lecture notes on basic differential topologylerario/antonio_lerario/...lecture notes on basic...

LECTURE NOTES ON BASIC DIFFERENTIAL TOPOLOGY

ANTONIO LERARIO

These are notes for the course “Advanced Geometry 2” for the Master Diploma in Mathematicsat the University of Trieste and at SISSA. These notes are by no means complete: excellentreferences for the subject are the books [3, 5, 7, 8, 9], from which in fact many proofs are takenor adapted. The notes contain some exercises, which the reader is warmly encouraged to solve(sometimes part of a proof is left as an exercise). I apologize in advance for the many mistakes andimprecisions that the reader might find: I will greatly appreciate if he/she could point out any ofthem.

Date: November 4, 2020.

1

2 ANTONIO LERARIO

1. Differentiable manifolds

Definition 1 (m–dimensional Ck manifold). Let m ∈ N and k ∈ N ∪ ∞, ω. A manifold M ofdimension m and class Ck is a paracompact, Hausdorff topological space, with countably manyconnected components1 and such that:

(1) for every point x ∈ M there exists a neighborhood U of x and a continuous functionψ : U → Rm which is a homeomorphism onto an open subset of Rm (the pair (U,ψ) iscalled a chart);

(2) for every pairs of charts (U1, ψ1) and (U2, ψ2) such that U1 ∩ U2 6= ∅ the map

ψ2 ψ1|−1U1∩U2

: ψ1(U1 ∩ U2)→ Rm

is a Ck map (for k = 0 we obtain topological manifolds, for k ≥ 1 differentiable manifolds,for k =∞ smooth manifolds and for k = ω analytic manifolds).

A collection of charts (Uj , ψj)j∈J as above such that⋃j∈J Uj = M is called a Ck atlas.

Remark 2. In the second condition above, given two charts (U1, ψ1) and (U2, ψ2) such thatU1 ∩ U2 6= ∅ it is not enough to check that the map from ψ2 ψ1|−1

U1∩U2is Ck, but also that its

inverse is Ck. For instance, let M = R with the two charts (U1, ψ1) = (R, x) and (U2, ψ2) = (R, x3).Then ψ2 ψ−1

1 = x3 is Cω but ψ1 ψ−12 = x1/3 is only C0.

Remark 3 (Atlases and differential structures). Two Ck atlases A = (Uα, ψα)α∈A and B =(Vβ , ϕβ)β∈B for M , are said to be equivalent if their union A ∪ B is still a Ck atlas. A Ck-differential structure on M is the choice of an equivalence class of Ck atlases. If we take the unionof all atlases belonging to a Ck-differential structure, we obtain a maximal Ck atlas. This atlascontains every chart that is compatible with the chosen differentiable structure. (There is a naturalone-to-one correspondence between differentiable structures and maximal differentiable atlases.)

From now on we will assume that the atlas we work with is maximal, so that we will have allpossible charts available.

A simple way to enrich a given atlas is as follows. Given a chart ψ : U → Rm around a pointx ∈ M (as in point (1) of Definition 1), and given a neighborhood V ⊂ U of x we can easilyconstruct a chart ϕ : V → Rm by simply taking ϕ = ψ|V . Note that in this way we can constructa chart (V, ϕ) around any point with V contractible: it is enough to take V = ψ−1(BRm(ψ(x), ε))for ε > 0 small enough.

Example 4. If ϕ : M → Rn is a homeomorphism, then M is an analytic manifold. In fact onecan cover M with the single chart (M,ϕ), and ϕ ϕ−1 = idRm : Rm → Rm is analytic.

Example 5. Given an open set U ⊆ Rm, the single chart given by the inclusion U → Rm turnsit into a smooth manifold. Why is the condition “with countably many connected components”in the above definition satisfied by U? (Try to prove it directly: an open set in Rm can have onlycountably many connected components).

Example 6 (Product manifolds). If M and N are smooth manifolds with respective atlases(Uα, ψα)α∈A and (Vβ , ϕβ)β∈B , then M × N is naturally a smooth manifold with the atlas(Uα × Vβ , ψα × ϕβ)(α,β)∈A×B .

Example 7 (Spheres). The unit sphere Sn = x20 + · · · + x2

n = 1 ⊂ Rn+1 can be endowed withthe structure of a smooth manifold as follows. Consider the point e0 = (1, 0, . . . , 0) ∈ Sn and the

1Recall that a paracompact space is a topological space X for which every open cover has a locally finiterefinement. More precisely: given an open cover U = Uαα∈A fo X, there exists another open cover V = Vββ∈Bsuch that (i) for every β ∈ B there exists α(β) ∈ A such that Vβ ⊂ Uα(β) (i.e. V refines U); (ii) for every x ∈ Xthere exists a neighborhood Vx of x which intersects only finitely many elements of V (i.e. V is locally finite).It is worth noticing that if a Hausdorff space is locally Euclidean (i.e. if it satisfies condition (1) of Definition1) and connected, then this space is paracompact if and only if it is second countable (i.e. its topology has acountable basis), see the discussion you can find at this webpage https://math.stackexchange.com/questions/

527642/the-equivalence-between-paracompactness-and-second-countablity-in-a-locally-eucl. There existlocally Euclidean spaces which are Hausdorff, paracompact but not second countable (for example R with thediscrete topology); that is why we also add the condition that a manifold should have countably many connected

components.

https://math.stackexchange.com/questions/527642/the-equivalence-between-paracompactness-and-second-countablity-in-a-locally-eucl

https://math.stackexchange.com/questions/527642/the-equivalence-between-paracompactness-and-second-countablity-in-a-locally-eucl

LECTURE NOTES ON BASIC DIFFERENTIAL TOPOLOGY 3

e0

p1

ψ1(p1)

p2

ψ1(p2)

Sn

Rn

b

bb

b

bc

Figure 1. The stereographic projection ψ1 : Sn\e0 → Rn.

two open sets of the the sphere defined by U1 = Sn\e0 and U2 = Sn\−e0. We produce explicitand nice homeomorphisms (i.e. charts) ψ1 : U1 → Rn and ψ2 : U2 → Rn, called stereographicprojections (see Figure 1), as follows:

ψ1(x0, . . . , xn) =1

1− x0(x1, . . . , xn) and ψ2(x0, . . . , xn) =

1

1 + x0(x1, . . . , xn).

It is easy to verify that ψ−11 : Rn → Sn is given by:

ψ−11 (y1, . . . , yn) =

1

1 + ‖y‖2(‖y‖2 − 1, 2y1, . . . , 2yn

).

This in particular allows one to write the explicit expression for ψ2 ψ1|−1U1∩U2

: Rn\0 → Rn\0:

ψ2 ψ−11 (y) =

y

‖y‖2 ,

which is a smooth map. Hence (U1, ψ1), (U2, ψ2) is a smooth atlas for Sn and turns it into asmooth manifold. This is called the standard differential structure on Sn.

Example 8 (Real projective spaces). The real projective space RPn can be endowed with thestructure of smooth manifold as follows. Recall that RPn = (Rn+1\0)/ ∼, where p1 ∼ p2 if andonly if there exists λ 6= 0 such that p1 = λp2. We denote by [x0, . . . , xn] the equivalence classof (x0, . . . , xn) ∈ Rn+1\0 (the xj are called homogeneous coordinates). For every j = 0, . . . , nconsider the open set Uj ⊂ RPn defined by:

Uj = [x0, . . . , xn] such that xj 6= 0,together with the homeomorphism ψj : Uj → Rn given by:

ψj([x0, . . . , xn]) =

(x0

xj, . . . ,

xjxj, . . . ,

xnxj

)(here the “hat” symbol denotes that this element has been removed from the list). The inverseψ−1j : Rn → RPn is given by:

ψ−1j (y0, . . . , yj , . . . , yn) = [y0, . . . , 1, . . . yn],

where the “1” is in position j. As a consequence, for every i 6= j we have:

ψi ψ−1j (y0, . . . , yj , . . . , yn) =

(y0

yi, . . . ,

yiyi, . . . ,

1

yi, . . .

ynyj

),

which is a diffeomorphism of Rn\0 to itself.

Exercise 9. Prove that RP1 and S1 are homeomorphic.

Example 10 (real Grassmannians). The real Grassmannian G(k, n) consists of the set of allk-dimensional vector subspaces of Rn, endowed with the quotient topology of the map:

q : M ∈ Rn×k such that rk(M) = k → G(k, n), q(M) = spancolumns of M.

4 ANTONIO LERARIO

In other words, G(k, n) (as a topological space) can be considered as the quotient of the set ofn× k real matrices of rank k (viewed as a subset of Rn×k) under the equivalence relation:

M1 ∼M2 ⇐⇒ there exists L ∈ GL(Rk) such that M1 = M2L.

Observe that G(1, n) = RPn−1 and that the above definition mimics the equivalence relationv1 ∼ v2 if and only if there exists λ ∈ GL(R) = R\0 such that v1 = λv2.

We want to endow G(k, n) with the structure of a smooth manifold. For every multi-indexJ = (j1, . . . , jk) ∈

nk

we denote by M |J the k× k submatrix of M obtained by selecting the rows

j1, . . . , jk (in this way M |Jc denotes the (n − k) × k submatrix of M obtained by selecting thecomplementary rows). For every such multi-index J we define the open set:

UJ = [M ] ∈ G(k, n) such that det(M |J) 6= 0.(Note that this set is well defined.) Mimicking again the definition for projective spaces, we definethe manifold charts ψJ : UJ → R(n−k)×k by:

ψJ([M ]) = (MM−1J )|Jc .

The expression of the inverse of a matrix in terms of its determinant and its cofactor sshows thatfor every pair of indices J1, J2 ∈

nk

the map ψJ2 ψ−1

J1is smooth. In this way (UJ , ψJ)J∈nk

is a smooth atlas for the k(n− k)-dimensional manifold G(k, n).

Exercise 11. Fill in all the details in the previous definition of the smooth structure on theGrassmannian.

Example 12 (The Complex projective line). Recall that the complex line CP1 is defined as thequotient space (C2\0)/ ∼ where (z0, z1) ∼ λ(z0, z1) for every λ ∈ C\0. As we did for realprojective spaces, we denote by [z0, z1] the homogeneous coordinates of a point on CP1. Considerthe two open sets U0 = z0 6= 0 and U1 = z1 6= 0 together with the charts ψj : Uj → C ' R2

for j = 0, 1 which are given by:

ψ1([z0, z1]) =z0

z1and ψ0([z0, z1]) =

z1

z0.

We have that ψ0 ψ−11 (z) = 1

z , which is a holomorphic map C\0 → C\0 and consequently,

using the identification C ' R2, a smooth map R2\0 → R2\0. If we wanted, we could also workwith ψj as a real map, as follows (however, as the reader will see, using the field structure of C ' R2

simplifies a lot the computations). Given (x0, y0, x1, y1) ∈ R4, let us denote by (x0 +iy0, x1 +iy1) =(z0, y0) ∈ C2. We can write ψ1 : U0 → C ' R2 as

ψ1([x0 + iy0, x1 + iy1]) =x0 + iy0

x1 + iy1=x0x1 + y0y1

x21 + y2

1

+ i · y0x1 − x0y1

x21 + y2

1

which means that the real map ψ1 : U1 → R2 is given by:

ψ1([x0 + iy0, x1 + iy1]) =

(x0x1 + y0y1

x21 + y2

1

,y0x1 − x0y1

x21 + y2

1

),

with inverse ψ−11 : R2 → CP1 given by:

ψ−11 (x, y) = [1, x+ iy].

In particular ψ0 ψ1|−1U0∩U1

is given by (x, y) 7→ 1x2+y2 (x,−y), which is indeed a smooth map

R2\0 → R2\0.The complex projective line CP1 is homeomorphic to S2 and in fact, as smooth manifolds, theyare indistinguishable (see Exercise 31).

Exercise 13. Generalize the previous example and prove that CPn can be endowed with thestructure of a smooth 2n-dimensional manifold.

Example 14 (A connected, non-paracompact “manifold”: the Prufer surface). LetH = (x, y), y >0 be the positive half-space and for every z ∈ R consider the set

Az = (x, y, z) ∈ R3 with y ≤ 0


x = z + ay x = z + by

z

a b

ǫH

Az

ǫ

bc

Figure 2. An open set of the type U3 (grey region) for the topology of the Prufer surface.

(each Az is a closed half space). The Prufer surface is the set

P = H ∪(⋃z∈R

Az

)endowed with the topology generated by open sets of the three following types:

(1) open sets U1 ⊂ H for the standard euclidean topology;(2) open sets U2 ⊂ Az ∩ y < 0 (so U2 is a euclidean open set in (x, y, z), y < 0 ' H);(3) open sets of the form U3 = Az(a, b, ε) ∪Hz(a, b, ε) where (see Figure 2):

Az(a, b, ε) = (x, y, z) ∈ R3, a < x < b,−ε < y ≤ 0Hz(a, b, ε) = (x, y) ∈ H, 0 < y < ε, z + ay < x < z + by.

The resulting topological space is Hausdorff and locally euclidean. To see this, for every z ∈ R letUz = Az ∪H (an open set) and define the map fz : R2 → Uz by:

fz(x, y) =

(x, y, z) y ≤ 0

(z + xy, y) y > 0

The map fz is continuous with continuous inverse (check it!) which we denote by ψz : Uz → R2.The “manifold” structure on P is then given by the atlas (Uz, ψz)z∈R. Notice now that the setZ = (0, 0, z), z ∈ R is an uncountable subset of P which inherits the discrete topology, hence Pcannot be second countable.

Example 15 (A non-Hausdorff “manifold”: the lines with two origins). The circle S1 can beseen as the one point compactification of the real line. We might as well define the n-pointscompactification Xn of R as follows. As a point set Xn = R ∪ ∞1, . . . ,∞n endowed with thetopology generated by all the open sets of the form:

U = ∞j ∪ A ⊂ R with R\A compact for some j = 1, . . . , n.

For n ≥ 2 the space Xn can be endowed with a structure satisfying the axioms for the definition ofsmooth manifold, except for the Hausdorff condition (see Figure 3). The space X2\0 is sometimescalled “the line with two origins”.

6 ANTONIO LERARIO

bb

∞1

∞2

X2 = R ∪ ∞1 ∪ ∞2

Figure 3. The two-points compactification of the real line: a compact, non-Hausdorff, C∞ manifold.

Example 16 (A paracompact, Hausdorff, “manifold” which is not second countable). ConsiderX = R with the discrete topology. Then, X is paracompact (given any cover U of X, the opencover V = xx∈R is a locally-finite refinement of U), Hausdorff and locally Euclidean (eachpoint x ∈ X is open and homeomorphic to R0). However X is not second countable, as it hasuncountably many components. More generally, let Y be a connected manifold (a honest manifold,in the sense of Definition 1) and consider X =

∐t∈R Yt, where each Yt is a copy of Y (just labeled

by a point in t ∈ R). Endow X with the disjoint union topology, so that it has uncountably manyconnected components and it is not second countable. However X is Hausdorff and paracompact.In fact let U = Uαα∈A be an open cover of X and for every t ∈ R consider the open coverUt = Uα ∩ Ytα∈A of the component Yt. Since Yt is paracompact, there exists a locally finiterefinement Vt = Vt,ββ∈Bt of Ut. Then V = Vt,β , t ∈ R, β ∈ Bt is a locally finite open cover ofX which refines U.

Exercise 17. Fill in all the details in the construction of the “manifold” structure for the abovenon-examples.

1.1. The tangent bundle.

Definition 18 (Tangent bundle). Let M be a smooth manifold of dimension m and (Uα, ψα)α∈Abe a smooth atlas for M . We define the tangent bundle of M as:

TM =

(∐α∈A

Uα × Rm)/ ∼,

where (x1, v1)α1∼ (x2, v2)α2

if and only if x1, x2 ∈ Uα1∩ Uα2

(the two open sets overlap andx1, x2 belong to their intersection), Jψα2

(x)(ψα1 ψ−1α2

)v2 = v1 (in other words, the Jacobian of the

coordinates change sends v2 to v1). We endow TM with the quotient topology from the definingequivalence relation “∼”. The tangent bundle is endowed with the structure of a smooth manifoldas follows (see Definition 1).

First, for every α ∈ A we call TM |Uα the set [(x, v)α] ⊂ TM and observe that (by construc-tion) TM |Uα = TUα is an open subset of TM which is homeomorphic to Uα × Rm (under theidentification map); we call ϕα : TM |Uα → Uα × Rm the inverse of this homeomorphism. Themanifold charts for TM are constructed as follows. We define the homeomorphism

φα : TM |Uα → ψα(Uα)× Rm ⊆ Rm × Rm

to be the composition φα = (ψα × idRm) ϕα, i.e. the map given by [(x, v)α] 7→ (ψα(x), v). Thefamily (Uα × Rm, φα)α∈A gives a family of charts for TM (called natural charts).

In order to prove that the above definition turns TM into a smooth manifold, we need to checkwhat is the regularity of the change of coordinates maps. For every pair of charts (U1 × Rm, φ1)and (U2 × Rm, φ2) such that (U1 × Rm) ∩ (U2 × Rm) 6= ∅, we consider therefore the map:

(1.1) φ2 φ−11 : ψ1(U1 ∩ U2)× Rm → ψ2(U1 ∩ U2)× Rm.


We see that this map is given by (y, v) 7→ (ψ2 ψ−11 (y), Jy(ψ2 ψ−1

1 )v) and it is smooth if M itselfwas smooth. Notice indeed that if M is a Ck manifold, the tangent bundle TM is only a Ck−1

manifold, since in (1.1) we are computing one more derivative (in particular the tangent bundle isnot defined if M is only C0).

The map p : TM →M defined by [(x, v)α] 7→ x is called the projection map and the fiber over apoint is denoted by TxM = p−1(x) and called the tangent space to M at x. An important propertyof the tangent space is that it carries the structure of an m-dimensional vector space. In fact,given a chart (U1, ψ1) containing x, calling by (TU1, φ1) the corresponding natural chart for TM ,the map φ1|TxM : TxM → Rm is a bijection and can be used to induce the vector space structureon TxM ; this structure is independent of the chart since, if (U2, ψ2) is another chart containing xthen φ1 (φ2|TxM )−1 : Rm → Rm is a linear automorphism.

Example 19. Since we can cover an open set U ⊂ Rm with the single chart ψ = idU : U → Rm,the tangent bundle to U is simply:

TU = U × Rm

with the projection map (x, v) 7→ x.

Exercise 20. Only using the definition of tangent bundle, try to visualize TS1 and to prove thatit is homeomorphic to S1 × R.

1.2. Smooth maps between manifolds and their differential.

Definition 21 (Smooth map). A continuous map f : M → N between two smooth manifolds issaid smooth if for every choice of charts (U,ψ) for M and (V, ϕ) for N such that f(U) ⊂ V (thesecharts will be called adapted) the function ϕ f ψ−1 : ψ(U) → Rn is smooth. If f is smooth,for every x ∈ M we pick a chart (U,ψ) = (Uα, ψα) containing x and a chart (V, ϕ) = (Vβ , ϕβ)containing f(x) (up to shrinking U we can assume these charts are adapted), and we define thedifferential of f to be the map df : TM → TN given by:

(1.2) df : [(x, v)α] 7→ [(f(x), Jψ(x)(ϕ f ψ−1)v)β ]

(applying the chain rule shows that the definition is independent of the charts). The restriction ofdf to TxM is denoted by:

dxf = df |TxM : TxM → Tf(x)N.

It is a linear map, called the differential of f at x.

Exercise 22. Prove that the projection map p : TM → M for the tangent bundle of a smoothmanifold is a smooth map.

Exercise 23. Prove that the map f : S1 → RP1 given by (x0, x1) 7→ [x0, x1] is smooth.

We immediately observe a couple of useful properties, which readily follows from the definition:

(1) if f : M → N and g : N → P are smooth maps between manifolds, then

dx(g f) = df(x)g dxf(“the differential of a composition is the composition of the differentials”);

(2) if f : Rm → Rn is smooth, then dxf : TxM ' Rm → Tf(x)N ' Rn is just the classicaldifferential of a smooth map, i.t. the unique linear map such that:

f(x+ v)− f(x) = dxfv +O(‖v‖2).

Remark 24 (Another definition of tangent space). For practical purposes it is often convenient tohave an alternative equivalent definition of tangent space and of (1.2). First, we can define TxMas the set of equivalence classes of smooth curves γ : (−ε, ε) → M such that γ(0) = x, with theequivalence relation γ1 ∼ γ2 if and only if for every chart (U,ψ) containing x we have:

d

dt(ψ(γ1(t)))

∣∣∣∣t=0

=d

dt(ψ(γ2(t)))

∣∣∣∣t=0

.

If (U,ψ) = (Uα, ψα) is a chart containing x, to every equivalence class [γ] we associate the element[(γ(0), ddt (ψ(γ(t)))|t=0)α]; viceversa, given [(x, v)α] we can construct the curve γ(t) = ψ−1

α (ψα(x)+tv), which is well defined on the interval (−ε, ε) for ε > 0 small enough. Since these two process

8 ANTONIO LERARIO

are inverse to each other, this shows that the two definitions are equivalent.This alternative definition is particularly convenient for “computing” the differential of a mapf : M → N . In fact, given v ∈ TxM we can write v = [γ] for some γ : (−ε, ε)→M with γ(0) = xand dxfv = [f γ].

Exercise 25. There is also a third equivalent definition of tangent space, viewing tangent vectorsas derivations. The reader is warmly encouraged to check out also this definition (for example asit is done in [8, Chapter 3]), and to prove that it is equivalent to the one we have given here.

Remark 26 (The differential in coordinates). Let f : M → N be a smooth map. Given a pointx ∈ M and adapted charts ψ : U → Rm and ϕ : V → Rn (with U neighborhood of x and Vneighborhood of f(x)) there is a natural way to construct bases for TxM and Tf(x)N which allowto write the matrix associated to the linear operator dxf in these bases. This is done as follows.First, let

TxRm = Rm = spane1, . . . , emand define for every i = 1, . . . ,m the vectors2

∂

∂xi= (dψ(x)ψ

−1)ej ∈ TxM.

Similarly, setting y = f(x), for j = 1, . . . ,m we define:

∂

∂yj= (dϕ(y)ϕ

−1)ei ∈ TyN.

Denoting byA = (aij) the matrix representing dxf in the coordinates given by the bases ∂∂xii=1,...,m

for TxM and ∂∂yjj=1,...,n for TyN , we have by definition:

dxf∂

∂xi=

n∑j=1

aij∂

∂yj.

Applying df(y)ϕ to both sides of the previous equation, and using the definitions, we have:

n∑j=1

aijej =

n∑j=1

aijdf(y)ϕ∂

∂yj= df(y)ϕ

n∑j=1

aij∂

∂yj

= df(y)ϕ

(dxf

∂

∂xi

)= df(y)ϕ

(dxf(dψ(x)ψ

−1)ej)

= dψ(x)(ϕ f ψ−1)ei.

In the last line we have the usual differential of a map ϕ f ψ−1 : Rm → Rn, whose representingmatrix in coordinates is the Jacobian matrix Jψ(x)(ϕf ψ−1); the above chain of equalities provesthat this Jacobiam matrix coincides with the matrix A.

Remark 27. Observe the following nice identity (which justifies the choice of the notation in theprevious Remark 26) for a smooth map f : Rm → R:

dxf∂

∂xi=

∂f

∂xi(x).

Definition 28 (Immersion, embedding and diffeomorphism). Let ϕ : M → N be a smooth map.We will say that ϕ is an immersion if for every x ∈M the differential dxϕ is injective. We will saythat ϕ is an embedding if it is an immersion and moreover it is a homeomorphism onto its image.We will say that a homeomorphism ϕ : M → N is a diffeomorphism if it is an embedding (notethat in particular the inverse ϕ−1 : N →M is also a diffeomorphism).

