Contents Introduction diﬀerential manifolds · forms and Milnor’s classic [Mil97] on which our section on Brouwer’s ﬁxed theorem will be based. 1. Differential Manifolds and

DIFFERENTIAL GEOMETRY

MARC BURGERSTEPHAN TORNIER

Abstract. These are notes of the course Differential Geometry I held at ETHZurich in 2015.

Disclaimer. This is a preliminary version. Please report any typos, mistakes,comments etc. to [email protected].

Acknowledgements. Thanks to those who pointed out typos and mistakesas these notes were written, in particular J. Allemann, H. Benner, P. Gantner,A. Hauswirth, S. Imfeld, C. Macho, P. Poklukar and H. Wu.

Contents

Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11. Differential Manifolds and Differentiable Maps . . . . . . . . . . . . . . 22. Tangent spaces, Differential and Whitney’s Embedding Theorem . . . 213. Differential Forms and Integration on Manifolds . . . . . . . . . . . . . 314. De Rham Cohomology . . . . . . . . . . . . . . . . . . . . . . . . . . . 545. De Rham’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . 65References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71

Introduction

Differential geometry is a synthesis of three different subjects: Analysis in Rn,topology and multilinear algebra. It precisely defines a class of “spaces” on which onecan do analysis, termed differential manifolds. These spaces as well as the associatednotion of differentiable functions are the central concept of this course. Differentialmanifolds look locally like Rn but are globally much less boring. Examples are thesphere as well as surfaces with holes:

Sphere Torus Surface with two holes

An example of what we mean by “do analysis” is the following: If D is a region inR2 contoured by a curve σ : [0, 1] → R2 and L,M : R2 → R are reasonably smoothfunctions, then by Green’s formula:

∮

c

L(x, y) dx+M(x, y) dy =

∫∫

D

(∂M

∂x− ∂L

∂y

)dx dy.

Note that whereas the left-hand side of the equation only takes into account thevalues of L andM on c it yet remembers something aboutD, namely the right-handside. Green’s formula, the divergence theorem and other well-known formulas are

Date: June 3, 2016.

1

2 MARC BURGER STEPHAN TORNIER

all incarnations of Stokes’ Theorem which is best understood in the framework ofmanifolds. In this context we will have to make precise what object makes sense tobe integrated over a manifold: differential forms. Stokes’ Theorem will also be usedto define invariants that can e.g. tell the above surfaces apart in the sense that onecannot be deformed into the other without tearing it apart.

Differential geometry forms a basis for many other subjects. Clearly Riemanniangeometry is one of them. However, so are Lie groups and physics. The relationshipbetween differential geometry in algebraic geometry is a special one. In a way,notions of one of the two fields echo in the other. For instance, highly abstractalgebraic geometric concepts are often easier to visualize in differential geometry.

References for this course include Boothby’s [Boo03] which is readable by stu-dents, Barden’s and Thomas’ [BT03] which will in particular be used for differentialforms and Milnor’s classic [Mil97] on which our section on Brouwer’s fixed theoremwill be based.

1. Differential Manifolds and Differentiable Maps

1.1. Differential Manifolds: Definitions and Examples. Differential mani-folds were first studied by Riemann in 1854 and later on by Poincaré. However,they were still thinking about manifolds being imbedded in some euclidean spaceand lacked a precise definition. Nevertheless, Stokes’ theorem and notions like cur-vature were already around. The first precise definition, however, was given in 1913by Weyl at ETH, see [RW13].

1.2. Differential Manifolds: Definitions and Examples. The first step to-wards defining differential manifolds is to introduce topological manifolds.

Definition 1.1. An n-dimensional topological manifold is a topological space which is(i) Hausdorff (T2),(ii) second countable, and(iii) locally homeomorphic to Rn.

Clearly, the last condition of 1.1 is the crucial one. Recall that a topologicalspace X is Hausdorff if for every pair of points x, y ∈ X with x 6= y there are opensets U, V ⊆ X containing x and y respectively such that U ∩ V = ∅. Moreover, Xis second countable if it its topology admits a countable basis B = Un | n ∈ N,i.e. any open set U ⊆ X can be written as U =

⋃Un⊆U

Un. Eventually, X is locallyhomeomorphic to Rn if every point in M admits an open neighbourhood which ishomeomorphic to an open subset of Rn.

We are going to discuss many examples and non-examples of differential mani-folds. Therefore our list of examples and non-examples of topological manifolds israther short.

Example 1.2.(i) Let M be a countable discrete space. Then M is a zero-dimensional topo-

logical manifold. The converse is true as well. However, the classification ofn-dimensional manifolds is much harder in larger n.

(ii) The circle S1 ⊆ R2 with the induced topology is clearly a topological mani-fold of dimension one. Every point p admits a neighbourhood that is home-omorphic to an interval:

b

p

DIFFERENTIAL GEOMETRY 3

(iii) The real line R is a one-dimensional topological manifold as well. This istautological.

If one restricts oneself to connected, one-dimensional topological manifolds then S1

and R are in fact the only examples up to homeomorphism.

(iv) In dimension two, the situation is already so rich that it defies any reason-able classification. Examples are, as before, the family of surfaces with gholes (g ∈ N0): These surfaces are all compact and connected and in facta classification of compact, connected, two-dimensional topological mani-folds is managable. To this end we will later on introduce the notion oforientability to distinguish between orientable examples as above and non-orientable examples like projective space. To get an idea of the wealth ofgeneral (connected) two-dimensional topological manifolds, note that thecomplement in R2 of a Cantor set is an example.

Non-Example 1.3.

(i) Consider M := [0, 1] ⊂ R with the induced topology. Every interior pointof M satisfies the third requirement of the definition of a topological man-ifold for n = 1 but the points 0, 1 ∈ M do not. For instance, a typicalneighbourhood of 0 ∈ M is given by U = [0, ε) for some 0 < ε < 1. Sup-pose ϕ : U → V ⊆ R is a homeomorphism onto an open subset V of R.Note U\0 is connected, but V \ϕ(0) is not. This contradicts ϕ beingcontinuous.

(ii) The set M := [0, 1]2 is not a (two-dimensional) topological manifold either.Here one may argue using the fundamental group instead of connectedness.

Both non-examples above are manifolds with boundary though, as defined later.

(iii) Let M = Q ⊂ R with the induced topology. Then every open set of Q iscountable and hence cannot be in bijection with an open subset of any Rn.

Remark 1.4. In order to show that the dimension of a topological manifold iswell-defined one has to show that if U ⊆ Rn and V ⊆ Rm are non-empty andhomeomorphic then n = m. If n = 1 or m = 1 one may argue in the above fashionusing connectedness. The other cases require e.g. some basic homology theory. If,however, one asks for U and V to be C1-diffeomorphic then basic linear algebraapplied to the derivative readily implies n = m. This argument is going to apply inthe context of differential manifolds.

The next definition constitutes the next important step towards the definitionof differential manifolds.

Definition 1.5. Let M be an n-dimensional topological manifold. A chart on Mis a pair (U,ϕ) consisting of an open subset U ⊆ M and a homoeomorphism ϕ :U → ϕ(U) =: V ⊆ Rn. The subset U is the coordinate neighbourhood and V is thecoordinate space.

We are now going to examine the case in which two charts intersect, see below.


Uα Uβ

ϕα(Uα) ϕβ(Uβ)

ϕα ϕβ

θβα

θαβ

The maps

θβα := ϕβ (ϕα|Uα∩Uβ)−1 : ϕα(Uα ∩ Uβ) → ϕβ(Uα ∩ Uβ),

θαβ := ϕα (ϕβ |Uβ∩Uα)−1 : ϕβ(Uα ∩ Uβ) → ϕα(Uα ∩ Uβ),

are coordinate transformations or change of charts. They are homeomorphismsbetween open subsets of Rn as θ−1

βα = θαβ .

Definition 1.6. Let M be a topological manifold. A C0-atlas A on M is a collectionof charts A = (Uα, ϕα) chart on M | α ∈ A such that

⋃α∈A Uα =M .

Note that our definition perfectly resembles real-life atlases, introduced by Mer-cator in 1585. By Definition 1.1, every topological manifold admits a C0-atlas. As anext step towards the definition of differential manifolds we now introduce smoothatlases.

Definition 1.7. Let A be a C0-atlas. Then A is a Ck-atlas if all coordinate trans-formations between members of A are Ck-maps. A C∞-atlas is smooth.

Recall that a map f : U → Rm from an open subset U ⊆ Rn to Rm given byx 7→ (f1(x), . . . , fm(x)) where fi := πi f for all i ∈ 1, . . . ,m is Ck if all partialderivatives of fi (i ∈ 1, . . . ,m) up to order k exist and are continuous. Also, notethat in a Ck-atlas all coordinate transformations are in fact Ck-diffeomorphisms bythe above, in contrast to the fact that there is a smooth homeomorphism f : R → Rwhich is not a diffeomorphism, for instance f : x 7→ x3. In particular, this mapcannot occur as a coordinate transformation in any smooth atlas.

Example 1.8. We now give examples of topological manifolds with smooth atlases.

(i) Let M := U be an open subset of Rn and A = (U, id). This example mayseem rather trivial but nevertheless includes an important example of a Liegroup, namely GL(n,R) := A ∈Mn,n(R) | detA 6= 0 ⊆Mn,n(R) ∼= Rn

2

.(ii) Let M := Sn ⊆ Rn+1 with the induced topology. If n ≥ 1 then an atlas of

M requires at least two charts since M is compact whereas an open subsetof Rn is not. Now, there are in fact many choices for an atlas with twoelements one of which may be constructed using stereographic projectionas follows: Let Sn = x = (x1, . . . , xn+1) ∈ Rn+1 | ∑n+1

i=1 x2i = 1, N :=

(0, . . . , 0, 1) ∈ Sn and S := (0, . . . , 0,−1). Further, set UN := Sn\N andUS := Sn\S. Then the pairs (UN , ϕN ) and (US , ϕS), where

ϕN,S : Sn → Rn, x 7→ (x1, . . . , xn)

1∓ xn+1


are given by stereographic projection, are charts for M .

bN

b

S

bϕN (x)

b

x

One checks that the associated coordinate transformation is given by

θSN : Rn \0 → Rn \0, y 7→ y

‖y‖2 .

Observe that θSN exchanges the inside and the outside of the unit spherein Rn. For n = 2 this is an example of a Möbius transformation.

Now, the first attempt to define a smooth manifold would be to say that asmooth manifold is a topological manifold together with a smooth atlas. In thiscase however, there would be as many smooth manifolds associated to e.g. thesphere as there are atlases on it, which is impractical. We therefore introcoduce thefollowing definitions to talk about maximal atlases only.

Definition 1.9. Let M be a topological manifold. Two charts (Uα, ϕα) and (Uβ, ϕβ)on M are Ck-compatible if both coordinate transformations are Ck. A Ck-atlas Aon M is maximal if each chart which is compatible with every chart in A is in A.

Retain the notation of Definition 1.9. If (U,ϕ) ∈ A and A is maximal then every(V, ϕ|V ) where V ⊆ U is open is contained in A as well. The following lemma isgoing to come in very handy when defining manifolds. Its proof is left as an exercise.

Lemma 1.10. Let M be a topological manifold. Every Ck-atlas A onM is containedin a unique maximal atlas.

Sketch of Proof. If a maximal atlas M containing A exists it contains the charts of

S := (U,ϕ) | (U,ϕ) is a chart on M compatible with every chart in A ⊇ A .

Conversely, if the above set is an atlas it is maximal among those containing A andnecessarily unique with this property since any other maximal atlas containing Aconsists of charts which are (in particular) compatible with all the charts of A andis thus contained in S. By maximality it then equals S. Therefore it only remainsto show that S is an atlas. Since A ⊆ S and A is an atlas of M we have

M =⋃

(U,ϕ)∈S

U.

As to coordinate transformations, let (Uα, ϕα), (Uβ , ϕβ) ∈ S. We show that thecoordinate transformation θβα : ϕα(Uα ∩ Uβ) → ϕβ(Uα ∩ Uβ) is Ck on a neigh-bourhood of every point ϕα(x) (x ∈ Uα ∩ Uβ) of its domain. Since A is an atlasthere is a chart (V, ψ) ∈ A with x ∈ V . Since further both (Uα, ϕα) and (Uβ, ϕβ)are compatible with (V, ψ) the coordinate transformations θβψ and θψα are smoothand hence so is θβα at ϕα(x) which agrees with θβψ θψα on Uα ∩ Uβ ∩ V ∋ x.

Finally, we are in a position to define the central concept of this course.

Definition 1.11. A Ck-differential manifold is a topological manifold with a maxi-mal Ck-atlas. A C∞-differential manifold is smooth.


We shall only concern ourselves with smooth manifolds in this course and notdiscuss subtleties arising from the exact value of k. One may wonder whether anytopological manifold admits a smooth atlas. The following remark answers thisquestion in the negative.

Remark 1.12. There is a compact topological manifold that does not admit asmooth atlas. However it does admit a system of pairwise compatible charts thatcover all but one point. See [Ker60].

As to uniqueness of maximal atlases, we remark the following.

Remark 1.13. Let M be a smooth manifold with maximal atlas A. If F :M →M isa homeomorphism then A′ := (F (U), ϕ F−1) | (U,ϕ) ∈ A is a maximal smoothatlas as well. This is readily checked by examining the coordinate transformations.They do not notice the changed point of view at all!

Even more remarkable is the fact that there are topological manifolds whichadmit incompatible maximal atlases that do not even arise from each other in theabove fashion.

Remark 1.14. The topological manifold S7 admits a maximal atlas which is notcompatible with the one defined in Example 1.8 and can even be given by polyno-mial equations. See [Mil56].

Example 1.15. We now describe three standard ways of producing new manifoldsout of old ones.

(i) (Regular value) Let f : Rn → Rm be a smooth map and let a ∈ Rm bea regular value, i.e. for every x ∈ f−1(a) the derivative Dxf : Rn → Rm

has maximal rank m. Then f−1(a) = x ∈ Rn | f(x) = a is a smoothmanifold in a natural way. Note that all manifolds produced in this way arenaturally subsets of Rn. Later on we give the details of this construction inmuch greater generality.

(ii) (Open subset) Let M be a smooth manifold with atlas A and let U ⊆M beopen. Then A |U := (U ∩ V, ϕU∩V ) | (V, ϕ) ∈ A is a smooth atlas on U .Note that A |U need not be maximal. However, it is contained in a uniquemaximal smooth atlas by Lemma 1.10.

(iii) (Products) Let M,N be smooth manifolds with atlases A = (Uα, ϕα | α ∈A and B = (Vβ , ψβ) | β ∈ B. Then P := (Uα × Vβ , ϕα × ψβ | (α, β) ∈A × B is a smooth atlas on M × N . We recall the definition of ϕα × ψβ((α, β) ∈ A×B): Let ϕα : Uα → Rm and ψβ : Vβ → Rn where m = dimMand n = dimN . Then

ϕα × ψβ : Uα × Vβ → Rm×Rn, (x, y) 7→ (ϕα(x), ψβ(x))

and Rm×Rn is identified with Rm+n. Note that P is rarely going to bemaximal as Uα ∩ Vβ typically has many open subsets that do not have aproduct structure. Again, Lemma 1.10 turns P into a maximal atlas.

Next, we describe a construction which does not yield smooth manifolds that arenaturally subsets of some euclidean space, namely quotients. As a matter of fact,the construction need not even begin with a smooth manifold. For instance considera square and glue two opposite sides together. The result is a cylinder. Continuingby gluing together its ends yields a torus which is a smooth manifold.


The reader is invited to think about what the following identification produces.For i ∈ 1, 2, the edge labelled ai (bi) is identified with the edge labelled a−1

i (b−1i )

in the direction shown.

a−11

b−11

b2

a2

a−12

b−12

b1

a1

The above construction can be generalized to every regular polygon whose num-ber of sides is divisible by four. Anyway, some work is needed to make the aboveconstructions precise. We begin by reviewing the quotient topology: Let X be atopological space and let ∼ be an equivalence relation on X . Further, let X/ ∼denote the set of equivalence classes, termed the quotient of X by ∼, and letπ : X → X/ ∼ be the map which to every x ∈ X associates its equivalenceclass. We turn X/ ∼ into a topological space as follows: A set U ⊆ X/ ∼ is open ifπ−1(U) is open. Using the fact that π respects all Boolean operations one verifiesthat this definition does indeed turn X/ ∼ into a topological space. Furthermore,π is continuous.

Example 1.16. (Torus). Let X := R2. For all x, y ∈ X , set x ∼ y if and onlyif x − y ∈ Z2. Since (Z2,+) is a group, this is indeed an equivalence relation.We are going to argue that X/ ∼ is homeomorphic to the torus. First of all wedetermine a suitable piece ofX which meets every equivalence class. The unit squareS := [0, 1]2 ⊂ R2 will serve. Indeed, let π : X → X/ ∼ denote the quotient map.Then π([0, 1]2) = X/ ∼. To see this, just floor the components of a vector x ∈ Xand subtract the resulting vector from x. However, it remains to be understoodwhich points of S are identified under the equivalence relation. Clearly, if x ∈ S isin the interior of S, no other point in S is equivalent to x. However, (x1, 0)T ∈ S isidentified with (x1, 1)

T ∈ S for all x1 ∈ [0, 1] and, similarly, (0, x2)T ∈ S is identifiedwith (1, x2)

T ∈ S for all x2 ∈ [0, 1]. As observed above, these identifications yield atorus. The details of a homeomorphism are to be worked out.

The following example observes that not all equivalence relations yield reasonablequotient spaces.

Example 1.17. (A pathology). Let X := R. For all x, y ∈ X , set x ∼ y if and onlyif x− y ∈ Q. We show that X/ ∼ is an uncountable space whose quotient topologyhas only the trivial open sets. Hence the topology is not suited at all to studythe quotient in this case, it does not give any shape. First of all, X/ ∼ is in factuncountable since otherwise R would be a countable union of countable equivalenceclasses and thus countable. Now, let U ⊆ X/ ∼ be open, i.e. π−1(U) is open. Henceπ−1(U) contains a non-empty open interval (a, b). As a result, (a, b) + Q = R ⊆π−1(U) since any real number can be approximated arbitrarily well by rationals.Hence the assertion.

The last example shows that in order to obtain quotient topologies with goodproperties one has to impose some conditions on the equivalence relation.


Definition 1.18. Let X be a topological space. An equivalence relation ∼ on X isopen if for every open set U ⊆ X , the image π(U) ⊆ X/ ∼ is open.

Retain the notation of Definition 1.18. The statement is equivalent to demandingthat the quotient map be open, or, by definition of the quotient topology, thatπ−1(π(U)) = y ∈ X | ∃x ∈ U : y ∼ x be open in X for all open U ⊆ X .

Reviewing Example 1.16 in this regard we indeed see that the associated equiv-alence relation is open: Let U ⊆ R2 be open. Then π−1(π(U)) =

⋃γ∈Z2(γ + U) is

open as a union of open sets.However, note that non-open equivalence relations need not exclusively yield

bad quotient spaces. In fact, consider S = [0, 1]2 ⊆ R2 with the induced topologyand identify opposite sides via ∼ as in Example 1.16. Then ∼ is not open but thequotient space is the same as above: The image of the following open set U ⊆ S isnot open. In a sense, one half of it is missing in the quotient.

U

In any case, open equivalence relations yield quotients which potentially sat-isfy the first two requirements on topological manifolds as follows. Recall that anequivalence relation ∼ on a set X is the subset R := (x, y) ∈ X ×X | x ∼ y.

Proposition 1.19. Let X be topological space and let ∼ be an open equivalencerelation on X .

(i) If X is second-countable then so is X/ ∼.(ii) The quotient X/ ∼ is Hausdorff if and only if R ⊆ X2 is closed.

Proof. As to (i), let B be a basis of open subsets of X and set B′ := π(U) | U ∈ B.Since ∼ is open, B′ consists of open sets. In addition, if V ⊆ X/ ∼ is open then bydefinition so is π−1(V ), hence π−1(V ) =

⋃U∈B, U⊆π−1(V ) U . Therefore

V = π(π−1(V )) =⋃

U ∈ BU ⊆ π−1(V )

π(U)

and hence B′ is a basis as well.Regarding (ii), first suppose that X/ ∼ is Hausdorff. We show that X ×X\R is

open. To this end, let (x, y) ∈ X × X\R, i.e. π(x) 6= π(y). By assumption, thereare open subsets Ux, Uy in X/ ∼ containing π(x) and π(y) respectively such thatUx ∩ Uy = ∅. Then π−1(Ux) and π−1(Uy) are open subsets of X containing x andy respectively. Hence π−1(Ux) × π−1(Uy) is an open subset of X × X containing(x, y). Finally, we observe that π−1(Ux) × π−1(Uy) does not intersect R. Indeed,suppose (z1, z2) ∈ R ∩ π−1(Ux) × π−1(Uy). Then π(z1) = π(z2) on the one handand π(z1) ∈ Ux and π(z2) ∈ Uy on the other hand which contradicts Ux ∩ Uy = ∅.

Conversely, assume that R ⊆ X ×X is closed and let x, y ∈ X such that π(x) 6=π(y). Then (x, y) /∈ R. Since R is closed there are open sets Vy and Vy containing xand y respectively such that (Vx × Vy)∩R = ∅. In other words, π(Vx) ∩ π(Vy) = ∅.Conclude by recalling that ∼ is open and hence so are π(Vx) and π(Vy).


The second part of Proposition 1.19 can be applied to Example 1.16. Indeed, theequivalence relation

R = (x, y) ∈ R2 ×R2 | x− y ∈ Z2= (x, y) ∈ R2 ×R2 | (e2πi(x1−x2), e2πi(y1−y2)) = (1, 1)

is a level set of a continuous function and as such closed.

Example 1.20. (Real projective space). In linear algebra, PnR is defined as thequotient of Rn+1 \0 by the equivalence relation x ∼ y if and only if x = λy forsome λ ∈ R∗. In fact, we could replace R with any other field. In this case, theequivalence relation is open and its graph R is closed. Hence PnR = Rn+1 \0/ ∼is second-countable and Hausdorff.

To see that ∼ is open, let A ⊆ Rn+1 \0 be open, then π−1(π(A)) =⋃λ∈R∗ λA

is open as well since multiplication by λ is a homeomorphism of Rn+1 \0.Now we show that R is closed. As before, we identify it as the level set of a

continuous function. Define

f : Rn+1 ×Rn+1 → R, (x, y) 7→n+1∑

i,j=1

(xiyj − xjyi)2.

If y = λx then yi = λxi for all i ∈ 1, . . . , n+1 and hence xiyj −xjyi = xi(λxj)−xj(λxi) = 0. Conversely, assume that x, y ∈ Rn+1 \0 and f(x, y) = 0. Thenxiyj−xjyi = 0 for all i, j ∈ 1, . . . , n+1. Without loss of generality we may assumethat x1 6= 0. Then x1yj −xjy1 = 0 implies yj = (y1/x1)xj for all j ∈ 1, . . . , n+1,i.e. y = λx where λ = y1/x1. Since y 6= 0 we conclude that λ 6= 0.

We now equip PnR with charts: Set Ui := x = (x1, . . . , xn+1) ∈ Rn+1 | xi = 1.

b

U3 ⊆ R3

Then π−1(π(Ui)) = (x1, . . . , xn+1) | xi 6= 0 is open and hence by definition so isUi := π(Ui) ⊆ PnR. Since π|Ui

: Ui → Ui is continuous by definition of the quotienttopology, as well as injective and open, it is a homeomorphism. To build a chart,let pri : R

n+1 → Rn, x 7→ (x1, . . . , xi, . . . , xn+1) and observe that pri |Ui: Ui → Rn

is a homeomorphism. Hence ϕi : Ui → Rn, y 7→ pri (π|Ui)−1(y) is a homeomor-

phism as well. Combining this with the fact that⋃n+1i=1 Ui = PnR shows that

(Ui, ϕi)i∈1,...,n+1 is an atlas on PnR.Next up, we compute the coordinate transformations. Let 1 ≤ i < j ≤ n+1. Then

ϕi(Ui ∩ Uj) = y = (y1, . . . , yn) | yj−1 6= 0 and ϕj(Ui ∩ Uj) = y = (y1, . . . , yn) |yi 6= 0. One then checks that θji : ϕi(Ui ∩ Uj) → ϕj(Ui ∩ Uj) is given by

(y1, . . . , yn) 7→1

yj−1(y1, . . . , yi−1, 1, yi, . . . , yj−1, . . . , yn)

and hence is smooth. Overall, we have shown that PnR admits the structure of asmooth n-manifold. Furthermore, it is compact: Simply note that π|Sn

: Sn → PnR


is surjective. (It is not injective, however; in fact every point in PnR has two pre-images.

Example 1.21. (Grassmann manifolds). Grassmann manifolds generalize projectivespace and play a major role in the theory of vector bundles which we study lateron. Let 1 ≤ k ≤ n and let G(k, n) be the set of all subspaces of Rn of dimension k.

First of all, we describe a topology on G(k, n) utilizing linear algebra: LetF (k, n) := (v1, . . . , vk) ∈ (Rn)k | v1, . . . , vk linearly independent. Since everyvector space admits a basis, we have a surjective map π : F (k, n) → G(k, n) givenby (v1, . . . , vk) 7→ R v1+ · · ·+R vk. This already puts us in a good position to definea topology: To k vectors w1, . . . , wk of Rn we associate the k × n-matrix

M(w1, . . . , wk) :=

w1

...wk

.

Then w1, . . . , wk are linearly independent if and only if rankM(w1, . . . , wk) = k. Inthis way, F (k, n) is identified with A ∈Mk,n(R) | rankA = k, that is

A ∈Mk,n(R)

∣∣∣∣∣∣∣∃1 ≤ j1 < · · · < jk ≤ n : det

A1j1 · · · A1jk

......

Akj1 · · · Akjk

6= 0

which is union over multi-indices of open subsets of Mk,n(R) and hence open. Fur-thermore, given w1, . . . , wk and w′

1, . . . , w′k one verifies that Rw1 + · · · + Rwk =

Rw′1 + · · ·Rw′

k if and only if there is B ∈ GL(k,R) such that M(w1, . . . , wk) =BM(w′

1, . . . , w′k). Using this, one can show that (w1, . . . , wk) ∼ (w′

1, . . . , w′k) if

and only if Rw1 + · · ·Rwk = Rw′1 + · · ·Rw′

k is an open equivalence relation:Indeed, given B ∈ GL(k,R), the map LB : Mk,n(R) → Mk,n(R), A 7→ BA isan invertible linear map with inverse LB−1 . In particular, LB is a homeomor-phism. It preserves the open subset F (k, n) and if Ω ⊆ Fk,n is open then so isπ−1(π(Ω)) =

⋃B∈GL(k,R) LB(Ω). One may also show that the graph of ∼ is a closed

subset of F (k, n)×F (k, n). We have thus equipped G(k, n) with a second-countableHausdorff topology.

We now define charts on G(k, n) which turn it into a smooth manifold. Notice theanalogy to the construction of Example 1.20. For any multi-index J = (j1, . . . , jk)where 1 ≤ j1 ≤ · · · ≤ jk ≤ n, set

UJ =

w1

...wk

∈Mk,n(R)

∣∣∣∣∣∣∣

w1j1 · · · w1jk

......

wkj1 · · · wkjk

= Id ∈Mk,k(R)

.

