Intro to Differ en Ti Able Manifolds

8/6/2019 Intro to Differ en Ti Able Manifolds

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LECTURE NOTES ON DIFFERENTIABLE MANIFOLDS

JIE WU

Contents

1. Tangent Spaces, Vector Fields in R n and the Inverse Mapping Theorem 21.1. Tangent Space to a Level Surface 21.2. Tangent Space and Vectors Fields on R n 21.3. Operator Representations of Vector Fields 31.4. Integral Curves 51.5. Implicit- and Inverse-Mapping Theorems 62. Topological and Differentiable Manifolds, Diffeomorphisms, Immersions, Submersions

and Submanifolds 92.1. Topological Spaces 92.2. Topological Manifolds 102.3. Differentiable Manifolds 112.4. Tangent Space 132.5. Immersions 152.6. Submersions 173. Examples of Manifolds 193.1. Open Stiefel Manifolds and Grassmann Manifolds 193.2. Stiefel Manifold 214. Fibre Bundles and Vector Bundles 224.1. Fibre Bundles 224.2. G-Spaces and Principal G-Bundles 254.3. The Associated Principal G-Bundles of Fibre Bundles 274.4. Vector Bundles 314.5. The Construction of Gauss Maps 315. Tangent Bundles and Vector Fields 386. Cotangent Bundles and Tensor Fields 38

7. Orientation of Manifolds 388. Tensor Algebras and Exterior Algebras 389. DeRham Cohomology 3810. Integration on Manifolds 3811. Stokes Theorem 38References 38

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1. Tangent Spaces, Vector Fields in R n and the Inverse Mapping Theorem

1.1. Tangent Space to a Level Surface. Let be a curve in R n : : t ( 1(t), 2(t), . . . , n (t)).

(A curve can be described as a vector-valued function. Converse a vector-valued function gives acurve in R n .) The tangent line at the point (t0) is given with the direction

d dt

(t0) =d 1

dt(t0), . . . ,

d n

t(t0) .

(Certainly we need to assume that the derivatives exist. We may talk about smooth curves , that is,the curves with all continuous higher derivatives.)

Consider the level surface f (x1 , x2 , . . . , x n ) = c of a differentiable function f , where x i refers toi-th coordinate. The gradient vector of f at a point P = ( x1(P ), x2(P ), . . . , x n (P )) is

f = (f x 1

, . . . ,f

x n).

Given a vector u = ( u1 , . . . , u n ), the directional derivative is

D u f = f u = f x 1u1 + + f

x nun .

The tangent space at the point P on the level surface f (x1 , . . . , x n ) = c is the (n 1)-dimensional(if f = 0) space through P normal to the gradient f . In other words, the tangent space is givenby the equation

f x 1

(P )(x1 x1(P )) + +f

x n(P )(xn xn (P )) = 0 .

From the geometric views, the tangent space should consist of all tangents to the smooth curveson the level surface through the point P . Assume that is a curve through P (when t = t0)that lies in the level surface f (x1 , . . . , x n ) = c, that is

f ( 1(t), 2(t), . . . , n (t)) = c.

By taking derivatives on both sides,f x 1

(P )( 1) (t0) + +f

x n(P )( n ) (t0) = 0

and so the tangent line of is really normal (orthogonal) to f . When runs over all possiblecurves on the level surface through the point P , then we obtain the tangent space at the point P .

Roughly speaking, a tangent space is a vector space attached to a point in the surface .How to obtain the tangent space: take all tangent lines of smooth curve through this point on the

surface.

1.2. Tangent Space and Vectors Fields on R n . Now consider the tangent space of R n . Accord-ing to the ideas in the previous subsection, rst we assume a given point P R n . Then we considerall smooth curves passes through P and then take the tangent lines from the smooth curves. The

obtained vector space at the point P is the n-dimensional space. But we can look at in a littledetail.Let be a smooth curve through P . We may assume that (0) = P . Let be another smooth

curve with (0) = P . is called to be equivalent to if the directives (0) = (0). The tangentspace of R n at P , denoted by T P (R n ), is then the set of equivalence class of all smooth curvesthrough P .


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LECTURE NOTES ON DIFFERENTIABLE MANIFOLDS 3

Let T (R n ) =P R n

T P (R n ), called the tangent bundle of R n . If S is a region of R n , let T (S ) =

P ST P (S ), called the tangent bundle of S .

Note. Each T P (R n ) is an n-dimensional vector space, but T (S ) is not a vector space. In otherwords, T (S ) is obtained by attaching a vector space T P (R n ) to each point P in S . Also S is assumedto be a region of R n , otherwise the tangent space of S (for instance S is a level surface) could be aproper subspace of T P (R n ).

If is a smooth curve from P to Q in R n , then the tangent space T P (R n ) moves along toT Q (R n ). The direction for this moving is given (t), which introduces the following importantconcept.

Denition 1.1. A vector eld V on a region S of R n is a smooth map (also called C -map)

V : S T (S ) P v(P ).

Let V : P v(P ) and W : P w(P ) be two vector elds and let f : S R be a smooth

function. Then V + W : P v(P ) + w(P ) and fV : P f (P ) v(P ) give (pointwise) addition andscalar multiplication structure on vector elds.

1.3. Operator Representations of Vector Fields. Let J be an open interval containing 0 andlet : J R n be a smooth curve with (0) = P . Let f = f (x1 , . . . , x n ) be a smooth functiondened on a neighborhood of P . Assume that the range of is contained in the domain of f . Byapplying the chain rule to the composite T = f : J R ,

D (f ) :=dT dt

=n

i =1

d i (t)dt

f x i

x i = i ( t )

Proposition 1.2.

D (af + bg) = aD (f ) + bD (g), where a, b are constant.

D (fg ) = D (f )g + fD (g).

