Notes-2012 Manifold w Excer

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INTRODUCTION TO

DIFFERENTIABLE

MANIFOLDS

spring 2012

Course by Prof. dr. R.C.A.M. van der Vorst

Solution manual Dr. G.J. Ridderbos

1


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Introduction to differentiable manifolds

Lecture notes version 2.1, November 5, 2012

This is a self contained set of lecture notes. The notes were written by Rob van

der Vorst. The solution manual is written by Guit-Jan Ridderbos. We follow the

book Introduction to Smooth Manifolds by John M. Lee as a reference text [1].

Additional reading and exercises are take from An introduction to manifolds by

Loring W. Tu [2].

This document was produced in LATEX and the pdf-file of these notes is available

on the following website

www.few.vu.nl/vdvorst

They are meant to be freely available for non-commercial use, in the sense that

free software is free. More precisely:

This work is licensed under the Creative Commons

Attribution-NonCommercial-ShareAlike License.

To view a copy of this license, visit

http://creativecommons.org/licenses/by-nc-sa/2.0/

or send a letter to Creative Commons, 559 Nathan Abbott Way, Stanford, California

94305, USA.

CONTENTS

References 5

I. Manifolds 6

1. Topological manifolds 6

2. Differentiable manifolds and differentiable structures 13

3. Immersions, submersions and embeddings 20

II. Tangent and cotangent spaces 32

4. Tangent spaces 32

5. Cotangent spaces 38

6. Vector bundles 41

6.1. The tangent bundle and vector fields 44

6.2. The cotangent bundle and differential 1-forms 46

III. Tensors and differential forms 50

7. Tensors and tensor products 50

8. Symmetric and alternating tensors 55


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9. Tensor bundles and tensor fields 61

10. Differential forms 66

11. Orientations 69

IV. Integration on manifolds 75

12. Integrating m-forms on Rm 75

13. Partitions of unity 76

14. Integration on of m-forms on m-dimensional manifolds. 81

15. The exterior derivative 87

16. Stokes Theorem 92

V. De Rham cohomology 97

17. Definition of De Rham cohomology 97

18. Homotopy invariance of cohomology 98

VI. Exercises 102

Manifolds

Topological Manifolds 102

Differentiable manifolds 103

Immersion, submersion and embeddings 104

Tangent and cotangent spaces

Tangent spaces 106

Cotangent spaces 107

Vector bundles 108

Tensors

Tensors and tensor products 109

Symmetric and alternating tensors 109

Tensor bundles and tensor fields 110

Differential forms 111

Orientations 111

Integration on manifolds

Integrating m-forms on Rm 112

Partitions of unity 112

Integration of m-forms on m-dimensional manifolds 112

The exterior derivative 113

Stokes Theorem 114


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Extras 115

De Rham cohomology

Definition of De Rham cohomology 117

VII. Solutions 118


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References

[1] John M. Lee.Introduction to smooth manifolds, volume 218 ofGraduate Texts in Mathematics.

Springer-Verlag, New York, 2003.

[2] Loring W. Tu.An introduction to manifolds. Universitext. Springer, New York, second edition,

2011.


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I. Manifolds

1. Topological manifolds

Basically an m-dimensional (topological) manifold is a topological space M

which is locally homeomorphic to Rm. A more precise definition is:

Definition 1.1. 1 A topological spaceMis called anm-dimensional (topological)

manifold, if the following conditions hold:

(i) Mis a Hausdorff space,

(ii) for anyp Mthere exists a neighborhood2 Uofp which is homeomorphicto an open subset VRm, and

(iii) Mhas a countable basis of open sets.

Axiom (ii) is equivalent to saying that p Mhas a open neighborhood U3 phomeomorphic to the open discDm in Rm. We say Mis locally homeomorphicto

Rm. Axiom (iii) says thatMcan be covered by countably many of such neighbor-

hoods.

FIGURE 1. Coordinate charts(U,).

1See Lee, pg. 3.2Usually an open neighborhood Uof a point p Mis an open set containing p. A neighborhood

of p is a set containing an open neighborhood containing p. Here we will always assume that a

neighborhood is an open set.


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FIGURE 2. The transition mapsi j.

Recall some notions from topology: A topological spaceMis calledHausdorff

if for any pair p,q M, there exist (open) neighborhoodsU3p, andU03 qsuchthatUU0= . For a topological spaceMwith topology, a collection isa basis if and only if each U can be written as union of sets in . A basis iscalled acountable basisif is a countable set.

Figure 1 displays coordinate charts(U,), whereU are coordinate neighbor-

hoods, or charts, and are (coordinate) homeomorphisms. Transitions between

different choices of coordinates are called transitions mapsi j= j 1i , whichare again homeomorphisms by definition. We usually write x= (p), : UVRn, as coordinatesfor U, see Figure 2, and p = 1(x), 1 : VUM, asa parametrization ofU. A collection A ={(i,Ui)}iIof coordinate charts withM= iUi, is called anatlasfor M.

The following Theorem gives a number of useful characteristics of topological

manifolds.

Theorem 1.4.

3

A manifold is locally connected, locally compact, and the union ofcountably many compact subsets. Moreover, a manifold is normal and metrizable.

J 1.5Example. M= Rm; the axioms (i) and (iii) are obviously satisfied. As for (ii)

we take U= Rm, andthe identity map. I

3Lee, 1.6, 1.8, and Boothby.


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J 1.6Example. M=S1 ={p= (p1,p2) R2 | p21+p22=1}; as before (i) and (iii)are satisfied. As for (ii) we can choose many different atlases.

(a): Consider the sets

U1={p S1 | p2>0}, U2={p S1 | p2


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clockwise rotation), and2(p) = (,) (complex argument). For example11 () = (cos(),sin()). I

J 1.8Example. M=Sn ={p= (p1, pn+1) Rn+1 | |p|2 =1}. Obvious exten-sion of the choices forS1. I

J 1.9Example.(see Lee) Let URn be an open set andg: URm a continuousfunction. Define

M= (g) ={(x,y) RnRm : x U, y=g(x)},endowed with the subspace topology, see 1.17. This topological space is an n-

dimensional manifold. Indeed, define : Rn Rm Rn as (x,y) = x, and= |(g), which is continuous onto U. The inverse

1(x) = (x,g(x)) is also

continuous. Therefore{((g),)} is an appropriate atlas. The manifold(g) ishomeomorphic to U. I

J 1.10 Example. M=PRn, the real projective spaces. Consider the following

equivalence relation on points on Rn+1\{0}: For any twox,y Rn+1\{0}definex yif there exists a 6=0, such thatx= y.

Define PRn ={[x] : x Rn+1\{0}} as the set of equivalence classes. One canthink ofPRn as the set of lines through the origin in Rn+1. Consider the natural

map :Rn+1\{0} PRn, via (x) = [x]. A set UPRn is open if1(U)is openin Rn+1\{0}. This makes a quotient map and is continuous. The equivalence

relation is open (see [2, Sect. 7.6]) and therefore PRn Hausdorff and secondcountable, see 1.17. Compactness of PRn can be proves by showing that there

exists a homeomorphism from PRn to Sn/, see [2, Sect. 7.6]. In order to verifythat we are dealing with an n-dimensional manifold we need to describe an atlas

forPRn. Fori=1, n + 1, defineVi Rn+1\{0}as the set of points x for whichxi6=0, and define Ui= (Vi). Furthermore, for any[x] Ui define

i([x]) =x1

xi, ,

xi1xi

,xi+1

xi, ,

xn+1

xi

,

which is continuous. This follows from the fact that is continuous. Thecontinuous inverse is given by

1i (z1, ,zn) =(z1, ,zi1,1,zi, ,zn).

These chartsUi coverPRn. In dimension n =1 we have that PR=S1, and in the

dimensionn=2 we obtain an immersed surfacePR2 as shown in Figure 4. I

The examples 1.6 and 1.8 above are special in the sense that they are subsets of

some Rm, which, as topological spaces, are given a manifold structure.


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FIGURE4. Identification of the curves indicated above yields an

immersion ofPR2 into R3.

Define

Hm ={(x1, ,xm)| xm 0},as the standard Euclidean half-space.

Definition 1.12. A topological spaceMis called an m-dimensional (topological)

manifold with boundaryMM, if the following conditions hold:(i) Mis a Hausdorff space,

(ii) for any pointp Mthere exists a neighborhoodU ofp, which is homeo-morphic to an open subset VHm, and

(iii) Mhas a countable basis of open sets.

Axiom (ii) can be rephrased as follows, any point p Mis contained in a neigh-borhood U, which either homeomorphic toDm, or toDmHm. The setMis locallyhomeomorphic to Rm, or Hm. TheboundaryMMis a subset ofMwhich con-sists of points pfor which any neighborhood cannot be homeomorphic to an open

subset of int (Hm). In other words point p Mare points that lie in the inverseimage ofV Hm for some chart(U,). Points p Mare mapped to points onHm andMis an(m1)-dimensional topological manifold.J 1.14Example.Consider the bounded cone

M= C={p= (p1,p2,p3) R3 | p21+p22= p23, 0 p3 1},

with boundary C= {pC | p3 = 1}. We can describe the cone via an atlasconsisting of three charts;

U1={p C| p3


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FIGURE5. Coordinate maps for boundary points.

The other charts are given by

U2={p C| 12< p3 1, (p1,p2) 6= (0,p3)},

U3={p C| 12< p3 1, (p1,p2) 6= (0,p3)}.

For instance2can be constructed as follows. Define

q= (p) =

p1

p3,p2

p3,p3

, (q) =

2q1

1

q2

,1q3

,

and x= 2(p) = ( )(p), 2(U2) = R [0, 12 ) H2. The map3 is definedsimilarly. The boundary is given by C=12 (R{0})13 (R{0}), see Figure6. I

J 1.16Example. The open cone M=C={p= (p1,p2,p3)| p21+p

22= p

23, 0

p3


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FIGURE6. Coordinate maps for the cone C.

to be homeomorphic. Using charts(U,), and(V,) for N and M respectively,

we can give a coordinate expression for f, i.e. f= f1.A(topological) embeddingis a continuous injective mapping f :XY, which

is a homeomorphism onto its image f(X) Y with respect to the subspace topol-ogy. Let f :NMbe an embedding, then its image f(N)Mis called asubman-ifoldofM. Notice that an open submanifold is the special case when f= i : UMis an inclusion mapping.J 1.17Facts.Recall the subspace topology. LetXbe a topological space and let

SXbe any subset, then thesubspace, or relative topologyon S(induced by thetopology onX) is defined as follows. A subset USis open if there exists an opensetVXsuch thatU= VS. In this case Sis called a (topological)subspaceofX.

