Download §8 De Rham cohomology

Differentiable Manifolds. Spring 2011
Theodore Voronov.
§8
De Rham cohomology
Last updated: May 21, 2011.
8.1
Examples, definitions, and basic properties
A k-form ω ∈ Ωk (M ) is closed if dω = 0. It is exact if there is a (k − 1)-form
σ ∈ Ωk−1 (M ) such that dσ = ω.
A form σ such that dσ = ω is called a primitive for ω. (Of course, when
it exists, it is not unique.)
The condition dω = 0 is local, in the sense that a form on M is closed if
it is closed near every point of M (in other words, if all the restrictions ω|U
are closed as forms on U ⊂ M ). To check that dω = 0, we simply check that
in local coordinates.
Compared to that, the property of being exact is not a local one. It is
necessary that a form σ such that dσ = ω is a form on the whole M , i.e., σ
should be everywhere defined.
We know that every exact form is closed, since d(dω) = 0. On the other
hand, there are closed but not exact forms.
Example 8.1. Consider the 1-form ω = dθ on R2 \ {0} where θ is the polar
angle. In standard Cartesian coordinates, as one can check,
ω=
xdy − ydx
.
x2 + y 2
This form is not exact, because for a cycle going around the origin once, say,
counterclockwise,
I
ω = 2π .
C
(The exact number is not important; that it is non-zero, is important.)
Remark 8.1. In the above example, dθ appear as exact (the differential of
the function θ), but it is not, because the function θ is ‘bad’. It cannot be
made everywhere defined, smooth and single-valued function on the whole of
R2 \ {0} at the same time. It can be made a ‘genuine’ function only locally,
in a sufficiently small domain, in particular where it is not possible to have a
cycle going around the origin. The case of dθ is typical. Every closed form on
a manifold can be regarded as exact if we surrender some of the properties:
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Theodore Voronov.
Differentiable Manifolds. Spring 2011
allow discontinuities or multi-valuedness, or consider the form only locally.
(The last claim is the content of the so-called Poincaré lemma, that every
closed form is locally exact, which will be discussed shortly.) Therefore, the
difference between closed forms and exact forms is a global feature of the
manifold in question and measures its ‘topological complexity’.
Definition 8.1. A closed k-form is cohomologous to zero if it is exact. Closed
k-forms ω and ω 0 on a manifold M are cohomologous if their difference is
exact: ω − ω 0 = dτ for some (k − 1)-form τ . (The form τ should be a
well-defined smooth form on the whole M .) This is an equivalence relation.
Equivalence classes w.r.t. this relation are called cohomology classes. The set
of all cohomology classes of degree k (sometimes people also say: in dimension
k), is denoted H k (M ). Notation for cohomology classes: [ω] ∈ H k (M ), if
ω ∈ Ωk (M ) is a closed form representing an element [ω].
(Check that it is indeed an equivalence relation!)
Theorem 8.1. For each k = 0, 1, , 2, . . . , the set of cohomology classes
H k (M ) is a vector space over R w.r.t. addition and multiplication by constants defined on representatives.
Proof. Consider the sum of two closed forms, ω + σ. It is closed, by the
linearity of d. If we replace one of the summands by a cohomologous form,
the sum will remain in the same cohomology class: ω + σ 0 = ω + (σ + dτ ) =
(ω + σ) + dτ . Thus addition of cohomology classes is well-defined and H k (M )
inherits the structure of an Abelian group. In a similar way we can show
that multiplication by real numbers is well-defined on cohomology classes.
Therefore H k (M ) is a vector space with respect to the operations
[ω1 ] + [ω2 ] := [ω1 + ω2 ] ,
k[ω] := [kω]
defined by the usual addition and multiplication by constants of closed forms
representing cohomology classes.
Hence, for each k, there is the cohomology space H k (M ). In particular,
it is an Abelian group and the traditional name for the vector space H k (M )
is the kth ‘cohomology group’.
Remark 8.2. Using the notion of the quotient space V /W of a vector space
V w.r.t. a subspace W ⊂ V , we can say, shortly, that
H k (M ) = Z k (M )/B k (M )
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Theodore Voronov.
Differentiable Manifolds. Spring 2011
where Z k (M ) := Ker(d : Ωk (M ) → Ωk+1(M )) stands for the space of all
closed k-form and B k (M ) := d Ωk−1 (M ) stands for the space of all exact
k-forms.
Remark 8.3. In the definition of cohomology groups we have used forms
with real coefficients. Sometimes it is convenient to consider complex-valued
functions and forms. The definition of cohomology applies to them as well