Exercise 29. Let f : M → N be a smooth map. Prove that the set of points x ∈ M such thatthe differential dxf has maximal rank is open in M .

Exercise 30. Prove that there is no immersion of S1 × S1 into R2. On the other hand there isan immersion of S1 × S1\one point → R2.

Exercise 31. Prove that S2 and CP1 are diffeomorphic.

2These are just symbols, but they have a useful and suggestive interpretation!


ψ

R

X

Figure 4. The set X = graph(x 7→ |x|) ⊂ R2 is an analytic manifold, becauseit can be covered with a single chart ψ : (x, |x|) 7→ x, but is is not a smoothsubmanifold of R2.

Remark 32. Consider on R the two atlases A1 = (R, ϕ1 = id) and A2 = (R, ϕ2 : x 7→ x3). Itfollows from Remark 2 above that the two atlases A1 and A2 are not C1-equivalent (but only C0

equivalent). However R with the differential structure induced by A1 and R with the differentialstructure induced by A2 are diffeomorphic, the diffeomorphism being the map x 7→ x1/3.

1.3. Submanifolds and how to produce them.

Definition 33 (Submanifold). Let M be a smooth manifold of dimension m. A subset X ⊂M iscalled a submanifold of dimension k if every x ∈ X belongs to the domain of a chart3 (U,ψ) for Msuch that U ∩X = ψ−1(Rk), where Rk ⊆ Rm is a linear subspace. (Observe that a submanifoldis itself a manifold with charts of the form (U ∩X,ψ|U∩X).) The difference dim(M)− dim(X) iscalled the codimension of X in M .

Example 34. For α ≥ 0 let Γα ⊂ R×R be the graph of the function x 7→ |x|α. For an integer k,if k < α < k+1 then Γα is a submanifold of R2 which is Ck but not Ck+1. In fact, let ϕ : R2 → R2

be the chart given by (x, y) 7→ (x, y − |x|α). Then ϕ−1(R× 0) = Γα.

Before proceeding we observe two important features of the notion of submanifold.

(1) It has a local character: X ⊆ M is a submanifold if and only if there is an open coverXii∈I of X and there are open sets Mii∈I in M such that for every i ∈ I the setXi ⊂Mi is a submanifold of Mi.

(2) It is invariant under diffeomorphism: if ϕ : M → N is a diffeomorphim, then X ⊆M is asubmanifold if and only if ϕ(X) ⊂ N is a submanifold.

Exercise 35. Work out the details showing that the notion of submanifold has local characterand is invariant under diffeomorphisms.

Theorem 36. A subset X ⊆M is a submanifold if and only if it is the image of an embedding.

Proof. If X ⊆ M is a submanifold, then the inclusion ι : X → M is an embedding: ι is ahomeomorphism onto its image ι(X) = X and the differential dxι : TxX → TxM is simply theinclusion (hence it is injective).

Suppose now that f : X → M is an embedding. We need to prove that f(X) is a submanifoldof M . For every x ∈ X let (U1, ψ) be a chart for M on a neighborhood of f(x) and (V1, ϕ) a chartfor X on a neighborhood of X. Since f is a homeomorphism onto its image, there exists an openset W1 ⊂M such that f(V1) = f(X) ∩W1. We set:

Uf(x) = U1 ∩W1 and Vx = f−1(Uf(x)).

Observe that f(X) ∩ Uf(x) = f(Vx) and consider the diagram of maps:

Rn

Vx Uf(x)

Rm

ϕ ψ

f

3Recall that we assumed that we have a maximal atlas at our disposal, see Remark 3.

10 ANTONIO LERARIO4 ANTONIO LERARIO

f

Figure 3. An injective immersion f : I → R2 of an open interval which is notan embedding (the image of f is a closed subset of R2).

Observe that f(X) ∩ Uf(x) = f(Vx) and consider the diagram of maps:

Rn

Vx Uf(x)

Rm

ϕ ψ

f

Because the definition of submanifold has a local character, it is enough to prove that f(Vx) ⊂U(fx) is a submanifold for every x ∈ X . Moreover, by the invariance under diffeomorphism,f(Vx) ⊂ U(fx) is a submanifold if and only if ψ(f(Vx)) ⊂ Rm is a submanifold. Note now that

ψ f ϕ−1 : ϕ(Vx) → Rm

is an embedding of ϕ(Vx) ⊂ Rn into an open subset of Rm and ψ f ϕ−1(ϕ(Vx)) = ψ(f(Vx)).Thus we are reduced to prove the statement in the special case X = ϕ(Vx) ⊂ Rn is an opensubset and h = ψ f ϕ−1 : X → Rm is an embedding.

We pick now a point p ∈ h(X) and we realize (locally around p) the image h(X) as the graphof a function. To this end we will assume that p is the origin (up to translations, which willnot change the property of being a submanifold) and we pick x0 ∈ X such that p = h(x0). Weconsider the splitting:

Rm = L1 ⊕ L2 where L1 = im(dx0h)

(L2 is any complementary space, for example L2 = L⊥1 ). Since h is an embedding, L1 ≃ Rm.

We also consider the projections on the two factors p1 : Rm → L1 and p2 : Rm → L2 and set:

h1 = p1 h and h2 = p2 h.

In this way, in the coordinates given by the splitting Rm = L1⊕L2 we have h = (h1, h2). Observethat, dx0h1 : Tx0X → Rn is invertible and by the Inverse Function Theorem there exists an openneighborhhod A ⊂ L1 and a smooth function α : A → X such that

h1(α(a)) = a for all a ∈ A.

We define now the function ϕ : A → L2 by:

ϕ(a) = h2(α(a)).

Figure 5. An injective immersion f : I → R2 of an open interval which is not anembedding (the image of f is a closed subset of R2).

Because the definition of submanifold has a local character, it is enough to prove that f(Vx) ⊂U(fx) is a submanifold for every x ∈ X. Moreover, by the invariance under diffeomorphisms,f(Vx) ⊂ U(fx) is a submanifold if and only if ψ(f(Vx)) ⊂ Rm is a submanifold. Note now that

ψ f ϕ−1 : ϕ(Vx)→ Rm

is an embedding of ϕ(Vx) ⊂ Rn into an open subset of Rm and ψ f ϕ−1(ϕ(Vx)) = ψ(f(Vx)).Thus we are reduced to prove the statement in the special case X = ϕ(Vx) ⊂ Rn is an open subsetand h = ψ f ϕ−1 : X → Rm is an embedding.

We pick now a point p ∈ h(X) and we realize (locally around p) the image h(X) as the graphof a function. To this end we will assume that p is the origin (up to translations, which will notchange the property of being a submanifold) and we pick x0 ∈ X such that p = h(x0). We considerthe splitting:

Rm = L1 ⊕ L2 where L1 = im(dx0h)

(L2 is any complementary space, for example L2 = L⊥1 ). Since h is an embedding, L1 ' Rn. Wealso consider the projections on the two factors p1 : Rm → L1 and p2 : Rm → L2 and set:

h1 = p1 h and h2 = p2 h.In this way, in the coordinates given by the splitting Rm = L1 ⊕L2 we have h = (h1, h2). Observethat dx0

h1 : Tx0X → Rn is invertible, and by the Inverse Function Theorem, there exist an open

neighborhood A ⊂ L1 and a smooth function α : A→ X such that:

h1(α(a)) = a for all a ∈ A.We define now the function ϕ : A→ L2 by:

ϕ(a) = h2(α(a)).

Since h is an embedding (in particular a homeomorphism onto its image), by possibly furthershrinking the neighborhood A, there exists a neighborhood B ⊂ L2 of zero such that:

h(X) ∩ (A×B) = h(α(A)).

In particular in the neighborhood A×B of p ∈ h(X) we have:

h(X) ∩ (A×B) = (a, ϕ(a)) | a ∈ A.Consider now the map ψ : A×B → Rm given by:

(a, b) 7→ (a, b− ϕ(a)).

It is easy to see that ψ is a diffeomorphism on a neighborhhod U of p (since its Jacobian isnonvanishing at this point); moreover, by construction:

h(X) ∩ U = (a, ψ(a)) | a ∈ A ∩ U = (a, b) ∈ U | b = ϕ(a) = ψ−1(0 × Rn−m) ∩ U.Let us summarize what we have proved: for every p ∈ h(X) there is a neighborhood U of p and amap ψ : U → Rm which is a diffeomorphism (a chart) such that h(X) ∩ U = ψ−1(Rn−m): this isexactly the requirement for h(X) to be a submanifold of Rn.

Exercise 37. Work out the details of the last step of the previous proof.


Exercise 38. LetG(2, 4) be the Grasmmanian of 2-planes in R4 and consider the map p : G(2, 4)→RP5 defined by:

p([M ]) = [det(M |j1,j2), . . . ,det(M |j3,j4)]

(the subscripts range over all possible pairs (i, j) ∈

42

, there are 6 such pairs.) Check that the map

p (called Plucker embedding) is an embedding. Denoting by [p12, p13, p14, p23, p24, p34] the homoge-neous coordinates of RP5, prove that the image of p is the quadric4 p12p34 +p13p24 +p14p23 = 0.More generally the Grassmannian G(k, n) embeds into RP(nk)−1 via an analogous Plucker embed-ding (though the description of the image of this embedding is more complicated). Can you figureout the general construction?

Definition 39 (Regular value). Let f : M → N be a smooth map. We will say that y ∈ N is aregular value for f if for every x ∈ f−1(y) the differential dxf (a linear map from TxM to TyN) issurjective.

Remark 40. Observe that a regular value does not need to be a value: if y /∈ im(f) then there isno x in f−1(y) and the condition in the above definition is automatically satisfied.

Theorem 41. Let y ∈ N be a regular value of a smooth map f : M → N . Then f−1(y) is asmooth submanifold of M ; if nonempty, it has dimension dim(M)− dim(N).

Proof. By using local character and invariance under diffeomorphisms of the definition of subman-ifold, we can reduce (as in the poof of Theorem 36) to the case M ⊂ Rm is an open subset andN = Rn. In this case the conclusion follows from the Implicit Function Theorem.

Whenever a manifold M is described as the preimage of a regular value M = f−1(y), we willsay that the equation f = y is regular.

Exercise 42. Work out the details of the last step in the previous proof.

Remark 43. Observe that if X = f−1(y) with y a regular value of f : M → N , then:

TxX = ker dxf.

One inclusion is simple: if v ∈ TxX, then there exists γ : (−ε, ε) such that γ(0) = x and [γ] = v.Then, since f(γ(t)) ≡ y is a constant curve, it follows that

dxfv = [f γ] = 0.

The other inclusion follows from the fact that, since the equation f = y is regular, TxX andker dxf have the same dimension!

Remark 44. We observe that the two ways we have to produce submanifolds (i.e. using embed-dings or giving their equations) are essentially the two ways we have to exhibit vector subspaces ofa vector space (i.e. as the span of some vectors, or as the set of solutions of a system of independentlinear equations).

Example 45 (Smooth projective hypersurfaces). Let F : Rn+1 → R be a homogeneous polynomialof degree d. Since F is homogeneous, the following set Z(F ) ⊂ RPn is well defined:

Z(F ) = [x0, . . . , xn] ∈ RPn such that F (x0, . . . , xn) = 0.If the vector ∇F =

(∂F∂x0

, . . . , ∂F∂xn

)(the gradient of F ) is nonzero at every non-zero point of

F = 0 ⊂ Rn+1 (a non-degeneracy condition), then Z(F ) is a smooth submanifold of RPn. To seethis we use the fact that being a submanifold is a local property, i.e. in order to prove that Z(F )is a submanifold, it is enough to cover RPn with the open sets RPn =

⋃nj= Uj defined in Example

8 and prove that Z(F )∩Uj is a submanifold of RPn for every j = 0, . . . , n. Moreover, since beinga submanifold is invariant under diffeomorphisms, it is enough to prove that ψj(Z(F ) ∩ Uj) is asubmanifold of Rn for every j = 0, . . . , n (the ψj : Uj → Rn are the manifold charts for RPn).Now, if we define the function fj : Rn → R by fj(y0, . . . , yj , . . . , yn) = F (y0, . . . , 1, . . . , yn), the setψj(Z(F ) ∩ Uj) is given by the equation:

ψj(Z(F ) ∩ Uj) = fj = 0.4This quadric has signature (3, 3) and is double-covered by S2 × S2; since χ(S2 × S2) = 4, then the Euler

characteristic of G(2, 4) is 2.

12 ANTONIO LERARIO

The condition ∇F |F=0\0 6= 0 tells that the equation fj = 0 is regular (i.e. 0 is a regularvalue of fj). In fact if for some y ∈ fj = 0 we had ∇fj(y) = 0, then:

∇F (y0, . . . , 1, . . . , yn) =

(∂F

∂x0, . . . ,

∂F

∂xj, . . . ,

∂F

∂xn

) ∣∣∣∣(y0,...,1,...,yn)

=

(0, . . . , 0,

∂F

∂xj

∣∣∣∣(y0,...,1,...,yn)

, 0, . . . , 0

)= (0, . . . , 0, d · F (y0, . . . , 1, . . . , yn), 0, . . . , 0)

= 0

contradicting the non-degeneracy condition (in the last line we have used Euler’s identity forhomogeneous functions d · F (x) =

∑nj=0 xj

∂F∂xj

(x).) We will review this example in Example 107

below.

Exercise 46. Let F : Rn+1 → R as in the previous example be a homogeneous polynomial ofdegree d and define the function f = F |Sn . Prove that if for every x ∈ Rn+1\0 such thatF (x) = 0 we have ∇F (x) 6= 0, then the equation f = 0 is a regular equation on Sn (i.e. f = 0is a smooth submanifold of Sn, of dimension n − 1 if nonempty). Prove that the covering mapq : Sn → RPn restricts to a smooth covering map p|f=0 : f = 0 → Z(F ) ⊂ RPn.

Exercise 47. Prove that if y is a regular value of a smooth map f : S1 → R, then f−1(y) haseven cardinality.

Exercise 48. Prove that there is no smooth function f : RP2 → R such that RP1 = f−1(y),where y ∈ R is a regular value.

Exercise 49. Let Z ⊂ RPn be a smooth algebraic hypersurface defined by the regular equationF = 0 with F : Rn+1 → R a homogeneous polynomial of even degree. Prove that there exists asmooth function ϕ : RPn → R whose zero set is Z and such that the equation ϕ = 0 is regularon RPn.

Exercise 50. Let M(r;n,m) ⊂ Rn×m be the set of matrices of rank r. Prove that M(r;n,m) isa smooth submanifold of codimension (n− r)(m− r).

1.4. Every compact manifold embeds into some Euclidean space.

Theorem 51. Let M be a compact manifold. There exists an embedding ϕ : M → Rn for some(possibly very large) n ≥ dim(M).

Proof. We first claim that if M is compact there exists a finite atlas (Uk, ψk)`k=1 such thatψk(Uk) ⊂ B(0, 2) ⊂ Rm (here m = dim(M)) and

(1.3) M =⋃k=1

int(ψ−1k (D(0, 1))

).

Let now λ : Rm → [0, 1] a bump function constructed as in Lemma 190 with the choice c1 = 1 andc2 = 2, and for every k = 1, . . . , ` define the function λk : M → [0, 1] by

λk(x) = χUk(x) · λ(ψk(x))

(we are “pulling back” the bump function λ to a bump function on M using the chart ψk). Observethat (1.3) ensures that the sets Dk = λ−1

k (1)`k=1 cover M .

Define now for k = 1, . . . , ` the functions fk : M → Rm by

fk(x) = χUk(x) · λk(x)ψk(x)

(we are using the λk to extend the charts ψk : Uk → Rm to the whole M).

We finally define the map ϕ : M → R`(m+1) by:

ϕ(x) = (f1(x), λ1(x), . . . , f`(x), λ`(x)).

We need to check that ϕ is a homeomorphism onto its image and that for every x ∈ M thedifferential dxϕ : TxM → Tϕ(x)R`(m+1) ' R`(m+1) is an injective linear map. Observe first thatϕ is injective: if x 6= y and y ∈ Dk, then either (1) x ∈ Dk, in which case ψk(x) 6= ψk(y), or (2)


x /∈ Dk and 1 = λk(y) 6= λk(x). Thus ϕ is injective and takes value in a Hausdorff space, M iscompact: hence ϕ is a homeomorphism onto its image. It remains to check that dxϕ is injectivefor every x ∈M . Note that

dxϕ = (dxf1, dxλ1, . . . , dxf`, dxλ`),

hence to prove that dxϕ is injective it is enough to prove that one of its components is injective.Since every x belongs to some Dk, then dxfk = dxψk is injective because ψk is a diffeomorphism.This concludes the proof.

Exercise 52. Prove that the product of spheres M =∏mi=1 S

ni embeds into R1+∑mi=1 ni .

1.5. The tangent bundle for embedded submanifolds. In the case M ⊆ Rn the tangentbundle can be defined in an equivalent way as follows (observe that if M is compact by Theorem51 we can always assume it is embedded in some Rn):

(1.4) EM = (x, v) ∈ Rn × Rn | v ∈ TxM.Exercise 53. Assuming that M ⊂ Rn is an embedded submanifold, using the definition (1.4),check that EM is a smooth submanifold of Rn × Rn. In other words, check that the “equations”v ∈ TxM defining EM ⊂ Rn × Rn are regular.

Exercise 54. Let ϕ : M → Rn be an embedding with image N = ϕ(M). Prove that the mapTM → TRn given by [γ] 7→ [ϕ γ] is an embedding with image TN. In particular if M ⊂ Rn, TMan EM can be naturally identified.

The previous definition is particularly useful because it allows to talk about the “length” of atangent vector without introducing the notion of Riemannian manifold.

Proposition 55. The set E1M = (x, v) ∈ EM | ‖v‖ = 1 (called the unit tangent bundle) is asmooth submanifold of EM of dimension 2 dim(M)− 1.

Proof. Consider the smooth function ρ : EM → R defined by:

ρ(x, v) = ‖v‖2.(This function is smooth because it is given by the composition of two smooth functions.) ThenE1M = ρ−1(1) and to see that E1M ⊂ EM is a submanifold it is enough to prove that 1 is aregular value for ρ and then use Theorem 41. To this end, let (x, v) ∈ TM such that ‖v‖2 = 1 andconsider the curve γ : (ε, ε)→ TM given by γ(t) = (x, (1 + t)v). Then γ(0) = (x, v) and

d

dtγ(t)

∣∣∣∣t=0

=d

dt‖tv‖2

∣∣∣∣t=0

=d

dt(1 + t)2

∣∣∣∣t=0

= 2.

Hence im(d(x,v)ρ) ⊃ spand(x,v)ρ(ddtγ(t)

∣∣t=0

) = R and d(x,v)ρ is surjective.

Remark 56. Observe that if M ⊂ Rn is compact, then T 1M is compact, because it is closed andbounded: M is compact, hence contained in a bounded set K ⊂ Rn and

TM ⊂ K ×B(0, 1)

which is bounded in Rn × Rn.

Example 57 (The group SO(3)). The group SO(3) is defined by:

SO(3) =A ∈ R3×3 such that AAT = 1 and det(A) = 1

.

This is an example of a Lie group, i.e. a group G which is also a smooth manifold and such thatthe group multiplication µ : G×G→ G, given by µ(g1, g2) = g1g2, and the inverse map η : G→ G,given by η(g) = g−1, are both smooth. The smooth structure on SO(3) is given as follows. Weconsider the map F : R3×3 → Sym(3,R) ' R6 given by:

F (A) = AAT ,

and we observe that 1 is a regular value of F (check it!), then we can use Theorem 41 and deducethat O(3) = F−1(1) is a smooth submanifold of R9. since the determinant function takes constantvalues on each components of O(3), we see that SO(3) is a union of connected components of O(3),

14 ANTONIO LERARIO

vx

TxS2

T 1xS2

S2

Figure 6. The (embedded) unit tangent bundle to S2. The vectors v and x areorthogonal and of norm one; the map (x, v) 7→ (x, v, x∧v) gives a homeomorphismT 1S2 → SO(3).

hence it is itself a smooth manifold. One important remark: O(3) consists actually of just twocomponents:

O(3) = SO(3) ∪ (−1 · SO(3)) .

The group SO(3) is homeomorphic to unit tangent bundle of S2 ⊂ R3. In fact we can define acontinuous, surjective and one-to-one map f : T 1S2 → SO(3) as follows (see Figure 6):

(x, v) 7→ (x, v, x ∧ v).

Since T 1S2 is compact, then f is a homeomorphism onto its image.

Exercise 58. Prove that SO(3) is also homeomorphic to RP3. (This implies that T 1S2 ' RP3.)

Exercise 59. Using Exercise 58 and Example 57, prove that every vector field on the sphere S2

(i.e. a continuous assignment v : S2 → TS2 with v(x) ∈ TxS2, see Section 3.1) must vanish atsome point.

Example 60 (The group SU(2)). The group SU(2) is defined by:

SU(2) =

A =

(z1 z2

z3 z4

)such that AA

T= 1 and det(A) = 1

.

We can endow SU(2) the structure of smooth manifold using again Theorem 41. As a smoothmanifold SU(2) is diffeomorphic to the sphere S3 and this can be seen as follows. Observe that

the condition AAT

= 1 can be written as:(z4 −z2

−z3 z1

)=

1

det(A)

(z4 −z2

−z3 z1

)= A−1 = A

T=

(z1 z3

z2 z4

),

which implies that A is of the form:

A =

(z1 z2

−z2 z1

).

In particular the group SU(2) can be described as:

SU(2) =

A =

(z1 z2

−z2 z1

)such that |z1|2 + |z2|2 = det(A) = 1

,

which is a sphere S3 ⊂ C2 = (z1, z2).The (embedded) tangent space to SU(2) at the identity is usually denoted by su(2) and it consists


of the vector space of matrices:

T1SU(2) = X ∈ C2×2 such that X = −XTand tr(X) = 0.

Writing a matrix X ∈ su(2) as

X =

(iu −v − iz

−v + iz −iu

), (u, v, z) ∈ R3,

we see that the determinant det(X) = u2 + v2 + z2 is a positive quadratic form on su(2) and itinduces on it a Euclidean structure. This allows to define a homomorphism of groups ϕ : SU(2)→O(3) (i.e. a group representation) via:

ϕ(A) = (RA : X 7→ AXAT

).

The image of ϕ is SO(3) and, recalling that SO(3) ' RP3 (Exercise 58), we therfeore have:

ϕ : SU(2) ' S3 → RP3 ' SO(3).

The map ϕ is a group homomorphism with kernel ±1 and topologically it is the double covermap.

Exercise 61. Try to prove the properties of ϕ : SU(2) → SO(3) rigorously (im(ϕ) = SO(3),ker(ϕ) = ±1, ϕ : S3 → RP3 is the covering space map), possibly referring to [2, Chapter 8].

1.6. Sets of measure zero. In this section we will define the notion of “sets of measure zero”,which introduces new ideas in the topic. We remark from the beginning that we will not definethe measure of a set, we will just define a special class of sets in a manifold: sentences like “theset A has measure zero” or “the measure of A is zero” mean that A belongs to this special class ofsets (those “of measure zero”).

Let D = [a1, b1]× · · · × [am, bm] be a cube in Rm. We will set µ(D) =∏mi=1 |bi − ai|.

Definition 62 (Sets of measure zero). We will say that A ⊂ Rm has measure zero if for everyε > 0 there exists a countable family of cubes Dkk∈J (countable means that the cardinality ofthe index set J is either finite or countable) such that

A ⊂⋃k∈J

Dk and∑k∈J

µ(Dk) ≤ ε.

If M is a smooth manifold, we will say that A ⊂M has measure zero if for every chart (U,ψ) forM the set ψ(A ∩ U) ⊂ Rm has measure zero.