For a subjective map : X Y we topologize Y via: U Y is open if andonly if1(U)is open in X. In this case is continuous and is called a quotient

map. For a given topological space Z the map f :Y Z is continuous if andonly if f

: X

Z is continuous. For an equivalence relation

, the mapping

:XX/ is a quotient map, see [1].If is a open equivalence relation, i.e. the image of an open set under :X

X/ is open, thenX/ is Hausdorff if and only if the set{(x,x0)XX| x x0}is closed in XX. IfXis second countable (countable basis of open sets), thenalsoX/ is second countable, see [2]. I


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2. Differentiable manifolds and differentiable structures

A topological manifoldMfor which the transition maps i j= j1i for all

pairsi,j in the atlas are diffeomorphisms is called a differentiable, or smooth

manifold. The transition maps are mappings between open subsets ofRm. Dif-

feomorphisms between open subsets ofRm areC-maps, whose inverses are also

C-maps. For two charts Ui and Uj the transitions maps are mappings:

i j= j 1i : i(UiUj) j(UiUj),and as such are homeomorphisms between these open subsets ofRm.

Definition 2.1. A C-atlas is a set of charts A={(U,i)}iIsuch that

(i) M= iIUi,(ii) the transition maps i j are diffeomorphisms between i(UiUj) and

j(UiUj), for alli 6= j(see Figure 2 ).

The charts in a C-atlas are said to be C-compatible. Two C-atlases A and A0

are equivalent ifAA0is again a C-atlas, which defines a equivalence relation onC-atlases. An equivalence class of this equivalence relation is called a differen-

tiable structure D on M. The collection of all atlases associated with D, denoted

AD, is called the maximal atlas for the differentiable structure. Figure 7 shows

why compatibility of atlases defines an equivalence relation.

Definition 2.3. Let Mbe a topological manifold, and let D be a differentiable

structure on Mwith maximal atlas AD. Then the pair(M,AD) is called a (C-

)differentiable manifold.

Basically, a manifold M with a C-atlas, defines a differentiable structure on

M. The notion of a differentiable manifold is intuitively a difficult concept. In

dimensions 1 through 3 all topological manifolds allow a differentiable structure

(only one up to diffeomorphisms). Dimension 4 is the first occurrence of manifolds

without a differentiable structure. Also in higher dimensions uniqueness of differ-

entiable structures is no longer the case as the famous example by Milnor shows;

S7 has 28 different (non-diffeomorphic) differentiable structures. The first example

of a manifold that allows many non-diffeomorphic differentiable structures occursin dimension 4; exotic R4s. One can also consider Cr-differentiable structures and

manifolds. Smoothness will be used here for the C-case.

Theorem 2.4. 4 Let M be a topological manifold with a C-atlas A. Then there

exists a unique differentiable structure D containing A, i.e. A AD.4Lee, Lemma 1.10.


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FIGURE 7. Differentiability of00 1 is achieved via the map-pings00 (0)1, and 0 1, which are diffeomorphisms sinceA A0, and A0 A00 by assumption. This establishes the equiva-lence A A00.

Proof:Let A be the collection of charts that are C-compatible with A. By the

same reasoning as in Figure 7 we prove that all charts in A are C-compatible with

eachother proving thatA

is a smooth altas. Now any chart that isC

-compatiblewith any chart in A is C-compatible with any chart in A and is thus in A and A is

a maximal atlas. Clearly any other maximal altas is contained in Aand therefore

A= AD.

J 2.5 Remark. In our definition of of n-dimensional differentiable manifold we

use the local model overstandardRn, i.e. on the level of coordinates we express

differentiability with respect to the standard differentiable structure on Rn, or dif-

feomorphic to the standard differentiable structure on Rn. I

J 2.6Example.The coneM= CR3 as in the previous section is a differentiablemanifold whose differentiable structure is generated by the one-chart atlas (C,)as described in the previous section. As we will see later on the cone is not a

smoothly embedded submanifold. I

J 2.7 Example. M= S1 (or more generally M= Sn). As in Section 1 consider

S1 with two charts via stereographic projection. We have the overlapU1 U2=S1\{(0,1)}, and the transition map 12=2 11 given by y= 12(x) = 4x ,


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x 6=0,, andy 6=0,. Clearly,12and its inverse are differentiable functionsfrom1(U1U2) = R\{0}to2(U1U2) = R\{0}. I

FIGURE8. The transition maps for the stereographic projection.

J 2.9Example.The real projective spaces PRn, see exercises Chapter VI. I

J 2.10Example.The generalizations of projective spacesPRn, the so-called(k,n)-

GrassmanniansGkRn are examples of smooth manifolds. I

Theorem 2.11. 5 Given a set M, a collection {U}A of subsets, and injectivemappings: URm, such that the following conditions are satisfied:

(i) (U) Rm is open for all;(ii) (UU)and(UU)are open in Rm for any pair, A;

(iii) for UU 6=, the mappings 1

: (UU) (UU)arediffeomorphisms for any pair, A;

(iv) countably many sets U cover M.

(v) for any pair p 6=q M, either p,q U, or there are disjoint sets U,Usuch that p Uand q U.

Then M has a unique differentiable manifold structure and(U,) are smooth

charts.

Proof:Let us give a sketch of the proof. Let sets1 (V),V Rn open, form abasis for a topology on M. Indeed, ifp 1 (V)1 (W)then the latter is againof the same form by (ii) and (iii). Combining this with (i) and (iv) we establish

a topological manifold over Rm. Finally from (iii) we derive that{(U,)} is a

smooth atlas.

Let N and Mbe smooth manifolds (dimensions n and m respectively). Let f :

NMbe a mapping from Nto M.

5See Lee, Lemma 1.23.


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Definition 2.12. A mapping f :N Mis said to be C, or smooth if for everyp N there exist charts(U,) of pand (V,)of f(p), with f(U) V, such that

f= f1 : (U) (V)is a C-mapping (from Rn to Rm).

FIGURE9. Coordinate representation for f, with f(U) V.The above definition also holds true for mappings defined on open subsets of

N, i.e. letW

Nis an open subset, and f : W

N

M, then smoothness onW

is defined as above by restricting to pionts p W. With this definition coordinatemaps : URm are smooth maps.

The definition of smooth mappings allows one to introduce the notion of dif-

ferentiable homeomorphism, ofdiffeomorphismbetween manifolds. LetNand M

be smooth manifolds. A C-mapping f :NM, is called a diffeomorphism ifit is a homeomorphism and also f1 is a smooth mapping, in which case N andMare said to be diffeomorphic. The associated differentiable structures are also

called diffeomorphic. Diffeomorphic manifolds define an equivalence relation. In

the definition of diffeomorphism is suffices require that f is a differentiable bijec-

tive mapping with smooth inverse (see Theorem 2.15). A mapping f :N

M is

called a local diffeomorphism if for every p Nthere exists a neighborhoodU,with f(U) open in M, such that f :U f(U) is a diffeomorphism. A mappingf :NMis a diffeomorphism if and only if it is a bijective local diffeomorphism.J 2.14Example. Consider N= Rwith atlas (R,(p) = p), and M= R with at-

las (R,(q) =q3). Clearly these define different differentiable structures (non-

compatible charts). Between N and M we consider the mapping f(p) = p1/3,


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which is a homeomorphism between Nand M. The claim is that fis also a diffeo-

morphism. Take U= V= R, then f1(p) = (p1/3)3 = p is the identity andthus C on R, and the same for f1 1(q) = (q3)1/3 = q. The associated dif-ferentiable structures are diffeomorphic. In fact the above described differentiable

structures correspond to defining the differential quotient via limh0f3(p+h)f3(p)

h .

I

Theorem 2.15. 6 Let N,M be smooth manifolds with atlases AN andAMrespec-

tively. The following properties hold:

(i) Given smooth maps f : U M, for all U AN, with f|UU =f|UU for all,. Then there exists a unique smooth map f :NMsuch that f|U= f.

(ii) A smooth map f :NM between smooth manifolds is continuous.(iii) Let f: NM be continuous, and suppose that the maps f= f1 ,

for charts(U,)AN, and(V,)AM, are smooth on their domainsfor all,. Then f is smooth.

Proof:Define f by f|U = f, which is well-defined by the overlap conditioins.

Given p M, there exists a chart U 3p and f= fwhich is smooth by definition,and thus fis smooth.

For any p U(chart) and choose f(p) V (chart), then fis a smooth map andf|U=

1 f : UV is continuous. Continuity holds for each neighborhood

ofp M.Let p U and f(p) V and set U= f1(V)UU which is open by

continuity of f. Now f= f on the charts (U,|U) and (V,) which proves

the differentiability of f.

With this theorem at hand we can verify differentiability locally. In order to

show that fis a diffeomorphism we need that fis a homeomorphism, or a bijection

that satisfies the above local smoothness conditions, as well as for the inverse.

J 2.17 Example. Let N=M= PR1 and points in PR1 are equivalence classes

[x] = [(x1,x2)]. Define the mapping f :PR1 PR1 as[(x1,x2)] 7 [(x21,x22)]. Con-

sider the charts U1 = {x16=0}, and U2 ={x2

6=0}, with for example 1(x) =

tan1(x2/x1), then 1(1) = [cos(1),sin(1)], with 1 V1= (/2,/2). Inlocal coordinates on V1we have

ef(1) =tan1sin2(1)/cos2(1),6See Lee Lemmas 2.1, 2.2 and 2.3.


