively oriented relative to those of nearby, overlapping charts. This is
captured by the Jacobian matrix of the transition maps.
Formally, an
n
-manifold
M
is
orientable
if it admits an
oriented atlas
:
an atlas such that any transition map of overlapping charts has an everywhere
positive Jacobian determinant on the intersection.
18
If we consider an
n
-form
ω
, then
ω
p
: T
p
M R
has the property that for any
ordered bases (
v
1
, . . . , v
n
) and (
w
1
, . . . , w
n
) of
T
p
M
, the map
T : T
p
M T
p
M
defined by T (v
i
) = w
i
satisfies:
ω
p
(w
1
, . . . , w
n
) = det(T ) · ω
p
(v
1
, . . . , v
n
).
This determines an orientation for these tangent spaces. Given that
ω
is smooth
(or simply continuous), the existence of such an
n
-form that never vanishes is
equivalent to having an oriented atlas:
Proposition 2.6.
An
n
-manifold
M
is orientable if and only if it admits a
never-vanishing (smooth) n-form, called a volume form.
Now we can discuss integration on manifolds. Let
M
be an orientable
n
-
manifold, and let
ω
n
(
M
) have compact support in a chart (
U, ϕ
) on
M
.
Then the integral of ω on U is
Z
U
ω :=
Z
ϕ(U)
(ϕ
1
)
ω,
which is an integral of an
n
-form on
ϕ
(
U
) and has the form of
R
ϕ(U)
f
(
x
)
dx
1
. . .dx
n
. It is integrated as
R
ϕ(U)
f
(
x
)
dx
1
. . . dx
n
, your typical Riemann integral.
Example 2.7
(Tu, 22.7)
.
Consider spherical coordinates. Let
r
denote the
distance of a point in
R
3
from the origin,
φ
be the polar angle from the positive
z
-axis, and
θ
be the azimuthal angle from the positive
x
-axis. Note that the
sphere (embedded in R
3
) has the volume form
ω = x dy dz y dx dz + z dx dy,
and hence is orientable. Consider
U = {(sin φ cos θ, sin φ sin θ, cos φ) S
2
| 0 < φ < π, 0 < θ < 2π},
which constitutes a chart (U, φ, θ) on S
2
. We wish to calculate
R
U
sin φ .
To this end, let f = (φ, θ) be the coordinate map on U and define:
u := (f
1
)
φ = φ f
1
, v := (f
1
)
θ = θ f
1
.
These correspond to the coordinate functions on
f(U) = {(u, v) R
2
| 0 < u < π, 0 < v < 2π}.
19
Now we just compute:
Z
U
sin φ =
Z
f(U)
(f
1
)
(sin φ )
=
Z
f(U)
sin(φ f
1
) d(f
1
)
φ d(f
1
)
θ
=
Z
f(U)
sin u du dv
=
Z
f(U)
sin u du dv
=
Z
2π
0
Z
π
0
sin u du dv
= 2π
Z
π
0
cos u du
= 4π.
So far all our manifolds have been “open” in the sense that they don’t
typically have an “edge”. The definition for a manifold can be extended to
include
manifolds with boundary
by requiring that any open subset of
M
be
homeomorphic to an open subset of H
n
,
H
n
:= {(x
1
, . . . , x
n
) R
n
| x
n
= 0},
the
Euclidean half-space
. In this way, some points
p M
are mapped by
charts to points in
H
n
with
x
n
= 0. We call these points
boundary points
,
and denote the set of them by
M
, the
boundary
of
M
. If
M
is an
n
-manifold,
then
M
constitutes an (
n
1)-manifold (by using the restriction of charts to
R
n1
H
n
). The theory of manifolds with boundary is developed analogously
to that of manifolds without boundary. We finish off this integration by stating
a general formulation of a famous theorem from multivariable calculus (and of
course, without proof!).
Theorem 2.8
(Stokes’)
.
Let
M
be an orientable
n
-manifold with boundary. If
ω is an (n 1)-form with compact support on M, then
Z
M
=
Z
M
ω.