Remark 63. Observe that Q ⊂ R has measure zero. In fact, let Q = rkk∈N and for every ε > 0consider the cubes Dk = [rk − ε

2k+1 , rk + ε2k+1 ]. Then:

Q ⊂⋃k∈N

Dk and∑k∈N

µ(Dk) =∑k∈N

ε

2k+1= ε.

Remark 64. If A ⊂M has measure zero, then it cannot contain any open set: in fact if V ⊂ A isopen, then for some chart (U,ψ) we have that ψ(U ∩V ) ⊂ Rm is open and nonempty. In particularψ(U∩V ) it contains a cube D with µ(D) = δ > 0. If we now try to cover ψ(U∩V ) with a countablecollection of cubes Dkk∈J , this collection must cover D and

∑k∈J µ(Dk) ≥ δ. In particular if

A ⊂M has measure zero, its complement is dense: let V ⊂M be any open set, then V ∩Ac 6= ∅,otherwise V would be contained in A.

On the other hand the reader should keep in mind that if B ⊂ M is dense, the complementdoesn’t necessarily have measure zero. For example let Q ∩ [0, 2π] = qkk∈N and set

B = (cos θ, sin θ) ∈ S1 | θ ∈ Q ∩ [0, 2π].Then B is dense in S1, but its complement does not have measure zero. (Can you prove thisrigorously?)

Exercise 65. For every ε > 0 construct an open and dense set A ⊂ Rm of Lebesgue measuresmaller than ε (this exercise requires measure theory).

Lemma 66. Let M be a smooth manifold and A ⊂M be a subset such that A ⊂ ⋃k∈J Ak with Jcountable and with each Ak ⊂M of measure zero; then A itself has measure zero.

16 ANTONIO LERARIO

Proof. Since J is countable, we may assume J = 1, 2, . . . ⊂ N. For simplicity let us just discussthe case A ⊂ Rm, leaving the manifold case to the reader.

Now, for every ε > 0 and for every k ∈ J cover Ak with cubes Dk,jj∈Jk with∑j∈Jk µ(Dk,j) ≤

ε2k+1 (such a cover exists because each Ak has measure zero). Then

A ⊂⋃k∈J

⋃j∈Jk

Dk,j and∑k∈J

∑j∈Jk

µ(Dk,j) ≤∑k∈J

ε

2k+1≤ ε.

Lemma 67. Let U ⊂ Rm be an open set and A ⊂ U be of measure zero. If f : U → Rn is a C1

function, then f(A) has measure zero.

Proof. Observe first that, since Rm is a second countable space, then U =⋃k∈J int(B(xk, rk))

where J is a countable set. Then U can also be written as:

U =⋃k∈J

⋃n∈N

B

(xk, rk −

1

n

),

i.e. U can be written as a countable union of closed (hence compact) balls. Relabeling, we canwrite:

U =⋃k∈N

Bk

where each Bk is a closed ball. Let us set Ak = A∩Bk; observe that Ak has measure zero (a subsetof a set of measure zero has itself measure zero). We will prove that f(Ak) has measure zero forevery k ∈ N, and the result will follow from Lemma 66, since f(A) =

⋃k∈N f(Ak).

Thus let us fix k. Observe that, since Bk is compact and f is C1, then f |Bk is Lispchitz andthere exists Lk > 0 such that for all x, y ∈ Bk:

‖f(x)− f(y)‖ ≤ Lk‖x− y‖.In particular, if C ⊂ Bk has diameter d, then f(C) has diameter at most d ·Lk; hence there existsCk > 0 (which also depends on m, but this is also fixed) such that if D ⊂ Bk is a cube, then f(D)is contained in a cube D′ with µ(D′) ≤ Ckµ(D).

Fix now ε > 0 and take a countable cover of Ak with balls Djj∈I such that∑j∈I µ(Dj) ≤ ε

Ck(this cover exists because we are assuming Ak has measure zero). By the above reasoning eachf(Dj) is contained in a D′j with µ(D′j) ≤ Ckµ(Dj). In particular:

f(Ak) ⊂⋃j∈I

D′j and∑j∈I

µ(D′j) ≤∑j∈I

Ckµ(Dj) ≤ Ck ·ε

Ck≤ ε,

which proves f(Ak) has measure zero.

Remark 68. Observe that the previous Lemma implies that for a set A ⊂ M to have measurezero it is enough to check ψ(A ∩ U) ⊂ Rm has measure zero for all charts (U,ψ) belonging to agiven atlas (i.e. we do not have to check it on all possible charts).

Lemma 69 (mini-Sard’s lemma). Let A and B be smooth manifold with dim(A) < dim(B) andg : A→ B be a C1 map. Then g(A) has measure zero in B and B\g(A) is dense.

Proof. Let (Uk, ψk)k∈I be a countable atlas for A and (Vj , ϕj)j∈J be a countable5 atlas for B.It is enough tot check that for every k ∈ I and j ∈ J the measure of ϕj(f(Uk) ∩ Vj) ⊂ Rb is zero.Let Ra be the codomain of ψk and Wk be its image. Consider the projection on the second factormap p : Rb−a×Wk →Wk. Then ϕj(f(Uk)∩ Vj) is the image of 0×Wk ⊂ Rb−a×Wk under theC1 map:

φ : Rb−a ×Wkp−→Wk

ψ−1

−→ Ukf−→ Vj

ψj−→ Rb

(this is a map between open sets in Rb.) Since 0×Wk has measure zero in Rb−a×Wk, then theresult follows from Lemma 67.

5The existence of such countable atlases follows from the assumption that both A and B are paracompact with

countably many components, hence second countable


Example 70. For every n there is a continuous surjective map f : S1 → Sn, which of course whenn > 1 cannot be C1 (this map is constructed using space-filling curves).

1.7. Whitney embedding theorem: weak form.

Theorem 71 (Weak Whitney embedding). Let M be a compact manifold of dimension m. Thereexists an embedding ϕ : M → R2m+1.

Proof. By Theorem 51 we can assume M is already embedded into some Rq. We will show thatif q > 2m + 1 then there exists a projection Rq → Rq−1 which restricts to an embedding ofM → Rq−1. The result will follow iterating the argument until q = 2m+ 1.

To make things more precise, let us put coordinates (x1, . . . , xq) on Rq and let us identifyRq−1 = xq = 0. For every unit vector v ∈ Sq−1\xq = 0 let us consider the linear mapfv : Rq → Rq−1 that projects in the direction of v. We need to find a vector v ∈ Sq−1\xq = 0such that fv|M is an embedding. To this end we need to verify:

(1) fv|M is injective;(2) for every x ∈M the differential dx(fv|M ) is injective.

The first condition is satisfied if there is no pair (x, y) ∈M ×M with x 6= y such that:

v =x− y‖x− y‖ .

We can rephrase this condition as follows. Let ∆ be the diagonal in M ×M ; then fv|M is injectiveif and only if v does not belong to the image of the map h1 : (M ×M)\∆→ Sq−1 given by:

h1(x, y) =x− y‖x− y‖ .

We apply now Lemma 69 to the smooth function h1. Since

dim(M ×M)\∆ = 2m < q − 1 = dim(Sq−1)

then im(h1) has measure zero in Sq−1 and there is a dense set D1 = Sq−1\im(h1) ⊂ Sq−1 of “good”choices for v (as long as we are concerned with injectivity of fv|M ).

The condition on the injectivity of the differential is handled as follows. Observe first that, sincefv is linear, the differential of fv at a point x ∈ Rq equals fv itself:

dxfv = fv.

Consequently the kernel of the linear map dxfv : TxRn ' Rn → Rn equals spanv (the linespanned by the direction of v, since by definition fv is the projection in the direction of v).Moreover, the differential of fv|M at a point x ∈M equals:

dx(fv|M ) = (dxfv)|TxMand ker dx(fv|M ) = TxM ∩ spanv. Thus the differential of fv|M at a point x ∈ M is injectiveif and only if v /∈ TxM . We can rephrase this condition as follows. Consider the smooth maph2 : T 1M → Sq−1 given by:

h2(x, v) = v.

Then fv|M is an immersion if and only if v does not belong to the image of h2. We can use nowLemma 69 again, since:

dim(T 1M) = 2m− 1 < q − 1 = dim(Sq−1).

As a consequence there is a dense subset D2 = Sq−1\im(h2) of “good” choices for v (as long as weare concerned with injectivity of dfv|M ).

Observe that since T 1M is compact (see Remark 56), then im(h2) is compact and D2 is notonly dense, but also open. In particular the intersection D1 ∩ D2 is nonempty (in fact dense inSq−1) and every v ∈ D1 ∩D2 produces a map fv which restricts to an embedding of M → Rq−1.

Exercise 72. Prove that if M is a compact smooth manifold of dimension m, then there existsan immersion of M in R2m.

18 ANTONIO LERARIO

Exercise 73. Let M be a compact manifold of dimension m and h : M → Rq be a smooth map,where q ≥ 2m+ 1. Prove that for every ε > 0 there exists an embedding hε : M → Rq such that:

maxx∈M

‖h(x)− hε(x)‖ ≤ ε.


2. Transversality and Sard’s lemma

2.1. Transversality.

Definition 74. Let f : M → N be a smooth map between two smooth manifolds and A ⊂ N bea smooth submanifold. We will say that f is transversal to A and write f t A if for every x ∈Msuch that f(x) ∈ A we have:

Tf(x)N = Tf(x)A+ im(dxf).

Moreover, if A,B are submanifolds of a given manifold N , we will say that A and B are transverseand write A t B if the inclusion ιA : A → N is transverse to B (or equivalently the inclusionιB : B → N is transverse to A), in other words for every y ∈ A ∩B we have:

TyN = TyA+ TyB.

(Note that A t B means that, if A and B intersect they do it in a transverse way; in particulartwo nonintersecting manifolds are transverse.)

Exercise 75. Let C ⊂ Rn be a closed set. Prove that there exists a smooth function f : Rn → Rwhose zero set coincides with C.

Exercise 76. Let C ⊂ Rn ⊂ Rn+1 be a closed set. Prove that there exists a submanifoldA ⊂ Rn+1 such that A ∩ Rn = C (in particular a nontransverse intersection of manifolds can bevery complicated, essentially as complicated as any closed set).

Exercise 77. Generalize Exercise 75: prove that if C is a closed set in a smooth manifold M ,there exists a smooth function f : M → R whose zero set coincides with C. (Hint: use partitionof unities.)

Observe that in the case A = y is a point f t A if and only if y is a regular value of f ;moreover in this case f−1(y) is a submanifold of M by Theorem 41. Next Theorem generalizesthis result.

Theorem 78. Let f : M → N a smooth map between two smooth manifolds and A ⊂ N be asubmanifold such that f t A. Then f−1(A) is a submanifold of M of codimension:

codimM (f−1(A)) = codimN (A).

Proof. Let x ∈ M such that f(x) = y ∈ A. Since A is a submanifold of N , for every y ∈ A thereexists a neighborhood Uy ⊂ N and a chart (Uy, ψ) such that:

A ∩ Uy = ψ−1(Ra).

Let us write now Rn = Ra × Rn−1 and let us consider the projection π : Rn → Rn−a. Denote byϕ the composite map ϕ = π ψ; then:

A ∩ Uy = ϕ−1(0).

We now consider the open set Wx = f−1(Uy) (a neighborhood of x in M) and the map:

φ = ϕ f : W → Rn−a.We claim that 0 is a regular value of φ. In fact, since f t A we have:

Rn = Tψ(y)ψ(A) + im(dx(ψ f)).

Applying dyπ on both sides we obtain:

Rn−a = Tφ(x)Rn−a = dyπ(Rn) = dyπ(Tψ(y)ψ(A)) + im(dxφ) = im(dxφ),

where we have used the fact that dyπ(Tψ(y)ψ(A)) = 0. This proves that 0 is a regular value for

φ and φ−1(0) = Wx ∩ f−1(A) is a submanifold of M . Repeating this for every x ∈ f−1(A) weget a cover of f−1(A) with open sets of M such that in each of these open sets f−1(A) ∩Wx is asubmanifold of M , hence f−1(A) itself is a submanifold of M .

The dimension of f−1(A) equals dim(f−1(A)∩Wx) = dim(M)−dim(N)−a, hence by Theorem41 its codimension is:

codimM (f−1(A)) = dim(M)− dim(f−1(A)) = dim(N)− a = codimN (A).

20 ANTONIO LERARIO

2.2. Sard’s Lemma.

Definition 79. Let f : M → N be a Ck function, k ≥ 1. We say that x ∈ M is a critical pointfor f is the differential dxf : TxM → Tf(x)N is not surjective. We will denote by Crit(f) ⊂M theset of critical points of f . A critical value is a point y ∈ N which is the image of a critical point.

We will need the following Lemma, whose proof is easy and is left as an exercise.

Lemma 80. Let I ⊂ R and K ⊂ I × Rn−1 ⊂ Rn be a closed set. Assume that for every t ∈ I theset Kt = K ∩ (t ×Rn−1) has measure zero in t ×Rn−1 ' Rn−1. Then K has measure zero inRn.

We will now prove the following result, which is fundamental in differential topology; the proofis taken from [9, Chapter 3].

Theorem 81 (Sard’s Lemma). Let f : M → N be a smooth map. Then f(Crit(f)) (i.e. the setof critical values of f) has measure zero in N . In particular the set of regular values of f is densein N (and open if M is compact).

Remark 82. The Theorem is not true if f is merely Ck with k <∞: in this case, to make it true,one has to add the requirement that k > max0,m− n.

Proof. First, let us observe that we can reduce to study the case of a map f : U → Rn, where U isan open subset of Rn. In fact we can find countable6 atlases (Uj , ψj)j∈N of M and (Vj , ϕj)j∈N ofN such that for every j ∈ N we have f(Uj) ⊂ Vj . Then, because we are only dealing with countablymany sets, thanks to Lemma 66 it will be enough to prove that f(Crit(f |Uj )) has measure zero for

every j ∈ N. But now observe that Crit(f |Uj ) = Crit(f ψ−1j ) (because ψj is a diffeomorphism);

moreover ϕj(Crit(f ψ−1j )) = Crit(ϕj f ψ−1

j ) and if we prove the statement for ϕj f ψ−1j ,

then it will also be proved for f ψ−1j (since ϕj is a diffeomorphism, it sends sets of measure zero

to sets of measure zero).

The proof is by induction on m, the case m = 0 being clearly true. Let us assume we haveproved it for all function f : U → Rn with U ⊂ Rm−1 and let’s prove it for a function f : U → Rnwith U ⊂ Rm.

We denote for simplicity by C the set of critical points of f and for every k ≥ 1 let us define:

Ck = x ∈ U | all partial derivatives of f of order up to k vanish at x.Note that each Ck is closed (it is given by the vanishing of finitely many continuous functions, thepartial derivatives of f up to order k) and:

C ⊃ C1 ⊃ C2 ⊃ C3 . . .

Note that this chain of inclusions is potentially infinite, for example for the function f(x) = e−1x2

the origin belongs to Ck for every k ≥ 1. The proof consists now of three steps.

(1) First we show that for k large enough the set f(Ck) has measure zero.(2) Then we show that f(C\C1) has measure zero.(3) Finally we prove that for every k ≥ 1 the set f(Ck\Ck+1) has measure zero.

From this the conclusion follows, since:

f(C) = f(C\C1) ∪ f(C1\C2) ∪ · · · ∪ f(Ck−1\Ck) ∪ f(Ck),

and each piece has zero measure by the above three steps.

Let us prove the first step. Let S ⊂ U be a cube whose sides have length δ > 0 (a fixed constant).The set Ck can be covered by countably many such cubes, again by Lemma 66 it is enough toprove that f(Ck ∩ S) has measure zero. If x ∈ Ck all partial derivatives of f up to order k vanishat x and we can write:

(2.1) f(x+ h) = f(x) + ‖h‖k+1R(x, h)

where R(x, h) is a continuous function, whose absolute value is consequently bounded by a constanta > 0 on S (which is compact).

6The existence of such countable atlases again follows from the assumption that manifolds are paracompact with

countably many components, hence second countable


Let us now cover S ⊂ Rm with rm many cubes of size δ/r. If S1 is a cube of this subdivisioncontaining a point x ∈ Ck, then every point in S1 is of the form x + h with ‖h‖ ≤ √mδ/r.Consequently (2.1) implies that f(S1) is contained in a cube centered at f(x) with sides of lengthb = 2a(

√mδ/r)k+1 (this is just a constant). It follows that f(Ck ∩ S) is contained in the union of

at most rm cubes whose total volume v is bounded by:

v ≤ rm(

b

rk+1

)n= bnrm−(k+1)n.

If k is large enough (say k ≥ m/n) then bnrm−(k+1)n → 0 as r → ∞ and for every ε > 0 the setf(Ck ∩ S) is contained in the union of finitely many cubes of total volume smaller than ε, i.e. it ishas measure zero. This proves point (1) above.

Let us now prove point (2). Let x ∈ C\C1: we will find an open neighborhood V of x suchthat f(V ∩ (C\C1)) has measure zero. Since C\C1 can be covered by countably many suchneighborhhoods (because Rm is second countable), this will prove point (2) (again because ofLemma 66).

Since x /∈ C1, then there is at least one partial derivative, say ∂f∂x1

, which is nonzero at x. Letus consider the map h : Rm → Rm given by:

h(x) = (f1(x), x2, . . . , xm).

Since ∂f∂x1

(x) 6= 0 this map is a diffeomorphism between two open sets V (a neighborhood of x)

and V ′ ⊂ Rm. We will assume V ′ is of the form V ′ = I × V ′′, with I ⊂ R. The critical values off |V are the same of g = f h−1. We denote by (g1, . . . , gn) the components of g. This map, byconstruction, can be written as:

g(t, x2, . . . , xn) = (t, g2(t, x), . . . , gn(t, x)).

In particular for every t ∈ I we obtain a smooth map gt : Rm−1 → Rn−1:

gt(x2, . . . , xm) = (g2(t, x), . . . , gn(t, x)).

Note that a point x ∈ V ′′ is critical for gt if and only if (x, t) is critical for g: in fact the Jacobianmatrix of g has the form:

Jg(t, x) =

1 0 · · · 0∗... Jgt(x)∗

.

In particular rk(Jg(t, x)) = rk(Jgt(x))+1 and g is a submersion at (t, x) if and only if rk(Jg(t, x)) =n, i.e. if and only if rk(Jgt(x)) = n− 1, which is equivalent to gt being a submersion at x. By theinductive hypothesis, the measure of the set of critical values of gt : V ′′ ⊂ Rm−1 → Rn−1 is zero.Let us denote now by K the set of critical values of g and by Kt ⊂ t × Rn−1 the set of criticalvalues of gt. Then K =

⋃t∈I Kt, and since we have proved that for every t ∈ I the measure of

Kt is zero, by Lemma 80 K itself has measure zero. Note that we can apply Lemma 80 because,up to shrinking the neighborhood V ′, we can assume K is closed: critical points are closed (theyare defined by the vanishing of continuous functions), and intersected with a small enough ballare compact; hence the image of their intersection with a small enough ball under a continuousfunction is compact, hence closed. Since C\C1 can be covered by countably many open sets likeV , then together with Lemma 66 this proves (2).

Let us finally prove point (3). Observe that Ck is defined by the vanishing of all partial deriva-tives of f up to order k. If x ∈ Ck\Ck+1, then there exists one such partial derivatives, call it u,such that u|Ck ≡ 0 but ∂u

∂x1(x) 6= 0. We argue now in a similar way as in the previous point and

define the diffeomorphism h : V → V ′ (V is a neighborhood of x):

h(x) = (u(x), x2, . . . , xm).

Since h(Ck ∩ V ) ⊂ 0 × Rm−1, the map g = f h−1 has all its critical points on 0 × Rm−1.Consider now g = g|0×Rm−1)). Each point in h(Ck ∩ V ) is critical point for g; by the inductivehypothesis, g(Crit(g)) has measure zero and consequently also f((Ck\Ck+1)∩V ) has measure zero.Since Ck\Ck+1 can be covered by countably many open sets like V , then the result follows fromLemma 66.

22 ANTONIO LERARIO

2.3. Parametric transversality.

Theorem 83 (Parametric transversality). Let F : V ×M → N be a smooth map between smoothmanifolds and A ⊂ N be a submanifold such that F t A. For every v ∈ V consider the mapfv : M → N defined by fv(x) = F (v, x). Then the set v ∈ V | fv t A is dense in V (in fact itscomplement has zero measure).

The meaning of this theorem is the following: if we interpret F : V ×M → N as a family ofmaps parametrized by V , under the assumption that F t A, “most” of the maps fv are transverseto A.

Remark 84. If a map F : M → N is a submersion, then it is automatically transverse to allsubmanifolds A → N. That is why sometimes in order to verify that the conditions of the previoustheorem are satisfied it is enough to check that F is a submersion.

Proof. Since F t A then W = F−1(A) is a submanifold of V × M by Theorem 78. Let p1 :V ×M → V the projection onto the first factor and π = p1|W . We claim that if v is a regular valueof π then fv is transversal to A.

To this end observe that if v is a regular value of π, then:

(2.2) T(v,x)(V ×M) = TxM + T(v,x)(F−1(A)).

We now write:

Tfv(x)N = TF (v,x)N

= TF (v,x)A+ d(v,x)F (T(v,x)(V ×M)) (since F t A)

= TF (v,x)A+ d(v,x)F (TxM) + d(v,x)(T(v,x)(F−1(A))) (because of (2.2))

= TF (v,x)A+ d(v,x)F (TxM) (because d(v,x)(T(v,x)(F−1(A))) ⊂ TF (v,x)A)

= Tfv(x)A+ dxfv(TxM),

which is the condition fv t A. The result follows now from Sard’s Lemma.

Exercise 85. Using the notation of the previous proof prove that indeed v is a regular value of πif and only if fv t A.

Exercise 86. Let A,B be two submanifolds of Rn. For every w ∈ Rn define the translate manifoldAw = a+ w | a ∈ A (the translation of A in the direction of w). Prove that there exists a denseset D ⊂ Rn such that for every w ∈ D the manifolds Aw and B are transverse to each other.

Exercise 87. Generalizing the previous exercise, let G be a Lie group and A,B be submanifolds.For every g ∈ G define Ag = Lg(A) (the left translation of A in the direction of g). Prove that theset of g ∈ G such that Ag t B is dense.

Exercise 88. Let f : Rn → Rn be a smooth function and K ⊂ Rn be a compact set. Assumethat n > 1. Prove that for every ε > 0 there exists a smooth function fε : Rn → Rn such that thedifferential dxfε is never zero (as a linear operator) and supx∈K ‖f(x)− fε(x)‖ ≤ ε.Exercise 89. Let f : Rn → Rm be a smooth function and K ⊂ Rn be a compact set. Assume thatm ≥ 2n. Prove that for every ε > 0 there exists an immersion fε : Rn → Rm (i.e. the differentialdxfε has rank n at every point x ∈ Rn) such that supx∈K ‖f(x)− fε(x)‖ ≤ ε.Exercise 90. A knot is a topological embedding ϕ : S1 → R3 (if ϕ is smooth, we call it a smoothknot). Let R2 ⊂ R3 be a linear subspace and ϕ : S1 → R3 be a smooth knot with image K (which,in particular, is a submanifold of R3). For every v ∈ R3\R2 denote by πv : R3 → R2 the orthogonalprojection in the direction of v. Prove that there exists v ∈∈ R3\R2 such that: (1) πv|K is animmersion; (2) every z ∈ πv(K) has at most two preimages (i.e. there are no triple-points) and (3)given z ∈ πv(K) with two preimages x1, x2, the images of the differentials dx1

(πv|K) and dx2(πv|K)

are transverse.


E

S1

π

Figure 7. The Moebius band has a structure of a vector bundle E → S1. Thetautological bundle τ1,2 → RP1 and the Moebius bundle are equivalent.

