manifolds notes

8/10/2019 Manifolds Notes

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Manifolds [7CCMMS18/ CM437Z - Semester 1, 2013]

Lecturer : George Papadopoulos

Department of Mathematics, Kings College London

Strand, London

WC2R 2LS

Email:

Notes most recently edited by Jan Gutowski and George Papadopoulos


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Contents

1. Manifolds 31.1 Elementary Topology and Definitions 3

1.2 Manifolds 4

2. The Tangent Space 112.1 Maps fromMto R 112.2 Tangent Vectors 12

2.3 Curves and their Tangents 17

3. Maps Between Manifolds 21

3.1 Diffeomorphisms 213.2 Push Forward of Tangent Vectors 21

4. Vector Fields 244.1 Vector Fields 24

4.2 Integral Curves and Local Flows 25

4.3 Lie Derivatives 28

5. Tensors 305.1 Co-Tangent Vectors 30

5.2 Pull-back and Lie Derivative of a co-vector 32

5.3 Tensors 34

6. Differential Forms 386.1 Forms 38

6.2 Exterior Derivative 41

6.3 Integration on Manifolds 47

7. Connections, Curvature and Metrics 547.1 Connections, Curvature and Torsion 54

7.2 Riemannian Manifolds 60

7.3 Symplectic Manifolds 657.4 Applications to classical mechanics 67

8. Spheres with different differentiable structures 68

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Recommended Books

C. Isham, Modern Differential Geometry for Physicists, World Scientific, 1989.

M. Nakahara, Geometry, Topology and Physics IOP, 1990.

I. Madsen and J. Tornehave, From Calculus to Cohomology, CUP, 1997. M. Gockler and T. Schuker, Differential Geometry, Gauge Thoeries and Gravity,

CUP, 1987.

S. Kobayashi and K. Nomizu, Foundations of Differential Geometry, vol. I, Wiley,1963.

A useful review of differential geometry (which includes much of the course material,

but also significant amounts of other material) can be found in

http://empg.maths.ed.ac.uk/Activities/GT/EGH.pdf

For an introduction to topological spaces, see

W. A. Sutherland, Introduction to Metric and Topological Spaces, OUP, 2009.

Course information can be found at:

http://keats.kcl.ac.uk/

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1. Manifolds

1.1 Elementary Topology and Definitions

This section should be a review of concepts (hence it is all definitions and no theorems).

Definition 1.1. A topological space X is a set equipped with a collectionU ={Ui} ofsubsets ofX, which are called open sets, that satisfy

(i)Uis closed under finite intersections and arbitrary unions(ii), X U

Definition 1.2. A set is closed if its complement inXis open.

Definition 1.3. An open cover ofX is a collection of open sets whose union isX.

A topology allows us to define notion of locality, convergence and continuity.Local properties are those of open sets, there is a definition of convergence of sequences

in a topological space and that of a continuous function between topological spaces. Some

other important concepts are:

Definition 1.4. X is connected if it is not the union of two disjoint open sets.

Definition 1.5. A map f : X Y between two topological spaces is continuous ifff1(V) ={xX :f(x)V} is open for any open setVY.Definition 1.6. If X Y, where Y is a topological space then X can be made into a

topological space too by considering the induced topology: open sets inXare generated byU XX whereUis an open set ofY.Problem 1.1. Show that the induced topology indeed satisfies the definition of a topology.

Definition 1.7. A Hausdorff space is a topological space with the additional property that

points can be separated: for any two distinct pointsx, yX, there exists open setsUx andUy such that x Ux, y Uy and UxUy =. Hausdorff spaces are also known as T2spaces (as there are also weaker notions of separability).

Non-Hausdorff spaces have various pathologies that we do not want to consider. There-

fore in what follows we will take all topological spaces to be Hausdorff unless otherwise

mentioned.

Definition 1.8. A functionf :X Y is one-to-one ifff(x) =f(y) impliesx= y.This guarantees the existence of a left inverse f1L : f(X)Y X such that f1L

f(x) =x for all xX, since every element in the image f(X) comes from a unique pointin X.

Definition 1.9. A function f : X Y is onto iff f(X) = Y, i.e. for all y Y thereexists anxX such thatf(x) =y.

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This guarantees the existence of a right inversef1R :Y Xsuch that ff1R (y) =yfor all yY, since every element in Y has some xX(not necessarily unique) which ismapped to it by f. Observe that f1R is not a map.

Definition 1.10. A bijection is a map which is both one-to-one and onto.

Definition 1.11. A homeomorphism is a bijection f : X Y which is continuous andwhose inverse is continuous.

Examples:

(X,U) withU={, X} (X,U) withU the set of all subsets ofX

We often have much more structure. For example if there is a notion of distance then

the usual topology is that defined by the open balls.

Definition 1.12. A metric space is a point set X together with a map d : X X Rsuch that

(i) d(x, y) =d(y, x)

(ii) d(x, y)0 with equality iffx= y(iii) d(x, y)d(x, z) + d(z, y)

U is an open subset ofX iff for every x

U there is an >0 such that the open ball

U(x) ={yX :d(x, y)< } (1.1)

is contained in U. Note that it follows that X is open since it is a union of open balls

X=xX

U(x) (1.2)

for any > 0 of your choosing. These spaces are always Hausdorff (property (ii) ensures

that any two distinct points have a finite distance between them and hence open balls with

taken to be half this distance will separate them).

In particular we will heavily use Rn

viewed as a metric topological space (with theusual Pythagorian definition of distance||x y||).

1.2 Manifolds

The use of topology allows one to define the notion of continuity on spaces while a manifold

structure deals with the notion of smoothness. Roughly speaking manifolds are topological

spaces which at suitably chosen open subsets look like Rn.

Definition 1.13.Ann-dimensional chart onMis a pair(U, )whereUis an open subsetofM and: U Rn is a homeomorphism onto its image(U) Rn

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Figure 1: Ann-dimensional chart (U, )

Definition 1.14. A n-dimensional differentiable structure onM is a collection of n-dimensional charts (Ui, i), iI such that

(i)M=iIUi(ii) For any pair of charts Ui, Uj with Ui Uj=, the map j 1i : i(Ui Uj)

j(Ui

Uj) isC

, i.e. all partial derivatives exist up to any order.

(iii) We always take a differentiable structure to be a maximal set of charts, i.e. the union

over all charts which satisfy (i) and (ii).

N. B. The functions j 1i are called transition functions.

Theorem 1.1. IfM is connected then n is well defined, i.e. all charts have the samevalue ofn.

Proof. Suppose that two charts had different values of n then it is clear that they cant

intersect because the map j 1i which takes a subset ofRni to a subset ofRnj is C

smooth and invertible. This in particular requires that the Jacobian ofj 1i is invertible

and so the map preserves the dimension. Since this is true for all charts we see thatMmust split into disjoint charts, at least one for each different value ofn. Since a chart is an

open set we can therefore writeMas a union over disjoint open sets, one for each value ofn. Thus ifMis connected it must have a unique value ofn.

Henceforth we will only consider connected topological spaces.

Definition 1.15. A differentiable manifoldM of dimension n is a connected Hausdorfftopological space equipped with ann-dimensional differentiable structure.

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Figure 2: Transition functions

N.B. One can also study differentiable structures where the transition functions are

only Ck for some k > 0. Alternatively one could replace Rn by Cn and demand that the

transition functions are holomorphic. These are therefore a special case of 2n-dimensional

real manifolds. This leads to the beautiful and rich subject of complex differential geometrywhich we will not have time to consider here.

Example: Trivially Rn is a n-dimensional manifold. A single chart that covers the

whole ofRn is (Rn, id) where id is the identity map id(x) =x

Example: Any open subset U Rn is a n-dimensional manifold. Again the singlechart (U,id) is sufficient. In fact any open subsetU of a manifoldM with charts (Ui, i)is also a manifold since one can use the charts (U Ui, i), where now i is restricted onU Ui.

Remark: Closed subsets may not be manifolds with respect to the induced charts and

transition functions fromM.

Example: The circle is a 1-dimensional manifold.

LetM={(x, y) R2 :x2 + y2 = 1}. We need at least two charts, say

U1 ={(x, y) R2 :x2 + y2 = 1, y > 12}

U2 ={(x, y) R2 :x2 + y2 = 1, y < 12} .

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We define i : Ui R by

1(x, y) =

4

,5

4

where (x, y) = (cos , sin )

2(x, y) = 54

,4

where (x, y) = (cos , sin )

Now

U1 U2={(x, y) R2 :x2 + y2 = 1, 12

< y < 1

2} =VL VR

where

VL={(x, y) R2 :x2 + y2 = 1, 12

< y < 1

2, x 0}

Figure 3: The open sets U1, U2

Now on 1(VL) = (34 ,

54 ), 2 11 () = 2, whereas on 1(VR) = (4 , 4 ),

211 () =. Similarly112 () =+2on 2(VL) = (54 , 34 ), and112 () = on 2(VR) = (4 , 4 ). These maps are C and hence we have a manifold.

Here on the circle we see that locally we can define a single coordinate, which we

think of as an angle. But is not defined over the whole circle, = 0 and = 2 are the

same.

This illustrates a key point: The maps i provide coordinates, just like a map in an

atlas provides coordinates in the form of longitude and latitude. However the coordinates

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will not in general work over the whole of the manifold. For example the surface of the

earth can be mapped in an atlas but the notion of latitude and longitude will break down

somewhere; at the poles longitude is not defined. Maps that one sees hanging on a wall

always break down somewhere (usually at both poles) but an atlas can smoothly cover the

whole earth.Often the Ui are called coordinate neighbourhoods, or patches, and the i are coordi-

nate maps. Ifp M is some point contained in a given patch Ui then the local coordinatesofp are

i(p) =

x1(p), x2(p), x3(p), . . . , xn(p)

Rn (1.3)

Figure 4: Local coordinates

Clearly there is a huge amount of choice of local coordinates. Typically in any given

patch Ui we could choose from an infinite number of different functions i. Furthermore

for a given manifold there will be infinitely many choices of open sets Ui which we use to

cover it with.

Differential geometry is the study of manifolds and uses tensorial objects which take

into account this huge redundancy in the actual way that we may choose to describe a given

manifold. This is the so-called coordinate free approach. Often, especially in older texts,

one fixes a covering and coordinate patches and writes any tensor in terms of its values in

some given local coordinate system. This may be convenient for some calculational purposes

but it obscures the true coordinate independent meaning of the important concepts. In

addition it should always be kept in mind when using explicit coordinates that they are

almost certainly not valid everywhere. One might often need to change coordinates, either

because we prefer to use a different choice of coordinates valid in the same patch, or because

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we need to transform to a new patch which covers a different portion of the manifold. In

this course we will use the coordinate free approach as much as possible.