Let C (R n ) denote the set of smooth functions on R n . An operation D on C (R n ) is called aderivation if D maps C (R n ) to C (Rn ) and satises the conditions

D (af + bg) = aD (f ) + bD(g), where a, b are constant.

D (fg ) = D (f )g + fD (g).Example: For 1 i n,

i : f f x i

is a derivation.

Proposition 1.3. Let D be any derivation on C (R n ). Given any point P in R n . Then thereexist real numbers a1 , a 2 , . . . , a n R such that

D(f )(P ) =n

i =1a i i (f )(P )

for any f C (R n ), where a i depends on D and P but is independent on f .


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Proof. Write x for (x1 , . . . , x n ). Dene

gi (x) =

1

0

f

xi (t(x P ) + P )dt.

Then

f (x) f (P ) = 1

0

ddt

f (t(x P ) + P )dt

= 1

0

n

i =1

f x i

(t(x P ) + P ) (x i x i (P ))dt

=n

i =1

(x i x i (P )) 1

0

f x i

(t(x P ) + P )dt =n

i =1

(x i x i (P ))gi (x).

Since D is a derivation, D(1) = D (1 1) = D(1) 1 + 1 D(1) and so D (1) = 0. It follows thatD (c) = 0 for any constant c. By applying D to the above equations,

D(f (x)) = D (f (x) f (P )) =n

i =1

D(x i x i (P ))gi (x) + ( x i x i (P ))D (gi (x))

=n

i =1

D (x i )gi (x) + ( x i x i (P ))D (gi (x))

because D (f (P )) = D (x i (P )) = 0. Let a i = D (x i )(P ) which only depends on D and P . Byevaluating at P ,

D (f )(P ) =n

i =1

D(x i )(P )gi (P ) + 0 =n

i =1

a i gi (P ).

Since

gi (P ) =

1

0

f

x i(t(P P ) + P )dt =

1

0

f

x i(P )dt =

f

x i(P ) = i (f )(P ),

D (f )(P ) =n

i =1a i i (f )(P ),

which is the conclusion.

From this proposition, we can give a new way to looking at vector elds:Given a vector elds P v(P ) = ( v1(P ), v2(P ), . . . , v n (P )) , a derivation

Dv =n

i =1

vi (P ) i

on C (R n ) is called an operator representation of the vector eld P v(P ).Note. The operation vi (x) i is given as follows: for any f C (R n ),

Dv (f )(P ) =n

i =1

vi (P ) i (f )(P )

for any P .From this new view, the tangent spaces T (R n ) admits a basis { 1 , 2 , . . . , n }.


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1.4. Integral Curves. Let V : x v(x ) be a (smooth) vector eld on an neighborhood U of P .An integral curve to V is a smooth curve s : ( , ) U , dened for suitable , > 0, such that

s (t) = v(s (t))

for < t < .

Theorem 1.4. Let V : x v(x ) be a (smooth) vector eld on an neighborhood U of P . Then thereexists an integral curve to V through P . Any two such curves agree on their common domain.

Proof. The proof is given by assuming the fundamental existence and uniqueness theorem for sys-tems of rst order differential equations.

The requirement for a curve s(t) = ( s1(t), . . . , s n (t)) to be an integral curve is:

ds 1 ( t )dt = v

1(s1(t), s 2(t), . . . , s n (t))ds 2 ( t )

dt = v2(s1(t), s 2(t), . . . , s n (t))

ds n ( t )dt = v

n (s1(t), s2(t), . . . , s n (t))

with the initial conditions

s(0) = P (s1(0) , s2(0) , . . . , s n (0)) = ( x1(P ), x2(P ), . . . , x n (P ))

s (0) = v(P )ds1

dt(0) , . . . ,

dsn

dt(0) = ( v1(P ), . . . , v n (P )) .

Thus the statement follows from the fundamental theorem of rst order ODE.

Example 1.5. Let n = 2 and let V : P v(P ) = ( v1(P ), v2(P )), where v1(x, y) = x and v2(x, y) =y. Given a point P = ( a1 , a2), the equation for the integral curve s(t) = ( x(t), y(t)) is

x (t) = v1(s(t)) = x(t)y (t) = v2(s(t)) = y(t)

with initial conditions ( x(0) , y(0)) = ( a1 , a 2) and ( x (0) , y (0)) = v(a1 , a 2) = ( a1 , a 2). Thus thesolution is

s (t) = ( a1et , a2et ).

Example 1.6. Let n = 2 and let V : P v(P ) = ( v1(P ), v2(P )), where v1(x, y) = x and v2(x, y) = y. Given a point P = ( a1 , a 2), the equation for the integral curve s(t) = ( x(t), y(t)) is

x (t) = v1(s(t)) = x(t)y (t) = v2(s (t)) = y(t)

with initial conditions ( x(0) , y(0)) = ( a1 , a 2) and ( x (0) , y (0)) = v(a1 , a 2) = ( a1 , a2). Thus thesolution is

s(t) = ( a1et , a 2e t ).


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1.5. Implicit- and Inverse-Mapping Theorems.

Theorem 1.7. Let D be an open region in R n +1 and let F be a function well-dened on D with

continuous partial derivatives. Let (x10 , x

20 , . . . , x

n0 , z0) be a point in D where

F (x10 , x20 , . . . , x

n0 , z0) = 0

F z

(x10 , x20 , . . . , x

n0 , z0) = 0 .

Then there is a neighborhood N (z0) R , a neighborhood N (x10 , . . . , x n0 ) R n , and a unique function z = g(x1 , x2 , . . . , x n ) dened for (x1 , . . . , x n ) N (x10 , . . . , x n0 ) with values z N (z0) such that

1) z0 = g(x10 , x20 , . . . , x n0 ) and

F (x1 , x2 , . . . , x n , g(x1 , . . . , x n )) = 0

for all (x1 , . . . , x n ) N (x10 , . . . , x n0 ).2) g has continuous partial derivatives with

gx i (x

1

, . . . , xn

) = F x i (x1 , . . . , x n , z)F z (x1 , . . . , x n , z)

for all (x1 , . . . , x n ) N (x10 , . . . , x n0 ) where z = g(x1 , . . . , x n ).3) If F is smooth on D , then z = g(x1 , . . . , x n ) is smooth on N (x10 , . . . , x n0 ).