3. Vector bundles

Definition 91 (Vector bundle). A vector bundle of rank k is a triple (π,E,M) where E and Mare smooth manifolds and

π : E −→M

is a smooth map such that:

(1) there is an open cover M =⋃α∈A Vα and diffeomorphisms ψα : π−1(Vα)→ Vα × Rk such

that the following diagram is commutative:

π−1(Vα) Vα × R

Vα

ψα

π p1

(p1 is the projection on the first factor). The family (Uα, ψα)α∈A is called a vector bundleatlas, and each ψα is called a trivialization.

(2) whenever Vα ∩ Vβ 6= ∅, there is a continuous map gαβ : Vα ∩ Vβ → GL(Rk) such that the

map ψα ψ−1β : (Vα ∩ Vβ)× Rk → (Vα ∩ Vβ)× Rk is given by:

(x, v) 7→ (x, gαβ(x) · v).

The fiber over x ∈ M is the vector space Ex = π−1(x). Given X ⊂ M the restriction of E isthe vector bundle (π|π−1(X), π

−1(X), X) (the vector space structure is well defined by condition(2) above). The family gαβα,β∈A is called the cocycle of the bundle and satisfies the interestingproperties:

(3.1) gαβ(x)gβγ(x) = gαγ(x) ∀x ∈ Vα ∩ Vβ ∩ Vγ , gαα(x) = 1.

Example 92 (The trivial bundle). The trivial rank-k bundle on M is simply E = M × Rk withπ : M ×Rk →M given by the projection on the first factor (in other words there is a single bundletrivialization: Vα = M and ψα = idM ).

Example 93 (The Moebius band). The Moebius band is the vector bundle on S1 = U1 ∪U2 (theopen cover defined in Example 7) constructed as follows. We have U1 ∩ U2 = I1

∐I2 (the disjoint

union of two intervals) and the cocycle for the Moebius band is:

g12|I1 ≡ 1 and g12|I2 ≡ −1.

Example 94 (How to construct a vector bundle). Let us consider an open cover U = Uαα∈Aof a smooth manifold M and for every α, β ∈ A such that Uα ∩ Uβ assume we are given a smooth

24 ANTONIO LERARIO

function gαβ : Uα ∩ Uβ → GL(Rk) such that the family gαβ satisfies properties (3.1). Then wecan construct a vector bundle E with cocylce gαβ by setting:

E =

(∐α∈A

Uα × Rk)/ ∼,

where (x1, v1)α1∼ (x2, v2)α2

if and only if x1 = x2 and v1 = gα1α2(x)v2. We endow E with the

quotient topology from the defining equivalence relation “∼” and the projection map π : E → Mis simply given by [(x, v)] 7→ x.

Example 95 (The tautological bundle). The tautological bundle τk,n is a vector bundle on theGrassmnannian G(k, n) which, as a topological space, is defined by:

τk,n = (W,x) ∈ G(k, n)× Rn |x ∈W ' Rk.The projection π : τk,n → G(k, n) equals the restriction to τk,n ⊂ G(k, n) × Rn of the projectionon the first factor G(k, n)× Rn → G(k, n).

Exercise 96. Let RPn = G(1, n+1) with the open cover U = Ujnj=0 as in Example 8. Prove that

τk,n|Uj is trivial and find the cocycle of τk,n associated to the open cover U. Prove that τ1,2 → RP1

is a Moebius band (see figure 7).

Exercise 97. Let f : RP2\[1 : 0 : 0] → RP1 be the smooth map:

f([x0 : x1 : x2]) = [x1 : x2].

Prove that RP2\[1 : 0 : 0] has the structure of a line bundle over RP1 with projection f . Is it atrivial bundle?

Exercise 98. Let Sym(2,R) ' R3 be the space of symmetric matrices and let Γ ⊂ Sym(2,R) bethe set:

Γ =

A =

(x yy z

)such that det(A) = −1, tr(A) = 0

.

Prove first that Γ is a smooth submanifold of Sym(2,R) which is diffeomorphic to S1. Consideralso the set:

E = (A, v) ∈ Γ× R2 such that Av = v.Prove that E has the structure of a line bundle over Γ with projection p : (A, v) 7→ A. Is it a trivialbundle?Hint: Observe the two identities:(

cos θ sin θsin θ − cos θ

)(1

sin θ1+cos θ

)=

(1

sin θ1+cos θ

), θ 6= π

2(cos θ sin θsin θ − cos θ

)(sin θ

1−cos θ

1

)=

(sin θ

1−cos θ

1

), θ 6= 0.

Definition 99 (Morphism of vector bundles). Let π1 : E1 →M1 and π2 : E2 →M2 be two vectorbundles. A morphism of vector bundles between E1 and E2 is a smooth map f : E1 → E2 whichis linear on every fiber and which covers a smooth map f : M1 →M2 (i.e. such that the followingdiagram is commutative):

E1 E2

M1 M2

π1

f

π2

f

If the linear map f |(E1)x : (E1)x → (E2)f(x) is injective for every x ∈ M1, then f is calleda monomoprhims; if it is surjective for every x ∈ M1, it is called an epimorphism. If for everyx ∈ M1 the linear map f |(E1)x : (E1)x → (E2)f(x) is an isomorphism, then f is called a bundlemap. An equivalence of vector bundles is a bundle map which covers a homeomorphism and anisomorphism of vector bundles is a bundle map which covers the identity.

A rank-k vector bundle π : E → M is called trivial if it is isomorphic to the trivial bundleM × Rk. An isomorphism E →M × Rk is called a trivialization of E. If the tangent bundle TMof a smooth manifold is trivial, M is said parallelizable.


Exercise 100. Prove that Lie groups are parallelizable. (In particular, for example, since bothRP3 and S3 are Lie groups, then they are parallelizable.)

Example 101 (Dual bundle). Given a vector bundle Eπ−→ M which trivializes over an open

cover U = Uαα∈A of M and with cocycle gEαβ, the dual bundle E∗p−→M is constructed as in

Example 94 by taking the cocyle gE∗αβ defined for x ∈ Uα ∩ Uβ by:

gE∗

αβ (x) = (gαβ(x)−1)T ∈ GL(Rk).

When E = TM , the dual bundle is denoted by T ∗M and called the cotangent bundle.

Example 102 (Determinant bundle). Given a vector bundle Eπ−→ M which trivializes over an

open cover U = Uαα∈A of M and with cocycle gEαβ, the determinant bundle det(E)p−→M is

the line bundle constructed as in Example 94 by taking the cocyle gdet(E)αβ defined for x ∈ Uα∩Uβ

by:

gdet(E)αβ (x) = det(gαβ(x)) ∈ GL(R).

Example 103 (Direct sum of vector bundles). Given two vector bundles Rk1 → E1π1−→ M and

Rk2 → E2π2−→M which trivialize over the same open cover7 U = Uαα∈A of M and with cocycles

gE1

αβ and gE2

αβ, the direct sum of E1 and E2 is the vector bundle E1 ⊕ E2 →M constructed asin Example 94 with the cocycle defined for x ∈ Uα ∩ Uβ by:

gE1⊕E2

αβ (x) =

(gE1

αβ(x) 0

0 gE2

αβ(x)

)∈ GL(Rk1+k2).

3.1. Sections of vector bundles.

Definition 104 (Section of a vector bundle). Let Eπ−→ M be a vector bundle. A section of E

is a smooth map s : M → E such that π(s(x)) = x for every x ∈ M , i.e. such that the followingdiagram is commutative:

E

M

π s

The zero section is the section that associates to every point x ∈M the zero vector in Ex.

Exercise 105. Let Rk → E → M be a rank-one bundle. Prove that E is trivial if and only ifadmits a nowhere vanishing section s : M → E.

Remark 106 (How to build a section of a vector bundle). Let Eπ−→M be a rank-k vector bundle

and U = Uαα∈A be an open cover of M such that ψα : E|Uα ' Uα×Rk for every α ∈ A. Let alsogαβ be the corresponding cocyle. Given a section s : M → E, for every α ∈ A we can considerthe map ψα s : Uα → Uα × Rk (the section in the trivialization), which has the following form:

(3.2) ψα(s(x)) = (x, σα(x)),

for some smooth function σα : Uα → Rk. Using the definition of the cocyle, we see that the familyof maps σα satisfies the relation:

(3.3) σα(x) = gαβ(x)σβ(x) ∀x ∈ Uα ∩ Uβ .Conversely, given a family of smooth functions σα : Uα → Rk such that (3.3) is satisfied, thisfamily defines a section of E (i.e. the unique section which in in the trivializaztions has theexpression (3.2)).

Example 107 (Polynomials and vector bundles over projective spaces). Let U = Ujnj=0 be theopen cover Uj = xj 6= 0 of RPn. For every d ∈ Z consider the cocycle:

gij([x0, . . . , xn]) =

(xjxi

)d, xi, xj 6= 0.

7Indeed, by Corollary 114 there is an open cover of M such that every vector bundle trivializes over this cover.

26 ANTONIO LERARIO

We denote by ORPn(d) the vector bundle constructed with the procedure of Example 94 corre-sponding to this cocyle. An interesting fact is that, when d ≥ 0, every homogeneous polynomialof degree d defines in a natural way a section of ORPn(d), as follows. Given p ∈ R[x0, . . . , xn](d)

define the family of functions (σp)j : Uj → Rnj=0 by

(σp)j([x0, . . . , xn]) = p

(x0

xj, . . . ,

xjxj, . . . ,

xnxj

).

Observe now that for every i, j and [x] ∈ Ui ∩ Uj we have:

gij([x])(σp)j([x]) =

(xjxi

)dp

(x0

xj, . . . ,

xjxj, . . . ,

xnxj

)= p

(x0

xi, . . . ,

xnxi

)= (σp)i([x])

(where in the second equality we have used the fact that p is homogeneous of degree d). In partic-ular, following the discussion of Remark 106, we see that the family (σp)jnj=0 defines a sectionσp of ORPn(d). (Note also that the map p 7→ σp is linear.) The procedure of de-homogenization ofa polynomial with respect to a variable can be geometrically described as the procedure of lookingat the polynomial as a section in the corresponding trivialization!

3.2. Metrics on vector bundles.

Definition 108 (Metric on a vector bundle). Let Rk → Eπ−→M be a vector bundle. A smooth

metric on E is a family hxx∈X where each hx is a symmetric, bilinear, positive definite quadraticform on Ex such that the map (x, y, z) 7→ hx(y, z) defined on (x, y, z) ∈M × E × E | p(y) = x =p(z) is smooth. (A smooth metric on TM is called a Riemannian metric.)

Proposition 109. Every vector bundles admits a smooth metric.

Proof. Let U = (Uα, ψα)α∈A be a vector bundle atlas and call p1 : Uα × Rk → Uα and p2 :Uα×Rk → Rk the projections on the two factors. Let ραα∈A be a partition of unity subordinatedto U (see Proposition 193) and denote by 〈·, ·〉 the standard scalar product on Rk. For every α ∈ Aand (x,w1, w2) ∈M × E|Uα × EUα such that π(w1) = x = p(w2), define:

hα,x(w1, w2) = ρα(x)〈p2(ψα(w1)), p2(ψα(w2))〉.Since the support of ρα is contained in Uα, this function extends to a smooth function on the setof (x, y, z) ∈ M × E × E such that π(y) = x = p(z) (when (x, y, z) /∈ M × E|Uα × EUα thenhα,x(y, z) = 0). The metric hx is defined by:

hx(y, z) =∑α

ρα(x)hα,x(y, z).

The fact that the family is smooth is clear by construction (it is the sum of smooth maps); thefact that for every x the quadratic form hx : Ex×Ex is positive definite follows from the fact thateach ρα ≥ 0 and that for every x ∈M there is at least one α ∈ A with ρα(x) 6= 0.

Exercise 110. Let π : E → M be a smooth vector bundle endowed with a smooth metric andE1 ⊆ E be a sub-bundle. Define the orthogonal complement bundle to E1 in E by:

E⊥1 = (x, v) ∈ X × E such that π(v) = x and v ∈ (E1)⊥x (where the orthogonal complement is taken with respect to the metric in Ex), together with theprojection p = p1|E⊥1 . Prove that p : E⊥1 → M is a smooth vector bundle of rank rk(E⊥1 ) =

rk(E)− rk(E1) and that E ' E1 ⊕ E⊥1 .

3.3. Parametric transversality for sections of vector bundles. Partition of unity can beused to produce “generic” sections of a vector bundle, as follows.

Theorem 111. Let Rk → Eπ−→ M be a vector bundle, s : M → E be a smooth section and Z

be the image of the zero section. We assume that E is endowed with a metric and we denote by‖v‖ the norm of an element v ∈ Ex. For every ε > 0 there exists a section sε : M → E such thats t Z and supx∈M ‖s(x)− sε(x)‖ < ε.


Proof. Let (Uj , ψj)`j=1 be a finite vector bundle atlas (such a finite atlas exists because M

is assumed to be compact) and ρj`j=1 be a partition of unity subordinated to U. Given the

trivialization ψj : E|Uj → Uj × Rk we define the smooth map φj : M × Rk → E by:

φj(x, v) =

ρj(x)ψ−1

j (x, v) if x ∈ Ujthe zero element of Ex otherwise

Using the family of maps φj we define the evaluation map Fs : Rk×` ×M → E by:

Fs(v1, . . . , v`, x) = s(x) +∑j=1

φj(x, vj).

Observe first that for every p = (v1, . . . , v`) ∈ Rk×` the map x 7→ F (p, x) is a section of E:

π(Fs(v1, . . . , v`, x)) = π

s(x) +∑j=1

φ(x, vj)

= x.

We claim that Fs t Z. Since ψj is a diffeomorphism, it is enough to check that for every j = 1, . . . , `the map ψj F is transversal to Uα × 0. To this end we write

ψj(Fs(v1, . . . , v`, x)) =

x, σ(x) +∑j=1

ρj(x)vj

,

from which we see that for every w ∈ Rk

d(0,...,0,x)Fs(w, . . . , w, 0) = (0, w),

and consequently (ψj Fs) t Uj × 0. The result follows now from Theorem 83: the set ofp = (v1, . . . , v`) such that x 7→ Fs(v1, . . . , v`, x) is transversal to Z is dense in Rk×`. Choose nowa sequence pnn∈N ⊂ Rk×` such that Fs(pn, ·) t Z and ‖pn‖ ≤ 1

n . Since Fs(pn, ·)n∈N convergesuniformly to s, it follows that there exists n large enough such that ‖Fs(pn, x)− s(x)‖ ≤ ε for allx ∈M .

3.4. The classification theorem.

Definition 112 (Pull-back bundle). Let π : E → Y be a smooth vector bundle and f : X → Ybe a smooth map. We define the total space of a new vector bundle, called the pull-back bundle:

f∗E = (x, e) ∈ X × E | f(x) = π(e).Denoting by p1 : X × E → X the projection on the first factor, the projection map for f∗E is

p = p1|f∗E : f∗E → X.

If (Uα, ψα)α∈A is a vector bundle atlas for E and denoting by p2 : Uα×Rk → Rk the projectionon the second factor, the vector bundle atlas for f∗E is (f−1(Uα), ϕα)α∈A, where:

ϕα(x, e) = (p1, p2 ψα)(x, e) = (x, p2(ψα(e))) ∀(x, e) ∈ f−1(Uα)× E|UαThe cocyle of the pull-back bundle f∗E is the family

hαβ : E|f−1(Uα)∩f−1(Uβ) → f−1(Uα) ∩ f−1(Uβ)× Rk

given by hαβ(x) = gαβ(f(x)).

Theorem 113. Let X be a smooth manifold and π : E → X × I be a vector bundle8. ThenE|X×0 and E|X×1 are isomorphic vector bundles.

Proof. The proof is based on the following two facts.

8Strictly speaking so fare we have only defined vector bundles when the base space is a smooth manifold, and

in this Theorem we are instead taking M × I. This is easily fixed by extending the notion of smooth manifold to

include manifolds with boundary, see Definition 186.

28 ANTONIO LERARIO

(1) A vector bundle π : E → X × [a, b] is trivial if there exists c ∈ [a, b] such that both therestriction E|X×[a,c] and E|X×[c,b] are trivial. In fact, given trivializations:

ψ1 : E|X×[a,c] → X × [a, c]× Rk and ψ2 : E|X×[c,b] → X × [c, b]× Rk,

we can replace ψ2 with the trivialization ψ3 : E|X×[c,b] → X × [c, b] × Rk defined bycomposing ψ2 with the map:

(x, t, v) 7→ (x, t, ψ1(ψ2)−1(x, c, v)).

(2) Given a vector bundle E → X × I, there exists an open cover Vαα∈A of X such thatE|Vα×I is trivial for every α ∈ A. In fact, for every x ∈ X and for every t ∈ [0, 1]we can find an open neighborhood Ux,t of (x, t) ∈ X × I such that E|Ux,t is trivial.Using the compactness of [0, 1], we can find open neighborhoods Ux,1, . . . , Ux,m of x and0 = t0 < t1 < · · · < tm = 1 such that E|Ux,j×[tj−1,tj ] is trivial for every j = 1, . . . ,m. Theprevious fact (1) implies that, after setting Ux = Ux,1∩· · ·∩Ux,m, then E|Ux×[0,1] is trivial.

Now the proof of the Theorem. In order to make the exposition more clear, let us first considerthe case when X is compact. Applying fact (2) above, we can find an open cover Uαα∈A of Xsuch that E|Uα×I is trivial. By compactness of X we can extract a finite open cover U = Ujkj=1

of X such that E|Uj×I is trivial. We consider now a partition of unity ρjkj=1 subordinated to U

and we define for j = 1, . . . , k:

θj = ρ1 + · · ·+ ρj .

Let Γj ⊂ X × [0, 1] be the graph of θj and Ej = E|Γj . We define for every j = 1, . . . , k a bundleisomorphism hj : Ej → Ej−1 as follows. Observe that

Γj ∩ (U cj × I) = Γj−1 ∩ (U cj × I),

we therefore set the map hj to be the identity outside π−1(Uj × I). Since over Uj × I the bundleis trivial, we have E|Uj×I ' (Uj × I)× Rk and we can set:

hj(x, θj(x), v) = (x, θj−1(x), v) ∀(x, θj(x), v) ∈ Ej .The composition h1h2 · · ·hk gives therefore an isomorphism between E|X×1 and E|X×0.

When X is only paracompact, we still get an open cover U = Uαα∈A such that E|Uα×I istrivial. Using paracompactness we refine U to a countable, locally finite V = Vjj∈N; for everyj ∈ N, since Vj ⊂ Uα for some α ∈ A, we still get that E|Vj×I is trivial. Let now ρjj∈N be apartition of unity subordinated to V and for every j ∈ N define as before:

θj = ρ1 + . . .+ ρj .

Again as before for every j ∈ N we get an isomorphism hj : Ej → Ej−1, where Ej = E|graph(θj).Now the delicate point would be the infinite composition h1h2 · · · , however near every point x ∈ Xonly finitely many ρj ’s are non zero, so there is a neighborhood where all but finitely many hj ’s arethe identity. Hence h is a well defined vector bundle isomorphism between E|X×1 and E|X×0.

An immediate consequence, which will be much useful is the following.

Corollary 114. Every vector bundle E over a contractible manifold M is trivial.

Proof. Since M is contractible, it means that the identity map f0 = idM is homotopic to a constantmap f1 ≡ x1. But then by the previous theorem:

E = id∗E ' f∗1E = M × Rk.

Proposition 115. Let π : E → M be a vector bundle. Then there exists another vector bundle

π : E →M such that:

E ⊕ E 'M × R`.


Proof. We only prove the statement under the additional assumption that M is compact – theproof is similar to the proof of Theorem 51.

Let U = Ujnj=1 be an open cover of M such that for every j = 1, . . . , n we have: (1) E|Uj istrivial; (2) there exists manifold charts ϕj : Uj → Rm with ϕj(Uj) = B(0, 1) and M is covered by

ϕ−1j (B(0, 1

2 ))nj=1. Let also ρjnj=1 be a family of bump functions with ρj supported on Uj .

For every j = 1, . . . , n define the map:

φj : E|Uj → Rm × Rk

which is the composition of the trivialization over Uj and the map ϕj × idRk . Out of these maps

we build φ : E 7→ Rn(m+k+1) defined by:

φ = ((ρ1 π)φ1, ρ1 π, . . . , (ρn π)φn, ρn π).

It is easy to see that φ : E → R` is an embedding (we have set ` = n(m+k+1)), and we can identifyE with a submanifold9 of R`. Consequently TE|M ' E⊕TM is a sub-bundle of M ×R` = TR`|Mand, after endowing TR` with a riemannian metric, we can finally set:

E = (TM ⊕ E)⊥ ⊕ TM.

(Here the orthgonal complement bundle (TM ⊕ E)⊥ is defined as in Exercise 110).

Theorem 116. Let p : τk,` → G(k, `) be the tautological vector bundle.

(1) Given a rank-k vector bundle π : E → M there exist ` > 0 and a smooth map f : M →G(k, `) such that E = f∗τk,`.

(2) Let E0 ' E1 be two isomorphic rank-k vector bundles over M . Then, by point (1), thereexist smooth maps f0 : M → G(k, `0) and f1 : M → G(k, `1) such that E0 = f∗0 τk,`0 andE1 = f∗1 τk,`1 . There exists ` > `0, `1 such that the two maps f0, f1 become homotopic aftercomposition with the natural inclusions G(k, ì) → G(k, `).

Proof. The proof of point (1) uses Proposition 115: let E such that E ⊕ E ' M × R`. Then Ecan be viewed as a sub-bundle of a trivial bundle and the map f : M → G(k, `) is simply:

f(x) = Ex ' Rk ⊂ R`.

For the second point we proceed as follows. For i = 0, 1, let gi : E → Rì be the mapEi → M × Rì → Rì (a linear injection on every fiber). Set ` = 2 max`0, `1 = 2`2 andconsider the two inclusions L0 : R`0 → R`2 and L1 : R`1 → R`2 as the first coordinates. Let nowjevent : R`0 → R` be the homotopy:

jevent (x1, . . . , x`2 , 0, . . . , 0) = (1− t)(x1, . . . , x`0 , 0, . . . , 0) + t(0, x1, . . . , 0, x`2).

Defining similarly joddt to be the homotopy to the odd coordinates, we see that both jeven

t and joddt

are homotopies of linear injections.

Let now g0 = jeven1 g0 and g1 = jodd

1 g1. Then for i = 0, 1 we have that gi : Ei → R` is alinear injection such that the map

fi : x 7→ gi(π−1i (x)) ∈ G(k, `)

has the property Ei = f∗i τk,`. Moreover, for i = 0, 1 we have that fi : M → G(k, `) is homotopic

to Mfi−→ G(k, ì) → G(k, `). The final homotopy is simply:

ft = (1− t)f0 + tf1.

Exercise 117. Prove that, up to isomorphism, there are only two line bundles on S1 (the tauto-logical τ1,2 and the trivial bundle).

9When M is not compact, one can use the stronger embedding theorem [7, Theorem 2.14] and directly embed

φ : E → R`. From this point the proofs, both in the compact and the non-compact case, are identical.

30 ANTONIO LERARIO

Remark 118. It is possible to provide a uniform estimate on the ` from point (2) of the previousTheorem, i.e. it is possible to prove that there exists ` > 0 (which depends on k and m = dim(M))such that if for two maps f0, f1 : M → G(k, `) we have f∗0 τk,` ' f∗1 τk,`, then the two maps f0

and f1 are homotopic. We give here a sketch of the proof for the case k = 1; this proof is missingmany details, which I warmly encourage the reader to fill-in, and has the only scope of giving aflavor of how nicely the techniques we have introduced so far combine together. We claim that iff0, f1 : M → G(1,m+ 2) ' RPm+1 pullback homotopic bundles, then they are homotopic. First,following the proof of the previous Theorem 116, if we compose these maps with the inclusionj : G(1,m+ 2) ' RPm+1 → G(m, 2m+ 4) ' RP2m+3, then they become homotopic:

h0 = j f0 ∼ j f1 = h1.