and to distinguish the two cases, we may use the notation such as H k (M, R)
and H k (M, C), which are vector spaces over R and C respectively. The
cohomology group H k (M ) defined using differential forms is also called the de
Rham cohomology group, after the Swiss mathematician George de Rham.
The name suggests that there are other cohomology theories, and this is
indeed true1 , but for manifolds they all give the same objects. This is exactly
the statement of the fundamental de Rham theorem conjectured by Élie
k
(M ) is used
Cartan and proved by G. de Rham. Sometimes the notation HDR
for de Rham cohomology.
Definition 8.2. The dimension dim H k (M ) is denoted bk (M ) and called,
the kth Betti number of M .
(One can see that we can use both real-valued or complex-valued forms
and this makes no difference for the dimension of cohomology: bk (M ) =
dimR H k (M, R) = dimC H k (M, C).)
As we know, forms
can be multiplied and they form a graded algebra
Ω∗ (M ) = Ωk (M ) , which is associative and graded-commutative.
Theorem 8.2. The exterior multiplication of forms induces a multiplication
of cohomology classes making the vectors spaces H k (M ) a graded-commutative
algebra. It is denoted H ∗ (M ) and is called the (de Rham) cohomology algebra of a manifold M .
Proof. Consider the product of closed forms: ω ∧ σ. The Leibniz rule implies
that it is closed: d(ω ∧ σ) = dω ∧ σ ± ω ∧ dσ = 0 + 0 = 0. If one of the factors
1
We may mention ‘ simplicial cohomology’ defined using a triangulation, ‘ singular cohomology’ based on the notion of ‘ singular simplex’ or, as a variant, ‘ singular cube’, and
‘ Čech cohomology’ defined using open covers. A feature distinguishing other cohomology
constructions from the Cartan – de Rham theory is the possibility to use the coefficients
more general than the real numbers. For example, using theories such as simplicial or Čech
it is possible to introduce certain cohomology groups H k (M, Z) or H k (M, Z2 ), which may
carry more information than the cohomology with real and complex coefficients.
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Theodore Voronov.
Differentiable Manifolds. Spring 2011
is replaced by a different representative of the same cohomology class, then
the cohomology class of the product will not change: ω 0 ∧ σ = (ω + dτ ) ∧ σ =
ω∧σ+dτ ∧σ = ω∧σ+d(τ ∧σ), by the same Leibniz rule, since dσ = 0. Hence,
the exterior multiplication of forms induces a well-defined multiplication on
cohomology classes:
[ω] [σ] := [ω ∧ σ] .
We arrive at the structure
of a graded algebra on the graded vector space
∗
k
H (M ) = H (M ) , called the cohomology algebra of M . The cohomology algebra H ∗ (M ) is associative and graded-commutative, inheriting these
properties from the algebra of forms Ω∗ (M ).
Consider a smooth map F : M → N . It induces the pull-back of forms:
F : Ωk (N ) → Ωk (M ). Let [ω] ∈ H k (N ) be a cohomology class represented
by a closed form ω ∈ Ωk (N ).
∗
Definition 8.3. The pull-back (or induced map) on cohomology classes is
defined by the formula:
F ∗ ([ω]) := [F ∗ (ω)] ,
where at the r.h.s. there is the pull-back of a differential form representing
a given cohomology class [ω].
Theorem 8.3. The pull-back of cohomology classes is well-defined. It is a
linear map
F ∗ : H k (N ) → H k (M ) ,
for each k = 0, 1, . . . . For the composition of smooth maps, we have
(G ◦ F )∗ = F ∗ ◦ G∗
(1)
on cohomology. The pull-back F ∗ is an algebra homomorphism of cohomology
algebras.
Proof. Suppose a form ω ∈ Ωk (N ) is closed and represents a cohomology class
[ω] ∈ H k (N ). We first need to check that the form F ∗ (ω) ∈ Ωk (M ) is also
closed and therefore gives a cohomology class in H k (M ). Indeed, dF ∗ (ω) =
F ∗ (dω) = 0. (The key fact here is that the exterior differential commutes
with pull-backs of forms.) Next, suppose a representative ω is replaced by
ω 0 = ω+dσ. Then F ∗ (ω 0 ) = F ∗ (ω+dσ) = F ∗ (ω)+F ∗ (dσ) = F ∗ (ω)+dF ∗ (σ),
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Theodore Voronov.
Differentiable Manifolds. Spring 2011
so [F ∗ (ω 0 )] = F ∗ (ω). Therefore the linear map F ∗ : H k (N ) → H k (M ) is welldefined. Equation (1) follows from the same property for the pull-backs of
forms. To see that F ∗ is an algebra homomorphism on cohomology, we write
F ∗ [ω][σ] = F ∗ [ω ∧ σ] = [F ∗ ω ∧ σ ] = [F ∗ ω ∧ F ∗ σ] =
[F ∗ ω][F ∗ σ] = F ∗ [ω] F ∗ [σ] .
so it is as claimed.
Corollary 8.1. If manifolds M and N are diffeomorphic, then the cohomology groups H k (M ) and H k (N ) are isomorphic (for all k).
Proof. Suppose F : M → N and G : N → M are mutually inverse smooth