Example: Let us consider RPn = (Rn+1{0})/ where is the equivalence relation(x1, x2, x3, . . . , xn+1)(x1, x2, x3, . . . , xn+1) (1.4)

for any R{0}. We denote an element of the equivalence class by [x1, x2, x3, . . . , xn+1].Let us choose for the charts

U1 ={[x1, x2, x3, . . . , xn+1] RPn :x1 = 0}U2 ={[x1, x2, x3, . . . , xn+1] RPn :x2 = 0}U3 ={[x1, x2, x3, . . . , xn+1] RPn :x3 = 0}

.

.

.

Un+1 ={[x1, x2, x3, . . . , xn+1] RPn :xn+1 = 0}with the functions

1([x1, x2, x3, . . . , xn+1]) =

x2

x1,x3

x1,x4

x1, . . . ,

xn+1

x1

Rn

2([x1, x2, x3, . . . , xn+1]) =

x1

x2,x3

x2,x4

x2, . . . ,

xn+1

x2

Rn

3([x1, x2, x3, . . . , xn+1]) = x

1

x3

,x2

x3

,x4

x3

, . . . ,xn+1

x3 R

n

.

.

.

n+1([x1, x2, x3, . . . , xn+1]) =

x1

xn+1,

x2

xn+1,

x3

xn+1, . . . ,

xn

xn+1

Rn

Clearlyn+1i=1Ui = RPn and each i is a homeomorphism (without loss of generalitywe can take xi = 1 in Ui). Consider the intersection ofU1 and U2, and consider

(u1, u2, . . . , un)1(U1 U2) (1.5)One must therefore take u1= 0, and hence

11 (u1, u2, u3, . . . , un) = [1, u1, u2, . . . , un] (1.6)

and hence

2 11 (u1, u2, u3, . . . , un) =

1

u1,u2u1

,u3u1

, . . . ,unu1

Rn . (1.7)

Since u1= 0 this map is C on 1(U1 U2). All the other intersections follow the sameway. Thus RPn is an n-dimensional manifold.

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Figure 5: Stereographic projection from US=S2 {(0, 0, 1)} R2

2. The Tangent Space

An important notion in geometry is that of a tangent vector. This is intuitively familiar for

a curve in Rn. But the elementary definition of a tangent vector, or indeed any vector, relies

on special properties ofRn such a fixed coordinate system and its vector space structure.

Once given a vector v= (v1, v2, v3)R3 for example, we can consider the derivative

in the direction ofv :

v1

x1+ v2

x2+ v3

x3 (2.1)

Here we view this expression as an operator acting on functions f : R3 R. Changingcoordinates, for example by performing a rotation, we also must change the coefficients

v1, v2, v3 in an appropriate way however the action on a function remains the same. We

need to generalise the notion of a tangent vector to manifolds in a coordinate free way.

There are several equivalent ways to do this but here we will use the identification of a

vector field with an operator acting on functions.

Our first step is to introduce differentiable functions on manifolds. We will then proceed

to understand vectors as operators on differentiable functions.

2.1 Maps fromM to RDefinition 2.1.A functionf :M R isC iff for each chart(Ui, i)in the differentiablestructure ofM

f 1i :i(Ui) R (2.2)

isC. The set of such functions on a manifoldM is denotedC(M).

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Note that iff 1i isC, and (Uj, j) is another chart with Ui Uj= 0, thenf 1jwill be C on Ui Uj. Thus we need only check thatf 1i is C over a set of chartsthat coversM.

Problem 2.1.Consider the circleS1

as above. Show thatf :S1

Rwithf(x, y) =x2

+yisC.

Definition 2.2.An algebraVis a real vector space along with an operation :V V V such that

(i) v 0= 0 v= 0

(ii) (v) u= v (u) =(v u)

(iii) v (u + w) =v u + v w

(iv) (u + w) v= u v+ w v

for allu, v, w V and R.Theorem 2.1. C(M) is an algebra with addition and multiplication defined pointwise

(f+ g)(p) = f(p) + g(p)

(f)(p) = f(p)

(f g)(p) = f(p)g(p) (2.3)

Proof. Let us show that f g is in C(M). Note that

(f g) 1i = (f 1i ).(g 1i ) (2.4)where f 1i and g1i are C. Therefore their pointwise product is too. Hence,(f g) 1i is C, which is what we needed to show.

The other properties can be shown in a similar manner.

Definition 2.3. For a pointp M, we letC(p) be the set of functions such that(i) f :U R wherep U MandU is an open set.

(ii) f

C(U) (recall that an open subset of a manifold is a manifold)

2.2 Tangent Vectors

We can now state our main definition.

Definition 2.4. A tangent vector at a point p M is a map Xp : C(p) R, whichsatisfies

(i) Xp(f+ g) =Xp(f) + Xp(g)

(ii) Xp(constant f unction) = 0

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(iii) Xp(f g) =f(p)Xp(g) + Xp(f)g(p) (Liebniz rule)

The set of tangent vectors at a pointp Mis called the tangent space toM atp andis denoted byTpM.

The union of all tangent spaces toM

is called the tangent bundle

TM=pMTpM . (2.5)This is an example of a fibre bundle and is itself a2n-dimensional manifold.

N.B.: In general objects which satisfy these properties are called derivations.

N.B.: With this definition a tangent vector is a linear map fromC(p) to R. SinceC(p)

is a vector space (it is an algebra) the tangent vectors are therefore elements of the dual

vector space to C(p). However C(p) and hence its dual are infinite dimensional. The

conditions (i), (ii) and (iii) restrict the possible linear maps that we identify as tangent

vectors and in fact we will see that they become a finite dimensional vector space.

Example: Consider Rn as a manifold with the obvious chart U = Rn, : U Rntaken to be the identity, then

X=n=1

x

is a tangent vector. In fact, we will learn that all tangent vectors have this form.

In what follows we denote

f

x =f,

2f

xx =f etc. (2.6)

wheref is defined on some open set in Rn. We can extend this to manifolds by the following

Definition 2.5. Let (x1, . . . xn) be local coordinates about a point p M. That is thereexists a chart(Ui, i) withpUi andi(q) = (x1(q), x2(q), . . . , xn(q)) for allqUi. Wedefine

x

p

: C(p) R (2.7)by

x p

f=(f 1i )(x1(p), . . . xn(p)) =(f 1i ) i(p) (2.8)

Theorem 2.2. xp is a tangent vector toM atp.

Proof. Letf , gC(p) be defined on an open set U inM that contains p, then

x

p

(f+ g) = ((f+ g) 1i )(x1(p), . . . xn(p))= (f 1i + g 1i )(x1(p), . . . xn(p))= (f 1i )(x1(p), . . . xn(p)) + (g 1i )(x1(p), . . . xn(p))=

xp

(f) +

xp

(g) (2.9)

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It is clear that xp

(constant map) = 0.

Also

x

p(f g) = ((f g) 1i )(x1(p), . . . xn(p))

= ((f 1i ).(g 1i ))(x1(p), . . . xn(p))= (f 1i )(x1(p), . . . xn(p))(g 1i )(x1(p), . . . xn(p))+ (f 1i )(x1(p), . . . xn(p))(g 1i )(x1(p), . . . xn(p))=

x

p

(f)g(p) +

x

p

(g)f(p) (2.10)

We will show that all tangent vectors arise in this way

Example: Consider RP1 = (R2 {0})/ where (x, y)(x, y), = 0.We have two charts

Ut ={(x, y) :x= 0}, t= t([x1, x2]) = x2

x1

Us ={(x, y) :y= 0}, s= s([x1, x2]) = x1

x2

thus on the intersection Us Ut, s = 1/t. Letp = [1, 3] and consider the tangent vector

X :C(p) R, X(f) = t

p

f= ddt

f 1t

(t= 3) . (2.11)

How does Xact in the other coordinate system (where they overlap)? Recall that

d

dt(f(t)) =

ds(t)

dt

d

ds(f(t(s)))

Nows(t) =s 1t and t(s) =t 1s so we have

X(f) = d

dt(f 1t )(t= 3)

= ds(t)

dt

d

ds(f 1t (t 1s ))(s= 13 )

= 1t2

d

ds(f 1s )

s= 13

=19

d

ds(f 1s )

s= 13

. (2.12)

Thus the tangent vector can look rather different, depending on the coordinate system one

chooses. However, its definition as a linear map from C(p) to R is independent of the

choice of coordinates, i.e. (2.11) and (2.12) will agree on any function in C(p).

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Lemma 2.1. Let(x1, . . . xn)be a coordinate system aboutp M. Then for every functionfC(p) there existn functionsf1, . . . f nC(p) such that

f(p) =

x pf (2.13)

and

f(q) =f(p) +

(x(q) x(p))f(q) (2.14)

Proof. To begin, let F =f1i which is defined onV =i(UUi) whereUis the domainoff. Let B be an open ball in V Rn centred on v = i(p), and take yB.

Then

F(y1, . . . , yn) = F(y1, . . . , yn)

F(y1, . . . yn1, vn)

+ F(y1, . . . yn1, vn) F(y1, . . . vn1, vn). . .

+ F(y1, v2, . . . vn) F(v1, . . . , vn)+ F(v1, . . . , vn)

= F(v1, . . . , vn) +

F(y1, . . . , y, v+1, . . . , vn) F(y1, . . . , v, v+1, . . . , vn)

= F(v1, . . . , vn) +

F(y1, . . . , v + t(y v), v+1, . . . , vn)t=1t=0

= F(v1, . . . , vn) +

10

ddtF(y1, . . . , v + t(y v), v+1, . . . , vn)dt

= F(v1, . . . , vn) +

10

F(y1, . . . , y1, v + t(y v), v+1, . . . vn)(y v)dt

(2.15)

If we define

F(y1, . . . yn) =

10

F(y1, . . . , v + t(y v), v+1, . . . , vn)dt (2.16)

then

F(y1, . . . , yn) =F(v1, . . . , vn) +

(y v)F(y1, . . . , yn) (2.17)

Finally, we recall that F =f1i so that if we let (y1, . . . , yn) =i(q) = (x1(q), . . . , xn(q))then this condition can be rewritten as

f 1i i(q) =f 1i i(p) +

(x(q) x(p))F i(q) (2.18)

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and so

f(q) =f(p) +

(x(q) x(p))f(q) (2.19)

where we identify f= F

i.

