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Chapter 1

Topology

We start by defining a topological space. A topological space is a set S together with a collection O ofsubsets called open sets such that the following are true:

i) the empty set and S are open, ,S O

ii) the intersection of a finite number of open sets is open; ifU1, U2 O, then U1 U2 O

iii) the union of any number of open sets is open, if Ui O, theniUi O irrespective of the range of i. 2

It is the pair {S,O} which is, precisely speaking, a topologicalspace, or a space with topology. But it is common to refer to S asa topological space which has been given a topology by specifyingO.

Example: S = R, the real line, with the open sets being openintervals ]a, b[ , i.e. the sets {x R | a < x < b} and their unions,plus and R itself. Then (i) above is true by definition.

For two such open sets U1 = ]a1, b1[ and U2 = ]a2, b2[ , we cansuppose a1 < a2. Then if b1 6 a2 , the intersection U1 U2 = O .Otherwise U1 U2 = ]a2, b1[ which is an open interval and thusU1 U2 O. So (ii) is true.

And (iii) is also true by definition. 2Similarly Rn can be given a topology via open rectangles, i.e.

via the sets {(x1, , xn) Rn | ai < xi < bi}. This is called thestandard or usual topology of Rn. The trivial topology on S consists of O = {,S}. 2

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The discrete topology on a set S is defined by O = {A |A S},i.e., O consists of all subsets of S. 2 A set A is closed if its complement in S, also written S\A oras A{, is open. 2

Closed rectangles in Rn are closed sets as are closed balls andsingle point sets.

A set can be neither open nor closed, or both open and closed.In a discrete topology, every set A S is both open and closed,whereas in a trivial topology, any set A 6= or S is neither open norclosed.

The collection C of closed sets in a topological space S satisfythe following:i) the empty set and S are open, ,S C

ii) the union of a finite number of open sets is open; if A1, A2 C ,then A1 A2 C

iii) the intersection of any number of open sets is open, if Ai C ,then

iAi C irrespective of the range of i.

Closed sets can also be used to define a topology. Given a set Swith a collection C of subsets satisfying the above three properties ofclosed sets, we can always define a topology, since the complementsof closed sets are open. (Exercise!) An open neighbourhood of a point P in a topological spaceS is an open set containing P . A neighbourhood of P is a setcontaining an open neighbourhood of P . Neighbourhoods can bedefined for sets as well in a similar fashion. 2

Examples: For a point x R, and for any > 0,]x , x+ [ is an open neighbourhood of x,[x , x+ [ is a neighbourhood of x,{x 6 y

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the same as the usual definition of continuity, which says that f : Rm Rn is continuous at x0 if given > 0, we can alwaysfind a > 0 such that |f(x) f(x0)| < whenever |x x0| < . 2

For the case of functions from a topological space to Rn, thisdefinition says that f : S Rn is continuous at s0 S if given > 0, we canalways find an open neighbourhoood U of s0 such that |f(s)f(s0)| < whenever s U. 2 If a map f : S1 S2 is one-to-one and onto, i.e. a bijection,and both f and f1 are continuous, f is called a homeomorphismand we say that S1 and cs2 are homeomorphic. 2

Proposition: The composition of two continuous maps is a con-tinuous map.

Proof: If f : S1 S2 and g : S2 S3 are continuous maps, andU is some open set in S3, then its pre-image g1(U) is open in S2.So f1(g1(U)), which is the pre-image of that, is open in S1. Thus(g f)1(U) = f1(g1(U)) is open in S1. Thus g f is continuous.2

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Chapter 2

Manifolds

Now that we have the notions of open sets and continuity, we areready to define the fundamental object that will hold our attentionduring this course. A manifold is a topological space which is locally like Rn. 2

That is, every point of a manifold has an open neighbourhoodwith a one-to-one map onto some open set of Rn. More precisely, a topological space M is a smooth n-dimensional manifold if the following are true:

i) We can cover the space with open sets U, i.e. every point ofM lies within some U.

ii) a map : U Rn, where is one-to-one and ontosome open set of Rn. is continuous, 1 is continuous, i.e. V Rn is a homeomorphism for V.(U, ) is called a chart (U is called the domain of thechart). The collection of charts is called an atlas.

iii) In any intersection U U, the maps 1, which arecalled transition functions and take open sets of Rn to opensets of Rn, i.e. 1 : (U U) (U U), aresmooth maps. 2

n is called the dimension of M. 2We have defined smooth manifolds. A more general definition is

that of a Ck manifold, in which the transition functions are Ck, i.e. ktimes differentiable. Smooth means k is large enough for the purposeat hand. In practice, k is taken to be as large as necessary, up toC. We get real analytic manifolds when the transition functions

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are real analytic, i.e. have a Taylor expansion at each point, whichconverges. Smoothness of a manifold is useful because then we cansay unambiguously if a function on the manifold is smooth as we willsee below. A complex analytic manifold is defined similarly by replac-ing Rn with Cn and assuming the transitions functions 1 tobe holomorphic (complex analytic). 2 Given a chart (U, ) for a neighbourhood of some point P, theimage (x1, , xn) Rn of P is called the coordinates of P in thechart (U, ). A chart is also called a local coordinate system.2

In this language, a manifold is a space on which a local coordi-nate system can be defined, and coordinate transformations betweendifferent local coordinate systems are smooth. Often we will suppressU and write only for a chart around some point in a manifold. Wewill always mean a smooth manifold when we mention a manifold.

Examples: Rn (with the usual topology) is a manifold. 2The typical example of a manifold is the sphere. Consider the

sphere Sn as a subset of Rn+1:

(x1)2 + + (xn+1)2 = 1 (2.1)

It is not possible to cover the sphere by a single chart, but it ispossible to do so by two charts.1

For the two charts, we will construct what is called the stereo-graphic projection. It is most convenient to draw this for a circlein the plane, i.e. S1 in R2, for which the equatorial plane is simplyan infinite straight line. Of course the construction works for anySn. consider the equatorial plane defined as the x1 = 0, i.e. theset {(0, x2, , xn+1)}, which is simply Rn when we ignore the firstzero. We will find homeomorphisms from open sets on Sn to opensets on this Rn. Let us start with the north pole N , defined as thepoint (1, 0, , 0).

We draw a straight line from N to any point on the sphere. Ifthat point is in the upper hemisphere (x1 > 0) the line is extended tillit hits the equatorial plane. The point where it hits the plane is the

1The reason that it is not possible to cover the sphere with a single chartis that the sphere is a compact space, and the image of a compact space un-der a continuous map is compact. Since Rn is non-compact, there cannot be ahomeomorphism between Sn and Rn.

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image of the point on the sphere which the line has passed through.For points on the lower hemisphere, the line first passes through theequatorial plane (image point) before reaching the sphere (sourcepoint). Then using similarity of triangles we find (Exercise!) thatthe coordinates on the equatorial plane Rn of the image of a pointon Sn\{N} is given by

N :(x1, x2, , xn+1) 7 ( x2

1 x1 , ,xn+1

1 x1). (2.2)

Similarly, the stereographic projection from the south pole is

S : Sn\{S} Rn,(x1, x2, , xn+1) 7 ( x2

1 + x1, , x

n+1

1 + x1

). (2.3)

If we write

z =

(x2

1 x1 , ,xn+1

1 x1), (2.4)

we find that

|z|2 (

x2

1 x1)2

+ +(xn+1

1 x1)2

=1 (x1)2(1 x1)2 =

1 + x1

1 x1 (2.5)

The overlap between the two sets is the sphere without the poles.Then the transition function between the two projections is

S N : Rn\{0} Rn\{0}, z 7 z|z|2 . (2.6)

These are differentiable functions of z in Rn\{0}. This shows thatthe sphere is an n-dimensional differentiable manifold. 2 A Lie group is a group G which is also a smooth (real analyticfor the cases we will consider) manifold such that group compositionwritten as a map (x, y) 7 xy1 is smooth. 2

Another way of defining a Lie group is to start with an n-parameter continuous group G which is a group that can beparametrized by n (and only n) real continuous variables. n is calledthe dimension of the group, n = dimG. (This is a different defi-nition of the dimension. The parameters are global, but do not ingeneral form a global coordinate system.)

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Then any element of the group can be written as g(a) wherea = (a1, , an) . Since the composition of two elements of G mustbe another element of G, we can write g(a)g(b) = g((a, b)) where = (1, , n) are n functions of a and b. Then for a Lie group,the functions are smooth (real analytic) functions of a and b.

These definitions of a Lie group are equivalent, i.e. define thesame objects, if we are talking about finite dimensional Lie groups.Further, it is sufficient to define them as smooth manifolds if we areinterested only in finite dimensions, because all such groups are alsoreal analytic manifolds. Apparently there is another definition ofa Lie group as a topological group (like n-parameter continuousgroup, but without an a priori restriction on n, in which the compo-sition map (x, y) 7 xy1 is continuous) in which it is always possibleto find an open neighbourhood of the identity which does not containa subgroup.