Proof. Step 1. We may assume that F z (x10 , x20 , . . . , x n0 , z0) > 0. Since F z is continuous, there

exists a neighborhood N (x10 , x20 , . . . , x n0 , z0) in which F z is continuous and positive. Thus for xed(x1 , . . . , x n ), F is strictly increasing on z in this neighborhood. It follows that there exists c > 0such that

F (x10 , x20 , . . . , x

n0 , z0 c) < 0 F (x10 , x20 , . . . , x

n0 , z0 + c) > 0

with(x10 , x

20 , . . . , x

n0 , z0 c), (x10 , x20 , . . . , x

n0 , z0 + c) N (x

10 , x

20 , . . . , x

n0 , z0).

Step 2. By the continuity of F , there exists a small > 0 such that

F (x1 , x2 , . . . , x n , z0 c) < 0 F (x1 , x2 , . . . , x n , z0 + c) > 0with

(x1 , x2 , . . . , x n , z0 c), (x1 , x2 , . . . , x n , z0 + c) N (x10 , x20 , . . . , xn0 , z0)

for (x1 , . . . , x n ) N (x10 , . . . , x n0 ).Step 3. Fixed ( x1 , . . . , x n ) N (x10 , . . . , x n0 ), F is continuous and strictly increasing on z. There isa unique z, z0 c < z < z 0 + c, such that

F (x1 , . . . , x n , z) = 0 .

This denes a function z = g(x1 , . . . , x n ) for (x1 , . . . , x n ) N (x10 , . . . , x n0 ) with values z (z0 c, z0 + c).Step 4. Prove that z = g(x1 , . . . , x n ) is continuous. Let ( x11 , . . . , x n1 ) N (x10 , . . . , x n0 ). Let(x1

1(k), . . . , x n

1(k)) be any sequence in N

(x1

0, . . . , x n

0) converging to ( x1

1, . . . , x n

1). Let A be any

subsequential limit of {zk = g(x11(k), . . . , x n1 (k))}, that is A = lims zk s . Then, by the continuity of F ,

0 = lims

F (x11(ks ), . . . , xn1 (ks ), zk s )

= F ( lims

x11(ks ), . . . , lims xn1 (ks ), lims zk s )


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= F (x11 , . . . , xn1 , A).

By the unique solution of the equation, A = g(x11 , . . . , x n1 ). Thus {zk } converges g(x11 , . . . , x n1 ) andso g is continuous.Step 5. Compute the partial derivatives zx i . Let h be small enough. Let

z + k = g(x1 , . . . , x i 1 , x i + h, x i +1 , . . . , x n ),

that isF (x1 , . . . , x i + h , . . . , x n , z + k) = 0

with z0 c < z + k < z 0 + c. Then

0 = F (x1 , . . . , x i + h , . . . , x n , z + k) F (x1 , . . . , x n , z)

= F x i (x1 , . . . , x i , . . . , x n , z)h + F z (x1 , . . . , x i , . . . , x n , z)kby the mean value theorem (Consider the function

(t) = F (x1 , . . . , x i + th , . . . , x n , z + tk )

for 0 t 1. Then (1) (0) = ()(1 0).), where x i is between x i and x i + h, and z is betweenz and z + k. Now

gx i

= limh 0

g(x1 , . . . , x i 1 , x i + h, x i +1 , . . . , x n ) zh

= limh 0

kh

= limh 0

F x i (x1 , . . . , x i , . . . , x n , z)F z (x1 , . . . , x i , . . . , x n , z)

. = F x iF z

,

where z z as h 0 because g is continuous (and so k 0 as h 0).Step 6. Since F z is not zero in this small neighborhood, gx i is continuous for each i. If F is smooth,then all higher derivatives of g are continuous and so g is also smooth.

Theorem 1.8 (Implicit Function Theorem) . Let D be an open region in R m + n and let F 1, F

2, . . . , F n

be functions well-dened on D with continuous partial derivatives. Let (x10 , x20 , . . . , x m0 , u10 , u20 , . . . , u n0 )be a point in D where

F 1(x10 , x20 , . . . , x m0 , u10 , u20 , . . . , u n0 ) = 0F 2(x10 , x20 , . . . , x m0 , u10 , u20 , . . . , u n0 ) = 0

F n (x10 , x20 , . . . , x m0 , u10 , u20 , . . . , u n0 ) = 0

and the Jacobian

J = (F 1 , F 2 , . . . , F n ) (u1 , u2 , . . . , u n )

= detF iu j

= 0

at the point (x10 , x20 , . . . , x m0 , u10 , u20 , . . . , u n0 ). Then there are neighborhoods N (x10 , . . . , x m0 ), N 1 (u10),N 2 (u20), . . . , N n (un0 ), and unique functions

u1 = g1(x1 , x2 , . . . , x m )u2 = g2(x1 , x2 , . . . , x m )

un = gn (x1 , x2 , . . . , x m )

dened for (x1 , . . . , x m ) N (x10 , . . . , x m0 ) with values u1 N 1 (u10), . . . , u n N n (un0 ) such that


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1) u i0 = gi (x10 , x20 , . . . , x m0 ) and

F i (x1 , x2 , . . . , x n , gi (x1 , . . . , x m )) = 0

for all 1 i n and all (x1 , . . . , x m ) N (x10 , . . . , x m0 ).2) Each gi has continuous partial derivatives with

g ix j

(x1 , . . . , x m ) = 1J

(F 1 , . . . , F n )

(u1 , u2 , . . . , u j 1 , x j , u j +1 , . . . , u n )

for all (x1 , . . . , x m ) N (x10 , . . . , x m0 ) where u i = gi (x1 , . . . , x m ).3) If each F i is smooth on D , then each u i = gi (x1 , . . . , x m ) is smooth on N (x10 , . . . , x m0 ).