It also follows that

x

p

f =

x(f 1i ) i(p)

= F

y i(p)

=

y

F(v1, . . . , vn) +

(y v)F(y1, . . . , yn)

y=i(p)

=

F(y

1, . . . , yn) +

(y v)F(y1, . . . , yn)y=i(p)

= F(i(p))= f(p) (2.20)

Theorem 2.3. Tp(M) is ann-dimensional real vector space and a set of basis vectors is

x

p

, = 1, . . . , n

(2.21)

i.e. a general element ofTp(M) can be written as

Xp=

x p (2.22)Proof. First we note that ifXp and Yp are two tangent vectors at a point p Mthen wecan add them or multiply them by a number R:

(i) (Xp+ Yp) :C(p) R, f Xpf+ Ypf

(ii) (Xp) :C(p) R, f (Xpf)

(Convince yourself of this).

Next, we show that the basis elements (2.21) span the vector space of tangent vectors.

To do this, note that the previous lemma implied

Xp(f) =Xp

f(p) +

(x

x

(p))f

(2.23)

NowXp(f(p)) = 0 and Xp(x(p)) = 0 as f(p) and x(p) are constants. Thus we have

Xp(f) =

((x(q) x(p))Xpf+ Xp(x)f(q)q=p

=

Xp(x)f(p)

=

Xp(x)

x

p

f (2.24)

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where we have used (2.20) in the last step. This shows that the elements (2.21) span

Tp(M). We must now show that they are also linearly independent. To this end, supposethat

xp = 0 (2.25)

The coordinate functions are in C(p), so we may consider

0 =

x

p

x =

x(x) =

= (2.26)

Thus = 0 for all .

So why are they called tangent vectors? First consider Rn, and a curveC: (0, 1)

Rn.

Recall from elementary geometry that a tangent vector to a pointp= C(t1) is a line through

p in the direction (i.e. with the slope)dC1

dt

t=t1

, . . . ,dCn

dt

t=t1

(2.27)

where C(t) = (C1(t), . . . , C n(t)) Rn is C.Now iffC(p), then by our definition, Xp: C(p) R defined by

Xp(f) = d

dtf(C(t))t=t1 (2.28)

is a tangent vector to Rn at p= C(t1), i.e. it acts linearly on the function f, vanishes on

constant functions, and satisfies the Liebniz rule. On the other hand, we also have

Xp(f) =dC

dt

t=t1

(f)C(t1)

(2.29)

dC

dt are the components ofXp in the basis x

p

.

2.3 Curves and their Tangents

We can now discuss curves on manifolds and their tangents.

Definition 2.6. Consider an open interval (a, b) R. A map C : (a, b) M to amanifoldM is called a smooth curve onM if i C is C where it is defined for anychart(Ui, i) ofM (i.e. withUi C((a, b))=).

Definition 2.7. For a point t1 (a, b) with C(t1) = p, we can define the tangent vectorTp(C)Tp(M) to the curveC atp by

Tp(C)(f) = d

dtf(C(t))

t=t1

(2.30)

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Figure 6: A smooth curve

It should be clear that Tp(C)Tp(M).Letp Mbe a point on a curve Catt = t1which is covered by a chart (Ui, i). Then

there is some > 0 such that

C((t1 , t1+ ))Ui (2.31)We can express the tangent to C atp as

Tp(C)(f) = d

dt(f(C(t)))

t=t1

= d

dt

(f 1i ) (i C)(t)

t=t1

=n=1

d

dt

(i C)(t))

t=t1

(f 1i )(i(C(t1))) (2.32)

Here we have split fC: (t1, t1+) R as the composition of a function f1i : R

n

Rwith i C : (t1 , t1+ ) Rn, and used the chain rule.

Thus we have

Tp(C)(f) =n=1

d

dt

(i C)(t))

t=t1

x

p

(f) (2.33)

so that

Tp(C) =n=1

d

dt

(i C)(t))

t=t1

x

p

(2.34)

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Conversely, suppose that Tp is a tangent vector toM at p. We will now show thatthere exists a curve through p such that Tp is its tangent atp. Let (x

1, . . . , xn) =i(q) be

local coordinates about p defined on an open set Ui. Hence we may write

Tp =

n=1

T

xp (2.35)

for some numbers T. We define C : (t1 , t1+ ) M by(i C(t)) =x(p) + (t t1)T (2.36)

where we picksufficiently small such that1i (x(p)+(tt1)T)Uifor t (t1, t1+).

It follows that

d

dt(f(C(t)))

t=t1

= d

dt

(f 1i ) ( C)(t)

t=t1

=n

=1

d

dt(i

C)

t=t1(f

1

i

)(i(C(t1)))

=n=1

T

xp

(f) (2.37)

We have therefore shown that all curves through p M define a tangent vector toM at p, and that conversely, all tangent vectors toMat p can be realized as the tangentvector to some curveC. However, this correspondence is not unique. Clearly many distinct

curves may have the same tangent vector at p, and conversely the construction of the curve

through p with tangentTp was not unique. But this does lead to the following theorem:

Theorem 2.4. TpM

is isomorphic to the set of curves throughp

Mmodulo the equiv-

alence relation

C(t) C(t) iff ddt

(f C)t=t1

= d

dt(f C)

t=t1(2.38)

for allfC(p) whereC(t1) = C(t1) =pProof. It is easy to check that the construction above provides a bijection between these

two spaces. The more difficult part , which we wont go into here, is to show that the

vector space structure is preserved. Indeed, to do this we would need to give a vector space

structure to the equivalence class of curves through p.

Theorem 2.5. Let (x1, . . . , xn) =

1, (y1, . . . , yn) =

2 be two coordinate systems at a

pointp M withU1 U2=, and supposeXpTpM. If

Xp =n=1

A

x

p

and Xp=n=1

B

y

p

(2.39)

then

B =n=1

A

xp

(y) (2.40)

wherey(x1, . . . , xn) =2 11 :1(U1 U2) 2(U1 U2) is a smooth function.

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Proof. We have in the second coordinate system that

Xp(y) =

n

=1B

yp

(y) =B (2.41)

but in the first coordinate system we also see that

Xp(y) =

n=1

A

x

p

(y) (2.42)

and as these two expressions must agree, we have proved the theorem.

N.B.: This formula is often simply written as

B =n

=1

Ay

x (2.43)

or even

A =Ax

x (2.44)

with a sum over and a prime denoting quantities in the new coordinate system.

We can see that

x

p

y, and

y px (2.45)

are inverses of each other (when viewed as matrices). To see this note that on interchanging

the coordinate systems, we must also have

A =n=1

B

yp

(x) (2.46)

It follows that

A =n=1

y

p

(x)n=1

A

x

p

(y) =n=1

A n=1

y

p

(x)

x

p

(y)

(2.47)

As this must be true for all possible A, it implies that

n=1

y

p

(x)

x

p

(y) = (2.48)

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3. Maps Between Manifolds

3.1 Diffeomorphisms

Definition 3.1.Suppose thatf :

M N, where

M, (Ui, i), i

I, and

N, (Va, a), a

A

are two manifolds. We say thatf isC iff

a f 1i :i(f1(Va)) Rn (3.1)

isC for alliI , aA (such thatUi f1(Va)=).

Problem 3.1. Show thatf :S1 S1 defined byf(e2i) =e2in isC for anyn.

Theorem 3.1. Suppose thatM,N andP are manifolds withf :M N andg :N PC, theng f :M P is also C.

Proof. This follows from the chain rule.

Definition 3.2. Iff :M N is a bijection with bothf andf1 C thenf is called adiffeomorphism.

Two manifolds which are diffeomorphic, i.e. for which there exists a diffeomorphism

between them, are equivalent geometrically.

Problem 3.2.Show that the charts of two diffeomorphic manifolds are in a 1-1 correspon-

dence.

Problem 3.3. Show that the set of diffeomorphisms from a manifold to itself forms agroup under composition.

3.2 Push Forward of Tangent Vectors

Theorem 3.2. Letf :M N beC, ifXpTpM then

fXf(p) : C(f(p)) R defined by g Xp(g f) (3.2)

is inTf(p)N, i.e. is a tangent vector toN atf(p)

Proof. Letg1, g2

C(f(p)),

R. Then

fXf(p)(g1+ g2) = Xp((g1+ g2) f)= Xp(g1 f+ g2 f)= Xp(g1 f) + Xp(g2 f)= fXf(p)(g1) + fXf(p)(g2) (3.3)

and

fXf(p)() =Xp( f) =Xp() = 0 . (3.4)

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Finally, we have

fXf(p)(g1.g2) = Xp((g1.g2) f)= Xp((g1

f).(g2

f))

= Xp(g1 f)(g2 f)(p) + (g1 f)(p)Xp(g2 f)= fXf(p)(g1)g2(f(p)) + g1(f(p))fXf(p)(g2) (3.5)

Definition 3.3. fXf(p) is called the push forward ofXp.

Figure 7: Push forward of tangent vector

Theorem 3.3. Suppose thatf :M N isC, (x1, . . . , xm) are local coordinates for apointp M and(y1, . . . , yn) are local coordinates for the imagef(p) N. If

Xp =m=1

A

x

p

(3.6)

is inTpM then

fXf(p)=m=1

n=1

A

xp

(y f). y

f(p)

(3.7)

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Proof. LetgC(f(p)). Then

fXf(p)(g) = Xp(g f)

=

m=1 A

xp(g f)

=m=1

A

x

g f 1i ) i(p)

=m=1

A

x

g 1a a f 1i i(p)

=m=1

n=1

A

x

(a f 1i )

(i(p)))

y(g 1a )(a(f(p)))

=

m=1

n=1

A

xp(y

f).

yf(p)(g) (3.8)

Corollory 3.1. The push forward acts linearly on vectors

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4. Vector Fields

4.1 Vector Fields

Next we consider vector fields.

Definition 4.1. A vector field is a map X :M TM such thatX(p) =XpTpM andfor allfC(M) the mapping

p X(f)(p) (4.1)

isC, whereX(f)(p) =Xp(f).

For vector fields we will drop the explicit subscript p. Thus a vector field assigns, in a

smooth way, a vector in TpM to each point p M.N.B.: As we defined it, a vector field is (globally) valid over allM, however it can

also be defined over an open subset U M (i.e. locally).Is the product of two vector fields also a vector field? To check this, take two vector

fields X, Y and f , gC(M). With multiplication of vectors taken to mean (X.Y)(f) =X(Y(f)), we see that

X(Y(f+ g)) = X(Y(f) + Y(g)) =X(Y(f)) + X(Y(g))

X(Y(constant map)) = X(0) = 0 (4.2)

However, we find

X(Y(f.g)) =X(Y(f)g+ f Y(g)) = g.X(Y(f)) + f.X(Y(g))

+ Y(f).X(g) + X(f).Y(g) (4.3)

and this is not a vector field, due to the presence of the terms on the second line. However,

note that the terms on the second line are symmetric in X, Y. Motivated by this, we

construct a vector field by taking

[X, Y](f) =X(Y(f)) Y(X(f)) (4.4)

so that

[X, Y](f.g) = g.X(Y(f)) + f.X(Y(g)) + Y(f).X(g) + X(f).Y(g)

g.Y(X(f)) + f.Y(X(g) + X(f).Y(g) + Y(f).X(g))

= [X, Y](f).g+ f.[X, Y](g) (4.5)

as required for a vector field.