Any of these definitions makes a Lie group a smooth manifold,an n-dimensional Lie group is an n-dimensional manifold. 2

The phase space of N particles is a 6N -dimensional manifold, 3Ncoordinates and 3N momenta. 2

The Mobius strip is a 2-dimensional manifold. 2The space of functions with some specified properties is of-

ten a manifold. For example, linear combinations of solutions ofSchrodinger equation which vanish outside some region form a man-ifold. 2

Finite dimensional vector spaces are manifolds. 2Infinite dimensional vector spaces with finite norm (e.g. Hilbert

spaces) are manifolds. 2 A connected manifold cannot be written as the disjoint unionof open sets. Alternatively, the only subsets of a connected manifoldwhich are both open and closed are and the manifold itself. 2

SO(3), the group of rotations in three dimensions, is a 3-dimensional connected manifold. O(3), the group of rotations plusreflections in three dimensions, is also a 3-dimensional manifold,but is not connected since it can be written as the disjoint unionSO(3)PSO(3) where P is reflection. 2L+, the group of proper (no space reflection) orthochronous (no

time reflection) Lorentz transformations, is a 6-dimensional con-nected manifold. The full Lorentz group is a 6-dimensional manifold,

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not connected. 2Rotations in three dimensions can be represented by 3 3 real

orthogonal matrices R satisfying RTR = I. Reflection is representedby the matrix P = I. The space of 3 3 real orthogonal matricesis a connected manifold. 2

The space of all n real non-singular matrices is called GL(n,R).This is an n2-dimensional Lie group and connected manifold. 2

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Chapter 3

Tangent vectors

Vectors on a manifold are to be thought in terms tangents to themanifold, which is a generalization of tangents to curves and sur-faces, and will be defined shortly. But a tangent to a curve is likethe velocity of a particle at that point, which of course comes frommotion along the curve, which is its trajectory. And motion meanscomparing things at nearby points along the trajectory. And com-paring functions at nearby points leads to differentiation. So in orderto get to vectors, let us first start with the definitions of these things. A function f : M R is differentiable at a point P M ifin a chart at P, the function f 1 : Rn R is differentiable at(P ). 2

This definition does not depend on the chart. If f 1 isdifferentiable at (P ) in a chart (U, ) at P , the f 1 isdifferentiable at (P ) for any chart (U, ) because

f 1 = (f 1) ( 1) (3.1)

and the transition functions ( 1) are differentiable.This should be thought of as a special case of functions from one

manifold to another. Consider two manifoldsM and N of dimensionm and n, and a mapping f :MN , P 7 Q. Consider local charts(U,) around P and (W,) around Q. Then f 1 is a mapfrom Rm Rn and represents f in these local charts. f is differentiable at P if f 1 is differentiable at (P ).In other words, f is differentiable at P if the coordinates yi = f i(x)of Q are differentiable functions of the coordinates x of P . 2 If f is a bijection (i.e. one-to-one and onto) and f and f1 are

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both differentiable, we say that f is a diffeomorphism and thatMand N are diffeomorphic. 2

In all of these definitions, differentiable can be replaced by Ck orsmooth. Two Lie groups are isomorphic if there is a diffeomorphismbetween them which is also a group homomorphism. 2 A curve in a manifold M is a map of a closed interval Rto M. (This definition can be given also when M is a topologicalspace.) 2

We will take this interval to be I = [0, 1] R. Then a curve is amap : I M. If (0) = P and (1) = P , for some , we say that joins P and P . A manifoldM is connected (actually arcwise connected)1 ifany two points in it can be joined by a continuous curve in M. 2

As for any map, a curve is called smooth iff its image in a chartis smooth in Rn, i.e., iff : I Rn is smooth in Rn.

Note that the definition of a curve implies that it is parametrized.So the same collection of points in M can stand for two differentcurves if they have different parametrizations.

We are now ready to define tangent vectors and the tangent spaceto a manifold. There are different ways of defining tangent vectors.i) Coordinate approach: Vectors are defined to be objects satis-

fying certain transformation rules under a change of chart, i.e.coordinate transformation, (U, ) (U, ).

ii) Derivation approach: A vector is defined as a derivation of func-tions on the manifold. This is thinking of a vector as defininga directional derivative.

iii) Curves approach: A vector tangent to a manifold is tangent toa curve on the manifold.

The approaches are equivalent in the sense that they end up defin-ing the same objects and the same space. We will follow the thirdapproach, or perhaps a mix of the second and the third approaches.Later we will briefly look at the derivation approach more carefullyand compare it with the way we have defined tangent vectors.

Consider a smooth function f : M R. Given a curve :I M, the map f : I R is well-defined, with a well-defined

1It can be shown that an arcwise connected space is connected.

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derivative. The rate of change of f along is written asdf

dt.

Suppose another curve another curve (s) meets (t) at somepoint P , where s = s0 and t = t0, such that

d

dt(f )

P

=d

ds(f )

P

f C(M) (3.2)That is, we are considering a situation where two curves are tangentto each other in geometric and parametric sense. Let us introduce aconvenient notation. In any chart containing the point P, let uswrite (P ) = (x1, , xn). Let us write f = (f 1) ( ),so that the maps are

f 1 : Rn R, x 7 f(x) or f(xi) (3.3) : I Rn, t 7 {xi((t))}. (3.4)

The last are the coordinates of the curve in Rn.Using the chain rule for differentiation, we find

d

dt(f ) = d

dtf(x((t))) =

f

xidxi((t))

dt. (3.5)

Similarly, for the curve we find

d

ds(f ) = d

dsf(x((s))) =

f

xidxi((s))

ds. (3.6)

Since f is arbitrary, we can say that two curves , have the sametangent vector at the point P M (where t = t0 and s = s0) iff

dxi((t))

dt

t=t0

=dxi((s))

ds

s=s0

. (3.7)

We can say that these numbers completely determine the rate ofchange of any function along the curve or at P. So we can definethe tangent to the curve. The tangent vector to a curve at a point P on it is definedas the map

P : C(M) R, f 7 P (f)

d

dt(f )|P . (3.8)

As we have already seen, in a chart with coordinates {xi} we canwrite using chain rule

P (f) =dxi((t))

dt

f

xi

(P )

(3.9)

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The numbersdxi((t))

dt

(P )

are thus the components of P . We will

often write a tangent vector at P as vP without referring to the curveit is tangent to.

We note here that there is another description of tangent vectorsbased on curves. Let us write if and are tangent to eachother at the point P . It is easy to see, using Eq. (3.7) for example,that this relation is transitive, reflexive, and symmetric. In otherwords, is an equivalence relation, for which the equivalence class[] contains all curves tangent to (as well as to one another) at P . A tangent vector at P M is an equivalence class of curvesunder the above equivalence relation. 2

The earlier definition is related to this by saying that if a vectorvP is tangent to some curve at P , i.e. if vP = P , we can writevP = [].

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Chapter 4

Tangent Space

The set of all tangent vectors (to all curves) at some point P M is the tangent space TPM at P. 2

Proposition: TPM is a vector space with the same dimension-ality n as the manifold M.

Proof: We need to show that TPM is a vector space, i.e.XP + YP TPM , (4.1)

aXP TPM , (4.2)XP , YP TPM, a R.

That is, given curves , passing through P such that XP =P , YP = P , we need a curve passing through P such thatP (f) = XP (f) + YP (f)f C(M).

Define : I Rn in some chart around P by = + (P ). Then is a curve in Rn, and

= : I M (4.3)is a curve with the desired property. 2

Note: we cannot define = + P because addition does notmake sense on the right hand side.

The proof of the other part works similarly. (Exercise!)To see that TPM has n basis vectors, we consider a chart with

coordinates xi. Then take n curves k such that

k(t) =(x1(P ), , xk(P ) + t, , xn(P )

), (4.4)

i.e., only the k-th coordinate varies along t. So k is like the axis ofthe k-th coordinate (but only in some open neighbourhood of P ).

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Now denote the tangent vector to k at P by

(

xk

)P

, i.e.,(

xk

)P

f = k(f)P

=d

dt(f k)

P

. (4.5)

This notation makes sense when we remember Eq. (3.9). Using it wecan write

k(f)P

=

(f

xk

)P

f C(M). (4.6)

Note that

(

xk

)P

is notation. We should understand this as(

xk

)P

f =

xk(f 1)

(P )

fxk

(P )

(4.7)

in a chart around P . The

(

xk

)P

are defined only when this chart

is given, but these are vectors on the manifold at P , not on Rn.Let us now show that the tangent space at P has k|P as a basis.

Take any vector vP TPM , which is the tangent vector to somecurve at P . (We may sometimes refer to P as (0) or as t = 0.)Then

vP (f) =d

dt(f )

t=0

(4.8)

=d

dt((f 1)

(P )

( ))t=0

. (4.9)

Note that : I Rn, t 7 (x1((t)), , xn((t))) are the coor-dinates of the curve , so we can use the chain rule of differentiationto write

vP (f) =

xi(f 1)

(P )

d

dt(xi )

t=0

(4.10)

=

xi(f 1)

(P )

vP (xi) . (4.11)

The first factor is exactly as shown in Eq. (4.7), so we can write

vP (f) =

(

xk

)P

fvP (xi) f C(M) (4.12)

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i.e., we can write

vP = viP

(

xk

)P

vP TPM (4.13)

where viP

= vP (xi). Thus the vectors

(xk

)P

span TPM . These areto be thought of as tangents to the coordinate curves in . Thesecan be shown to be linearly independent as well, so

(xk

)P

form a

basis of TPM and viP are the components of vP in that basis.The

(xk

)P

are called coordinate basis vectors and the set{(xk

)P

}is called the coordinate basis.

It can be shown quite easily that for any smooth (actually C1)function f a vector vP defines a derivation f 7 vP (f) , i.e., satisfieslinearity and Leibniz rule,

vP (f + g) = vP (f) + vP (g) (4.14)

vP (fg) = vP (f)g(P ) + f(P )vP (g) (4.15)

f, g C1(M) and R (4.16)

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Chapter 5

Dual space

The dual space T PM of TPM is the space of linear mappings : TPM R. 2

We will write the action of on vP TPM as (vP ) or sometimesas | vP .