Then π|UJ: UJ → G(k, n) is injective and

π−1(π(UJ )) =

v1...vk

∣∣∣∣∣∣∣det

v1j1 · · · v1jk

......

vkj1 · · · vkjk

6= 0

.

Therefore, π(UJ ) ⊆ G(k, n) is open and we have⋃J π(UJ ) = G(k, n). As a con-

sequence π|UJ: UJ → π(UJ ) is bijective, continuous and open, that is, a home-

omorphism. In addition, we have the projection maps pJ : UJ∼=−→Mk,n−k(R) and

one verifies that (π(UJ ), pJ (π|UJ)−1)J is a smooth atlas on G(k, n). We have

dimG(k, n) = k(n− k).


1.3. Differentiable Maps: Definition and Examples. From the categoricalviewpoint, we have so far introduced the objects. Morphisms, however, are stillmissing. In other words, without the means to compare smooth manifolds, nothingis going to happen.

Recall that in Rk we measure the length of vectors by ‖v‖ =∑k

i=1 v2i . This is

used in the definition of differentiability of a map defined on an open subset of Rn.

Definition 1.22. Let U ⊆ Rn be open and let p ∈ U . A map f : U → Rm isdifferentiable at p ∈ U if there exists a linear map L : Rn → Rm such that f(p+v) =f(p) + L(v) +R(p, v) for all v ∈ Rn such that p+ v ∈ U with

limv→0

‖R(p, v)‖‖v‖ = 0.

In the situation of Definition 1.22, we write Dpf = L. One of the most usefulcriteria for differentiability in Rn is the following.

Theorem 1.23. Let U ⊆ Rn be open. Further, let f : U → Rm, x 7→ (f1(x), . . . , fm(x))be C1 on U . Then f is differentiable at every point p ∈ U and the matrix of Dpfwith respect to the standard bases on Rn and Rm is given by

∂f1∂x1

· · · ∂f1∂xn

......

∂fm∂xn

· · · ∂fm∂xn

Recall that in fact the existence of partial derivatives alone does not imply dif-ferentiability. For instance, the function

f : R2 → R, (x, y) 7→

xyx2+y2 (x, y) 6= (0, 0)

0 (x, y) = (0, 0)

has partial derivatives everywhere but is not differentiable at (x, y) = (0, 0). In fact,it is not even continuous at this point.

We now define smooth functions on manifolds.

Definition 1.24. Let M be a smooth manifold and p ∈ M . A function f : M → Ris differentiable at p if for some chart (U,ϕ) at p, the function ϕ(U) → R, x 7→f ϕ−1(x) is differentiable at ϕ(p).

Definition 1.24 is in fact independent of the chart chosen. The argument whichshows this is an important and much used one: Let (U,ϕ) and (V, ψ) be chartsincluding p ∈M . For x ∈ ψ(U ∩V ) we compute f ψ−1(x) = f ϕ−1 (ϕψ−1)(x)which is a composition of the smooth map ϕ ψ−1 defined on ψ(U ∩ V ) and f ϕ−1 which is differentiable at ϕ(p). The same argument shows that the followingdefinitions make sense.

Definition 1.25. Let M and N be smooth manifolds. Then(i) f ∈ Ck(M,R) if for every chart (U,ϕ) ofM the function f ϕ−1 : ϕ(U) → R

is Ck,(ii) f : M → N is differentiable at p ∈ M if there are charts (U,ϕ) at p and

(W,ψ) at f(p) such that (1) f(U) ⊆W and (2) ψ f ϕ−1 : ϕ(U) → ψ(W )is differentiable at ϕ(p),

(iii) f ∈ Ck(M,N) if f : M → N is continuous and for every pair (U,ϕ),(W,ψ) of charts respectively on M and N such that f(U) ⊆ W the mapϕ(U) → ψ(W ), x 7→ ψ f ϕ−1(x) is Ck, and

(iv) f : M → N is a Ck-diffeomorphism if f is a homeomorphism and both fand f−1 are Ck-maps.


Retain Definition 1.25. Note that in (ii) we do not simply require (2) becausewithout (1) it might be satisfied automatically for rather bad choices of f . Also,the reader is invited to check that Definition 1.25 is consistent, i.e. for instance thatpart (iii) for N = R yields the same notion as part (i). The following is an exampleof how results of calculus, namely Theorem 1.23 generalize to manifolds.

Theorem 1.26. Let M and N be manifolds and let f :M → N be a C1-map. Thenf is differentiable at every point p ∈M .

Remark 1.27. We now collect several examples to illustrate Definition 1.25.(i) Given a manifold M , the set Ck(M,R) of Ck-functions on M is an R-

algebra for pointwise addition and multiplication of functions. This followsfrom the according statement for M = R.

(ii) LetM,N and R be manifolds and let f ∈ Ck(M,N) as well as g ∈ Ck(N,R)be given. Then g f ∈ Ck(M,R) by the chain rule.

(iii) Let M and N be a manifolds and let M =⋃α∈A Uα be an open covering

of M . Then f ∈ Ck(M,N) if and only if f |Uα∈ Ck(Uα, N) for all α ∈ A,

i.e. being Ck is a local condition.(iv) Let M = R and N = S1 = z ∈ C | |z| = 1. Then exp : R → S1, t 7→ e2πit

is smooth.(v) Consider GL(n,R) with its smooth structure defined in Example 1.8. Then

the map GL(n,R) × GL(n,R) → GL(n,R), (A,B) 7→ AB is smooth. Asa consequence, given g ∈ GL(n,R) the map Lg : GL(n,R) → GL(n,R),a 7→ ga is a diffeomorphism. In fact, a two-sided inverse of Lg is Lg−1 .

Remark 1.28. We remark further that any homeomorphism of a manifold can beturned into a smooth map: Let M be a smooth manifold with maximal atlas A andlet F : M → M be a homeomorphism. Set A′ = (F (U), ϕ F−1) | (U,ϕ) ∈ A.Then A′ is a (maximal) smooth atlas on M and F : (M,A) → (M,A′) is a smoothdiffeomorphism.

We are now working towards the rank theorem for which we recall that the rankof a linear map T : Rn → Rm is given by the dimension of its image which equalsthe row and column rank of any coordinate matrix of T .

Definition 1.29. Let M and N be manifolds, p ∈ M and let f : M → N bedifferentiable at p. Further, let (U,ϕ) be a chart of M at p and (W,ψ) a chart of Nat f(p). The rank of f at p is the rank at ϕ(p) of the linear map Dϕ(p)(ψ f ϕ−1) :Rn → Rm.

The notion of rank of a map will be exploited a lot in the sequel. It leads tothe concept of immersion, submersion, immersed manifolds etc. The key ingredientcomes from the rank theorem of calculus to which we now turn.

1.4. The Rank Theorem. First of all, we recall the inverse function theorem.

Theorem 1.30. Let W ⊆ Rn be open and let F :W → Rn be a Cr-map (r ≥ 1) suchthat at a ∈ W , the derivative DaF : Rn → Rn is invertible. Then there are opensubsets U and V of Rn containing a and F (a) respectively such that F |U : U → Vis a Cr-diffeomorphism.

This theorem is adaptable to the setting of smooth manifolds. However, fornow we apply it to the following calculus result which says that up to smoothchange of coordinates a map with constant rank k is the projection onto the firstk components. To this end, recall the following fact from linear algebra: Let L :Rn → Rm be a linear map of rank k. Note that necessarily k ≤ min(m,n). Thenthere are invertible linear maps G : Rn → Rn and H : Rm → Rm such that(H L G−1)(x1, . . . , xn) = (x1, . . . , xk, 0, . . . , 0).


Theorem 1.31. Let A0 ⊆ Rn and B0 ⊆ Rm be open and let F : A0 → B0 be aCr-map (r ≥ 1). Assume that F has constant rank k on A0. Then given a0 ∈ A0

there are open subsets A and B of A0 and B0 respectively, containing a0 and F (a0)respectively as well as Cr-diffeomorphisms G : A → U ⊆ Rn and H : B → V ⊆Rm to open subsets U and V of Rn and Rm respectively such that G(a0) = 0,H(F (a0)) = 0 and HFG−1(x1, . . . , xn) = (x1, . . . , xk, 0, . . . , 0).

The assertion of Theorem 1.31 may be visualized as follows.

A0

b

A

B0

b

BF

U

b0

V

b 0HFG−1

G H

We state that the change of variable maps G and H of Theorem 1.31 are notgenerally defined on the whole of A0 and B0.

Proof. Altering G and H with translations if necessary we may without loss ofgenerality assume that a0 = 0 as well as F (a0) = 0. Now D0F = (∂Fj/∂xi)i,jwhere 1 ≤ j ≤ m and 1 ≤ i ≤ n has rank k. Modulo permuting coordinates in Rn

and Rm we may assume that the first principal k × k-minor of D0F has non-zerodeterminant, that is

det

(∂Fj∂xi

)

1≤i≤k1≤j≤k

6= 0.

Define G(x) := (F1(x), . . . , Fk(x), xk+1, . . . , xn) where F (x) = (F1(x), . . . , Fm(x)).Observe that G(0) = 0 Furthermore, we have x ∈ A0:

DxG =

∂F1

∂x1· · · ∂F1

∂xk

∂F1

∂xk+1· · · ∂F1

∂xn

......

......

∂Fk

∂x1· · · ∂Fk

∂xk

∂Fk

∂xk+1· · · ∂Fk

∂xn

0 · · · 0 1...

.... . .

0 · · · 0 1

=

((∂Fi

∂xj

)∗

0 Idn−k

).

That is, detDxG = det(∂Fi/∂xj) 6= 0. Hence we may apply the Inverse FunctionTheorem to G. There are open subsets A1 ⊂ A and U1 ⊆ Rn, containing 0 respec-tively such that G : A1 → U1 is a Cr-diffeomorphism. Now, let us compute F G−1

on U1: Let y ∈ U1 and write y = G(x) with x ∈ A1. Then

F G−1(y) = F G−1(F1(x), . . . , Fk(x), xk+1, . . . , xn) = F (x1, . . . , xn)

= (F1(x), . . . , Fk(x), Fk+1(x), . . . , Fm(x))

= (y1, . . . , yk, fk+1(y), . . . , fm(y))

for some fl(y) = Fl(x) with l ≥ k+1. Now observe that F G−1 has constant rankk on U1 and that for y ∈ U1 we have Dy(F G−1) = DG−1(y)F DyG. Hence the


lower right block in the matrix

Dy(F G−1) =

Idk 0 · · · 0

∗ ∂fk+1

∂yk+1· · · ∂fk+1

∂yn...

......

∗ ∂fm∂yk+1

· · · ∂fm∂yn

vanishes. Modulo replacing U1 with a smaller connected ε-ball we may thus assumethat the functions fk+1, . . . , fm only depend on y1, . . . , yk. Thus

F G−1(y1, . . . , yn) = (y1, . . . , yk, fk+1(y1, . . . , yk), . . . , fm(y1, . . . , yk))

Now define T (z) := (z1, . . . , zk, zk+1+fk+1(z1, . . . , zk), . . . , zm+fm(z1, . . . , zk)) forz ∈ Rm with (z1, . . . , zk) ∈ prk1(U1) where prk1 projects onto the first k coordinates.Then T (0) = 0 and (

Idk 0∗ Idn−k

)

is invertible. By the Inverse Function Theorem there is an open set B1 ⊆ Rm

containing 0 such that T : B1 → T (B1) ⊆ Rm is a Cr-diffeomorphism onto its openimage. Furthermore,

T (z1, . . . , zk, 0, . . . , 0) = (z1, . . . , zk, fk+1(z1, . . . , zk), . . . , fm(z1, . . . , zk))

which implies the assertion: T−1FG−1(y1, . . . , yn) = (y1, . . . , yk, 0, . . . , 0). Now setH = T−1.

We remark that in Mm,n(R) the condition of being of rank k is neither opennor closed and hence constitutes a strong assumption in the above theorem whichutterly fails without it. As an immediate corollary of Theorem 1.31, we recordthat for a smooth map between manifolds there are always charts “adapted” toit. To this end we introduce the following notation: Given a ∈ Rn and ε > 0 setCnε (a) := x ∈ Rn | ∀i ∈ 1, . . . , n : |xi−ai| < ε. The sets Cnε (a) are hypercubes,for instance Cnε (0) = (−ε, ε)n.Corollary 1.32. Let N and M be manifolds of dimension n and m respectively andp ∈ M . Further, let F : M → N be a smooth map of constant rank k. Then thereare charts (U,ϕ) at p and (W,ψ) at F (p) such that ϕ(p) = 0, ϕ(U) = Cnε (0) ⊆ Rn

as well as ψ(F (p)) = 0, ψ(W ) = Cmε (0) and

ψFϕ−1(x1, . . . , xn) = (x1, . . . , xk, 0, . . . , 0)

for some ε > 0.

1.5. Submanifolds, Immersions, Embeddings etc. We have defined the cat-egory of smooth manifolds and smooth maps. In some categories the image andkernel of a morphism are naturally identified as subobjects of the codomain anddomain of the morphism. For instance, if L : V → W is a linear map betweenvector spaces then L(V ) is a subspace of W and L−1(0) ⊆ V is subspace of V . Ananalogous statement holds true in the category of groups. For smooth manifolds,however, images and level sets of smooth maps can be pretty bad. In the sequelwe address both of these problems. First we establish conditions under which levelsets of smooth maps are “nice”. This leads us to introduce the notion of a subman-ifold. For instance, S1 is a submanifold of the torus in e.g. the way depicted below.However, a real line wrapped around the torus at an irrational angle α ∈ R \Q asin ι : R → S1 × S1, t 7→ (e2πit, e2πiαt) is not going to be a submanifold because aneighbourhood of a point in the image contains a dense set of image points. In fact,the image of ι is dense in S1 × S1.


Definition 1.33. Let M be an m-manifold. A subset N ⊆ M is an n-submanifoldif for any p ∈ N there is a chart (U,ϕ) at p such that ϕ(p) = 0, ϕ(U) = Cmε (0) =(−ε, ε)m and ϕ(U ∩N) = x ∈ Cmε (0) | xn+1 = · · · = xm = 0.

We remark that in the literature, Definition 1.33 often defines a regular subman-ifold. Also, one verifies easily that the restriction to N of all charts as in Definition1.33 gives a smooth atlas on N and hence a structure of smooth n-manifold. Strictlyspeaking, we have the charts

ϕ|U∩N : U ∩N → x ∈ Cmε (0) | xn+1 = · · · = xm = 0 prk1−−→ Cnε (0) ⊆ Rn

Showing that the associated coordinate transformations are smooth amounts tosaying that the restriction of a smooth map defined on Rn remains smooth whenrestricted to a coordinate plane.

Our results enable us to introduce the following mechanism to construct smoothmanifolds.

Theorem 1.34. Let N and M be manifolds of dimension n and m respectively.Further, let F : N → M be a smooth map of constant rank k and let q ∈ F (N).Then F−1(q) ⊆ N is a submanifold of N of dimension n− k.

Proof. Let p ∈ F−1(q) = x ∈ N | F (x) = q. By Corollary 1.32 there are charts(U,ϕ) and (W,ψ) at p and F (p) respectively such that ϕ(p) = 0, ϕ(U) = Cnε (0),ψ(W ) = Cmε (0) and ψ(q) = 0 as well as ψFϕ−1(x1, . . . , xn) = (x1, . . . , xk, 0, . . . , 0):

UF |U

//

ϕ

W

ψ

Cnε (0)prk1

// Cmε (0)

In particular, (F |U )−1(q) = U ∩ F−1(q) and ϕ(U ∩ F−1(q)) = (prk1)−1(0) = x ∈

Cnε (0) | x1 = · · · = xk = 0 which is a hypercube. Up to swapping coordinates thisshows that F−1(q) is a submanifold in the sense of Definition 1.33.

Using the following definitions we broaden the applicability of Theorem 1.34. Inthe sequel, Nn and Mm denote manifolds of dimension n and m respectively.

Definition 1.35. Let Nn and Mm be smooth manifolds and let f : N → M be asmooth map.

(i) A point x ∈ Nn is a critical point if the rank of f at x is strictly less thanm. The corresponding value f(x) is critical.

(ii) A value y ∈M is regular if f has rank m at every point of f−1(y)

Retain the notation of Definition 1.35. Notice that if m n then every x ∈ N isa critical point. We now have the following.

Theorem 1.36. Let Nn and Mm be smooth manifolds and let f : N → M be asmooth map. Further, let y ∈M be a regular value of f . Then f−1(y) is a regular(n−m)-dimensional submanifold of N .


Note that in Theorem 1.36, the set f−1(y) on which f has constant rank is aclosed subset of N . Therefore, a priori, Theorem 1.34 is useless. However, we willexploit the fact that the rank at any point in f−1(y) is the maximal possible oneusing the following lemma.

Lemma 1.37. Let n ≥ m be integers. Then the set Rm := A ∈Mm,n(R) | rankA =m is open in Mm,n(R).

Proof. For every A ∈ Rm we construct an open neighbourhood of A contained inRm. Let 1 ≤ j1 < · · · < jm ≤ n be a multi-index such that the associated minor ofA has non-zero determinant:

det

A1j1 · · · A1jm

......

Amj1 · · · Amjm

6= 0.

Then

VA :=

B ∈Mm,n(R)

∣∣∣∣∣det(Bijl)1≤l≤m

1≤i≤m

6= 0

is an open subset of Mm,n(R) containing A which is in fact contained in Rm bymaximality of m since rankB ≥ m for all B ∈ VA.

We now prove Theorem 1.36.

Proof. Consider the subset R := x ∈ N | rankx f = m of N . By hypothesis,R contains f−1(y). Furthermore, R is open in N by Lemma 1.37: Let x ∈ R andlet (U,ϕ) and (W,ψ) be charts at x and f(x) respectively such that f(U) ⊆ W .Consider ϕ(U) → ψ(W ), z 7→ ψfϕ−1(z). We know that rankDϕ(x)(ψfϕ

−1) = m.In addition, the map ϕ(U) → Mn,n(R) which to z ∈ ϕ(U) associates Dz(ψfϕ

−1)is continuous. Hence, by Lemma 1.37, the set z ∈ ϕ(U) | rankDz(ψfϕ

−1) = mis open and contains ϕ(x). Therefore, there is an open subset of U containingx on which f has constant rank m which implies that R is open. Thus R is asmooth n-manifold on which f has constant rank. Applying Theorem 1.34 yieldsthe assertion.

Given that individual level sets of a smooth function f : Rn → R may bearbitrary closed subsets of Rn, one would not expect regular values to be abundant.However, we shall see in Theorem 1.43 that in fact they are in a very precise sense.First of all, though, we give a sample application of Theorem 1.36.

Theorem 1.38. The orthogonal group O(n) := X ∈ GL(n,R) | XTX = Id is aregular submanifold of GL(n,R) of dimension n(n− 1)/2: To see this, consider themap F : Mn,n(R) → Mn,n(R) given by X 7→ XTX . We show that F has constantrank n(n− 1)/2 on GL(n,R). We compute

DXF (Y ) = limh→0

F (X + hY )− F (X)

h= Y TX +XTY

and note that the map

DXF : Mn,n(R) →Mn,n(R), Y 7→ Y TX +XTY

ranges in symmetric matrices. We therfore reconsider

DXF :Mn,n(R) → Symn(R), Y 7→ Y TX +XTY

and note that if X is invertible then the image of DXF is Symn(R). Hence F hasconstant rank n(n+ 1)/2 on GL(n,R) which implies the assertion.


Having discussed level sets of smooth functions we now turn to the questionunder which circumstances the image of a smooth map between manifolds is amanifold.

Definition 1.39. Let Nn and Mm be smooth manifolds and let f : N → M be asmooth map. Then f is an immersion if the rank of f at every point is n.

Observe that an immersion f : Nn → Mm is only possible for n ≤ m. In localcoordinates, an immersion looks nice by Theorem 1.34. Also, the inverse image ofa point under f is a 0-dimensional submanifold of N and hence discrete.

Example 1.40. We give two examples of an immersion.

(i) Let f : R → R2, t 7→ (2 cos t, sin 2t). Since f(t) = (−2 sin t, 2 cos 2t) 6= 0 forall t ∈ R this map is in fact an immersion, of constant rank one. Its graphlooks as follows:

b

Notice that f(π/2) = (0, 0)T = f(3π/2). In particular, immersions need notbe injective. However, the tangent vectors of f at these two points differ.

(ii) As before, consider the irrationally imbedded real line into the torus givenby f : R → S1 × S1, t 7→ (e2πit, e2πiαt) where α ∈ R \Q. Whereas f isinjective, its image is dense which is considered bad for reasons that willbecome clear later.

The following is a stronger notion than immersion.

Definition 1.41. Let Nn and Mm be smooth manifolds and let f : N → M be asmooth map. Then f is an embedding if f(N) ⊆M is a regular submanifold and iff : N → f(N) is a diffeomorphism.

Proposition 1.42. Let Nn and Mm be smooth manifolds and let f : N → M be asmooth map. If f is an immersion and a homeomorphism onto its image then f isan embedding.

This follows essentially from Theorem 1.34.

Proof. Since f : N → M is a homeomorphism onto its image we know that forevery open subset V ⊆ N there is an open subset W ⊆M with f(V ) ⊆W ∩ f(N).Now choose charts (V, ϕ) and (W,ψ) at p and f(p) respectively such that f(V ) ⊆W ∩f(N) and ϕ(V ) = Cnε (0), ϕ(p) = 0 as well as ψ(W ) = Cmε (0), ψ(f(p)) = 0 and

ψfϕ−1 : Cnε (0) → Cmε (0), x 7→ (x, 0).

ψ(W ∩ f(N)) = y ∈ Cmε (0) | yn+1 = · · · = ym = 0which shows that f(N) is a regular submanifold of M .


1.6. Sard’s Theorem. We now turn to the arguably most important theorem inDifferential Topology. While individual level sets of smooth maps can be very bad,there are quite general situations in which a smooth map has lots of regular values;this follows from Sard’s Theorem which says that in any case the image of the setof critical points is always small.

Theorem 1.43 (Sard, 1942). Let U ⊆ Rn be open and let f : U → Rm be smooth.Let C := x ∈ U | rankDxf < m be the set of critical points of f . Then f(C) ⊆Rm has Lebesgue measure zero.

A few remarks about this theorem are in order. First of all, recall that a subsetE ⊆ Rm has measure zero if for every ε > 0 there is a covering E ⊆ ⋃∞

i=1 Ci bycountably many hypercubes such that

∑∞i=1 vol(Ci) < ε. We will use several times

the fact that a countable union of sets of measure zero has measure zero. Thisfollows from the above definition by choosing geometrically decaying values for therespective ε.

For instance, this implies the disturbing fact that Q ⊆ R has measure zero:Indeed, let f : N → Q be a bijection. Then f(n) ∈ (f(n) − ε/2n, f(n) + ε/2n) andthe sum of the volumes of these intervals is 2ε.

Also, note the following: Let f : [0, 1] → R be monotonically increasing, i.e.f(t1) ≤ f(t2) whenever t1 ≤ t2. Then the set of points at which f is not differen-tiable is of measure zero. It therefore seems like one could deduce a lot of informationabout f using its derivative; however, there is a monotonically increasing f whosederivative is zero whenever it exists. See [RSN72] for a nice introduction to examplesof this sort.

Remark 1.44. We now discuss Theorem 1.43 in the two cases n < m and n ≥ mbecause the flavour is somewhat different.

(i) (n < m). In this case, C = U and hence the statement is that f(U) is ofmeasure zero in Rm. As a warm-up to this, we prove that if U ⊆ R is openand f : U → Rm for m ≥ 2 is C1 then f(U) has measure zero. First of all,note that the C1 assumption is indeed necessary as there is a continuoussurjective map f : [0, 1] → [0, 1]2. By reparameterization, assume that[0, 1] ⊆ U . Then there is a constant C > 0 such that ‖f(x) − f(y)‖ ≤C|x − y| for all x, y ∈ [0, 1]. Given k ∈ N, subdivide the interval [0, 1]into the subintervals [j/k, (j + 1)/k]. Then f([j/k, (j + 1)/k]) is containedin a hypercube of side length at most C/k and f(C) can be covered byk hypercubes of this side length. The volume of each such hypercube is(C/k)m and hence the sum of their volumes is at most k(C/k)m = Cmk1−m.Since m ≥ 2, this tends to zero as k tends to infinity.

(ii) (n ≥ m). Assume that f : U → Rm is defined on an open subset U ⊆ Rn.Also, let us assume that C 6= U , that is, there is a regular point x ∈ U .Then there is an open subset V in U containing x on which f has maximalrank. Hence f(V ) ⊆ Rm is open and as a consequence cannot have measurezero. Hence f(U)\f(C) is non-empty. In particular, there are tons of regularvalues! In fact, Sard’s Theorem 1.43 is often applied by stating the existenceof a regular value.

(iii) We remark that Sard’s Theorem remains true under the weaker assumptionthat f be Ck for some k ≥ max1, n−m+1. Thus for n = 1 it suffices infact to ask for f to be C1 whereas when m = 1 the function f needs to beCn. For instance, there is an example of a C1-function f : R2 → R whoseimage contains an interval of singular values.

We now turn to the proof of Theorem 1.43. It contains two major ideas.


Proof. The proof proceeds by induction on n. Note that the statement makes sensefor n ≥ 0 and m ≥ 1. For n = 0 and m ≥ 1 the theorem is true since the imageof the one-point space R0 has measure zero in Rm for m ≥ 1. Now, let n ≥ 1 andwrite f = (f1, . . . , fm). Recall that C = x ∈ U | rankDxf < n. Set

C1 := x ∈ U | Dxf ≡ 0 =

x ∈ U

∣∣∣∣∂fr∂xi

(x) = 0 ∀r ∈ 1, . . . ,m ∀i ∈ 1, . . . , n.

More generally, for i ≥ 1, let Ci denote the set of all x ∈ U for which all partialderivatives of f up to order i vanish at x. Then

· · · ⊆ Ci+1 ⊆ Ci ⊆ · · · ⊆ C1 ⊆ C.

The proof is now divided into the follwing three steps.(i) Show that f(C\C1) has measure zero.(ii) Show that f(Ci\Ci+1) has measure zero for all i ≥ 1.

These two steps are not yet enough to conclude the assertion as we do not knowabout the measure of

⋂i≥1 Ci. We therefore end with the following step.

(iii) Show that f(Ck) has measure zero for large enough k. In fact, k > n/m−1.The first two steps are based on the same idea whereas the third one introduces anew one.

Step (i). We consider C\C1. Observe that for m = 1 we have C = C1. Indeed, for afunction f : U → R the condition that rankDxf < 1 is equivalent to the vanishingof all first partial derivative at x. Hence f(C\C1) is empty. We may thus assumem ≥ 2. In this situation, let x ∈ C\C1. By definition and without loss of generalitywe may thus assume (∂f1/∂x1)(x) 6= 0. Define a change of variables h : U → Rn

by x 7→ (f1(x), x2, . . . , xn) and observe that

Dxh =

∂f1∂x1

· · · · · · ∂f1∂xm

1. . .