Let us call H : M×I → RP2m+3 the (smooth, by construction) homotopy between h0 = H(·, 0) andh1 = H(·, 1). We use now the fact that if RPb ⊂ RPa is a subspace which is disjont from RPa−b−1,then RPa\RPb deformation retracts to RPa−b−1. We apply this with the choice a = 2m + 3 andb = m + 1 (we call B ' RPm+1 = RPb a complementary subspace in RP2m+3 to our originalRPm+1): this says that there exists a retraction

r : RP2m+3\RPm+1 → RPm+1,

which is a homotopy equivalence. Now we apply Theorem 83 (Parametric Transversality) for themap F : O(2m+ 4)× RP2m+3 → RP2m+3 given by:

F (g, [v]) = [gv].

It is not difficult to verify that F is a submersion and consequently that there are elements gεarbitrarily close to the identity 1 ∈ O(2m+ 4) such that the map

gε H : (x, t) 7→ [gεH(x, t)]

is transversal to B. Since codimRP2m+3(B) = m+2 and dim(M ×I) = m+1, transversality meansthat:

(gε H)(M × I) ∩B = ∅.In particular gε H gives a homotopy between gε h0 and gε h1, which misses B. Since both f0, f1

have image in RPm+1, for ε > 0 small enough and for i = 0, 1 the maps j fi : M → RPm+1 ⊂RP2m+3\B and gε hi : M → RP2m+3\B are homotopic as maps with values in RP2m+3\B (thisis because “close enough” maps are homotopic – a statement that has to be justified carefully).Thus so far we have obtained homotopy of maps:

j f0 ∼ gε h0 ∼ gε h1 ∼ j f1 : M → RP2m+3\B,where the homotopies are as maps with value in RP2m+3\B. In particular this gives a homotopyf t between f0 = j f0 : M → RP2m+3\B and f1 = j f1 : M → RP2m+3\B. We finally composenow this homotopy with the retraction r : RP2m+3\B → RPm+1 and we get a homotopy of maps:

r f t : M → RPm+1

such that r f0 = r j f0 = f0 and r j f1 = r f1 = f1.

3.5. Vector bundles on spheres. Consider a rank-k vector bundle on the sphere

Rk E

Sn

π

and let D+ and D− be, respectively, the upper and the lower hemisphere. Then, by Corollary 114both E|D+ and E|D− are trivial and there is a well defined map:

g : D+ ∩D− = Sn−1 → GL(Rk).

(This is simply the cocyle of the cover Sn = D+∪D−; this is not an “open” cover, but the cocycle

is still defined.) The map g : Sn−1 → GL(Rk) is called a clutching function. Denoting by Vectk(X)the set of isomorphism classes of rank-k vector bundles over X and by [Y : Z] the set of homotopyclasses of maps f : Y → Z, there is a well defined map:

γ : [Sn−1 : GL(Rk)]→ Vectk(Sn),


which associates to every clutching function (i.e. cocycle) g the vector bundle Eg constructed asin Example 94.

Exercise 119. Prove that γ : [Sn−1 : GL(Rk)] → Vectk(Sn) is well defined, i.e. that homotopicclutching functions define isomorphic vector bundles. (Hint: use Theorem 113.)

Since GL(Rk) is not connected, in order to classify vector bundles on Sn we restrict to theso called oriented bundles, i.e. those vector bundles E → Sn such that the determinant bundledet(E) is trivial and a trivialization has been chosen (this last condition is equivalent to choosingan orientation of one fiber Ex0

over a fixed point x0).

When n = 1, there are just two homotopy classes of maps g : S0 → GL(Rk) such that g(x0) ∈GL+(Rk). These correspond to the trivial bundle and the direct sum of (k − 1) many trivial linebundles with the Moebius bundle.

Exercise 120. Consider a vector bundle of rank k > 1:

Rk E

S1

π .

Prove that there exists a section s : S1 → E of this vector bundle that is never zero, and inparticular that E ' E1 ⊕ spans where E1 is a rank-(k − 1) vector bundle and spans is trivialrank-one bundle. (Hint: use Theorem 111.)

When n ≥ 2, the sphere Sn−1 is path connected and if for a clutching function we have g(x0) ∈GL+(Rk), then g(Sn−1) ⊂ GL+(Rk)10. In particular all rank-k bundles over Sn (with n ≥ 2) areorientable and every vector bundle has precisely two orientations.

We denote by Vectk+(Sn) the set of isomorphism classes of oriented, rank-k vector bundles onSn, where the isomorphisms must preserve the orientation of the chosen fiber. For general k, n, themap Vectk+(Sn)→ Vectk(Sn) that forgets the orientation is at most two-to-one, in fact exactly two-to-one except on those vector bundles that have an automorphism which reverts the orientation;this can only happen when Sn−1 is not connected, i.e. when n = 1, for which Exercise 120 applies.When n > 1 the map Vectk+(Sn) → Vectk(Sn) is two-to-one. In light of this observation, theclassification of vector bundles on Sn follows from next Theorem 121.

Theorem 121. The map γ+ : [Sn−1 : GL+(Rk)] → Vectk+(Sn), that associates to a map g :

Sn−1 → GL+(Rk) the bundle Eg with clutching function g, is a bijection.

Proof. Observe first that if we take g : Sn−1 → GL+(Rk) then the corresponding bundle is oriented.The map γ+ : [g] 7→ [Eg] is well defined by Exercise 119 and we wish to find an inverse for it.The obvious candidate is the map [E] 7→ [gE ] which associates to a vector bundle E over Sn theclutching function that we obtain by trivializing the vector bundle over D+ and D−. The fact thatthis map is well defined is left as an exercise (the proof uses the properties of a cocycle and thefact that GL+(Rk) is path connected).

Exercise 122. Prove that the inclusion SO(k) → GL+(Rk) is a homotopy equivalence, andconsequently that [Sn−1 : GL+(Rk)] = πn−1(SO(k)). (Hint: use the Gram-Schmidt procedure.)

Exercise 123. Prove that there is a natural bijection θ : Vect2+(S2) → Z = π1(S1). Under this

identification we have θ(TS2) = 2 and θ(τC1,2) = −1 (here C ' R2 → τC1,2 → S2 ' CP1 denotes thecomplex tautological bundle, seen as a real, rank-2 oriented vector bundle).

Exercise 124. Prove that every rank-2 bundle over Sn is trivial when n > 2.

10Here GL+(Rk) denotes the set of invertible k × k matrices with positive determinant, i.e. one of the two

components of GL(Rk).

32 ANTONIO LERARIO

3.6. The normal bundle.

Definition 125 (Normal bundle). Let X ⊆ Y be an embedded submanifold. Endow the vector

bundle TYp−→ Y with a Riemannian metric (see Definition 108). The normal bundle NX of X

in Y is the vector bundle NX = TX⊥ → X (see Exercise 110). Note that NX ⊂ TY |X .

Exercise 126. Let X ⊂ Y be a submanifold given by regular equations X = f1 = · · · = fc = 0.Prove that the normal bundle NX is trivial. (We assume a metric on TY has been chosen so thatit makes sense to speak about normal bundle of a submanifold.) Use this result to give anotherproof that RP1 ⊂ RP2 cannot be the regular zero set of a function on RP2.

Proposition 127. Let M ⊂ Rn be a smooth embedded manifold and endow TRn = Rn ×Rn withthe standard vector bundle metric. Denote by N εM = (x, v) ∈ NM | ‖v‖ < ε (an open set inNM). If M is compact, there exists ε > 0 such that the map τ : N εM → Rn given by:

τ : (x, v) 7→ x+ v

is an embedding.

Proof. Let us first observe that for every x ∈M we have:

(3.4) rk(d(x,0)τ) = n.

First, clearly the image of d(x,0) contains TxM . Moreover for every v ∈ TxM⊥ let us consider thecurve γ : (−δ, δ)→ NM given by γ(t) = (x, tv). Then γ(0) = x and

d

dtτ(γ(t))

∣∣∣∣t=0

=d

dt(x+ tv)

∣∣∣∣t=0

= v.

Then the image of d(x,0) also contains TxM⊥ and since Rn = TxM + TxM

⊥, then (3.4) follows.

It is easy to see that the map τ is smooth and for every x ∈ M there exists a neighborhoodU ′x ⊂M and ε′x > 0 such that:

rk(d(x,v)τ) = n for all (x, v) ∈ U ′xM such that x ∈ Ux and ‖v‖ ≤ εx.As a consequence, by the Inverse Function Theorem, there exist Ux ⊂ U ′x and εx < ε′x such that:

(3.5) τ |NεxUx : N εxUx → Rn is an embedding.

By compactness of M , we can extract a finite cover Uxi`i=1 of M out of Uxx∈M and thereexists 0 < ε0 < min` εxi2 such that

(3.6) τ |N2εUxi: N2εUxi → Rn

is an embedding for every i = 1, . . . , `.

Let us now consider for every 0 < ε < ε0 the compact set:

N32 εM =

(x, v) ∈ NM

∣∣∣∣ ‖v‖ ≤ 3

2ε

⊂ N2εM.

We claim that for ε > 0 small enough τ |N

32εM

: N32 εM → Rn is a homeomorphism onto its

image. Since N32 εM is compact, Rn is Hausdorff and τ is continuous, it is enough to show that

for ε > 0 small enough τ |N

32εM

: N32 εM → Rn is injective. Suppose by contradiction that there

are sequences (xn, vn)n∈N and (yn, wn)n∈N with

(3.7) (xn, vn) 6= (yn, wn)

and (xn, vn), (yn, wn) ∈ N32nM such that

τ(xn, vn) = τ(yn, wn).

Up to subsequences, we can assume xn → x ∈M and yn → y ∈M . Then

x = limn→∞

τ(xn, vn) = limn→∞

τ(yn, wn) = y,

which means that (again, up to subsequences) the two sequences (xn, vn)n∈N and (yn, wn)n∈Neventually lie in the same N εxUx, on which τ is injective by (3.5), contradicting (3.7).


Thus there exists ε > 0 such that τ : N32 εM → Rn is a homeomorphism onto its image. In

particular, since N εM ⊂ N32 εM , then:

τ |NεM : N εM → Rn

is a homeomorphism onto its image. Consequently, because of (3.6), τ |NεM is an embedding.

Corollary 128. Let M ⊂ Rn be a compact manifold. Then there is an open set U ⊂ Rn containingM and a smooth retraction r : U →M which is a homotopy equivalence.

Proof. Let ε > 0 such that τ |NεM : N εM → Rn is an embedding. Observe that

U = τ(N εM)

is an open set containing M . Let p : NM → NM be the smooth map (x, v) 7→ (x, 0) and define:

r = τ p τ−1 : U →M.

Observe that the map p : N εM → N εM is homotopic to idNεM , through the homotopy:

pt(x, v) = (x, tv), p0 = p, p1 = idNεM .

Consequently, denoting by i : M → U the inclusion we have:

r i = idM and i r = τ p τ−1 ∼ τ idNεM τ−1 = idU ,

which proves both the inclusion i : M → U and the retraction r : U → M are homotopy equiva-lences.

3.7. An approximation result.

Theorem 129. Let A and B be smooth compact manifold and f : A→ B be a continuous function.There is a smooth map g : A→ B which is homotopic to f .

Proof. Let B → Rn be an embedding given by Theorem 51; we identify B with its image underthis embedding. Proposition 127 says that there exists an ε0 > 0 such τ embeds N ε0B onto aneighborhood U = τ(N ε0B) of B; Corollary 128 says that the retraction r : U → B is a homotopyequivalence.

We claim that we there exists ε > 0 small enough such that for every y ∈ B the ball B(y, ε) (aball in Rn) is entirely contained in U . In fact, since U is open, then for every y ∈ B there is εy > 0such that B(y, 3εy) ⊂ U . Since M is compact, there are y1, . . . , y` ∈M such that:

M ⊂⋃k=1

B(yk, 3εyk).

Let ε = minεy1 , . . . , εy`. Then every y ∈M belongs to some B(yk, 3εyk) and

B(y, ε) ⊂ B(yk, 3εyk) ⊂ U.

We can think at f as a continuous function f : A→ Rn with image in B. Using Theorem 195,we produce a smooth map fε : A→ Rn such that

(3.8) maxx∈A‖f(x)− fε(x)‖ ≤ ε.

Of course the image of the map fε is not in B, but (3.8) implies fε(x) ∈ B(f(x), ε), hence theimage of fε lies in U . Using the retraction r we define:

g = r fε : A→ B.

To see that g is homotopic to f we simply define the homotopy:

gt(x) = r(tfε(x) + (1− t)f(x)), g0 = f, g1 = g.

(The homotopy is well defined because for every t ∈ [0, 1] the point tfε(x) + (1− t)f(x) belongs tothe convex set B(f(x), ε) ⊂ U , i.e. tfε(x) + (1− t)f(x) belongs to the domain of the map r.)

34 ANTONIO LERARIO

Exercise 130. Let A and B be two compact manifold. Let us assume that B ⊂ Rn and let usput a metric on the set of continuous functions C0(A,B) by:

d(f, g) = maxx∈A‖f(x)− g(x)‖ f, g ∈ C0(A,B).

Under this assumptions, refine the conclusion of Theorem 129 and prove that for every continuousfunction f : A→ B and every ε > 0 there is a smooth map fε : A→ B such that fε is homotopicto f and d(fε, f) ≤ ε.

As a corollary we can deduce the following important theorem.

Theorem 131. If k < n then πk(Sn) = 0.

Proof. Let f : Sk → Sn be a continuous map. Then, by Theorem 129, f is homotopic to a smoothmap g : Sk → Sn. Since k < n, by Lemma 69, the map g is not surjective; consequently it missesa point and factors through a map Sk → Rn → Sn, hence g is homotopic to a constant map andthe same is true for f .

Exercise 132. Use the previous Theorem together with the results from Section 3.5 to prove thatevery vector bundle on S3 is trivial.


4. Differential forms and cohomology

A note to the reader: the subject of this section requires a bit more to be digested, because theconstruction of differential forms is a bit involved. To complement this chapter, I suggest to alsoread [1, Section 7], which is an excellent source and might help developing the “intuition” on theseobjects.

4.1. Alternating forms. Let V be a n-dimensional vector space. We define the k-th exteriorpower of V ∗ (the dual space of V ) as:

∧k(V ∗) = α : V × · · · × V︸︷︷︸k times

→ R multilinear and alternating.

Elements of ∧k(V ∗) are called k-forms.

Let now η1, . . . , ηn be a basis for V ∗ and I = i1, . . . , ik ∈nk

be a multi-index. We define11

ηI = ηi1 ∧ · · · ∧ ηik ∈ ∧k(V ∗) as the k-form such that:

(4.1) ηI(v1, . . . , vk) = det

ηi1(v1) · · · ηi1(vk)...

...ηik(v1) · · · ηik(vk)

.

We observe that to the same multi-index I we can associated different forms: for example, inprinciple, given I = 1, 2 we could consider η1 ∧ η2 and η2 ∧ η1 (note that in virtue of definition(4.1) these two multilinear maps are different). To resolve this ambiguity we adopt the conventionthat multi-indices are always ordered and I = i1 < · · · < ik.

The following fact is elementary and is left as an exercise...

Proposition 133. Letnk

denote the set of all possible multi-indices I = i1 < · · · < ik. The

set ηII∈nk is a basis for ∧k(V ∗). In particular the dimension of ∧k(V ∗) is(nk

).

Proof. Try to prove it yourself!

We now make clear the meaning of the symbol “∧”.

Definition 134 (Wedge product). Let γ ∈ ∧k(V ∗) and ω ∈ ∧`(V ∗) be the forms (written in theabove basis) as:

γ =∑I∈nk

γI ηi1 ∧ · · · ∧ ηik and ω =∑J∈n`

ωJ ηj1 ∧ · · · ∧ ηj` .

We define the form γ ∧ ω ∈ ∧k+`(V ∗) (called “the wedge of γ and ω”) as:

γ ∧ ω =∑I,J

γIωJ ηi1 ∧ · · · ∧ ηik ∧ ηj1 ∧ · · · ∧ ηj` .

The wedge product of forms has the following properties, which eventually follows from (4.1):

(1) (bilinearity) For every a1, a2, b1, b2 ∈ R and forms γ1, γ2 ∈ ∧k(V ∗) and ω1, ω2 ∈ ∧`(V ∗) wehave:

(a1γ1 + a2γ2) ∧ (b1ω1 + b2ω2) = a1b1γ1 ∧ ω1 + a1b2γ1 ∧ ω2 + a2b1γ2 ∧ ω1 + a2b2γ2 ∧ ω2.

(2) (associativity) For every γ ∈ ∧k(V ∗) , ω ∈ ∧`(V ∗) and θ ∈ ∧m(V ∗) we have:

γ ∧ (ω ∧ θ) = (γ ∧ ω) ∧ θ.(3) (anticommutativity) For every γ ∈ ∧k(V ∗) , ω ∈ ∧`(V ∗) we have:

γ ∧ ω = (−1)k`ω ∧ γ.

11At this point ηi1 ∧ · · · ∧ ηik is just a symbol for denoting a special k-form.

36 ANTONIO LERARIO

(Notice in particular that from the last property it follows that γ ∧ γ = 0.) The wedge productturns

∧∗(V ∗) =

dim(V )⊕k=0

∧k(V ∗)

into a graded algebra.

Exercise 135. Check the details for the previous properties, using (4.1).

Definition 136 (Pull-back). Let WL−→ V be a linear map between two finite-dimensional vector

spaces. We define the pull-back map L∗ : ∧k(V ∗)→ ∧k(W ∗) by:

(L∗η)(w1, . . . , wk) = η(Lw1, . . . ,Lwk), η ∈ ∧k(V ∗), w1, . . . , wk ∈W.

Note that a simple computation applying the above definition shows that:

(4.2) L∗(η1 ∧ · · · ∧ ηk) = L∗η1 ∧ · · · ∧ L∗ηk.

Let now e1, . . . , en be a basis for V , f1, . . . , fm be a basis for W and let η1, . . . , ηn andφ1, . . . , φm be the corresponding dual bases for V ∗ and W ∗ respectively. Let L ∈ Rn×m be thematrix representing L : W → V in these bases and for every pair of indices I ∈

nk

and J ∈

mk

let LI,J be the submatrix of L obtained by extracting the rows with index in I and the columnswith index in J .

Proposition 137. With the above notations, for every I ∈nk

we have:

(4.3) L∗(ηI) =∑

J∈mkdet(LI,J)φJ .

Proof. Given α = α1, . . . , αk, β = β1, . . . , βk ∈mk

observe that (4.1) implies:

φα1 ∧ · · · ∧ φαk(fβ1 , . . . , fβk) = δαβ ,

consequently in order to prove (4.3) it is enough to prove that L∗ηI(fj1 , . . . , fjk) = det(LI,J). Tothis end observe that:

L∗ηI(fj1 , . . . , fjk) = ηi1 ∧ · · · ∧ ηik(Lfj1 , . . . ,Lfjk)

= ηi1 ∧ · · · ∧ ηik

(n∑b=1

Lbj1eb, . . . ,

n∑b=1

Lbjkeb

)

= det

ηi1 (∑nb=1 Lbj1eb) · · · ηi1 (

∑nb=1 Lbjkeb)

......

ηik (∑nb=1 Lbj1eb) · · · ηik (

∑nb=1 Lbjkeb)

= det

ηi1(Li1j1ei1) · · · ηi1(Li1jkei1)...

...ηik(Likj1eik) · · · ηik(Likjkeik)

= det(LI,J).

4.2. Differential forms. Let U ⊂ Rn be an open set and consider the linear coordinate functionsx1, . . . , xn : U → R. The differentials dx1, . . . , dxn of these functions are smooth sections ofT ∗U ' U × (Rn)∗ → U . Given a function f ∈ C∞(U,R) and a multi-index i1 < . . . < ik, wecan consider:

ηx = f(x)dxi1 ∧ · · · ∧ dxikwhich is a section of the trivial vector bundle U ×∧k(Rn)∗ → U . In other words, for every x ∈ U ,ηx defines an element of ∧k(TxU)∗. A section of U × ∧k(Rn)∗ → U is called a k-differential form


on U and the space of all sections of this bundle is denoted by:

Ω∗(U) =

η =∑I∈nk

fI dxi1 ∧ · · · ∧ dxik , f ∈ C∞(U,R)

=C∞ sections of U × ∧k(Rn)∗ → U

.

Let now W ⊂ Rm and V ⊂ Rn be open sets, ψ : W → V be a smooth map and η ∈ Ωk(V ). Wedefine the pull-back of η under ψ, denoted ψ∗η ∈ Ωk(W ) as the section of W × ∧k(Rm)∗ → Wgiven by x 7→ (x, dxψ

∗ηψ(x)). In other words:

ψ∗ηx(w1, . . . , wk) = ηψ(x)(dxψw1, . . . , dxψwk).

Observe that, by Proposition 137, given I = i1 < · · · < ik we have:

ψ∗(dxi1 ∧ · · · ∧ dxik) =∑

J∈mkdet(Jψ(x)I,J)dxj1 ∧ · · · ∧ dxjk .

When ψ : W → V is the inclusion, ψ∗η is denoted by η|W and called the restriction of η to W .

Example 138. Let ψ : Rm → Rn be a smooth map with components ψ = (ψ1, . . . , ψn) andconsider dxi ∈ Ω1(Rn). Then we have:

(4.4) ψ∗(dxi) =

m∑j=1

∂ψi∂xj

(x)dxj .

In practice combining (4.2) and (4.4) gives the simplest way to compute the pull-back of a differ-ential form.

We turn now to the definition of the main object of this section.

Definition 139 (Differential form). Let M be a smooth manifold with atlas (Uα, ψα)α∈A. Adifferential k-form η12 on M is a collection ηαα∈A of differential forms ηα ∈ Ωk(ψα(Uα)) suchthat for every α, β ∈ A with Uα ∩ Uβ 6= ∅ we have:

(4.5) (ψα ψ−1β )∗ηα = ηβ .

We want to show now that there exists a vector bundle

∧k(Rm)∗ → ∧k(T ∗M) −→M

such that differential k-forms are exactly sections of this bundle. In fact, given the atlas (Uα, ψα)α∈Afor the m-dimensional manifold M , we define the cocyle gαβ : Uα ∩ Uβ → GL(∧k(Rm)∗) as themap:

gαβ(x) = the

(m

k

)×(m

k

)matrix with entries det

(J(ψα ψ−1

β )(x))I,J

, I, J ∈m

k

.

We define ∧k(T ∗M) → M to be the vector bundle with cocycle gαβ, using the construction ofExample 94 . Recalling the discussion from Remark 106, it follows now from (4.3) and (4.5) thata differential k-form is a section of ∧k(T ∗M)→M . We denote by Ωk(M) the space of all smoothsections of this bundle:

Ωk(M) = C∞ sections of ∧kT ∗M.A differential k-form η ∈ Ωk(M), evaluated at x ∈ M is therefore a multilinear, alternating mapηx ∈ ∧k(T ∗xM) :

ηx : TxM × · · · × TxM︸︷︷︸k times

→ R multilinear and alternating.

12k is called the degree of the form η and denoted by deg(η).

38 ANTONIO LERARIO

Example 140. We have the following identities:

Ω0(M) = C∞(M,R)

Ω1(M) = sections of T ∗M

Ω2(M) =

objects which in coordinates look like∑i<j

fij(x)dxi ∧ dxj

...