Linearity of the mapping means

(uP + avP ) = (uP ) + a(vP ) , (5.1)

uP , vP TPM and a R .

The dual space is a vector space under the operations of vector ad-dition and scalar multiplication defined by

a11 + a22 : vP 7 a11(vP ) + a22(vP ) . (5.2)

The elements of T PM are called dual vectors, covectors,cotangent vectors etc. 2

A dual space can be defined for any vector space V as the space oflinear mappings V R (or V C if V is a complex vector space).

Example:

Vector Dual vectorcolumn vectors row vectorkets | bras |functions linear functionals, etc.2

Given a function on a manifold f : M R , every vector atP produces a number, vP (f) R vP TPM . Thus f defines a

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covector df , given by df(vP ) = vP (f) called the differential orgradient of f . 2

Since vP is linear, so is df ,

df(vP + awP ) = (vP + awP )(f)

= vP (f) + awP (f) (5.3)

vP , wP TPM, a R .Thus df T PM .

Proposition: T PM is also n-dimensional.Proof: Consider a chart with coordinate functions xi . Each xi

is a smooth function xi :M R . then the differentials dxi satisfy

dxi(

xj

)P

=

(

xj

)P

(xi) =

xj(xi 1)

(P )

= ij . (5.4)

The differentials dxi are covectors, as we already know. So wehave constructed n covectors in T PM . Next consider a linear combi-nation of these covectors, = idx

i. If this vanishes, it must vanishon every one of the basis vectors. In other words,

= 0 (

xj

)P

= 0

idxi(

xj

)P

= 0

iij = 0 i.e. j = 0 . (5.5)So the dxi are linearly independent.

Finally, given any covector , consider the covector = (

xi

)P

dxi . Then letting this act on a coordinate basis vector, weget

(

xj

)P

=

(

xj

)P

(

xi

)P

dxi(

xj

)P

=

(

xj

)P

(

xi

)P

ij = 0j (5.6)

So vanishes on all vectors, since the

(

xj

)P

form a basis. Thus

the dxi span T PM , so T PM is n-dimensional.

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Also, as we have just seen, any covector T PM can be writtenas

= idxi where i =

(

xi

)P

, (5.7)

so in particular for = df , we get

i (df)i = df(

xi

)P

=

(f

xi

)(P )

(5.8)

This justifies the name gradient.It is straightforward to calculate the effect of switching to another

overlapping chart, i.e. a coordinate transformation. In a new chart where the coordinates are yi (and the transition functions are thusyi(x)) we can use Eq. (5.8) to write the gradient of yi as

dyi =

(yi

xj

)P

dxj (5.9)

This is the result of coordinate transformations on a basis of covec-tors.

Since

{(

xi

)P

}is the dual basis in TPM to {dxi}, in order

for

{(

yi

)P

}to be the dual basis to {dyi} we must have(

yi

)P

=

(xj

yi

)P

(

xj

)P

(5.10)

These formulae can be generalized to arbitrary bases.Given a vector v, it is not meaningful to talk about its dual, but

given a basis {ea}, we can define its dual basis {a} by a(eb) = ab .We can make a change of bases by a linear transformation,

a 7 a = Aabb , ea 7 ea = (A1)baeb , (5.11)

with A a non-singular matrix, so that a(eb) = ab .

Given a 1-form we can write it in both bases,

= aa = a

a = aAab

a , (5.12)

from which it follows that a = (A1)bab .

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Similarly, if v is a vector, we can write

v = vaea = vaea = v

a(A1)baeb , (5.13)

and it follows that va = Aabvb .

Quantities which transform like a are called covariant, whilethose transforming like va are called contravariant. 2

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Chapter 6

Vector fields

Consider the (disjoint) union of tangent spaces at all points,TM =

PM

TPM . (6.1)

This is called the tangent bundle of M. 2 A vector field v chooses an element of TPM for every P , i.e.v : P 7 v(P ) vP TPM. 2

We will often write v(f)|P = vP (f).Given a chart, v has components vi in the chart,

vP = vi

(

xi

)P

, (vi)P = vP (xi) . (6.2)

The vector field v is smooth if the functions vi = v(xi) aresmooth for any chart (and thus for all charts). 2 A rule that selects a covector from T PM for each P is called aoneform (often written as a 1form). 2 Given a smooth vector field v (actually C1 is sufficient) we candefine an integral curve of v, which is a curve in M such that(t)|P = vP at every P . (One curve need not pass through allP M.) 2

Suppose is an integral curve of a given vector field v, with(0) = P. Then in a chart containing P , we can write

(t) = v ddtxi((t)) = vi(x(t)) , (6.3)

with initial condition xi(0) = xi|P . This is a set of ordinary first orderdifferential equations. If vi are smooth, the theory of differential

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equations guarantees, at least for small t (i.e. locally), the existenceof exactly one solution. The uniqueness of these solutions impliesthat the integral curves of a vector field do not cross.

One use of integral curves is that they can be thought of as co-ordinate lines. Given a smooth vector field v such that v|P 6= 0, itis possible to define a coordinate system {xi} in a neighbourhoodaround P such that v =

xi.

A vector field v is said to be complete if at every point P Mthe integral curve the integral curve (t) of v passing through P canbe extended to all t R . 2

The tangent bundle TM is a product manifold, i.e., a point inTM is an ordered pair (P, v) where P M and v TPM. The topo-logical structure and differential structure are given appropriately. The map pi : TMM, (P, v) 7 P (where v TPM) is calledthe canonical projection (or simply projection). 2 For each P M, the pre-image pi1(P ) is TPM. It is called thefiber over P . Then a vector field can be thought of as a section ofthe tangent bundle. 2

Given a smooth vector field v, we can define an integral curve through any point P by (t) = v , i.e.,

d

dtxi((t)) = vi((t)) v(xi ((t))) , (6.4)

(0) = P . (6.5)

We could also choose (t0) = P.Then in any neighbourhood U of P we also have Q , the integral

curve through Q. So we can define a map : I U M given by(t, Q) = Q(t) where Q(t) satisfies

d

dtxi(Q(t)) = v(x

i(Q(t))

), (6.6)

Q(0) = Q . (6.7)

This defines a map t : U M at each t by t(Q) = (t, Q) =Q(t) , i.e. for given t, t takes a point by a parameter distance t alongthe curve Q(t). This t is called the local flow of v. 2

The local flow has the following properties:

i) 0 is the identity map of U ;

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ii) s t = s+t for all s, t, s+ t U ;iii) each flow is a diffeomorphism with t

1 = t .

The first property is obvious, while the second property followsfrom the uniqueness of integral curves, i.e. of solutions to first orderdifferential equations. Then the integral curve passing through thepoint Q(s) is the same as the integral curve passing through Q, sothat moving a parameter distance t from Q(s) finds the same pointon M as by moving a parameter distance s+ t from Q(0) Q .

A vector field can also be thought of as a map from the spaceof differentiable functions to itself v : C(M) C(M), f 7v(f) , with v(f) :M R, P 7 vP (f) . Often v(f) is called the Liederivative of f along v and denoted vf .

The map v : f 7 v(f) has the following properties:v(f + g) = v(f) + v(g) (6.8)

v(fg) = fv(g) + v(f)g (6.9)

f, g C(M), RThe set of all (real) vector fields V (M) on a manifold M has thestructure of a (real) vector space under vector addition defined by

(u+ v)(f) = u(f) + v(f), u, v V (M), R .(6.10)

It is possible to replace by some function on C(M). If u, vare vector fields onM and is now a smooth function onM, defineu+ v by

(u+ v)P (f) = uP (f) + (P )vP (f) f C(M), P M .(6.11)

This looks like a vector space but actually it is what is called amodule. A ring R is a set or space with addition and multiplicationdefined on it, satisfying (xy)z = x(yz) , x(y + z) = xy + xz , (x +y)z = xz + yz , and two special elements 0 and 1, the additive andmultiplicative identity elements, 0 + x = x + 0 = x , 1x = x1 =x . A module X is an Abelian group under addition, with scalarmultiplication by elements of a ring defined on it.

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A module becomes a vector space when this ring is a commutative division ring, i.e. when the ring multiplication is commu-tative, xy = yx, and an inverse exists for every element except 0.Given a smooth function , in general 1 / C(M), so the spaceof vector fields on M is in general a module, not a vector space.

Given a vector field v, in an open neighbourhood of some P Mand in a chart, and for any f C(M) , we have

v(f)P

= vP (f) = viP

(f

xi

)P

, where viP

= vP (xi) . (6.12)

Thus we can write

v = vi

xiwith vi = v(xi) , (6.13)

as an obvious generalization of vector space expansion to the moduleV (M).

The vi are now the components of the vector field v, and xi

arenow vector fields, which we will call the coordinate vector fields.Note that this is correct only in some open neighbourhood on which achart can be defined. In particular, it may not be possible in generalto define the coordinate vector fields globally, i.e. everywhere onM,and thus the components vi may not be defined globally either.