1

which has non-zero determinant. Hence, by the Inverse Function Theorem, there isan open subset V of U containing x and an open subset V ′ ⊆ Rn containing h(x)such that h|V : V → V ′ is a diffeomorphism. Now define g := f(h|V )−1 : V ′ → Rm.Let C′ ⊆ V ′ be the set of critical points of g. Then the chain rule implies thath(V ∩ C) = C′. Hence f(V ∩ C) = g(C′). It now suffices to show that g(C′) hasmeasure zero since we can cover U\C1 with countably many such V ’s and deducethat f(C\C1) has measure zero.

As in Theorem 1.34, we show that g has a special form. Let (t, x2, . . . , xn) =(f1(x), x2, . . . , xn) = h(x) ∈ V ′. Then

g(t, x2, . . . , xn) = gh(x) = f(x)

= (f1(x), . . . , fm(x))

= (t, f2h−1(t, x2, . . . , xn), . . . , fmh

−1(t, x2, . . . , xn))

= (t, gt(x2, . . . , xn)

where gt : Rn−1 → Rm−1 is given by

gt(x2, . . . , xn) := (f2h−1(t, x2, . . . , xn), . . . , fmh

−1(t, x2, . . . , xn).

Observe that n−1 ≥ 0 and m−1 ≥ 1 which is covered by the induction hypothesis.Compute

D(t,x2,...,xn)g =

(1 0∗ D(x2,...,xn)g

t

).


Therefore, (t, x2, . . . , xn) is critical for g if and only if (x2, . . . , xn) is critical for gt.Let Ct denote the critical points of gt. Then g(C′) = (t, z) | z ∈ gt(Ct and henceg(C′) ∩ (t × Rm−1) = t × gt(Ct). By the induction hypothesis, the Lebesguemeasure of gt(Ct) is zero for all t and hence by Fubini’s Theorem g(C′) has measurezero as well. For an efficient introduction to measure theory, see [Rud87].

This completes step one.

Step (ii). Let i ≥ 1 and x ∈ Ci\Ci+1. Then for some 1 ≤ r ≤ m and some multi-index 1 ≤ j1 ≤ · · · ≤ ji+1 ≤ n we have

w(x) :=∂ifr(x)

∂xj2 · · · ∂xji+1

= 0

but ∂w∂xj1

6= 0. Without loss of generality, assume j1 = 1, that is ∂w∂x1

6= 0. Again,consider the map h : U → Rn given by x 7→ (w(x), x2, . . . , xn) which is a diffeomor-phism from some open set V ⊆ U containing x onto some open set V ′ ⊆ Rn. Bydefinition, h(Ci ∩ V ) ⊆ (0 × Rn−1). Now, consider again the map g := f h−1 :V ′ → Rm. Then gh(Ci ∩V ) = f(Ci ∩V ) but since h(Ci ∩V ) ⊆ 0×Rn−1 we mayconsider g := g|0×Rn−1 . Observe that any point in h(Ci ∩V ) is certainly a criticalpoint of g. Hence we are done by recurrence.

Step (iii). We now show for large enough k, the set f(Ck) has measure zero. To thisend, let Iδ ⊆ U be a closed hypercube whose edges have length δ. We show thatf(Ck∩Iδ) has measure zero for k > n/m−1. By Taylor, we have for all x ∈ Ck∩Iδand h with x+ h ∈ Iδ that

f(x+ h) = f(x) +R(x, h)

where ‖R(x, h)‖ ≤ C‖h‖k+1 and C only depends on f and Iδ. More precisely, thisis a consequence of the integral form of the remainder. Now, pick N to subdivideIδ into cubes of side length δ/N . Let I be one of these small cubes containing apoint x ∈ Ck ∩ I.

Iδ

Ck

b

Then for all h such that x+ h ∈ I we have

‖f(x+ h)− f(x)‖ ≤ C‖h‖k+1 ≤ C

(√nδ

N

)k+1

.

Hence f(I) is contained in a hypercube with edges of the above length. There areat most Nn such little cubes and hence f(Ck ∩Iδ) is contained in a union of hyper-cubes whose sum of volumes is less than Nn(C

√nδ/N)(k+1)m = (C

√nδ)(k+1)m ·

Nn−(k+1)m. For k > n/m− 1 we can make this sum arbitrarily small by choosingN large enough. Hence the assertion.


2. Tangent spaces, Differential and Whitney’s Embedding Theorem

2.1. Tangent Spaces. The naive idea of a tangent “plane” approximating a surfaceup to order one cannot be implemented in general for lack of an ambient space.However, there are at least two ways to define a tangent space and recover the aboveintuition. We will start from the following viewpoint: Let M and N be manifoldsand p ∈ M . The tangent space TpM should be equal to Rm if M = Rm, and suchthat the construction is functorial, meaning that the derivative Dpf is a linear mapfrom TpN → Tf(p)M .

b

Rn

b

0

v

bp

The mental picture of course is that a tangent vectorv at a point p is attached

to p as it represents a small increment of the position vector of p. Now, let A be anatlas on M and define

Ap := (U,ϕ, ξ) | (U,ϕ) is a chart at p, ξ ∈ RmOn Ap we define a relation ∼p as follows. Set (U,ϕ, ξ) ∼p (V, ψ, η) if and only ifDϕ(p)(ψ ϕ−1)ξ = η.

U V

b

ξb

ηϕ(U) ψ(V )

ϕ ψ

ψ ϕ−1

Lemma 2.1. Retain the above notation. The relation ∼p is an equivalence relation.

Proof. Clearly, ∼p is reflexive. As to symmetry, suppose (U,ϕ, ξ) ∼= (V, ψ, η), thatis Dϕ(p)(ψϕ

−1)ξ = η. Hence ξ = (Dϕ(p)(ψϕ−1))−1(η). However, by the chain

rule, (Dϕ(p)(ψϕ−1))−1 = Dψ(p)(ϕψ

−1) and hence ξ = Dψ(p)(ϕψ−1)η which is the

assertion. For transitivity, suppose (U1, ϕ1, ξ1) ∼= (U2, ϕ2, ξ2) and (U2, ϕ2, ξ2) ∼=(U3, ϕ3, ξ3). Then Dϕ1(p)(ϕ2ϕ

−11 )ξ1 = ξ2 and Dϕ2(p)(ϕ3ϕ

−12 )ξ2 = ξ3. Now observe

that on ϕ1(U1 ∩U2 ∩U3), which contains ϕ1(p) we have (ϕ3ϕ−12 )(ϕ2ϕ

−11 ) = ϕ3ϕ

−11 .

Hence the chain rule implies that

Dϕ1(p)(ϕ3ϕ−11 ) = Dϕ2(p)(ϕ3ϕ

−12 ) Dϕ1(p)(ϕ3ϕ

−11 )

maps ξ1 to ξ3 which is the assertion.


Given the above lemma, we now define TpM := Ap / ∼p. Hence a tangent vectorat p is a consistent choice of tangent vectors in the charts. Next, we record thatTpM does indeed admit a natural vector space structure.

Lemma 2.2. Let M be a manifold and let (U,ϕ) be a chart at p ∈ M . Then themap Rm → TpM, ξ 7→ [(U,ϕ, ξ)] is a bijection. The resulting vector space structureon TpM is independent of the choice of the chart (U,ϕ).

Proof. First, we show injectivity: Let ξ1, ξ2 ∈ Rm and suppose that (U,ϕ, ξ1) ∼p(U,ϕ, ξ2), i.e. Dϕ(p)(ϕ ϕ−1)ξ1 = ξ2 which implies ξ1 = ξ2. As to surjectivity: Letv = [(V, ψ, η)] ∈ TpM . Then (U,ϕ,Dψ(p)(ϕ ψ−1)η) is equivalent to v.

Regarding the vector space structure, we need to show that addition and scalarmultiplication defined with respect to different choices of (U,ϕ) coincide. To thisend, let v1, v2 ∈ TpM . Assume that vi (i ∈ 1, 2) is represented by both (U,ϕ, ξi)and (V, ψ, ηi). We need to show that (U,ϕ, ξ1 + ξ2) ∼p (V, ψ, η1 + η2). Indeed, wehaveDϕ(p)(ψϕ

−1)ξ1 = η1 and Dϕ(p)(ψϕ−1)ξ2 = η2 and therefore Dϕ(p)(ψϕ

−1)(ξ1+ξ2) = η1 + η2 which amounts to the assertion. A similar argument works for scalarmultiplication.

Remark 2.3. With this definition of tangent space we indeed recover the following:Let U ⊆ Rm be an open subset, considered as a smooth m-manifold. Then usingthe chart (U, Id) at any point p ∈ U leads to the identification TpU = Rm.

Now, let f : N → M be a map which is differentiable at p. We define thedifferential Dpf : TpN → Tf(p)M of f at p in the following way: Pick local charts(U,ϕ) at p and (V, ψ) at f(p) such that f(U) ⊆ V . Given v := [(U,ϕ, ξ)] ∈ TpN wedefine Dpf(v) := [(V, ψ,Dϕ(p)(ψfϕ

−1)ξ)]. Recall that by definition of smoothnessof f at p the map ψfϕ−1 is indeed smooth at ϕ(p). The proof of the followinglemma is now left as an exercise.

Lemma 2.4. Retain the above notation. The map Dpf : TpN → Tf(p)M is awell-defined linear map. Further, the rank of f at p equals dimDpf(TpN).

Also, the chain rule is generalized to the setting of manifolds.

Lemma 2.5. Let N,M and R be manifolds and f : N → M as well as g : M → Rbe smooth maps. If f is differentiable at p ∈ N and g is differentiable at f(p) ∈Mthen gf :M → R is differentiable at p ∈ N and Dpgf = Df(p)g Dpf .

The following proposition underlines once more that our definition captures theintuition: Let N,M be manifolds and f : N → M a smooth map of constant rankk. Then for y ∈ f(N), the set f−1(y) ⊆ N is a regular (n− k)-submanifold. Let usconsider f−1(y) as an (n−k)-manifold and the injection i : f−1(y) → N which is ofrank n− k. Then for all x ∈ f−1(y) we have the map Dxi : Tx(f

−1(y)) → Ti(x)N .

Proposition 2.6. Retain the above notation. For every x ∈ f−1(y) we have

Dxi(Txf−1(y)) = kerDxf.

The reader is encouraged to verify that this captures the intuition from above inthe case N = R3 \0, M = (0,∞), f(x) = ‖x‖2, y = 1 and f−1(y) = S2:


b

Proof. This can be proven directly by unravelling the definitions but there is alsoa way to avoid that. Consider the constant map f i : f−1(y) →M, x 7→ y. ThenDx(f i) = 0. Using the chain rule we deduce Di(x)f Dxi = 0. In particular,Dxi(Txf

−1(y)) ⊆ ker(Di(x)f). For the other inclusion, observe that rank of i atany point x ∈ f−1(y) is n − k. Hence dimDxi(Txf

−1(y)) = n − k. Since therank of f at any point is k we conclude that dimkerDi(x)f = n − k becausedimTi(x)N − dimkerDi(x)f = dimDi(x)f = k and dim Ti(x)N = n. This impliesthe converse inclusion.

2.2. Tangent Vectors and Derivations. In this section, we discuss an equiva-lent, more algebraic definition of the tangent space which is particularly useful e.g.in the setting of Lie groups. Let N be an n-manifold and let p ∈ N . Recall thatC∞(N) := C∞(N,R) is an R-algebra.

Definition 2.7. Let N be a manifold and p ∈ N . A derivation of C∞(N) at p is amap δ : C∞(N) → R such that

(i) δ is an R-linear map, and(ii) (Leibniz rule) for all f, g ∈ C∞(N) we have δ(fg) = δ(f)g(p) + f(p)δ(g).

Let Derp C∞(N) be the vector space of derivations at p

Theorem 2.8. Let N be a manifold and p ∈ N . The map

TpN → Derp C∞(N), v 7→ (δv : f 7→ Dpf(v))

is an isomorphism of vector spaces.

Note that Definition 2.7 is much easier to write down as a definition of thetangent space but is also a lot less transparent. Nevertheless, as noted above, it isan important description of the tangent space. We remark that the C∞-assumptionin Definition 2.7 is crucial. There are much more derivations of Ck(N) than thosewhich come from tangent vectors.

The strategy of proof of Theorem 2.8 is to translate the problem to Rn usingcharts, solve it there and translate back the solution. The following lemmas areconcerned with the Euclidean setting.

In particular, the following function is needed inthe proof: Let σ : R → R be defined by

σ(x) :=

e−

1x2 x > 0

0 x ≤ 0.

Then σ ∈ C∞(R).

b

0

Lemma 2.9. Let K ⊆ Rn be compact and let F ⊆ Rn be closed with K ∩ F = ∅.Then there is f ∈ C∞(Rn) such that 0 ≤ f(x) ≤ 1 for all x ∈ Rn, f |K ≡ 1 andf |F ≡ 0.


Proof. We utilize the following more precise statement which will also be used inthe proof of Whitney’s embedding theorem: Let a ∈ Rn and ε > 0. Then there isf ∈ C∞(Rn) such that 0 ≤ f(x) ≤ 1 for all x ∈ Rn as well as f−1(1) = Cnε/2(a)and f−1(0) = Rn \Cnε (a).

First, recall the function σ from above and consider the map R → R which sendst to σ(t)σ(1/2 − t) and whose graph looks as follows:

b b

0 12

Next, consider the map R → R, x 7→∫ x−∞ σ(t)σ(1/2− t) dt. One easily determines

that its graph looks as follows:

b b

0 12

Finally, normalize the above function so that the constant value which it assumeson [1/2,∞) is one, i.e. define g : R → R by

g(x) :=

∫ x−∞ σ(t)σ(12 − t) dt∫∞

−∞ σ(t)σ(12 − t) dt

Now, set β(x) := g(1 + x)g(1− x) whose graph is

b b b b b

− 12

0 12

−1 1

This essentially solves the problem: It only remains to set

f(x) :=

n∏

i=1

β

(xi − aiε

).

We are now in a position to prove the actual statement of the Lemma: Letε > 0 and a1, . . . , al ∈ K such that K ⊆ ⋃l

i=1 Cnε/2(ai) ⊆ ⋃l

i=1 Cnε (ai) ⊆ Rn \F .

For each ai (i ∈ 1, . . . , l) pick βi as above for the hypercube Cnε/2(ai) and setf(x) = 1 −∏l

i=1(1 − βi(x)). This function has all the right properties: Firstly, itranges betweeen zero and one because the βi do. Secondly, for x ∈ K there is somei0 ∈ 1, . . . , l such that x ∈ Cnε/2(ai0); hence βi0(x) = 1 whence f(x) = 1. Thirdly,if x ∈ F then x /∈ Cnε (ai) and hence βi(x) = 0 for all i ∈ 1, . . . , l and thereforef(x) = 0. This proves the assertion.

A good account for analytical aspects of the theory as the one above is [dR56].The next lemma states that the image of a derivation on a function is determinedby the local behaviour of that function.

Lemma 2.10. Let N be a manifold and p ∈ N . Further, let δ ∈ Derp(C∞(N)) and

g ∈ C∞(N) such that g ≡ 1 in a neighbourhood of p. Then δ(f) = δ(fg) for allf ∈ C∞(N).


Proof. By linearity of δ it suffices to show that δ(h) = 0 where h := f − fg. Since hvanishes in a neighbourhood V of p ∈ N there is some ψ ∈ C∞(N) whose supportis contained in V and which satisfies ψ(p) 6= 0: Indeed, let (U,ϕ) be a chart at pwhose domain is contained in V . Let C be a hypercube centered at ϕ(p) containedin ϕ(U) and let β ∈ C∞(Rn) be a function adapted to C as in the previous lemma.Then set ψ := β ϕ, naturally extended to the whole of N .

Then h · ψ = 0 and hence 0 = δ(h · ψ) = h(p)δ(ψ) + δ(h)ψ(p). Since h(p) = 0and ψ(p) 6= 0 this implies δ(h) = 0 and hence δ(f) = δ(fg).

We now get to the C∞-part of the argument without which Theorem 2.8 fails.

Lemma 2.11. Let a ∈ Rn and U ⊆ Rn a star-shaped neighbourhood of a. Further,let f ∈ C∞(U). Then there are g1, . . . , gn ∈ C∞(U) with

(i) (g1(a), . . . , gn(a)) =(∂f∂x1

(a), . . . , ∂f∂xn(a)), and

(ii) f(x) = f(a) +∑n

i=1(xi − ai)gi(x)

In a way, the above lemma formulates a factorization property: For instance, ifn = 1, it asserts f(x) − f(a) = (x − a)g(x) where g is again a C∞-function. If fwas only Ck for finite k then g would be at most Ck−1.

Proof. Since U is star-shaped with respect to a, we may write

f(x) = f(a) +

∫ 1

0

∂f

∂t(a+ t(x− a)) dt

Furthermore, by the chain rule

∂f

∂t(a+ t(x− a)) =

n∑

i=1

(xi − ai)∂f

∂xi(a+ t(x− a))

and hence

f(x) = f(a) +

n∑

i=1

(xi − ai)

∫ 1

0

∂f

∂xi(a+ t(x − a)) dt

︸︷︷︸=:gi(x)

.

In particular, gi(a) =∫ 1

0 (∂f/∂xi)(a) dt = (∂f/∂xi)(a).

Despite this factorization property, C∞-functions can be pretty “bad”. In fact itis a result of Borel that given a sequence (an)n of real numbers there is a functionf ∈ C∞(R) with f (n)(0) = an. This, however, is far from true for real analyticfunctions.

We are now in a position to prove Theorem 2.8. As announced, we translate theproblem to one in Rn and solve it there.

Proof. (Theorem 2.8). Let (U,ϕ) be a chart at p. Pick ε > 0 such that Cn2ε(ϕ(p)) ⊆ϕ(U). Now let g ∈ C∞(Rn) be as f in the proof of Lemma 2.9, i.e. g equals one onCnε/2(ϕ(p)) and g equals zero outside Cnε (ϕ(p)).

For f ∈ C∞(N), we have δ(f) = δ(f · (g ϕ)) by Lemma 2.10 where we agreethat g ϕ is extended to N\U by zero. This in turn equals δ(((f ϕ−1) · g) ϕ)where now (f ϕ−1) · g is a smooth function on Rn with support in Cnε (ϕ(p)).

Now, for every F ∈ C∞(Cn2ε(ϕ(p))) define α(F ) := δ((F · g) ϕ) which is aderivation of C∞(Cn2ε(ϕ(p))) at ϕ(p). For ease of notation, set a := ϕ(p). ApplyLemma 2.11 to F to get functions Gi ∈ C∞(Cn2ε(a)) such that

F (x) = F (a) +

n∑

i=1

(xi − ai)Gi(x).


Applying α yields

α(F ) = α(F (a) · 1) +n∑

i=1

α(x 7→ (xi − ai))Gi(a) +

n∑

i=1

0 · α(Gi)

=

n∑

i=1

α(x 7→ (xi − ai))︸︷︷︸ci

∂F

∂xi(a).

We therefore have

δ(f) = α(f ϕ−1) =

n∑

i=1

ci∂(f ϕ−1)

∂xiϕ(p)

= Dϕ(p)(f ϕ−1)(v) = (Dpf)(Dϕ(p)ϕ−1)(v)

which is the assertion.

Now, let N and M be manifolds and let f : N → M be a smooth map. Definef∗ : C∞(M) → C∞(N) by g 7→ g f . This is an R-algebra homomorphism. Givenp ∈ N we further define

f∗ : Derp(C∞(N)) → Derf(p)(C

∞(M)), δ 7→ δ f∗.

The map f∗ does indeed range in Derp(C∞(M)): Given δ ∈ Derp(C

∞(N)) andg1, g2 ∈ C∞(M), we have

f∗δ(g1g2) = δ(f∗(g1g2)) = δ((f∗g1)(f∗g2)) = δ((g1 f)(g2 f))

= (g1 f)(p)δ(g2 f) + (g2 f)(p)δ(g1 f)= g1(f(p))f∗(δ)(g2) + g2(f(p))f∗(δ)(g1).

Also, the following diagram commutes:

TpNDpf

//

∼=

Tf(p)M

∼=

Derp C∞(N)

f∗

// Derf(p) C∞(M)

In particular, the derivative of f can be defined without ever writing down an actualderivative in Rn. This algebraization of the derivative is used a lot in algebraicgeometry. It is also very useful when working with vector fields as we shall see later.However, both versions of the tangent space and the derivative are important.

2.3. The Tangent Bundle. In this section, we want to show how one can organizethe set of tangent vectors of a manifold M into a manifold TM on its own. Forinstance, this is essential to define smooth vector fields and plays a major rule inWhitney’s embedding theorem which we will see later. As a set, we have

TM =⋃

x∈M

x × TxM = (x, v) | x ∈M, v ∈ TxM.

The tangent bundle TM of a manifold M comes with the natural projection mapπ : TM →M, (x, v) 7→ x. For any subset U ⊆M we define

TU := π−1(U) = (x, v) | x ∈ U, v ∈ TxMand use these subsets of TM to define a topology as follows: Let (U,ϕ) be any chartof M . Then Dϕ : TU → ϕ(U)×Rm given by (x, v) 7→ (ϕ(x), Dxϕ(v)) is a bijection.We now declare a subset E ⊆ TM to be open if and only if for every chart (U,ϕ)of M the set Dϕ(E ∩ TU) ⊆ ϕ(U)× Rm is open.


Lemma 2.12. Retain the above notation. Then TM is a Hausdorff and secondcountable topological space. Also, given a chart (U,ϕ) of M the map Dϕ from TUto ϕ(U)× Rm is a homeomorphism, and π : TM →M is continuous and open.

Proof. First of all, we show that the set

E ⊆ TM | ∀(U,ϕ) chart of M : Dϕ(E ∩ TU) ⊆ ϕ(U)× Rm is opendefines a topology on TM . Note that for a chart (U,ϕ) of M the subset TU ⊆ TMis open. Hence so is TM since TM ∩TU = TU . Also, the empty set is open. As tofinite intersections, suppose that E1, . . . , En are open subsets of TM . Then

n⋂

i=1

Ei ∩ TU =

n⋂

i=1

(Ei ∩TU)

is open. Also, if (Eα)α∈A is a family of open subsets of TM then⋃

α∈A

Eα ∩ TU =⋃

α∈A

(Eα ∩ TU)

is open. The definition also readily implies that Dϕ : TU → ϕ(U) × Rm is ahomeomorphism for every chart (U,ϕ) of M .

With the above topology, TM is a Hausdorff space: Indeed, let (p1, v1) 6= (p2, v2)be distinct points in TM . If p1 6= p2, choose charts (U1, ϕ1) and (U2, ϕ2) at p1 andp2 respectively such that U1∩U2 = ∅. Then TU1 and TU2 are non-intersecting openneighbourhoods of (p1, v1) and (p2, v2) respectively. If p1 = p2 then v1 6= v2. Hence,if (U,ϕ) is a chart at p := p1 = p2 then (p, v1), (p, v2) ∈ TU ∼= ϕ(U) × Rm andwe may use the fact that Rm is a Hausdorff space to separate (p1, v1) and (p2, v2).Second countability is left to the reader as an exercise.

It remains to show that π : TM → M is continuous and open. Given an atlasA = (Uα, ϕα | α ∈ A of M we have M =

⋃α∈A Uα and it suffices to show that

π|Uα: TUα → Uα is continuous and open for every α ∈ A. This follows from the

fact the map pr1 in the following diagram is continous and open,

TUαDϕα

∼=//

π

ϕα(Uα)× Rm

pr1

Uα ϕα

∼= // ϕα(Uα),

because Dϕα and ϕα are homeomorphisms.

The proof of Lemma 2.12 in fact shows that if A = (Uα, ϕα | α ∈ A is anatlas on M then (TUα, Dϕα) | α ∈ A is a C0-atlas on TM . This atlas is in factsmooth.

Lemma 2.13. Retain the above notation. The atlas (TUα, Dϕα) | α ∈ A on TMis smooth.

Proof. We determine the coordinate transformations. Let α, β ∈ A. Then we haveTUα ∩ TUβ = T (Uα ∩ Uβ) and

T (Uα ∩ Uβ)Dϕα

vv

Dϕβ

((

ϕα(Uα ∩ Uβ)× Rm(ϕβϕ

−1α ,D(ϕβϕ

−1α ))

// ϕβ(Uα ∩ Uβ)× Rm .

In particular, the coordinate transformations are smooth.


As a consequence of the proof of Lemma 2.12 we also record that π : TM →Mhas constant rank m as in local coordinates it is given by a projection in Euclideanspace.

The smooth structure on TM allows to define smooth vector fields.

Definition 2.14. Let M be a manifold. A (continuous, Ck, smooth) tangent vectorfield on M is a (continuous, Ck, smooth) map X :M → TM with π X = idM .

It is an important question whether or not a manifold admits a nowhere vanishingcontinuous or smooth vector field as this potentially allows us to distinguish betweengiven manifolds. For instance, S1 clearly admits such a vector field. We shall seelater that in fact any smooth vector field on S2 has to have a zero somewhere. Asa third example, note that SO(3) admits a nowhere vanishing smooth vector field:Simply choose a non-zero tangent vector v at Id ∈ SO(3) and translate it aroundusing the derivatives of the smooth maps Lg : SO(3) → SO(3), x 7→ gx whereg ∈ SO(3). The reader is invited to think about the case of S3.

2.4. Whitney’s Embedding Theorem. In this section we prove that every (com-pact) manifold can be embedded into some Euclidean space. The strong version ofthis Theorem due to Whitney reads as follows.

Theorem. Let M be an m-manifold. Then M embeds into R2m.

In full generality, the dimension of the Euclidean target space cannot be reducedany further: For instance, the two-dimensional manifold P2 R cannot be embeddedinto R3 for orientability reasons. We shall prove the following version of the abovetheorem.

Theorem 2.15. Let M be a compact m-manifold. Then M embeds into R2m+1.

A byproduct of the proof of Theorem 2.15 is that in fact M can be immersedinto R2m. However, it requires a new idea to get rid of potential double points.

Proof. The first step of the proof constitutes in constructing an embedding of Minto a Euclidean space of some large dimension. To this end, recall Proposition 1.42by which it suffices to construct an immersion of M into Euclidean space which isa homeomorphism onto its image. We then work on reducing the dimension usingSard’s Theorem 1.43.

Given p ∈ M , let (Up, ϕp) be a chart at p with ϕp(p) = 0 and pick εp > 0 suchthat Cm3εp(0) ⊆ ϕp(Up). Further, given any ε > 0, let ψε : Rm → [0, 1] be a smoothfunction with ψ−1

ε (1) = Cmε (0) and ψε = 0 outside of Cm2ε(0). Now define

fp :M → R, x 7→ψεp ϕp(x) x ∈ Up

0 x 6∈ Up

which is smooth for every p ∈M . Using fp we further define

Fp :M → Rm, x 7→fp(x) · ϕp(x) x ∈ Up

0 x 6∈ Up.

Let Vp := ϕ−1p (Cmεp(0)) ⊆ Up which is open and contains p ∈M . SinceM is compact,

there are finitely many points p1, . . . , pl ∈ M such that M =⋃li=1 Vpi . Then the


following map is the announced injective immersion:

Ψ :M → Rml+l, x 7→ (Fp1 (x), . . . , Fpl(x), fp1 (x), . . . , fpl(x)).