Sketch of Proof. The proof is given by induction on n. Assume that the statement holds for n 1with n > 1. (We already prove that the statement holds for n = 1.) Since the matrix

F iu j

is invertible at the point P = ( x10 , x

20 , . . . , x

m0 , u

10 , u

20 , . . . , u

n0 ) (because the determinant is not zero),we may assume that

F nu n

(P ) = 0 .

(The entries in the last column can not be all 0 and so, if F iu n (P ) = 0, we can interchange F i andF n .)

From the previous theorem, there is a solution

un = gn (x1 , . . . , x m , u1 , . . . , u n 1)

to the last equation. ConsiderG1 = F 1(x1 , . . . , x m , u1 , . . . , u n 1 , gn )G2 = F 2(x1 , . . . , x m , u1 , . . . , u n 1 , gn )

Gn 1 = F n 1(x1 , . . . , x m , u1 , . . . , u n 1 , gn ).Then

G iu j

=F iu j

+F iu n

gnu j

for 1 i, j n 1, whereF nu j

+F nu n

gnu j

= 0 .

Let

B =

1 0 0 0 00 1 0 0 00 0 1 0 0

0 0 0 1 0g n

u 1g nu 2

g nu 3

g nu n 1 1

ThenF iu j

B =G iu j n 1,n 1

0 F nu n.


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By taking the determinant,

J = (F 1 , . . . , F n )

(u1

, . . . , un

)=

F n

un

(G1 , . . . , G n 1)

(u1

, . . . , un 1

).

Thus (G 1 ,...,G n 1 ) (u 1 ,...,u n 1 ) = 0 at P and, by induction, there are solutions

u i = gi (x1 , . . . , x m )

for 1 i n 1.

Theorem 1.9 (Inverse Mapping Theorem) . Let D be an open region in R n . Let

x1 = f 1(u1 , . . . , u n )x2 = f 2(u1 , . . . , u n )

xn = f n (u1 , . . . , u n )

be functions dened on D with continuous partial derivatives. Let (u10, . . . , u n

0) D satisfy x i

0=

f i (u10 , . . . , u n0 ) and the Jacobian

(x1 , . . . , x n ) (u1 , . . . , u n )

= 0 at ( u10 , . . . , un0 ).

Then there are neighborhood N (x10 , . . . , x n0 ) and N (u10 , . . . , u n0 ) such that

u1 = f 11 (x1 , . . . , x n )u2 = f 12 (x1 , . . . , x

n )

un = f 1n (x1 , . . . , x n )

is well-dened and has continuous partial derivatives on N (x10 , . . . , x n0 ) with values in N (u10 , . . . , u n0 ).Moreover if each f i is smooth, then each f 1i is smooth.

Proof. Let F i = f i (u1 , . . . , u n ) x i . The assertion follows from the Implicit Function Theorem.

2. Topological and Differentiable Manifolds, Diffeomorphisms, Immersions,Submersions and Submanifolds

2.1. Topological Spaces.

Denition 2.1. Let X be a set. A topology U for X is a collection of subsets of X satisfyingi) and X are in U ;

ii) the intersection of two members of U is in U ;iii) the union of any number of members of U is in U .

The set X with U is called a topological space . The members U U are called the open sets .

Let X be a topological space. A subset N X with x N is called a neighborhood of x if thereis an open set U with x U N . For example, if X is a metric space, then the closed ball D (x)and the open ball B (x) are neighborhoods of x. A subset C is said to closed if X \ C is open.

Denition 2.2. A function f : X Y between two topological spaces is said to be continuous if for every open set U of Y the pre-image f 1(U ) is open in X .


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A continuous function from a topological space to a topological space is often simply called amap. A space means a Hausdorff space , that is, a topological spaces where any two points hasdisjoint neighborhoods.

Denition 2.3. Let X and Y be topological spaces. We say that X and Y are homeomorphic if there exist continuous functions f : X Y, g : Y X such that f g = id Y and g f = id X . Wewrite X = Y and say that f and g are homeomorphisms between X and Y .

By the denition, a function f : X Y is a homeomorphism if and only if i) f is a bijective;

ii) f is continuous andiii) f 1 is also continuous.

Equivalently f is a homeomorphism if and only if 1) f is a bijective, 2) f is continuous and 3) f isan open map, that is f sends open sets to open sets. Thus a homeomorphism between X and Y isa bijective between the points and the open sets of X and Y .

A very general question in topology is how to classify topological spaces under homeomorphisms.For example, we know (from complex analysis and others) that any simple closed loop is homeo-morphic to the unit circle S 1 . Roughly speaking topological classication of curves is known. Thetopological classication of (two-dimensional) surfaces is known as well. However the topologicalclassication of 3-dimensional manifolds (we will learn manifolds later.) is quite open.

The famous Poicare conjecture is related to this problem, which states that any simply connected3-dimensional (topological) manifold is homeomorphic to the 3-sphere S 3 . A space X is called simply connected if (1) X is path-connected (that is, given any two points, there is a continuous path joiningthem) and (2) the fundamental group 1(X ) is trivial (roughly speaking, any loop can be deformedto be the constant loop in X ). The manifolds are the objects that we are going to discuss in thiscourse.

2.2. Topological Manifolds. A Hausdorff space M is called a (topological) n-manifold if eachpoint of M has a neighborhood homeomorphic to an open set in R n . Roughly speaking, an n-manifold is locally R n . Sometimes M is denoted as M n for mentioning the dimension of M .