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FIGURE 10. Coordinate diffeomorphisms and their local repre-

sentation as smooth transition maps.

and a similar expression forV2. Clearly, f is continuous and the local expressions

also prove that fis a differentiable mapping using Theorem 2.15. I

The coordinate maps in a chart (U,) are diffeomorphisms. Indeed, : UMRn, then if(V,)is any other chart with UV6=, then by the definition thetransition maps1 and 1 are smooth. For we then have = 1and 1 = 1, which proves that is a diffeomorphism, see Figure 10. This

justifies the terminologysmooth charts.J 2.18 Remark. For arbitrary subsets U Rm and V Rn a map f :U Rm V Rn is said to be smooth at pU, if there exists an open set U Rmcontaining p, and a smooth map f :U Rm, such that f and f coincide onUU. The latter is called an extension of f. A mapping f :U V betweenarbitrary subsetsU Rm and V Rn is called a diffeomorphism if f maps Uhomeomorphically onto V and both f and f1 are is smooth (as just describedabove). In this case Uand Vare said to be diffeomorphic. I

J 2.19Example. An important example of a class of differentiable manifolds are

appropriately chosen subsets of Euclidean space that can be given a smooth mani-

fold structure. Let M R` be a subset such that every p Mhas a neighborhoodU3pin M(open in the subspace topology, see Section 3) which is diffeomorphicto an open subset VRm (or, equivalently an open discDm Rm). In this case thesetMis a smoothm-dimensional manifold. Its topology is the subsapce topology

and the smooth structure is inherited from the standard smooth structure on R`,

which can be described as follows. By definition a coordinate map is a diffeo-


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FIGURE 11. The transitions maps1 and 1 smoothmappings establishing the smooth structure onM.

morphisms which means that : UV= (U)is a smooth mapping (smoothnessvia a smooth map), and for which also1 is a smooth map. This then directlyimplies that the transition maps 1 and 1 are smooth mappings, seeFigure 11. In some books this construction is used as the definition of a smooth

manifold. I

J 2.21Example.Let us consider the coneM= Cdescribed in Example 1 (see also

Example 2 in Section 1). We already established that Cis manifold homeomorphic

to R2, and moreover C is a differentiable manifold, whose smooth structure is

defined via a one-chart atlas. However, Cis not a smooth manifold with respect tothe induced smooth structure as subset ofR3. Indeed, following the definition in the

above remark, we have U= C, and coordinate homeomorphism (p) = (p1,p2) =

x. By the definition of smooth maps it easily follows thatis smooth. The inverse

is given by1(x) =x1,x2,

qx21+x

22

, which is clearly not differentiable at the

cone-top (0,0,0). The cone C is not a smoothly embedded submainfold ofR3

(topological embedding). I

Let U,Vand Wbe open subsets ofRn, Rk andRm respectively, and let f: UVandg : VWbe smooth maps withy = f(x), andz =g(y). Then the Jacobiansare

J f|x=

f1x1

f1xn...

. . . ...

fkx1

fkxn

, Jg|y=

g1y1

g1yk...

. . . ...

gmy1

gmyk

,and J(g f)|x = J g|y=f(x)J f|x (chain-rule). The commutative diagram for themaps f,g and g fyields a commutative diagram for the Jacobians:


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V

U W

AAAAAAU

g

f

-gf

Rk

Rn

Rm

AAAAAAU

Jg|y

J f|x

-J(gf)|x

For diffeomorphisms between open sets in Rn we have a number of important

properties. Let URn and VRm be open sets, and let f : UVis a diffeomor-phism, thenn=m, andJ f|xis invertible for anyx U. Using the above describedcommutative diagrams we have that f1

fyields that J(f1)|y=f(x)J f|x is the

identity on Rn, and f f1 yields that J f|xJ(f1)|y=f(x) is the identity on Rm.Thus J fx has an inverse and consequently n= m. Conversely, if f :U Rn,URn, open, then we have the Inverse Function Theorem;Theorem 2.22. If J f|x: R

n Rn is invertible, then f is a diffeomorphism betweensufficiently small neighborhoods U0 and f(U0)of x and y respectively.

3. Immersions, submersions and embeddings

Let NandMbe smooth manifolds of dimensions n and m respectively, and let

f :NMbe a smooth mapping. In local coordinates f= f

1

:(U) (V), with respects to charts(U,)and (V,). Therankof f at p N is definedas the rank of f at(p), i.e. rk(f)|p= rk(Jf)|(p), whereJf|(p) is the Jacobian

of f at p:

J f|x=(p)=

f1x1

f1xn

... . . .

...fmx1

fmxn

This definition is independent of the chosen charts, see Figure 12. Via the commu-

tative diagram in Figure 12 we see that f= (0

1)

f

(0

1)1, and by the

chain ruleJf|x0= J(0 1)|y Jf|x J(0 1)1|x0 . Since 0 1 and 0 1are diffeomorphisms it easily follows that rk(Jf)|x= rk(J

f)|x0 , which shows that

our notion of rank is well-defined. If a map has constant rank for all p Nwe sim-ply write rk(f). These are called constant rank mappings. Let us now consider

the various types of constant rank mappings between manifolds.


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FIGURE 12. Representations of fvia different coordinate charts.

Definition 3.2. A mapping f :NMis called an immersionif rk(f) = n, and asubmersionif rk(f) = m. An immersion that is injective,7 or 1-1, and is a home-

omorphism onto (surjective mapping8) its image f(N)M, with respect to thesubspace topology, is called asmooth embedding.

A smooth embedding is an (injective) immersion that is a topological embed-

ding.

Theorem 3.3. 9

Let f :NM be smooth with constant rankrk(f) =k. Then foreach pN, and f(p)M, there exist coordinates(U,)for p and(V,)for f(p),with f(U) V, such that

( f1)(x1, xk,xk+1, xn) = (x1, xk,0, ,0).J 3.4 Example. Let N= (

2, 3

2), and M= R2, and the mapping f is given by

f(t) = (sin(2t),cos(t)).In Figure 13 we displayed the image of f. The Jacobian is

given by

J f|t=

2cos(2t)

sin(t)

!Clearly, rk(J f|t) =1 for all tN, and f is an injective immersion. SinceN isan open manifold and f(N)M is a compact set with respect to the subspace

7A mapping f: NMis injective if for all p,p0 Nfor which f(p) = f(p0)it holds that p =p0.8A mapping f :NM is surjective if f(N) =M, or equivalently for everyq Mthere exists a

p Nsuch that q= f(p).9See Lee, Theorem 7.8 and 7.13.


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topology, it follows fis not a homeomorphism onto f(N), and therefore is not an

embedding. I

FIGURE13 . Injective parametrization of the figure eight.

J 3.6Example.Let N= S1 be defined via the atlas A={(Ui,

i)}, with

1

1 (t) =

((cos(t),sin(t)), and12 (t) = ((sin(t),cos(t)), andt (2 , 32). Furthermore, letM= R2, and the mapping f :NM is given in local coordinates; in U1 as inExample 1. Restricted toS1 R2 the map fcan also be described by

f(x,y) = (2xy,x).

This then yields for U2 that f(t) = (sin(2t),sin(t)). As before rk(f) =1, which

shows that f is an immersion ofS1. However, this immersion is not injective at

the origin in R2, indicating the subtle differences between these two examples, see

Figures 13 and 14. I

FIGURE14 . Non-injective immersion of the circle.

J 3.8Example.LetN= R,M= R2, and f :NMdefined by f(t) = (t2,t3). Wecan see in Figure 15 that the origin is a special point. Indeed, rk(J f)|t= 1 for all

t6=0, and rk(J f)|t= 0 fort=0, and therefore f is not an immersion. IJ 3.10Example. Consider M= PRn. We establishedPRn as smooth manifolds.

Forn=1 we can construct an embedding f :PR1 R2 as depicted in Figure 16.Forn=2 we find an immersion f :PR2 R3 as depicted in Figure 17. I


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FIGURE 15. The map f fails to be an immersion at the origin.

FIGURE1 6. PR is diffeomorphic toS1.

FIGURE 17 . Identifications for PR2 giving an immersed non-

orientable surface in R3.

J 3.13Example.Let N= R2 andM= R, then the projection mapping f(x,y) =x

is a submersion. Indeed,J f|(x,y)= (1 0), and rk(J f|(x,y)) =1. I


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J 3.14Example.LetN=M= R2 and consider the mapping f(x,y) = (x2,y)t. The

Jacobian is

J f|(x,y)= 2x 00 1

! ,and rk(J f)|(x,y)=2 forx 6=0, and rk(J f)|(x,y)=1 forx=0. See Figure 18. I

FIGURE18. The projection (submersion), and the folding of the

plane (not a submersion).

J 3.16Example.We alter Example 2 slightly, i.e. N= S1 R2, andM= R2, andagain use the atlas A. We now consider a different map f, which, for instance on

U1, is given by f(t) = (2cos(t),sin(t))(or globally by f(x,y) = (2x,y)). It is clear

that fis an injective immersion, and since S1 is compact it follows from Lemma

3.18 below that S1 is homeomorphic to it image f(S1), which show that f is a

smooth embedding (see also Appendix in Lee). I

J 3.17 Example. Let N = R, M = R2, and consider the mapping f(t) =

(2cos(t),sin(t)). As in the previous example fis an immersion, not injective how-

ever. Also f(R) = f(S1)in the previous example. The manifoldNis the universal

covering ofS1 and the immersion f descends to a smooth embedding ofS1. I

Lemma 3.18. 10 Let f :NM be an injective immersion. If(i) N is compact, or if

(ii) f is a proper map,11

then f is a smooth embedding.Proof:For (i) we argue as follows. Since Nis compact any closed subset XN

is compact, and therefore f(X)Mis compact and thus closed; f is a closed10Lee, Prop. 7.4, and pg. 47.11A(ny) map f :NMis called proper if for any compactKM, it holds that also f1(K)N

is compact. In particular, when Nis compact, continuous maps are proper.


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mapping.12 We are done if we show that f is a topological embedding. By the

assumption of injectivity, f :N f(N) is a bijection. Let XNbe closed, then(f1)1(X) = f(X) f(N) is closed with respect to the subspace topology, andthus f1 : f(N)Nis continuous, which proves that fis a homeomorphism ontoits image and thus a topological embedding.

A straightforward limiting argument shows that proper continuous mappings

bewteen manifolds are closed mappings (see Exercises).

Let us start with defining the notion of embedded submanifolds.

Definition 3.19. A subset NM is called a smooth embedded n-dimensionalsubmanifoldin M if for every p N, there exists a chart(U,)for M, with p U ,such that

(UN) = (U) Rn{0}={x (U) : xn+1= =xm=0}.The co-dimensionof N is defined as codimN=dimMdimN.

The set W=UN in Nis called a n-dimensional slice, or n-sliceofU, seeFigure 19, and(U,)aslice chart. The associated coordinatesx= (x1, ,xn)are

calledslice coordinates. Embedded submanifolds can be characterized in terms of

embeddings.

FIGURE19. Take ak-sliceW. On the left the image(W)corre-

sponds to the Eucliden subspace RkRm.

Theorem 3.21. 13 Let NM be a smooth embedded n-submanifold. Endowedwith the subspace topology, N is a n-dimensional manifold with a unique (induced)

smooth structure such that the inclusion map i :NM is an embedding (smooth).12 A mapping f : N M is called closed if f(X) is closed in M for any closed set X N.

Similarly, fis calledopenif f(X)is open in Mfor every open setXN.13See Lee, Thms 8.2


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To get an idea let us show that N is a topological manifold. Axioms (i) and

(iii) are of course satisfied. Consider the projection : Rm Rn, and an inclusionj: Rn Rm defined by

(x1, ,xn,xn+1, ,xm) = (x1, ,xn),

j(x1, ,xn) = (x1, ,xn,0, ,0).

Now set Z= ( )(W) Rn, and = ( )|W, then 1 = (1 j)|Z, and: W Zis a homeomorphism. Therefore, pairs(W,) are charts for N, whichform an atlas forN. The inclusioni :NMis a topological embedding.