maps giving a diffeomorphism M ∼
= N . Consider the vector spaces H k (M )
k
and H (N ) for some given k. There are linear maps
F ∗ : H k (N ) → H k (M )
and
G∗ : H k (M ) → H k (N ) .
The identities F ◦ G = id and G ◦ F = id imply (F ◦ G)∗ = G∗ ◦ F ∗ = id and
(G ◦ F )∗ = F ∗ ◦ G∗ = id. Hence the linear maps F ∗ and G∗ are mutually
inverse, so the vector spaces H k (M ) and H k (N ) are isomorphic (in particular,
have the same dimension).
Remark 8.4. Moreover, if M and N are diffeomorphic, the cohomology
algebras H ∗ (M ) and H ∗ (N ) are isomorphic. Indeed, we have isomorphism
of vector spaces (in each degree) that also preserves the multiplication.
Corollary 8.1 means that the de Rham cohomology is a ‘diffeomorphism
invariant’ of a smooth manifold.
We shall see shortly that de Rham cohomology is, in fact, an invariant
under a much coarser equivalence relation. Namely, it is a ‘homotopy invariant’.
Proposition 8.1. For any manifold M , the dimension of the space H 0 (M )
is the number of connected components of M (if it is finite).
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Theodore Voronov.
Differentiable Manifolds. Spring 2011
Proof. The space of 0-forms is the space of smooth functions C ∞ (M ). A
function f ∈ Ω0 (M ) = C ∞ (M ) is closed if df = 0. What does it mean?
Near each point f must be a constant (indeed, we may introduce coordinates
and write df = 0 in coordinates). Hence f is a local constant. It need not
be a constant on the whole M (a global constant), which is demonstrated
by the example of a manifold consisting of two connected components such
as two disjoint copies of Rn . The function may be zero on one component
and 1, on another. Notice that there no exact 0-forms (because there are no
−1-forms). Therefore H 0 (M ) is the space of all local constants on M . It is
intuitively clear and can be proved by a simple topological argument that on
a connected topological space any local constant is a constant. Therefore,
if a topological space, in particular, a manifold M , is the disjoint union of
its connected components (maximal connected subspaces), then any local
constant is defined by its values of the components (constants); thus it is a
function on the set of components. If the number of components is finite,
the space of functions on this set is finite-dimensional. As a basis one take
functions that are identically 1 on one component and identically 0 on all
other components. Hence the dimension is the number of components.
In most of our examples, the manifolds in question are connected. Then,
automatically, H 0 (M ) ∼
= R for them.
Consider examples of cohomology classes in degrees higher than zero.
Example 8.2. We saw above that [dθ] is a non-zero class in H 1 (R2 \ {0}).
Example 8.3. If we consider θ as a parameter on the circle S 1 (defined up
to 2π), the 1-form dθ is well-defined and gives a non-zero class [dθ] in H 1 (S 1 ).
We shall shortly see that all other cohomology classes of the circle in degree
1 are proportional to [dθ], i.e., dim H 1 (S 1 ) = 1 and [dθ] can be taken as a
basis (consisting of a single element).
8.2
Homotopy invariance and the Poincaré Lemma
Definition 8.4. Two smooth maps f0 , f1 : M → N are homotopic if there is
a smooth map F : M × [0, 1] → N such that F (x, 0) = f0 (x) and F (x, 1) =
f1 (x) for all x ∈ M . Notation: f0 ∼ f1 . (In other words, there is a family
ft = F (·, t) giving ‘smooth interpolation’ between f0 and f1 .) The map F is
known as a homotopy between f0 and f1 .
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Theodore Voronov.
Differentiable Manifolds. Spring 2011
Of course, homotopy also makes sense just for topological spaces, not
manifolds. In such a case instead of smoothness we impose continuity. What
we have defined above is known as ‘smooth homotopy’, but we shall skip the
adjective.
Theorem 8.4 (Homotopy invariance). Homotopic maps of smooth manifolds
induce the same linear map of their de Rham cohomology.
A proof is given below. An immediate application of this property is as
follows.
Definition 8.5. Two manifolds (more generally, two topological spaces) are
homotopy equivalent if there are maps in the opposite directions such that
their compositions are homotopic to identities. That is, for X and Y , there
are f : X → Y and g : Y → X such that f ◦ g ∼ idY and g ◦ f ∼ idX .
Notation: X ∼ Y . A manifold (more generally, topological space) homotopy
equivalent to a point is called contractible.
Example 8.4 (Contractibility of Rn ). Let us show that Rn ∼ {∗} (a single
point). We can identify the point with 0 ∈ Rn . Consider the obvious maps