Indeed, Ψ is of rank m everywhere: Let x ∈M and i ∈ 1, . . . , l such that x ∈ Vpi .Consider

Ψ :M → Rml+l = Rm× · · · × Rm× · · · × Rm︸︷︷︸l

× R× · · · × R

and denote by pi the projection of Rml+l onto the i-th Rm-factor. Then pi Ψagrees with ϕpi on a neighbourhood of x. As a consequence Ψ has (maximal) rankm at x ∈ M since ϕpi does. Note that in order to get an immersion from M toa Euclidean space we could have dropped the coordinates fp1 , . . . , fpl from thedefinition of Ψ. However, we need them to ensure injectivity: Let x, y ∈ M andi ∈ 1, . . . , l be such that x ∈ V pi = ϕ−1

pi (Cmεp(0)). First suppose that y ∈ V pi as

well and assume that Ψ(x) = Ψ(y). Since Fpi and ϕpi agree on V pi we concludethat ϕpi(x) = ϕpi(y) and hence x = y. Now suppose that y 6∈ V pi . Then fpi(y) < 1.But since fpi(x) = 1 this implies that Ψ(x) 6= Ψ(y).

The strong version of Whitney’s embedding theorem requires an additional ideain this step to deal with non-compactness.

As a second step, we now reduce the dimension of the target Euclidean spaceutilizing Sard’s Theorem 1.43. We have an embedding of M into some large Rn andconsider TM as a regular submanifold of TRn = Rn×Rn, and in fact

σ(M) = (p, v) ∈ TM | ‖v‖ = 1as a (2m−1)-dimensional regular submanifold of Rn×Rn. We now identify Rn−1 asx ∈ Rn | xn = 0 and let v ∈ Rn \Rn−1 with ‖v‖ = 1. Further, let pv : Rn → Rn−1

be the projection parallel to v given by the decomposition Rn = Rn−1 ⊕R v. Nowconsider the map f : σ(M) → Sn−1 given by (p, w) 7→ w and observe that pv|M isan immersion if and only if v 6∈ f(σM). If dim(σ(M)) = 2m−1 < n−1 = dimSn−1,then by Sard’s Theorem f(σM) has measure zero in Sn−1. Hence there is v ∈ Sn−1

with v 6∈ f(σM) for which consequently pv :M → Rn−1 is an immersion. Repeatingthe argument, one eventually gets an immersion of M to R2m.

As a third step, we adress the injectivity issue. In doing so, we loose one di-mension. The strong version of Whitney’s Theorem avoids with using a deeperunderstanding about how to get rid of potentially introduced double points. Con-sider

g : (M ×M)\∆(M) → Sn−1, (x, y) 7→ x− y

‖x− y‖which, using the immersion above, is smooth as the restriction of a smooth mapfrom Rn×Rn. Observe that pv|M is injective if and only if v 6∈ im(g): We areonly intersted in the following direction: Suppose that pv|M is not injective. Thenthere are x, y ∈ M with pv(x) = pv(y) and hence pv(x − y) = 0, i.e. x − y ∈ R v.Therefore g(x, y) ∈ ±v and hence v ∈ im(g). By Sard’s Theorem, the image ofg has measure zero in Sn−1 if 2m < n − 1. Putting everything together, we havethat if 2m < n− 1, the set im(f) ∪ im(g) has measure zero in Sn−1. Therefore wecan find v 6∈ im(f) ∪ im(g) for which pv is an injective immersion and hence anembedding by Proposition 1.42 since M is compact.

Whitney’s Theorem and Sard’s Theorem can be applied to obtain considerableinformation about a given manifold. For instance, suppose that M is a compactmanifold and that f : M → R is a smooth map all of whose values are regular.Then f−1(y) for y ∈ im(f) is a regular submanifold of M of lower dimension and


in fact M is built up from such lower-dimensional submanifolds in a certain way.However, a function f : M → R does not exist as for instance a point x ∈ M atwhich f assumes its maximum is critical. However, one can still insist on using thismethod and try to obtain more precise information about these critical values usingthe second-derivative of f . A Morse function is one all of whose second derivativesare non-degenerate. Combining Whitney’s and Sard’s Theorem one can show theexistence of Morse functions which can be used to analyze the structure of M . Forinstance, this yields a classification of all compact surfaces.

As a second remark, we state that although the dimension of the Euclideanspace in the strong version of Whitney’s theorem is optimal for some m, for instancepowers of two, for which projective space realizes the worst case, it is not in general.However, the general relation is a rather complicated one.

2.5. The Cotangent Bundle. In this section we introduce the cotangent bundleof a manifold in analogy to its tangent bundle and it turns out to be even more im-portant. First, we fix some notation. Given an R-vector space V , let V ∗ = Lin(V,R)denote its dual space. If T : V → W is a linear map of R-vector spaces V and W ,then T ∗ :W ∗ → V ∗ denotes its adjoint map, given by (T ∗λ)(v) = λ(Tv). Now, letM be a smooth manifold of dimension m. As in the case of the tangent bundle,cf. Section 2.3 we define a smooth structure on T∗M =

⋃x∈Mx × TxM

∗. Thisleads to the cotangent bundle T∗M

π−→ M with smooth, submersive π. The topol-ogy and the smooth structure on T∗M are defined using charts (U,ϕ) on M : LetT∗U = π−1(U). Then we have the map

T∗U → ϕ(U)× (Rm)∗, (x, λ) 7→ (ϕ(x), (Dϕ(x)ϕ−1)∗(λ)).

From here on, one proceeds as in Section 2.3.

Definition 2.16. Let M be a manifold. A (continuous, Ck, smooth) differential 1-form on M is a (continuous, Ck, smooth) map ω :M → T∗M such that πω = idM .

In a sense, differential 1-forms are more natural then vector fields: Let Ω1(M)be the vector space of smooth 1-forms on M . The derivative of smooth functionsgives rise to a natural map

d : C∞(M) → Ω1(M), f 7→ (df : x 7→ Dxf)

where by “naturality” we mean the following: Recall that a smooth map ψ : N →Mbetween manifolds induces an algebra homomorphism ψ∗ : C∞(M) → C∞(N) anda map ψ∗ : Ω1(M) → Ω1(N) given by (ψ∗ω)x(v) = ωψ(x)Dxψ(v). The constructiond is now a natural transformation between the functors C∞(−) and Ω1(−), i.e. thefollowing diagram commutes.

C∞(M)dM //

ψ∗

Ω1(M)

ψ∗

C∞(N)dN

// Ω1(N).

Also, note that Ω1(M) is not only a vector space but in fact a C∞(M)-module withmultiplicative structure given by (fω)x := f(x)ωx.

We now start to generalize calculus on Rm to smooth manifolds. For instance,what kind of object can be integrated over a manifold? Looking at the case of Rm

it seems like the answer should be “functions” and indeed one could try to definean integral of a function over a manifold by patching together integrals in chartcodomains but this would not be well-defined as the integral in Rm is not invariantunder diffeomorphisms. The right objects to integrate will turn out to be k-forms.


We can already see that a 1-form ω could be integrated over a one-dimensionalregular submanifold σ : [0, 1] →M of a manifold by defining

∫σω :=

∫ 1

0ωc(t)c(t) dt.

σ

2.6. Differential 1-forms in Local Coordinates. In order to give a meaningto the expressions that occur in e.g. Green’s formula in the introduction we nowexpress 1-forms in local coordinates. To this end, letM be a smoothm-manifold andω ∈ Ω1(M) a smooth 1-form. Further, let (U,ϕ) be any coordinate chart of M . OnRm we have the coordinate functions πi : Rm → R given by x = (x1, . . . , xm)T 7→ xi.These functions give rise to the 1-forms dπ1, . . . , dπm on Rm which at every pointx ∈ Rm give a basis of (TxRm)∗ = (Rm)∗. In fact, for x ∈ Rm and v ∈ Rm we have(dπi)xv = vi which is essentially due to the fact that the differential of a linear mapis the map itself. Now, since ϕ : U → ϕ(U) is a diffeomorphism onto its image, thepullback forms ϕ∗(dπ1), . . . , ϕ

∗(dπm) form a basis of TpM∗ at every point p ∈ U .To shorten the notation, we shall simply write dxi := ϕ∗(dπi) ∈ Ω1(U). We then

have ωx =∑n

i=1 ai(x)(dxi)x and the functions ai : U → R are smooth. We will alsowrite ω =

∑ni=1 ai dxi in the C∞(U)-module Ω1(U). Furthermore, if V ⊆ Rm is

open we will also consider V as a smooth manifold with the single chart (V, id) anduse the notation dxi ∈ Ω1(V ) for the 1-form obtained by using the chart (V, id).

3. Differential Forms and Integration on Manifolds

We have seen above that 1-forms can be integrated over one-dimensional man-ifolds in an invariant way. In this section we introduce k-forms which constitutethe natural objects to be integrated over k-dimensional manifolds. As a motivation,consider again Green’s formula from the introduction:

σ

D

∫

σ

ω =

∮

∂D

P dx+Q dy =

∫

D

(∂Q

∂x− ∂P

∂y

)dx dy.

If ω ∈ Ω1(R2) then ω = P dx+Q dy for smooth functions P,Q : R2 → R by theabove. Note that the expressions dx and dy now have a precise meaning, namelythey are 1-forms on R2, which is typically vague in calculus courses. The first twoterms in Green’s formula are now well-defined.

The remainder of this section in particular identifies (∂Q/∂x− ∂P/∂y) dx dyas a 2-form, the exterior derivative dω of ω, and defines intregration on manifoldsM with boundary of such forms. The pinnacle will be Stokes’ Theorem∫

∂M

ω =

∫

M

dω

which is arguably one of the most wonderful formulas in mathematics.

3.1. Alternating Forms on Vector Spaces. To implement the above programwe require several multilinear algebra notions. Let V be a finite-dimensional realvector space (neither finite-dimensionality nor real coefficients are required every-where but we shall not worry about these things here).

Definition 3.1. Retain the above notation. A multilinear k-form on V is a functionµ : V × · · · × V → R which is linear in each argument.


Lemma 3.2. Retain the above notation and let µ be a multilinear k-form on V .Then the following are equivalent.

(i) The form µ is zero whenever two arguments coincide.(ii) The form µ changes its sign whenever two arguments are interchanged.(iii) For all v1, . . . , vk ∈ V and σ ∈ Sk: µ(vσ(1), . . . , vσ(k)) = sign(σ)µ(v1, . . . , vk).

Based on Lemma 3.2 we make the following definition.

Definition 3.3. Retain the above notation. A multilinear, alternating k-form on Vis a multilinear form on V satisfying one of the equivalent properties of Lemma 3.2.

Proof. (Lemma 3.2). We show that (i) implies (ii): Let v1, . . . , vk ∈ V . Then wehave for all 1 ≤ i < j ≤ n:

0 = µ(v1, . . . , vi−1, vi + vj , vi+1, . . . , vj−1, vi + vj , vj+1, . . . , vn)

= µ(. . . , vi, . . . , vi, . . .) + µ(. . . , vi, . . . , vj , . . .)+

+ µ(. . . , vj , . . . , vi, . . .) + µ(. . . , vj , . . . , vj , . . .)

= µ(. . . , vi, . . . , vj , . . .) + µ(. . . , vj , . . . , vi, . . .).

Hence the assertion. Also, (ii) implies (iii): Rephrasing (ii), we have

µ(vτ(1), . . . , vτ(k)) = (−1) · µ(v1, . . . , vk)for every transposition τ ∈ Sk. Now use the fact that every permutation σ ∈ Skcan be written as a product of transpositions and that sign : Sk → ±1 is a grouphomomorphism. The implications (iii)⇒(ii) and (ii)⇒(i) are immediate.

We now organize the alternating forms on a vector space into a graded algebra:First of all, given k ∈ N0, let Λk(V ∗) denote the vector space of alternating k-formson V where we define Λ0(V ∗) := R. In particular, we have Λ1(V ∗) = V ∗. Nowconsider the graded vector space

Λ∗(V ∗) =⊕

k≥0

Λk(V ∗),

termed the exterior algebra of V ∗ because it admits the following multiplication,termed wedge product : Let α ∈ Λp(V ∗) and β ∈ Λq(V ∗). Define

(α ∧ β)(v1, . . . , vp+q) :=1

p!q!

∑

σ∈Sp+q

sign(σ)α(vσ(1) , . . . , vσ(p))β(vσ(p+1) , . . . , vσ(p+q))

=∑

σ∈Sp+q

σ(1)<···<σ(p)

σ(p+1)<···<σ(p+q)

sign(σ)α(vσ(1) , . . . , vσ(p))β(vσ(p+1), . . . , vσ(p+q))

where equality is left as an exercise. The special kind of permutation of 1, . . . , p+qthat occurs in the second expression is called (p, q)-shuffle for apparent reasons.

Proposition 3.4. Retain the above notation. Then Λ∗(V ∗) is an associative, gradedcommutative R-algebra.

Before we proceed to the proof of Proposition 3.4 several remarks are in order.

Remark 3.5. Let α ∈ Λp(V ∗) and β ∈ Λq(V ∗).

(i) If p = 0 then α ∧ β = α · β.(ii) The term graded-commutative means that β∧α = (−1)pqα∧β. In particular,

even order forms commute with any other form.


(iii) The definition of the wedge product may be motivated as follows: Given anmultilinear k-form µ on V there is a natural alternating k-form Aµ on Vassociated to µ, namely

Aµ(v1, . . . , vk) :=1

k!

∑

σ∈Sk

sign(σ)µ(vσ(1), . . . , vσ(k)).

It is immediate that Aµ is multilinear. To see that it is also alternating welet v1, . . . , vk ∈ V , τ ∈ Sk and compute

Aµ(vτ(1), . . . , vτ(k)) =1

k!

∑

σ

sign(σ) sign(σ)µ(vτσ(1), . . . , vτσ(k))

=1

k!

∑

σ∈Sk

sign(τ−1σ)µ(vσ(1), . . . , vσ(k))

= sign(τ)1

k!

∑

σ∈Sk

sign(σ)µ(vσ(1) , . . . , vσ(k))

= sign(τ)Aµ(v1, . . . , vk)

Now consider µ(v1, . . . , vp+a) := α(v1, . . . , vp)β(vp+1, . . . , vp+q) which ismultilinear. Then α ∧ β = (p+ q)!/(p!q!)Aµ.

(iv) So far, we have only defined the multiplication within Λ∗(V ∗) on pairs of pand q-forms. However, we may simply extend this definition linearly to thewhole of Λ∗(V ∗) realizing distributivity: Suppose α, β ∈ Λ∗(V ∗) are givenby α =

∑i αi and β =

∑j βj where αi ∈ Λi(V ∗) and βj ∈ Λj(V ∗) then

α ∧ β =∑

i,j αi ∧ βj . However, mixed products are rare in practice.

Proof. (Proposition 3.4). First consider associativity: Let α ∈ Λp(V ∗), β ∈ Λq(V ∗),γ ∈ Λr(V ∗) and v1, . . . , vp+q+r ∈ V . Then ((α ∧ β) ∧ γ)(v1, . . . , vp+q, . . . , vp+q+r)equals

1

(p+ q)!r!

∑

σ∈Sp+q+r

sign(σ)(α ∧ β)(vσ(1), . . . , vσ(p+q))γ(vσ(p+q+1), . . . , vσ(p+q+r))

which in turn equals

1

(p+ q)!p!q!r!

∑

σ∈Sp+q+r

∑

τ∈Sp+q

sign(σ) sign(τ) α(vστ(1), . . . , vστ(p))··β(vστ(p+1), . . . , vστ(p+q))·

·γ(vσ(p+q+1), . . . , vσ(p+q+r)).

Now, to every τ ∈ Sp+q we associate τ ∈ Sp+q+r by fixing p+ q + 1, . . . , p+ q + r:

τ(j) :=

τ(j) j ∈ 1, . . . , p+ qj j ∈ p+ q + 1, . . . , p+ q + r .

Then the above expression can be rewritten as

1

(p+ q)!p!q!r!

∑

σ∈Sp+q+r

∑

τ∈Sp+q

sign(στ )α(vστ (1), . . . , vστ(p))··β(vστ(p+1), . . . , vστ(p+q))·

·γ(vστ(p+q+1), . . . , vστ (p+q+r)).

Since the map Sp+q+r → Sp+q+r, σ 7→ στ is a bijection we may continue with

1

(p+ q)!p!q!r!

∑

τ∈Sp+q

∑

σ∈Sp+q+r

sign(σ)α(vσ(1) , . . . , vσ(p))··β(vσ(p+1), . . . , vσ(p+q))·

·γ(vσ(p+q+1), . . . , vσ(p+q+r)).


which equals1

p!q!r!

∑

σ∈Sp+q+r

sign(σ)α(vσ(1), . . . , vσ(p)) β(vσ(p+1), . . . , vσ(p+q))··γ(vσ(p+q+1), . . . , vσ(p+q+r)).

Computing α ∧ (β ∧ γ) in a similar way yields the same. This proves associativity.We now turn to proving that Λ∗(V ∗) is graded commutative: Let α ∈ Λp(V ∗),

β ∈ Λq(V ∗) and v1, . . . , vp+q ∈ V . Then

(α ∧ β)(v1, . . . , vp+q) =1

p!q!

∑

σ∈Sp+q

sign(σ)α(vσ(1) , . . . , vσ(p))β(vσ(p+1), . . . , vσ(p+q))

Again, we use that the map Sp+q → Sp+q, σ 7→ στ is a bijection for any τ ∈ Sp+q.Applying this to

τ :=

(1 · · · p p+ 1 · · · p+ q

q + 1 · · · q + p 1 · · · q

)∈ Sp+q

which has sign (−1)pq yields1

p!q!

∑

σ∈Sp+q

sign(στ)α(vστ(1) , . . . , vστ(p))β(vστ(p+1), . . . , vστ(p+q))

= (−1)pq1

p!q!

∑

σ∈Sp+q

sign(σ)β(vσ(1), . . . , vσ(q))α(vσ(q+1), . . . , vσ(q+p))

which is (−1)pq(β ∧ α)(v1, . . . , vp+q).

In order to get an impression of how to use graded commutativity, consider thefollowing. Let V be a real three-dimensional vector space with basis (e1, e2, e3)and let (e∗1, e

∗2, e

∗3) be the associated dual basis of V ∗ = Λ1(V ∗). Given ai, bi ∈ R

(i ∈ 1, 2, 3) we have the 1-forms a1e∗1 + a2e∗2 + a3e

∗3 and b1e

∗1 + b2e

∗2 + b3e

∗3. To

compute their wedge product, which is a 2-form, note that for two f, g ∈ V ∗ we havef ∧ f = 0 since f ∧ f = (−1)1·1f ∧ f and f ∧ g = −g ∧ f by graded commutativity.Therefore (a1e

∗1 + a2e

∗2 + a3e

∗3) ∧ (b1e

∗1 + b2e

∗2 + b3e

∗3) equals

a1b2e∗1 ∧ e∗2 + a1b3e

∗1 ∧ e∗3 + a2b1e

∗2 ∧ e∗1 + a2b3e

∗2 ∧ e∗3 + a3b1e

∗3 ∧ e∗1 + a3b2e

∗3 ∧ e∗2

= (a1b2 − a2b1)e∗1 ∧ e∗2 + (a1b3 − a3b1)e

∗1 ∧ e∗3 + (a2b3 − b2a3)e

∗2 ∧ e∗3.

In general, the evaluation of a k-form on a k-tuple of vectors is a determinant.

Proposition 3.6. Let V be a finite-dimensional real vector space. Furthermore, letφ1, . . . , φk ∈ V ∗ and v1, . . . , vk ∈ V . Then (φ1∧· · ·∧φk)(v1, . . . , vk) = det(φi(vj))i,j .

Proof. We argue by induction on k. If k = 1, then φ(v) = det(φ(v)). Now, assumek ≥ 2. We expand the determinant det(φi(vj))

ki,j=1 along the last row as

k∑

j=1

(−1)k−1(−1)j−1φk(vj) det

φ1(v1) · · · φ1(vj) · · · φ1(vk)...

......

φk−1(v1) · · · φk−1(vj) · · · φk−1(vk)

.

By the induction hypothesis, we thus have

det(φi(vj))ki,j=1 =

k∑

j=1

(−1)k−j(φ1 ∧ · · · ∧ φk−1)(v1, . . . , vj , . . . , vk)φk(vj)

which is exactly the definition of ((φ1 ∧ · · · ∧ φk−1) ∧ φk)(v1, . . . , vk) in terms of(k − 1, 1)-shuffles.

We shall now use Proposition 3.6 to determine bases of spaces of k-forms and inparticular the dimensions of the latter.


Corollary 3.7. Let V be an n-dimensional real vector space with basis (e1, . . . , en)and let (e∗1, . . . , e

∗n) denote the dual basis. Then (e∗j1 ∧ · · · ∧ e∗jk)1≤j1<···<jk≤n is a

basis of Λk(V ∗). In particular

dimΛk(V ∗) =

(nk

).

Proof. We show that the vectors of said tuple are linearly independent and thatΛk(V ∗) has at most the asserted dimension. To this end, let 1 ≤ j1 < · · · < jk ≤ nand 1 ≤ l1 < · · · < lk ≤ n. Then by Proposition 3.6,

(e∗j1 ∧ · · · ∧ e∗jk)(el1 , . . . , elk) = det

e∗j1(el1) · · · e∗j1(elk)

......

e∗jk(el1) · · · e∗jk(elk)

.

Assume that the above expression is non-zero. Then in particular the first rowis non-zero which implies j1 ∈ l1, . . . , lk by definition of the dual basis. Sim-ilarly, ji ∈ l1, . . . , lk for all i ∈ 1, . . . , k and therefore ji = li for all i ∈1, . . . , k by the ordering of the multi-indices. As a consequence, we have (e∗j1 ∧· · · ∧ e∗jk)(el1 , . . . , elk) = δj1l1 · · · δjklk . Suppose now that there is a linear relation

∑

j1<···<jk

aj1···jke∗j1 ∧ · · · ∧ e∗jk = 0 (aj1···jk ∈ R)

among the asserted basis vectors. Then evaluation on (el1 , . . . , elk) shows al1···lk = 0by the above.

Now we show that Λk(V ∗) has at most the asserted dimension. For this, it sufficesto show that the linear map

e : Λk(V ∗) → R(n

k), ω 7→ (ω(ej1 , . . . , ejk))1≤j1<···<jk≤n

is injective: Suppose e(ω) = 0 and let v1, . . . , vk ∈ V . It suffices to show thatω(v1, . . . , vk) = 0. Indeed, we compute

ω(v1, . . . , vk) = ω

∑

j

v1jej, . . . ,∑

j

vkjej

=∑

j1,...,jk

v1j1 · · · vkjkω(ej1 · · · ejk) =∑

j1,...,jkdistinct

v1j1 · · · vkjkω(ej1 · · · ejk)

The last sum vanishes since for every k-tuple j1, . . . , jk ∈ 1, . . . , n of distinctnumbers there is a permutation σ ∈ Sk such that jσ(1) < · · · < jσ(k) and thereforeω(ej1 , . . . , ejk) = sign(σ)ω(ejσ(1)

, . . . , ejσ(k)) which vanishes by assumption.

Recall that binomial coefficients are the entries of Pascal’s triangle. In particular,Λk(V ∗) = 0 for k > dim V and Λk(V ∗) = ΛdimV−k(V ∗) for k ∈ 0, . . . , dimV .

1

1 2 1

1 3 3 1

...

Remark 3.8. Before organizing k-forms on a manifold into a bundle we remark oncontravariance and coordinate-free definitions of the determinant of a matrix andthe cross-ratio of four lines.


(i) Let V,W be vector spaces and let T : V →W be a linear map. Further, letT ∗ : W ∗ → V ∗ denote the dual map of T . Then there is an induced mapΛkT ∗ : Λk(W ∗) → Λk(V ∗) defined by

ΛkT ∗(α)(v1, . . . , vk) := α(Tv1, . . . , T vk)

where α ∈ Λk(W ∗) and v1, . . . , vk ∈ V .(ii) Let V be an n-dimensional real vector space and let T ∈ End(V ). Then

ΛnT ∗ ∈ End(Λn(V ∗)). Since Λn(V ∗) is one-dimensional, so is its endomor-phism algebra. Furthermore, the endomorphism algebra of a one-dimensionalreal vector space E is, in contrast to the space itself, canonically isomorphicto R via R → End(E), λ 7→ λ Id. Therefore ΛnT ∗ is canonically associatedto a real number. It is an exercise to show that this number is detT .

(iii) To further illustrate the importance of the fact that the endomorphismalgebra of a one-dimensional real vector space is canonically isomorphic toR consider the following: In geometry, one defines the cross-ratio of fournon-zero complex numbers z1, z2, z3, z4 by

(z1, z2; z3, z4) :=(z1 − z3)(z2 − z4)

(z2 − z3)(z2 − z4)

Now, identify C as a two-dimensional real vector space and let l1, l2, l3 andl4 be the lines through the origin associated to z1, z2, z3 and z4 respectively.It is a fact that for a two-dimensional vector space V its general linear groupGL(V ) acts transitively on the set of lines through the origin. It even actstransitively on the set of triples of lines through the origin. Hence thereis no non-trivial linear invariant of such lines. However, it does not acttransitively on quadruples of lines through the origin and the cross-ratioprovides a linear invariant for such quadruples. Now, consider the mapϕ3 : l1 → l2 given by moving parallel to l2 towards l3 and then parallel tol1 towards l2. Similarly, define ϕ4 : l1 → l2. Then ϕ−1

4 ϕ3 ∈ GL(l1) ∼= R∗ isthe cross-ratio.

l1

l4l3l2

b

3.2. Differential Forms on Manifolds. Previously we have introduced the spaceof k-forms Λk(V ∗) of a real finite-dimensional vector space V . We now apply thisdefinition in the context of manifolds to organize the k-forms on the tangent spacesof a manifold M into a bundle: Define

Λk(M) :=⋃

x∈M

x × Λk(TxM∗).

Let π : Λk(M) → M , (x, ω) 7→ x denote the projection onto the first factor ofΛk(M). As in the case of the tangent and cotangent bundle we equip Λk(M) with thestructure of a smooth manifold such that π becomes a smooth map which we outlinenow: Given a coordinate chart (U,ϕ) on M and x ∈ M , we have an isomorphism


of vector spaces Dϕ(x)ϕ−1 : Rm → TxM which induces an isomorphism

Λk((Dϕ(x)ϕ−1)∗) : Λk((TxM)∗) → Λk((Rm)∗).

Define the topology and smooth structure on Λk as before using the bijections

π−1(U) → ϕ(U)× Λk((Rm)∗), (x, ω) 7→ (ϕ(x),Λk(Dϕ(x)ϕ−1)∗(ω)).

Definition 3.9. Let M be a manifold. A differential k-form on M is a map ω :M →Λk(M) such that π ω = idM .

In other words, a differential k-form assigns an alternating k-form ωx on thetangent space TxM to every point x ∈M , denoted by ωx : TxM ×· · ·×TxM → R.Notice that despite the name, there is no regularity assumption in the definitionof differential k-form. As a map between smooth manifolds, it can be measurable,continuous, Ck, smooth and so on. We denote by Γ(Λk(M)) the set of differentialk-forms on M and by Ωk(M) the set of smooth k-forms on M . The latter is goingto play a fundamental role in what is to follow. Now, we generalize the structure ofsets of alternating forms to the context of manifolds:

(i) (Vector space). Let α, β ∈ Γ(Λk(M)) and let µ, ν ∈ R. Define µα + νβ ∈Γ(Λk(M)) by (µα+ νβ)x := µαx + νβx for all x ∈M .