Given slice charts(U,)and (U0,0)and associated charts(W,)and (W0,0)for N. For the transitions maps it holds that 1 = 0 1 j, which are

diffeomorphisms, which defines a smooth atlas for N. The inclusioni :NMcan

be expressed in local coordinates;

i= i1, (x1, ,xn) 7 (x1, ,xn,0, ,0),

which is an injective immersion. Sincei is also a topological embedding it is thus

a smooth embedding. It remains to prove that the smooth structure is unique, see

Lee Theorem 8.2.

Theorem 3.22. 14 The image of an embedding is a smooth embedded submanifold.

Proof:By assumption rk(f) = nand thus for any p Nit follows from Theorem3.3 that ef(x1, ,xn) = (x1, ,xn,0, ,0),for appropriate coordinates(U,)for pand(V,)for f(p), with f(U) V. Con-sequently, f(U)is a slice in V, since (f(U)satisfies Definition 3.19. By assump-

tion f(U)is open in f(N)and thus f(U) =Af(N)for some open setA M. Byreplacing V by V0=AVand restricting to V0,(V0,)is a slice chart with sliceV0f(N) = V0f(U).

Summarizing we conclude that embedded submanifolds are the images of

smooth embeddings.

A famous result by Whitney says that considering embeddings intoRm

is notnot really a restriction for defining smooth manifolds.

Theorem 3.23. 15 Any smooth n-dimensional manifold M can be (smoothly) em-

bedded into R2n+1.

14See Lee, Thms 8.315See Lee, Ch. 10.


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A subset N M is called an immersed submanifold if N is a smooth n-dimensional manifold, and the mapping i :N Mis a (smooth) immersion. Thismeans that we can endow Nwith an appropriate manifold topology and smoothstructure, such that the natural inclusion ofNintoMis an immersion. If f :NMis an injective immersion we can endow f(N)with a topology and unique smooth

structure; a setU f(N) is open if and only if f1(U)N is open, and the(smooth) coordinate maps are taken to be f1, wheres are coordinate mapsforN. This way f :N f(N)is a diffeomorphism, andi : f(N)Man injectiveimmersion via the composition f(N) NM. This proves:Theorem 3.24. 16 Immersed submanifolds are exactly the images of injective im-

mersions.

We should point out that embedded submanifolds are examples of immersed

submanifolds, but not the other way around. For any immersion f :NM, theimage f(N)is called animmersed manifoldin M.

In this setting Whitney established some improvements of Theorem 3.23.

Namely, for dimension n> 0, any smooth n-dimensional manifold can be em-

bedded into R2n (e.g. the embedding of curves in R2). Also, forn>1 any smooth

n-dimensional manifold can be immersed into R2n1 (e.g. the Klein bottle). Inthis course we will often think of smooth manifolds as embedded submanifolds of

Rm. An important tool thereby is the general version of Inverse Function Theorem,

FIGURE 20. An embedding ofR [left], and an immersion ofS2

[right] called the Klein bottle.

which can easily be derived from the Euclidean version 2.22.

Theorem 3.26. 17 Let N,M be smooth manifolds, and f : N M is a smoothmapping. If, at some point p N, it holds that Jf|(p)=x is an invertible matrix,

16See Lee, Theorem 8.16.17See Lee, Thms 7.6 and 7.10.


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then there exist sufficiently small neighborhoods U03p, and V0 f(p)such thatf|U0 : U0V0 is a diffeomorphism.

As a direct consequence of this result we have that if f :NM, with dimN=dimM, is an immersion, or submersion, then f is a local diffeomorphism. If f is a

bijection, then fis a (global) diffeomorphism.

Theorem 3.27. Let f :NM be a constant rank mapping withrk(f) = k. Thenfor each q f(N), the level set S= f1(q)is an embedded submanifold in N withco-dimension equal to k.

Proof: Clearly, by continuitySis closed inN. By Theorem 3.3 there are coor-

dinates(U,)ofp Sand(V,)of f(p) =q, such that

ef(x1, ,xk,xk+1, ,xn) = (x1, ,xk,0, ,0) = (q) =0.

The points inSUare characterized by(SU) ={x|(x1, ,xk) =0}. There-fore,SU is a (nk)-slice, and thusSis a smooth submanifold inMof codimen-sionk.

In particular, when f : NM is a submersion, then for each q f(N), thelevel setS= f1(q)is an embedded submanifold of co-dimension codim S=m=dimM. In the case of maximal rank this statement can be restricted to just one

level. A point p Nis called aregular point if rk(f)|p= m =dimM, otherwisea point is called a critical point. A valueq f(N)is called aregular value if allpoints p f1(q)are regular points, otherwise a value is called a critical value. Ifqis a regular value, then f1(q)is called a regular level set.Theorem 3.28. Let f :NM be a smooth map. If q f(N) is a regular value,then S= f1(q)is an embedded submanifold of co-dimension equal todimM.

Proof:Let us illustrate the last result for the important caseN=Rn andM=Rm.

For any p S= f1(q)the JacobianJ f|p is surjective by assumption. Denote thekernel ofJ f|pby kerJ f|pRn, which has dimensionnm. Define

g:N= Rn RnmRm= Rn,by g() = (L,f() q)t, where L: N= Rn Rnm is any linear map whichis invertible on the subspace kerJ f|pRn. Clearly, Jg|p= L J f|p, which, byconstruction, is an invertible (linear) map on Rn. Applying Theorem 2.22 (Inverse

Function Theorem) tog we conclude that a sufficiently small neighborhood ofU

of p maps diffeomorphically onto a neighborhood V of(L(p),0). Since g is a

diffeomorphism it holds thatg1 mapsRnm{0}V onto f1(q)U(the 0

Rm corresponds toq). This exactly says that every point p Sallows an(nm)-slice and is therefore an(nm)-dimensional submanifold in N= Rn (codim S=m).


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FIGURE 21. The mapg yields 1-slices for the set S.

J 3.30Example. Let us start with an explicit illustration of the above proof. Let

N= R2,M= R, and f(p1,p2) = p21+p

22. Consider the regular value q =2, then

J f|p= (2p1 2p2), and f1

(2) = {p : p2

1+p2

2= 2}, the circle with radius

2.

We have kerJ f|p= span{(p1,p2)t}, and is always isomorphic to R. For examplefix the point(1,1) S, then kerJ f|p=span{(1,1)t}and define

g() =

L()

f()2

!=

1221+

222

!, Jg|p=

1 12 2

!where the linear map is L = (1 1). The map g is a local diffeomorphism andonSUthis map is given by

g(1,

q221) =

1

q221

0

!,

with1 (1 ,1 + ). The first component has derivative 1 + 1221

, and there-

foreSUis mapped onto a set of the form R{0}V. This procedure can becarried out for any point p S, see Figure 21. IJ 3.31Example.LetN=R2\{(0,0)}R=R3\{(0,0,)},M= (1,)(0,),and

f(x,y,z) =

x2 +y21

1

!, with J f|(x,y,z)=

2x 2y 0

0 0 0

!We see immediately that rk(f) = 1 onN. This map is not a submersion, but is of

constant rank, and therefore for any value q f(N) Mit holds that S= f1(q)is a embedded submanifold, see Figure 22. I

J 3.33Example.Let N,Mas before, but now take

f(x,y,z) =

x2 +y21

z

!, with J f|(x,y,z)=

2x 2y 0

0 0 1

!


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FIGURE22. An embedding of a cylinder via a constant rank map-

ping.

Now rk(f) = 2, and f is a submersion. For everyq

M, the set S= f1(q)is a

embedded submanifold, see Figure 23. I

FIGURE23 . An embedding circle of an submersion.

J 3.35Example.LetN=R2,M=R, and f(x,y) = 14

(x21)2 +12y2. The Jacobian

isJ f|(x,y)= (x(x21) y). This is not a constant rank, nor a submersion. However,

any q> 0, q 6= 14

, is a regular value since then the rank is equal to 1. Figure 24

shows the level set f1(0)(not an embedded manifold), and f1(1)(an embeddedcircle) I

Theorem 3.37. 18 Let SM be a subset of a smooth m-dimensional manifold M.Then S is a k-dimensional smooth embedded submanifold if and if for every p

S

there exists a neighborhood U3p such that US is the level set of a submersionf : URmk.

Proof: Assume that SM is a smooth embedded manifold. Then for eachp Sthere exists a chart (U,), p U, such that(SU) = {x : xk+1= =

18See Lee, Prop.8.12.


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FIGURE24 . A regular and critical level set.

xm= 0}. Clearly,SUis the sublevel set of f : URmk at 0, given by

ef(x) =

(xk+1, ,xm), which is a submersion.

Conversely, if SU= f1(0), for some submersion f :U Rmk, then byTheorem 3.28,SUis an embedded submanifold ofU. This clearly shows that Sis an embedded submanifold in M.


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II. Tangent and cotangent

spaces

4. Tangent spaces

For a(n) (embedded) manifold M R` the tangent space TpM at a point pM can be pictured as a hyperplane tangent to M. In Figure 25 we consider the

parametrizationsx+teiin Rm. These parametrizations yield curves i(t) =

1(x +

tei)on Mwhose velocity vectors are given by

0(0) = d

dt1(x + te i)

t=0

=J1|x(ei).

The vectors p +0(0)are tangent toMat p and span anm-dimensional affine linearsubspace Mp ofR

`. Since the vectors J1|x(ei) span TpMthe affine subspace isgiven by

Mp:=p + TpM R`,

which is tangent to Mat p.

FIGURE 25 . Velocity vectors of curves of M span the tangent

space.


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The considerations are rather intuitive in the sense that we consider only em-

bedded manifolds, and so the tangent spaces are tangent m-dimensional affine sub-

spaces ofR`. One can also define the notion of tangent space for abstract smoothmanifolds. There are many ways to do this. Let us describe one possible way (see

e.g. Lee, or Abraham, Marsden and Ratiu) which is based on the above considera-

tions.

Leta


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The above definition does not depend on the choice of charts at pM. Let(U0,0)be another chart at p M. Then, using that( )0(0) = ( )0(0), for(0 )0(0)we have

(0 )0(0) =h

(0 1) ( )i0

(0)

= J(0 1)x

( )0(0)= J(0 1)

x( )0(0)

=h

(0 1) ( )i0

(0) = (0 )0(0),

which proves that the equivalence relation does not depend on the particular choice

of charts at p

M.

One can prove that TpM= Rm. Indeed, TpMcan be given a linear structure asfollows; given two equivalence classes[1]and[2], then

[1] + [2] := : ( )0(0) = ( 1)0(0) + ( 2)0(0)

,

[1] := : ( )0(0) = ( 1)0(0)

.