(Note. If you are not familiar with topological spaces, you just think that M is a subspace of R N for a large N .)

For example, R n and the n-sphere S n is an n-manifold. A 2-dimensional manifold is calleda surface . The objects traditionally called surfaces in 3-space can be made into manifolds ina standard way. The compact surfaces have been classied as spheres or projective planes withvarious numbers of handles attached.

By the denition of manifold, the closed n-disk D n is not an n-manifold because it has theboundary S n 1 . D n is an example of manifolds with boundary. We give the denition of manifold with boundary as follows.

A Hausdorff space M is called an n-manifold with boundary (n 1) if each point in M has aneighborhood homeomorphic to an open set in the half space

R n+ = {(x1 , , xn ) R n |xn 0}.Manifold is one of models that we can do calculus locally. By means of calculus, we need local

coordinate systems. Let x M . By the denition, there is a an open neighborhood U (x) of x anda homeomorphism x from U (x) onto an open set in R n+ . The collection {(U (x), x ) |x M } hasthe property that 1) {U (x)|x M } is an open cover and 2) x is a homeomorphism from U (x)


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onto an open set in R n+ . The subspace x (U x ) in R n+ plays a role as a local coordinate system. Thecollection {(U (x), x )|x M } is somewhat too large and we may like less local coordinate systems.This can be done as follows.

Let M be a space. A chart of M is a pair ( U, ) such that 1) U is an open set in M and 2) isa homeomorphism from U onto an open set in R n+ . The map

: U R n+can be given by n coordinate functions 1 , . . . , n . If P denotes a point of U , these functions areoften written as

x1(P ), x2(P ), . . . , x n (P )or simply x1 , x2 , . . . , x n . They are called local coordinates on the manifold.

An atlas for M means a collection of charts {(U , )| J } such that {U | J } is an opencover of M .

Proposition 2.4. A Hausdorff space M is a manifold (with boundary) if and only if M has an

atlas.Proof. Suppose that M is a manifold. Then the collection {(U (x), x )|x M } is an atlas. Con-versely suppose that M has an atlas. For any x M there exists such that x U and so U isan open neighborhood of x that is homeomorphic to an open set in R n+ . Thus M is a manifold.

We dene a subset M as follows: x M if there is a chart ( U , ) such that x U and (x) R n 1 = {x R n |xn = 0 }. M is called the boundary of M . For example the boundary of D n is S n 1 .

Proposition 2.5. Let M be a n-manifold with boundary. Then M is an (n 1)-manifold without boundary.

Proof. Let {(U , )| J } be an atlas for M . Let J J be the set of indices such thatU M = if J . Then Clearly

{(U M, |U M | J }

can be made into an atlas for M .

Note. The key point here is that if U is open in R n+ , then U R n 1 is also open because: Since U is open in R n+ , there is an open subset V of R n such that U = V R n+ . Now if x U R n 1 , thereis an open disk E (x) V and so

E (x) R n 1 V R n 1 = U R n 1

is an open (n 1)-dimensional -disk in R n 1 centered at x.

2.3. Differentiable Manifolds.

Denition 2.6. A Hausdorff space M is called a differential manifold of class C k (with boundary)if there is an atlas of M

{(U , ) | J }such that


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For any , J , the composites

1 : (U U ) Rn+

is differentiable of class C k .The atlas {(U , | J } is called a differential atlas of class C k on M .

(Note. Assume that M is a subspace of R N with N >> 0. If M has an atlas {(U , ) | J }such that each : U R n+ is differentiable of class C k , then M is a differentiable manifold of class C k . This is the denition of differentiable (smooth) manifolds in [6] as in the beginning theyalready assume that M is a subspace of R N with N large. In our denition (the usual denitionof differentiable manifolds using charts), we only assume that M is a (Hausdorff) topological spaceand so is only an identication of an abstract U with an open subset of R n+ . In this case we cannot talk differentiability of unless U is regarded as a subspace of a (large dimensional) Euclidianspace.)

Two differential atlases of class C k {(U , )| I } and {(V , )| J } are called equivalent

if {(U , )| I } {(V , )| J }is again a differential atlas of class C k (this is an equivalence relation). A differential structure of class C k on M is an equivalence class of differential atlases of class C k on M . Thus a differentialmanifold of class C k means a manifold with a differential structure of class C k . A smooth manifoldmeans a differential manifold of class C .Note: A general manifold is also called topological manifold . Kervaire and Milnor [4] have shownthat the topological sphere S 7 has 28 distinct oriented smooth structures.

Denition 2.7. let M and N be smooth manifolds (with boundary) of dimensions m and n re-spectively. A map f : M N is called smooth if for some smooth atlases {(U , | I } for M and {(V , ) J } for N the functions

f 1 | ( f 1 (V )U ) : (f 1(V ) U ) R n+are of class C .

Proposition 2.8. If f : M N is smooth with respect to atlases

{(U , | I }, {(V , | J }

for M , N then it is smooth with respect to equivalent atlases

{(U , | I }, {(V , | J }

Proof. Since f is smooth with respect with the atlases

{(U , | I }, {(V , | J },

f is smooth with respect to the smooth atlases

{(U , | I } {(U , | I }, {(V , | J } {(V , | J }

by look at the local coordinate systems. Thus f is smooth with respect to the atlases

{(U , | I }, {(V , | J }.


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Thus the denition of smooth maps between two smooth manifolds is independent of choice of atlas.

Denition 2.9. A smooth map f : M N is called a diffeomorphism if f is one-to-one and onto,and if the inverse f 1 : N M is also smooth.

Denition 2.10. Let M be a smooth n-manifold, possibly with boundary. A subset X is called aproperly embedded submanifold of dimension k n if X is a closed in M and, for each P X , thereexists a chart ( U, ) about P in M such that

(U X ) = (U ) R k+ ,

where R k+ R n+ is the standard inclusion.