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Chapter 7

Pull back and push forward

Two important concepts are those of pull back (or pull-back or pull-back) and push forward (or push-forward or pushforward) of mapsbetween manifolds. Given manifolds M1,M2,M3 and maps f : M1 M2 , g :M2 M3 , the pullback of g under f is the map fg :M1 M3defined by

fg = g f . (7.1)

2 So in particular, if M1 and M2 are two manifolds with a mapf : M1 M2 and g : M2 R is a function on M2 , the pullbackof g under f is a function on M1 ,

fg = g f . (7.2)

While this looks utterly trivial at this point, this concept will becomeincreasingly useful later on. Given two manifoldsM1 andM2 with a smooth map f :M1 M2, P 7 Q the pushforward of a vector v TPM1 is a vectorfv TQM2 defined by

fv(g) = v(g f) (7.3)

for all smooth functions g :M2 R . 2Thus we can write

fv(g) = v(fg) . (7.4)

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The pushforward is linear,

f(v1 + v2) = fv1 + fv2 (7.5)f(v) = fv . (7.6)

And if M1,M2,M3 are manifolds with maps f : M1 M2 , g :M2 M3, it follows that

(g f) = gf , i.e.(g f)v = gfv v TPM1 . (7.7)

Remember that we can think of a vector v as an equivalence class ofcurves []. The pushforward of an equivalence class of curves is

fv = f[] = [f ] (7.8)Note that for this pushforward to be defined, we do not need the

original maps to be 1-1 or onto. In particular, the two manifolds mayhave different dimensions.

SupposeM1 andM2 are two manifolds with dimension m and nrespectively. So in the respective tangent spaces TPM1 and TQM2are also of dimension m and n respectively. So for a map f :M1 M2, P 7 Q , the pushforward f will not have an inverse if m 6= n .

Let us find the components of the pushforward fv in terms ofthe components of v for any vector v. Let us in fact consider, givencharts : P 7 (x1, , xm) , : Q 7 (y1, , yn) the pushforwardof the basis vectors.

For the basis vector(xi

)P

, we want the pushforward f(xi

)P,

which is a vector in TQM2 , so we can expand it in the basis(

yi

)Q,

f(

xi

)P

=

(f

(

xi

)P

)( y

)Q

(7.9)

In any coordinate basis, the components of a vector are given by theaction of the vector on the coordinates as in Chap. 4,

vP

= vP (y) (7.10)

Thus we can write(f

(

xi

)P

)= f

(

xi

)P

(y) (7.11)

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But

fv(g) = v(g f) , (7.12)so

f(

xi

)P

(y) =

(

xi

)P

(y f) . (7.13)

But yf are the coordinate functions of the map f , i.e., coordinatesaround the point f(P ) = Q . So we can write y f as y(x) , whichis what we understand by this. Thus(

f(

xi

)P

)=

(

xi

)P

(y f) = y(x)

xi

P

. (7.14)

Because we are talking about derivatives of coordinates, these areactually done in charts around P and Q = f(P ) , so the chart mapsare hidden in this equation. The right hand side is called the Jacobian matrix (of y(x) =y f with respect to xi). Note that since m and n may be unequal,this matrix need not be invertible and a determinant may not bedefined for it. 2

For the basis vectors, we can then write

f(

xi

)P

=y(x)

xi

P

(

y

)f(P )

(7.15)

Since f is linear, we can use this to find the components of (fv)Qfor any vector vP ,

fvP = f[viP

(

xi

)P

]= vi

Pf

(

xi

)P

= viP

y(x)

xi

P

(

y

)f(P )

(7.16)

(fvP ) = viPy(x)

xi

P

. (7.17)

Note that since f is linear, we know that the components of fvshould be linear combinations of the components of v , so we can

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already guess that (fvP ) = Ai v

iP

for some matrix Ai . The matrixis made of first derivatives because vectors are first derivatives.

Another example of the pushforward map is the following. Re-member that tangent vectors are derivatives along curves. SupposevP TPM is the derivative along . Since : I M is a map,we can consider pushforwards under , of derivatives on I . Thus for : I M , t 7 (t) = P , and for some g :M R ,

(d

dt

)t=0

g =d

dt(g )|t=0

= P (g)|t=0 = vP (g) , (7.18)

so

(d

dt

)t=0

= vP (7.19)

We can use this to give another definition of integral curves.Suppose we have a vector field v onM . Then the integral curve of vpassing through P M is a curve : t 7 (t) such that (0) = Pand

(d

dt

)t

= v|(t) (7.20)

for all t in some interval containing P . 2Even though in order to define the pushforward of a vector v

under a map f , we do not need f to be invertible, the pushforwardof a vector field can be defined only if f is both one-to-one and onto.

If f is not one-to-one, different points P and P may have thesame image, f(P ) = Q = f(P ) . Then for the same vector field v wemust have

fv|Q = f(vP ) = f(vP ) , (7.21)

which may not be true. And if f :M N is not onto, fv will bemeaningless outside some region f(M) , so fv will not be a vectorfield on N .

If f is one-to-one and onto, it is a diffeomorphism, in which casevector fields can be pushed forward, by the rule

(fv)f(P ) = f (vP ) . (7.22)

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Chapter 8

Lie brackets

A vector field v is a linear map C(M) C(M) since it is basi-cally a derivation at each point, v : f 7 v(f) . In other words, givena smooth function f , v(f) is a smooth function on M . Suppose weconsider two vector fields u , v . Then u(v(f)) is also a smooth func-tion, linear in f . But is uv u v a vector field? To find out, weconsider

u(v(fg)) = u(fv(g) + v(f)g)

= u(f)v(g) + fu(v(g)) + u(v(f))g + v(f)u(g) . (8.1)

We reorder the terms to write this as

uv(fg) = fuv(g) + uv(f)g + u(f)v(g) + v(f)u(g) , (8.2)

so Leibniz rule is not satisfied by uv . But if we also consider thecombination vu , we get

vu(fg) = f(vu(g) + vu(f)g + v(f)u(g) + u(f)v(g) . (8.3)

Thus

(uv vu)(fg) = f(uv vu)(g) + (uv vu)(f)g , (8.4)which means that the combination

[u , v] := uv vu (8.5)

is a vector field on M , with the product uv signifying successiveoperation on any smooth function on M .

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This combination is called the commutator or Lie bracket ofthe vector fields u and v . 2

In any chart around the point P M , we can write a vectorfield in local coordinates

v(f) = vif

xi, (8.6)

so that

u(v(f)) = uj

xj

(vif

xi

)= uj

vi

xjf

xi+ ujvi

2f

xjxi,

v(u(f)) = vjui

xjf

xi+ ujvi

2f

xjxi. (8.7)

Subtracting, we get

u(v(f)) v(u(f)) = uj vi

xjf

xi vj u

i

xjf

xi, (8.8)

from which we can read off the components of the commutator,

[u , v]i = ujvi

xj vj u

i

xj(8.9)

The commutator is antisymmetric, [u , v] = [v , u] , and satisfiesthe Jacobi identity

[[u , v] , w] + [[v , w] , u] + [[w , u] , v] = 0 . (8.10)

The commutator is useful for the following reason: Once we have a

chart, we can use

{

xi

}as a basis for vector fields in a neighbour-

hood.Any set of n linearly independent vector fields may be chosen as

a basis, but they need not form a coordinate system. In a coordinatesystem, [

xi,

xj

]= 0 , (8.11)

because partial derivatives commute. So n vector fields will form acoordinate system only if they commute, i.e., have vanishing commu-tators with one another. Then the coordinate lines are the integral

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30 Chapter 8. Lie brackets

curves of the vector fields. For analytic manifolds, this condition issufficient as well.

A simple example is the polar coordinate system in R2 . The unitvectors are

er = ex cos + ey sin

e = ex sin + ey cos , (8.12)

with ex =

xand ey =

ybeing the Cartesian coordinate basis

vectors, and

cos =x

r, sin =

y

r, r =

x2 + y2 (8.13)

Using these expressions, it is easy to show that [er , e] 6= 0 , so{er, e} do not form a coordinate basis.

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Chapter 9

Lie algebra

An (real) algebra is a (real) vector space equipped with a bi-linear operation (product) under which the algebra is closed, i.e., foran algebra A

i) x y A x, y Aii) (x+ y) z = x z + y z

x (y + z) = x y + x z x, y, z A , , R .If , are complex numbers and A is a complex vector space, weget a complex algebra. 2 A Lie algebra is an algebra in which the operation isi) antisymmetric, x y = y x , andii) satisfies the Jacobi identity ,

(x y) z + (y z) x+ (z x) y = 0 . (9.1)

2

The Jacobi identity is not really an identity it does not holdfor an arbitrary algebra but it must be satisfied by an algebra forit to be called a Lie algebra.

Example:

i) The space Mn = {alln n matrices} under matrix multipli-cation, A B = AB . This is an associative algebra sincematrix multiplication is associative, (AB)C = A(BC) .

ii) The same space Mn of all n n matrices as above, but now

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32 Chapter 9. Lie algebra

with matrix commutator as the product,

A B = [A ,B] = AB BA . (9.2)

This product is antisymmetric and satisfies Jacobi identity, soMn with this product is a Lie algebra.

iii) The angular momentum algebra in quantum mechanics.If Li are the angular momentum operators with [Li , Lj ] =iijkLk , we can write the elements of this algebra as

L =

{a =

i

iLi|i C}

(9.3)

If a =aiLi and b =

biLi , their product is

a b [a , b] =

aibj [Li , Lj ] = i

ijkaibjLk . (9.4)

This is a Lie algebra because it [a , a] = 0 and the Jacobi iden-tity is satisfied.

iv) The Poisson bracket algebra of a classical dynamical sys-tem consists of functions on the phase space, with the productdefined by the Poisson bracket,

f g = [f , g]P.B. . (9.5)

This is a Lie algebra. As a vector space it is infinite-dimensional.

v) Vector fields on a manifold form a real Lie algebra under thecommutator bracket, since the Jacobi identity is a genuine iden-tity, i.e. automatically satisfied, as we have seen in the previ-ous chapter. This algebra is infinite-dimensional. (It can bethought of as the Lie algebra of the group of diffeomorphisms,Diff(M)) .