The above argument shows that these operation are well-defined, i.e. indepen-

dent of the chosen chart at p M, and the operations yield non-empty equivalenceclasses. The mapping

: TpMRm,

[]

= ( )0(0),

is a linear isomorphism and 0 = J(0 1)|x . Indeed, by considering curves

i(x) = 1(x + te i),i=1,...,m, it follows that[i] 6= [j],i 6= j, since

( i)0(0) =ei6=ej= ( j)0(0).

This proves the surjectivity of. As for injectivity one argues as follows. Suppose,

(

)0(0) = (

)0(0), then by definition[] = [], proving injectivity.

Given a smooth mapping f :NMwe can define how tangent vectors in TpNare mapped to tangent vectors in TqM, with q= f(p). Choose charts(U,) for

p N, and(V,) for q M. We define the tangent map or pushforward of f asfollows, see Figure 27. For a given tangent vectorXp= [] TpN,

d fp= f: TpNTqM, f([]) = [f].


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FIGURE 27 . Tangent vectors in XpTpNyield tangent vectorsfXp TqMunder the pushforward of f.

The following commutative diagram shows that fis a linear map and its definitiondoes not depend on the charts chosen at p N, or q M.

TpN

f

TqM

y yRn

J(f1)|x Rm

Indeed, a velocity vector( )0(0)is mapped to(f())0(0), and(f())0(0) = (f1 )0(0) =J f

x( )0(0).

Clearly, this mapping is linear and independent of the charts chosen.

If we apply the definition of pushforward to the coordinate mapping :NRn,thencan be identified with, andJ(

f

1)|x with(

f

1). Indeed,

([]) = ( )0(0) and ([]) = [ ], and in Rn the equivalence class can belabeled by( )0(0). The labeling map is given as follows

id([ ]) = ( )0(0),and is an isomorphism, and satisfies the relations

id = , = 1id .


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From now one we identify TxRn with Rn by identifying and. This justifies

the notation

([]) = [ ]:= ( )0(0).Properties of the pushforward can be summarized as follows:

Lemma 4.5. 20Let f :NM, and g :M P be smooth mappings, and let pM,then

(i) f : TpN Tf(p)M, and g : Tf(p)M T(gf)(p)P are linear maps (homo-morphisms),

(ii) (g f)=g f: TpNT(gf)(p)P,(iii) (id)=id : TpNTpN,(iv) if f is a diffeomorphism, then the pushforward f is a isomorphism from

TpN to Tf(p)M.

Proof:We have that f([]) = [f], andg([f]) = [g f], which definesthe mapping(g f)([]) = [gf]. Now

[g f] = [g (f)] =g([f]) =g(f([])),

which shows that(gf)=g f.

A parametrization 1 : Rm Mcoming from a chart (U,) is a local dif-feomorphism, and can be used to find a canonical basis for TpM. Choosing local

coordinates x= (x1, ,xn) = (p), and the standard basis vectors e i for Rm, we

define

xi

p

:= 1 (ei).

By definition xi

p TpM, and since the vectorsei form a basis for Rm, the vectors

xi

p

form a basis for TpM. An arbitrary tangent vector XpTpM can now bewritten with respect to the basis{ xi

p

}:

Xp= 1 (Xiei) =Xi

xi

p.

where the notationXixi p=iXi

xi pdenotes the Einstein summation convention,and(Xi)is a vector in Rm!

We now define the directional derivative of a smooth function h :M R in thedirection ofXp TpMby

Xph:=hXp= [h ].

20Lee, Lemma 3.5.


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In fact we have that Xph= (h )0(0), withXp= []. For the basis vectors ofTpMthis yields the following. Leti(t) =

1(x + tei), andXp= xi p, then(2) Xph=

xi

p

h=

(h1) ( i)0

(0) = h

xi,

in local coordinates, which explains the notation for tangent vectors. In particular,

for general tangent vectors Xp,Xph=Xihxi

.

Let go back to the curve :N= (a,b) Mand express velocity vectors. Con-sider the chart(N, id)for N, with coordinatest, then

d

dt

t=0:=id1 (1).

We have the following commuting diagrams:

N M

id

y yR

= Rm

TtN TpM

id

y yR

RmWe now define

0(0) = d

dt

t=0

= 1

( )0(0) TpM,

by using the second commuting diagram.

If we take a closer look at Figure 27 we can see that using different charts at

p M, or equivalently, considering a change of coordinates leads to the followingrelation. For the charts(U,) and (U0,0) we have local coordinates x= (p)and x0= 0(p), and p UU0. This yields two different basis for TpM, namely xi

p

, and

x0i

p

. Consider the identity mapping f= id :MM, and the push-

forward yields the identity on TpM. If we use the different choices of coordinates

as described above we obtainid

xi

p

h= (0 1) h

x0j=x0jxi

h

x0j.

In terms of the different basis forTpMthis gives

xi

p

=x0jxi

x0j

p.

Let us prove this formula by looking a slightly more general situation. Let N,M

be smooth manifolds and f :NMa smooth mapping. Let q = f(p)and (V,)


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is a chart forMcontainingq. The vectors yj

qform a basis for TqM. If we write

Xp=Xixi p, then for the basis vectors

xi p we have

f

xi

p

h =

xi

p

(hf) = xi

ghf= xi

h f1

=

xi

hf1=

xi

h f

= h

yj

fjxi

=fjxi

yj

q

h

which implies that TpNis mapped to TqMunder the map f. In general a tangentvectorXi

xi

p

is pushed forward to a tangent vector

(3) Yj yj

q= "fj

xiXi#

yj

q TqM,

expressed in local coordinates. By taking N= Mand f=id we obtain the above

change of variables formula.

5. Cotangent spaces

In linear algebra it is often useful to study the space of linear functions on a

given vector space V. This space is denoted by Vand called thedual vector spaceto V again a linear vector space. So the elements ofV are linear functions : V R. As opposed to vectors v V, the elements, or vectors inV are called

covectors.

Lemma 5.1. Let V be a n-dimensional vector space with basis {v1, ,vn}, then

the there exist covectors{1, ,n}such that

i vj := i(vj) =

ij, Kronecker delta,

and the covectors{1, ,n}form a basis for V.This procedure can also be applied to the tangent spaces TpMdescribed in the

previous chapter.

Definition 5.2. Let M be a smooth m-dimensional manifold, and let TpMbe the

tangent space at some p M. The thecotangent spaceTpMis defined as the dualvector space ofTpM, i.e.

TpM:= (TpM).


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By definition the cotangent spaceTpMis alsom-dimensional and it has a canon-ical basis as described in Lemma 5.1. As before have the canonical basis vectors

xi |p for TpM, the associated basis vectors for TpMare denoted d xi|p. Let us nowdescribe this dual basis forTpMand explain the notation.

The covectors dx i|p are called differentials and we show now that these are

indeed related dhp. Let h : M R be a smooth function, then h : TpM Rand hTpM. Since the differentials dxi|p form a basis of TpM we have thath= idxi|p, and therefore

h

xi

p

= jdxj|p

xi

p

= jji =

i = h

xi,

and thus

(4) dhp=h= h

xidx ip.

Chooseh such thath satisfies the identity in Lemma 5.1, i.e. leth=xi ( h=xi = h,eii). These linear functionsh of course spanTpM, and

h= (xi )=d(xi )p=dxi|p.

Cotangent vectors are of the form

p= idxip.

The pairing between a tangent vector Xp and a cotangent vector p is expressed

component wise as follows:

p Xp= iXj

ij=

iXi.

In the case of tangent spaces a mapping f : NM pushes forward to a linearf; TpN TqMfor each p N. For cotangent spaces one expects a similar con-struction. Letq= f(p), then for a given cotangent vectorq TqMdefine

(f

q) X

p=

q (fX

p) TpN,

for any tangent vectorXp TpN. The homomorphism f: TqM TpN, defined byq 7 fqis called thepullbackof f atp. It is a straightforward consequence fromlinear algebra that f defined above is indeed the dual homomorphism of f, alsocalled the adjoint, or transpose (see Lee, Ch. 6, for more details, and compare the

the definition of the transpose of a matrix). If we expand the definition of pullback


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in local coordinates, using (3), we obtain

fdyjq Xi xi p = dyjqfXi xi p= dyj

q

"fjxi

Xi

#

yj

q

=fjxi

Xi

Using this relation we obtain thatfjdyj

q

Xi

xi

p

= jfjxi

Xi= j

fj

xidxip

Xi

xi

p,

which produces the local formula

(5) fjdyjq= "j fjxi #dxip= "j ffjxi #x

dxip.Lemma 5.3. Let f :NM, and g:M P be smooth mappings, and let p M,then

(i) f: Tf(p)M TpN, and g: T(gf)(p)P Tf(p)M are linear maps (homo-

morphisms),

(ii) (g f)= f g: T(gf)(p)P TpN,(iii) (id)=id : TpNTpN,(iv) if f is a diffeomorphism, then the pullback fis a isomorphism from T

f(p)M

to TpN.Proof:By definition of the pullbacks of f andg we have

fq Xp= q f(Xp), gg(q)Yq= g(q) g(Yq).

For the compositiong f it holds that(g f)(gf)(p) Xp=(gf)(p) (g f)(Xp).Using Lemma 4.5 we obtain

(g f)(gf)(p) Xp = (gf)(p) g

f(Xp)

= g(gf)(p) f(Xp) = f

g(gf)(p)

Xp,

which proves that(gf)= f g.

Now consider the coordinate mapping : UM Rm, which is local diffeo-morphism. Using Lemma 5.3 we then obtain an isomorphism : Rm TpM,which justifies the notation

dx i|p= (ei).


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6. Vector bundles

The abstract notion of vector bundle consists of topological spaces E (total

space) and M (the base) and a projection : E M (surjective). To be moreprecise:

Definition 6.1. A triple(E,M,)is called areal vector bundleof rankkoverMif

(i) for each p M, the set Ep= 1(p)is a real k-dimensional linear vectorspace, called thefiber over p, such that

(ii) for everyp Mthere exists a open neighborhood U3p, and a homeomor-phism : 1(U) URk;(a)

1

(p,) = p, for all Rk;

(b) 7

1(p,)is a vector space isomorphism between Rk andEp.

The homeomorphismis called a local trivializationof the bundle.

If there is no ambiguity about the base spaceMwe often denote a vector bundle

by E for short. Another way to denote a vector bundle that is common in the

literature is :EM. It is clear from the above definition that ifMis a topologicalmanifold then so is E. Indeed, via the homeomorphisms it follows that E

FIGURE28. Charts in a vector bundleEoverM.

is Hausdorff and has a countable basis of open set. Definee= IdRk ,ande : 1(U) VRk is a homeomorphism. Figure 28 explains the choice ofbundle charts forErelated to charts forM.