(ii) (Module). Let f : M → R be a function and α ∈ Γ(Λk(M)). We definefα ∈ Γ(Λk(M)) by (fα)x := f(x)αx.

(iii) (Algebra). Let α ∈ Γ(Λp(M)) and β ∈ Γ(Λq(M)). Then we define α ∧ β ∈Γ(Λp+q(M)) by (α ∧ β)x := αx ∧ βx for all x ∈ M to the effect thatΓ(Λ∗(M)) :=

⊕k≥0 Γ(Λ

k(M)) becomes an associative graded-commutativeR-algebra.

A crucial feature of k-forms on a manifold is that they can be pulled back fromone manifold to another via a smooth map. This is the basis of the “invariantcalculus” on manifolds which is to follow.

Definition 3.10. Let M and N be manifolds and let f : N →M be a smooth map.Further, let ω ∈ Γ(Λk(M)). Define f∗ω ∈ Γ(Λk(N)) by

f∗(ω)x(v1, . . . , vk) := ωf(x)(Dxf(v1), . . . , Dxf(vk))

for all x ∈ N and v1, . . . , vk ∈ TxN .

The pullback operation behaves well with respect to the algebra structures.

Proposition 3.11. Let M,N and P be manifolds and let f : N →M and g :M → Pbe smooth maps. Then

(i) f∗ : Γ(Λk(M)) → Γ(Λk(N)) is R-linear,(ii) for ω1 ∈ Γ(Λp(M)) and ω2 ∈ Γ(Λq(M)): f∗(ω1 ∧ ω2) = f∗(ω1) ∧ f∗(ω2),(iii) for g :M → R and ω ∈ Γ(Λk(M)) we have f∗(gω) = (g f)f∗ω, and(iv) (g f)∗(ω) = f∗g∗(ω).

The last part of Proposition 3.11 implies that the functor which to a manifoldassociates its space of k-forms is contravariant.


Proof. (Proposition 3.11). We only prove (ii) and (iv). For (ii), let x ∈ N andv1, . . . , vp+q ∈ TxN . Then

f∗(ω1 ∧ ω2)x(v1, . . . , vp+q) = (ω1 ∧ ω2)f(x)(Dxfv1, . . . , Dxfvp+q)

= (ω1,f(x) ∧ ω2,f(x))(Dxfv1, . . . , Dxfvp+q)

=1

p!q!

∑

σ∈Sp+q

sign(σ)ω1,f(x)(Dxfvσ(1), . . . , Dxfvσ(p))

·ω2,f(x)(Dxfvσ(p+1), . . . , Dxfvσ(p+q))

= (f∗ω1)x ∧ (f∗ω2)x(v1, . . . , vp+q)

= (f∗ω1 ∧ f∗ω2)x(v1, . . . , vp+q)

which is the assertion of part (ii). Part (iv) is yet another incarnation of the chainrule. Given ω ∈ Γ(Λk(P )) and v1, . . . , vk ∈ TxN we compute

(g f)∗(ω)x(v1, . . . , vk) = ωgf(x)(Dx(gf)v1, . . . , Dx(gf)vk)

= ωgf(x)(Df(x)gDxf(v1), . . . , Df(x)gDxf(vk))

= g∗ωf(x)(Dxfv1, . . . , Dxfvk)

= f∗g∗ωx(v1, . . . , vk)

In order to deal with smooth k-forms we are going to need to express them in localcoordinates. To this end, recall that on Rm we have the global coordinate functionsπi : R

m → R given by x 7→ xi which provide globally defined 1-forms dπ1, . . . , dπm.For a coordinate chart (U,ϕ) on a manifold M we define dxi := ϕ∗(dπi). Thenfor every p ∈ U , the 1-forms (dx1)p, . . . , (dxm)p form a basis of (TpM)∗. Let nowω ∈ Γ(Λk(M)). Then by Corollary 3.7 we have for every p ∈ U that ωp can bewritten as

ωp =∑

i1<···<ik

ai1,...,ik(p)(dxi1 )p ∧ · · · ∧ (dxik )p

in a unique way. In terms of the algebra structure defined above, this reads

ω|U =∑

i1<···<ik

ai1,...,ik dxi1 ∧ · · · ∧ dxik .

One often writes I to refer to a multi-index i1 < · · · < ik and defines aI := ai1,...,ikas well as dxI := dxi1 ∧ · · · ∧ dxik in which case the above is condensed to

ω|U =∑

I

aI dxI

This expression of a differential k-form in local coordinates allows one to checkthe smoothness of it in the following way.

Lemma 3.12. Let M be a manifold and let ω ∈ Γ(Λk(M)). Then ω is smooth if andonly if in every coordinate chart (U,ϕ) all the functions aI : U → R in the localcoordinate expression of ω with respect to (U,ϕ) are smooth.

The proof of Lemma 3.12 is left as an exercise. It requires unravelling the defi-nitions of smoothness and smooth structure on Λk(M); however, once this is done,the statement is rather tautological. Using Lemma 3.12 we can prove the following.

Proposition 3.13. Let M and N be a manifolds and let f : N → M be a smoothmap. Then the following hold.

(i) The exterior product of smooth k-forms on M is smooth, and(ii) the pullback via f of a smooth k-form on M is a smooth k-form on N .


Proof. For (i), let α ∈ Ωp(M) and β ∈ Ωq(M). Further, let (U,ϕ) be a coordinatechart on M . Writing α and β in local coordinates with respect to (U,ϕ) we haveα =

∑I aIdxI and β =

∑J bJdxJ for smooth functions aI , bJ :M → R. Therefore

α ∧ β =∑

I,J

aIbJ dxI ∧ dxJ =∑

I∩J 6=∅

aIbJ(±1) dxQ

where Q is the ordered multi-index corresponding to I ∪ J . The claim now followsfrom the fact that the functions U → R given by x 7→ aI(x)bJ (x) are smooth.

For part (ii), let ω ∈ Ωk(M). Further, let (V, ψ) and (U,ϕ) be charts on N andM respectively such that f(V ) ⊆ U . Then in (U,ϕ)-local coordinates we have

ω|U =∑

I

aI dxI =∑

i1<···<ik

ai1,...,ik dxi1 ∧ · · · ∧ dxik .

Therefore f∗ω =∑

i1<···<ii(ai1,...,ik f)f∗(dxi1 ) ∧ · · · ∧ f∗(dxik ). Since the forms

f∗(dxi1 ) are smooth as pullbacks of 1-forms, part (i) implies the assertion.We could also have determined a local coordinate expression for the f∗(dxi).

Namely,

f∗(dxi) = f∗(ϕ∗(dπi)) = (ϕf)∗(dπi) = (ϕfψ−1ψ)∗(dπi) = ψ∗(ϕfψ−1)∗(dπi)

Now, let F : ψ(V ) → ϕ(U) denote the map ϕ f ψ−1. If further we denote bypj : Rn → R the coordinate functions on Rn and by dp1, . . . , dpn the associated1-forms then with dyi := ψ∗(dpj) we have

F ∗(dπi)y =n∑

j=1

∂Fi∂yj

(y) dpj

and therefore

f∗(dxi)p =n∑

j=1

∂Fi∂yj

(ψ(p)) dyj

which implies the assertion.

We now compute the pullback of a differential form in a particularly importantcase which will reappear later on: Let U, V ⊆ Rm be open and let f : U → V besmooth. A top form ω ∈ Ωm(V ) can be written as ω = adx1 ∧ · · · ∧ dxm wherea : V → R is a smooth function. To compute f∗ω, note that for dimension reasonswe already know that f∗ω = g dx1∧· · ·∧dxm for some smooth function g : U → R.To determine g(p) for p ∈ U we just need to evaluate (f∗ω)p on the standard basis(e1, . . . , em) since dx1 ∧ · · · ∧ dxm(e1, . . . , em) = 1. By definition:

(f∗ω)p(e1, . . . , em) = ωf(p)(Dpf(e1), . . . , Dpf(em))

= a(f(p))(dx1 ∧ · · · ∧ dxm)f(p)(Dpf(e1), . . . , Dpf(em))

= a(f(p))((dx1)f(p) ∧ · · · ∧ (dxm)f(p))(Dpf(e1), . . . , Dpf(em))

= a(f(p)) det((dxi)f(p)(Dpf(ej)))i,j

= a(f(p)) det(e∗i (Dpf(ej))i,j

= a(f(p)) det(Dpf).

Overall, we have (f∗ω)p = a(f(p)) det(Dpf) dx1 ∧ · · · ∧ dxm. This computationcan be generalized to k-forms in which case the coefficients appearing in f∗ω aredeterminants of minors of Dpf .


3.3. Partition of Unity. In this section we construct the most powerful tool forstudying smooth manifolds, namely partition of unity. It allows one to constructvarious global objects out of locally defined ones, such as a Riemannian metricand the integral of a suitable differential form over a manifold, denotes

∫M ω. We

demonstrate the idea in the case of a Riemannian metric which is a smooth choice ofscalar products (〈−,−〉x)x∈M on the tangent spaces TxM of a manifold M . Usingthese we can define the length of a C1-path c : [0, 1] →M by

l(c) :=

∫ 1

0

‖c(t)‖c(t) dtc(t)b

bb

c(t)

yx

This in turn allows us to define the Riemannian distance of points x, y ∈ Massociated to the Riemannian metric by

d(x, y) := infl(c) | c : [0, 1] → M C1-path with c(0) = x and c(1) = y.The problem lies in defining the Riemannian metric in the first place. For instance,if (U,ϕ) is a chart of M at x ∈ M one could define 〈−,−〉x as the pullback of thestandard scalar product on Rm = DxϕTxM . This works well as long as there isonly one chart involved. However, if a point x ∈ M lies in two charts domains Uand V we cannot guarantee that the scalar products so defined on TxM coincide.

U V

bx

However, we may take a convex linear combination. For instance, if ϕU : M → Rand ϕV : M → R are functions which vanish outside U and V respectively andwhose sum at x is one, then ϕU (x)〈−,−〉x,U + ϕV 〈−,−〉x,V is a scalar product onTxM . This method works as long as there are only finitely many chart domainscontaining x. In the following we make these difficulties precise and resolve them.

Lemma 3.14. Let M be a topological manifold with a maximal atlas. Further, letVα | α ∈ A be a cover of M by open sets. Then there is an at most countablefamily of charts (Ui, ϕi) | i ∈ S with

(i) Ui | i ∈ S being a locally finite cover of M refining Vα | α ∈ A, and(ii) ϕi(Ui) ⊆ Cm3ε(0) as well as

⋃i∈S Vi =M where Vi = ϕ−1

i (Cmε (0)).If M is a smooth manifold with atlas the then the charts can be chosen to bediffeomorphisms.

We recall the terminology appearing in Lemma 3.14. A cover Ui | i ∈ S of atopological space M is locally finite if every point in M has a neighbourhood whichintersects only finitely many Ui (i ∈ S). It refines another cover Vα | α ∈ A if forevery i ∈ S there is α ∈ A such that Ui ⊆ Vα.

As an example, consider the real line R and the cover (−∞, n) | n ∈ Z.Then (n − 2, n) | n ∈ Z is a locally finite subcover refining the former since(n− 2, n) ⊂ (−∞, n) for all n ∈ Z and every point in R is contained in at most twoelements of (n− 2, n) | n ∈ Z. Note, however, that the refining step is necessary:There is no locally finite subcover of (−∞, n) | n ∈ Z itself.

Proof. (Lemma 3.14). We may assume that M is connected and non-compact. LetPi | i ∈ N be a countable basis of open sets such that P i is compact for everyi ∈ N which exists for the following reason: If B = Bi | i ∈ N is a basis thenBc := B ∈ B | B is compact works: It is indeed a basis: Since M is locally


compact Hausdorff, every Bi is a union of open sets with compact closure andevery open set with compact closure is a union of elements of B which hence liein Bc. Now, we define an exhaustion of M by compact sets (Kn)n∈N with nicelybehaved interiors and closures which will aid in definining the asserted locally finiterefinement of Vα | α ∈ A:

Set K1 := P 1. By compactness and the fact that Pi | i ∈ N is a basis there isj ∈ N such that P1∪· · ·∪Pj ⊇ K1 = P 1. In fact, we must have j ≥ 2 since there areno non-empty open compact subsets M by the assumption that it be connected andnon-compact. Let r1 := minP 1 ⊆ P1 ∪ · · · ∪Pj ≥ 2 and set K2 := P 1 ∪ · · · ∪P r1 .Iterate this process to defineKj := P 1∪· · ·∪P rj−1 . ThenKj−1 ⊆ P1∪· · ·∪Prj−1 butKj−1 6⊆ P1 ∪ · · · ∪Pl for any l < rj−1. Also, observe Kj ⊇ P1 ∪ · · · ∪ Prj−1 ⊇ Kj−1.Therefore, Ki+2\Ki−1 is an open set which contains the compact set Ki+1\Ki.

Ki−1

Ki

Ki+1

Ki+2

Now consider Ki+2\Ki−1 ∩ Vα: For every p ∈ Ki+2\Ki−1 ∩ Vα let (U ip, ϕip) be a

chart at p contained in Ki+2\Ki−1 ∩ Vα and such that ϕip(Uip) = Cm3ε(0). Then

set V ip := (ϕip)−1(Cmε (0)). For fixed i ∈ N, consider the set of all charts (U ip, ϕ

ip)

obtained in this way by varying α over A. Since⋃α∈A Vα =M we have in particular

that the sets V ip so obtained cover Ki+1\Ki. By compactness we can thereforefind a finite set Si ⊂ Ki+2\Ki−1 such that

⋃p∈Si

V ip ⊇ Ki+1\Ki. Now consider thecollection (U ip, ϕip) | i ∈ N, p ∈ Si. First of all, the U ip of this collection form acountable refinement of Vα | α ∈ A. Also,

⋃i∈N,p∈Si

V ip =M by construction. Asto local finiteness, let p ∈ M and i ∈ N such that p ∈ Ki−1. Observe that for allj ≥ i we have Kj+2\Kj−1 ∩ Ki−1 = ∅ and conclude by noting that for all j ≥ i andall q ∈ Sj we have U jq ⊆ Kj+2\Kj−1 whence U jq ∩ Ki−1 = ∅. That is, only finitelymany domains of (U ip, ϕip) | i ∈ N, p ∈ Si intersect Ki−1.

The statement about the smooth case is immediate from the above.

As a corollary to Lemma 3.14 we have the following.

Theorem 3.15. Let M be a smooth manifold and let Vα | α ∈ A be an open coverof M . Then there is a countable family fi | i ∈ F of functions on M such that

(i) fi ≥ 0, fi ∈ C∞(M) for all i ∈ F with compact support,(ii) supp fi | i ∈ F is a locally finite cover of M refining Vα | α ∈ A, and(iii)

∑i∈F fi(x) = 1 for all x ∈M .

On the real line, a partition of unity may look as follows.

. . . . . .

Proof. (Theorem 3.15). Let (Ui, ϕ)i | i ∈ I be a covering as in Lemma 3.14. Now,let g : Rm → R with 0 ≤ g(x) ≤ 1 for all x ∈ Rm be smooth such that g ≡ 1 on


Cmε (0) and g ≡ 0 outside Cm2ε(0). Then set

gi(x) :=

g(ϕi(x)) x ∈ Ui

0 x 6∈ Ui

Then∑i∈I gi : M → R is well-defined, smooth and strictly positive everywhere.

We may thus set fi := gi/(∑

j∈I gj).

3.4. Orientation. In this section we the introduce the last notion needed to defineexpressions of the form

∫Mω where M is a manifold and ω is a top-dimensional

form on M , i.e. ω ∈ Ωm(M) where m = dimM . Namely, M has to be orientable.First of all, we recall the notion of orientation on a finite-dimensional real vec-tor space V . We remark that interestingly this notion does depend on the fieldof scalars. Two bases (e1, . . . , en) and (f1, . . . , fn) of V are defined to be equiva-lent if the change of basis matrix A ∈ GL(n,R) from (e1, . . . , en) to (f1, . . . , fn)whose coefficients are given by fi =

∑nj=1 ajiei (i ∈ 1, . . . , n) has positive de-

terminant. This defines an equivalence relation on the set of bases of V sinceGL+(n,R) := A ∈ GL(n,R) | detA > 0 is a subgroup of GL(n,R). An orienta-tion on V is a choice of an equivalence class of bases. Hence there are two possibleorientations on a given finite-dimensional real vector space. For instance, on R2,the standard basis (e1, e2) is typically said to define the positive orientation and(e1,−e2) is said to define the negative orientation. The fact that these bases definedifferent orientations is also reflected by the fact that they cannot be transformedinto each other continuously while maintaing the property of forming a basis. Thisin turn comes from the fact that GL(n,R) has exactly two connected components.

For a smooth manifold M , an orientation would be a consistent, continuouschoice of an orientation on TpM for every p ∈M . However, this is not always pos-sible. The lowest dimension in which it is not is two. Indeed, P2 R is not orientable:Think of P2 R as the northern hemisphere of a sphere with opposite points on theequator identified. Then the image in P2 R of the strip depicted in the image belowis diffeomorphic to a Möbius strip within P2 R.

b r

r b

Hence P2 R cannot be orientable. In fact, using the theory to be developed in thissection, it is an exercise to show that PnR is orientable if and only if n is odd.

Definition 3.16. Let M be a smooth manifold.(i) An atlas A on M is oriented if for any two overlapping charts (Uα, ϕα)

and (Uβ , ϕβ) in A the coordinate transformation θβα : ϕα(Uα ∩ Uβ) →ϕβ(Uα ∩Uβ) has the property that detDxθβα > 0 for all x ∈ ϕα(Uα ∩Uβ).

(ii) The manifold M is orientable if it admits an oriented atlas.

Given an oriented atlas A on a manifold M we obtain for every p ∈ M a well-defined orientation on TpM by stipulating that for every (U,ϕ) ∈ A with p ∈ Uthe map Dpϕ : TpM → Rm is orientation-preserving; here we put the standardpositive orientation on Rm.

There is a characterization of orientability of manifolds in terms of certain dif-ferential forms which better allows for computations. For this, recall that for anm-dimensional real vector space V its space of alternating m-forms ΛmV ∗ is one-dimensional. However, the bundle of alternating m-forms over a manifold behavesvery differently from the trivial bundle with fiber R whose smooth sections are justthe smooth functions on M in the sense that the latter admits constants whereasthe first one not necessarily does. Hence the following definition makes sense.


Definition 3.17. Let M be a smooth manifold. A volume form on M is a smoothdifferential m-form ω on M which is nowhere vanishing, i.e. ωx 6= 0 for all x ∈M .

From the remark above it is clear that if ω is a volume form on M then any othervolume form on M is of the form f · ω where f ∈ C∞(M) is nowhere vanishing.The term volume form is due to the fact that ωx(e1, . . . , en) may be interpreted asthe volume of the parallelepiped spanned by the vectors (e1, . . . , en).

Proposition 3.18. Let M be a smooth manifold. Then M is orientable if and onlyif it admits a volume form.

Proof. Suppose first that M admits a volume form ω. Let A be an atlas of M . Wemodify A into a new atlas A′ which is oriented. First of all, passing to connectedcomponents and restricting charts we may assume that all chart domains of A areconnected. Now, given (U,ϕ) ∈ A consider (ϕ−1)∗(ω) = a dx1 ∧ · · · ∧ dxm wherea : ϕ(U) → R is a nowhere vanishing smooth function. Then by connectedness ofU , the function a has constant sign. If it has positive sign we declare (U,ϕ) to bepart of the new atlas A′. If not, set (U, s1 ϕ) to be in A′ where s1 : Rm → Rm

is given by (x1, . . . , xm) 7→ (−x1, x2, . . . , xm); in fact any function whose derivativehas negative determinant everywhere would do: If we set ϕ′ := s1 ϕ then

(ϕ′−1)∗(ω) = (s−11 )∗(ϕ−1)∗(ω) = (s−1

1 )∗(a dx1 ∧ · · · ∧ dxm) =

= −a(−y1, y2, . . . , ym) dy1 ∧ · · · ∧ dym = b dy1 ∧ · · · ∧ dymon s1(ϕ(U)) where b is of positive sign. We claim that A′ so defined is an orientedatlas on M . It is an atlas simply because its chart domains cover M . To see thatit is oriented, let (U,ϕ) and (V, ψ) in A′ be overlapping charts. Then (ϕ−1)∗(ω) =a dx1 ∧ · · · ∧ dxm and (ψ−1)∗(ω) = b dx1 ∧ · · · ∧ dxm where a and b are strictlypositive functions on ϕ(U ∩ V ) and ψ(U ∩ V ) respectively. Furthermore, since thetriangle

U ∩ V

ϕ(U ∩ V )

ϕ−1

99rrrrrrrrrr

ψ(U ∩ V )ϕψ−1

oo

ψ−1

ee

commutes we have (ϕ ψ−1)∗(ϕ−1)∗ω = (ψ−1)∗ω and hence

(ϕ ψ−1)∗(a dx1 ∧ · · · ∧ dxm) = b dx1 ∧ · · · ∧ dxm.However, we know from before that b(p) = a(ϕ ψ−1(p)) detDp(ϕ ψ−1) for allp ∈ ψ(U ∩ V ). Hence, since a and b are positive, we deduce that detDp(ϕ ψ−1) ispositive forall p ∈ ψ(U ∩ V ). That is, A′ is oriented.

Conversely, suppose that A is an oriented atlas on M . The idea on how toconstruct a volume form is to pullback volume forms from chart codomains and tosum the forms so obtained using a partition of unity: Let (Ui, ϕi)i∈I be a locallyfinite refinement of A and let fi | i ∈ I be a partition of unity subordinate to it,i.e. supp fi ⊂ Ui, fi ∈ C∞(M), 0 ≤ fi ≤ 1 and

∑i∈I fi(x) = 1 for all x ∈ M . For

every i ∈ I, define ωi ∈ Ωm(M) by

(ωi)p :=

fi(p)ϕ

∗i (dx1 ∧ · · · ∧ dxm) x ∈ Ui

0 x 6∈ Ui

Now define ω :=∑

i∈I ωi ∈ Ωm(M). Then ω is a volume form on M : Given p ∈M ,let k ∈ I be such that p ∈ Uk. Then ωp equals the finite sum

∑p∈Ui∩Uk

(ωi)p. OnUi ∩ Uk we can express ωi in terms of ωk as follows: As before, we have

(ϕi)∗(dx1 ∧ · · · ∧ dxm)p = (ϕk)

∗((ϕiϕ−1k )∗(dx1 ∧ · · · ∧ dxm))p =

= detDϕk(p)(ϕiϕ−1k )(ϕ∗

k(dx1 ∧ · · · ∧ dxm))p.


Therefore we have

ωp =∑

p∈Ui∩Uk

fi(p) detDϕk(p)(ϕi ϕ−1k )(ϕ∗

k(dx1 ∧ · · · ∧ dxm))p

Now observe that since ϕ∗k(dx1 ∧ · · · ∧ dxm)p 6= 0 we only need to show that the

coefficients in the above sum are non-zero. Indeed, since A is oriented we have∑

p∈Ui∩Uk

fi(p) detDϕk(p)(ϕi ϕ−1k ) ≥ min

p∈Ui∩Uk

detDϕk(p)(ϕi ϕ−1k )

∑

p∈Ui∩Uk

fi(p)

︸︷︷︸1

.

Hence the assertion.

Note that the argument of the second part of the proof of Proposition 3.18 worksfor any manifold up to the point of proving that the coefficients do not vanish, butthey will for non-orientable manifolds.

Given a volume form ω on a manifold M , we get a consistent choice of orientationof each TpM (p ∈ M) by saying that a tuple (e1, . . . , em) of vectors in TpM ispositively oriented if ωp(e1, . . . , em) > 0.

3.5. Integrating Smooth Compactly Supported Forms And More. Let Mbe a smooth manifold with an oriented atlas A and let Ωpc(M) denote the spanin Ωp(M) of all smooth p-forms with compact support; recall that for ω ∈ Ωp(M)

we set supp(ω) = x ∈M | ωx 6= 0. Now, let (U,ϕ) ∈ A. We define a linear formI(U,ϕ) : Ω

mc (U) → R, where m = dimM , by

I(U,ϕ)(ω) :=

∫

Rm

a(x1, . . . , xm) dµ(x1) · · · dµ(xm)

using a dx1∧· · ·∧dxm = (ϕ−1)∗(ω) ∈ Ωmc (ϕ(U)), where a ∈ C∞c (Rm) has compact

support contained in ϕ(U), and the Lebesgue measure, which can be replaced bythe Jordan content as long as ω is smooth. The orientability on M is key in provingthe following compatibility lemma.

Lemma 3.19. Let M be a manifold with an oriented atlas A. Further, let (U,ϕ)and (V, ψ) be charts, and ω ∈ Ωmc (U ∩ V ). Then I(U,ϕ)(ω) = I(V,ψ)(ω).

Proof. Write (ϕ−1)∗(ω) = a dx1 ∧ · · · ∧ dxm and (ψ−1)∗(ω) = b dx1 ∧ · · · ∧ dxm.Utilizing that ϕ ψ−1 : ψ(U ∩ V ) → ϕ(U ∩ V ) is a diffeomorphism between opensubsets of Rm and the calculation at the end of Section 3.2 we conclude b(x) =a(ϕ ψ−1(x)) detDx(ϕ ψ−1). We therefore have

I(V,ψ)(ω) =

∫

Rm

b(x) dµ(x) =

∫

Rm

a(ϕ ψ−1(x)) detDx(ϕ ψ−1) dµ(x).

Since A is oriented we have detDx(ϕ ψ−1) > 0 and hence, by the classical changeof variables formula with y := ϕ ψ−1(x), we may continue the above with

=

∫

Rm

a(y) dµ(y) = I(U,ϕ)(w).

We now extend the definition of integral to the whole manifold: Let (Ui, ϕi, fi)i∈N

be such that the (Ui, ϕi) (i ∈ N) are contained in the oriented atlas A and forma locally finite covering, and (fi)i∈N is a partition of unity subordinate to (Ui)i∈N.Now given ω ∈ Ωmc (M) we set

∫

M

ω :=∑

i∈N

I(Ui,ϕi)(fiω).


Observe that for every i ∈ N the form fiω lies in Ωmc (Ui) and that the above sumis finite as only finitely many Ui intersect the compact set supp(ω). We proceed byshowing that the above sum is independent of the involved choices: If (Vj , ψj , gj)j∈N

is another choice as above we may write fiω =∑j∈N gjfiω and therefore

I(Ui,ϕi)(fiω) =∑

j

I(Ui,ϕi)(gjfiω) =∑

j∈N

I(Vj ,ψj)(gjfiω)

where we have used that supp(gjfiω) ⊆ Ui ∩ Vj . Consequently, we have∑

i∈N

I(Ui,ϕi)(fiω) =∑

i∈N

∑

j∈N

I(Vj ,ψj)(gjfiω)

=∑

j∈N

∑

i∈N

I(Vj ,ψj)(gjfiω)

=∑

j∈N

I(Vj ,ψj)

(∑

i∈N

gjfiω

)=∑

j∈N

I(Vj ,ψj)(gjω).