Definition 4.2. [X, Y] is called the commutator of two vector fields

Problem 4.1. What goes wrong if we try to make a vector field using the definition

(X.Y)(f) =X(f).Y(f)?

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Problem 4.2. Show that, if in a particular coordinate system

X=

X(x)

x

p

, Y =

Y(x)

x

p

(4.6)

then

[X, Y] =

XY

YX

xp

(4.7)

Theorem 4.1. With the product of two vector fields defined as the commutator, the space

of vector fields is an algebra

Proof. We have already seen that TpM is a vector space for a particularp M. this ensuresthat the space of vector fields is also a vector space, with addition and scalar multiplication

defined pointwise. We need to check the conditions (i)

(iv) in the definition of an algebra.

Conditions (i), (ii) are obviously satisfied. In addition since [X, Y] =[Y, X], we needonly check that

[X, Y + Z](f) = X(Y + Z)(f) (Y + Z)X(f)= X(Y(f)) + X(Z(f)) Y(X(f)) Z(X(f))= [X, Y](f) + [X, Z](f) (4.8)

as required

Problem 4.3. Show that for three vector fieldsX, Y,Z onMtheJacobi Identityholds:

[X, [Y, Z]] + [Y, [Z, X]] + [Z, [X, Y]] = 0 (4.9)

Problem 4.4. Consider a manifold with a local coordinate system Ui, i = (x1, . . . xn).

InUi we can simply write

x

p

=

x (4.10)

(i) Show that x ,

x

= 0.

(ii) Evaluate x1 , (x1, x2) x2 where(x1, x2) is aC function ofx1, x2.4.2 Integral Curves and Local Flows

Given a vector field X, we can construct curves that pass through p M for which thetangent vector atp is X.

Definition 4.3. LetXbe a vector field onM and consider a point p M. An integralcurve ofXpassing throughp is a curveC(t) inM such thatC(0) =p and

C

d

dt

=XC(t) (4.11)

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for allt in some open interval(, ) R. Here we are viewing ddt as a vector field onR,so that the push forward is defined by

C ddt(f) = d

dt

(f(C(t))) =Tp(C)(f) (4.12)

is just the tangent vector to C(t) atp.

If we introduce a local coordinate system so that in an open set about p M,

X=

X(x)

x

1i (x)

(4.13)

then we find the integral curve is

C d

dt

(f) = d

dtf(C(t))

= d

dt

f 1i i C

(t)

=

dC

dt (t)

x

C(t)

(f) (4.14)

where we have set C = (i C). On the other hand we have

XC(t)(f) =

X(C(t))

x C(t)

(f). (4.15)

Thus we see that the condition for an integral curve is a first order differential equation

for the coordinates of the curve C(t):

d

dtC(t) =X(C(t)) (4.16)

together with the initial condition x(C(0)) = x(p). This is a first order differential

equation, and as such it has (at least locally) a unique solution with the given initial

condition (Picards theorem). However it is by no means clear whether or not the solution

can be extended to all values oft. In particular, even if there is a solution to the differential

equation (4.16) for all t, one must also worry about patching solutions together over thedifferent coordinate patches. This leads to

Definition 4.4. A vector fieldXon M is complete if for every pointp M the integralcurve ofXcan be extended to a curve onM for all values of t

Theorem 4.2. IfMis compact (i.e. all open covers have a finite subcover) then all vectorfields onM are complete.

Proof. We will not prove this

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Let(t, p) be an integral curve of a vector field Xthat passes throughp at t = 0. This

therefore satisfies

d

dt(t, p) =X((t, p)) (4.17)

along with the initial condition (0, p) = p. Thus we have found the following, at least

locally (and globally for complete vector fields):

Definition 4.5. The flow generated by the vector fieldXis a differentiable map

: R M M (4.18)

such that

(i) At each pointp M, the tangent to the curveCp(t) =(t, p) atp isX.

(ii) (0, p) =p

(iii) (t + s, p) =(t, (s, p))

We must show that the third property holds. To do this we dimply note that since

(4.17) is a first order differential equation it has a unique solution for a fixed initial condi-

tion. So consider

d

dt(t + s, p) =

d(t + s, p)

d(t + s) =X((t + s, p)) (4.19)

which satisfies the initial condition(0 + s, p) =(s, p). On the other hand we also have

d

dt(t, (s, p)) =X((t, (s, p))) (4.20)

with the initial condition (0, (s, p)) = (s, p). Thus both (t+ s, p) and (t, (s, p))

satisfy the same equation (4.17) with the same initial condition, and so they must be

equal.

We see that each point p M defines a curve Cp(t) =(t, p) inM whose tangent isX, and such thatCp(0) =p.

Another way of looking at this is that, for each t the flow defines a map t:M M

such thatt+s= t s. Now for t = small, we have, from (4.17),(p) =

(, p) =x(p) + X(p) + O(2) (4.21)

Thus, at least for a sufficiently small value oft, t is 1-1 and C. Hence it is a diffeomor-

phism onto its image (at least for some open set inM). This leads to another definitionDefinition 4.6. A 1-parameter family of diffeomorphisms ofM is a collection of diffeo-morphismst :M M witht R, such that

(i) 0= id where id is the identity map.

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(ii) t s = t+s(iii) t t= id

So in effect, we can think of vector fields as generating infinitessimal diffeomorphisms

through their flows. In this sense, vector fields can be identified with the Lie algebra of thediffeomorphism group.

4.3 Lie Derivatives

A vector field allows us to introduce the notion of a derivative on a manifold. The problem

with the usual derivative,

f

p= lim

0

f(p + ) f(p)

(4.22)

is that we dont know how to add two points on a manifold, i.e. what isp+? However,

we saw that at least locally, a vector field generates a unique integral flow about any givenpoint p. Therefore we can use the flow to take us to a nearby point and hence form a

derivative. This is the notion of a Lie derivative.

Definition 4.7. The X be a vector field onM and f C(M). We define the Liederivative off alongX to be

LXf(p) = lim0

f((, p)) f(p)

(4.23)

where(, p) is the flow generated byX atp.

Theorem 4.3.

LXf =X(f) (4.24)Proof. We have from the definition

LXf(p) = lim0

f((, p)) f((0, p))

= d

dtf((t, p))

t=0

= (t, p)

d

dt

t=0

(f)

= Xp(f) (4.25)

where we have used the defining property of the flow that the tangent at p is Xp.

We can also define the Lie derivative of a vector field Y along X;

Definition 4.8.

LXYp = lim0

Yp Yp

(4.26)

where again () is the flow generated by X, and we have suppressed the dependence on

the pointp.

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Theorem 4.4.

LXY = [X, Y] (4.27)Proof. Let us introduce coordinates (x1, . . . , xn) about p

Msuch that

X=

X(x)

x

1(x)

, Y =

Y(x)

x

1(x)

(4.28)

Recall that

(, p) =x(p) + X(p) + O(2) (4.29)

and note for any f,

Y()(f()) =

Y(())

x()

(f())

=

(Y + XY) x

(f()) + O(

2)

=

(Y + XY)(f() + X

f()) + O(2)

=

Y + XY

(f Xf) + Xf

+ O(2)

=

(Y+ XY

YX)f+ O(2)= =

(Y+ [X, Y]

)f+ O(2) (4.30)

N.B. In the 3rd line, we have suppressed the 1i term (inf 1i ) for convenience.

We also adopt the Einstein summation convention, whereby repeated indices aresummed over.

It follows that

1

()Yp(f) Yp(f)

= (XY

YX)f(p) + O()= [X, Y]p(f) + O() (4.31)

and we are done.

Remark: Sometimes the push forward of a vector field is defined asXp(f) =Xp(f ) (4.32)

ie although Xp is a vector at (p) the subscript of X is the point p rather than

the point (p) that the vector is defined, compare with (3.2). In the above case, the Lie

derivative is defined as

LXYp= lim0

Y(,p) Yp

(4.33)

compare with (4.26). Of course, there is no difference, it is a matter of notation.

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5. Tensors

5.1 Co-Tangent Vectors

Recall that the tangent space Tp

M at a point p

M is a vector space. For any vector

space there is a natural notion of a dual vector space which is defined at the space of linear

maps from the vector space to R.

Problem 5.1. Show that the dual space is a vector space, of the same dimension as the

original vector space (provided it is finite-dimensional).

Thus we have the

Definition 5.1. The co-tangent space toM atp M is the dual vector space to Tp(M)and is denoted byTpM.

In other words, pT

pMiffp: TpM R

is a linear map. We denote the action ofp on a vector XpTpMby

p(Xp) =p, Xp (5.1)

Since p is a linear map we have

p, Xp+ Yp= p(Xp+ Yp) =p(Xp) + p(Yp) =p, Xp + p, Yp (5.2)

Furthermore, as the dual space is a vector space,

p+ p, Xp

= (p+ p)(Xp) =p(Xp) + p(Xp) =

p, Xp

+

p, Xp

(5.3)

Thus, is linear in each of its entries.Now the dual of the dual of a vector space is just the original space itself. To see this,

note that for a fixed vector Xp, we can construct the map

p p(Xp) R (5.4)

The properties of dual vectors ensure that this is a linear map. Thus we can view vectors

Xp as linear maps acting on co-vectors p via

Xp(p) =

p, Xp

(5.5)

Just as for the tangent bundle, we can define the co-tangent bundle to be

TM=pM

TpM (5.6)

which is a 2n-dimensional manifold.

Definition 5.2. A smooth co-vector field is a map :M TM such that(i) pTpM.

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(ii) (X) :M R, defined by (X)(p) = p(Xp) is in CM for all smooth vectorfieldsX

Recall that a chart (Ui, i) defines a natural basis ofTpM for p Ui:

x

p (5.7)

where i(p) = (x1(p), . . . , xn(p)). This allows us to define a natural basis for TpM by

dxp

,

xp= (5.8)

i.e. dxp

is a linear map from TpM to R that maps the vectorn=1

v

xp

(5.9)

to v.