For two trivializations : 1(U)URk and : 1(V)VRk, we havethat the transition map 1 :(UV)Rk (UV)Rk has the following


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form

1(p,) = (p,(p))),where :UV Gl(k,R) is continuous. It is clear from the assumptions that

FIGURE29. Transition mappings for local trivializations.

1

(p,) = (p,(p,)). By assumption we also have that 7 1

(p,) =A(p), and 71(p,) =B(p). Now,

1(p,) = (A(p)) = (p,(p,)),and 1(p,) =A(p)=B(p). Therefore (p,) =B1A=: (p). Continuityis clear from the assumptions in Definition 6.1.

If bothEand Mare smooth manifolds and is a smooth projection, such that the

local trivializations can be chosen to be diffeomorphisms, then(E,M,)is called

asmooth vector bundle. In this case the maps are smooth. The following result

allows us to construct smooth vector bundles and is important for the special vector

bundles used in this course.Theorem 6.4. 21 Let M be a smooth manifold, and let{Ep}pMbe a family of k-dimensional real vector spaces parametrized by p M. Define E= FpMEp, and :EM as the mapping that maps Ep to p M. Assume there exists

(i) an open covering{U}A for M;

21See Lee, Lemma 5.5.


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(ii) for each A, a bijection :1(U)U Rk, such that 71 (p,)is a vector space isomorphism between Rk and Ep;

(iii) for each ,I, with U U6= , smooth mappings : U UGl(k,R)such that

1 (p,) = (p,(p)).Then E has a unique differentiable (manifold) structure making (E,M,)a smooth

vector bundle of rank k over M.

Proof: The proof of this theorem goes by verifying the hypotheses in Theorem

2.11. Let us propose charts for E. Let(Vp,p),VpU, be a smooth chart forM containing p. As before define

ep :

1(Vp)

eVp Rk. We can show, using

Theorem 2.11, that(1(Vp),ep), p Mare smooth charts, which gives a uniquedifferentiable manifold structure. For details of the proof see Lee, Lemma 5.5.J 6.5Example.Consider the setEdefined as

E=

(p1,p2,1,2)| p1=cos(),p2=sin(), cos(/2)1+ sin(/2)2=0.

Clearly, E is a smooth vector bundle over S1 of rank 1, embedded in R4. For

example ifU= S1\{(1,0)}( (0,2)), then(1(U)) = (p1,p2,1)is a localtrivialization. This bundle is called the Mobius strip.

To show that E is a smooth vector bundle we need to verify the conditions in

Theorem 6.4. Consider a second chartU0= S1\{(1,0)} ( (,)), and thetrivialization(1(U0)) = (p1,p2,2). For thewe have that

1(p1,p2,1) =

p1,p2,1,cos(/2)sin(/2)

1

,

and thus

1(p1,p2,1) =

p1,p2,cos(/2)sin(/2)

1

,

which gives that(p)is represented by = cos(/2)sin(/2)

, for (0,), which invert-ible and smooth in p. I

Mappings between smooth vector bundles are calledbundle maps, and are de-

fined as follows. Given vector bundles : E M and0 : E0M0 and smoothmappingsF: EE0 and f :MM0, such that the following diagram commutes

E F E0

y y0M

f M0


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and F|Ep :EpE0f(p) is linear, then the pair (F,f) is a called a smooth bundlemap.

Asection, orcross section of a bundleEis a continuous mapping : ME,such that =IdM. A section is smooth if is a smooth mapping. The space of

FIGURE30. A cross section in a bundleE.

smooth section inEis denoted by E(M). Thezero sectionis a mapping :MEsuch that(p) =0 Ep for all p M.J 6.7Remark.By identifyingMwith the trivial bundleE0=M{0}, a section isa bundle map from E0= M{0}to E. Indeed, define0 : E0 E by0(p,0) =(p), and let f=IdM. Then 0= =IdM. I

6.1. The tangent bundle and vector fields

The disjoint union of tangent spaces

T M:=G

pMTpM

is called thetangent bundleofM. We show now thatT Mis a fact a smooth vector

bundle overM.

Theorem 6.8. The tangent bundle T M is smooth vector bundle over M of rank m,

and as such T M is a smooth2m-dimensional manifold.

Proof:Let (U,)be a smooth chart for p M. Define

Xi

xi p= p,(Xi),which clearly is a bijective map between 1(U) andURm. Moreover, Xp=(Xi) 71(p,Xp) =Xi xi

p

is vector space isomorphism. Let(U,)be a cov-

ering of smooth charts ofM. Let x=(p), and x0= (p). From our previous

considerations we easily see that

1 (p,Xp) = (p,(p)Xp),


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In Section 4 we introduced the notion of push forward of a mapping f :NM.Under submersions and immersion a vector fieldX: N T Ndoes not necessarilypush forward to a vector field on M. If fis a diffeomorphism, then fX= Y is avector field on M. We remark that the definition of f also allows use to restatedefinitions about the rank of a map in terms of the differential, or pushforward fat a point p N.

6.2. The cotangent bundle and differential 1-forms

The disjoint union

TM:=G

pMTpM

is calledcotangent bundleofM.

Theorem 6.12. 22 The cotangent bundle TM is a smooth vector bundle over M ofrank m, and as such TM is a smooth2m-dimensional manifold.

Proof:The proof is more identical to the proof for T M, and is left to the reader

as an exercise.

The differentiald h:M TMis an example of a smooth function. The aboveconsideration give the coordinate wise expression for d h.

Definition 6.13. Asmooth covector fieldis a smooth mapping

: M TM,with the property that =idM. In order words is a smooth section in TM.The space of smooth covector fields on Mis denoted by F(M). Usually the spaceF(M)is denoted by1(M), and covector fields are referred to a(smooth) differ-ential 1-formon M.

For a chart(U,)a covector field can be expressed as follows

= idxip,

wherei :U R. Smoothness of a covector fields can be described in terms ofthe component functionsi.

Lemma 6.14. A covector field is smooth at p U if and only if the coordinatefunctionsi : U

R are smooth.

Proof:See proof of Lemma 6.11.

We saw in the previous section that for arbitrary mappings f :NM, a vectorfieldX F(N)does not necessarily push forward to a vector on Munder f. Thereason is that surjectivity and injectivity are both needed to guarantee this, which

22See Lee, Proposition 6.5.


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requires fto be a diffeomorphism. In the case of covector fields or differential 1-

forms the situation is completely opposite, because the pullback acts in the opposite

direction and is therefore onto the target space and uniquely defined. To be moreprecise; given a 1-form 1(M)we define a 1-form f 1(N)as follows

(f)p= ff(p).

Theorem 6.15. 23 The above defined pullback fof under a smooth mappingf :NM is a smooth covector field, or differential 1-form on N.

Proof: Write 1(M) is local coordinates; =idyiq

. By Formula (5) we

then obtain

f

idyif(p)= "iy fi

xj#dxjp= "i ffi

xj #xdx jp,which proves that f 1(N).

Given a mapping g : MR, the differential, or push forward g= dg definesan element in TpM. In local coordinates we haved g

p

= egxi

dx ip

, and thus defines

a smooth covector field onM;d g 1(M). Given a smooth mapping f :NM,it follows from (5) that

fdg= f

eg

xidx i

f(p)=

eg

xief

efi

xjdyj

p.

By applying the same to the 1-formd(g f)we obtain the following identities

(6) fdg=d(g f), f(g) = (g f)f.

Using the formulas in (6) we can also obtain (5) in a rather straightforward way.

Letg= j= h,eji =yj, and =dg=dyj|q in local coordinates, then

f

(j )= j ffdg = j fd(gf) = hj ffjxi

ix

dxip,

where the last step follows from (4).

Definition 6.16. A differential 1-form 1(N)is called anexact1-form if thereexists a smooth functiong :N R, such that =dg.

23Lee, Proposition 6.13.


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The notation for tangent vectors was motivated by the fact that functions on a

manifold can be differentiated in tangent directions. The notation for the cotangent

vectors was partly motivated as the reciprocal of the partial derivative. The in-troduction of line integral will give an even better motivation for the notation for

cotangent vectors. Let N= R, and a 1-form on Ngiven in local coordinates by

t=h(t)dt, which can be identified with a function h. The notation makes sense

becausecan be integrated over any interval [a,b] R:Z

[a,b] :=

Z ba

h(t)dt.

LetM=R, and consider a mapping f :M= RN= R, which satisfies f0(t) > 0.Thent= f(s)is an appropriate change of variables. Let[c,d] = f([a,b]), then

Z[c,d]

f =Z d

ch(f(s))f0(s)ds =

Z ba

h(t)dt=Z

[a,b],

which is the change of variables formula for integrals. We can use this now to

define the line integral over a curveon a manifoldN.

Definition 6.17. Let: [a,b] R N, and let be a 1-form on N and thepullback of, which is a 1-form on R. Denote the image ofin Nalso by, thenZ

:=

Z[a,b]

=Z

[a,b]i((t))0i(t)dt,

24

the expression in local coordinates.

The latter can be seen by combining some of the notion introduced above:

d

dt= ()t= (t)

d

dt= (t)

0(t).

Therefore, = ()tdt= (t) 0(t)dt= i((t))0i(t)dt, andZ =

Z[a,b]

=Z

[a,b](t)

0(t)dt=Z

[a,b]i((t))0i(t)dt.

If0 is nowhere zero then the map:[a,b] Nis either an immersion or em-bedding. For example in the embedded case this gives an embedded submanifold

Nwith boundary={(a),(b)}. Let =dgbe an exact 1-form, thenZ

dg =g=g((b))g((a)).Indeed,

Z

dg =Z

[a,b]dg=

Z[a,b]

d(g ) =Z b

a(g )0(t)dt= g((b))g((a)).

24The expressionsi(t) R are the components of0(t) T(t)M.


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This identity is called theFundamental Theorem for Line Integrals and is a spe-

cial case of the Stokes Theorem (see Section 16).


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III. Tensors and differential

forms

7. Tensors and tensor products

In the previous chapter we encountered linear functions on vector spaces, linear

functions on tangent spaces to be precise. In this chapter we extend to notion of

linear functions on vector spaces to multilinear functions.

Definition 7.1. Let V1, ,Vr, and W be real vector spaces. A mapping T :V1 VrW is called amultilinear mappingif

T(v1, ,vi+v0i, ,vr) = T(v1, ,vi, vr) +T(v1, ,v

0i, vr), i,

and for all, R i.e. f is linear in each variable vi separately.