Overall,∫M

: Ωmc (M) → R is a well-defined linear form.

Remark 3.20. Using the Lebesgue integral, the linear form∫M

can be extendedto forms ω which are Borel measurable, have compact support and are bounded.Here, Borel measurability refers to ω as a map from M to Λk(M) and boundednessmeans the following: Let η ∈ Ωm(M) be a volume form. A differential m-form αon M with compact support is bounded if there is a constant c > 0 such that|αx(v1, . . . , vm)| ≤ c|ηx(v1, . . . , vm)| for all x ∈ M and v1, . . . , vm ∈ TxM . Thisdefinition is independent of the chosen volume form.

This extension of∫M is important for instance to be able to integrate forms of

the type χD · ω where ω is a smooth form on M and D is some domain.

3.6. Exterior Derivative. Having introduced integration of forms we now turnto a differentiation type operation on forms. We have seen that given a smoothmanifold M one can define a natural derivative d : Ω0(M) → Ω1(M). We are nowgoing to construct natural maps Ωk(M) → Ωk+1(M) for every value of k whichplay the role of a derivative. First of all, we define these maps in the case whereM = U is an open subset of Rm and then transplant the result back into manifolds.For f ∈ Ω0(U) = C∞(U) we have already defined df ∈ Ω1(U); in our case:

df =m∑

i=1

∂f

∂xidxi.

Now, let k ≥ 1 and ω ∈ Ωk(U). Then ω can be written in a unique way as ω =∑I aI dxI where the sum is taken over all ordered multi-indices of length k ranging

between 1 and n. For I = (i1, . . . , ik) recall that dxI := dxi1 ∧ · · · ∧ dxik .

Definition 3.21. Let U ⊆ Rm be open and ω ∈ Ωk(U). Define the exterior derivativedω ∈ Ωk+1(U) of ω by

dω :=∑

I

daI ∧ dxI

where ω =∑

I aI dxI .

Example 3.22. To illustrate that the above formula is far from arbitrary, consideran open subset U ⊆ R2 and a 1-form ω = P dx + Q dy where P,Q : U → R are


smooth functions. Then

dω = dP ∧ dx+ dQ ∧ dy =

(∂P

∂xdx+

∂P

∂ydy

)∧ dx+

(∂Q

∂xdx+

∂Q

∂ydy

)∧ dy

=∂P

∂xdx ∧ dx+

∂P

∂ydy ∧ dx +

∂Q

∂xdx ∧ dy + ∂Q

∂ydy ∧ dy

=

(∂Q

∂x− ∂P

∂y

)dx ∧ dy

In this case we recover Green’s theorem as∫∂D ω =

∫D dω which we later identify

as an incarnation of Stokes’ theorem.

Example 3.23. As a second example, consider an open subset U ⊆ R3 and the2-form ω := F1 dy ∧ dz − F2 dx ∧ dz + F3 dx ∧ dy. As above, we compute

dω = dF1 ∧ dy ∧ dz − dF2 ∧ dx ∧ dz + F3 ∧ dx ∧ dy

=∂F1

∂xdx ∧ dy ∧ dz + ∂F2

∂ydx ∧ dy ∧ dz + ∂F3

∂zdx ∧ dy ∧ dz

=

(∂F1

∂x+∂F2

∂y+∂F3

∂z

)dx ∧ dy ∧ dz.

This resembles the divergence theorem which is yet another incarnation of Stokes’theorem: Indeed, if we consider the vector field F (x) := (F1(x), F2(x), F3(x)) on R3

then dω = divF dx ∧ dy ∧ dz.Next, we collect some fundamental properties of the exterior derivative.

Proposition 3.24. Let d be as in Definition 3.21. Then(i) d is R-linear,(ii) d(ω1 ∧ ω2) = dω1 ∧ ω2 + (−1)degω1ω1 ∧ dω2 for any two differential forms

ω1 and ω2 on U ,(iii) d2 = 0, and(iv) if f : V → U is a smooth map between open subsets V ⊆ Rn and U ⊆ Rm

then df∗ = f∗d.

Note that part (iii) of Proposition 3.24 means that d is not a derivative in thenaive sense. Rather it resembles the geometric intuition that the boundary ∂D ofa region D does not have a boundary itself, i.e. “∂2 = 0”.

Proof. Part (i) and (ii) are left as exercises but are used. For (iii), suppose first thatf ∈ Ω0(U). We show that d(df) vanishes and then consider the general case. Thisis in fact in incarnation of Schwarz’s Theorem:

d(df)(i)=

m∑

i=1

d

(∂f

∂xidxi

)=

m∑

i=1

d

(∂f

∂xi

)∧ dxi =

m∑

i=1

m∑

j=1

∂2f

∂xj∂xidxj

∧ dxi

=

m∑

i,j=1

∂2f

∂xj∂xidxj ∧ dxi =

∑

i6=j

∂2f

∂xj∂xidxj ∧ dxi

=∑

i<j

∂2f

∂xj∂xidxj ∧ dxi +

∑

i>j

∂2f

∂xj∂xidxj ∧ dxi

=∑

i<j

∂2f

∂xj∂xidxj ∧ dxi +

∑

j>i

∂2f

∂xi∂xjdxi ∧ dxj

=∑

i<j

(− ∂2f

∂xj∂xi+

∂2f

∂xi∂xj

)dxi ∧ dxj = 0.


Now, if ω ∈ Ωk(U) for k ≥ 1, write ω =∑I aI dxI . Then

d(dω) = d

(∑

I

daI ∧ dxI)

(i)=∑

I

d(daI ∧ dxI)(ii)=∑

I

d(daI)︸︷︷︸=0

∧ dxI − daI ∧ d(dxI)︸︷︷︸=0

where d(daI) = 0 by the above and the term d(dxI) vanishes by recurrence: Recallthat dxI = dxi1 ∧ · · · ∧ dxik ; therefore d(dxI )

(i)= d(dxi1 ∧ · · · ∧ dxik ) which equals

d(dxi1 ) ∧ (dxi2 ∧ · · · ∧ dxik )− dxi ∧ d(dxi2 ∧ · · · ∧ dxik). Here, d(dxi1 ) vanishes bythe discussion for functions and d(dxi2 ∧ · · · ∧dxik ) by recurrence. Thus d(dω) = 0.

For part (iv) we again first consider the case g ∈ Ω0(U). Then

f∗(dg) = f∗

(m∑

i=1

∂g

∂xidxi

)=

m∑

i=1

f∗

(∂g

∂xidxi

)=

m∑

i=1

(∂g

∂xi f)f∗(dxi).

If f = (f1, . . . , fm) then we have f∗(dxi) =∑n

i=1∂fi∂yj

dyj . Therefore:

f∗(dg) =

m∑

i=1

(∂g

∂xi f) n∑

j=1

∂fi∂yj

dyj

Now observe thatm∑

i=1

(∂g

∂xi f)∂fi∂yj

by the chain rule, computing the partial derivative with respect to yj of the functiony 7→ g(f1(y), . . . , fm(y)). Hence

f∗(dg) =

m∑

j=1

∂g f∂yj

dyj = d(g f) = d(f∗g)

which proves the statement for functions g ∈ Ω0(U). Now, if ω ∈ Ωk(U) and k ≥ 1then ω =

∑I aI dxI and we compute

d(f∗ω) = d

(∑

I

f∗(aI)f∗(dxI)

)=∑

I

d(f∗(aI) · f∗(dxI))

(ii)=∑

I

d(f∗(aI)) ∧ f∗(dxI) +∑

I

f∗(aI)d(f∗(dxI))

=∑

I

f∗(daI) ∧ f∗(dxI) +∑

I


= f∗

(∑

I

daI ∧ dxI)

+∑

I


= f∗(dω) +∑

I

f∗(aI)d(f∗(dxI))︸︷︷︸=0

where the term d(f∗(dxI)) vanishes by recurrence as before: If dxI = dxi1∧· · ·∧dxikthen f∗(dxI) = f∗dπi1 ∧ · · · ∧ f∗dπik and applying d yields

d(f∗(dxI)) = d(f∗d(πi1 )) ∧ f∗(dπi2 ) ∧ · · · ∧ f∗(dπik )

− f∗(dπi1) ∧ d(f∗(dπi2 ) ∧ · · · ∧ f∗(dπik ))

in which d(f∗(dπi1 )) = f∗(ddπi1 ) = 0 and d(f∗(dπi2 ) ∧ · · · ∧ f∗(dπik )) vanishes byrecurrence. This proves the assertion.

We now have an understanding of how the exterior derivate behaves with respectto the various operations for forms defined on open subsets of Euclidean space.Suppose next that M is a manifold and ω ∈ Ωk(M). We claim that there is a


well-defined (k + 1)-form dω ∈ Ωk+1(M) such that if∑

I aI dxI is the expressionof ω in a chart (U,ϕ) then the expression of dω in the same chart is

∑I daI ∧ dxI .

For this, recall that aI is a smooth function on U and that dxI = ϕ∗(dπI) wheredπI = dπi1 ∧ · · · ∧ dπik . To see that dω ∈ Ωk+1(M) is well-defined we rewrite∑

I

daI ∧ dxI =∑

I

d(aI ϕ−1 ϕ) ∧ ϕ∗(dπI) =∑

I

d(ϕ∗(aI ϕ−1)) ∧ ϕ∗(dπI)

= ϕ∗

(∑

I

d(aI ϕ−1) ∧ dπI)

= ϕ∗d

(∑

I

aI ϕ−1 dπI

)= ϕ∗d((ϕ−1)∗ω)

Note that the exterior derivative d in the last expression is indeed applied to aform (ϕ−1)∗ω defined on Euclidean space. In order to show that d is well-definedon manifolds we now show that given any two charts (Uα, ϕα) and (Uβ , ϕβ) on Mwe have

ϕ∗α(d(ϕ

−1α )∗(ω)) = ϕ∗

β(d(ϕ−1β )∗(ω)).

Let θαβ : ϕβ(Uα ∩ Uβ) → ϕα(Uα ∩ Uβ) denote the coordinate transformation fromα to β and set ωα := (ϕ−1

α )∗ω. Then

ϕ∗α(dωα) = (θαβϕβ)

∗(dωα) = ϕ∗βθ

∗αβ(dωα) =

= ϕ∗βd(θ

∗αβωα) = ϕ∗

β(d(θ∗αβ(ϕ

−1α )∗ω)) = ϕ∗

β(d(ϕ−1β )∗ω)


Uα ∩ Uβ

ϕα(Uα ∩ Uβ)

ϕ−1α

77♣♣♣♣♣♣♣♣♣♣♣

ϕβ(Uα ∩ Uβ)θαβ :=ϕαϕ

−1β

oo

ϕ−1β

gg

Proposition 3.24 now carries over to the setting of manifolds.

Theorem 3.25. Let M be a manifold and let d : Ωk(M) → Ωk+1(M) be defined asabove. Then

(i) d : Ω0(M) → Ω1(M) is the usual differential,(ii) d(ω1 ∧ ω2) = dω1 ∧ ω2 + (−1)degω1ω1 ∧ dω2 for any two differential forms

ω1 and ω2 on M ,(iii) d2 = 0, and(iv) if F : N → M is smooth map between manifolds then dF ∗ω = F ∗(dω) for

all ω ∈ Ωk(M).

It is an exercise to convice oneself that these properties of the exterior deriva-tive on manifolds do indeed follow from their counterparts in Euclidean spaces. Amore interesting exercise is to show that a collection of maps Ωk(M) → Ωk+1(M)satisfying the above properties is necessarily given by d.

Before finally turning to Stokes’ Theorem we remark on a few invariants ofmanifolds related to the exterior derivative. Note that given a manifold M we havea sequence

0d−1

// Ω0(M)d0 // Ω1(M)

d1 // · · · // Ωk−1(M)dk−1

// Ωk(M)dk // Ωk+1(M) // · · ·

Since dk dk−1 = 0 we have im dk−1 ⊆ ker dk. We define the k-th de Rham coho-mology of M as the quotient vector space Hk

dR(M) := ker dk/ imdk−1. For instance,we have H0

dR(M) = ker d0 ∼= Rπ0(M) where π0(M) is the number of connectedcomponents of M . Thus, in a sense, de Rham cohomology spaces measure higherconnectivity problems. Straight from the definition one may also view them as ob-structions to solving certain differential equations: Given ω ∈ Ωk(M), is there a


form η ∈ Ωk−1(M) such that dη = ω? A necessary condition is that dω be zero.However, this is not sufficient in general.

As another fact we state that if M is compact then all its de Rham cohomol-ogy spaces are finite-dimensional despite its constituents being uncountable dimen-sional. For instance, if Sg denotes the surface with g holes then H0

dR(Sg)∼= R

since Sg is connected, H2dR(Sg)

∼= R by Poincaré duality since Sg is orientable andH1

dR(Sg)∼= R2g. In this case, de Rham cohomology detects the number of holes.

If M is a compact manifold and Sk(M) = σ : ∆k → M smooth denotes theset of smooth k-simplices in M then integration against a k-form provides a naturalmap Sk(M) → R, σ 7→

∫σω. This induces a duality map Hk(M,R)×Hk

dR(M) → Rwhere Hk(M,R) denotes the singular homology of M over R computed with respectto smooth k-simplices.

3.7. Stokes’ Theorem. The general form of Stokes’ Theorem requires introducingthe concept of smooth manifolds M with boundary ∂M and to develop the conceptsof smoothness for maps, differential forms etc. in this context. Whereas this doesnot present any major difficulties, it takes time. A shortcut that we will take is toprove Stokes’ Theorem so called regular domains in smooth manifold which is lessgeneral only in appearance.

Definition 3.26. Let M be a manifold. A regular domain in M is a subset D ⊆Msuch that

(i) D is closed,(ii) D is not empty, and(iii) for every p ∈ ∂D = D\D there is a chart (U,ϕ) of M at p such that

ϕ(U) = Cmε (0), ϕ(p) = 0 and ϕ(U ∩D) = x ∈ Cmε (0) | xm ≥ 0.

D

U

bϕ

b

ϕ(U)

Note that Definition 3.26 in particular implies that ∂D is a regular (m − 1)-submanifold of M .

Example 3.27. The reader is invited to convince himself that the following areexamples of regular domains.

(i) D = x ∈ Rm | ‖x‖ ≤ 1.(ii) Let N be a smooth manifold and set M := N × R. Then N × [0, 1] is a

regular domain in M with boundary ∂D = N × 0 ∪N × 1.

0 1

Theorem 3.28. Let M be an oriented manifold of dimension at least two and letD be a regular domain in M . Then ∂D is orientable and the orientation of Mcanonically determines an orientation on ∂D.

Proof. Let A be the set of all charts as in the definition of a regular domain.Using a volume form on M we can orient A as in the proof of Proposition 3.18


without changing property (iii) of Definition 3.26: Indeed, either Dpϕ : TpM → Rm

preserves the orientation in which case we leave it or it reverses the orientation inwhich case we compose it with the map Rm → Rm which sends (x1, x2, . . . , xm)to (−x1, x2, . . . , xm); observe that this coordinate transformation preserves x ∈Cmε (0) | xm ≥ 0. Now, let p ∈ ∂D, and (U,ϕ) and (V, ψ) in A be charts at p. Then

ψ ϕ−1(x ∈ ϕ(U ∩ V ) | xm ≥ 0) = y ∈ ψ(U ∩ V ) | ym ≥ 0and

ψ ϕ−1(x ∈ ϕ(U ∩ V ) | xm = 0) = y ∈ ψ(U ∩ V ) | ym = 0.In particular, D0(ψ ϕ−1(x))(Rm−1 ×0) = Rm−1 ×0. Writing ψ ϕ−1(x) =(F1(x), . . . , Fm(x)) for x ∈ ϕ(U ∩ V ) we therefore have

D0(ψ ϕ−1) =

∂F1

∂x1(0) · · · ∂F1

∂xm−1(0) ∂F1

∂xm(0)

......

...∂Fm−1

∂x1(0) · · · ∂Fm−1

∂xm−1(0) ∂Fm−1

∂xm(0)

0 · · · 0 ∂Fm

∂xm(0)

=

(Mm−1 ∗

0 ∂Fm

∂xm(0)

).

Since A is oriented we have detMm−1 · ∂Fm

∂xm(0) = detD0(ψ ϕ−1) > 0. From the

fact ψ ϕ−1(x ∈ ϕ(U ∩ V ) | xm > 0) = y ∈ ψ(U ∩ V ) | ym > 0 we deduce that∂Fm

∂xm(0) > 0 which implies detMm−1 > 0 by the above. Now, for every (U,ϕ) ∈ A

define ϕ := ϕ|U∩∂D. Since Mm−1 = D0(ϕ ψ−1) we conclude that (U ∩ ∂D, ϕ)is an oriented atlas of ∂D.

In other words, we take the following from Theorem 3.28.

Scholium. Retain the notation of the proof of Theorem 3.28. Given p ∈ ∂D and(U,ϕ) ∈ A the space

TpM>0 := v ∈ TpM | (Dpϕ(v))m > 0is a well-defined half-space, independent of the choice of chart. It contains the“inward” directions.

b

Now recall that if M is orientable and A an oriented atlas on M , then an orien-tation on M is a consistent choice of orientation on TpM for all p ∈ M . If M isconnected there are two possible orientations: the one for which for every choice ofp ∈ M and (U,ϕ) ∈ A at p the map Dpϕ : TpM → Rm is orientation-preservingand the one for which all these maps are orientation-reversing; here, we equip Rm

with the canonical orientation in which (e1, . . . , em) is positively oriented. Givenan orientation on M , the induced orientation on ∂D is defined as follows: Givenp ∈ ∂D, a basis (f1, . . . , fm−1) of Tp∂D is positively oriented if for all v ∈ TpM>0

the basis (f1, . . . , fm−1, v) of TpM is positively oriented.

Theorem 3.29. Let M be a smooth oriented manifold and let D ⊆M be a regulardomain. Further, let ω ∈ Ωm−1(M) be a smooth (m− 1)-form. Assume that eitherD is compact or that ω has compact support. Then∫

D

dω =

∫

∂D

i∗ω

where i : ∂D → M denotes the inclusion and ∂D denotes the boundary of D withthe induced orientation if m is even and the opposite of the induced orientation ifm is odd.


The proof of Theorem 3.29 reduces to the fundamental theorem of calculus. Thereduction however, employs all our knowledge on differential forms.

Proof. Let (Ui, ϕi)i∈I be a countable, locally finite covering of M taken from anoriented atlas and such that (Ui, ϕi) is as in Definition 3.26 whenever Ui ∩ ∂D 6= ∅.Furthermore, let fi | i ∈ I be a partition of unity subordinate to (Ui, ϕi)i∈I .We now give a proof in the case where ω has compact support. It constitutes incomputing both sides of the stated equality. First of all ω =

∑i∈I fiω contains only

finitely many non-zero terms since suppω is compact. Hence∫

D

dω =∑

j∈I

∫

D

d(fjω) and∫

∂D

i∗ω =∑

j∈I

∫

∂D

i∗(fjω)

It therefore suffices to show∫D d(fjω) =

∫∂D

i∗(fjω) for all j ∈ I which amountsto showing that ∫

D∩Uj

d(fjω) =

∫

˜∂D∩Uj

i∗(fjω).

By definition, the left hand side of the above is computed as follows. Considerϕj : Uj → Cmε (0). Then

(ϕ−1j )∗(fjω) =

m∑

l=1

gl dx1 ∧ · · · ∧ dxl ∧ · · · ∧ dxm

for smooth functions gj : Rm → R with compact support contained in Cmε (0). Usingthe fact that pulling back commutes with taking the exterior derivative we compute

(ϕ−1j )∗(dfjω) = d((ϕ−1

j )∗(fjω))

=

m∑

l=1

dgl ∧ dx1 ∧ · · · ∧ dxl ∧ · · · ∧ dxl ∧ · · · ∧ dxm

=

m∑

l=1

(m∑

s=1

∂gl∂xs

dxs

)∧ dx1 ∧ · · · ∧ dxl ∧ · · · ∧ dxm

=

m∑

l=1

∂gl∂xl

dxl ∧ dx1 ∧ · · · ∧ dxl ∧ · · · ∧ dxm

=

m∑

l=1

(−1)l−1 ∂gl∂xl

dx1 ∧ · · · ∧ dxm.

We therefore obtain∫

D∩Uj

d(fjω) =

∫

ϕj(D∩Uj)

m∑

l=1

(−1)l−1 ∂gl∂xl

dx1 · · · dxm

=

∫

xm≥0

dxm

∫

Rm−1

dx1 · · · dxm−1

(m∑

l=1

(−1)l−1 ∂gl∂xl

)

of which we compute each summand individually: If l = m we have

(−1)m−1

∫

xm≥0

dxm

∫

Rm−1

dx1 · · · dxm∂gm∂xm

= (−1)m−1

∫

Rm−1

dx1 · · · dxm−1

∫

xm≥0

∂gm∂xm︸︷︷︸

dxm

=−gm(x1,...,xm−1,0)

= (−1)m∫

Rm−1

gm(x1, . . . , xm−1, 0) dx1 · · · dxm−1.


On the other hand, for l 6= m we get∫

xm≥0

dxm

∫

Rm−2

dx1 · · · dxl · · · dxm−1

∫

R

dxl∂gl∂xl︸︷︷︸

=0

= 0

Overall, we have∫

D∩Uj

d(fjω) = (−1)m∫

Rm−1

gm(x1, . . . , xm−1, 0) dx1 · · · dxm−1.

We now turn to computing∫

˜∂D∩Uji∗(fjω): Consider the following diagram

Ujϕj

// Cmε (0)

∂D ∩ Uj

i

OO

pmϕj// Cm−1ε (0)

I

OO

where pm(x1, . . . , xm) := (x1, . . . , xm−1) and I(x1, . . . , xm−1) := (x1, . . . , xm−1, 0).Then ((pm ϕj)−1)∗(i∗(fjω)) = I∗((ϕ−1

j )∗(fjω)). Recall that

(ϕ−1j )∗(fjω) =

m∑

l=1

gl dx1 ∧ · · · ∧ dxl ∧ · · · ∧ dxm

and note that

I∗(dx1 ∧ · · · ∧ dxl ∧ · · · ∧ dxm) = I∗(dx1) ∧ · · · ∧ I∗(dxl) ∧ · · · ∧ I∗(dxm)

where I∗(dxm) = 0 and I∗(dxl) = dxl for l 6= m. This implies

I∗((ϕ−1j )∗(fjω)) = gm(. . . , 0) dx1 ∧ · · · ∧ dxm−1

and hence the result; note that if ∂D ∩ Uj = ∅ then clearly∫

˜∂D∩Uji∗(fjω) = 0.

Also,∫D∩Uj

d(fjω) vanishes in this case since integration will be over the whole ofRm without restriction on the xm-coordinate.

Corollary 3.30. Let M be an oriented manifold of dimension at least two and letD be a compact regular domain in M. Further, let Ω ⊇ D be open. Then there isno smooth map f : Ω → ∂D which is the identity on ∂D.

Proof. Assume that f : D → ∂D is as asserted. Pick a volume form on ω ∈Ωm−1(∂D) on ∂D. Then

∫∂D ω 6= 0. On the other hand, note that dΩ(f∗ω) =

f∗(d∂Dω) = 0 since by d∂Dω is an m-form on ∂D and hence identically zero.Therefore Theorem 3.29 yields a contradiction:

0 =

∫

D

d(f∗ω) =

∫

∂D

i∗(f∗ω) =

∫

∂D

(f i)∗(ω) 6= 0.

Corollary 3.31 (Brouwer’s Fixed Point Theorem). Let m ≥ 2 and B := x ∈ Rm |‖x‖2 ≤ 1. Then any continuous map f : B → B has a fixed point.

Proof. As a first step, we prove a slightly different statement in a smooth setting:Let δ > 0 and G : B1+δ → B a smooth map. We claim that G has a fixed point. IfG(x) 6= x for all x ∈ B(0, 1 + δ) then we may consider the ray G(x) + t(x −G(x))(t ≥ 0) and define f(x) to be the intersection of this ray with S(0, 1). Then f :B(0, 1 + δ) → B is smooth and restricts to the identity on ∂B = S(0, 1). This,however, contradicts Corollary 3.30.

As a second step, we pass from the smooth to the continuous setting usingWeierstrass approximation. Assume that there is a continuos map F : B → B withF (x) 6= x for all x ∈ B. Then there is ε > 0 such that 0 < 2ε < minx∈B ‖F (x)−x‖.


By Weierstrass approximation there is a polynomial map P : Rm → Rm such that‖F (x)−P (x)‖ < ε for all x ∈ B. Since ‖F (x)‖ ≤ 1 we get maxx∈B ‖P (x)‖ < 1+ ε.Now define G : Rm → Rm by G(x) = P (x)/(1 + ε). Then maxx∈B ‖G(x)‖ < 1.Hence there is δ > 0 such that G(B(0, 1 + δ)) ⊆ B and we claim that G(x) 6= xfor all x ∈ B(0, 1 + δ). This is due to the fact that G approximates F well: Ifx ∈ B(0, 1 + δ)\B then G(x) 6= x since G(B(0, 1 + δ)) ⊆ B. For x ∈ B we have

‖G(x)− F (x)‖ ≤ ‖G(x) − P (x)‖+ ‖P (x)− F (x)‖ < εP (x)

1 + ε< 2ε

and therefore ‖G(x)− x‖ ≥ ‖F (x)− x‖ − 2ε > 0.

3.8. De Rham Cohomology in Top Degree. Let M be a smooth manifold. Wehave already seen that H0

dR(M) consists of locally constant functions and hence canbe identified with Rπ0 where π0 denotes the number of connected components of M .In this section, we determine Hm

dR(M) in the case where M is connected, compactand oriented of dimension m. We will see that integration over M , which can beviewed as a linear map from Ωm(M) → R, induces an isomorphism Hm

dR(M) ∼= R.The following two lemmas work towards this statement.

Lemma 3.32. Let M be a manifold and let (U,ϕ) be a chart of M . Further, supposethat F ⊆ U is such that ϕ(F ) = [0, 1]m. If ω ∈ Ωm(M) satisfies

∫Mω = 0 and

suppω ⊆ F then there is η ∈ Ωm−1(M) such that supp η ⊆ F and ω = dη.

Proof. Consider the form (ϕ−1)∗ω ∈ Ωm(ϕ(U)) with support in ϕ(F ) = [0, 1]m.Then we can write (ϕ−1)∗ω = f dx1 ∧ · · · ∧ dxm for some f ∈ C∞(Rm) withsupport in ϕ(F ). In this situation, it is an exercise to show that there are fi ∈C∞(Rm) (i ∈ 1, . . . ,m) with support in ϕ() such that f =

∑mi=1

∂fi∂xi

. Now defineα ∈ Ωm−1(Rm) by

α :=

m∑

i=1

(−1)i−1fi dx1 ∧ · · · ∧ dxi ∧ · · · ∧ dxm

which is supported in ϕ(F ). One readily computes that dα = (ϕ−1)∗ω and hencewe may put η := ϕ∗α.

The second lemma deals with adjusting the support of a general m-form to becontained in a set F as in Lemma 3.32.