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Chapter 10

Local flows

We met local flows and integral curves in Chapter 6. Given a vectorfield v , write its local flow as t . The collection t for t < (for some > 0, or alternatively fort < 1) is a oneparameter group of local diffeomorphisms . 2

Consider the vector field in a neighbourhood U of a point Q M . Since t : U M, Q 7 Q(t) is local diffeomorphism , i.e.diffeomorphism for sufficiently small values of t , we can use t to pushforward vector fields. At some point P we have the curve t(P ) . Wepush forward a vector field at t = to t = 0 and compare with thevector field at t = 0 .

We recall that for a map :M1 M2 the pullback of a functionf C(M2) is defined as

f = f :M1 R , (10.1)

and f C(M1) if is C .The pushforward of a vector vP is defined by

vP (f) = vP (f ) = vP (f) (10.2)vP TPM1 , vP T(P )M2 . (10.3)

If is a diffeomorphism, we can define the pushforward of a vectorfield v by

v(f)|(P ) = v (f )|Pi.e. v(f)|Q = v (f )|1Q

= v (f)|1Q . (10.4)

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We can rewrite this definition in several different ways,

(v)(f) = v (f ) 1= (1) (v (f ))= (1) (v (f)) . (10.5)

If : M1 M2 is not invertible, v is not a vector field onM2 . If 1 exists but is not differentiable, v is not differentiable.But there are some and some v such that v is a differentiablevector field, even if is not invertible or 1 is not differentiable.Then v and v are said to be related. 2

Proposition: Given a diffeomorphism :M1 M2 (say bothC manifolds) the pushforward is an isomorphism on the Liealgebra of vector fields, i.e.

[u , v] = [u , v] . (10.6)

Proof:

[u , v](f) = [u , v] (f ) 1= u (v (f )) 1 u v , (10.7)

while [u , v](f) = u (v (f)) u v= u (v (f) ) 1 u v= u

((v (f ) 1) ) 1 u v

= u (v (f )) 1 u v . (10.8)

2

A vector field v is said to be invariant under a diffeomorphism :MM if v = v , i.e. if (vP ) = v(P ) for all P M . 2

We can write for any f C(M)

(v) (f) =(1

)(v (f))

((v) (f)) = v (f) , v = v . (10.9)

So if v is an invariant vector field, we can write

v = v . (10.10)

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This expresses invariance under , and is satisfied by all differentialoperators invariant under .

Consider a vector field u , and the local flow (or one-parameterdiffeomorphism group) t corresponding to u ,

t(Q) = Q(t) , Q(t) = u(Q(t)) . (10.11)

But for any f C(M) ,

Q(f) =d

dt

(f Q(t)

)=

d

dt(f t(Q))

=d

dt(t (f)) = uQ (t)(f) u(f)

Q(t)

(10.12)

At t = 0 we get the equation

d

dt(t (f))

t=0

= u(f)Q

(10.13)

We can also write

d

dt(t f) (Q) = u(f) (t(Q)) =

tu(f)(Q) . (10.14)

This formula can be used to solve linear partial differential equationsof the form

tf (x, t) =

ni=1

vi(x)

xif(x, t) (10.15)

with initial condition f(x, 0) = g(x) and everything smooth. This isan equation on Rn+1 , so it can be on a chart for a manifold as well.

We can treat vi(x) as components of a vector field v . Then asolution to this equation is

f(x, t) = t g(x) g (t(x)) g t(x) , (10.16)

where t is the flow of v .Proof:

tf(x, t) =

d

dt(t g) = v(f) vi

f

xi, (10.17)

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using Eq. (10.13) . 2Thus the partial differential equation can be solved by finding the

integral curves of v (the flow of v) and then by pushing (also calleddragging) g along those curves. It can be shown, using well-knowntheorems about the uniqueness of solutions to first order partial dif-ferential equations, that this solution is also unique.

Example: Consider the equation in 2+1 dimensions

tf(x, t) = (x y)

(f

x fy

)(10.18)

with initial condition f(x, 0) = x2 + y2 . The corresponding vectorfield is v(x) = (x y,x + y) . The integral curve passing throughthe point P = (x0, y0) is given by the coordinates

(t) = (vx(P )t+ x0, vy(P )t+ y0) , (10.19)

so the integral curve passing through (x, y) in our example is givenby

(t) = ((x y)t+ x, (x+ y)t+ y) (10.20)= t(x, y) ,

the flow of v. So the solution is

f(x, t) = t f(x, 0) = f(x, 0) t(x, y)= [(x y)t+ x]2 + [(x+ y)t+ y]2= (x y)2t2 + x2 + 2(x y)xt+ (x y)2t2 + y2 2(x y)yt= 2(x y)2t2 + (x2 + y2)(1 + 2t) 4xyt . (10.21)

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Chapter 11

Lie derivative

Given some diffeomorphism , we have Eq. (10.5) for pushforwardsand pullbacks,

(v)f =(1

)v ((f)) . (11.1)

We will apply this to the flow t of a vector field u , defined by

d

dt(t f)

t=0

= u(f)Q

. (11.2)

Applying this at t , we gettv(f) =

(1t

)v(t(f)

)= t v

(t(f)

), (11.3)

where we have used the relation 1t = t . Let us differentiate thisequation with t ,

d

dt(tv)(f)

t=0

=d

dtt v

(t(f)

) t=0

(11.4)

On the right hand side, t acts linearly on vectors and v acts linearlyon functions, so we can imagine At =

t v as a kind of linear operator

acting on the function ft =(tf

). Then the right hand side is of

the form

d

dtAtft

t=0

=

(d

dtAt

)ft

t=0

+Atd

dtft

t=0

=

(d

dtt v)ft

t=0

+ At

(d

dtt(f)

)t=0

= u (v(f))t=0 v (u(f))

t=0

= [u , v](f)t=0

. (11.5)

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The things in the numerator are numbers, so they can be comparedat different points, unlike vectors which may be compared only onthe same space. We can also write this as

limt0

tvt(P ) vPt

= [u , v] . (11.6)

This has the look of a derivative, and it can be shown to havethe properties of a derivation on the module of vector fields, appro-priately defined. So the Lie bracket is also called the Lie derivative,and written as

uv = [u , v] . (11.7)

The derivation on functions by a vector field u : C(M) C(M) , f 7 u(f) , can be defined similarly as

u(f) = limt0

t f ft

. (11.8)

So this can also be called the Lie derivative of f with respectto u , and written as uf . 2

Then it is easy to see that

u(fg) = (uf) g + f (ug) ,

and u(f + ag) = uf + aug . (11.9)

So u is a derivation on the space C(M) . Also,

u(v + aw) = uv + auw ,

and u(fv) = (uf) v + fuv f C(M) .(11.10)

So u is a derivation on the module of vector fields. Also, usingJacobi identity, we see that

u(v w) = (uv) w + v (uw) , (11.11)

where v w = [v , w] , so u is a derivation on the Lie algebra ofvector fields.

Lie derivatives are useful in physics because they describe invari-ances. For functions, uf = 0 means

t f = f , so the function does

not change along the flow of u . So the flow of u preserves f , or leavesf invariant.

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If there are two vector fields u and v which leave f invariant,uf = 0 = vf . But we know from the Eq. (11.8), which definesthe Lie derivative of a function that

u+avf = uf + av = 0 a Rand [u ,v]f = [u ,v]f = 0 . (11.12)

So the vector fields which preserve f form a Lie algebra.Similarly, a vector field is invariant under a diffeomorphism if

v = v , as mentioned earlier. Using the flow of u , we find that avector field v is invariant under the flow of u if

tv = v uv = v . (11.13)

So if a vector field w is invariant under the flows of u and v , i.e. ifuw = 0 = vw , we find that

0 = uvw vuw = [u ,v]w . (11.14)Thus again the vector fields leaving w invariant form a Lie algebra. Let us also define the corresponding operations for 1-forms. Aswe mentioned in Chap. 6, a 1form is a section of the cotangentbundle

T M =P

T PM . (11.15)

Alternatively, a 1-form is a smooth linear map from the space ofvector fields on M to the space of smooth functions on M , : v 7 (v) C(M), (u+ av) = (U) + a(v) . (11.16)

A 1-form is a rule that (smoothly) selects a cotangent vector at eachpoint. 2 Given a smooth map M1 M2 (say a diffeomorphism, forconvenience), the pullback is defined by

() (v) = () . (11.17)

We have already seen the gradient 1-form for a function f :M R , which is a linear map from the space of vector fields tofunctions,

df(u+ av) = u(f) + av(f) , (11.18)

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and which can be written as

df =f

xidxi (11.19)

in some chart. 2For an arbitrary 1-form , we can write in a chart and for any

vector field v ,

= idxi , v = vi

xi, (v) = iv

i . (11.20)

All the components i , vi are smooth functions, so is Iv

i . Thespace of 1-forms is a module. Since the function (v) is chart-independent, we can find the components i of in a new chartby noting that

(v) = ivi = iv

i . (11.21)

Note that the notation is somewhat ambiguous here i also runsfrom 1 to n , and the prime actually distinguished the chart, or thecoordinate system, rather than the index i .