Now consider the special case that W= R, thenTbecomes amultilinear func-

tion, or form, and a generalization of linear functions. If in additionV1= =

Vr= V, then

T : V VR,

is a multilinear function on V, and is called acovariantr-tensoron V. The number

of copiesris called the rank ofT. The space of covariant r-tensors on Vis denoted

byTr(V), which clearly is a real vector space using the multilinearity property in

Definition 7.1. In particular we have thatT0(V) = R, T1(V) = V, and T2(V) isthe space of bilinear forms on V. If we consider the case V1= = Vr= V

, then

T : V VR,

is a multilinear function on V, and is called a contravariant r-tensoron V. Thespace of contravariant r-tensors on V is denoted by Tr(V). Here we have that

T0(V) = R, andT1(V) = (V)= V.J 7.2Example. The cross product on R3 is an example of a multilinear (bilinear)

function mapping not to R to R3. Letx,y R3, then

T(x,y) =xy R3,

which clearly is a bilinear function on R3. I


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Since multilinear functions on Vcan be multiplied, i.e. given vector spaces V,W

and tensorsT Tr(V), andS Ts(W), the multilinear function

R(v1, vr,w1, ,ws) =T(v1, vr)S(w1, ,ws)

is well defined and is a multilnear function on VrWs. This brings us to thefollowing definition. LetT Tr(V), andS Ts(W), then

TS: VrWs R,

is given by

TS(v1, ,vr,w1, ws) =T(v1, ,vr)S(w1, ws).

This product is called the tensor product. By takingV= W, TSis a covariant(r+ s)-tensor on V, which is a element of the space Tr+s(V) and : Tr(V)Ts(V) Tr+s(V).Lemma 7.3. Let T Tr(V), S,S0 Ts(V), and R Tt(V), then

(i) (TS)R=T (SR)(associative),(ii) T (S+ S0) =TS+ TS0 (distributive),

(iii) TS6=ST (non-commutative).The tensor product is also defined for contravariant tensors and mixed tensors.

As a special case of the latter we also have the product between covariant and

contravariant tensors.J 7.4Example.The last property can easily be seen by the following example. Let

V= R2, andT,S T1(R2), given byT(v) =v1+ v2, andS(w) =w1w2, then

TS(1,1,1,0) =2 6=0=ST(1,1,1,0),

which shows that is not commutative in general. I

The following theorem shows that the tensor product can be used to build the

tensor spaceTr(V)from elementary building blocks.

Theorem 7.5. 25Let{v1, ,vn}be a basis for V , and let{1, ,n}be the dual

basis for V. Then the set

B=i1 ir : 1 i1, , ir n

,

is a basis for the nr-dimensional vector space Tr(V).

25See Lee, Prop. 11.2.


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Proof:Compute

Ti1ir

i1

ir

(vj1 , ,vjr) = Ti1iri1

(vj1 ) ir

(vjr)= Ti1ir

j1j1

jrjr =Tj1jr

= T(vj1 , ,vjr),

which shows by using the multilinearity of tensors that Tcan be expanded in the

basis B as follows;

T=Ti1iri1 ir,

where Ti1ir= T(vi1 , ,vir), the components of the tensor T. Linear independence

follows form the same calculation.

J 7.6Example.Consider the the 2-tensorsT(x,y) =x1y1+x2y2,T0(x,y) =x1y2+

x2y2andT00=x1y1 +x2y2 +x1y2on R2. With respect to the standard bases 1(x) =

x1,2(x) =x1, and

11(x,y) =x1y1, 12(x,y) =x1y2,12(x,y) =x2y1, and 22(x,y) =x2y2.

Using this the components ofTare given byT11= 1,T12= 0,T21= 0, andT22= 1.

Also notice that T0= S S0, where S(x) = x1+x2, and S0(y) = y2. Observe thatnot every tensorT

T2(R2)is of the form T=S

S0. For exampleT00

6=S

S0,

for anyS,S0 T1(R2). I

In Lee, Ch. 11, the notion of tensor product between arbitrary vector spaces is

explained. Here we will discuss a simplified version of the abstract theory. Let

V and W be two (finite dimensional) real vector spaces, with bases {v1, vn}

and {w1, wm} respectively, and for their dual spaces V and W we have the

dual bases{1, ,n}and{1, ,m}respectively. If we use the identification

{V}= V, and{W}= Wwe can define VWas follows:

Definition 7.7. The tensor product ofV andWis the real vector space of (finite)

linear combinations

VW :=ni jviwj : i j R

o=h

viwj

i,j

i,

whereviwj(v,w) := v(vi)w(wj), using the identificationvi(v) := v(vi), andwj(w

):=w(wj), with(v,w) VW.


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To get a feeling of what the tensor product of two vector spaces represents con-

sider the tensor product of the dual spaces V andW. We obtain the real vector

space of (finite) linear combinations

VW:=ni j

ij : i jRo

=hij

i,j

i,

whereij(v,w) = i(v)j(w)for any(v,w) VW. One can show thatVWis isomorphic to space of bilinear maps from VWto R. In particular elementsvw all lie inVW, but not all elements in VW are of this form. Theisomorphism is easily seen as follows. Let v= ivi, andw= jwj, then for a given

bilinear form b it holds that b(v,w) =ijb(vi,wj). By definition of dual basis

we have thatij= i(v)j(w) = ij(v,w), which shows the isomorphism by

settingi j=b(vi,wj).

In the case VW the tensors represent linear maps from V to W. Indeed,from the previous we know that elements in VW represent bilinear maps fromVWto R. For an elementb VWthis means thatb(v, ) : WR, and thusb(v, ) (W)= W.J 7.8Example.Consider vectors a V andbW, thena (b) can be iden-tified with a matrix, i.e a (b)(v, ) = a(v)(b)() =a(v)b. For example leta(v) =a1v1+ a2v2+ a3v3, and

Av = a(v)b = a1b1v1+ a2b1v2+ a3b1v3

a1b2v1+ a2b2v2+ a3b2v3 != a1b1 a2b1 a3b1

a1b2 a2b2 a3b2 !

v1

v2

v3

.

Symbolically we can write

A=ab=

a1b1 a2b1 a3b1

a1b2 a2b2 a3b2

!=

a1 a2 a3

b1

b2

!,

which shows how a vector and covector can be tensored to become a matrix. Note

that it also holds that A= (a b)=b a. I

Lemma 7.9. We have that

(i) VW and WV are isomorphic;(ii) (UV)W and U (VW)are isomorphic.

With the notion of tensor product of vector spaces at hand we now conclude that

the above describe tensor spaces Tr(V)and Vr(V)are given as follows;

Tr(V) = V V| {z }r times

, Tr(V) = V V| {z }r times

.


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By considering tensor products ofVs andVs we obtain the tensor space ofmixed tensors;

Trs(V):= V V| {z }r times

V V| {z }s times

.

Elements in this space are called(r,s)-mixed tensorson V rcopies ofV, andscopies ofV. Of course the tensor product described above is defined in general for

tensorsT Trs(V), andS Tr0

s0 (V):

: Trs(V)Tr0

s0(V) Tr+r0

s+s0 (V).

The analogue of Theorem 7.5 can also be established for mixed tensors. In the next

sections we will see various special classes of covariant, contravariant and mixed

tensors.J 7.10Example. The inner product on a vector space Vis an example of a covariant

2-tensor. This is also an example of a symmetric tensor. I

J 7.11Example.The determinant ofn vectors in Rn is an example of covariant n-

tensor onRn. The determinant is skew-symmetric, and an example of an alternating

tensor. I

If f :V Wis a linear mapping between vector spaces and Tis an covarianttensor onWwe can define concept of pullback ofT. LetT Tr(W), then fTTr(V)is defined as follows:

fT(v1, vr) =T(f(v1), ,f(vr)),

and f: Tr(W) Tr(V)is a linear mapping. Indeed, f(T+ S) =T f+ S f=fT+ fS, and fT =T f =fT. If we represent f by a matrix A withrespect to bases{vi} and {wj} for V andWrespectively, then the matrix for the

linear f is given by

A A| {z }rtimes

,

with respect to the bases{i1 ir}and{j1 jr}forTr(W)andTr(V)respectively.

J 7.12Remark.The direct sum

T(V) =M

r=0

Tr(V),

consisting of finite sums of covariant tensors is called the covariant tensor algebra

ofV with multiplication given by the tensor product. Similarly, one defines the


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contravariant tensor algebra

T(V) =

Mr=0

Tr(V).

For mixed tensors we have

T(V) =M

r,s=0

Trs(V),

which is called thetensor algebra of mixed tensorofV. Clearly,T(V)andT(V)subalgebras ofT(V). I

8. Symmetric and alternating tensors

There are two special classes of tensors which play an important role in the

analysis of differentiable manifolds. The first class we describe are symmetric

tensors. We restrict here to covariant tensors.

Definition 8.1. A covariantr-tensorTon a vector space V is calledsymmetricif

T(v1, ,vi, ,vj, ,vr) =T(v1, ,vj, ,vi, ,vr),

for any pair of indices i j. The set of symmetric covariant r-tensors on V isdenoted byr(V) Tr(V), which is a (vector) subspace ofTr(V).

Ifa Sris a permutation, then defineaT(v1, ,vr) =T(va(1), ,va(r)),

where a({1, , r}) = {a(1), ,a(r)}. From this notation we have that for two

permutationsa,b Sr, b(aT) = ba T. Define

SymT= 1

r!aSr

aT.

It is straightforward to see that for any tensor T Tr(V), SymTis a symmetric.Moreover, a tensor T is symmetric if and only if Sym T = T. For that reason

SymTis called the(tensor) symmetrization.J 8.2 Example. Let T,T0T2(R2) be defined as follows: T(x,y) =x1y2, andT0(x,y) =x1y1. Clearly,T is not symmetric andT0 is. We have that

SymT(x,y) = 1

2T(x,y) +

1

2T(y,x)

= 1

2x1y2+

1

2y1x2,


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which clearly is symmetric. If we do the same thing forT0 we obtain:

SymT0(x,y) = 1

2 T0(x,y) +1

2 T0(y,x)

= 1

2x1y1+

1

2y1x1=T

0(x,y),

showing that operation Sym applied to symmetric tensors produces the same ten-

sor again. I

Using symmetrization we can define the symmetric product. LetS r(V)andT s(V)be symmetric tensors, then

S T=Sym(ST).

The symmetric product of symmetric tensors is commutative which follows di-rectly from the definition:

S T(v1, ,vr+s) = 1

(r+ s)! aSr+s

S(va(1), ,va(r))T(va(r+1), ,va(r+s)).

J 8.3 Example. Consider the 2-tensors S(x) = x1+x2, and T(y) = y2. Now ST(x,y) = x1y2+x2y2, and TS(x,y) = x2y1+x2y2, which clearly gives that ST6=TS. Now compute

Sym(ST)(x,y) = 12

x1y2+1

2x2y2+

1

2y1x2+

1

2x2y2

= 1

2x1y2+

1

2x2y1+x2y2=S T(x,y).