Thus, if in a local coordinate chart we have a vector

V(p) =n=1

V(p)

xp

(5.10)

and a co-vector

(p) =n=1

(p)dxp

(5.11)

Then

(p), V(p) =

n=1 (p)dx

p,

n=1 V

(p)

xp

=n=1

n=1

(p)V(p)dx

p,

x

p

=n=1

n=1

(p)V(p)

=n=1

(p)V(p) (5.12)

Theorem 5.1. Let(x1, . . . , xn) =1 , and(y1, . . . , yn) =2 be two coordinate systems at

a pointp M withp U1 U2, and suppose thatpTpM If

p=n=1

Adxp

, and p =n=1

Bdyp

(5.13)

then

A =n=1

B

xp

y (5.14)

wherey(x1, . . . , xn) =2 11 is a smooth function1(U1 U2) 2(U1 U2).

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Proof. In the first coordinate system

p

x

p

=

n=1

Adxp

,

x

p= A (5.15)

In the second coordinate system, we see that

p

xp

=

n=1

Bdyp

,

xp (5.16)

Also recall that

xp

=

xp

(y)

yp

(5.17)

Thus we find

p x

p =n

=1

B

dyp,

x p(y)

y p

=n=1

B

x

p

(y)

dy

p,

x

p

=n=1

B

xp

y (5.18)

and since these must agree the theorem is proved.

N.B.: This formula is often simply written as

A =n=1

By

x (5.19)

or even

A= Ax

x (5.20)

with a sum over understood and a prime denoting quantities in the new coordinate

system. Note the different positions of the prime and unprimed coordinates as compared

to the analogous formula for a vector.

5.2 Pull-back and Lie Derivative of a co-vector

Suppose we have a smooth mapf :M N. We saw that we could push-forward a vectorXpTpM to a vector fXf(p)Tf(p)N by

fXf(p)(g) =Xp(g f) (5.21)We therefore can consider the dual map f :Tf(p)N TpMdefined by

f(f(p))(Xp) =f(f(p)), Xp=f(p), fXf(p)= f(p)(fXf(p)) (5.22)It is clear that iff(p)Tf(p)N thenf(f(p)) is a linear map TpM R, this follows fromthe linearity of the push-forward f.

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Theorem 5.2. Letf :M N be C, (y1, . . . , yn) be local coordinates onV N, and(x1, . . . , xm) be local coordinates onU f1(V) M (assumingU f1(V)=). If

=n

=1

dyf(p)

(5.23)

then

f=m=1

n=1

x(y f)dx

p (5.24)

Proof. We have

, fXf(p) =n

=1

(f Xf(p))

=n=1

m=1

X

x

p

y f) (5.25)

where we used out earlier result for the components of the push forward of a vector.

However we also have

f, Xp=m=1

(f)X (5.26)

by definition. As these two expressions must be equal for all possible choices ofX, the

result follows.

We can also extend the definition of the Lie derivative to co-vector fields (and in fact

all tensor fields).

Definition 5.3. IfX is a smooth vector field and a smooth co-vector field onM then

LX= lim0

1

()

(5.27)

Theorem 5.3. If, in a local coordinate system(x1, . . . , xn),

X=

X(x)

x

p

, =

(x)dxp

(5.28)

then

LX=

X

+ X

dxp

(5.29)

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Proof. Suppose YTpM, consider

()Y = ()Y()

=

()(Y())

=

+ X

(

Y()) + O(2) (5.30)

where

(Y())

= Y(())

= Y(x + X) + O(2)

= Y + YX + O(2) (5.31)

as a consequence of Theorem 3.3.

Combining these expressions one obtains

()Y =

+ X

Y + YX

+ O(2)

=

Y +

X+ X

Y + O(2) (5.32)

It follows that

lim0

1

() Y = X+ Xdx (Y) (5.33)and the theorem follows.

5.3 Tensors

We can now construct the definition of a (r, s) tensor. First we need to recall the definition

of the tensor product. IfV,W are two vector spaces with basis{vi : i = 1, . . . , n} and{wa: a = 1, . . . , m} respectively, then the vector space sum is an + m-dimensional vectorspace with basis

{vi, wa : i = 1, . . . , n; a= 1, . . . , m

}, i.e.

V W= Spani,a{vi, wa} (5.34)

so that a general element is

ni=1

aivi+ma=1

bawa (5.35)

where the sum is interpreted as a formal sum. HenceV W is an + m-dimensional vectorspace.

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We can also construct the tensor product ofV andW, which is spanned by{vi wa:i= 1, . . . , n; a= 1, . . . , m}, i.e.

V W= Spani,a{vi wa} (5.36)

where vi wa is a formal product. A general element isni=1

ma=1

ciavi wa . (5.37)

and we have the following identities:

(v1+ v2) w = (v1 w) + (v2 w)v (w1+ w2) = (v w1) + (v w2)

(v) w = v (w) =(v w) (5.38)for v , v1, v2

V, w, w1, w2

W and

R.

V W is a nm-dimensional vector space.

Definition 5.4. A (r, s)-tensorT at a pointp M is an element ofT(r,s)p M=

r TpM s TpM (5.39)wherer denotes ther-th tensor product.

It follows that, given a local coordinate system, a local basis for (r, s)-tensors is given

by

x1

p

xr

p

dx1p

dxsp

(5.40)

and if we write

T =

T1...r1...s

x1

p

xr

p

dx1p

dxsp

(5.41)

then the components can be computed as

T1...r1...s =T

dx1

p

, . . . , d xrp

,

x1

p

, . . . ,

xs

p

(5.42)

Problem 5.2. Show that if in some local coordinate system

T =

A1...r1...s

x1

p

xr

p

dx1p

dxsp

(5.43)

is a(r, s)-tensor, and

Xi =

Xi

x

p

, i= 1, . . . , s (5.44)

ares vectors and

I =

Idxp

, I= 1, . . . , r (5.45)

arer co-vectors then

T(1, . . . , r, X1, . . . , X s) =

A1...r1...s11. . .

rrX

11 . . . X

ss (5.46)

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Problem 5.3. Suppose that T is a (r, s)-tensor, (U1, 1 = (x1, . . . , xn)) and (U1, 1 =

(y1, . . . , yn)) are two local coordinate charts withU1 U2=, such that

T = A1...r1...s

x1 p

xr p dx1p dxsp (5.47)and

T =

B1...r1...s

y1

p

yr

p

dy1p

dysp

(5.48)

Then

B1...r1...s =n

1=1

n

s=1

n1=1

n

r=1

A1...r1...s

x1 py1

. . .

xr pyr

y1

p

x1

. . .

ys

p

xs

(5.49)

where y(x) is the -th component of the transition function 2 11 , and x(y) is the-th component of the transition function1 12 .

Definition 5.5. A (r, s)-tensor field is a map

T :M (rTM) (sTM) such that T(p)(rTpM) (sTpM)(5.50)

which is smooth in the sense that for any choice of r smooth co-vector fields 1, . . . , r,

ands smooth vector fieldsV1, . . . , V s, the map T(1, . . . , r, V1, . . . , V s) :M RisC.Some tensor fields have special names.

(i) A (0, 0)-tensor is a scalar; as a field it assigns a number to each point inM(ii) A (1, 0)-tensor is a vector; as a field it assigns a tangent vector to each point in M.

(iii) A (0, 1)-tensor is a 1-form; as a field it assigns a co-vector to each point inM

Sometimes, especially in older books, (r, 0)-tensors are called covariant, and (0, s)

tensors are called contravariant.

A (r, 0) tensor is called symmetric if

T(P(1), . . . , P(r)) =T(1, . . . , r) (5.51)

and similarly a (0, s) tensor is called symmetric if

T(VP(1), . . . , V P(s)) =T(V1, . . . , V s) (5.52)

On the other hand, they are called anti-symmetric if

T(P(1), . . . , P(r)) = sgn(P)T(1, . . . , r) (5.53)

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or

T(VP(1), . . . , V P(s)) = sgn(P)T(V1, . . . , V s) (5.54)

HerePis a permutation, and sgn(P) is its sign. Recall thatPis a bijectionP :{1, . . . , k} {1, . . . , k}, and can always be written in terms of either an even (sgn(P) = 1) or odd(sgn(P) =1) of interchanges (where two neighbouring integers are swapped, and every-thing else is unaltered).

Similarly, one can talk about symmetry properties of the (r, 0) and (0, s) components

of a mixed (r, s)-tensor separately.

It is straightforward to extend the Lie derivative to act on (r, s) tensors by requiring

that

(i)LX(T1+ T2) =LXT1+ LXT2, where T1, T2 are (r, s) tensors(ii)LX(T) =(LXT), where T is a (r, s) tensor and R is constant.

(iii)LX(T1 T2) = (LXT1) T2+ T1 (LXT2) whereT1, T2 are (r, s) and (r, s) tensors.

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6. Differential Forms

N.B.: Conventions about forms can vary from book to book (by various factors

ofp! and minus signs), so be careful when comparing different sources.

6.1 Forms

Definition 6.1. Ap-form on a manifoldMis a smooth anti-symmetric(0, p)-tensor fieldonM. In particular, if is ap-form then

(XP(1), . . . , X P(p)) = sgn(P)(X1, . . . , X p) (6.1)

Definition 6.2. A 0-form onM is a function inC(M).

Theorem 6.1. IfM isn-dimensional then allp-forms withp > n vanish.Proof. First note that a p-form acting on a set of vectors with the same vector appearing

twice vanishes, because

(X1, . . . , Y , X 2, . . . , Y , X 3, . . . ) = sgn(P)(Y , Y , X 1 . . . )

= (1)sgn(P)(Y , Y , X 1, . . . )= (1)(sgn(P))2(X1, . . . , Y , X 2, . . . , Y , X 3, . . . )=(X1, . . . , Y , X 2, . . . , Y , X 3, . . . ) (6.2)

where in the first line we used a permutation P to place the two Y vectors next to eachother, in the second line we used an interchange to swap the two Y-vectors, which introduces

an extra coefficient of1, and in the third line we apply the inverse permutation to P torestore the original order of the vector fields (recall that the sign of P1 is the same as

that ofP) which introduces another factor of sgn(P).

Now, if is a p form with p > n, then in any collection of p basis vectors, at least

two must be the same. So evaluated on any such collection of basis vector must vanish.

Since it vanishes on any set of basis vectors, it must vanish identically.

The space ofp-forms onMis denoted p

(M,R), and we let(M) = 0(M,R) 1(M,R) + + n(M,R) (6.3)

Here we have included an explicit reference to the field R over which the manifold is defined.

Note that if and are ap-form andq-form respectively, then will be a (0, p+q)-tensor field, but not a (p + q)-form, since it will not be antisymmetric.