If the components of v in the new chart are related to those inthe old one by vi

= Ai

j v

j , it follows that

iAij v

j = jvj iAij = j (11.22)

Since coordinate transformations are invertible, we can multiply bothsides of the last equation by A1 and write

i =(A1

)ji j . (11.23)

For coordinate transformations from a chart {xi} to a chart {xi} ,

Aij =

xi

xj,

(A1

)ji =

xj

xi(11.24)

so vi

=xi

xjvj , i =

xj

xij . (11.25)

We can define the Lie derivative of a 1-form very conveniently bygoing to a chart, and treating the components of 1-forms and vectorfields as functions,

u(v) = u(iv

i)

= uj

xj(iv

i)

= ujixj

vi + ujivi

xj. (11.26)

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But we want to define things such that

u(v) = (u) (v) + (uv) . (11.27)

We already know the left hand side of this equation from Eq. (11.26),and the right hand side can be calculated in a chart as

(u) (v) + (uv) = (u)i vi + i (uv)

i

= (u)i vi + i[u , v]

i

= (u)i vi + i

(ujvi

xj vj u

i

xj

).(11.28)

Equating the right hand side of this with the right hand side ofEq. (11.26), we can write

(u)i = uj ixj

+ juj

xi. (11.29)

These are the components of u in a given chart {xi} .For the sake of convenience, let us write down the Lie deriva-

tives of the coordinate basis vector fields and basis 1-forms. Thecoordinate basis vector corresponding to the i-th coordinate is

v =

xi vj = ji . (11.30)

Putting this into the formula for Lie derivatives, we get

u

xi= [u , v]j

xj

=

(ukvj

xk vk u

j

xk

)

xj

=

(0 ki

uj

xk

)

xj

= (uj

xi

)

xj. (11.31)

Similarly, the 1-form corresponding to the i-th basis coordinate is

dxi = ijdxj , i.e.

(dxi)j

= ij . (11.32)

Using this in the formula Eq. (11.29) we get

udxi = ik

uk

xjdxj =

ui

xjdxj . (11.33)

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There is also a geometric description of the Lie derivative of 1-forms,

u|P = limt01

t

[t|t(P ) P

]=

d

dtt

P

. (11.34)

We will not discuss this in detail, but only mention that it leads tothe same Leibniz rule as in Eq. (11.27), and the same description interms of components as in Eq. (11.29).

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Chapter 12

Tensors

So far, we have defined tangent vectors, cotangent vectors, and alsovector fields and 1-forms. We will now define tensors. We will dothis by starting with the example of a specific type of tensor. A (1, 2) tensor AP at P M is a map

AP : TPM TPM T PM R (12.1)which is linear in every argument. 2

So given two vectors uP , vP and a covector P ,

AP : (uP , vP , P ) 7 AP (uP , vP ;P ) R . (12.2)Suppose {ea}, {a} are bases for TPM, T PM . Write

Acab = AP (ea, eb;c) . (12.3)

Then for arbitrary vectors uP = uaea , vP = v

aea , and covector P =a

a we get using linearity of the tensor map,

AP (uP , vP ;P ) = AP

(uaea, v

beb;cc)

= uavbcAcab . (12.4)

It is a matter of convention whether A as written above shouldbe called a (1, 2) tensor or a (2, 1) tensor, and the convention variesbetween books. So it is best to specify the tensor by writing indicesas there is no confusion about Acab .

A tensor of type (p, q) can be defined in the same way,

Ap,qP

: TPM TPM q times

T PM T PM Rp times (12.5)

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in such a way that the map is linear in every argument. Alternatively, AP is an element of the tensor product space

AP TPM TPM p times

T PM T PM q times (12.6)

We can define the components of this tensor in the same way thatwe did for the (1, 2) tensor. Then a (p, q) tensor has componentswhich can be written as

Aa1apb1bq .

Some special types of (p, q) tensors have special names. A (1, 0)tensor is a linear map AP : T

PM R , so it is a tangent vector. A

(0, 1) tensor is a cotangent vector. A (p, 0) tensor has componentswith p upper indices. It is called a contravariant ptensor. A(0, q) tensor has components with q lower indices. It is called acovariant qtensor. 2

It is possible to add tensors of the same type, but not of differenttypes,

Aa1apb1bq +B

a1apb1bq = (A+B)

a1apb1bq . (12.7)

A tensor field is a rule giving a tensor at each point. 2We can now define the Lie derivative of a tensor field by using

Leibniz rule in a chart. Let us first consider the components of atensor field in a chart. For a (1, 2) tensor field A , the componentsin a chart are

Akij = A(

xi,

xj; dxk) . (12.8)

The components are functions of x in a chart. Thus we can writethis tensor field as

A = Akijdxi dxj

xk, (12.9)

where the indicates a product, in the sense that its action on twovectors and a 1-form is a product of the respective components,(

dxi dxj xk

)(u, v;) = uivjk . (12.10)

Thus we find, in agreement with the earlier definition,

A(u, v;) = Akijuivjk . (12.11)

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Under a change of charts, i.e. coordinate system xi xi , thecomponents of the tensor field change according to

A = Akij dxi dxj

xk= Ak

ij dx

i dxj xk

(12.12)

Since

dxi

=xi

xidxi ,

xi=

xi

xi

xi(12.13)

(i and i are not equal in general), we get

Akij dxi dxj

xk= Ak

ij

xi

xidxi x

j

xjdxj x

k

xk

xk.

(12.14)

Equating components, we can write

Akij = Akij

xi

xixj

xjxk

xk(12.15)

Akij = A

kij

xi

xixj

xjxk

xk. (12.16)

From now on, we will use the notation i for

xiand if for

f

xiunless there is a possibility of confusion. This will save some spaceand make the formulae more readable.

We can calculate the Lie derivative of a tensor field (with respectto a vector field u, say) by using the fact that u is a derivative onthe modules of vector fields and 1-forms, and by assuming Leibnizrule for tensor products. Consider a tensor field

T = Tmnab m n dxa dxb . (12.17)Then

uT = (uTmnab ) m n dxa dxb

+Tmnab (um) n dxa dxb + +Tmnab m n (udxa) dxb ,+

(12.18)

where the dots stand for the terms involving all the remaining up-per and lower indices. Since the components of a tensor field arefunctions on the manifold, we have

uTmnab = u

iiTmnab , (12.19)

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and we also know that

um = ui

xmi , udx

a =ua

xidxi . (12.20)

Putting these into the expression for the Lie derivative for T andrelabeling the dummy indices, we find the components of the Liederivative,

(uT )mnab = u

i iTmnab

T inab ium Tmiab iun+Tmnib au

i + + Tmnai bui . (12.21)

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Chapter 13

Differential forms

There is a special class of tensor fields, which is so useful as to havea separate treatment. There are called differential pforms orpforms for short. A pform is a (0, p) tensor which is completely antisymmetric,i.e., given vector fields v1 , , vp , (v1 , , vi , , vj , , vp) = (v1 , , vj , , vi , , vp)

(13.1)for any pair i, j . 2

A 0-form is defined to be a function, i.e. an element of C(M) ,and a 1-form is as defined earlier.

The antisymmetry of any p-form implies that it will give a non-zero result only when the p vectors are linearly independent. On theother hand, no more than n vectors can be linearly independent inan n-dimensional manifold. So p 6 n .

Consider a 2-form A . Given any two vector fields v1 , v2 , we haveA(v1 , v2) = A(v2 , v1) . Then the components of A in a chart are

Aij = A (i , j) = Aji . (13.2)Similarly, for a p-form , the components are i1ip , and compo-nents are multiplied by (1) whenever any two indices are inter-changed.

It follows that a p-form has

(n

p

)independent components in n-

dimensions.Any 1-form produces a function when acting on a vector field. So

given a pair of 1-forms A,B, it is possible to construct a 2-form

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48 Chapter 13. Differential forms

by defining

(u, v) = A(u)B(v)B(u)A(v), u, v . (13.3)

This is usually written as = ABBA , where is calledthe outer product. 2 Then the above construction defines a product written as

= A B = B A , (13.4)

and called the wedge product . Clearly, is a 2-form. 2Let us work in a coordinate basis, but the results we find can be

generalized to any basis. The coordinate bases for the vector fields,{i} , and 1-forms, {dxi} , satisfy dxi(j) = ij . A 1-form A can bewritten as A = Aidx

i , and a vector field v can be written as v = vii ,so that A(v) = Aiv

i . Then for the defined above and for any pairof vector fields u, v,

(u, v) = A(u)B(v)B(u)A(v)= Aiu

iBjvj BiuiAjvj

= (AiBj BiAj)uivj . (13.5)

The components of are ij = (i , j) , so that

(u, v) = (uii , vjj) = iju

ivj . (13.6)

Then ij = AiBj BiAj for the 2-form defined above. We can nowconstruct a basis for 2-forms, which we write as dxi dxj ,

dxi dxj = dxi dxj dxj dxi . (13.7)

Then a 2-form can be expanded in this basis as

=1

2!ijdx

i dxj , (13.8)

because then

(u, v) =1

2!ij(dxi dxj dxj dxi) (u, v)

=1

2!ij(uivj ujvi) = ijuivj . (13.9)

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Similarly, a basis for pforms is

dxi1 dxip = dx[i1 dxip] , (13.10)

where the square brackets stand for total antisymmetrization: alleven permutations of the indices are added and all the odd permu-tations are subtracted. (Caution: some books define the squarebrackets as antisymmetrization with a factor 1/p! .) For example,for a 3-form, a basis is

dxi dxj dxk = dxi dxj dxk dxj dxi dxk+dxj dxk dxi dxk dxj dxi+dxk dxi dxj dxi dxk dxj . (13.11)

Then an arbitrary 3-form can be written as

=1

3!ijkdx

i dxj dxk . (13.12)

Note that there is a sum over indices, so that the factorial goes away ifwe write each basis 3-form up to permutations, i.e. treating differentpermutations as equivalent. Thus a pform can be written interms of its components as

=1

p!i1ip dx

i1 dxip . (13.13)

Examples: A 2-form in two dimensions can be written as

=1

2!ij dx

i dxj

=1

2!