Note. In the above denition, the collection {(U X, |U X )} is an atlas for making X to a smoothk-manifold with boundary X = X M .

If M = , by dropping the requirement that X is a closed subset but keeping the requirementon local charts, X is called simply a submanifold of M .

2.4. Tangent Space. Let S be an open region of R n . Recall that, for P S , the tangent spaceT P (S ) is just the n-dimensional vector space by putting the origin at P . Let T be an open regionof R m and let f = ( f 1 , . . . , f m ) : S T be a smooth map. Then f induces a linear transformation

T f : T P (S ) T f (P ) (T )

given by

T f ( v) =f ix j m n

v1v2 vn n 1

=

v1 1(f 1) + v2 2(f 1) + + vn n (f 1)v1 1(f 2) + v2 2(f 2) + + vn n (f 2)

v1 1(f m ) + v2 2(f m ) + + vn n (f m )

,

namely T f is obtained by taking directional derivatives of ( f 1 , . . . , f m ) along vector v for anyvT P (S ).

Now we are going to dene the tangent space to a (differentiable) manifold M at a point P asfollows:

First we consider the set

T P = {(U, , v) | P U, (U, ) is a chart vT ((P ))( (U ))}.

The point is that there are possibly many charts around P . Each chart creates an n-dimension vectorspace. So we need to dene an equivalence relation in T P such that, T P modulo these relations isonly one copy of n-dimensional vector space which is also independent on the choice of charts.

Let (U, , v) and ( V,, w) be two elements in T P . That is ( U, ) and ( V, ) are two charts withP U and P V . By the denition,

1 : (U V ) - (U V )is diffeomorphism and so it induces an isomorphism of vector spaces

T ( 1) : T (P ) ((U V )) - T (P ) ((U V )) .Now (U, , v) is called equivalent to ( V,, w), denoted by ( U, , v) (V,, w), if

T ( 1)( v) = w.


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Dene T P (M ) to be the quotientT P (M ) = T P / .

Exercise 2.1. Let M be a differentiable n-manifold and let P be any point in M . Prove thatT P (M ) is an n-dimensional vector space. [Hint: Fixed a chart ( U, ) and dened

a(U, , v) + b(U,, w) := ( U, , av + b w).

Now given any ( V, , x), (V ,, y) T P , consider the map

1 : (U V ) (U V ) 1 : (U V ) (U V )and dene

a(V, , x) + b(V ,, y) = ( U,,aT ( 1)( x) + bT ( 1)( y)) .Then prove that this operation gives a well-dened vector space structure on T P , that is, independenton the equivalence relation.]

The tangent space T P (M ), as a vector space, can be described as follows: given any chart ( U, )

with P U , there is a unique isomorphismT : T P (M ) T (P ) ((U )) .

by choosing ( U, , v) as representatives for its equivalence class. If ( V, ) is another chart withP V , then there is a commutative diagram

(1)

T P (M )T =- T (P ) ((U V ))

T P (M )T =- T (P ) ((U V )) ,

T ( 1)

?

where T ( 1

) is the linear isomorphism induced by the Jacobian matrix of the differentiablemap 1 : (U V ) (U V ).

Exercise 2.2. Let f : M N be a smooth map, where M and N need not to have the samedimension. Prove that there is a unique linear transformation

T f : T P (M ) - T f (P ) (N )such that the diagram

T P (M )T =

- T (P ) ((U ))

T f (P ) (N )

T f

?T =- T ( f (P )) ((V ))

T ( f 1)

?

commutes for any chart ( U, ) with P U and any chart ( V, ) with f (P ) V . [First x a choiceof (U, ) with P U and ( V, ) with f (P ) V , the linear transformation T f is uniquely denedby the above diagram. Then use Diagram (1) to check that T f is independent on choices of charts.


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2.5. Immersions. A smooth map f : M N is called immersion at P if the linear transformation

T f : T P (M ) T f (P ) (M )

is injective.

Theorem 2.11 (Local Immersion Theorem) . Suppose that f : M m N n is immersion at P . Then there exist charts (U, ) about P and (V, ) about f (P ) such that the diagram

U f |U - V

R m

(P ) = 0

?

canonical coordinate inclusion - R n

(f (P )) = 0

?

commutes.

Proof. We may assume that (P ) = 0 and (f (P )) = 0. (Otherwise replacing and by (P )and (f (P )), respectively.)

Consider the commutative diagram

U f |U - V

(U )

? g = f 1- (V )

?

R m ?

R n . ?

By the assumption,

T g: T 0((U ))-

T 0((V ))

is an injective linear transformation and so

rank( T g) = m


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at the origin. The matric for T g is

(2)

g1

x 1 g1

x 2 g1

x m

g2

x 1g2

x 2 g

2

x m

gm

x 1gm

x 2 g

m

x m

gn

x 1g

n

x 2 g

n

x m

.

By changing basis of R n (corresponding to change the rows), we may assume that the rst m rowsforms an invertible matrix Am m at the origin.

Dene a function

h = ( h1 , h2 , . . . , h n ) : (U ) R n m - R n

by setting

h i (x1 , . . . , x m , xm +1 , . . . , x n ) = gi (x1 , . . . , x m )

for 1 i m and

h i (x1 , . . . , x m , xm +1 , . . . , x n ) = gi (x1 , . . . , x m ) + x i

for m + 1 i n. Then Jacobian matrix of h is

Am m 0m (n m )

B (n m ) m I n m

,

where B is taken from ( m + 1)-st row to n-th row in the matrix (2). Thus the Jacobian of h is notzero at the origin. By the Inverse Mapping Theorem, h is an diffeomorphism in a small neighborhoodof the origin. It follows that there exist open neighborhoods U U of P and V V of f (P ) such


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that the following diagram commutes

U f |U - V

(U )

|U ?

g = f 1 - (V )

|V ?

(U ) 0 - (U ) U 2

= h 1

?