Lemma 3.33. Let M be a connected, compact manifold and let (U,ϕ) be a chart ofM such that F ⊂ U satisfies ϕ(F ) = [0, 1]m. If ω ∈ Ωm(M) there is η ∈ Ωm−1(M)such that supp(ω + dη) ⊂ F ,

Proof. The proof proceedes in two steps.

(i) Claim: Let (Vi, ϕi, Fi)ni=0 be such that for every i ∈ 0, . . . , n the tuple

(Vi, ϕi) is a chart with Fi ⊂ Vi satisfying ϕi(Fi) = [0, 1]m. Furthermore,assume that Fi−1 ∩ Fi 6= ∅ for 1 ≤ i ≤ n and let ω0 ∈ Ωm(M) be supportedin F0. Then there is η ∈ Ωm−1(M) with supp(ω0 + dη) ⊆ Fn.

b b b

V0, F0 V1, F1 V2, F2 Vn, Fn


Choose m-forms ω1, . . . , ωn with suppωi ⊆ Fi−1 ∩ Fi and such that∫Fj−1

ωj =∫Fj−1

ωj−1 for all j ∈ 1, . . . , n. Then supp(ωi+1 − ωi) ⊆ Fifor 0 ≤ i ≤ n − 1 and

∫Viωi+1 − ωi = 0. Then by Lemma 3.32 there are

ηi ∈ Ωm−1(M) such that dηi = ωi+1 − ωi for all 0 ≤ i ≤ n− 1. Hence

ωn = (ωn − ωn−1) + (ωn−1 − ωn−2) + · · ·+ (ω1 − ω0) + ω0

=

n−1∑

i=0

dηi + ω0

Hence the assertion.(ii) We now finish the proof of the lemma. Let (U1, ϕ1, F1), . . . , (Un, ϕn, Fn)

be charts as in step (i) and suppose (U1, ϕ1, F1) = (U,ϕ, F ). Furthermore,assume by compactness that

⋃ni=1 Fi =M and let (fi)

ni=1 be a partition of

unity subordinate to (Fi)ni=1. Then we may write ω =

∑ni=1 fiω. Now, for

i ≥ 2 there is, by connectedness and step (i), a form ηi ∈ Ωm−1(M) suchthat supp(fiω + dηi) ⊆ F1 = F . Then ω +

∑ni=1 dηi has support in F .

We are now in a position to prove the following theorem.

Theorem 3.34. Let M be a compact, connected and oriented manifold. Then thesequence

Ωm−1(M)dm−1

// Ωm(M)

∫M // R

is exact, i.e. ker∫M

= im dm−1.

Proof. First, we show that im dm−1 ⊆ ker∫M . Let ω ∈ Ωm−1(M). Take any chart

(U,ϕ) of M such that the closed unit ball B is contained in ϕ(U) and define D :=

ϕ−1(B). Then both D and M\D are regular domains. Furthermore, they sharethe same boundary ∂D = ∂(M\D) but induce opposite orientations on it. As aconsequence, Stokes’ Theorem implies∫

M

dω =

∫

D

dω +

∫

M\D

dω =

∫

∂D

ω +

∫˜

∂(M\D)

ω = 0.

We now turn to proving that ker∫M

⊆ im dm−1. To this end, let ω ∈ ker∫M

andlet (U,ϕ) be a chart such that F ⊆ U satisfies ϕ(F ) = [0, 1]m. Then by Lemma3.33 there is η ∈ Ωm−1(M) with supp(ω + dη) ⊂ F . Also,

∫Mω + dη = 0 by the

first step. Hence Lemma 3.32 implies that there is α ∈ Ωm−1(M) with ω+dη = dα.Thus ω = d(α− η).

4. De Rham Cohomology

4.1. Basic Definitions. Let M be a a smooth manifold of dimension m. Recallthe sequence

0 // Ω0(M)d0 // Ω1(M)

d1 // · · · dm−2// Ωm−1(M)

dm−1// Ωm(M)

0 // · · · .Since d2 = 0, we have im dk−1 ⊆ ker dk for every k ≥ 0 and we may therefore define

HkdR(M) := ker dk/ im dk−1

Elements of im dk−1 are called exact forms and elements of ker dk are called closedforms. The de Rham cohomology spaces are fundamental invariants of M . So farwe know that

(i) H0dR(M) ∼= Rπ0 , and

(ii) HmdR(M) ∼= R via integration if M is compact, connected and oriented.


Any differential form ω ∈ Ωm(M) with∫M ω = 1 is said to represent the fundamen-

tal class of M . Next, we describe the functorial properties of de Rham cohomology:Let M,N be manifolds and let f : M → N be smooth. Combining pullback via fand exterior differentiation we obtain the diagram

0 // Ω0(N)d0 //

f∗

Ω1(N)d1 //

f∗

· · · dk−1// Ωk(N)

dk //

f∗

· · ·

0 // Ω0(M)d0 // Ω1(M)

d1 // · · · dk−1// Ωk(M)

dk // · · ·Since pullback and exterior differentiation commute, so does the above diagram.As a consequence, we have

f∗(ker dNk ) ⊆ ker dMk and f∗(im dNk−1) ⊆ im dMk−1.

Indeed, if for instance ω ∈ ker dNk then dMk f∗ω = f∗dNk ω = 0 and similarly for the

assertion about images. Therefore, f induces for every k ≥ 0 a map f∗ : HkdR(N) →

HkdR(M). If the context is clear, we drop the superscript.

4.2. The Degree of a Map. Let M and N be compact, connected and orientedmanifolds of dimension m. Further, let f :M → N be a smooth map. By Theorem3.34, integration induces isomorphisms

HmdR(N)

f∗m //

I

HmdR(M)

I

R //❴❴❴❴❴❴ R

and hence there is a unique linear map R → R which makes the diagram commute.

Definition 4.1. Let M and N be compact, connected and oriented manifolds ofdimension m. Further, let f : M → N be a smooth map. The degree of f is theunique real number deg f ∈ R such that the linear map R → R, x 7→ (deg f) · xmakes the above diagram commute.

In the following, we show that the degree of a smooth map when defined isactually an integer. First, we record the following lemma which states that f isclose to being a covering map.

Lemma 4.2. LetM andN be compact manifolds of dimensionm and let f :M → Nbe a smooth map. Let q ∈ f(M) be a regular value. Then f−1(q) is finite and thereis an open connected set Vq ⊆ N containing q and for each p ∈ f−1(q) an openconnected set Up ⊆M containing p such that f |Up

: Up → Vq is a diffeomorphism.

Proof. Since q is regular, for all p ∈ f−1(q) the differential Dpf : TpM → TqN hasfull rank and hence, in our case, is an isomorphism. Hence, by the inverse functiontheorem, there are open neighbourhoods U ′

p of p ∈ M and V ′q of q ∈ N such that

f |U ′p: U ′

p → Vq is a diffeomorphism. In particular, U ′p ∩ f−1(q) = p. Therefore

f−1(q) is discrete in M and hence finite by compactness of M . Now set Vq to bethe connected component of

⋂p∈f−1(q) V

′q and Up := (f |U ′

p)−1(Vq).

Retain the situation of Lemma 4.2 and assume in addition that M and N are ori-ented. For each p ∈ f−1(q), the differential Dpf : TpM → TqN is an isomorphism.We define the index I(f, p) of f at p by

I(f, p) :=

1 if Dpf is orientation-preserving−1 if Dpf is orientation-reversing


The degree of f can be expressed in terms of indices:

Proposition 4.3. Let M and N be compact, connected, oriented manifolds of di-mension m. Further, let f : M → N be a smooth map. Then either there is noregular value in which case deg f = 0 or deg f =

∑p∈f−1(q) I(f, p) for any regular

value q of f .

As an immediate consequence of Proposition 4.3 we record the following.

Corollary 4.4. Let M and N be compact, connected, oriented manifolds of di-mension m. Further, let f : M → N be a smooth map. Then deg f ∈ Z and| deg f | ≤ |f−1(q)|.

Given a compact, connected, oriented manifold M , it is an interesting questionto determine which integers occur as degrees of smooth maps f : M → M . Forinstance, all integers occur as degrees of smooth maps from the torus to itself butonly the degrees −1, 0 and 1 occur for smooth maps from a genus two surfaceto itself which is deeply related to the fact that the latter admits non-Euclideangeometries whereas the torus does not.

Proof. (Proposition 4.3). If there is no regular value then f(M) has measure zeroin N by Sard’s Theorem 1.43. In addition, f(M) is compact and hence closed. Asa reslt, f(M) 6= N and N\f(M) is non-empty and open. Now, let ω ∈ Ωm(N) be arepresentative of the fundamental class of N such that the support of ω is containedin N\f(M). Then f∗ω = 0 and hence deg f =

∫M f∗(ω) = 0.

Now, assume that a regular value q ∈ N of f exists. Choose open connectedneighbourhoods Up for each p ∈ f−1(q) and Vq of q as in Lemma 4.2. Since the setsUp are connected, f |Up

is either orientation-preserving or orientation-reversing onthe whole of Up. We therefore have

deg f =

∫

M

f∗(ω) =∑

p∈f−1(q)

∫

Up

(f |Up)∗ω =

∑

p∈f−1(q)

I(f, p)

∫

Vq

ω =∑

p∈f−1(q)

I(f, p)

by the change of variables formula. This proves the assertion.

4.3. Poincaré Lemma. Suppose that M and N are smooth manifolds and thatf : N →M is a smooth map. Then f induces linear maps f∗ : Hk

dR(M) → HkdR(N)

in cohomology in every degree. In this section we work towards the statement thatif one smooth map f can be smoothly deformed into another smooth map g thenthe two induce the same maps in cohomology. For instance, if M = Rn and f = idthen f can be smoothly deformed via the family ft := t · id (t ∈ [0, 1]) to the zeromap g = 0. As a consequence, all de Rham cohomology spaces of Rn except thedegree zero one vanish.

The Poincaré lemma will also play a role in proving the fact that the de Rham co-homology of a manifold can be recovered from combinatorial intersection propertiesof the elements of certain open covers of the manifold.

We begin with the following easy fact from linear algebra.

Lemma 4.5. Let V be a real finite-dimensional vector space and let p : R×V → Vbe the projection onto the second factor. Define λ ∈ (R×V )∗ by λ(t, v) = t. Then

Λk((R×V )∗) = p∗(Λk(V ∗))⊕ λ ∧ p∗(Λk−1(V ∗))

Proof. Let e1, . . . , em be a basis of V and e∗1, . . . , e∗m the dual basis. Then the

elements p∗(e∗1), . . . , p∗(e∗m), λ is a basis of (R×V )∗. Any k-form on R×V is a sum

over wedge products of length k of these 1-forms. Each such product either containsλ in which case it is a k-form in the first summand or it does not contain λ in whichcase it is a k-form in the second summand.


Now, let M be a smooth manifold. Consider R×M and let π : R×M → Mdenote the projection onto M . Define dt ∈ Ω1(R×M) by (dt)t,m(s, u) = s for alls ∈ R = TtR and for all u ∈ TmM . We want to decompose a k-form on R×M intosomething which genuinely comes from M and something with the dt-part. Theprevious lemma implies

Λk((T(t,m))∗) = (D(t,m)π)

∗(Λk(Tm(M)∗)⊕ dt(t,m) ∧ (D(t,m)π)∗(Λk−1(Tm(M)∗))

Therefore, every ω ∈ Ωk(R×M) can be written as ω = ω1 + dt ∧ ω2 where ω1 ∈Ωk(R×M), ω2 ∈ Ωk−1(R×M) and

(ω1)(t,m) ∈ (D(t,m)π)∗(Λk(Tm(M)∗)) and (ω2)(t,m) ∈ (D(t,m)π)

∗(Λk−1(Tm(M)∗)).

For every t ∈ R let Ω2(t) ∈ Ωk−1(M) be such that

(ω2)(t,m) = (D(t,m)π)∗(Ω2(t)m).

In this way, we get a well-defined map from R to Ωk−1(M) given by t 7→ Ω2(t). Wedefine

I(ω) =

∫ 1

0

Ω2(t) dt.

More precisely, for every m ∈M and all v1, . . . , vk−1 ∈ TmM we have

I(ω)m(v1, . . . , vk−1) =

∫ 1

0

Ω2(t)m(v1, . . . , vk−1) dt.

The Poincaré lemma now reads as follows.

Lemma 4.6. Retain the above notation. Let it : M → R×M, (m 7→ (t,m)). Thenfor every ω ∈ Ωk(R×M) we have

i∗1(ω)− i∗0(ω) = dM (I(ω)) + I(dR×Mω).

Proof. We compute both sides of the asserted equality in local coordinates. Let(U,ϕ) be a chart on M . Then (R×U, id×ϕ) is a chart on R×M . Recall that insuch local coordinates we express differential forms on M in terms of dxI where Iis a multi-index and dxI is a form on U . Lifting them to R×U via π : R×M →Mwe can express a k-form ω ∈ Ωk(R×M) on R×U as follows:

ω =∑

|I|=k

aI(t, x)π∗(dxI)

︸︷︷︸ω1

+ dt ∧∑

|J|=k−1

bJ(t, x)π∗(dxJ )

︸︷︷︸ω2

.

As a consequence, Ω2(t) =∑

|J|=k−1 bJ(t, x)dxJ . We now look for the ω2-part indω = dω1 − dt ∧ dω2. To this end, compute

dω1 =∑

|I|=k

(∂aI∂t

dt+

m∑

i=1

∂aI∂xi

π∗(dxi)

)∧ π∗(dxI)

=∑

|I|=k

∂aI∂t

dt ∧ π∗(dxI) + “stuff not containing dt”

and

dω2 =∑

|J|=k−1

∂bJ

∂tdt+

m∑

j=1

∂bJ∂xj

π∗(dxj)

∧ π∗(dxJ )

= “stuff containing dt” +∑

|J|=k−1

m∑

j=1

∂bJ∂xj

π∗(dxj ∧ dxJ ).


We therefore have

dω = “terms in dx” + dt ∧

∑

|I|=k

∂aI∂t

π∗(dxI)−∑

|J|=k−1

n∑

j=1

∂bJ∂xj

π∗(dxj ∧ dxJ

and

I(dω) =

∫ 1

0

∑

|I|=k

∂aI∂t

dxI dt−∫ 1

0

dt∑

|J|=k−1

m∑

j=1

∂bJ∂xj

dxj ∧ dxJ

whence

d(I(ω)) = d

∫ 1

0

∑

|J|=k−1

bJ(t, x) dxJ

dt =

∫ 1

0

∑

|J|=k−1

m∑

j=1

∂bJ∂xj

dxj

∧ dxJ

dt.

Overall, we conclude

I(dω) + d(Iω) =∑

|I|=k

aI(1, x) dxI −∑

|I|=k

aI(0, x) dxI = i∗1(ω)− i∗0(ω).

As mentioned in the beginning, one important application of Poincaré’s Lemmais that homotopic maps induce the same map in cohomology. This is made precisein the following.

Definition 4.7. Let M and N be manifolds and let f0, f1 : N → M be smoothmaps. A homotopy between f0 and f1 is a smooth map h : R×N → M such thath(0,m) = f0(m) and h(1,m) = f1(m) for all m ∈ N . In this situation, f0 and f1are homotopic.

Example 4.8. Let N = M = Rn, f1 = id and f0 = 0. Then h : R×N → M givenby h(t, x) = t · x is a homotopy between f0 and f1.

Proposition 4.9. Let M and N be manifolds and let f0, f1 : N →M be homotopicmaps. Then f∗

0 , f∗1 : H∗

dR(M) → H∗dR(N) coincide.

Proof. Let h : R×N → M be a homotopy between f0 and f1 and consider thefollowing diagram

Ωk−1(M)h∗

// Ωk+1(R×N)I // Ωk(N)

Ωk(M)h∗

//

d

OO

Ωk(R×N)

d

OO

I// Ωk−1(N)

d

OO

Let α ∈ ker dk. We need to show that f∗1 (α) and f∗

0 (α) differ only by an exact form.To this end, we apply Poincaré’s Lemma 4.6 to ω := h∗(α):

i∗1(ω)− i∗0(ω) = dI(ω) + I(dω).

This reads

f∗1α−f∗

0α = (hi1)∗α−(hi0)∗α = dI(h∗α)+I(dh∗α) = dI(h∗α)+Ih∗dα = dIh∗α


Definition 4.10. Let M and N be manifolds. Then M and N are homotopy equiv-alent if there are smooth maps f : M → N and g : N → M such that g f ishomotopic to idM and f g is homotopic to idN .

Corollary 4.11. Let M and N be manifolds. If M and N are homotopy equivalentthen Hk

dR(M) ∼= HkdR(N) for all degrees k.


Definition 4.12. Let M be a manifold. If M is connected and homotopy equivalentto a point then M is contractible.

Corollary 4.13.(i) Hk

dR(Rn) = 0 if k ≥ 1 and H0

dR(Rn) = R.

(ii) Let M be a manifold. Then HkdR(R

n×M) ∼= HkdR(M) for all degrees k.

For later use we also recall the following consequence of the proof of Lemma 4.6.

Corollary 4.14. Let M be contractible to m0 ∈ M . Then there are linear mapsHk : Ωk(M) → Ωk−1(M) for all degrees k such that

(i) α = dHk(α) +Hk+1(dα) for all α ∈ Ωk(M) with k ≥ 1, and(ii) α− α(m0) = H1(dα) for all α ∈ Ω0(M).

Proof. Let h : R×M →M be a homotopy between f1 = idM and the map f0 withconstant value m0, and go through the proof of Lemma 4.6.

Retain the notation of Corollary 4.14. If α ∈ Ωk(M) is closed, i.e. dα = 0 thenHk(α) is a primitive for α.

4.4. Mayer-Vietoris Sequence. Let M be a manifold and write M = U ∪ V foropen sets U, V ⊆ M . In this section we develop Mayer-Vietoris sequences whichrelate the de Rham cohomologies of M , U , V and U ∩ V . For instance consider thesphere S2 and let U and V be small open enlargements of the northern and south-ern hemisphere respectively. Then both U and V are diffeomorphic to R2, hencecontractible and therefore cohomologically inexistent. The intersection U∩V on theother hand is diffeomorphic to R×S1 whence has the same de Rham cohomologyas S1. The Mayer-Vietoris sequence will enable us to turn this observation into aprecise induction argument to compute the de Rham cohmology of spheres.

S2 U V U ∩ V

Definition 4.15. Let A, B and C be vector spaces and let f : A→ B and g : B → Cbe linear maps. The sequence

Af

// Bg

// C

is exact at B if im f = ker g. An infinite sequence

· · · // Vn−1

fn−1// Vn

fn// Vn+1

fn+1// · · ·

of vector spaces and linear maps is exact if it is exact at every Vn

As an example of exact sequences, we remark that the sequence

0 // Vf

// W ( Vf

// W // 0 )

is exact if and only if f is injective (surjective).

Now, let M be a manifold and write M = U ∪ V for open sets U, V ⊆ M . Wehave the canonical injections iU,M of U into M , iV,M of V into M , iU∩V,U of U ∩Vinto U and iU∩V,V of U ∩ V into V . To avoid an overload of notation we oftenabbreviate e.g. i∗U,M (ω) =: ω|U for a form on M .


Theorem 4.16. Now, let M be a manifold and write M = U ∪ V for open setsU, V ⊆M . For every k ≥ 0, the sequence

0 // Ωk(M)i∗U⊕V

// Ωk(U)⊕ Ωk(V )i∗V ⊖U

// Ωk(U ∩ V ) // 0,

where i∗U⊕V (ω) := (i∗U,M (ω), i∗V,M (ω)) for all ω ∈ Ωk(M), and i∗V⊖U (α, β) :=

i∗U∩V,V (β)− i∗U∩V,U (α) for all α ∈ Ωk(U) and β ∈ Ωk(V ), is exact.

Proof. We need to show exactness at the three slots Ωk(M), Ωk(U) ⊕ Ωk(V ) andΩk(U ∩ V ). As to the first one, it is immediate that i∗U⊕V is injective: If a k-formon M vanishes on both U and V then it vanishes identically as U and V cover M .

Consider now the slot Ωk(U) ⊕ Ωk(V ). It is clear that im i∗U⊕V ⊆ ker i∗V⊖U . Inorder to see the converse inclusion, let (α, β) ∈ Ωk(U) ⊕ Ωk(V ) such that (α, β) ∈ker i∗V⊖U , i.e. α|U∩V = β|U∩V . Thus we may define ω ∈ Ωk(M) by setting ω|U = αand ω|V = β to the effect that i∗U⊕V (ω) = (α, β).

Finally, we show that i∗V⊖U is surjective: Given ω ∈ Ωk(U ∩ V ), we construct(ωU , ωV ) ∈ Ωk(U) ⊕ Ωk(V ) such that i∗V⊖U (ωU , ωV ) = ω. Let fU and fV be apartition of unity subordinate to the cover (U, V ) of M and define ωV ∈ Ωk(V ) by

ωV |U∩V := fU · ωωV |V \(U∩V ) = 0

.

In order to verify that ωV is smooth, it suffices to cover its domain V by open setsand show that ωV is smooth on each of these open sets. In this case, we have V =(U ∩V )∪ (V ∩ (supp fU )

c) and indeed both ωV |U∩V = fUω and ωV |V ∩(supp fU )c = 0

are smooth. Similarly, we may define a smooth ωU ∈ Ωk(U) byωU |U∩V = −fV · ωωU |U\(U∩V ) = 0

.

We obtain ωV |U∩V − ωU |U∩V = ω.

For ease of notation we abbreviate

0 // Ωk(M)i∗ // Ωk(U)⊕ Ωk(V )

p∗// Ωk(U ∩ V ) // 0

and denote the induced maps in cohomology by

HkdR(M)

i∗ // HkdR(U)⊕ Hk

dR(V )p∗

// HkdR(U ∩ V )

The reason we do not include the zero maps in the sequence in cohomology is thati∗ need not be injective and p∗ need not be surjective: For instance, H2

dR(S2) = R

but H2dR(U) and H2

dR(V ) = 0. Also, in degree one, we have H1dR(U) and H1

dR(V ) = 0but H1

dR(U ∩ V ) ∼= H1dR(S

1) ∼= R.Nonetheless, the sequence in cohomology is exact at the middle slot.

Lemma 4.17. Retain the above notation. Then im i∗ = ker p∗.

Proof. We know that p∗i∗ = 0. Hence im i∗ ⊆ ker p∗. For the opposite inclusion, let(α, β) ∈ Ωk(U)⊕Ωk(V ) with dα = 0 = dβ. The assumption that the class ([α], [β]) ∈Hk

dR(U)⊕ HkdR(V ) is in the kernel of p∗ means that β|U∩V − α|U∩V = dγ for some

γ ∈ Ωk−1(U ∩ V ). Let (α′, β′) ∈ Ωk−1(U) ⊕ Ωk−1(V ) with β′|U∩V − α′|U∩V = γ.Hence dβ′|U∩V − dα′|U∩V = dγ = β|U∩V − α|U∩V . Now let ω ∈ Ωk(M) be suchthat ω|U = α− dα′ and ω|V = β − dβ′ and finish by observing that

dω|U = d(α− dα′) = dα = 0

dω|V = d(β − dβ′) = dβ = 0


whence dω = 0. Hence ω defines a cohomology class in degree k which by construc-tion satisfies i∗([ω]) = ([α], [β]).

We now concern ourselves with the question to which extent the above sequencein cohomology fails to be exact in the first and the third slot. For instance, letω ∈ Ωk(U ∩V ) represent an element of Hk

dR(U ∩V ), i.e. dω = 0. Since the sequence

0 // Ωk(M)i∗ // Ωk(U)⊕ Ωk(V )

p∗// Ωk(U ∩ V ) // 0

is exact, there is (α, β) ∈ Ωk(U) ⊕ Ωk(V ) such that p∗(α, β) = ω. However, thereis no reason for α and β to be exact which would imply that p∗ is surjective. Theobstruction to this is captured by the following diagram

Ωk(U)⊕ Ωk(V )

(d,d)

p∗// Ωk(U ∩ V )

d

// 0

0 // Ωk+1(M)i∗ // Ωk+1(U)⊕ Ωk+1(V )

p∗// Ωk+1(U ∩ V )

We have p∗ (d, d)(α, β) = d p∗(α, β) = dω = 0. Hence there is ω ∈ Ωk+1(M)such that i∗ω = (dα, dβ). Note that the choice of the pre-image (α, β) of ω is notunique. However, once a choice of (α, β) has been made, ω is determined uniquelyby injectivity of i∗ in the second row.

Furthermore, observe that dω = 0 since dω|U = d(ω|U ) = d(dα) = 0 and ω|V =d(ω|V ) = d(dβ) = 0. This is important as our aim is the construction of a well-defined map Hk

dR(U ∩ V ) → Hk+1dR (M) that captures the amount to which the

map p∗ : HdR(U) ⊕ HdR(V ) → HdR(U ∩ V ) fails to be exact. To this end, wedirectly analyze what happens in case we pick ω′ ∈ Ωk(U ∩ V ) representing thesame cohomology class as ω, i.e. assuming that there is η ∈ Ωk−1(U ∩ V ) withω′ = ω + dη. Thus consider the extended diagram

Ωk−1(U)⊕ Ωk−1(V )

(d,d)

p∗// Ωk−1(U ∩ V )

d

// 0

Ωk(U)⊕ Ωk(V )

(d,d)

p∗// Ωk(U ∩ V )

d

// 0

0 // Ωk+1(M)i∗ // Ωk+1(U)⊕ Ωk+1(V )

p∗// Ωk+1(U ∩ V )

Since its first row is exact there is (a, b) ∈ Ωk−1(U) ⊕ Ωk−1(V ) with p∗(a, b) = η.For the same reason, we may choose pre-images (α, β) of ω and (α′, β′) of ω′.Then (α, β) + (da, db) is a pre-image of ω′ by commutativity. Therefore, (α′ −α − da, β′ − β − db) is in the kernel of p∗. Hence there is a unique element c ∈Ωk(M) such that i∗c = (α′ − α− da, β′ − β − db). This implies that dc and ω′ − ωmap to (dα′ − dα, dβ′ − dβ). This shows that we obtain a well-defined connectinghomomorphism δ : Hk

dR(U ∩ V ) → Hk+1dR (M). Overall, the short exact sequences


from above now fit into a long Mayer-Vietoris sequence

0 // H0dR(M)

i∗ // H0dR(U)⊕ H0

dR(V )p∗

// H0dR(U ∩ V ) 5423 δ

7601//❫❫❫❫❫ H1

dR(M)i∗ //

−−−

H1dR(U)⊕ H1

dR(V )p∗

// H1dR(U ∩ V )

−−−

5423 δ

HkdR(M)

01//

76i∗ // Hk

dR(U)⊕ HkdR(V )

p∗// Hk

dR(U ∩ V ) 5423 δ,

which is actually exact.

Theorem 4.18. Let M be a manifold and assume that U, V are open subsets of Msuch that M = U ∪ V . Then the Mayer-Vietoris sequence above is exact.

The proof of Theorem 4.18 is not difficult but requires to “deconfuse” oneself andhence constitutes a good exercise which is left to the reader.

Corollary 4.19. Let m ≥ 1 and k ∈ N0. Then

HkdR(S

m) =

R k ∈ 0,m0 otherwise

.