However, one can construct a (p + q)-form from ap-form and aq-form using thewedge

product:

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Definition 6.3. Ifp(M) andq(M) then

( )(X1, . . . , X p+q) =P

sgn(P)( )(XP(1), . . . , X P(p+q)) (6.4)

where

( )(XP(1), . . . , X P(p+q)) =(XP(1), . . . , X P(p)) . (XP(p+1), . . . , X P(p+q)) (6.5)

Note that is clearly a (0, p+q) tensor, because each term in the above sum is.Also, ifQ is some permutation, then

( )(XQ(1), . . . , X Q(p+q)) =P

sgn(P)( )(XQP(1), . . . , X QP(p+q))

=P

sgn(Q1P)( )(XQQ1P(1), . . . , X QQ1P(p+q))

=P

sgn(Q1P)( )(XP(1), . . . , X P(p+q))

= sgn(Q)P

sgn(P)( )(XP(1), . . . , X P(p+q))

= sgn(Q)( )(X1, . . . , X p+q) (6.6)

where in going from the first to the second line, the permutation P in the sum is replaced

with Q1P(this leaves the sum unaffected). So is indeed a (p + q)-form.Example:

dxp dxp= dxp dxp dxp dxp (6.7)

Example:

dxp

(dxp

dxp

) = dxp

dxp

dxp

dxp

dxp

dxp

+ dxp

dxp

dxp

dxp

dxp

dxp

+ dxp

dxp

dxp

dxp

dxp

dxp

(6.8)

Note: by convention iff0(M,R) and k(M,R) then

f = f=f . (6.9)Theorem 6.2. The wedge product satisfies:

(i) (1+ 2) = (1 ) + (2 )(ii) (1+ 2) = ( 1) + ( 2)

(iii) () = () =( )for, 1, 2p(M), , 1, 2q(M) and R

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Proof. This follows directly from the definition of the wedge product, and is left as an

exercise

Theorem 6.3. Ifp(M) andq(M) then = (1)pq .Proof. LetQ be the permutation that maps the ordered list

{1, . . . , p , p + 1, . . . , p + q} {q+ 1, . . . , q +p, 1, . . . , q } (6.10)Note that sgn(Q) = (1)pq.Then

( )(X1, . . . , X p, Xp+1, . . . , X p+q)=P

sgn(P)( )(XP(1), . . . , X P(p), XP(p+1), . . . , X P(p+q))

=

Psgn(P Q)( )(XPQ(1), . . . , X PQ(p), XPQ(p+1), . . . , X PQ(p+q))

=P

sgn(P)sgn(Q)( )(XP(q+1), . . . X P(q+p), XP(1), . . . X P(q))

= (1)pqP

sgn(P)( )(XP(q+1), . . . X P(q+p), XP(1), . . . X P(q))

= (1)pqP

sgn(P)( )(XP(1), . . . , X P(q), XP(q+1), . . . , X P(p+q))

= (1)pq( )(X1, . . . , X p, Xp+1, . . . , X p+q) (6.11)

Problem 6.1. If

= A12dx1 dx2 + A34dx3 dx4

= B123dx1 dx2 dx3 + B125dx1 dx2 dx5 (6.12)

then what is ? (N.B. here we have dropped the subscriptp for convenience)Theorem 6.4. A basis fork(M) at a pointp M is given by

dx1p

dxkp

(6.13)

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for any permutation P. It then follows that

1

k!

i

1,...kdx1p

dxk p

= 1

k!

i

P

sgn(P)1...kdxP(1)

p

dxP(k)

= 1k!i

P

P(1),...P(k)dxP(1)p dxP(k)

=i

1,...,kdx1p

dxkp

= (6.16)

This shows that dx1p

dxkp

span the space ofk-forms.

Now consider the k-form dx1p dxk

p. This either vanishes, because not all

the i are distinct, or there exists a permutation P of 1, . . . , k such that

dx1p dxkp= sgn(P)dxP(1)p dx

P(k)p (6.17)with P(1)


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Theorem 6.5. Ifk(M) thendk+1(M)Proof. To start with, we shall prove thatd is a (0, k + 1) tensor. Having established that,

we will prove that it is antisymmetric by evaluating its components in a local basis.

First, fix such that 1k + 1. Note that from the construction given in (6.21),one findsd(X1, . . . , X + Y, . . . , X k+1) =d(X1, . . . , X , . . . , X k+1) + d(X1, . . . , Y , . . . , X k+1)

(6.22)

Next, suppose that fC(M).To simplify the expressions in what follows, it will be convenient to denote by{X}i

the (ordered)k-tuple obtained by removingXi from the (ordered) list ofk + 1 vector fields

X1, . . . , X k+1; and by{X}i,j the (ordered) (k 1)-tuple obtained by removing Xi and Xjfrom X1, . . . , X k+1 (i=j).

We can proceed to evaluate

d(X1, . . . , f X , . . . , X k+1) =i=

(1)i+1Xi

f ({X}i)

+ (1)+1f X

({X})

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On adding together the contribution from (6.25) and (6.27), one finds that the terms

involving Xi(f) cancel, and one finds

d(X1, . . . , f X , . . . , X k+1) =f d(X1, . . . , X , . . . , X k+1) (6.28)

Having established linearity, we compute the components ofd in the usual basis of

vectors

d

x1

p

, . . . ,

xk+1

p

=

k+1i=1

(1)i+1 xi

p

x1

p

, . . . ,

xi1

p

,

xi+1

p

, . . . ,

xk+1

p

(6.29)

where we have used the fact that

x p,

xp = 0 (6.30)so that the second term in the definition ofd (6.21) vanishes. So, on defining the compo-

nents ofd as

d1,...,k+1 =d

x1

p

, . . . ,

xk+1

p

(6.31)

one obtains

d1,...,k+1 =k+1i=1

(1)i+1i1,...,i1,i+1,...k+1= (k+ 1)[12,...,k+1] (6.32)

where the square brackets are defined as

Y[1,...k] = 1

k!

P

sgn(P)YP(1),...,P(k) (6.33)

It is clear that the components ofd are antisymmetric.

Note that in this coordinate basis

d = 1

(k+ 1)!(d)1,...,k+1dx1p dx

k+1p=

1

k!

[12,...,k+1]dx

1p

dxk+1p

= 1

k!

12,...,k+1dx

1p

dxk+1p

(6.34)

where the last line follows due to the antisymmetry of the wedge product terms.

Before proceeding further, we shall establish a correspondence between the exterior

derivative acting on the co-ordinate functions (which although not globally defined, are

nevertheless locally defined), and the dual operators.

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In particular we have denoted by dxp

the dual vectors to the tangent vectors xp

at p M, defined in equation (5.8).Suppose that (x1, . . . , xn) is some local co-ordinate system associated with the chart

(Ui, i). Note that x(p) = (i(p))

are C(Ui) functions, and hence we can (locally)

define their exterior derivative dx where

dx(X) =X(x) (6.35)

for any (locally defined) smooth vector field X. Sodx is a (smooth) 1-form locally defined

on the patch Ui. To evaluate this 1-form at p Ui note thatdx

p

(

xp

) =

x

p

(x) = (6.36)

where on the LHS of this expression dxp denotes the evaluation ofdx at p. Fromthe above equation it is clear that dx evaluated at p is identical to the dual vector dx

p

as defined in equation (5.8), and so our choice of notation in section 5 is consistent with

the exterior derivative as defined in this section.

Example: Consider R3. A 0-form f is just a function, and

df=

f dx (6.37)

is just the gradient off. A 1-form = dx has

d =

dx dx= (12 21)dx1 dx2 + (23 32)dx2 dx3 + (31 13)dx1 dx3

(6.38)

whose components are those of curl(). Lastly, for a 2-form = 12

dx dx, we

have

d=1

2

dx

dx dx = (123+ 231+ 312)dx1 dx2 dx3 (6.39)

and these are the components of div() where

= 23dx1 + 31dx

2 + 12dx3 (6.40)

Next we prove the most important property of the exterior derivative

Theorem 6.6. d2 = 0

Proof. Let us choose a coordinate system as previously, so that

d= 1

k!

1,...,kdx

dx1 dxk (6.41)

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Then it follows that

d2= 1

k!

1,...,kdx

dx dx1 dxk (6.42)However this vanishes, because

1,...,k =1,...,k (6.43)

Theorem 6.7. Ifp(M) andq(M), thend( ) = (d) + (1)p d (6.44)

Proof. In components we have

= 1

p!

1...pdx

1 dxp (6.45)

and

= 1

q!

1...qdx

1 dxq (6.46)

and hence

d( ) = 1p!q!

1...p1...q

dx dx1 ... dxp dx1 ... dxq

= 1

p!q!

1...p1...q + 1...p1...q

dx dx1 ... dxp dx1 ... dxq

= 1

p!q! 1...p1...qdx

dx1

...

dxp

dx1

...

dxq

+(1)p 1p!q!

1...p1...qdx

1 ... dxp dx dx1 ... dxq

= (d) + (1)p d (6.47)where the factor of (1)p appears because

dx dx1 dxp = (1)pdx1 dxp dx (6.48)

Let us return to the notion of a pull-back. This can be easily extended to any p-form.

Consider a C map f :M N

between two manifolds, and let be a p-form onN

.

Then we define

(f)p(X1, . . . , X p) =f(p)(fX1, . . . , f Xp) (6.49)

for tangent vectors Xi Tp(M). Clearly, f is antisymmetric, because is, so ifp(N,R) then fp(M,R). We can also define the pull-back of a 0-form or functiong:N R by the rule

fg= g f (6.50)so thatfg:M R.

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Theorem 6.8.Letf :M N,(y1, . . . yn)be local coordinates onV N, and(x1, . . . , xm)local coordinates onU f1(V) M. If

= 1

k!

n

=1

1,...,k(q)dy1

q dyk

q

(6.51)

then

(f)p = 1

k!

m=1

n=1

1,...,k(f(p))

x1(y1 f) . . .

xk(yk f)dx1

p dxk

p

(6.52)

forq N, p M.Proof. It is sufficient to compute

(f)p

x1

p

, . . . ,

xk

p

= f(p)

f

x1

p

, . . . , f

xk

p

(6.53)

We have already proven that

f

xi

p

=i

xi(yi f)

yi

f(p)

(6.54)

and hence on substituting this back into (6.53), one obtains the result.

Theorem 6.9. The exterior derivative and the pull-back commute: d(f

) =f

d.Proof. In a local coordinate system we have proven that

f= 1

k!

1,...,k(f(x))

f1

x1. . .

fk

xkdx1 dxk (6.55)

where we have defined f =y f. Therefore,

df = 1

k!

x

1,...,k(f(x))

f1

x1. . .

fk

xk

dx dx1 dxk

= 1

k! f

x

y1,...,k(f(x))

f1

x1. . .

fk

xkdx dx1 dxk

+ 1

k!