Ωk(M) =

objects which in coordinates look like

∑i1<···<ik

fi1···ik(x)dxi1 ∧ dxik

Given η ∈ Ωk(M) and θ ∈ Ω`(M) the wedge η ∧ θ ∈ Ωk+`(M) is defined by the pointwiseoperation (η ∧ θ)x = ηx ∧ θx. The wedge product of differential forms satisfies the followingproperties:

(1) (C∞(M,R)-bilinearity) For every a1, a2, b1, b2 ∈ C∞(M,R) and forms γ1, γ2 ∈ Ωk(M) andω1, ω2 ∈ Ω`(M) we have:

(a1γ1 + a2γ2) ∧ (b1ω1 + b2ω2) = a1b1γ1 ∧ ω1 + a1b2γ1 ∧ ω2 + a2b1γ2 ∧ ω1 + a2b2γ2 ∧ ω2.

(2) (associativity) For every γ ∈ Ωk(M) , ω ∈ Ω`(M) and θ ∈ Ωn(M) we have:

γ ∧ (ω ∧ θ) = (γ ∧ ω) ∧ θ.

(3) (anticommutativity) For every γ ∈ Ωk(M) , ω ∈ Ω`(M) we have:

γ ∧ ω = (−1)k`ω ∧ γ.

The vector space

Ω∗(M) =

dim(M)⊕k=0

Ωk(M)

together with the multiplication given by the wedge of forms is a graded algebra over C∞(M,R).

The pull-back of a differential form ω ∈ Ωk(N) under a smooth map ψ : M → N is defined tobe the form ψ∗ω such that:

(ψ∗ω)x(w1, . . . , wk) = ηψ(x)(dxψw1, . . . , dxψwk).

4.3. The exterior differential. Let U ⊂ Rm be an open set. We define now a linear operatord : Ωk(U)→ Ωk+1(U) by the following two conditions:

(1) if f ∈ Ω0(U) = C∞(U,R), then df coincides with the usual differential of functions;(2) if ω =

∑I∈mk fIdxi1 ∧ · · · ∧ dxik then dω =

∑I∈mk dfI ∧ dxi1 ∧ · · · ∧ dxik .

Example 141. If ω = f(x, y)dx+ g(x, y)dy ∈ Ω1(R2), then

dω =∂f

∂xdx ∧ dx+

∂f

∂ydy ∧ dx+

∂g

∂xdx ∧ dy +

∂g

∂ydy ∧ dy

=∂f

∂ydy ∧ dx+

∂g

∂xdx ∧ dy

=

(∂g

∂x− ∂f

∂y

)dx ∧ dy.

The exterior differential satisfies the following properties.


(1) d(γ ∧ ω) = dγ ∧ ω + (−1)deg(γ)γ ∧ dω.To see that this is true let13 γ = fIdxI and ω = gJdxJ and write:

d(γ ∧ ω) = d(fIgJdxI ∧ dxJ)

= dfIdxI ∧ gJdxJ + fIdgJ ∧ dxI ∧ dxJ= dγ ∧ ω − fIdxi1 ∧ dgJ ∧ dxi2 ∧ · · · ∧ dxik ∧ dxJ= dγ ∧ ω + fIdxi1 ∧ dxi2 ∧ dgJ ∧ dxi3 ∧ · · · ∧ dxik ∧ dxJ= dγ ∧ ω + (−1)deg(γ)γ ∧ dω.

(2) d2 = 0

To see this let first f ∈ Ω0(U), such that df =∑mj=1

∂f∂xj

dxj . Then:

d(df) = d

m∑j=1

∂f

∂xjdxj

=

m∑j=1

d

(∂f

∂xjdxj

)

=

m∑j=1

(m∑i=1

∂2f

∂xi∂xjdxj

)∧ dxi

=

m∑i,j=1

∂2f

∂xi∂xjdxj ∧ dxi = (∗)

In the last sum each pair of indices i, j ∈ 1, . . . ,m appears two times: one time we see∂2f

∂xi∂xjdxj ∧ dxi and the other ∂2f

∂xj∂xidxi ∧ dxj . However dxi ∧ dxj = −dxj ∧ dxi and

∂2f∂xj∂xi

= ∂2f∂xi∂xj

, consequently (∗) = 0.

For a general ω ∈ Ωk(U) with ω =∑I∈mk fIdxI , using the previous point we get:

d(dω) = d

∑I∈mk

dfI ∧ dxI

=

∑I∈mk

d(dfI)︸︷︷︸0

∧dxI + dfI ∧ d(dxI)︸︷︷︸0

= 0.

(3) Given ψ ∈ C∞(W,V ) with W ⊂ Rm and V ⊂ Rn, we have d ψ∗ = ψ∗ d.Starting again with the case of functions we have:

(d ψ∗)f = d(f ψ) =

m∑j=1

∂(f ψ)

∂xjdxj

=

m∑j=1

(n∑i=1

∂ψi∂xj

∂f

∂yj ψ)dxj

=

n∑i=1

(∂f

∂yi ψ) m∑j=1

∂ψi∂xj

dxj︸︷︷︸ψ∗dyi

=

n∑i=1

(ψ∗

∂f

∂yi· ψ∗dyi

)= ψ∗df.

13Here and below we use the convention that for a multi-index I = i1 < · · · < ik ∈mk

the symbol dxI

denotes the form dxi1 ∧ · · · ∧ dxik .

40 ANTONIO LERARIO

Now, for a general form ω =∑I∈nk fIdyI :

(d ψ∗)ω = d

∑I∈nk

ψ∗fI · ψ∗dyi1 ∧ · · · ∧ ψ∗dyik

= d

∑I∈nk

ψ∗fI · d(ψ∗yi1) ∧ · · · ∧ d(ψ∗yik)

=∑I∈nk

dψ∗fI ∧ ψ∗dyi1 ∧ · · · ∧ ψ∗dyik

= ψ∗

∑I∈nk

dfI ∧ dxi1 ∧ · · · ∧ dxik

= (ψ∗d)ω.

We want to extend now the exterior differential to forms on a manifold. Let M be a smoothmanifold and η ∈ Ωk(M). We define dη ∈ Ωk+1(M) to be the section of ∧kT ∗M which in thetrivialization given by an atlas (Uα, ψα)α∈A is defined by:

(dη)α = d(ηα) ∈ Ωk+1(ψα(Uα)).

In order to check that this definition is well posed, i.e. that dηαα∈A defines a section of ∧kT ∗Mwe need to verify that (3.3) is verified for the cocycle of this bundle. Using the above propertieswe see that:

gαβdηα =(ψα ψ−1

β

)∗dηα = d

((ψα ψ−1

β

)∗ηα

)= dηβ ,

which proves dη ∈ Ωk+1(M) is well defined.

Exercise 142. Prove that the exterior differential has the following important property (whichfollows from the previous local properties): given f ∈ C∞(M,N) and ω ∈ Ωk(N), then:

d(f∗ω) = f∗dω.

4.4. De Rham cohomology. We introduce now the concept of De Rham cohomology of a smoothmanifold M . This concept combines the differential structure of the manifold with its topologyand gives a homotopy invariant of the manifold M which is of fundamental importance.

Definition 143 (De Rham cohomology). The differential complex (Ω∗(M), d) is called the DeRham complex fo M . A form ω ∈ Ωk(M) is called closed if dω = 0 and exact if there existsη ∈ Ωk−1(M) such that dη = ω. The fact that d2 = 0 implies that every exact form is closed andconsequently we can define the k-th De Rham cohomology of M as the vector space:

Hk(M) =closed k-forms on Mexact k-forms on M .

Example 144 (The cohomology of R). The De Rham complex in this case is simply:

0 −→ Ω0(R)d0−→ Ω1(R)

d1−→ 0.

The kernel of d0 equals the set of constant functions and consequently:

H0(R) = ker(d0) ' R.

We have ker(d1) = Ω1(R) and we claim that also im(d0) = Ω1(R): in fact, given ω = g(x)dx ∈Ω1(R), if we define the function:

f(x) =

∫ x

0

g(t)dt,

we see that df = ω and consequently:

H1(R) = 0.


Remark 145. Notice that it follows from the definition that if two manifolds are diffeomorphic,then they have the same De Rham cohomology (prove it with details!). We will prove indeedthat they have the same cohomology even if they are just homotopy equivalent (this will be aconsequence of Corollary 147.

4.5. Poincare Lemma. In this section we introduce a basic tools for computing the cohomologyof a manifold. Consider the two commutative diagrams:

(4.6)

M × R Ω∗(M × R)

M Ω∗(M)

π s∗s π∗

where the map π is the projection on the first factor and s(x) = (x, 0) is the “zero section”. NextTheorem implies that the both π∗ and s∗ induce isomorphisms in cohomology.

Theorem 146 (Poincare Lemma). In the following diagram, induced by the diagrams in (4.6):

H∗(M × R)

H∗(M)

s∗π∗

both the maps π∗ and s∗ are isomorphims.

Proof. We will prove that:

s∗π∗ = 1H∗(M) and π∗s∗ = 1H∗(M×R),

which will imply the statement. The first identity is simple, since:

s∗π∗ = (π s)∗ = (idM )∗ = 1H∗(M).

For the second identity we will construct a linear map K : Ω∗(M × R) → H∗−1(M × R) (calledchain homotopy) such that:

(4.7) π∗s∗ − 1 = ±(Kd± dK).

If (4.7) is true, then for every [ω] ∈ H∗(M × R) we have:

π∗s∗[ω]− [ω] = ±[(Kd± dK)ω]

=dω=0±[dKω]

= [d(Kω)]

=d(Kω) is exact

0.

For the proof of (4.7) we observe that every form ω ∈ Ω∗(M × R) can be written in a unique wayas ω = ω1 + ω2 with:

ω1 = f1(x, t)π∗β1 and ω2 = f2(x, t)π∗β2 ∧ dt,where f1, f2 ∈ C∞(M × R) and β1, β2 ∈ Ω∗(M). We define:

Kω1 = 0 and Kω2 =

(∫ t

0

f(x, s)ds

)︸︷︷︸∈C∞(M×R)

π∗β2.

We check now that (4.7) is true for both forms ω1 (“of type (1)”) and ω2 (“of type (2)”).

(1) Let ω1 = fπ∗β1 of the first type. Then:

π∗s∗ω1 − ω1 = π∗s∗(f1(x, t)π∗β1)− f1(x, t)π∗β1

= (f1(x, 0)− f1(x, t))π∗β1

= (∗1)

42 ANTONIO LERARIO

(Kd− dK)ω1 = (Kd− dK)(f1π∗β1)

=Kω1=0

Kd(f1π∗β1)

= K (∂xf1 ∧ π∗β1 + ∂tf1dt ∧ π∗β1 + f1π∗dβ1)

=K(type (1))=0

K(∂tf1dt ∧ π∗β1)

= (−1)deg(ω1)K(∂tf1π∗β1 ∧ dt)

=by def. of K(type (2))

(−1)deg(ω1)

(∫ t

0

∂tf1(x, s)ds

)π∗β1

= (−1)deg(ω1)(f1(x, t)− f1(x, 0))π∗β1

= (−1)deg(ω)−1(∗1).

This proves (4.7) holds true on forms of the first type.(2) Let now ω2 = f2π

∗β2 ∧ dt. We have:

π∗s∗ω2 − ω2 = −f2(x, t)π∗β2 ∧ dt = (∗2),

because s∗dt = 0.

(Kd− dK)ω2 =K (∂xf2 ∧ π∗β2 ∧ dt+ f2π∗dβ2 ∧ dt) +

− d((∫ t

0

f2(x, s)ds

)π∗β2

)=

(∫ t

0

∂xf2(x, s)ds

)π∗β2 +

(∫ t

0

f2(x, s)ds

)π∗dβ2

−(∫ t

0

∂xf2(x, s)ds

)∧ π∗β2 − f2(x, t)dt ∧ π∗β2+

−(∫ t

0

f2(x, s)ds

)π∗dβ2

=− f2(x, t)dt ∧ π∗β2

=(−1)deg(β2)(∗2)

=(−1)deg(ω)−1(∗2).

Corollary 147. The De Rham cohomology of Rn vanishes except in dimension zero, whereH0(Rn) = R.

Proof. This follows immediately using Theorem 146 with the choice M = Rn−1:

H∗(Rn) = H∗(Rn × R) ' H∗(Rn−1) ' · · · ' H∗(point).

Corollary 148 (Homotopy axiom for De Rham cohomology). Homotopic maps induce the samemap in cohomology.

Proof. Let f0, f1 ∈ C∞(M,N) be homotopic maps between smooth manifolds, i.e. f0 = F (·, 0)and f1 = F (·, 1) for a smooth map F : M ×R→ N (here we have extended the homotopy in sucha way that F (x, t) ≡ f0(x) for t ≤ 0 and F (x, t) ≡ f1(x) for t ≥ 1).

For every t ∈ R consider the commutative diagram

M × R N

M

F

π stft ,


where st(x) = (x, t). Since for every t ∈ R the maps s∗t and π∗ are inverse to each other incohomology, it follows that s∗1 = s∗0 and consequently

f∗0 = s∗0F∗ = s∗1F

∗ = f∗1 .

As a corollary we immediately see that if two smooth manifolds are homotopy equivalent, theyhave the same De Rham cohomology.

Corollary 149. Homotopy equivalent manifolds have the same cohomology.

Proof. If two manifolds are homotopy equivalent in the C0-sense, then they are also homotopyequivalent in the C∞ sense (by a variation of the argument in Theorem 129 for noncompactmanifolds, if two maps are continuously homotopic, they are homotopic to smooth maps that aresmoothly homotopic) and we can apply the previous Corollary 148.

Exercise 150. Using the fact that every manifold (even non-compact) embeds in some Rn, workout all the details for the proof Corollary 149. (This is a long, but rewarding, exercise...)

Exercise 151. Compute the De Rham cohomology of R2\point.Exercise 152. Compute the De Rham cohomolgy of the Moebius band M (hint: use the fact thatπ : M → S1 is a line bundle...)

4.6. Mayer-Vietoris sequence. Let M be the union of two open sets A ∪ B. Then we have adiagram of maps:

(4.8) M AqB A ∩BiB

iA

,

where iA and iB are the inclusions in A and B respectively. The Mayer-Vietoris sequence of formsis the sequence obtained by taking the induced map at the level of differential forms (and takingthe difference of the two maps on the right):

0 Ω∗(M) Ω∗(A)⊕ Ω∗(B) Ω∗(A ∩B) 0

ω (ω|A, ω|B)

(α, β) β|A∩B − α|A∩B

(1)

j i∗B−i∗A(2) (3)

Proposition 153. The Mayer-Vietoris sequence of forms is exact.

Proof. We check exactness at every step.

(1) j is injective since a form ω ∈ Ω∗(M) is zero if and only if both its restrictions to A andB are zero.

(2) clearly (i∗B − i∗A)(ω|A, ω|B) = 0; moreover if α|A ∩B = β|A∩B , then there exists ω ∈Ω∗(A ∪B) such that ω|A = α and ω|B = β.

(3) to check that every element η ∈ Ω∗(A∩B) is the difference of two forms we simply take apartition of unity ρA, ρB subordinated to the cover A,B and define:

α = −ρBη ∈ Ω∗(A) and β = ρAη ∈ Ω∗(B).

With this choice we have (i∗B − i∗A)(α, β) = η.

As a corollary we derive the following Theorem, which is a second basic tool for computing thecohomology of a manifold.

44 ANTONIO LERARIO

Theorem 154. Let M = A ∪ B be an m-dimensional manifold with A,B two open sets. Thereexists a long exact sequence (called the Mayer-Vietoris long exact sequence):

· · · Hq(A ∪B) Hq(A)⊕Hq(B) Hq(A ∩B) Hq+1(A ∪B) · · ·d∗

which starts with 0→ H0(A ∪B)→ · · · and ends with · · · → Hm(A ∩B)→ 0.

Proof. This follows from Proposition 153 and Proposition 196

Example 155 (Cohomology of spheres). The sphere S0 consists of two points and consequently:

H0(S0) = R⊕ R and Hq(S0) = 0 ∀q > 0.

We want to show now that for n > 0 we have:

Hq(Sn) =

R for q = 0, n0 otherwise

Let us start with the case of S1. Considering the open cover U1, U2 from Example 7 (i.e. thecover given by the upper and the lower hemispheres). We have:

U1 ' R ' U2 and U1 ∩ U2 ' Rq R.Consequently the Mayer-Vietoris long exact sequence for this open cover is:

0 H0(S1) H0(U1)⊕H0(U2) H0(U1 ∩ U2) H1(S1) 0 .

Since S1 is connected it follows that H0(S1) ' R, hence the above sequence equals:

0 R R⊕ R R⊕ R H1(S1) 0

and the only unknown element is H1(S1). However the alternating sum between two zeroes of thedimensions of vector spaces in an exact sequences is zero (see Exercise 197) and:

dim(H1(S1))− 2 + 2− 1 = 0

which gives H1(S1) = R.For the sphere Sn we argue as follows. First, since Sn is connected (n > 0), again H0(Sn) = R.

Moreover, since both U1 and U2 are diffeomorphic to Rn, for n > 0 their cohomology vanishes byCorollary 147. Conseuqently for n > 1 the only nonzero piece of the long Mayer-Vietoris exactsequence for the open cover given by the upper and the lower hemispheres is:

Hn−1(U1)⊕Hn−1(U2) = 0 Hn−1(U1 ∩ U2) Hn(Sn) 0' .

Since U1 ∩ U2 is homotopy equivalent to a sphere Sn−1, Corollary 148 implies that:

Hn−1(U1 ∩ U2) ' Hn−1(Sn−1)

and we conclude by induction.

Exercise 156. Using the Mayer-Vietoris sequence, compute the cohomology of R2\p1, p2.Exercise 157. Using the Mayer-Vietoris sequence, prove that:

H0(RPn) = R and Hn(RPn) =

R n odd0 n even

Remark 158 (The generalized Mayer-Vietoris principle). There is a far generalization of theMayer-Vietoris sequence to countably many open sets, which works as follows. Let U = Ujj∈Jbe a finite open cover (possibly with infinitely many open sets, but countably). For every k ∈ Nwe define the topological space:

Σk =∐

j0<···<jkUj0 ∩ · · · ∩ Ujk

Consider now the following diagram, which generalizes (4.8):

M Σ0 Σ1 Σ2 · · ·∂0

∂1∂2

∂0

∂1


where ∂i is the inclusion which ignores the i-th open set (e.g. ∂0 : Uj0∩Uj1∩Uj2 → Uj1∩Uj2). Usingthe maps from this diagram we can form an exact sequence (which generalizes the Mayer-Vietorissequence of forms):

(4.9) 0 Ω∗(M) Ω∗(Σ0) Ω∗(Σ1) · · ·r δ δ

Here δ : Ω∗(Σk)→ Ω∗(Σk+1) is defined as follows. First observe that:

Ω∗(Σk) =∏

j0<···<jkΩ∗(Vj0 ∩ · · · ∩ Vjk)

and consequently we can write η ∈ Ω∗(Σk) as:

η = (ηj0···jk) with ηj0···jk ∈ Ω∗(Vj0 ∩ · · · ∩ Vjk).

We define δη ∈ Ω∗(Σk+1) as:

(δη)j0···jk+1=

k+1∑i=0

(−1)iηj0···ji···jk .

The sequence (4.9) is called the generalized Mayer-Vietoris sequence and it is exact (see [3,Proposition 8.5]). The trick that we used in point (3) of Propostion 153 for proving exact-ness here generalizes as follows: pick η = (ηj0···jk) ∈ Ω∗(Σk) such that δ(η) = 0 and defineτ = (τj0···jk−1

) ∈ Ω∗(Σk−1) by:

τj0···jk−1=∑j=1

ρjτjj0···jk−1,

where ρj`j=1 is a partition of unity subordinated to the open cover U. (Check that in thecase of two open sets this gives the same as we already did!). Out of the generalized Mayer-Vietoris sequence one can construct the so-called Mayer-Vietoris spectral sequence, which is afar generalization of the long exact Mayer-Vietoris sequence. I encourage the willing reader toelaborate on this by reading [3, Chapter 8].

4.7. The Mayer-Vietoris argument. The Mayer-Vietoris sequence can be used as a tool forproving statements by induction on the cardinality of “good” open covers of a manifold. Thisidea (combining the Mayer-Vietoris sequence with inductive proofs using “good” open covers) issometimes called the Mayer-Vietoris argument. What is meant by “good” is the following.

Definition 159 (Good cover). An open cover of an n-dimensional manifold is called a good coverif all finite nonempty intersections of open sets from it are diffeomorphic to Rn

Remark 160. Every manifold has a good cover – this is proved using Riemannian geometry. Wewill take this statement for granted. Moreover the generalized Mayer-Vietoris sequence techniquecan be used to prove that the De Rham cohomology ofM is isomorphic to the simplicial cohomologyof the nerve of a good cover [3, Theorem 8.9].

Theorem 161. If a manifold has a finite good cover, then its cohomology is finite dimensional.

Proof. The proof is an interesting application of the above principle. First observe that from theexactness of the long Mayer-Vietoris sequence for the open cover M = A ∪ B it follows that thecohomology of M has dimension bounded by:

(4.10) dim(H∗(M)) ≤ dim(H∗(A)) + dim(H∗(B)) + dim(H∗(A ∩B)),

and in particular if A, B and A ∩ B have finite-dimensional cohomology, then also A ∪ B does.Now we prove the statement of the theorem by induction on the cardinality k of the good cover.For k = 1 the manifold is diffeomorphic to Rn and the theorem is true. Assume now the theoremis proved for all manifolds with a good cover of cardinality k and let

M = U1 ∪ · · · ∪ Uk+1

such that Ujk+1j=1 is a good cover. Then we can take the cover of M given by the two open sets:

A = U1 ∪ · · · ∪ Uk and B = Uk+1.

46 ANTONIO LERARIO

Both A and B have finite dimensional cohomology by induction; moreover

U1 ∩ Uk+1, . . . , Uk ∩ Uk+1is a good cover of A∩B of cardinality k, so A∩B also has finite dimensional cohomology and thestatement follows from (4.10).

4.8. Compact support. Given a smooth manifold M of dimension m it is also possible to restrictour attention only to forms with compact supports, i.e. we can consider

Ωkc (M) = sections of ∧kT ∗M which have compact support.Observe that the exterior differential preserves the property of having compact support and hencethere exists a well-defined subcomplex:

0 −→ Ω0c(M)

d0−→ Ω1c(M)

d1−→ · · · dm−2−→ Ωm−1c (M)

dm−1−→ Ωmc (M)dm−→ 0.

The cohomology of this complex is called the compactly supported De Rham cohomology and de-noted by:

Hkc (M) =

compactly supported, closed k-forms on Mcompactly supported, exact k-forms on M .

Example 162 (The compactly supported cohomology of R). In this case the complex for com-pactly supported forms is:

0 −→ Ω0c(R)

d0−→ Ω1c(R)

d1−→ 0.

We see that ker(d0) equals the set of constant functions with compact supports, i.e. ker(d0) = 0and consequently:

H0c (R) = 0.

For the cohomology H1c (R) the situation is already a bit more interesting. Consider the linear

map: ∫R

: Ω1c(R)→ R

which takes a 1-form ω = g(x)dx and gives∫R g(x)dx. This map is clearly surjective (being nonzero)

and:

(4.11)Ω1c(R)

ker(∫R ·)' R.

We claim that:

(4.12) im(d0) = ker

(∫R·).

In fact given f ∈ Ω0c(R), then for a < 0 < b with absolute value large enough we have∫

Rdf =

∫R

∂f

∂x(x)dx = f(b)− f(a) = 0

and df ∈ ker(∫R ·). Viceversa, let ω = g(x)dx ∈ ker(

∫R ·). Then we can define the function

f(x) =∫ x

0g(t)dt which has the property of being compactly supported (because ω is compactly

supported) and satisfies df = ω.From this, using (4.11) and (4.12) we see that:

H1c (R) =

ker(d1|Ω1c(R))

im(d0|Ω0c(R))

=Ω1c(R)

im(d0|Ω0c(R))

' Ω1c(R)

ker(∫R ·)

= R.