(12dx

1 dx2 + 21dx2 dx1)

=1

2!(12 21) dx1 dx2

= 12 dx1 dx2 . (13.14)

2

A 2-form in three dimensions can be written as

=1

2!ijdx

i dxj

= 12 dx1 dx2 + 23 dx2 dx3 + 31 dx3 dx1 (13.15)

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2

In three dimensions, consider two 1-forms = idxi , = idx

i .Then

= (ij ji) 12!dxi dxj

= ijdxi dxj

= (12 21) dx1 dx2+ (23 32) dx2 dx3+ (31 13) dx3 dx1 . (13.16)

The components are like the cross product of vectors in three dimen-sions. So we can think of the wedge product as a generalization ofthe cross product. We can also define the wedge product of a pform and aqform as a (p + q)form satisfying, for any p + q vector fieldsv1, , vp+q , (v1, , vp+q) = 1

p!q!

P

(1)degP (P (v1, , vp+q)) .(13.17)

Here P stands for a permutation of the vector fields, and degP is 0 or1 for even and odd permutations, respectively. In the outer producton the right hand side, acts on the first p vector fields in a givenpermutation P , and acts on the remaining q vector fields. 2

The wedge product above can also be defined in terms of thecomponents of and in a chart as follows.

=1

p!i1ip dx

i1 dxip

=1

q!j1jq dx

j1 dxjq

= 1p!q!

i1ip j1jq(dxi1 dxip) (dxj1 dxjq) .

(13.18)

Note that = 0 if p + q > n , and that a term in which some iis equal to some j must vanish because of the antisymmetry of thewedge product.

It can be shown by explicit calculation that wedge products areassociative,

( ) = ( ) . (13.19)

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Cross-products are not associative, so there is a distinction betweencross-products and wedge products. In fact, for 1-forms in threedimensions, the above equation is analogous to the identity for thetriple product of vectors,

a (b c) = (a b) c . (13.20)For a p-form and q-form , we find

= (1)pq . (13.21)Proof: Consider the wedge product written in terms of the com-ponents. We can ignore the parentheses separating the basis formssince the wedge product is associative. Then we exchange the basis1-forms. One exchange gives a factor of 1 ,

dxip dxj1 = dxj1 dxip . (13.22)Continuing this process, we get

dxi1 dxip dxj1 dxjq= (1)pdxj1 dxi1 dxip dxj2 dxjq= = (1)pqdxj1 dxjq dxi1 dxip . (13.23)

Putting back the components, we find

= (1)pq (13.24)as wanted. 2 The wedge product defines an algebra on the space of differentialforms. It is called a graded commutative algebra . 2 Given a vector field v , we can define its contraction with ap-form by

v = (v, ) (13.25)with p1 empty slots. This is a (p1)-form. Note that the positionof v only affects the sign of the contracted form. 2

Example: Consider a 2-form made of the wedge product of two1-forms, = = . Then contraction by v gives

v = (v, ) = (v) (v) = ( , v) . (13.26)

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If we have a p-form = 1p!i1ip dxi1 dxip , its contraction

with a vector field v = vii is

v =1

(p 1)! ii2ipvidxi2 dxip . (13.27)

Note the sum over indices. To see how the factor becomes 1(p1)! , wewrite the contraction as

v =1

p!i1ipdx

i1 dxip (vii) . (13.28)Since the contraction is done in the first slot, so we consider theaction of each basis 1-form dxik on i by carrying dx

ik to the firstposition and then writing a iki . This gives a factor of (1) for eachexchange, but we get the same factor by rearranging the indices of , thus getting a +1 for each index. This leads to an overall factorof p . given a diffeomorphism : M1 M2 , the pullback of a 1-form (on M2) is , defined by

(v) = (v) (13.29)

for any vector field v on M1 . 2Then we can consider the pullback dxi of a basis 1-form dxi .

For a general 1-form = idxi , we have = (idxi) . But

(v) = (v) = i dxi(v) . (13.30)

Now, dxi(v) = dxi(v) and the thing on the right hand side is afunction on M1 , so we can write this as

(v) = (i)dxi(v) , (13.31)

where i are now functions on M1 , i.e.(i)|P = i|(P ) (13.32)

So we can write = (i)dxi . For the wedge product of two1-forms,

( )(u, v) = ( )(u , v)= (u , v) (u , v)= (u)(v) (u)(v)= (u)(v) (u)(v)= ( )(u , v) . (13.33)

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Since u, v are arbitrary vector fields it follows that

( ) = (dxi dxj) = dxi dxj . (13.34)

Since the wedge product is associative, we can write (by assumingan obvious generalization of the above formula)

(dxi dxj dxk

)=

((dxi dxj) dxk)

= (dxi dxj) dxk

= dxi dxj dxk , (13.35)

and we can continue this for any number of basis 1-forms. So for anyp-form , let us define the pullback by

(v1 , , vp) = (v1 , , vp) , (13.36)

and in terms of components, by

=1

p!

(i1ip

)dxi1 dxip . (13.37)

We assumed above that the pullback of the wedge product of a2-form and a 1-form is the wedge product of the pullbacks of therespective forms, but it is not necessary to make that assumption it can be shown explicitly by taking three vector fields and followingthe arguments used earlier for the wedge product of two 1-forms.

Then for any p-form and q-form we can calculate from thisthat

( ) = . (13.38)Thus pullbacks commute with (are distributive over) wedge products.

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Chapter 14

Exterior derivative

The exterior derivative is a generalization of the gradient of a func-tion. It is a map from p-forms to (p + 1)-forms. This should be aderivation, so it should be linear,

d(+ ) = d+ d p-forms , . (14.1)

This should also satisfy Leibniz rule, but the algebra of p-forms isnot a commutative algebra but a graded commutator algebra, i.e.,involves a factor of (1)pq for exchanges. So we need

d( ) = d + (1)pqd , (14.2)

or alternatively,

d( ) = d + (1)p d . (14.3)

This will be the Leibniz rule for wedge products. Note that it givesthe correct result when one or both of , are 0-forms, i.e., functions.The two formulas are identical by virtue of the fact that d is a(q + 1)-form, so that

d = (1)p(q+1)d . (14.4)

We will try to define the exterior derivative in a way such that it hasthese properties.

Let us define the exterior derivative of a p-form in a chart as

d =1

p!ii1ipdx

i1 dxip (14.5)

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This clearly has the first property of linearity. To check the (graded)Leibniz rule, let us write in components. Then

d( ) = 1p!q!

i(i1ipj1jq

)dxi dxi1 dxjq

=1

p!q!

[(ii1ip

)j1jq + i1ip

(ij1jq

)]dxi dxi1 dxjq

=1

p!q!

(ii1ip

)j1jq dx

i dxi1 dxip dxj1 dxjq

+1

p!q!(1)p i1ip

(ij1jq

)dxi1 dxip dxi dxj1 dxjq

= d + (1)p d . (14.6)

A third property of the exterior derivative immediately followsfrom here,

d2 = 0 . (14.7)

To see this, we write

d(d) =1

p!d(ii1ipdx

i dxi1 dxip)=

1

p!jii1ipdx

j dxi dxi1 dxip . (14.8)

But the wedge product is antisymmetric, dxj dxi = dxi dxj ,and the indices are summed over, so the above object must be anti-symmetric in j , i . But that vanishes. So d

2 = 0 on all forms.Note that we can also write

d =1

p!

(di1ip

) dxi1 dxip , (14.9)where the object in parentheses is a gradient 1-form correspondingto the gradient of the component.

Consider a 1-form A = Adx where A are smooth functions on

M . Then using this definition we can write

dA = (dA) dx= Adx

dx

=1

2(A A) dx dx

(dA) = A A . (14.10)

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We can generalize this result to write for a p-form,

=1

p!1pdx

1 dxp (14.11)

d =1

p!

(1p

)dx1 dxp

=1

(p+ 1)![1p]dx

dx1 dxp

(d)1p = [1p] (14.12)Example: For p = 1 i.e. for a 1-form A we get from this formula

(dA) = A A , in agreement with our previous calculation.For p = 2 we have a 2-form, call it . Then using this formula

we get

(d) = []

= + + .(14.13)

Note that d is not defined on arbitrary tensors, but only on forms.2

By definition, d2 = 0 on any p-form. So if = d , it follows thatd = 0 . But given a p-form for which d = 0 , can we say thatthere must be some (p 1)-form such that = d ? This is a good place to introduce some terminology. Any form such that d = 0 is called closed, whereas any form such that = d is called exact. 2

So every exact form is closed. Is every closed form exact? Theanswer is yes, in a sufficiently small neighbourhood. We say thatevery closed form is locally exact. Note that if a p-form = d , wecannot uniquely specify the (p 1)-form since for any (p 2)-form , we can always write = d , where = + d .