i : {0} → Rn (inclusion) and p : Rn → {0} (projection). We have p ◦ i = id,
but i ◦ p is not the identity, it is the map that send every vector to zero.
However, it is homotopic to the identity, the family ft : x 7→ tx, where
t ∈ [0, 1], giving the desired homotopy.
Example 8.5. Open and closed cylinders, S 1 × R and S 1 × [−1, 1], are both
homotopy equivalent to the circle S 1 . (The same is true if we replace S 1 by
any space X.) (Construct the maps and homotopies following the previous
example.)
The previous example shows that homotopy equivalence is much coarser
relation than diffeomorphism or homeomorphism: it does not have to preserve dimension (the cylinders are 2-dimensional and S 1 is 1-dimensional)
and a non-compact space (such as the open cylinder) may be homotopy
equivalent to a compact space (such as the closed cylinder).
Corollary 8.2 (from Theorem 8.4). Homotopy equivalent manifolds M and
N have isomorphic cohomology groups H k (M ) and H k (N ) for all k.
Proof. Let f : M → N and g : N → M be such maps that f ◦ g ∼ idN and
g ◦ f ∼ idM . Then by Theorem 8.4, for each k, we have linear maps of vector
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Theodore Voronov.
Differentiable Manifolds. Spring 2011
spaces f ∗ : H k (N ) → H k (M ) and g ∗ : H k (M ) → H k (N ) with the property
f ∗ ◦ g ∗ = id on H k (M ) and g ∗ ◦ f ∗ = id on H k (N ) (‘homotopic to identity’
becomes ‘equal to identity’ for the maps of cohomology). That mean that
these vector spaces are isomorphic.
Example 8.6. All ‘generalized cylinders’ over a manifold M such as M × R,
M × (−1, 1), and M × [−1, 1] have the same cohomology as M . For example,
H 1 (S 1 × (−1, 1)) ∼
= R and H 2 (S 1 × (−1, 1)) = 0 .
Applying Corollary 8.2 to the special case of Rn (or any star-shaped
domain in Rn , see below) we obtain the following.
Theorem 8.5 (“The Poincaré Lemma”). On Rn every closed k-form, k ≥ 1,
is exact. Or:
(
R
for k = 0
H k (Rn ) ∼
=
0
for k > 0 .
Proof. Indeed, Rn is contractible. Therefore, by Corollary 8.2, H k (Rn ) ∼
=
H k ({∗}) for all k = 0, 1, 2, . . . . The claim follows.
Although the Poincaré lemma is the statement about vanishing of certain
cohomology, namely, H k (Rn ) = 0 for k ≥ 0, it plays a key role in calculating
the cohomology for manifolds when it is possibly non-zero. Indeed, since in a
neighborhood of a point every manifold M n is undistinguishable from Rn or
an open ball in Rn , the Poincaré lemma immediately implies that there is no
local obstruction for a closed form on M n to be exact, but only locally. See
the discussion at the beginning of this section. So if there is an open cover of
a manifold by contractible domains Uα (for example, diffeomorphic to Rn ), it
is possible to find a local primitive for each closed k-form: i.e., write ω = dσα
for σα ∈ Ωk−1 (Uα ). Then the task of finding a global primitive for ω (or an
obstruction for that) is reduced to considering the differences σα − σβ on the
intersections and whether or not, by a choice of each σα , these differences can
be made zero. We shall see an application of this idea in the next subsection2 .
Let us now give a proof of the homotopy invariance of de Rham cohomology.
2
Moreover, although this remains beyond the scope of this course, provided all the
pairwise and higher intersections such as Uα ∩ Uβ , Uα ∩ Uβ ∩ Uγ , etc., are also contractible,
it is possible to reduce the description of the cohomology classes of M n to a purely combinatorial data in terms of the open cover U = (Uα ), ‘encoding’ in this case the topology
of M n .
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Theodore Voronov.
Differentiable Manifolds. Spring 2011
Proof of Theorem 8.4. Suppose f0 and f1 are two homotopic maps M → N .
Let F : M ×[0, 1] → N be a homotopy. We wish to show that the linear maps
f0∗ : H k (N ) → H k (M ) and f1∗ : H k (N ) → H k (M ) coincide for all k ≥ 0.
That means that if we take a cohomology class [ω] ∈ H k (N ) represented by
a closed k-form ω ∈ Ωk (N ), the pull-backs f0∗ (ω) and f1∗ (ω) not necessarily
coincide, but they must differ only by an exact form. To this end, we shall
introduce a linear transformation
K : Ωk (N ) → Ωk−1 (M ) ,
decreasing degrees, such that for any form ω ∈ Ωk (N ),
f1∗ ω − f0∗ ω = (d ◦ K + K ◦ d)ω .
(2)
Thus for closed forms there will be f1∗ ω − f0∗ ω = d(Kω). We define K, using
the homotopy F . Consider F ∗ ω ∈ Ωk (M ×[0, 1]). Each k-form σ on M ×[0, 1]
(0)
(1)
(0)
(1)
can be uniquely written as σ = σt + dt ∧ σt where σt and σt are forms
on M depending on the parameter t. (Use some local coordinates on M to
(0)
(1)