1,...,k

x

f1

x1. . .

fk

xk

dx dx1 dxk

= 1

k!

fx

y1,...,k(f(x))

f1

x1. . .

fk

xkdx dx1 dxk

+ 1

k!

1,...,k

x

f1

x1. . .

fk

xk

dx dx1 dxk

= 1

k!

y

1,...,k(f(x))f

xf1

x1. . .

fk

xkdx dx1 dxk

= fd (6.56)

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Note that line 4 of this expression is obtained by using the chain rule. The contribution

from line 5 vanishes - this is because 2f1

xx1 is symmetric in , 1, whereas dx

dx1 dxk is antisymmetric in , 1; similarly all other contributions from 2f

j

xxj terms

vanish.

There is also a particularly elegant relationship between the exterior derivatived, and

the Lie derivative.

Definition 6.5. Given a vector fieldY, the interior product iY is a map iY : p(M)

p1(M) defined by(iY)(X1, . . . , X p1) =(Y, X1, . . . , X p1) (6.57)

forp(M) and vector fieldsX1, . . . , X p1.

It is clear that if is a p-form then iY is a p 1-form, as it is linear in the Xi, andinterchange of any pairXi, Xjchanges the sign. Note that ifis a 1-form theniY = (Y).

Theorem 6.10. IfX is a vector field and a 1-form then

LX= d(iX) + iXd (6.58)Proof. Working in local co-ordinates x, note that

d(iX) + iXd

=

(X) + Xd

=

()X+ X

+ X( )

=

X+ X

= (LX) (6.59)

In fact, one can show thatLX= d(iX) + iXd for anyp-form , however we shallnot do this here.

6.3 Integration on Manifolds

Let us first recall how we would integrate = p(x, y)dx+ q(x, y)dy along a curve C :[0, 1] R2 in R2. A natural prescription is

Cp(x, y)dx + q(x, y)dy=

10

p(C(t))

dCx

dt + q(C(t))

dCy

dt

dt (6.60)

Here we are thinking ofdx and dy as the infinitessimal change in x and y along the curve

C; (dx, dy) =dCx

dt dt,dCy

dt dt

. We can rewrite this asC

=

10

C (6.61)

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Problem 6.2. Show that

C=

p(C(t))

d

dt(x C) + q(C(t))d

dt(y C)

dt (6.62)

This definition clearly extends to the integral of a 1-form along a curve in an arbitrary

manifold. To define the integral of a general p-form over a manifold, we need to generalize

a curve to a p-dimensional surface.

Definition 6.6. LetIp = [0, 1]p ={(x1, . . . , xp) Rp : 0 x 1} be thep-cube inRp.

(i) A p-simplex onM is a C map C : J M where J is an open set in Rp thatcontainsIp.

(ii) A 0-simplex is a map from{0} M i.e. it is just a point inM.(iii) The support|C| of ap-simplex is the setC(Ip) M.

Next we can consider sums of such surfaces.Definition 6.7. A p-chain onM is a finite formal linear combination of p-simplices onM with real coefficients, i.e. a generalp-chain is

p= r1C1+ + rkCk (6.63)whereri R andCi arep-simplexes. The support of ap-chainp= r1C1+ + rkCk is

|p|=

i: ri=0

|Ci| (6.64)

Definition 6.8. We define the maps(1)i :Ip1 Ip and(0)i :Ip1 Ip by

(1)i (t1, . . . , tp1) = (t1, . . . , ti1, 1, ti, . . . , tp1)

(0)i (t1, . . . , tp1) = (t1, . . . , ti1, 0, ti, . . . , tp1) (6.65)

i.e. these project a(p 1)-cube inRp1 onto the side of ap-cube inRp.The boundary of ap-cube can then be constructed as the sum of all sides weighted with

a plus sign for front sides and a minus sign for the back sides, in other words a (p1)-chain.Example: If we look atI2 then

(0)1 (t) = (0, t)

(0)2 = (t, 0)

(1)1 (t) = (1, t)

(1)2 = (t, 1) (6.66)

As a point set the boundary is

{(1, t) :t I1} {(t, 1) :t I1} {(0, t) :t I1} {(t, 0) :t I1} (6.67)but this does not take into account the fact that some sides are oriented differently to

others. This is achieved by considering the 1-chain

(1)1 (0)1 (1)2 + (0)2 (6.68)

whose support is the point set consisting of the boundary ofI2.

This allows us to define the boundary of a p-simplex inM.

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Figure 8: The (oriented) boundary ofI2

Definition 6.9. IfC is ap-simplex inM then the boundary of C is denoted byC andis defined as the(p 1)-chain

C=

pi=1

(1)i+1(C (1)i C (0)i ) (6.69)

For ap-chain= r1C1+ + rkCk we defineC=

riCi (6.70)

The next theorem summarizes the notion that boundaries have no boundaries.

Theorem 6.11. 2 = 0

Proof. It suffices to show this for a p-simplexC inM. As

C=

pi=1

(1)i+1(C (1)i C

(0)i ) (6.71)

it follows that

2C =

pi=1

(1)i+1(C (1)i ) (C (0)i )

=

p1j=1

pi=1

(1)i+jC (1)i (1)j C (1)i (0)j C (0)i (1)j + C (0)i (0)j (6.72)

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Now, ifj < i then

()i ()j (t1, . . . , tp2) =()i (t1, . . . , , . . . , tp2) = (t1, . . . , , . . . , , . . . , tp2) (6.73)

and also (as j

i

1)

()j ()i1(t1, . . . , tp2) =()j (t1, . . . , , . . . , tp2) = (t1, . . . , , . . . , , . . . , tp2) (6.74)

where the final expression has in the j-th position and in the i-th position. Thus if

j < i then

()i ()j =()j ()i1 (6.75)

This shift ofi i 1 introduces a minus sign into the sum due to the (1)i+j factor.Hence, we see that the first term and the last term in 2Ceach sum to zero, and the

middle two terms together sum to zero.

Finally, we can define the integral of a p-form over a p-chain.

Definition 6.10. LetC be ap-simplex inM and ap-form. ThenC

=

Ip

C (6.76)

where ifC= f(t1, . . . , tp)dt1 dtp, the RHS is understood to mean the usual integral

Ip

C= 10

. . . 10

f(t1

, . . . , tp

)dt1

. . . d tp

(6.77)

If=

riCi is ap-chain then

=i

ri

Ci

(6.78)

Example: Consider the manifold R2 {(0, 0)}, the 1-form

= ydx

x2 + y2 xdy

x2 + y2 (6.79)

and the curve C(t) = (cos(2t), sin(2t)). ThenC

=

10

Cdt

=

10

sin(2t)

d

dt(cos(2t)) cos(2t)d

dt(sin(2t))

dt

=2 10

(sin2(2t) + cos2(2t))dt

=2 (6.80)

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Problem 6.3. Consider the manifoldR2 {(0, 0)}, the 1-form

= ydx

x2 + y2 xdy

x2 + y2 (6.81)

What is C

(6.82)

along the curveC(t) = (2 + cos(2t), 2 + sin(2t)). Next consider the 2-form

=dx dyx2 + y2

(6.83)

What is C

(6.84)

whereC :I2 R2 {(0, 0)} is given byC(t1, t2) = (t1+ 1)(cos(2t2), sin(2t2)).Problem 6.4. Consider the manifoldR2 {(0, 0)} and the 1-form

= ydx

x2 + y2 xdy

x2 + y2 (6.85)

Show thatd= 0. Is there a smooth functionf such that= df?

The exterior derivative has one very important property:

d2 = 0 (6.86)

Thus if = d then it follows that d = 0. This motivates two definitions

Definition 6.11. A p-form is closed ifd= 0. We denote the set of closedp-forms on

M byZp(M,R).Definition 6.12.Ap-form is exact if = d for some(p1)form onM. We denotethe set of exactp-forms onM byBp(M,R).Theorem 6.12. Bp(M,R)Zp(M,R)Proof. If Bp(M,R) then = d for some (p 1)-form , hence d = d2 = 0, soZ

p

(M,R).

Since the space ofp-forms is a vector space over R, we can define the following:

Definition 6.13. Thep-th de Rahm cohomology group Hp(M,R) is the quotient space

Hp(M,R) = Zp(M,R)

Bp(M,R) (6.87)

where two p-forms are viewed as equivalent iff their difference is an exact form. The

dimension ofHp(M,R) is called thep-th Betti numberbp

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Theorem 6.13. IfM andNare two manifolds, andf : M N is a diffeomorphismthenHp(M,R)=Hp(N,R).

Proof. Recall that we proved that the pull-back and exterior derivative commute (Theorem

6.9). Therefore if is a closed p-form onN then f

is a closed form onM:df= fd= 0 . (6.88)

Furthermore, if = d is an exact form onN then

f= fd= d(f) (6.89)

is an exactp-form onM. Similarly closed forms onMare pulled back using f1 to closedforms onNand exact forms onMare pulled back to exact forms onN.

Thus the de Rahm cohomology groups are capable of distinguishing between two dis-tinct manifolds. By distinct, we mean that two manifolds are equivalent if there is a

diffeomorphism between them. Note though that the converse is not true. There are

plenty of examples of inequivalent manifolds that have the same de Rahm cohomology

groups Hk(M,R). The general idea of cohomology can be applied to any operator whichis nilpotent, i.e. whose action squares to zero, and is a central element of modern algebraic

and geometric topology.

We finally arrive at a central theorem in differential geometry

Theorem 6.14 (Stokess Theorem). If(p1)(M,R) and is ap chain then

d=

(6.90)

Proof. By linearity, it suffices to show this is true for p-simplexes C. Recall that as a

consequence of Theorem 6.9 C

d=

Ip

Cd=

Ip

dC (6.91)

and by definition

C

=Ip

C (6.92)

So to prove that these two quantities are equal, it is sufficient to show thatIp

d =

Ip

(6.93)

for any (p 1)-form on Rp. As this condition is linear in it is sufficient to consider

= f(x)dx1 dxp1 (6.94)

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so

d = (1)p1pf dx1 dxp (6.95)

and hence we evaluateIp

d = (1)p1Ip

pf dx1 dxp

= (1)p1 10

dx1 . . . d xp1 10

pf dxp

= (1)p1 10

dx1 . . . d xp1

f(x1, . . . , xp1, 1) f(x1, . . . , xp1, 0) (6.96)On the other hand, the boundary ofIp is

Ip =

pi=1

(1)i+1

((1)i

(0)i )

=

pi=1

(1)i+1

{(x1, . . . , xi1, 1, xi+1, . . . xp) :xj I1}

{(x1, . . . , xi1, 0, xi+1, . . . xp) :xj I1}

(6.97)

Now = f dx1 dxp1 will only have a non-vanishing contribution to the totalintegral on those boundary components with xp constant and x1, . . . , xp1 varying. HenceIp

= (1)p+1{(x1,...,xp1,1:xiI1}

(1)p+1{(x1,...,xp1,0:xiI1}

= (1)p+1 10

dx1 . . . d xp1f(x1, . . . , xp1, 1) (1)p+1 10

dx1 . . . d xp1f(x1, . . . , xp1, 0)

(6.98)

and this establishes the proof.