2.6 Cohomology
In algebra, a cochain of vector spaces is a family {C
k
}
kZ
of vector spaces,
· · ·
d
1
C
0
d
0
C
1
d
1
C
2
d
2
· · ·
where {d
k
: C
k
C
k+1
}
kZ
is a matching family of linear maps satisfying
im d
k
ker d
k+1
.
20
We denote the cochain by
C
. If instead of
we have equality, then the cochain
is said to be
exact
. Because the
C
k
are vector spaces we can take the quotients,
H
k
(C
) := ker d
k+1
/ im d
k
,
which we call the cohomology groups (they are vector spaces which are “abelian
groups” under addition). If
C
is exact, then we notice that the
H
k
(
C
) are
trivial, and so we tend to view the homology groups as measuring to what degree
C
fails to be exact.
Now define the de Rham complex Ω(M) of M:
Ω(M) :=
dim M
M
i=0
k
(M),
which is a graded algebra. Since d
2
= 0, we have that
d(Ω
k1
(M)) ker d|
k
(M)
, ()
which gives us a cochain (implicitly, every space preceding
0
(M) is {0})
0
(M)
d
1
(M)
d
2
(M)
d
3
(M)
d
. . .
where taking any two successive arrows is 0 as a result of (
). Define
Z
k
(
M
) :=
ker d|
k
(M)
as the set of
closed k-forms
(that is, forms
ω
k
(
M
) such that
= 0), and
B
k
(
M
) :=
d
(Ω
k1
(
M
)) as the set of
exact k-forms
(that is,
forms
ω
k
(
M
) such that there exists a (
k
1)-form
µ
such that
=
ω
). In
other words, () says that B
k
(M) Z
k
(M) and so we can take the quotient
H
k
dR
(M) := Z
k
(M)/B
k
(M)
which measures to what degree closed forms fail to be exact forms. That is, we
say two
k
-forms are “cohomologous” (in the same equivalence class) if they differ
by an exact
k
-form. In particular, exact forms are cohomologous to 0. Because
the de Rham complex Ω(
M
) has a product structure given by the wedge product,
H
dR
(M) inherits this product structure, given by:
[ω] [η] = [ω η],
and so
H
dR
(
M
) :=
L
n
k=0
H
k
dR
(
M
) forms a graded-commutative algebra, called
the
de Rham cohomology of M
. The significance of the de Rham cohomology
is that it is an algebraic invariant:
Theorem 2.9.
Suppose that
M
and
N
are diffeomorphic smooth manifolds.
Then
H
dR
(
M
)
=
H
dR
(
N
) as vector spaces (and hence have the same dimension).
Differential topology is often summarized as the use of topology in order to
determine manifolds up to diffeomorphism, and so with respect to this, the de
Rham cohomology of
M
serves as an important tool to this end. The dimension
21
b
k
(
M
) :=
dim H
k
dR
(
M
) is called the
k
th
Betti number
of
M
, where we note
by a dimension argument on
k
(
M
) that for
k > dim M
,
b
k
(
M
) = 0. Using the
Betti numbers, we can calculate the
Euler characteristic
of
M
as the (finite)
alternating sum:
χ(M) := b
0
(M) b
1
(M) + b
2
(M) b
3
(M) + . . . ,
another (rather famous) algebraic topological invariant that is commonly used in
many areas of mathematics. In the context of polyhedra, the Euler characteristic
is
χ
=
V E
+
F
where
V, E,
and
F
are respectively the number of vertices,
edges, and faces of the polyhedron in question. It’s well known that any convex
polyhedron has an Euler characteristic of 2.
Example 2.10
(de Rham cohomology of
S
1
)
.
First we consider
H
0
dR
(
S
1
) and
then finally H
1
dR
(S
1
). We know that H
k
dR
(S
1
) = {0} for k > dim S
1
= 1.