see how it works.) The degree of σt is k, while the degree of σt is k − 1.
(1)
Indeed, the term with σt contains an extra factor of dt, which raises the
degree by 1. We define the operation K by Kω = (K̃ ◦ F ∗ )ω, where K̃ sends
R1
(0)
(1)
(1)
any σ = σt + dt ∧ σt on M × [0, 1] to 0 dt(σt ) (integration over t as
a parameter). The result is a (k − 1)-form. Let us check that K has the
desired property (2). Firstly, d ◦ F ∗ = F ∗ ◦ d, so we need to calculate the
commutator of d and K̃. On M × [0, 1],
d = dx + dt ∧
∂
∂t
where dx can be considered as the differential on M . In local coordinates,
dx = dxi ∧ ∂x∂ i . If we apply it to σ ∈ Ωk (M × [0, 1]) written as above, we
arrive at
!
(0)
∂ (0)
∂σ
(1)
(0)
(1)
dσ = dx + dt ∧
σt + dt ∧ σt
= dx σt + dt ∧ −dx σt + t
.
∂t
∂t
Hence
Z
K̃dσ =
0
1
(0)
∂σ
(1)
dt −dx σt + t
∂t
!
Z
=−
0
9
1
(1)
(0)
(0)
dt dx σt
+ σ1 − σ0 .
Differentiable Manifolds. Spring 2011
Theodore Voronov.
On the other hand, clearly
Z
dK̃σ = d
1
(1)
dt(σt )
=
1
(1)
dt dx σt
.
0
0
Therefore
Z
(0)
(0)
d ◦ K̃ + K̃ ◦ d σ = σ1 − σ0 .
(3)
Now the desired relation (2) follows from Equation (3) applied to σ = F ∗ ω.
We have, for K = K̃ ◦ F ∗ ,
∗
∗
(d ◦ K + K ◦ d) ω = d ◦ K̃ ◦ F + K̃ ◦ F ◦ d ω
= d ◦ K̃ ◦ F ∗ + K̃ ◦ d ◦ F ∗ ω = d ◦ K̃ + K̃ ◦ d (F ∗ ω)
(0)
∗
∗
∗
= (F ∗ ω)(0)
1 − (F ω)0 = f0 ω − f1 ω .
Indeed, when we view F as the family ft = F (·, t), we see that ft∗ ω is
(0)
obtained from F ∗ ω by setting dt to zero, i.e., by passing to (F ∗ ω)t in the
above notation. Hence f0∗ ω and f1∗ ω are obtained by additionally setting t
to 0 and to 1, respectively, which gives the desired formula.
Remark 8.5. The method used in the proof above is a prototype of a general construction called ‘algebraic homotopy’. More precisely, the ‘algebraic
homotopy’ in the proof is the operator K : Ωk (N ) → Ωk−1 (M ) satisfying
the property f1∗ − f0∗ = [d, K] (the commutator understood in the sense of
graded algebras). We deduced the existence of such an operator K from the
existence of a homotopy of smooth maps F = (ft ). From the existence of K
immediately follows that the maps given by f0∗ and f1∗ coincide on the level
of cohomology. These constructions can be axiomatized and generalized as
it is done in the area of mathematics called ‘homological algebra’.
The Poincaré Lemma is often given a direct proof independent from the homotopy invariance of de Rham cohomology. The underlying idea is very simple and
is related with integration of forms.
Example 8.7. On the real line R, any 1-form ω = f (x)dx (which is automatically
closed), is exact. Indeed, consider the function
Z x
F (x) =
f (y)dy .
0
It is defined on the whole line R and we have dF = ω by the Newton–Leibniz
theorem (the “main theorem of calculus”).
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Theodore Voronov.
Differentiable Manifolds. Spring 2011
This can be seen as a prototype of the general statement concerning Rn or the
so-called ‘star-shaped’ domains in Rn . Let us define them
Definition 8.6. An open domain U ⊂ Rn is called star-shaped if there is a point
x0 ∈ U such that for each x ∈ U , all points of the segment [x0 x] are in U . (The
segment [x0 x] consists of all points of the form (1 − t)x0 + tx where t ∈ [0, 1].)
The point x0 is often referred to as the center of the star-shaped domain U .
For example, any convex domain (i.e., such that for any two points the segment
joining them is also contained in the domain) is star-shaped. However, a starshaped domain does not have to be convex.
Example 8.8. An open ball in Rn is convex, therefore star-shaped.
Example 8.9. The interior of any “star” in R2 is star-shaped. (Notice that a star
is not convex.)
A version of the Poincaré Lemma is as follows.
Theorem 8.6 (Poincaré Lemma). On Rn or any star-shaped domain in Rn , every
closed k-form is exact, if k ≥ 1.
We can give an independent direct proof for 1-forms. (It is in fact generalizes
to k-forms as well, but we shall not go into that here.)
Direct proof of the Poincaré Lemma for 1-forms. Let A = Ai (x)dxi be a closed 1form on a star-shaped domain U ⊂ Rn . That means that ∂i Aj −∂j Ai = 0. Without
loss of generality we may assume that the center x0 of U is the origin 0 ∈ Rn . For
any point x ∈ U , the segment [0x] consists of points tx where t ∈ [0, 1]. Define a
function f by the formula
Z 1
f (x) =
xi Ai (tx)dt
0
for all x ∈ U . It makes sense because tx ∈ U . We have
Z 1
Z 1
∂
∂ i
df = dxj j
xi Ai (tx)dt = dxj
x
A
(tx)
dt
i
j
∂x 0
0 ∂x
Z 1
j
= dx
δji Ai (tx) + xi t ∂j Ai (tx) dt
| {z }
0
=∂i Aj (tx)
= dx
j
Z 1
i
Aj (tx) + tx ∂i Aj (tx) dt = dx
0
j
Z
0
1
d
tAj (tx) dt
dt
t=1
= dxj tAj (tx)
= A,
t=0
therefore A = df is exact, as claimed.