Exercise 163. Let A,B be two open sets in an m-dimensional manifold. There is a Mayer-Vietorissequence of forms with compact support:

0 Ω∗c(A ∩B) Ω∗c(A)⊕ Ω∗c(B) Ω∗c(A ∪B) 0ι∗ σ ,

where ι∗(α) = (α,−α) and σ(α, β) = α + β. Prove that this sequence is exact and consequentlythat there exists a long exact sequence:

· · · Hqc (A ∩B) Hq

c (A)⊕Hqc (B) Hq

c (A ∪B) Hq+1c (A ∩B) · · ·d∗


which starts with 0 −→ H0c (A ∩B) −→ · · · and ends with · · · −→ Hm

c (M) −→ 0. This long exactsequence is called the Mayer-Vietoris long exact sequence with compact support.

Exercise 164 (The compactly supported cohomology of Rn). The scope of this guided exerciseis to generalize Example 162 and to prove that:

Hqc (Rn) '

R q = n0 q 6= n

We will mimic the construction of Theorem 146. First we consider the diagram of maps:

Ωq+1c (M × R)

Ωqc(M)

π∗ e∗ ,

where π∗ and e∗ are constructed as follows. Every form in Ωqc(M × R) can be written (using thenotation of the proof of Theorem 146) as a unique linear combinations of forms of two types:

ω1 = f1(x, t)π∗β1 and ω2 = f2(x, t)π∗β2 ∧ dt,where now the involved forms have compact support. We define π∗ to be zero on forms of the firsttype and we set:

π∗ (f2(x, t)π∗β2 ∧ dt) =

(∫ ∞−∞

f2(x, t)dt

)· π∗β2.

Check that this defines π∗ (this amounts to verify that the decomposition into sum of forms ofthe two types is unique and that π∗ maps to the space fo forms with compact support). For thedefinition of e∗ we pick a form e = e(t)dt ∈ Ω1

c(R) such that∫R e(t)dt = 1 and we set:

e∗(η) = π∗η ∧ e, η ∈ Ωqc(M).

Using the strategy of Theorem 146 prove that π∗ and e∗ induce isomorphisms in cohomology. (Hereone has to find an appropriate K : Ωqc(M×R)→ Ωq−1

c (M×R) such that e∗π∗−1 = ±(Kd±dK)...)

4.9. Orientation and integration. Pick coordinate functions x1, . . . , xn : Rn → R, which weassume to be ordered. Let ω ∈ Ωnc (Rn) and write it as:

ω = fdx1 ∧ · · · ∧ dxn,for some function f ∈ C∞c (Rn,R). We define the integral of ω as:∫

[Rn]

f(x)dx1 ∧ · · · ∧ dxn =

∫Rnf(x)dx1 · · · dxn.

On the right hand side we have the standard Riemann integral of a smooth, compactly supportedfunction, while on the left hand side we have introduced the symbol “[Rn]” to stress that a orderingof the coordinates has been chosen (this notation will become more transparent soon).

Let now ϕ ∈ C∞(Rn,Rn) be a diffeomorphism and consider the form

ϕ∗ω = (f ϕ) · det(Jϕ) · dy1 ∧ · · · ∧ dyn.(This identity follows from Proposition 137.) From this we see that the integral of a top differentialform ω ∈ Ωnc (Rn) is not invariant under diffeomorphisms, but only under diffeomorphisms thatpreserve the orientation, i.e. which have everywhere positive Jacobian. In fact, using the abovedefinition we have:∫

[Rn]

ϕ∗ω =

∫Rn

(f ϕ) · det(Jϕ) · dy1 · · · dyn

= sign(det(Jϕ))

∫Rn

(f ϕ) · | det(Jϕ)| · dy1 · · · dyn

= sign(det(Jϕ))

∫Rnfdx1 · · · dxn

= sign(det(Jϕ))

∫[Rn]

ω.(4.13)

This discussion justifies the following definition.

48 ANTONIO LERARIO

Definition 165 (Orientable manifold). LetM be a smooth manifold with an atlas A = (Uα, ψα)α∈A.We say that M is orientable if for every α, β ∈ A such that Vα ∩ Vβ 6= ∅ we have:

(4.14) det(J(ψα ψ−1β )) > 0.

In this case A is called an oriented atlas. We say that a smooth manifold is orientable14 if it admitsan oriented atlas.

The manifolds Rn and Sn are both orientable: Rn because it can be endowed with a single chart(the identity), while for spheres one has to pay attention because the atlas constructed in Example7 is not oriented..

Exercise 166. Prove that the sphere Sn is orientable, by building an oriented atlas.

Observe that the condition (4.14) can be reinterpreted in the language of vector bundles: since(ψα ψ−1

β ) is the cocycle of the tangent bundles, a manifold is orientable if and only if its tangent

bundle is orientable (as a vector bundle15, in the sense of Section 3.5). Using the language of vectorbundles we can rephrase the orientability condition as follows.

Proposition 167. A smooth manifold M of dimension m is orientable if and only if the bundle∧m(T ∗M)→M is trivial.

Proof. Assume that M is orientable and let (Uα, ψα)α∈A be an oriented atlas. Let also ραα∈Abe a partition of unity subordinated to this atlas. Fix coordinates x1, . . . , xm : Rm → R and set:

σα = ραψ∗α(dx1 ∧ · · · ∧ dxm) ∈ Ωm(M).

This is a form on the whole M because we are multiplying ψ∗α(dx1 ∧ · · · ∧ dxm) (which is a formon Uα) by the function ρα which is zero outside Uα. Consider now the form:

ω =∑α∈A

σα ∈ Ωm(M).

We claim that this form is never zero, i.e. it is a nowhere zero section of the line bundle ∧m(T ∗M),which will therefore be trivial by Exercise 105. In order to see that ω is never zero, it is enough

to show that for every β ∈ A the form ωβ =(ψ−1β

)∗ω ∈ Ω∗(ψβ(Uβ)) is never zero. To this end

observe that:

ωβ =(ψ−1β

)∗ω

=∑α∈A

(ψ−1β

)∗σα

=∑α∈A

(ψ−1β

)∗ραψ

∗α(dx1 ∧ · · · ∧ dxm)

=∑α∈A

(ρα ψ−1

β

)·(ψα ψ−1

β

)∗(dx1 ∧ · · · ∧ dxm)

=∑α∈A

(ρα ψ−1

β

)· det

(J(ψα ψ−1

β ))dx1 ∧ · · · ∧ dxm

= gβ(x)dx1 ∧ · · · ∧ dxm,

where we have set gβ =∑α∈A

(ρα ψ−1

β

)· det

(J(ψα ψ−1

β ))

. Since this function is never zero,

it follows that ωβ is never zero.

Assume now that the bundle ∧m(T ∗M) is trivial. Then there exists ω ∈ Ωm(M) which is neverzero – we will use this form to build an oriented atlas for M . Let U = (Uα, ψα)α∈A be an atlasfor M (it might not be oriented yet). For every α ∈ A there exists a never vanishing function fαsuch that:

ψ∗α(dx1 ∧ · · · ∧ dxm) = fαω.

14As we will see, not all manifolds are orientable (for example RP2 is not orientable).15We say orientable “as a vector bundle”, because TM as a smooth manifold is always orientable (i.e. TTM is

orientable as a vector bundle). Try to prove this!


Now, if fα > 0 we set ψα = ψα, otherwise if fα < 0 we replace with the chart:

ψα = ϕ ψα : Uα → Rn

where ϕ(x1, . . . , xm) = (−x1, x2, . . . , xm) is a diffeomorphism which reverses the orientation. It is

now easy to check that the atlas (Uα, ψα)α∈A is an oriented atlas.

Let now ω1, ω2 ∈ Ωm(M) be two nowhere zero forms. We say that they are equivalent ω1 ∼ ω2

if ω1 = fω2 with f > 0. If M is orientable, an orientation for M is the choice of an equivalenceclass of nowhere vanishing forms (notice that on a connected manifold there are only two possibleorientations). Once this choice has been made, then one can build an oriented atlas for M as inthe proof of Proposition 167. The choice of the orientation class is denoted by [M ].

Example 168 (RP2 is not orientable). We show in this example that RP2 is not orientableby proving that ∧2(T ∗RP2) is not a trivial bundle. To this end let us first consider the formω ∈ Ω2(R3) given by:

ω = x1dx2 ∧ dx3 − x2dx1 ∧ dx3 + x3dx1 ∧ dx2.

We restrict this form to the two sphere S2, i.e. we consider ω ∈ Ω2(S2) given by:

ω = i∗ω

where i : S2 → R3 is the inclusion. Notice that, denoting by α : S2 → S2 the antipodal map, wehave:

(4.15) α∗ω = −ω,because α = −1|S2 and −1∗ω = −ω (here −1 : R3 → R3 is the map x 7→ −x). Observe thatω ∈ Ω2(S2) is never zero, hence it defines a trivialization of ∧2(T ∗S2). In other words, themap (x, t) 7→ tωx gives a trivializing diffeomorphism S2 × R → ∧2(T ∗S2) (we already knew thisbundle is trivial since S2 is orientable). Assume now that RP2 is orientable. Then there existsa nowhere vanishing form η ∈ Ω2(RP2) and we can pull-back this form to S2 using the quotientmap q : S2 → RP2. The resulting form q∗τ is also never vanishing and consequently there existsa nowhere zero function f : S2 → R such that:

q∗η = fω.

From the commutativity of the diagram

S2 S2

RP2

q

α

q

we deduce that α∗q∗η = q∗η. However we have:

fω = q∗η = α∗(q∗η) = α∗(fω) = −fω,(the last identity by (4.15)), which contradicts the fact that f is never zero.

Definition 169 (Integral of a form of degree m on an oriented m-dimensional manifold). LetM be an orientable and oriented manifold, with oriented atlas (Uα, ψα)α∈A, and consider thecorresponding subordinated partition of unity ραα∈A. We define the integral of an m-formτ ∈ Ωmc (M) as the number: ∫

[M ]

τ =∑α∈A

∫ψα(Uα)

(ψ−1α

)∗(ρατ).

In order to see that this definition does not depend on the choice of the partition of unity orthe oriented atlas, we argue as follows. First observe that, once an orientation has been fixed andan oriented atlas (Uα, ψα)α∈A has been chosen, then for every form σ ∈ Ωm(Uα) the integral:∫

[Uα]

σ =

∫Rn

(ψ−1α

)∗σ

is well defined, i.e. it does not depend on the choice of the oriented chart. This follows from (4.13),since the transition function for two oriented charts belonging to the same orientation class have

50 ANTONIO LERARIO

positive Jacobian. As a consequence, if now (Vβ , ϕβ)β∈B is another oriented atlas positivelyoriented with (Uα, ψα)α∈A, with corresponding partition of unity γββ∈B , then:∫

[M ]

τ =∑α∈A

∫ψα(Uα)

(ψ−1α )∗(ρατ)

=∑α∈A

∫[Uα]

ρατ

=∑γβ=1

∑α∈A,β∈B

∫[Uα]

ραγβτ

=supp(ραγβ)⊂Uα∩Vβ

∑α∈A,β∈B

∫[Uα∩Vβ ]

ραγβτ

=by symmetry

∑β∈B

∫[Vβ ]

γβτ.

4.10. Stokes’ theorem. In this section we work with orientable manifolds that can possibly haveboundary (see Definition 186 below). We start with the following Lemma.

Lemma 170. Let M be orientable with ∂M 6= ∅. Then ∂M is orientable.

Proof. Let (Vα, ϕα)α∈A be an oriented atlas for M :

det(J(ϕα ϕ−1

β ))> 0 whenever Vα ∩ Vβ 6= ∅.

We claim that (Uα ∩ ∂M,ϕα|Uα∩∂M )α∈A is an oriented atlas for ∂M (here A ⊆ A denotes theindex set labeling only the charts that intersect ∂M).

Whenever Vα∩Vβ ∩∂M 6= ∅ we have that the map ϕαβ = ϕα ϕβ : W → H is a diffeomorphismbetween an open set W ⊆ H and its image such that:

φαβ.= ϕαβ |W∩∂H : W ∩ ∂H −→ ∂H,

and we want to prove that det(Jφαβ) > 0 (note that φαβ is a map between open subsets of Rm−1,where m = dim(M)). Let us put coordinates (x, y) ∈ H = Rm−1 × [0,∞) such that H = y ≥ 0and ∂H = y = 0. Observe that the condition ϕαβ(W ∩ ∂H) ⊆ ∂H implies that, writing the

components ϕαβ = (ϕ(1)αβ , ϕ

(2)αβ) ∈ Rm−1 × R, then for all j = 1, . . . ,m− 1 we have:

∂ϕ(2)αβ

∂xj(x, 0) = 0 and

∂ϕ(2)αβ

∂y(x, 0) > 0.

In particular the Jacobian matrix of ϕαβ at points of the form (x, 0) has the form:

Jϕαβ(x, 0) =

(∂φαβ∂x (x) ∗

0∂ϕ

(2)αβ

∂y (x, 0)

).

Since∂ϕ

(2)αβ

∂y (x, 0) > 0 and det(Jϕαβ) > 0, then it must also be det(Jφαβ) > 0.

Definition 171 (Induced orientation on the boundary). Let M be an orientable and orientedmanifold of dimension m with boundary ∂M . The induced orientation on ∂M is (−1)m theorientation that one obtains considering the restriction to ∂M of an oriented atlas for M . If [M ]is the orientation on M , we denote by [∂M ] the induced orientation on the boundary ∂M .

In other words, if the orientation on H is given by the class (in the sense of Section 4.9) of theform dx1∧· · ·∧dxm−1∧dy, the induced orientation on ∂H is given by the class of (−1)mdx1∧· · ·∧dxm−1. The reason to introduce the (−1)m is to have a formula with no signs in Stokes’ Theorem.

Theorem 172 (Stokes’ Theorem). Let M be an orientable and oriented manifold of dimensionm, possibly with boundary. For every ω ∈ Ωm−1

c (M) we have:

(4.16)

∫[M ]

dω =

∫[∂M ]

ω.


Proof. Let (Uα, ϕα)α∈A be an oriented atlas for M and ραα∈A be a subordinated partition ofunity. We will prove that for every α ∈ A we have:

(4.17)

∫[Uα]

d(ραω) =

∫[∂Uα]

ραω,

where ∂Uα is the oriented manifold Uα ∩ ∂M (and not the boundary of Uα in M !) with theorientation obtained by restricting the induced orientation [∂M ]. From (4.17) the conclusion(4.16) follows, since: ∫

[M ]

dω =

∫[M ]

d

(∑α∈A

ραω

)

=

∫[M ]

∑α∈A

d (ραω)

=∑α∈A

∫[M ]

d (ραω)

=supp(d(ραω))⊂Uα

∑α∈A

∫[Uα]

d (ραω)

=(4.17)

∑α∈A

∫[∂Uα]

ραω

=supp(ραω)⊂Uα

∑α∈A

∫[M ]

ραω

=

∫[∂M ]

ω.

In order to prove (4.17) we observe that there are two types of charts in an altas (Uα, ϕα)α∈A fora manifold with boundary: (i) charts such that Uα∩∂M = ∅ and (ii) charts such that Uα∩∂M 6= ∅;we prove (4.17) separately for these two types.

Let ϕ : Uα → Rm be a chart of the first type. Then, by definition:∫[Uα]

d (ραω) =

∫[ϕ(Uα)]

(ϕ−1α

)∗d(ραω) =

∫[Rm]

d((ϕ−1α

)∗ραω

).

We can write the form(ϕ−1α

)∗ραω ∈ Ωm−1

c (ϕα(Uα)) as:

(4.18)(ϕ−1α

)∗ραω =

m∑j=1

fα,j(x)dx1 ∧ · · · ∧ dxj ∧ · · · dxm,

and consequently we have:

(4.19) d((ϕ−1α

)∗ραω

)=

m∑j=1

(−1)j+1 ∂fα,j∂xj

(x)dx1 ∧ · · · ∧ dxm.

We integrate the summands of (4.19) one by one:∫[Rn]

(−1)j+1 ∂fα,j∂xj

(x)dx1 ∧ · · · ∧ dxm =

∫Rm

(−1)j+1 ∂fα,j∂xj

(x)dx1 · · · dxm

=

∫Rm−1

(∫ ∞−∞

∂fα,j∂xj

(x)dxj

)dx1 · · · dxj · · · dxm∫

Rm−1

(limR→∞

fα,j(R)− fα,j(−R))dx1 · · · dxj · · · dxm

=

∫Rm−1

0 · dx1 · · · dxj · · · dxm = 0,

where fα,j(±R) = fα,j(x1, . . . ,±R, . . . , xm) and we have used the fact that∂fα,j∂xj

(x) has compact

support. Since for an open set Uα of the first type we have ∂Uα = ∅, this proves (4.17) in this case.

52 ANTONIO LERARIO

Let us consider now the case of a chart ϕα : Uα → H such that ϕα(Uα)∩ ∂H 6= ∅. Observe first

that, writing(ϕ−1α

)∗ραω as in (4.18), we have:

(4.20)

∫[∂Uα]

ραω = (−1)m∫∂H

fα,m(x, 0)dx1 · · · dxm,

where the (−1)m comes from the definition of induced orientation. Proceeding as before we get:

(4.21)

∫[Uα]

d (ραω) =

∫H

m∑j=1

(−1)j+1 ∂fα,j∂xj

(x)dx1 · · · dxm−1dy

(notice that we have used a different symbol “y” to denote the last coordinate) and again weintegrate the summands one by one, i.e. we consider:

(4.22)

∫H

(−1)j+1 ∂fα,j∂xj

(x)dx1 · · · dxm−1dy.

For j 6= m we can write the integral in (4.22) (up to its sign) as:∫H

∂fα,j∂xj

(x, y)dx1 · · · dxm−1dy =

∫Rm−1×[0,∞)

∂fα,j∂xj

(x, y)dx1 · · · dxm−1dy

=

∫Rm−2×[0,∞)

(∫ ∞−∞

∂fα,j∂xj

(x, y)dxj

)dx1 · · · dxj · · · dxm−1dy

=

∫Rm−2×[0,∞)

0 · dx1 · · · dxj · · · dxm−1dy = 0(4.23)

where we have used again the fact that∂fα,j∂xj

has compact support. For the last summand we have

instead:

(−1)m+1

∫H

∂fα,m∂y

(x, y)dx1 · · · dxm−1dy = (−1)m+1

∫Rm−1

(∫ ∞0

∂fα,m∂y

(x, y)dy

)dx1 · · · dxm−1

= (−1)m+1

∫Rm−1

−fα,m(x, 0)dx1 · · · dxm

=(4.20)

∫[∂Uα]

ραω.

Together with (4.21) and (4.23), this proves (4.17) also in this second case and concludes the proof.

Exercise 173 (The volume form of Sn). Let us denote by ω ∈ Ωn(Sn) the form:

(4.24) ω = i∗

n+1∑j=1

(−1)j+1xjdx1 ∧ · · · ∧ dxj ∧ · · · ∧ dxn+1

where i : Sn → Rn+1 denotes the inclusion. This is called the volume form of Sn. Compute theintegral of ω on Sn, with respect to the orientation induced on Sn by viewing it as the boundaryof the disk Dn+1 and orienting Dn+1 in the standard way as a subset of Rn+1 and show that:∫

[Sn]

ω =2π

n+12

Γ(n+1

2

) .Example 174. Let us consider again the form ω ∈ Ωn(Sn) defined in (4.24). We claim thatHn(Sn) is generated by [ω]. In fact we know that dω = 0 (because it is a top form), hence itscohomology class is well defined. Moreover, by the previous Exercise 173, we have:∫

[Sn]

ω > 0.

Assume that [ω] = 0 in Hn(Sn). Then ω = dη for some η ∈ Ωn−1(Sn) and consequently, usingStokes’

0 <

∫[Sn]

ω =

∫[Sn]

dη =Stokes

∫[∂Sn]

η =∂Sn=∅

0,

which is a contradiction.


4.11. Poincare duality. Let M be an orientable and oriented manifold without boundary and ofdimension m. Then we can define a linear map P : Hq(M) −→ (Hn−q

c (M))∗

as follows: given anelement [α] ∈ Hq(M) the linear functional P ([α]) : Hn−q

c (M)→ R is defined by:

β 7−→∫

[M ]

α ∧ β.

The fact that the map P is well defined follows from Stokes’ Theorem. Poincare duality assertsthat, when the cohomology of M has finite dimension, P is an isomorphism.

Theorem 175 (Poincare duality). Let M be an orientable and oriented manifold of dimension m,with no boundary and with finite-dimensional cohomology. Then P : Hq(M) −→ (Hn−q

c (M))∗

isan isomorphism for every q ∈ N.

Proof. We will prove the statement under the additional assumption that M has a finite goodcover. Under this assumption the strategy of the proof is another example of application of theMayer-Vietoris argument: we proceed by induction on the cardinality k of the good cover. (Inother words: we will assume Poincare duality is true for all manifolds with a good cover consistingof at most k elements and we prove it for manifolds with a good cover of cardinality k + 1.)

The case k = 0 (the base of the induction) follows immediately from Exercise 164: becauseH∗c (Rn) = Hn

c (Rn) ' R, this vector space is generated by [f(x)dx1 ∧ · · · ∧ dxn], where f ∈C∞c (Rn,R) (the form f(x)dx1 ∧ · · · ∧ dxn on [Rn] cannot be exact by Stokes’ theorem, since itsintegral does not vanish).

For the inductive step we consider a manifold M with a good cover Vjk+1j=1 and write it as

M = A ∪B where:

A = V1 ∪ · · · ∪ Vk and B = Vk+1.

We now “pair together” the long exact Mayer-Vietoris sequence for A,B and the dual of theMayer-Vietoris sequence with compact support for A,B from Exercise 163:

Hq(A ∩B) Hq+1(A ∪B) Hq+1(A)⊕Hq+1(B)

(Hn−qc (A ∩B))

∗ (Hn−q−1c (A ∪B)

)∗ (Hn−q−1c (A)

)∗ ⊕ (Hn−q−1c (B)

)∗PA∩B PA∪B PA⊕PB

The bottom sequence (the dual of the sequence from Exercise 163) is exact since the dual of anexact sequence is still exact. These two long exact sequences give rise to diagram of maps whichare all commutative, up to a sign. The only tricky part is to verify commutativity of the square:

Hq(A ∩B) Hq+1(A ∪B)

(Hn−qc (A ∩B))

∗ (Hn−q−1c (A ∪B)

)∗d∗

PA∩B PA∪B

(d∗)∗

,

where the map d∗ is the connecting homomorphism of the Mayer-Vietoris long exact sequence and(d∗)∗ is the dual of the connecting homomorphism of the long exact Mayer-Vietoris sequence forcompact supports. Given [ω] ∈ Hq(A ∩B) we want to prove that the two linear maps

(d∗)∗PA∩B [ω] : Hn−q−1

c (A ∪B) −→ R and PA∪Bd∗[ω] : Hn−q−1

c (A ∪B) −→ R

are the same (up to a sign). In other words we need to prove that for every [η] ∈ Hn−q−1c (A ∪B)

we have:

(4.25) ((d∗)∗PA∩B [ω]) [η] =

∫[A∩B]

ω ∧ d∗η = ±∫

[A∪B]

d∗ω ∧ η = ± (PA∪Bd∗[ω]) [η].

Let us recall now the definition of the involved connecting homomorphisms. Let ρA, ρB be apartition of unity subordinated to A,B. A representative d∗ω of d∗[ω] is a form in Ωq+1(A∪B)such that (see Section 4.6):

(4.26) d∗ω|A = −d (ρBω) and d∗ω|B = d (ρAω) .

54 ANTONIO LERARIO

Since ω ∈ Ωq(A ∩ B), it follows from (4.26) that d∗ω has support in A ∩ B. A representative d∗ηof d∗[η] is a form in Ωn−qc (A ∩B) such that (see Exercise 163):

(4.27) (−d∗η, d∗η) = (d (ρAη) , d (ρBη)) .