R m = R m 0 ?

- R n , ?

where U 2 is a small neighborhood of the origin in R n m .

Theorem 2.12. Let f : M N be a smooth map. Suppose that

1) f is immersion at every point P M ,2) f is one-to-one and 3) f : M f (M ) is a homeomorphism.

Then f (M ) is a smooth submanifold of M and f : M f (M ) is a diffeomorphism.

Note. In Condition 3, we need that if U is an open subset of M , then there is an open subset V of N such that V f (M ) = f (U ).

Proof. For any point P in M , we can choose the charts as in Theorem 2.11. By Condition 3, f (U )is an open subset of f (M ). The charts {(f (U ), | f (U ) )} gives an atlas for f (M ) such that f (M ) isa submanifold of M . Now f : M f (M ) is a diffeomorphism because it is locally diffeomorphismand the inverse exists.

Condition 3 is important in this theorem, namely an injective immersion need not give a dif-feomorphism with its image. (Construct an example for this.) An injective immersion satisfyingcondition 3 is called an embedding .

2.6. Submersions. A smooth map f : M N is called submersion at P if the linear transforma-tion

T f : T P (M ) T f (P ) (M )

is surjective.


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Theorem 2.13 (Local Submersion Theorem) . Suppose that f : M m N n is submersion at P .Then there exist charts (U, ) about P and (V, ) about f (P ) such that the diagram

U f |U - V

R m

(P ) = 0

?

canonical coordinate proj . - R n

(f (P )) = 0

?

commutes.

For a smooth map of manifolds f : M N , a point Q N is called regular if T f : T P (M ) T Q (N ) is surjective for every P f 1(Q), the pre-image of Q.

Theorem 2.14 (Pre-image Theorem) . Let f : M N be a smooth map and let Q N such that f 1(Q) is not empty. Suppose that Q is regular. Then f 1(Q) is a submanifold of M with dim f 1(Q) = dim M dim N .Proof. From the above theorem, for any P f 1(Q),

| f 1 (Q ) : f 1(Q) U - R m n

gives a chart about P .

Let Z be a submanifold of N . A smooth map f : M N is said to be transversal to Z if

Im( T f : T P (M ) T f (P ) (N )) + T f (P ) (Z ) = T f (P ) (N )

for every x f 1(Z ).

Theorem 2.15. If a smooth map f : M N is transversal to a submanifold Z N , then f 1(Z )

is a submanifold of M . Moreover the codimension of f 1

(Z ) in M equals to the codimension of Z in N .

Proof. Given P f 1(Z ), since Z is a submanifold, there is a chart ( V, ) of N about f (P ) suchthat V = V 1 V 2 with V 1 = V Z and ( V 1 , |V 1 ) is a chart of Z about f (P ). By the assumption,the composite

f 1(V )f | f 1 ( V )- V proj.- V 2

is submersion. By the Pre-image Theorem, f 1(V ) f 1(Z ) is a submanifold of the open subsetf 1(V ) of M and so there is a chart about P such that Z is a submanifold of M .

With respect to the assertion about the codimensions,

codim( f 1(Z )) = dim V 2 = codim( Z ).

Consider the special case that both M and Z are submanifolds of N . Then the transversalcondition is

T P (M ) + T P (Z ) = T P (N )for any P M Z .


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Corollary 2.16. The intersection of two transversal submanifolds of N is again a submanifold.Moreover

codim( M Z ) = codim( M ) + codim( Z )in N .

3. Examples of Manifolds

3.1. Open Stiefel Manifolds and Grassmann Manifolds. The open Stiefel manifold is thespace of k-tuples of linearly independent vectors in R n :

V k,n = {( v1 , . . . , vk )T | vi R n , {v1 , . . . , vk } linearly independent },

where V k,n is considered as the subspace of k n matrixes M (k, n ) = R kn . Since V k,n is an opensubset of M (k, n ) = R kn , V k,n is an open submanifold of R kn .

The Grassmann manifold Gk,n is the set of k-dimensional subspaces of R n , that is, all k-planesthrough the origin. Let

: V k,n Gk,nbe the quotient by sending k-tuples of linearly independent vectors to the k-planes spanned by kvectors. The topology in Gk,n is given by quotient topology of , namely, U is an open set of Gk,nif and only if 1(U ) is open in V k,n .

For ( v1 , . . . , vk )T V k,n , write v1 , . . . , vk for the k-plane spanned by v1 , . . . , vk . Observe thattwo k-tuples ( v1 , . . . , vk )T and ( w1 , . . . , wk )T spanned the same k-plane if and only if each of themis basis for the common plane, if and only if there is nonsingular k k matrix P such that

P ( v1 , . . . , vk )T = ( w1 , . . . , wk )T .

This gives the identication rule for the Grassmann manifold Gk,n . Let GL k (R ) be the space of general linear groups on R k , that is, GL k (R ) consists of k k nonsingular matrixes, which is anopen subset of M (k, k) = R k

2. Then Gk,n is the quotient of V k,n by the action of GL k (R ).

First we prove that Gk,n is Hausdorff. If k = n, then Gn,n is only one point. So we assume thatk < n . Given an k-plane X and w R n , let w be the square of the Euclidian distance from w toX . Let {e1 , . . . , e k } be the orthogonal basis for X , then

w (X ) = w w k

j =1

( w ej )2 .

Fixing any w R n , we obtain the continuous map

w : Gk,n - R

because w : V k,n R is continuous and Gk,n given by the quotient topology. (Here we usethe property of quotient topology that any function f from the quotient space Gk,n to any spaceis continuous if and only if f from V

k,nto that space is continuous.) Given any two distinct

points X and Y in Gk,n , we can choose a w such that w (X ) = w (Y ). Let V 1 and V 2 be disjointopen subsets of R such that w (X ) V 1 and w (Y ) V 2 . Then 1 w (V 1) and

1 w (V 2) are two open

subset of Gk,n that separate X and Y , and so Gk,n is Hausdorff.Now we check that Gk,n is manifold of dimension k(n k) by showing that, for any X in Gk,n ,

there is an open neighborhood U X of such that U X = R k (n k ) .