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Differentiable Manifolds. Spring 2011
Theodore Voronov.
RemarkR 8.6. In the proof for 1-forms above, the function f is nothing but the
integral [x0 x] A over the straight line segment [x0 x]. A similar proof for k > 1,
which we do not give, also uses integration, but in a more sophisticated way (since
the integral of a k-form over a k-surface would give a number, but we need to
obtain a (k − 1)-form).
8.3
Further examples of calculation
Consider some examples of calculating cohomology.
We know that for any connected manifold M , the space H 0 (M ) is onedimensional. Consider the opposite extreme. What can be said of the n-th
cohomology of a manifold M n , where dim M n = n? All n-forms on M n are
automatically closed. Suppose M n is compact and orientable. It is not
difficult to see that in such case on a manifold M n there exists an n-form ω
such that for all charts of an oriented atlas, ω = n1! ω12...n dx1 ∧. . .∧dxn where
ω12...n > 0. (For surfaces in R3 , one can take the area element; similarly, a
‘volume element’ can be constructed for higher dimensional manifolds, for
example, using an embedding into RN .) For the form ω,
Z
ω>0
Mn
due to the positivity of ω12...n . Suppose M n is also without boundary.
Then, by the Stokes theorem, the integral of any exact form dσ vanishes:
I
Z
Z
dσ =
σ=
σ = 0.
Mn
∂M
∅
Therefore the form ω cannot be exact, and we conclude that the space
H n (M n ) for an arbitrary closed manifold is non-zero. (Recall that a manifold
is called ‘closed’ if it is compact, orientable and without boundary3 .)
In fact, more can be said:
Theorem 8.7. For a connected closed manifold M n , the top cohomology
space is one-dimensional:
H n (M n ) ∼
= R.
3
Hopefully the terminology — in the context of de Rham cohomology — will not confuse
the reader. Yes, there are manifolds that are ‘closed’ and there are differential forms that
are ‘closed’. What is in common, is that ∂M = ∅ for a closed manifold and dω = 0 for a
closed form, so the two notions are not so far apart.
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Theodore Voronov.
Differentiable Manifolds. Spring 2011
A proof of that is beyond the scope of this course. We can only note that
a map
H n (M n ) → R
giving the desired isomorphism is nothing but the integration
I
[ω] 7→
ω
Mn
for ω ∈ Ωn (M n ).
Remark 8.7. The condition for M n to be connected is important. If M n has
several connected components, then one can consider integrals over each of
them, arriving therefore at several real numbers instead of one. Theorem 8.7
generalizes to the statement that for an arbitrary closed manifold, the vector
space H n (M n ) is isomorphic to the dual of the vector space H 0 (M n ) and
therefore there is the equality of the Betti numbers:
bn (M n ) = b0 (M n ) .
More generally, for a closed k-form ω and a closed (n − k)-form σ on a closed
n-dimensional manifold M n , the map
I
(ω, σ) 7→
ω∧σ ∈R
Mn
∗
defines a linear transformation H k (M n ) → H n−k (M n ) . (Check that it is
well-defined!) A generalization of Theorem 8.7 tells that this linear transformation is an isomorphism, so for an arbitrary closed manifold M n ,
∗
H k (M n ) ∼
= H n−k (M n )
and therefore
bk (M n ) = bn−k (M n ) ,
for all k = 0, 1, . . . , n. This is known as the Poincaré Duality Theorem.
Again, its proof is outside our course.
Let us now calculate the de Rham cohomology in some simple examples
where it can be done “by hands” (without developing special technique).
13
Differentiable Manifolds. Spring 2011
Theodore Voronov.
Example 8.10 (Cohomology of the circle S 1 ). Consider S 1 as the unit circle
with center at the origin in R2 . We already know that [dθ],
H where θ is the
polar angle, is a non-zero class in H 1 (S 1 ). This is because S 1 dθ 6= 0. Claim:
dim H 1 (S 1 ) = 1 and one can take [dθ] as a basis. To prove it, note that
arbitrary 1-forms on the circle can be written as f (θ)θ where f (θ+2π) = f (θ),
i.e., as “periodic” 1-forms on the real line. Any such form is the differential
of a function g on R, which is possibly not periodic: f (θ)dθ = dg where
Z
θ
f (s)ds .
g(θ) =
0
It turns out that the condition for g(θ + 2π) = g(θ) is exactly the vanishing
of the integral
Z 2π
I
f (s)ds =
f (θ)θ .
S1
0
1
Therefore a 1-form on S is exact if and only if it has zero integral over S 1 .
Any 1-form ω ∈ Ω1 (S 1 ) (automatically closed because all 2-forms are zero)
is cohomologous to c · [dθ] where
I
1
ω.
c=
2π S 1
Thus the cohomology classes in H 1 (S 1 ) are in 1-1-correspondence with numbers c ∈ R and H 1 (S 1 ) ∼
= R. Conclusion:


for k = 0
R
k
1 ∼
H (S ) = R
for k = 1


0
for k > 1 .
(It is also zero for k < 0, but we shall always omit such obvious equality.)
Example 8.11 (Cohomology of the cylinder S 1 × R). A similar direct argument can be applied to the cylinder S 1 × R (or a finite open cylinder
S 1 × (−1, 1), diffeomorphic to it) to show that H 1 (S 1 × R) ∼
= R (a representative of a non-zero class is again [dθ] where θ is the angular coordinate on S 1 )
and that H 2 (S 1 × R) = 0 . It is also possible to apply a homotopy argument,
that S 1 × R is homotopy equivalent to S 1 , therefore H k (S 1 × R) ∼
= H k (S 1 )
for all k.
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Theodore Voronov.
Differentiable Manifolds. Spring 2011
Example 8.12 (Cohomology of the sphere S 2 ). To find the cohomology of S 2
we may argue in the following way. Consider an open cover of S 2 by two open
sets, open neighborhoods of the north and south poles respectively, such that
each of them is diffeomorphic to a disk and the intersection is diffeomorphic
to a cylinder S 1 × (−1, 1). For example, one can take the coordinate domains
for the stereographic atlas, where U0 = S 2 \ {N } and U1 = S 2 \ {S}. Then
it is possible to compute the cohomology of S 2 by reducing it to the known
cohomology of U0 , U1 and U01 = U0 ∩ U1 .
Suppose ω ∈ Ω1 (S 2 ) is closed. Consider ω|U0 and ω|U1 . By the Poincaré
lemma, ω|U0 = df0 and ω|U1 = df1 where f0 ∈ C ∞ (U0 ) and f1 ∈ C ∞ (U1 ),
respectively. On the intersection U01 , d(f0 − f1 ) = 0, so f0 − f1 is a constant
(since U01 is connected), say, f0 − f1 = c ∈ R. Consider f10 := f1 + c on U1 ;
for it, ω = df10 still holds, and f0 and f10 agree on U01 . Therefore there is a
globally defined function f such that f |U0 = f0 and fU1 = f10 , and ω = df .
We can conclude that every closed 1-form on S 2 is exact, i.e., H 1 (S 2 ) = 0.
Consider now ω ∈ Ω2 (S 2 ). It is automatically closed. We know that
there are 2-forms whose cohomology classes are non-zero. For example, the
area element dS = sin θ dθ ∧ dϕ on S 2 . We shall show that H 2 (S 2 ) is onedimensional. In a similar way to the above, ω|U0 = dσ0 and ω|U1 = dσ1
where σ0 ∈ Ω1 (U0 ) and σ1 ∈ Ω1 (U1 ), respectively. On the intersection,
d(σ0 − σ1 ) = 0, hence σ01 := σ0 − σ1 represents a cohomology class in
H 1 (U01 ). As U01 is diffeomorphic to the cylinder S 1 × (−1, 1), we know that
H 1 (U01 ) ∼
= R. If we choose σ0 or σ1 differently, say, take σ00 = σ0 +df0 instead
of σ0 , the cohomology class of σ01 = σ0 − σ1 on U01 will not change. If we
replace the original ω by a cohomologous form on S 2 , say, take, ω 0 = ω + dσ
for some global σ ∈ Ω1 (S 2 ), then for the new form it is possible to take
0
= σ01 . Hence we arrive at a wellσ00 = σ0 + σ and σ10 = σ1 + σ, so σ01
2
2
1
defined map H (S ) → H (U01 ). It is linear (check!). It is an isomorphism.
Indeed, consider a closed 1-form σ01 on U01 . Using a partition of unity on S 2
subordinate to the cover (U0 , U1 ), g0 +g1 = 1, Supp g0 ⊂ U0 and Supp g1 ⊂ U1 ,
we can define σ0 := g1 σ01 as a 1-form on U0 (by extending by zero from U01 )
and, similarly, σ1 := −g0 σ01 as a 1-form on U1 . Note the minus sign. Then
σ0 − σ1 = g0 σ01 − (−g1 σ01 ) = (g0 + g1 )σ01 = σ01 on U01 . Applying d, we
see that dσ0 − dσ1 = 0 on U01 . Therefore there is a well-defined 2-form ω
on S 2 = U0 ∪ U1 such that ω|U0 = dσ0 and ω|U0 = dσ0 respectively. In this
way we have constructed the inverse map for H 2 (S 2 ) → H 1 (U01 ). So it is an
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Theodore Voronov.
Differentiable Manifolds. Spring 2011
isomorphism and we conclude that H 2 (S 2 ) ∼
= H 1 (U01 ) ∼
= R. To summarize:


for k = 0
R
k
2 ∼
H (S ) = 0
for k = 1


R
for k = 2 .
(It is of course zero also for k < 0 or k > 2.)
Example 8.13 (Cohomology of S n for n > 2). The cohomology of S n for
n > 2 can be calculated by the same method as for S 2 . It is clear that
H 0 (S n ) ∼
= R because the sphere is connected. By arguing as above, one can
see that H 1 (S n ) = 0. In the same way one can establish an isomorphism
H k (S n ) → H k−1 (S n−1 )
for all k > 1. Using induction on n, one can arrive at the following answer:


for k = 0
R
k
n ∼
H (S ) = 0
for 0 < k < n


R
for k = n .
(Question: what is the cohomology of S 0 ?)
We shall give more examples — without proofs, which would require developing some technical tools. They are not particularly difficult, but we do
not have the time.
Example 8.14 (Real projective space). The cohomology of RP n is as follows. The answer depends on the parity of n. For the even-dimensional real
projective spaces,
(
R
for k = 0
H k (RP 2m ) ∼
=
0
for k > 0 .
For the odd-dimensional real projective spaces,


for k = 0
R
k
2m+1 ∼
H (RP
)= 0
for 0 < k < 2m + 1


R
for k = 2m + 1 .
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Theodore Voronov.
Differentiable Manifolds. Spring 2011
Here m = 0, 1, 2, . . . . (For example, RP 1 ∼
= S 1 and H ∗ (RP 1 ) ∼
= H ∗ (S 1 ).)
The difference of the two cases (and the whole answer) can be explained by
the fact that RP n is obtained from S n by the identification of the antipodal
points. So the k-forms on RP n can be naturally considered as the k-forms on
the sphere S n invariant under the antipodal map x 7→ −x. With some more
care, one can see that for all k the vector space H k (RP n ) can be identified
with a subspace in H k (S n ). The question is therefore about the cohomology
in degree n. Here the difference between n = 2m and n = 2m + 1 arises:
in the first case the manifold is non-orientable and there is no volume form
giving a non-trivial class in H n (RP n ), while in the second case the volume
form on S n is invariant, thus giving a volume form for RP n and a non-trivial
class in H n (RP n ).
Example 8.15 (Complex projective space). The answer is as follows:


R
for k = 0




0
for k = 1





R
for k = 2



...
H k (CP n ) ∼
=

R
for k = 2p





0
for k = 2p + 1





...



R
for k = 2n
(recall that dim CP n = 2n). For example, when n = 1, we know that
CP 1 ∼
= S 2 and the answer above agrees with the results for the sphere. We
have a one-dimensional space in all even degrees 0 ≤ 2p ≤ 2n and zero
otherwise. In the previous examples there was no need to consider the multiplicative structure in cohomology (since by dimensional considerations, the
multiplication was always zero). Here the algebra structure of H ∗ (CP n ) can
help to describe the answer better. The following is true. Take a non-zero
cohomology class x ∈ H 2 (CP n ). (It spans the one-dimensional vector space
H 2 (CP n ).) Then all its powers: x, x2 , . . . , xn (up to the nth power), are nonzero and therefore span the corresponding cohomology spaces H 2p (CP n ).
The cohomology algebra of CP n can be described as the “truncated polynomial algebra”
H ∗ (CP n ) ∼
= R[x]/(xn+1 ) ,
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Theodore Voronov.
Differentiable Manifolds. Spring 2011
i.e., the factor-algebra of the algebra of polynomials R[x] w.r.t. the relation
xn+1 = 0. There is a nice choice of a basis class x ∈ H 2 (CP n ) given by a certain closed 2-form ω ∈ Ω2 (CP n ). (We skip the formula for ω in coordinates.)
In particular, this form satisfies
I
ω=1
(S 2 , i)
for a natural embedding i of S 2 = CP 1 into CP n .
(When n grows, the relation xn+1 = 0 in the algebra H ∗ (CP n ) “goes
away to infinity”. The limit CP ∞ makes sense, which is called infinite complex projective space. It is not a finite-dimensional manifold, but cohomology
still makes sense. For it, the cohomology algebra is just the algebra of polynomials: H ∗ (CP ∞ ) ∼
= R[x].)
Example 8.16 (the n-torus). By definition, T n is the product S 1 × . . . × S 1 .
Denote the angular coordinate on each circle as θi , where i = 1 . . . n. The 1forms dθi are obviously closed and define the non-trivial cohomology classes
yi = [dθi ] ∈ H 1 (T n ). (Indeed, the forms dθi give non-zero integrals over the
corresponding embedded circles.) They can be multiplied and we have
yi yj = −yj yi
(as represented by the wedge product of 1-forms, dθi ∧ dθj = −dθj ∧ dθi ).
The statement is that there are no other relations between the products of
the classes yi and that the products yi1 . . . yik with i1 < . . . < ik span the
cohomology spaces H k (T n ). So the Betti numbers are given by the binomial
coefficients:
n!
,
bk (T n ) = Cnk =
k!(n − k)!
and the cohomology algebra H ∗ (T n ) can be described as the exterior algebra
on n generators y1 , . . . , yn . In particular, for the 2-torus we have


for k = 0
R
k
2 ∼
2
H (T ) = R
for k = 1


R
for k = 2 ,
with [dθ1 ], [dθ2 ] giving a basis of H 1 (T 2 ), and [dθ1 ∧ dθ2 ], the basis of H 2 (T 2 ).
18