Using this information, we can compute:∫[A∩B]

ω ∧ d∗η =(4.27)

∫[A∩B]

ω ∧ d (ρBη)

=dη=0

∫[A∩B]

ω ∧ dρB ∧ η

= (−1)degω

∫[A∩B]

dρB ∧ ω ∧ η

=dω=0

(−1)degω

∫[A∩B]

d (ρBω) ∧ η = (∗).

Similarly we have: ∫[A∪B]

d∗ω ∧ η =supp(d∗ω)⊂A∩B

∫[A∩B]

−d (ρBω) ∧ η

=suppdω=0

−∫

[A∩B]

dρBω ∧ η = ±(∗).

This proves (4.25) and we have established that the diagrams are commutative. Poincare dualityfollows from the five’s lemma: in fact in the next commutative diagram (this is just the previousdiagram where we have denoted by V the cohomology of A ∩ B and by W the direct sum of thecohomology of A and the one of B)

V q W q Hq(A ∪B) V q+1 W q+1

(V n−qc )∗

(Wn−qc )

∗(Hn−q

c (A ∪B))∗ (

V n−q−1c

)∗ (Wn−q−1c

)∗PA∩B PA⊕PB PA∪B PA∩B PA⊕PB

PA ⊕PB and PA∩B are isomorphisms (by the inductive hypothesis, because A, B and A∩B havea good cover of cardinality smaller than k), hence PA∪B is an isomorphism as well.

4.12. Kunneth theorem. Kunneth theorem allows to compute the cohomology of a product oftwo manifolds as a tensor product of the two cohomologies. Observe first that we have a diagram:

M ×N

M N

π1 π2

which allows to pull back forms on M and N to forms on M ×N .

Theorem 176 (Kunneth). Let M and N be two smooth manifolds. There is an isomorphism

φ : H∗(M)⊗H∗(N) −→ H∗(M ×N)

which is given by:

(4.28) [α]⊗ [β] 7→ [π∗1α ∧ π∗2β].

In particular for every q ≥ 0 we have:

(4.29) Hq(M ×N) =⊕p≥0

Hp(M)⊗Hq−p(N)

(where we adopt the convention Hp = 0 for p < 0).


Proof. (sketch) We will prove the theorem under the additional assumption that M has a finitegood cover, again using the Mayer-Vietoris argument. Since the reader should at this point befamiliar with this technique we only sketch an outline, leaving the details to him/her.

The base of the induction is just Theorem 146: M has a good cover consisting of just one openset, hence M ' Rm and consequently:

H∗(Rm ×N) ' H∗(N) = H∗(Rm)⊗H∗(N) = R⊗H∗(N).

For the inductive step, as before, we consider a manifold M with a good cover Vjk+1j=1 and

write it as M = A ∪B where:

A = V1 ∪ · · · ∪ Vk and B = Vk+1.

Then we take the Mayer-Vietoris exact sequence of the pair A,B and, for every p we tensor itwith Hp(N), obtaining many exact sequences (tensoring an exact sequence with a vector spacekeeps exactness):

· · · Hq(A ∪B)⊗Hp(N) (Hq(A)⊕Hq(B))⊗Hp(N) · · ·We then take the direct sum over all p and combine all these sequences together into a singlesequence which, after relabeling the indices, looks like:

· · · ⊕qp=0 (Hp(A ∪B)⊗Hp−q(N))

⊕qp=0 ((Hp(A)⊕Hp(B))⊗Hp−q(N)) · · ·

This exact sequence is naturally paired with the long Mayer-Vietoris exact sequence of the opencover A×N,B ×N for M ×N :

· · · ⊕qp=0 (Hp(A ∪B)⊗Hq−p(N))

⊕qp=0 ((Hp(A)⊕Hq−p(B))⊗Hn−q(N))

· · · Hq((A ∪B)×N) Hq(A×N)⊕Hq(B ×N)

φA∪B φA⊕φB

where the vertical arrows are the corresponding Kunneth maps, as in (4.28). By the inductivehypothesis we know that all these maps – except φA∪B , are isomorphisms: if we prove that theprevious diagram is commutative, the theorem follows again by the five lemma.

The only tricky part is again the commutativity of the piece of the diagram which contains theconnecting homomorphisms:⊕q

p=0 (Hp(A ∩B)⊗Hp−q(N))⊕q+1

p=0

(Hp+1(A ∪B)⊗Hq−p(N)

)Hq((A ∩B)×N) Hq+1((A ∪B)×N)

d∗

φA∩B φA∪B

d∗

We leave to the reader the verification that also this piece of the diagram is commutative (oneshould take, in the definition of the bottom connecting homomorphism, the partition of unityρAπ1, ρBπ1...)

Example 177 (The cohomology of Sn×Sn). The cohomology of Sn×Sn can be computed usingKunneth Theorem. In fact, using (4.29), we have:

(4.30) Hq(Sn × Sn) 'n⊕p=0

Hq(Sn)⊗Hp−q(Sn)

and, using the fact that H∗(Sn) = H0(Sn)⊕Hn(Sn), the only non-zero cohomology in (4.30) is:

H0(Sn × Sn) ' H0(Sn)⊗H0(Sn)

Hn(Sn × Sn) '(H0(Sn)⊗Hn(Sn)

)⊕(Hn(Sn)⊗H0(Sn)

)H2n(Sn × Sn) ' Hn(Sn)⊗Hn(Sn).

Using the explicit formula (4.28) of the Kunneth isomorphism, if we denote by [ω] ∈ Hn(Sn) thegenerator from Example 174, we have:

H0(Sn × Sn) = 〈[f ≡ 1]〉, Hn(Sn × Sn) = 〈[π∗2ω], [π∗1ω]〉, H2n(Sn × Sn) = 〈[π∗1ω ∧ π∗2ω]〉.

56 ANTONIO LERARIO

Exercise 178. Using Kunneth formula, compute the cohomology of S1×· · ·×S1 (k many factors).

Exercise 179. Compute the cohomology of S1 × S1 minus a point.

4.13. Invariant forms. Let M be a smooth manifold and G be a finite group acting freely on it.Then the quotient space M/G can be endowed the structure of a smooth manifold in such a waythat the quotient map π : M → M/G is a normal covering space and a smooth submersion. TheDe Rham cohomology of M/G can be computed using the following theorem.

Theorem 180. Let π : M →M/G be the quotient under a free action of a finite group. Then themap π∗ : H∗(M/G)→ H∗(M) is injective with image the vector space:

H∗(M)G = [ω] ∈ H∗(M) such that g∗[ω] = [ω] for every g ∈ G.

Proof. We first observe a preliminary fact: if a form ω ∈ Ω∗(M) satisfies g∗ω for all g ∈ G (sucha form is called “invariant”), then it descends to a form ω ∈ Ω∗(M/G) such that π∗ω = ω (tryto make this rigorous using the definition of a form as a bunch of local sections gluing togethernicely...)

We start by proving injectivity of π∗. Let [α] ∈ H∗(M/G) such that π∗[α] = 0, i.e. such thatπ∗α = dη for some η ∈ Ω∗(M). Then we can consider the form:

η =1

|G|∑g∈G

g∗η,

which is readily seen to be invariant and descends to a form η on M/G. The form η satisfiesdα = η, hence [α] = 0.

Let us prove now that the image of π∗ equals H∗(M)G. One inclusion is clear, since g∗π∗[α] =π∗[α] (for every g ∈ G we have π g = π). For the other inclusion, let [ω] ∈ H∗(M) such thatg∗[ω] = [ω] for every g ∈ G. Then:

g∗ω = ω + dηg

for some form ηg. Consider the invariant form ω = 1|G|∑g g∗ω. Then there is a closed form ω on

M/G such that π∗ω = ω and [ω] = π∗[ω]. It remains to show that [ω] = [ω]. To this end observethat:

1

|G|∑g

g∗ω − ω =1

|G|∑g

(g∗ω − ω) =1

|G|∑g

dηg = d

1

|G|∑g∈G

ηg

,

and ω and ω are co-homologous.

Example 181 (The cohomology of RPn revised). Let G = idSn , α ' Z2 where α : Sn → Sn isthe antipodal map x 7→ −x. Then RPn = Sn/Z2 and

H∗(RPn) ' H∗(Sn)Z2 .

We know from Example 174 that Hn(Sn) is generated by the volume form [ω] and consequently:

H∗(Sn) = 〈[f ≡ 1], [ω]〉.On the other hand an explicit computation shows that:

α∗[f ≡ 1] = [f ≡ 1] and α∗[ω] = (−1)n+1[ω].

Thus, using Theorem 180, we see that the cohomology of RPn can be identified with:

H∗(RPn) '⟨

[f ≡ 1],

(1 + (−1)n+1

)2

[ω]

⟩Example 182 (The cohomology of G(2, 4)). Recall from Exercise 38 that G(2, 4) can be embeddedinto RP5 through a map p : G(2, 4)→ RP5 whose image is:

p(G(2, 4)) = p12p34 + p13p24 + p14p23 = 0(here [p12, p13, p14, p23, p24, p34] denote the homogeneous coordinates of RP5). The quadric definingp(G(2, 4)) has signature (3, 3) and putting it in normal form, we can find coordinates [y0, . . . , y5]in RP5 such that:

G(2, 4) ' q(y) = y20 + y2

1 + y22 − y2

3 − y24 − y2

5 = 0.


Let us look at the manifold defined by the same equation in S5:

M = (y0, . . . , y5) ∈ S5 | q(y) = y20 + y2

1 + y22 − y2

3 − y24 − y2

5 = 0.It is easy to see that M is a smooth submanifold of S5 (because the equation is regular) whichdouble covers G(2, 4). The antipodal map α : S5 → S5 restricts to a map τ : M → M whosequotient is G(2, 4):

π : M → G(2, 4) is a normal covering space.

It follows from Theorem 180 that:

H∗(G(2, 4)) ' H∗(M)Z2

where Z2 = idM , τ. The manifold M is given in R6 by the equations ‖y‖2 = 2 (we take the sphere

of radius√

2 to make the notation simpler) and q(y) = 0. Using coordinates (w, z) ∈ R3 × R3,these two equations are equivalent to:

M =‖w‖2 = 1, ‖z‖2 = 1

,

hence M ' S2×S2 ⊂ R3×R3. The cohomology of M is given by Kunneth Theorem: denoting byπ1, π2 : S2 × S2 → S2 the two projections and by [ω] ∈ H2(S2) the generator from Example 174,

H∗(M) = H∗(S2 × S2) = 〈[f ≡ 1], [π∗1ω], [π∗2ω], [π∗1ω ∧ π∗2ω]〉.The action of Z2 on M is given by:

τ(z, w) = (−z,−w),

and consequently:

τ∗[f ≡ 1] = [f ≡ 1], τ∗[π∗1ω] = −[π∗1ω], τ∗[π∗2ω] = −[π∗2ω], τ∗[π∗1ω ∧ π∗2ω] = [π∗1ω ∧ π∗2ω].

From this, using Theorem 180, we see that the cohomology of G(2, 4) is isomorphic to:

H∗(G(2, 4)) ' 〈[f ≡ 1], [π∗1ω ∧ π∗2ω]〉.(Notice in particular that G(2, 4) is orientable.)

Exercise 183 (The cohomology of a projective quadric). Generalizing the previous example, letq : Rn+1 → R be a nondegenerate quadratic form of signature (a, b), a+ b = n+ 1. Compute thecohomology of q = 0 ⊂ RPn and q = 0 ⊂ Sn.Exercise 184. The Klein bottle K can be seen as (S1 × S1)/Z2, where Z2 = idS1 , τ andτ(z, w) = (1/z,−w). Compute the cohomology of K. Is K orientable?

Exercise 185. Prove that a compact and connected manifold M of dimension n is orientable ifand only if Hn(M) ' R and that M is not orientable if and only of Hn(M) = 0. (Hint: use theorientation double cover, as defined in [6, pag 234]...)

58 ANTONIO LERARIO

5. Appendix

5.1. Manifolds with boundary. In this section we introduce the notion of a manifold withboundary. To start with, define a half-space H ⊂ Rn as:

H = x ∈ Rn |λ(x) ≥ 0where λ : Rn → R is a linear map. The half-space is called proper if λ 6= 0, in which case theboundary of H is the set:

∂H = x ∈ H |λ(x) = 0.The definition of a manifold chart can now be extended to be a map ϕ : U → Rn which is ahomeomorphism to an open subset of a half-space.

Definition 186 (Manifold with boundary). A manifold with boundary, of dimension m and classCk is a paracompact Hausdorff space M such that:

(1) for every point x ∈ M there exists a neighborhood U of x and a continuous functionψ : U → H which is a homeomorphism onto an open subset of a half-space H (the pair(U,ψ) is still called a chart);

(2) for every pairs of charts (U1, ψ1) and (U2, ψ2) such that U1 ∩ U2 6= ∅ the map

ψ2 ψ1|−1U1∩U2

: ψ1(U1 ∩ U2)→ Rm

is a Ck map16.

By the Invariance of Domain Theorem [6, Theorem 2B.3], a coordinate change cannot map aninterior point of a half-space into a boundary point; consequently the set of boundary points of M(points which in some chart are mapped the boundary of a half-space), is well defined and calledthe boundary of M .

5.2. The inverse and the implicit function theorem. These are basic tools in DifferentialTopology, but we will not prove them here. There are many excellent references for these theorems,for instance a complete discussion can be found online at [4].

Theorem 187 (Inverse function theorem). Let U ⊆ Rn be an open set and f : U → Rn be adifferentiable function of class Ck, k ≥ 1. If the Jacobian matrix Jf(x) is nonsingular at x0 thereexists a neighborhood A of x0 such that f(A) is open and f |A : A → f(A) ⊆ Rn is invertible;moreover the inverse function g : f(A)→ A is also Ck.

In the above statement, note that the conclusion “f(A) is open” follows from the Invarianceof Domain Theorem; moreover, using the language of Definition 28 below, the function f |A is adiffeomorphism.

Theorem 188 (Implicit function theorem). Let U ⊆ Ra+b be an open set with coordinates (x, y) ∈Ra × Rb and f : U → Rb be a differentiable function of class Ck, k ≥ 1. If the matrix ∂f

∂y (x, y)

(consisting of the partial derivatives of f with respect to the y variables) is nonsingular at (x0, y0),there exist neighborhoods A of x0 and B of y0 and a function ϕ : A→ B of class Ck such that:

f = f(x0, y0) ∩ (A×B) = graph(ϕ).

5.3. Bump functions.

Definition 189 (Bumb function). Let 0 < c1 < c2 be positive distinct numbers. A bumb functionis a C∞ function λ : Rn → [0, 1] such that (see Figure 8):

(1) λ(x) = 1 for ‖x‖ ≤ c1;(2) 0 < λ(x) < 1 for c1 < ‖x‖ < c2;(3) λ(x) = 0 for ‖x‖ ≥ c2.

Lemma 190. Bumb functions exist for every 0 < c1 < c2.

16A map f : W1 → W2 between open sets in half-spaces W1 ⊆ H1 and W2 ⊆ H2 is called Ck if it extends to a

Ck map on a open neighborhood W1 ⊂ W1 ⊂ Rn.


b b b b

−c2 −c1 c1 c2

Figure 8. A bump function

Proof. Let α : R→ R be the C∞ function

α(x) = e−1xχx>0(x),

where χx>0(x) denotes the characteristic function of x > 0. Out of this function we constructγ : R→ R by:

γ(x) =

∫ c2xα(t− c1)α(c2 − t)dt∫ c2

c1α(t− c1)α(c2 − t)dt

.

Finally we set λ(x) = γ(‖x‖). (The reader can see [7, Section 2.2] for pictures explaining theconstruction step by step.)

5.4. Partitions of unity.

Definition 191 (Partition of unity). Let M by a smooth manifold and U = Ujj∈J be an opencover of M . A partition of unity subordinated to U is a collection of smooth maps ρj : M →[0, 1]j ∈ J such that:

(1) for every j ∈ J the support supp(ρj) ⊂ Uj ;(2) the family supp(ρj)j∈J is locally finite;(3)

∑j∈J ρj ≡ 1 (the sum is well defined because of the previous condition).

In order to prove the existence of partitions of unity, we will need the following Lemma.

Lemma 192. Let U = Ujj∈J be an open cover for M . Then there is a locally finite atlas

(Vα, ψα)α∈A such that V αα∈A refines U, each ψα(Vα) ⊂ Rm is bounded and each V α ⊂ M iscompact. Moreover the open cover V = Vαα∈A has a shrinking W = Wαα∈A, i.e. W is itselfa cover and for every α ∈ A we have Wα ⊂ Vα.

Proof. The proof is left as an exercise. Work out the details carefully!

Proposition 193. Let M be a smooth manifold. Every open cover U of M admits a subordinatepartition of unity.

Proof. Let U = Ujj∈J be the open cover. We observe first that if we can construct a partitionof unity λii∈I subordinated to a refinement V = Vii∈I of U, then we can also construct apartition of unity subordinated to U. In fact let g : I → J be such that for every i ∈ I we havethe inclusion Vi ⊂ Ug(i) (the existence of such g is in the definition of refinement) and define forj ∈ J the function:

ρj(x) =∑

i∈g−1(j)

λi(x).

It is easy to check that ρjj∈J is a partition of unity subordinated to U (essentially to obtain theρj we collate together some of the λi).

Let now V be the open cover produced by Lemma 192 and W its shrinking. For every α ∈ Awe cover the compact set ψα(Wα) ⊂ Rm by finitely many closed balls contained in ψα(Vα):

ψα(Wα) ⊂kα⋃k=1

Bα,k,

where Bα,k = B(xα,k, rα,k) for some points xα,k ∈ Rm and radii rα,k > 0. For every ball Bα,klet λα,k : Rm → [0, 1] be a bump function centered at xα,k such that λα,k(x) > 0 if and only if

60 ANTONIO LERARIO

x ∈ int(Bα,k) (it suffices to take a bumb function λ, as in Lemma 190, with c1 < rα,k and c2 = rα,xand then translate it at xα,k, i.e. λα,k(x) = λ(x − xα,k)). Put now λα =

∑kαk=1 λα,k and define

µα : M → [0,∞) by:

µα(x) = χVα(x) · λα(ψα(x)).

Each µα is a smooth function, strictly positive on Wα and with supp(µα) ⊂ Vα. We define thedesired partition of unity ραα∈A subordinated to V by:

ρα(x) =µα(x)∑β∈A µβ(x)

.

Exercise 194 (Smooth Urysohn’s Lemma). Let A and B be two closed and disjoint submanifoldsof a manifold M . Prove that there is a smooth function f : M → [0, 1] such that f |A ≡ 0 andf |B ≡ 1.

5.5. A first approximation result. Partitions of unity are used to glue together locally definedobject on a manifold (for example vector fields, Riemannian metrics...). Here is an importantapplication of the existence of partitions of unity: we can approximate continuous functions f :M → Rn with smooth functions (we refine this result in Theorem 129).

Theorem 195. Let M be a compact, smooth manifold and f : M → Rn be a continuous function.Then, for every ε > 0 there is a smooth function fε : M → Rn such that:

maxx∈M

‖f(x)− fε(x)‖ ≤ ε.

Proof. Let ε > 0 and for every z ∈M let Uz be an obpen neighborhood of z such that

(5.1) ‖f(z)− f(y)‖ ≤ ε for all y ∈ Uz.consider the open cover Uzz∈M and let ρzz∈M be a partition of unity subordinated to thiscover. Define the smooth function:

fε(x) =∑z∈M

ρz(x)f(z).

Let us evaluate

‖f(x)− fε(x)‖ =

∥∥∥∥∥∑z∈M

ρz(x)f(z)−∑z∈M

ρz(x)f(x)

∥∥∥∥∥=

∥∥∥∥∥∑z∈M

ρz(x)(f(z)− f(x))

∥∥∥∥∥≤∑z∈M

ρz(x)‖f(z)− f(x)‖ = (∗) (since ρz ≥ 0).

Now, either (1) x ∈ supp(ρz) ⊂ Uz, in which case ρz(x)‖f(z) − f(x)‖ ≤ ρz(x)ε by (5.1) or (2)ρz(x) = 0, in which case also ρz(x)‖f(z)− f(x)‖ = 0 ≤ ρz(x)ε. Hence

(∗) ≤∑z∈M

ρz(x)ε = ε.

This concludes the proof.

5.6. Some homological algebra. We recall here some basic facts on homological algebra. Adifferential complex is a sequence of vector spaces with a differential d such that d2 = 0:

· · · Cq−1 Cq Cq+1 · · ·d

d2=0

d

The condition that d2 = 0 implies that we can define the cohomology of the complex by

Hq(C) =ker(d|Cq )

im(d|Cq−1).


Given two differential complexes (A, dA) and (B, dB), a chain map is a linear map

f : A∗ −→ B∗ such that fdA = dBf.

The condition fdA = dBf implies that there is a well defined map:

f∗ : H∗(A) −→ H∗(B)

given by f∗[α] = [f(α)]. An exact sequence of differential complexes

0 A B C 0f g

is called a short exact sequence.

Proposition 196. A short exact sequence of differential complexes:

0 A B C 0f g

in which f and g are chain maps, induces a long exact sequence between the cohomologies of thesecomplexes:

· · · Hq(A) Hq(B) Hq(C) Hq+1(A) · · ·f∗ g∗ d∗

Proof. The interesting part of the proof is the construction of the map d∗. This is done as follows.Pick [c] ∈ Hq(C), with c ∈ Cq. such that dCc = 0 and consider the commutative diagram:

......

...

0 Aq+1 Bq+1 Cc+1 0

0 Aq Bq Cq 0

......

...

f g

f

dA

g

dB dC.

Since g is surjective, there exists b ∈ Bq such that g(b) = c. Consider db. By commutativity ofthe diagram, 0 = dg(b) = g(db) and consequently db is in the kernel of g and, by exactness, thereexists a ∈ Aq+1 such that f(a) = db. We set:

d∗[c] = [a].

The rest of the proof is left as an exercise to the reader.

Exercise 197. Consider the exact sequence of finite dimensional vector spaces:

0 V1 V2 · · · Vm−1 Vm 0

Prove that∑mk=1(−1)k dim(Vk) = 0.

References

[1] V. I. Arnold. Mathematical methods of classical mechanics, volume 60 of Graduate Texts in Mathematics.

Springer-Verlag, New York, [1989?]. Translated from the 1974 Russian original by K. Vogtmann and A. Weinstein,Corrected reprint of the second (1989) edition.

[2] M. Artin. Algebra. Prentice Hall, Inc., Englewood Cliffs, NJ, 1991.[3] R. Bott and L. W. Tu. Differential forms in algebraic topology, volume 82 of Graduate Texts in Mathematics.

Springer-Verlag, New York-Berlin, 1982.

[4] M. Cornalba. Note di geometria differeziale. http://mate.unipv.it/cornalba/dispense/geodiff.pdf.

[5] V. Guillemin and A. Pollack. Differential topology. AMS Chelsea Publishing, Providence, RI, 2010. Reprint ofthe 1974 original.

[6] A. Hatcher. Algebraic topology. Cambridge University Press, Cambridge, 2002.[7] M. W. Hirsch. Differential topology, volume 33 of Graduate Texts in Mathematics. Springer-Verlag, New York,

1994. Corrected reprint of the 1976 original.

[8] J. M. Lee. Introduction to smooth manifolds, volume 218 of Graduate Texts in Mathematics. Springer, NewYork, second edition, 2013.

[9] J. W. Milnor. Topology from the differentiable viewpoint. Princeton Landmarks in Mathematics. Princeton

University Press, Princeton, NJ, 1997. Based on notes by David W. Weaver, Revised reprint of the 1965 original.

http://mate.unipv.it/cornalba/dispense/geodiff.pdf

62 ANTONIO LERARIO

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E-mail address: [email protected]

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