(i) H
0
dR
(
S
1
): A 0-form
f
is closed if
df
= 0, so locally
f
is constant. Since
S
1
is connected, if it is locally constant, it is globally constant on
S
1
. Hence
H
1
dR
(S
1
)
=
R.
(ii) H
1
dR
(S
1
): First note that any 1-form on S
1
is closed (otherwise
2
(S
1
) is
non-zero) and so H
1
dR
(S
1
) = Z
1
(S
1
). Consider the maps:
R
e
S
1
i
R
2
,
where
e
(
t
) = (
cos t, sin t
) and
i
is the inclusion. Consider the non-vanishing
1-form
˜α
:=
y dx
+
x dy
on
R
2
which is closed (exercise) but not exact
14
.
Pull back this form along i to a 1-form α := i
˜α on S
1
.
One can calculate that
i
e
d
dt
= y
x
+ x
y
,
and so:
1 = ˜α
i
e
d
dt
= i
˜α
e
d
dt
= α
e
d
dt
= e
α
d
dt
,
where we conclude that e
α = dt.
The map
f 7→ e
f
:=
f e
is an isomorphism of
C
(
S
1
) to the set
of smooth, 2
π
-periodic functions on
R
. The map
e
:
1
(
S
1
)
1
(
R
)
has for each
β
1
(
S
1
) that
e
β
=
g dt
for some
g C
(
R
) which
is 2
π
-periodic. So there exists a
f C
(
S
1
) with
g
=
e
f
so that
e
β = g dt = e
f e
α = e
(fα). So e
gives us a correspondence β fα.
14
If there were such an
f C
(
R
2
) such that
df
=
˜α
, then
f
would attain a maxi-
mum/minimum on a compact subset such as
S
1
. In this case,
df
would vanish, however it is
non-vanishing on S
1
.
22
Define a new map:
φ:
1
(S
1
) R : β 7→
Z
S
1
β.
This map φ is surjective onto R because
φ(α) =
Z
S
1
α =
Z
[0,2π]
e
α =
Z
2π
0
dt = 2π 6= 0.
Hence by linearity, it must be surjective onto
R
. Stokes’ theorem says
that any exact form is in the kernel of
φ
. On the other hand, suppose
that
fα ker φ
. Then
R
S
1
fα
= 0 and so
˜g
(
t
) :=
R
t
0
e
f
is a 2
π
-periodic
function and hence is identified with some
g C
(
S
1
) so that
˜g
=
e
g
.
We note that
e
(dg) = d(e
g) = d˜g = e
f dt = e
fe
α = e
(fα),
and hence (since e
is an isomorphism), dg = fα is exact.
Hence by the first isomorphism theorem
15
:
1
(S
1
)/ ker φ = H
1
dR
(S
1
)
=
R.
In conclusion, we have found that:
H
k
dR
(S
1
) =
(
R, k = 0, 1;
0, k = 2,
and the Euler characteristic of S
1
is χ(S
1
) = 1 1 = 0.
Intuitively, the Betti numbers give the number of
k
-dimensional “holes” in
the manifold. So the common explanation is that
b
0
(
M
) counts the connected
components,
b
1
(
M
) counts the number of “circular” holes, and
b
2
(
M
) counts
the number of hollow cavities. In our example of
S
1
, we have one connected
component and one circular hole. As one can see, actually computing the de
Rham cohomology is non-trivial, and because of this, it is rare that the entirety
of the cohomology of a manifold
M
is calculated. It is only in special cases is it
easily done.
A Further Reading
For more on differential geometry, see:
Introduction to manifolds, Loring W. Tu.
15
If
T : V W
is a linear map of vector spaces, then the first isomorphism theorem says
that V / ker T
=
im T .
23
Differential Manifolds, Antoni A. Kosinski.
An Introduction to Differential Manifolds, Jacques Lafontaine.
For more on differential geometry and applications to algebraic topology, see:
Differential Forms in Algebraic Topology, Raoul Bott & Loring W. Tu.
For more on differential geometry as it relates to physics, see:
Gauge Fields, Knots and Gravity, John Baez & Javier P. Muniain.
24