Similarly,

Sym(TS)(x,y) = 12

x2y1+1

2x2y2+

1

2y2x1+

1

2x2y2

= 1

2x1y2+

1

2x2y1+x2y2=T S(x,y),

which gives thatS T=T S. I

Lemma 8.4. Let{v1, ,vn} be a basis for V, and let{1, ,n} be the dual

basis for V. Then the set

B=i1 ir : 1 i1 ir n

,

is a basis for the (sub)space r(V)of symmetric r-tensors. Moreover, dimr(V) = n + r1

r

!= (n+r1)!

r!(n1)!.


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Proof: Proving that B is a basis follows from Theorem 7.5, see also Lemma

8.10. It remains to establish the dimension ofr(V). Note that the elements in the

basis are given by multi-indices (i1, , ir), satisfying the property that 1 i1 ir n. This means choosing rintegers satisfying this restriction. To do thisredefine jk:= ik+k1. The integers jrange from 1 throughn + r1. Now chooserintegers out ofn + r1, i.e.

n + r1

r

!combinations(j1, ,jr), which are

in one-to-one correspondence with(i1, , ir).

Another important class of tensors are alternating tensors and are defined as

follows.

Definition 8.5. A covariantr-tensorTon a vector space V is calledalternatingif

T(v1, ,vi, ,vj, ,vr) = T(v1, ,vj, ,vi, ,vr),

for any pair of indices i j. The set of alternating covariant r-tensors on V isdenoted byr(V) Tr(V), which is a (vector) subspace ofTr(V).

As before we define

AltT= 1

r!aSr

(1)a aT,

where(1)a is+1 for even permutations, and1 for odd permutations. We saythat AltT is the alternating projection of a tensor T, and AltT is of course aalternating tensor.

J 8.6 Example. Let T,T0T2(R2) be defined as follows: T(x,y) =x1y2, andT0(x,y) =x1y2 x2y1. Clearly, T is not alternating and T0(x,y) =T0(y,x) isalternating. We have that

AltT(x,y) = 1

2T(x,y) 1

2T(y,x)

= 1

2x1y2 1

2y1x2=

1

2T0(x,y),

which clearly is alternating. If we do the same thing for T0we obtain:

AltT0(x,y) = 1

2T0(x,y) 1

2T0(y,x)

= 1

2x1y2 1

2x2y1 1

2y1x2+

1

2y2x1=T

0(x,y),

showing that operation Alt applied to alternating tensors produces the same tensor

again. Notice thatT0(x,y) =det(x,y). I


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This brings us to the fundamental product of alternating tensors called thewedge

product. LetS r(V)and T s(V)be symmetric tensors, then

ST=(r+ s)!r!s!

Alt(ST).The wedge product of alternating tensors is anti-commutative which follows di-

rectly from the definition:

ST(v1, ,vr+s) = 1r!s! aSr+s

(1)aS(va(1), ,va(r))T(va(r+1), ,va(r+s)).

In the special case of the wedge of two covectors, V gives = .

In particular we have that

(i) (TS)R=T (SR);(ii) (T+ T0)S=TS+ T0S;

(iii) TS= (1)rsST, forT r(V)and S s(V);(iv) TT=0.

The latter is a direct consequence of the definition of Alt . In order to prove these

properties we have the following lemma.

Lemma 8.7. Let T Tr(V)and S Ts(V), thenAlt(T

S) =Alt((AltT)

S) =Alt(T

AltS).

Proof: Let G =Srbe the subgroup ofSr+s consisting of permutations that onlypermute the element {1, , r}. For aG, we have a0Sr. Now a(TS) =a0TS, and thus

1

r!aG

(1)a a(TS) = (AltT)S.

For the right cosets{ba : a G}we have

aG

(1)ba ba(TS) = (1)b b

aG(1)a a(TS)

= r!(

1)b b(AltT)S.

Taking the sum over all right cosets with the factor 1(r+s)! gives

Alt(TS) = 1(r+ s)!

b

aG(1)ba ba(TS)

= r!

(r+ s)!b

(1)b b

(AltT)S

=Alt((AltT)S),


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where the latter equality is due to the fact that r! terms are identical under the

definition of Alt((AltT)S).Property (i) can now be proved as follows. Clearly Alt(Alt(TS)TS) = 0,

and thus from Lemma 8.7 we have that

0=Alt((Alt(TS)TS)R) =Alt(Alt(TS)R)Alt(TSR).By definition

(TS)R = (r+ s + t)!(r+ s)!t!

Alt((TS)R)

= (r+ s + t)!

(r+ s)!t! Alt

(r+ s)!r!s!

Alt(TS)R

=

(r+ s + t)!

r!s!t! Alt(TSR).The same formula holds forT(SR), which prove associativity. More generallyit holds that forTi ri (V)

T1 Tk= (r1+ + rk)!r1! rk!

Alt(T1 Tk).Property (iii) can be seen as follows. Each term in TScan be found inST.

This can be done by linking the permutations a and a0. To be more precise, howmany permutation of two elements are needed to change

a (i1, , ir,jr+1, , jr+s) into a0 (jr+1, ,jr+s, i1, , ir).This clearly requiresrs permutations of two elements, which shows Property (iii).

J 8.8Example. Consider the 2-tensorsS(x) = x1+x2, and T(y) = y2. As before

ST(x,y) =x1y2+x2y2, andTS(x,y) =x2y1+x2y2. Now compute

Alt(ST)(x,y) = 12

x1y2+1

2x2y2 1

2y1x2 1

2x2y2

= 1

2x1y2 1

2x2y1=2

ST(x,y)

.

Similarly,

Alt(TS)(x,y) = 12

x2y1+1

2x2y2 1

2y2x1 1

2x2y2

= 12

x1y2+12

x2y1= 2TS(x,y),which gives that ST= T S. Note that ifT=e1, i.e. T(x) = x1, andS=e2,i.e. S(x) =x2, then

TS(x,y) =x1y2x2y1=det(x,y).I


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J 8.9Remark.Some authors use the more logical definition

S

T=Alt(S

T),

which is in accordance with the definition of the symmetric product. This definition

is usually called the alt convention for the wedge product, and our definition is

usually referred to as the determinant convention. For computational purposes the

determinant convention is more appropriate. I

If{e1, ,en}is the standard dual basis for (R

n), then for vectorsa1, ,anRn,

det(a1, ,an) =e1 en(a1, ,an).

Using the multilinearity the more general statement reads

(7) 1 n(a1, ,an) =deti(aj),wherei are co-vectors.

The alternating tensor det = e1 enis called the determinant function onRn.If f : VWis a linear map between vector spaces then the pullback fTr(V)of any alternating tensorT r(W)is given via the relation:

fT(v1, ,vr) =T

f(v1), ,f(vr), f: r(W) r(V).

In particular, f(TS) = (fT)f(S). As a special case we have that if f : VV, linear, and dimV=n, then

(8) fT=det(f)T,for any alternating tensor Tn(V). This can be seen as follows. By multilinearitywe verify the above relation for the vectors{ei}. We have that

fT(e1, ,en) = T(f(e1), ,f(en))

= T(f1, ,fn) =c det(f1, , fn) =c det(f),

where we use the fact thatn(V) = R (see below). On the other handdet(f)T(e1, en) = det(f)c det(e1, ,en)

= c det(f),

which proves (8).

Lemma 8.10. Let{1, ,n}be a basis for V, then the set

B=i1 ir : 1 i1


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Proof: From Theorem 7.5 we know that any alternating tensor T r(V) canbe written as

T=Tj1jrj1 jr.

We have that AltT=T, and so

T=Tj1jrAlt(j1 jr) = 1

r!Tj1jr

j1 jr

In the expansion the terms with jk= j`are zero since jkj` =0. If we order the

indices in increasing order we obtain

T=

1

r! Ti1iri1

ir

,

which show that Bspansr(V).

Linear independence can be proved as follows. Let 0= i1iri1 ir, and

thusi1ir= i1 ir(vi1 , ,vir) =0, which proves linear independence.

It is immediately clear that Bconsists of

n

r

!elements.

As we mentioned before the operation Alt is called the alternating projection.

As a matter of fact Sym is also a projection.

Lemma 8.11. Some of the basic properties can be listed as follows;

(i) SymandAlt are projections on Tr(V), i.e. Sym2 =Sym, andAlt2 =Alt;

(ii) T is symmetric if and only ifSym T=T , and T is alternating if and only

ifAlt T=T ;

(iii) Sym(Tr(V)) = r(V), andAlt(Tr(V)) = r(V);

(iv) SymAlt= Alt Sym= 0, i.e. if T r(V), then SymT =0, and ifT r(V), thenAlt T=0;

(v) let f : VW, thenSym andAlt commute with f : Tr(W) Tr(V), i.e.Sym f= f Sym, andAlt f= f Alt.

9. Tensor bundles and tensor fields

Generalizations of tangent spaces and cotangent spaces are given by the tensor

spaces

Tr(TpM), Ts(TpM), and Tr

s(TpM),


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whereTr(TpM) =Tr(T

pM). As before we can introduce the tensor bundles:

TrM= GpM

Tr(TpM),

TsM=G

pMTs(TpM),

TrsM=G

pMTrs(TpM),

which are called the covariant r-tensor bundle, contravariant s-tensor bundle,

and themixed(r,s)-tensor bundleon M. As for the tangent and cotangent bundle

the tensor bundles are also smooth manifolds. In particular, T1M=TM, andT1M= T M. Recalling the symmetric and alternating tensors as introduced in the

previous section we also define the tensor bundlesrMand rM.

Theorem 9.1. The tensor bundles TrM,TrM and TrsM are smooth vector bundles.

Proof: Using Section 6 the theorem is proved by choosing appropriate local

trivializations. For coordinates x= (p)and x0= (p)we recall that

xi

p

=x0jxi

x0j

p, dx 0j|p=

x0jxi

dx i|p.

For a covariant tensor T TrM this implies the following. The components aredefined in local coordinates by

T=Ti1irdxi1 dx ir, Ti1ir= T

xi1, ,

xir.

The change of coordinatesx x0then gives

Ti1ir= Tx0j1xi1

x0j1, ,

x0jrxir

x0jr

=T0j1jr

x0j1xi1

x0jrxir

.

Define

(Ti1irdxi1 dx ir) = p,(Ti1ir),

then

01p,(T0j1jr)=

p,

T0j1jrx0j1xi1

x0jrxir

!.

The products of the partial derivatives are smooth functions. We can now apply

Theorem 6.4 as we did in Theorems 6.8 and 6.12.

On tensor bundles we also have the natural projection

:TrsMM,defined by(p,T) = p. A smooth sectionin TrsMis a smooth mapping

:M TrsM,


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such that =idM. The space of smooth sections in TrsMis denoted by F

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