This is a beautiful generalization of the following well-known result for 1-forms on R:

ba

df =f(b) f(a) (6.99)

In particular, Stokess theorem enables us to see quite explicitly the connection between

the identities d2 = 0 and 2 = 0, because if is a p-chain and is a (p 2)-form then

0 =

d2=

d =

2

= 0 (6.100)

where the first equality (reading from left to right) follows from d2 = 0, the next two

equalities follow from Stokess theorem, and the last equality follows from 2 = 0.

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7. Connections, Curvature and Metrics

So far, everything that we have discussed about manifolds has been intrinsic to the man-

ifold, defined as a topological space with a differentiable structure, and has not required

introducing any additional structures. However, it is very common and intuitive to intro-duce some additional structures.

7.1 Connections, Curvature and Torsion

The first additional structure that we can introduce is that of a connnection. We have

been emphasising that we cannot just differentiate a generic object such as a tensor on a

manifold because we dont know how to construct something like x + where x Mand is a small parameter.

Problem 7.1. Show that if

X=

X x

p

(7.1)

is a vector field expanded in some local coordinate charti(p) = (x1(p), . . . , xn(p)) and we

define

dX=

Xdx

p

x

p

(7.2)

thendXis not a tensor. [N.B. dis to be distinguished from the exterior derivatived; hered is simply an operator defined by the above equation!]However, check that if=

dx

p

is a co-vector (1-form) then

d=

( )dxp

dxp

(7.3)

is a tensor. Hint: consider how things look in a different coordinate chart i(p) =

(y1, . . . , yn).

However we can simply declare that there is a suitable derivative

Definition 7.1.A connection (or more accurately an affine connection) on a manifoldMis an operatorD which assigns to each vector fieldX onM a mappingDX :TM TMsuch that for allX, YTM andfCM:

(i) DX(Y + Z) =DXY + DXZ

(ii) DX+YZ=DXZ+ DYZ

(iii) DfXY =f DXY

(iv) DX(f Y) =X(f)Y + f DXY

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DXY is called the covariant derivative ofY alongX.

N.B. The commutator (Lie derivative) obeys all but condition (iii).

In other words, DXacts as a directional derivative along the direction determined by

X. However the existence of a connection does not follow from the definition of a manifold

but requires us to add it in. In particular a typical manifold can be endowed with infinitely

many different connections.

It is convenient to introduce a new notation. If x1, . . . , xn are local coordinates on

some chart ofM, we let

D = D x

p

(7.4)

It is straightforward to see that DX is entirely determined by its action on a set of basis

vectors, hence we introduce

D

xp

=

xp

(7.5)

and the are known as the connection coefficients. Thus if

X=

X

x

p

, and Y =

Y

x

p

(7.6)

then it follows from the definition ofDX that

DXY = DX

xp

Y

xp

=

XD

Y

x

p

=

X

xp

(Y)

xp

+ YD

xp

=

XY + X

Y

xp

(7.7)

Theorem 7.1. Let be the connection coefficients in a coordinate system(x1, . . . , xn),

then in another overlapping coordinate system(y1, . . . , yn), we have

= x

yx

yy

x+

yx

2x

yy (7.8)

where we think ofx(y) as the transition functions.

Proof. We have seen that the relationship between the standard vector field basis elements

in two overlapping coordinate systems is

x

p

= y(x)

x

y

p

and

y

p

= x(y)

y

x

p

(7.9)

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By definition, we have

D y p

yp

=

yp

(7.10)

However, we also have

D y

p

yp

= D x(y)

y

x

p

x(y)y

xp

= x(y)

y D

x

p

x(y)y

x

p

= x(y)

yx(y)

y

x

p

+ 2x(y)

yy

x

p

=

x(y)

yx(y)

y

x p

+

2x(y)

yy

x p

=

x(y)y

x(y)y

y(x)

x +

2x(y)yy

y(x)x

y

p

On comparing the coefficients of y

p

in these two expressions, one establishes the proof.

Thus the connection coefficients cannot be thought of as the components of a tensor,

because they do not transform in the appropriate way. However, from the connection we

can construct two associated tensors.

Definition 7.2. The torsion tensor is a(1, 2) tensor defined by

T(X , Y , ) =(DXY DYX [X, Y]) (7.11)

forX, YTM, TM.Theorem 7.2. In a local coordinate system, the torsion tensor is

T =

Tdxp

dxp

x

p

(7.12)

where

T=

(7.13)

Proof. We evaluate

T = dx

D

x D

x

= dx

x

x

= (7.14)

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Secondly, we have the curvature (1, 3)-tensor R:

Definition 7.3. The curvature(1, 3) tensor is

R(X ,Y,Z ,) = DX(DYZ) + DY(DXZ) + D[X,Y]Z (7.15)

forX, Y,Z TM andTM.Problem 7.2. Prove thatR is a tensor.

Theorem 7.3. In a local coordinate system the curvature tensor is

R=

Rdx

p

dxp

dxp

x

p

(7.16)

where

R = + + (7.17)

Problem 7.3. Prove this.

The important property of these tensors is that they contain coordinate independent

information. In particular, if a tensor, such as torsion or curvature, vanishes in one coor-

dinate system, then it vanishes in all. This cannot be said of things like the connection

coefficients or other quantities that one might encounter when working with a particular

coordinate system.

Definition 7.4. A vector fieldXis parallel transported along a curveC if

DTCX= 0 (7.18)

at each point onC, whereTCis the tangent to C.

This means that we think ofXas being transported in such a way that it points in

the same direction along the curve. This is possible because we have a connection which

tells us how to compare vectors in the tangent spaces at different points.

We can give a geometric meaning to both the torsion and curvature tensors. First the

torsion: consider an infinitessimal displacement of the coordinate x by a vector X:

Xx =X (7.19)

Then parallel transport this displacement along a direction Y. Parallel transport means

that

0 =Y

X+

X

(7.20)

so that

YX =YX

=

YX (7.21)

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Thus

YXx =2

YX

(7.22)

On changing the order of displacement first along Y and then Xone similarly finds

XYx =2XY (7.23)Therefore the difference between these two is measured by the torsion

[X, Y]x =2

XYT (7.24)

To understand the curvature, we first parallel transport a vector Z around a curve

with tangent Xby an infinitessimal amount. Using the above formula one finds

Z Z

XZ (7.25)

Let us now parallel transport this around Y by an infinitessimal amount:

Z

Z

XZ (Y)(x + X)(Z XZ)= Z

XZ

Y( + X

)(Z

XZ)= Z

(X + Y)Z

2

YXZ

+ 2

Y

XZ + O(3)

(7.26)

If we first transport around Y and then Xwe find

Z Z

(Y + X) 2

XYZ

+ 2

X

YZ + O(3)

(7.27)

Then, with a little work, one sees that the difference between parallel transport along X

then Y, minus the parallel transport along Y and then X is expressed in terms of thecurvature:

[X, Y]Z =2

R

XYZ (7.28)

Figure 9: Parallel transport along great circle arcs on S2

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In the above, the vector field at A can be taken toCby parallel transport A B Cor by A D C. The resulting vector fields at C are notequal.

Figure 10: Parallel transport of vector field Z around a small parallelogram

The connection can be extended to define a covariant derivative on any tensor field.

We start by defining it on a co-vector by

(DX)(Y) =X((Y))

(DX(Y)) (7.29)

for any vector field Y.

Clearly,

(DX)(Y + Z) = (DX)(Y) + (DX)(Z) (7.30)

for vector fields X , Y , Z , and iffC(M) then

(DX)(f Y) = X((f Y)) (DX(f Y))= X(f (Y)) (X(f)Y + f DX(Y))= X(f)(Y) + f X((Y)) X(f)(Y) f (DX(Y))= f(DX)(Y) (7.31)

and hence DX is a co-vector as claimed. In coordinates this is

D=

(7.32)

where we have taken X= xp

, Y = xp

so that

(Y) = and (DXY) = (7.33)

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This implies that

D(dx) =

dx

(7.34)

The extension to a (r, s)-tensor is

DXT(1, . . . , r, Y1, . . . , Y s) = X(T(

1, . . . , r, Y1, . . . , Y s))

i

T(1, . . . , DXi, . . . r, Y1, . . . , Y s)

i

T(1, . . . , r, Y1, . . . , DXYi, . . . , Y s) (7.35)

Problem 7.4. Convince yourself that DXT is a (r, s)-tensor and in a local coordinate

system

DT1,...,r1,...,s = T1,...,r1,...,s

+i

iT1,...,,...,r

1,...,s

i

iT1,...,r

1,...,,...,s (7.36)

7.2 Riemannian Manifolds

Another object that is frequently discussed is a metric. This allows one to measure distances

and angles on manifolds. Again, this is not implicit to a manifold, and typically there are

infinitely many possible metrics for a given manifold. For example, General Relativity is a

theory of gravity which postulates that spacetime is a manifold. The dynamical equationsof General Relativity (Einsteins equations) then determine the metric.

Definition 7.5.A metricg on a manifoldMis a non-degenerate symmetric(0, 2)tensordefined at each pointp M; that is a map gp : TpM TpM R such that

g(X, Y) =g(Y, X), g(X, Y + f Z) =g(X, Y) + f g(X, Z) (7.37)

for X, Y,Z TM, f C(M). We will assume thatg is a smooth (0, 2) tensor field onM.

The non-degeneracy condition is equivalent to requiring that in any local co-ordinate

system(x1, . . . , xn), the components of the metric

g=g

x,

x

(7.38)

form a matrix with nonzero determinant.

Definition 7.6. A Riemannian manifold is a manifold with a positive definite metric

tensor field (positive definite meansg(X, X)0 with equality iffX= 0). If the metric isnot positive definite it is called a pseudo-Riemannian manifold.

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From elementary linear algebra, an inner product allows us to define the lengths and

angles of vectors. Thus, with a (positive definite) metric we can define the length and

angles of tangent vectors. For example, we can define the angle between two intersecting

curves as

arccos

g(T1, T2)g(T1, T1)g(T2, T2)

(7.39)

where T1, T2 are the tangent vectors to the two curves where they intersect. We can also

define the length of a curve C with tangent vector T to beC

g(T, T)d (7.40)

So we simply integrate the length of the tangent vector at each point of t

manifolds notes

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