Thus a more precise statement is that given any p-form suchthat d = 0 in a neighbourhood of some point P , there is someneighbourhood of this point and some (p1)-form such that = din that neighbourhood. But this may not be true globally. Thisstatement is known as the Poincare lemma. 2

Example: In R2 remove the origin. Consider the 1-form

=xdy ydxx2 + y2

. (14.14)

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Then

d =

(1

x2 + y2 2x

2

(x2 + y2)2

)dx dy

(1

x2 + y2 2y

2

(x2 + y2)2

)dy dx

=2

x2 + y2dx dy 2 x

2 + y2

(x2 + y2)2dx dy = 0 . (14.15)

Introduce polar coordinates r, with x = r cos , y = r sin .Then

dx = dr cos r sin d dy = dr sin + r cos d

=r cos (sin dr + r cos d)

r2 r sin (cos dr r sin d)

r2

=r2

(cos2 + sin2

)d

r2= d . (14.16)

Thus is exact, but is multivalued so there is no function fsuch that = df everywhere. In other words, = d is exact onlyin a neighbourhood small enough that remains single-valued.

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Chapter 15

Volume form

The space of p-forms in n dimensions is

(n

p

)dimensional. So the

space of n-forms in n dimensions is 1-dimensiona, i.e., there is onlyone independent component, and all n-forms are scalar multiples ofone another.

Choose an n-form field. Call it . Suppose 6= 0 at some pointP . Then given any basis {e} of TPM , we have (e1, en) 6= 0since 6= 0 . Thus all vector bases at P fall into two classes, one forwhich (e1, en) > 0 and the other for which it is < 0 .

Once we have identified these two classes, they are independent of . That is, if is another n-form which is non-zero at P , there mustbe some function f 6= 0 such that = f . Two bases which gavepositive numbers under will give the same sign both positive orboth negative under and therefore will be in the same class. So every basis (set of n linearly independent vectors) is a mem-ber of one of the two classes. These are called righthanded andlefthanded. 2 A manifold is called orientable if it is possible to define a con-tinuous n-form field which is non-zero everywhere on the manifold.Then it is possible to choose a basis with the same handedness ev-erywhere on the manifold continuously. 2

Euclidean space is orientable, the Mobius band is not. An orientable manifold is called oriented once an orientationhas been chosen, i.e. once we have decided to choose basis vectorswith the same handedness everywhere on the manifold. 2 It is necessary to choose an oriented manifold when we discussthe integration of forms. On an n-dimensional manifold, a set of n

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59

linearly independent vectors define an n-dimensional parallelepiped.If we define an n-form 6= 0 we can think of the value of these vectorsas the volume of this parallelepiped. This is called a volume form.2

Once a volume form has been chosen, any set of n linearly inde-pendent vectors will define a positive or negative volume.

The integral of a function f on Rn is the sum of the values of f ,multiplied by infinitesimal volumes of coordinate elements. Similarly,we define the integral of a function f on an oriented manifold as thesum of the values of f , multiplied by infinitesimal volumes. The wayto do that is the following.

Given a function f , define an n-form in a chart by = fdx1 dxn . To integrate over an open set U , divide it up into infinitesimalcells, spanned by vectors{

x1

x1,x2

x2, ,xn

xn

},

where the xi are small numbers.Then the integral of f over one such cell is approximately

fx1x2 xn = fdx1 dxn (x11, ,xnn)= (cell) . (15.1)

Adding up the contributions from all cells and taking the limit of cellsize going to zero, we find

U

=

(U)

fdnx . (15.2)

The right hand side is the usual integration in calculus of n variables,and the left hand side is our notation which we are defining.

The right hand side can be seen to be independent of the choiceof coordinate system. If we choose a different coordinate system, weget a Jacobian, but also a redefinition of the region (U) . Let uscheck that the left hand side is also invariant of the choice of thecoordinates. We will do this in two dimensions with = fdx1dx2 .

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In another coordinate system (y1, y2) corresponding to (U)

dx1 =x1

y1dy1 +

x1

y2dy2

dx2 =x2

y1dy1 +

x2

y2dy2

dx1 dx2 =(x1

y1x2

y1 x

1

y2x2

y2

)dy1 dy2

= Jdy1 dy2 , (15.3)

and J is the Jacobian.So what we have here is

U

=

U

f(x1, x2)dx1 dx2

=

U

f(y1, y2)Jdy1 dy2

=

(U)

f(y1, y2)Jd2y , (15.4)

so we get the same result both ways.Given the same f , if we choose a basis with the opposite orien-

tation, the integral of will have the opposite sign. This is why thechoice of orientation has to be made before integration.

Manifolds become even more interesting if we define a metric.

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Chapter 16

Metric tensor

A metric on a vector space V is a function g : V V Rwhich is

i) bilinear:

g(av1 + v2, w) = ag(v1, w) + g(v2, w)

g(v, w1 + aw2) = g(v, w1) + ag(v, w2) , (16.1)

i.e., g is a (0,2) tensor;

ii) symmetric:g(v, w) = g(w, v); (16.2)

iii) non-degenerate:

g(v, w) = 0 w v = 0 . (16.3)

2

If for some v, w 6= 0 , we find that g(v, w) = 0 , we say that v, ware orthogonal. 2 Given a metric g on V , we can always find an orthonormalbasis {e} such that g(e, e) = 0 if 6= and 1 if = . 2 If the number of (+1)s is p and the number of (1)s is q , wesay that the metric has signature (p, q) .

We have defined a metric for a vector space. We can generalizethis definition to a manifold M by the following. A metric g on a manifoldM is a (0, 2) tensor field such that if(v, w) are smooth vector fields, g(v, w) is a smooth function on M ,and has the properties (16.1), (16.2) and (16.3) mentioned earlier. 2

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62 Chapter 16. Metric tensor

It is possible to show that smoothness implies that the signatureis constant on any connected component of M , and we will assumethat it is constant on all of M .

A vector space becomes related to its dual space by the metric.Given a vector space V with metric g , and vector v defines a linearmap g(v, ) : V R , w 7 g(v, w) R . Thus g(v, ) V where V is the dual space of V . But g(v, ) is itself linear in v , so the mapV V defined by g(v, ) is linear. Since g is non-degenerate, thismap is an isomorphism. It then follows that on a manifold we canuse the metric to define a linear isomorphism between vectors and1-forms.

In a basis, the components of the metric are g = g(e , e) . Thisis an n n matrix in an n-dimensional manifold. We can thus writeg(v, w) = gv

w in terms of the components. Non-degeneracyimplies that this matrix is invertible. Let g denote the inversematrix. Then, by definition of an inverse matrix, we have

gg = = g

g . (16.4)

Then the linear isomorphism takes the following form.

i) If v = ve is a vector field in a chart, and {} is the dualbasis to {e} ,

g(v, ) = v , (16.5)where v = gv

.

ii) If A = A is a 1-form written in a basis {} , the corre-

sponding vector field is Ae , where A = gA .

This is the isomorphism between vector fields and 1-forms. (Wecould of course define a similar isomorphism between vectors andcovectors without referring to a manifold.) A similar isomorphismholds for tensors, e.g. in terms of components,

T T T T (16.6)T T T T (16.7)

These correspondences are not equalities the components are notequal. What it means is that, if we know one set of components, sayT , and the metric, we also know every other set of components.

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Using the fact that a non-degenerate metric defines a 1-1 linearmap between vectors and 1-forms, we can define an inner productof 1forms, by

A | B = gAB (16.8)for 1-forms A,B . This result is independent of the choice of basis, i.e.independent of the coordinate system, just like the inner productof vector fields,

v | w = g(v, w) = gvw . (16.9)

2

Given a manifold with metric, there is a canonical volume formdV (sometimes written as vol) , which in a coordinate chart reads

dV =|det g |dx1 dxn . (16.10)

Note that despite the notation, this is not a 1-form, nor the gradientof some function V . This is clearly a volume form because it is ann-form which is non-zero everywhere, as g is non-degenerate.

We need to show that this definition is independent of the chart.Take an overlapping chart. Then in the new chart, the correspondingvolume form is

dV =|det g |dx1 dxn . (16.11)

We wish to show that dV = dV . In the overlap,

dx =x

xdx = Adx

(say) (16.12)

Then dx1 dxn = (detA)dx1 dxn .On the other hand, if we look at the components of the metric

tensor in the new chart,

g = g(, )

=

(x

x ,

x

x

)= g

((A1

) ,

(A1

)

)=(A1

)

(A1

)g . (16.13)

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Taking determinants, we find

det g = (detA)2 (det g) . (16.14)

Thus | det g | = |detA|1

| det g | , (16.15)

and so dV = dV . This is called the metric volume form and written as

dV =|g|dx1 dxn (16.16)

in a chart. 2When we write dV , sometimes we mean the n-form as defined

above, and sometimes we mean|g|dnx , the measure for the usual

integral. Another way of writing the volume form in a chart is interms of its components,

dV =

|g|n!

1ndx1 dxn (16.17)

where is the totally antisymmetric Levi-Civita symbol, with12n = +1 . Thus

|g| 1n are the components of the volumeform.

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Chapter 17

Hodge duality

We will next define the Hodge star operator. We will defineit in achart rather than abstractly. The Hodge star operator, denoted ? in an n-dimensionalmanifold is a map from p-forms to (n p)-forms given by

(?)1np |g|p!

1n gnp+11 gnp 1p , (17.1)

where is a p-form. 2The ? operator acts on forms, not on components.Example: Consider R3 with metric +++, i.e. g =

diag(1, 1, 1) . Then |g| g = 1 , gdiag(1, 1, 1) . Write the coordi-nate basis 1-forms as dx, dy, dz . Their components are clearly

(dx)i = 1i , (dy)i =

2i , (dz)i =

3i , (17.2)

the s on the right hand sid