Proof. Clearly, we have HdR(Sm) = 0 for k ≥ m + 1. For the remainder, we may

assume m ≥ 2 and write Sm = U ∪ V where U = Sm\S and V = Sm\N.Then U ∩ V ∼= R×Sm−1. Now, consider the first part of the long exact sequencefor M = Sm. It reads

0 // H0dR(S

m)

∼=

// H0dR(U)⊕ H0

dR(V )

∼=

// H0dR(U ∩ V )

∼=

// H1dR(S

m) // H1dR(U)⊕ H1

dR(V )

0 // Ri∗ // R⊕R

p∗// R

δ // H1dR(S

m) // 0

Indeed, U and V are both contractible, hence the assertion about their cohomologyin degree zero and one. Furthermore, U ∩ V ∼= R×Sm−1 has the same cohomologyas Sm−1 which is connected since m ≥ 2. Now, the image of i∗ equals the kernelof p∗ which is hence one-dimensional. Therefore, the image of p∗ is of dimension2 − 1 = 1, i.e. p∗ is surjective. Therefore δ is the zero map. However, δ is alsosurjective by exactness. This implies H1

dR(Sm) = 0.

Now, for m ≥ 2 and j ≥ 1 we have

HjdR(U)⊕ Hj

dR(V ) // HjdR(U ∩ V )

∼=

δ // Hj+1dR (Sm) // Hj+1

dR (U)⊕ Hj+1dR (V )

0 // HjdR(S

m−1)δ // Hj+1

dR (Sm) // 0

which implies HjdR(S

m−1) ∼= Hj+1dR (Sm). We now have all the information we need

to fill in the following tableau by recurrence.

H0dR H1

dR H2dR H3

dR H4dR · · ·

S1 R R 0 0 0 · · ·S2 R 0 R 0 0 · · ·S3 R 0 0 R 0 · · ·S4 R 0 0 0 R · · ·...

......

. . .. . .

. . .. . .



4.5. De Rham cohomology of T n. In this section we compute the de Rhamcohomology of the T n := (S1)n. This can be done using the long exact sequenceintroduced in the previous section, see [BT03] for an account of this. We shallpresent a different approach pertaining to the geometry and topology of homoge-neous spaces of Lie groups. These methods could have been applied to compute thede Rham cohomology of spheres as well.

Theorem 4.20. There is a natural isomorphism of HkdR(T

n) and Λk((Rn)∗).

As a consequence of Theorem 4.20 we obtain dimHkdR(T

n) =(nk

). For the re-

mainder of this section shall think of S1 as the unit circle S1 = z ∈ C | |z| =1 = e2πit | t ∈ R. Then S1 and T n = (S1)n are abelian groups. In addition, forevery ξ ∈ T n the map Lξ : T n → T n, η 7→ ξη is a diffeomorphism. In this context,a k-form ω ∈ Ωk(T n) is invariant if L∗

ξω = ω for all ξ ∈ T n. Let Ωkinv(Tn) denote

the subspace of Ωk(T n) of invariant forms and let 1 = (1, . . . , 1) ∈ T n denote theidentity element.

Lemma 4.21. The map Ωkinv(Tn) → Λk((T

1

T n)∗), ω 7→ ω1

is an isomorphism.

Proof. We construct an inverse to the given map. Let ϕ ∈ Λk((T1Tn)∗) and ξ ∈ T n.

We define ω ∈ Ωkinv(Tn) by ωξ(w1, . . . , wk) := ϕ(DξLξ−1(w1), . . . , DξLξ−1(wk)) for

all w1, . . . , wk ∈ TξT n. One verifies that the k-form ω so defined is invariant andsatisfies ω

1

= ϕ.

As a first step towards Theorem 4.20 we show that invariant forms are closed.Then can be checked using local coordinates or with the help of the followingtrick due to É. Cartan: Consider the exponential map E : Rn → T n given byx 7→ (e2πix1 , . . . , e2πixn). This map is a group homomorphism and a local diffeo-morphism. In fact, it induces an isomorphism Rn /Zn → T n. Furthermore, lets : T n → T n, ξ 7→ ξ−1 denote the inversion which is a diffeomorphism of T n fixing1. We have D

1

s = − Id: Indeed, note that s(E(x)) = E(x)−1 = E(−x) for allx ∈ Rn and hence (D0E)−1 D

1

s D0E = − Id which implies D1

s = − Id.

Lemma 4.22. Let ω ∈ Ωkinv(Tn). Then dω = 0.

Proof. First, observe that s∗ preserves Ωkinv(Tn): For all ξ ∈ T n we have s Lξ =

Lξ−1 s and therefore

L∗ξ(s

∗ω) = (sLξ)∗ω = (Lξ−1s)∗ω = s∗L∗

ξ−1ω = s∗ω.

for all ξ ∈ T n, i.e. s∗ω is invariant.Next, we compute s∗ω. By invariance, it suffices to determine s∗ω at 1 ∈ T n.

Let v1, . . . , vk ∈ T1

T n. Then

(s∗ω)1

(v1, . . . , vk) = ω1

(D1

sv1, . . . , D1

svk) = (−1)kω1

(v1, . . . , vk)

Hence the invariant forms s∗ω and (−1)kω conincide everywhere. On the otherhand, dω ∈ Ωk+1

inv (T n) since differential and pullback commute, therefore

(−1)k+1dω = s∗(dω) = d(s∗ω) = d((−1)kω = (−1)kdω

which implies dω = 0.

We aim to show that the combinded map

Ωkinv(Tn) → kerdk → kerdk/ im dk−1 = Hk

dR(Tn)

is an isomorphism. To this end, we introduce an averaging operator in k-formswhich ranges in invariant k-forms: Let L denote the Lebesgue measure on Rn with


L([0, 1]n) = 1. Given ω ∈ Ωk(T n), we define Pkω ∈ Ωk(T n) by “∫[0,1]n L

∗E(x)ω dL(x)”,

that is, for ζ ∈ T n and v1, . . . , vk ∈ TζT n we set

(Pkω)ζ(v1, . . . , vk) :=

∫

[0,1]n(L∗

E(x)ω)ζ(v1, . . . , vk) dL(x).

Lemma 4.23. The operator Pk on Ωk(T n) is a projection ranging in Ωkinv(Tn) and

commutes with the differential.

The proof of Lemma 4.23 is left as an exercise. It readily implies that the map

Ωkinv(Tn) → kerdk → kerdk/ im dk−1 = Hk

dR(Tn)

is injective: Assume that ω ∈ Ωkinv(Tn) maps to the trivial class in Hk

dR(Tn), i.e.

there is η ∈ Ωk−1(T n) such that dη = ω. Then

ω = Pkω = Pk(dη) = dPk−1η = 0

since Pk−1η is invariant and hence closed.In order to see surjectivity, note that Pkω is defined as a “sum” of pullbacks

via diffeomorphisms that are all homotopic to the identity, hence Pkω should notchange the cohomology class of ω. To make this precise, let x ∈ Rn and considerthe homotopy Hx : R×T n → T n given by (t, ξ) 7→ E(tx)ξ. Then LE(x) = Hx i1and Id = Hx i0 where it : T n → R×T n maps ξ to (t, ξ). Furthermore, letI : Ωk(R×T n) → Ωk−1(T n) be the operator that occurs in the Poincaré Lemma.Then for every α ∈ Ωk(R×T n) we have i∗1α − i∗0α = dI(α) + I(dα). Applying thisto α = (Hx)

∗ω with ω ∈ Ωk(T n) we get

i∗1(H∗1 )ω − i∗0(Hx)

∗ω = dIH∗xω + IdH∗

xω

and hence L∗E(x)ω − ω = dIH∗

xω + IH∗xω. Integrating over [0, 1]n we finally get

Pkω − ω = dKω −Kdω

where K(ω) :=∫[0,1]n

(IH∗x)(ω) d(x).

Now, let ω ∈ Ωk(T n) be such that dω = 0, i.e. representing a class of HkdR(T

n).Then

Pkω − ω = dKω +Kdω = dKω,

that is, Pkω ∈ Ωkinv(Tn) and ω represent the same cohomology class. This completes

the proof of Theorem 4.20.

We remark that the methods employed to prove Theorem 4.20 can be vastlygeneralized to the setting of smooth actions of compact connected Lie groups onmanifolds: A Lie group G is a smooth manifold G endowed with a group structuresuch that the multiplication map G×G→ G and the inversion map i : G→ G aresmooth. Now, pick any non-zero alternating form λ on TeG where n = dimG ande ∈ G denotes the identity element. Then ω ∈ Ωn(G) defined by ωg(v1, . . . , vn) :=λ(DgLg−1v1, . . . , DgLg−1vn) for all v1, . . . , vn ∈ TgG is a nowhere vanishing top-form, i.e. a volume form. In particular, G is orientable and admits integration: IfE ⊆ G is any relatively compact Borel set, we define L(E) :=

∫GχEω. Then L

is an invariant regular Borel measure on G, called Haar measure. This measurefacilitates averaging arguments as in the computation of H∗

dR(Tn): For instance, if

K is a compact connected Lie group and K×M →M is a smooth action of K on amanifold M then the set of closed K-invariant k-forms on M surjects onto Hk

dR(M),i.e. the de Rham cohomology of M can be computed using invariant forms only.For instance, this could have been used to compute H∗

dR(Sm) using the action of

SO(m+ 1) on Sm. As another exercise, the reader is invited to show that for anysmooth manifold M one has

HkdR(S

1 ×M) ∼= HkdR(M)⊕ Hk−1

dR (M)


using the action of S1 on S1 ×M which acts from the left on S1 and fixes M andthe details of the Poincaré Lemma.

5. De Rham’s Theorem

Historically, de Rham’s theorem is the first instance of a comparison of differentcohomology theories. He showed that for a manifold, de Rham cohomology andsingular cohomology, which is defined in the more general context of topologicalspaces, coincide. The approach to proving this theorem is via Čech cohomologywhich is combinatorial in nature and can be shown to be equal to singular and deRham cohomology for any manifold. A classical account of this is [Wei52]. A niceconsequence of this is the finite-dimensionality of the de Rham cohomology of acompact manifold in every degree which follows easily from Čech cohomology.

5.1. Čech Cohomology. Let X be a set and let U = Ui | i ∈ I ⊆ P(X) bea collection of subsets of X ; think of X being a manifold and the Ui being openand covering M . The nerve of U is the set N (U) of subsets J ⊆ I such thatUJ :=

⋂j∈J Uj 6= ∅. For q ≥ 0, a q-simplex is an ordered (q + 1)-tuple of indices

σ = (i0, . . . , iq) such that |σ| := i0, . . . , iq ∈ N (U). The j-th face of a q-simplexσ (q ≥ 1) is the q − 1-simplex σj = (i0, . . . , ij, . . . , iq).

The nerve N (U) is an instance of what is called a simplicial complex, a combi-natorial object for which there is a topological realization. For instance, considerthe covering U1, U2, U3 of S1 indicated below.

U1

U2U3

U4

b b

b b

3 2

1 4

2, 3

1, 21, 3

1, 2, 4

If the set U4 is added to the covering, then a 2-simplex is introduced in thesimplicial complex. However, note that this simplex does not change the homotopytype. Now, let Sq(U) denote the set of all q-simplices of N (U) and let Cq(U ,R)denote the vector space of all functions f : Sq(U) → R, termed q-cochains. Wedefine the coboundary operator

δq : Cq(U ,R) → Cq+1(U ,R), (δqf)(σ) =

q+1∑

j=0

(−1)jf(σj)

For instance, consider the following simplicial complex.

b b

b

1 2

3

(1, 2)

(2, 3)(1, 3)

(1, 2, 3)


Then (δ1f)((1, 2, 3)) = f((2, 3))− f((1, 3)) + f((1, 2)). Thus, in a sense, δ1 yields acombinatorial boundary of the 2-simplex (1, 2, 3) with built-in orientation.

Lemma 5.1. Retain the above notation. The map δq is linear and δq+1 δq = 0.

As a consequence of the above lemma, we obtain a complex

C0(U ,R) δ0 // C1(U ,R) δ1 // C2(U ,R) δ2 // · · ·

whose cohomology is the Čech cohomology: Hq(U ,R) := ker δq/ im δq−1. Note that

if I is finite then all Cq(U ,R) and hence all Hq(U ,R) are finite-dimensional.

5.2. Statements. This section collects the main statements about Čech cohomol-ogy. They are proven in the next one. First of all we apply the previous section inthe context of manifolds.

Definition 5.2. Let M be a smooth manifold and let U = (Ui)i∈I be a covering ofM by subsets of M . Then U is admissible if

(i) Ui is open for every i ∈ I,(ii) U is locally finite, and(iii) for all J ∈ N (U ,M), the intersection UJ :=

⋂i∈J Ui is contractible.

De Rham’s Theorem now reads as follows.

Theorem 5.3. Let M be a smooth manifold and let U be an admissible covering ofM . Then for all k ≥ 0:

HkdR(M) ∼= H

k(U ,R).

Concerning the notation Hk(U ,R) we remark that instead of taking the real

numbers as the target of functions on simplices we could have taken any abeliangroup, e.g. Z. In some cases, this yields more refined invariants. In this case however,it makes sense to compare the R-vector spaces H

k(U ,R) and Hk

dR(M).Note that the statement of Theorem 5.3 is empty without having proven the

existence of admissible coverings. This is a theorem on its own.

Theorem 5.4. LetM be a smooth manifold. Then M admits an admissible covering.

One strategy to prove the existence of admissible coverings is to apply Whitney’sEmbedding Theorem 2.15 first and then argue within Euclidean space. Anotherone is based on Riemannian geometry and the existence of convex neighbourhoods.Anyway, as a Corollary to Theorem 5.3 we immediately record the following.

Corollary 5.5. Let M be a compact manifold. Then HkdR(M) is finite-dimensional

for all k ≥ 0.

Proof. Let U be an admissible covering of M and extract a finite subcover. Observethat since U is finite, so is Sq(U) for every q ≥ 0 and hence Cq(U ,R) is finite-

dimensional whence Hk(U ,R) is finite-dimensional for all k ≥ 0.

We remark that in the case of compact manifolds, the above implies that deRham cohomology can be computed by a machine if an admissible cover and itsintersection pattern are provided. This is in sharp contrast to e.g. the fundamentalgroup in which case not even triviality can be decided computationally.

For the next statement recall that we computed

dimHkdR(T

n) =

(n

k

)=

(n

n− k

)= dimHn−k

dR (T n).

This can be generalized to the following extent.


Theorem 5.6 (Poincaré Duality). Let M be a compact oriented manifold. Then thepairing

Ωk(M)× Ωn−k(M)∧−→ Ωn(M)

I−→ R, (α, β) 7→∫

M

α ∧ β.

is non-degenerate and hence HkdR(M) ∼= (Hn−k

dR (M))∗.

Often, dimHkdR(M) is referred to as the k-th Betti number of M , denoted bk(M).

5.3. Proofs. We now turn to proving de Rham’s theorem. Throughout, M denotesa manifold and U = (Ui)i∈I an admissible covering of M . Recall that Sp(U) denotesthe set of all p-simplices, i.e. (p + 1)-tuples σ = (i0, . . . , ip) ∈ Ip+1 such that⋂pν=0 Uiν 6= ∅. In this case, |σ| = i0, . . . , ip denotes the set of vertices.

Definition 5.7. A Čech form of bidegree (k, p) (k, p ≥ 0) is an Sp(U)-tuple Ω =(ωσ)σ∈Sp(U) with ωσ ∈ Ωk(U|σ|). We denote by Ω(k,p) the real vector space of Čechforms of bidegree (k, p).

For example, a Čech form of bidegree (k, 0) is Ω = (ωi)i∈I with ωi ∈ Ωk(Ui).Also, a Čech form of bidegree (0, p) is a tuple (fσ)σ∈Sp(U) where fσ : U|σ| → R aresmooth functions. We define

d : Ω(k,p) → Ω(k+1,p), (dΩ)σ = d(ωσ)

Consider in particular the map d : Ω(0,p) → Ω(1,p). In the following we identify thekernel of this map: Suppose Ω = (fσ)σ∈Sp(U) ∈ kerd, i.e. dfσ = 0 for all σ ∈ Sp(U).Since U|σ| is connected, this implies that fσ takes a constant value depending onlyon σ. Therefore, an element Ω ∈ ker(d) can be viewed as an element of Cp(U ,R).In the following we introduce differentials δ : Ω(k,p) → Ω(k,p+1) which together withthe differentials d fit into the following complex

......

......

......

Ω2(M)

OO

// Ω(2,0)

OO

// Ω(2,1)

OO

// Ω(2,2)

OO

// Ω(2,3)

OO

// . . .

Ω1(M)

OO

// Ω(1,0)

OO

// Ω(1,1)

OO

// Ω(1,2)

OO

// Ω(1,3)

OO

// . . .

Ω0(M)

OO

// Ω(0,0)

OO

// Ω(0,1)

OO

// Ω(0,2)

OO

// Ω(0,3)

OO

// . . .

C0(U ,R)

OO

// C1(U ,R)

OO

// C2(U ,R)

OO

// C3(U ,R)

OO

// · · ·

The horizontal differentials δ are defined as follows. First, δ : Ωk(M) → Ω(k,0) isdefined by ω 7→ (ω|Ui

)i∈I . For p > 0 we mimic the boundary operator in the Čechcomplex:

δ : Ω(k,p) → Ω(k,p+1), Ω = (ωσ)σ∈Sp(U) 7→ δΩ = (ω′η)η∈Sp+1(U) ∈ Ω(k,p+1)

where η = (i0, . . . , ip+1) ∈ Sp+1(M) and ω′η =

∑p+1ν=0(−1)νω(i0,...,iν ,...,ip+1)

|U|η|. Re-

call that U|η| =⋂p+1l=1 Uil which is indeed contained in the support of ω(i0,...,iν ,...,ip+1)

for all ν. As an example, consider Ω(k,0) = Ω = (ωi)i∈I | ωi ∈ Ωk(Ui) whereI = S0(U). Then δΩ = (ω′

i0,i1)(i0,i1)∈S1(U) where

ω′(i0,i1)

= ωi1 |Ui0∩Ui1− ωi0 |Ui0∩Ui1

.

This resembles the restriction map in the Mayer-Vietoris sequence.


Lemma 5.8. Retain the above notation. All squares in the above diagram commute.

Proof. Let Ω = (ωσ)σ∈Sp(U) ∈ Ω(k,p) and consider δΩ := (ω′η)η∈Sp+1(U). By defini-

tion,

ω′η =

p+1∑

ν=0

(−1)νωην |U|η|

where ην is the ν-th face of η, i.e. ην = (i0, . . . , iν , . . . , ip+1). Now applying d to δΩyields dδΩ = (dω′

η)η∈Sp+1(U) where

dω′η =

p+1∑

ν=0

(−1)νd(ωην |U|η|) =

p+1∑

ν=0

(−1)ν(dωην )|U|η|

On the other hand, δdΩ = δ(dωσ)σ∈Sp(U) = (ω′′η )η∈Sp+1(U) where

ω′′η =

p+1∑

ν=0

(−1)ν(dωην )|U|η|.


5.3.1. Homotopies. All vertical and horizontal complexes in the above diagram arein fact exact. This is the content of the two lemmas in this section. Since for everyσ ∈ Sp(U) the set U|σ| is contractible, the Poincaré Lemma 4.6 yields a linearmap I|σ| : Ω

m(U|σ|) → Ωm−1(U|σ|) (m ≥ 1) such that ω = dI|σ|ω + I|σ|dω for allω ∈ Ωm(U|σ|). Now define

I : Ωm,p → Ωm−1,p, (ωσ)σ∈Sp(U) → (I|σ|ωσ)σ∈Sp(U).

Then the following is immediate.

Lemma 5.9. Retain the above notation. Then Ω = dIΩ + IdΩ for all Ω ∈ Ω(m,p)

with m ≥ 1 and p ≥ 0.

In particular, all vertical complexes are exact.

Now we define maps K : Ω(m,p) → Ω(m,p−1) (p ≥ 1) and K : Ω(m,0) → Ωm(M),playing a role analogous to the maps I above for horizontal complexes. To this end,let (fi)i∈I be a partition of unity subordinate to U . First, we define K : Ω(m,0) →Ωm(M) by (ωi)i∈I 7→ ∑

i∈I(fiωi) where fiωi is extended to the whole of M byzero. Similarly, for p ≥ 1 and Ω = (ωσ)σ∈Sp(U) we set KΩ := (ζη)η∈Sp−1(U) where

ζi0,...,ip−1 =∑

k∈I

fkωk,i0,...,ip−1

in which fkωk,i0,...,ip−1 = 0 if Uk ∩ Ui0,...,ip−1 = ∅, and if Uk ∩ Ui0,...,ip−1 6= ∅ then

fkωk,i0,...,ip−1 :=

fkωk,i0,...,ip−1 on Uk ∩ Ui0,...,ip−1

0 on Ui0,...,ip−1\Uk.

This is an extension similar to the one in the Mayer-Vietoris sequence.

Lemma 5.10. Retain the above notation. Then Ω = δKΩ+KδΩ for all Ω ∈ Ω(m,p)

with m ≥ 1 and p ≥ 0.

In particular, all horizontal complexes are exact. We are now in a position toprove Theorem 5.3, i.e. to show that

HkdR(M) ∼= H

k(U ,R)

for every k ≥ 0 whenever U is an admissible covering of M . Building on the above,our strategy is as follows. Let ZmdR(M) = ker(d : Ωm(M) → Ωm+1(M)). Given ω ∈ZmdR(M) we have d(δω) = δ(dω) = 0. Hence there is ω1 ∈ Ω(m−1,0) with dω1 = δω.


Next, since d(δω1) = δ(dω1) = δ2ω = 0, there is ω2 ∈ Ω(m−2,1) with dω2 = δω1.Continuing in this fashion we reach δωm ∈ Ω(0,m) which satisfies dδωm = 0 andhence δωm ∈ Cm(U ,R). In fact, δ(δωm) = 0 and hence δωm ∈ Zm(U ,R) = ker(δ).

ZmdR(M)δ // Ω(m,0)

Ω(m−1,0)

d

OO

δ // Ω(m−1,1)

Ω(m−2,1)

d

OO

This can be made into a well-defined map from ZmdR(M) → Zm(U ,R) and we willshow that it induces the sought-after isomorphism in cohomology. To this end,define for 0 ≤ n ≤ m− 1:

Fm,n := Ω ∈ Ω(m−1−n,n) | dδΩ = 0and

Hm,n := Ω ∈ Ω(m−1−n,n) | Ω = X + Y with dX = 0 and δY = 0.Lemma 5.11. Retain the above notation. The map

Iδ : ZmdR(M)δ−→ Ω(m,0) I−→ Ω(m−1,0)

takes values in F (m,0) and induces an isomorphism

HmdR(M) → Fm,0 /Hm,0 .

Proof. For ω ∈ ZmdR(M) we have d(δω) = δ(dω) = 0 and hence δω = dI(δω). Fromthis we deduce 0 = δ2ω = δd(Iδ(ω)) and hence Iδ(ω) ∈ Fm,0.

Now, let Ω0 ∈ Fm,0. Then δ(dΩ0) = 0 and hence dΩ0 ∈ Ω(m,0) is in the imageof δ: Let ω ∈ Ωm(M) with δω = dΩ0. Then we have

δ(dω) = dδω = d2Ω0 = 0

and since δ : Ωm+1(M) → Ω(m+1,0) is injective we get dω = 0, that is ω ∈ ZmdR(M).Thus dIδω = dΩ0 and hence Ω0 − Iδω ∈ Hm,0. This shows that the composition ofIδ with the projection ZmdR(M) → Fm,0 /Hm,0 is surjective.

Finally, let ω = dα for some α ∈ Ωm−1(M). Then δω = δdα = d(δα). Therefore,d(Iδω) = δω = d(δα) and hence Iδω = δα+ β with dβ = 0, i.e. Iδω ∈ Hm,0.

Conversely, if ω ∈ ZmdR(M) and Iδω = α + β with dα = 0 and δβ = 0 thenδω = d(Iδω) = dβ. But since δβ = 0 there is β′ ∈ Ωm−1(M) with δβ′ = β andhence δω = dδβ′ = δdβ′ whence ω = dβ′.

Lemma 5.12. Retain the above notation. The map

δ : Fm,m−1 → Ω(0,m)

takes values in Zm(U ,R) and induces an isomorphism

Fm,m−1 /Hm,m−1 → Hm(U ,R).

Proof. Let Ω ∈ Fm,m−1 ⊆ Ω(0,m−1). Then dδΩ = 0 and δΩ = (fσ)σ∈Sm(U) with fσconstant on U|σ|. Thus δΩ ∈ Cm(U ,R). In addition δΩ ∈ Zm(U ,R) since δ2Ω = 0.

Now, let c ∈ Zm(U ,R). Then dc = 0 and δc = 0. Hence there is Ω ∈ Ω(0,m−1)

with δΩ = c. Clearly, dδΩ = dc = 0 and hence Ω ∈ Fm,m−1. Therefore the mapδ : Fm,m−1 → Zm(U ,R) is surjective.

Finally, let Ω ∈ Hm,m−1, i.e. Ω = α+ β with dα = 0 and δβ = 0. Then δΩ = δαand since dα = 0 we have α ∈ Č

m−1(U ,R). Hence δΩ = δα ∈ Č

m(U ,R). Conversely,


assume that δΩ = δγ with γ ∈ Čm−1

(U ,R). Then dγ = 0 and δ(Ω − γ) = 0 andtherefore Ω ∈ Hm,m−1.

Now, for 0 ≤ n ≤ m− 2 consider the following diagram:

Ω(m−1−n,n) δ // Ω(m−1−n,n+1)

Ω(m−1−(n+1),n+1).

d

OO

Lemma 5.13. Retain the above notation. Then the map

δ : Fm,n → Ω(m−1−(n+1)),n+1)

takes values in Fm,n+1 and induces an isomorphism

Fm,n /Hm,n → Fm,n+1 /Hm,n+1 .

The proof of Lemma 5.13 is similar to the proofs of Lemmas 5.11 and 5.12 andis left to the reader. Overall, this proves Theorem 5.3.


References

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[dR56] G. de Rham, Variétés différentiables, 1956.[Ker60] M. A. Kervaire, A manifold which does not admit any differentiable structure, Commen-

tarii Mathematici Helvetici 34 (1960), no. 1, 257–270.[Mil56] John Milnor, On manifolds homeomorphic to the 7-sphere, Annals of Mathematics 64

(1956), no. 2, pp. 399–405.[Mil97] J. W. Milnor, Topology from the Differentiable Viewpoint, Princeton University Press,

1997.[RSN72] F. Riesz and B. Szökefalvi-Nagy, Leçons d’analyse fonctionnelle, Gauthier-Villars 1

(1972).[Rud87] W. Rudin, Real and complex analysis, Tata McGraw-Hill Education, 1987.[RW13] R. Remmert and H. Weyl, Die Idee der Riemannschen Fläche, vol. 5, Springer-Verlag,

2013.[Wei52] André Weil, Sur les théorèmes de de Rham, Commentarii Mathematici Helvetici 26

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Contents Introduction diﬀerential manifolds · forms and Milnor’s classic [Mil97] on which our section on Brouwer’s ﬁxed theorem will be based. 1. Differential Manifolds and