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Let X Gk,n be spanned by ( v1 , . . . , vk )T . There exists a nonsingular n n matrix Q such that

( v1 , . . . , vk )T = ( I k , 0)Q,

where I k is the unit k k-matrix. Fixing Q, deneX = {(P k , B k,n k )Q | det( P k ) = 0 , B k,n k M (k, n k)} V k,n .

Then E X is an open subset of V k,n . Let U X = (E X ) Gk,n . Since

1(U X ) = E Xis open in V k,n , U X is open in Gk,n with X U X . From the commutative diagram

GLk (R ) M (k, n k)(P, A ) (P,PA )Q

=- E X

M (k, n k)

proj .

?A (I k , A)Q

1X- U X ,

?

U X is homeomorphic to M (k, n k) = R k (n k ) and so Gk,n is a (topological) manifold.For checking that Gk,n is a smooth manifold, let X and Y Gk,n be spanned by ( v1 , . . . , vk )T

and ( w1 , . . . , wk )T , respectively. There exists nonsingular n n matrixes Q and Q such that

( v1 , . . . , vk )T = ( I k , 0)Q, ( w1 , . . . , wk )T = ( I k , 0)Q.

Consider the maps:M (k, n k)

1X- U X A (I k , A)Q

M (k, n k) 1

Y

- U Y A (I k , A)Q .

If Z U X U Y , thenZ = (I k , AZ )Q = (I k , B Z )Q

for unique A, B M (k, n k). It follows that there is a nonsingular k k matrix P such that

(I k , B Z )Q = P (I k , AZ )Q (I k , B Z ) = P (I k , AZ )QQ 1 .

Let

T = QQ 1 =T 11 T 12

T 21 T 22.

Then(I k , B Z ) = ( P,PA Z )T = ( P T 11 + PA Z T 21 , P T 12 + PA Z T 22 )

I k = P (T 11 + AZ T 21 )

BZ = P (T 12 + AZ T 22 ).It follows that

Z U X U Y if and only if det( T 11 + AZ T 21 ) = 0 (that is, T 11 + AZ T 21 is invertible).


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From the above, the composite

X (U X U Y ) 1X- U X U Y

Y - M (k, n )is given by

A (T 11 + AT 21 ) 1 (T 12 + AT 22 ) ,which is smooth. Thus Gk,n is a smooth manifold.

As a special case, G1,n is the space of lines (through the origin) of R n , which is also calledprojective space denoted by RP n 1 . From the above, R P n 1 is a manifold of dimension n 1.

3.2. Stiefel Manifold. The Stiefel manifold , denoted by V k,n , is dened to be the set of k orthog-onal unit vectors in R n with topology given as a subspace of V k,n M (k, n ). Thus

V k,n = {A M (k, n ) | A AT = I k }.We prove that V k,n is a smooth submanifold of M (k, n ) by using Pre-image Theorem.

Let S (k) be the space of symmetric matrixes. Then S (k) = R( k +1) k

2 is a smooth manifold of

dimension. Consider the mapf : M (k, n ) S (k) A AAT .

For any A M (k, n ), T f A : T A (M (k, n )) T f (A ) (S (k)) is given by setting T f A (B ) is the directionalderivative along B for any B T A (M (k, n )), that is,

T f A (B ) = lims 0

f (A + sB ) f (A)s

= lims 0

(A + sB )(A + sB )T AAT

s

= lims 0

AAT + sAB T + sBA T + s2BB T AAT

s= AB T + BA T .

We check that T f A : T A (M (k, n )) T f (A ) (S (k)) is surjective for any A f 1(I k ).

By the identication of M (k, n ) and S (k) with Euclidian spaces, T A (M (k, n )) = M (k, n ) andT f (A ) (S (k)) = S (k)). Let A f 1(I k ) and let C T f (A ) (S (k)). Dene

B =12

CA T A (M (k, n )) .

ThenT f A (B ) = AB T + BA T =

12

AAT C T +12

CAA T =======AA T = I k 1

2C T +

12

C =====C = C T

C.

Thus T f : T A (M (k, n )) T f (A ) (S (k)) is onto and so I k is a regular value of f . Thus, by Pre-imageTheorem, V k,n = f 1(I k ) is a smooth submanifold of M (k, n ) of dimension

kn (k + 1) k

2=

k(2n k 1)2

.

Special Cases: When k = n, then V n,n = O(n) the orthogonal group. From the above, O(n) is a

(smooth) manifold of dimension n (n 1)2 . (Note. O(n) is a Lie group, namely, a smooth manifoldplus a topological group such that the multiplication and inverse are smooth.)

When k = 1, then V 1,n = S n 1 which is manifold of dimension n 1.When k = n 1, then V n 1,n is a manifold of dimension (n 1) n2 . One can check that

V n 1,n = SO (n)


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the subgroup of O(n) with determinant 1. In general case, V k,n = O(n)/O (n k).As a space, V k,n is compact. This follows from that V k,n is a closed subspace of the k-fold

Cartesian product of S n 1 because V k,n is given by k unit vectors ( v1 , . . . , vk )T in R n that aresolutions to vi vj = 0 for i = j , and the fact that any closed subspace of compact Hausdorff spaceis compact. The composite

V k,n - V k,n-- Gk,n

is onto and so the Grassmann manifold Gk,n is also compact. Moreover the above composite is asmooth map because is smooth and V k,n is a submanifold. This gives the diagram

V k,n submanifold- M (k, n ) submersion at I k

A AAT -- S (k)

Gk,n

smooth ? ?

Note. By the construction, Gk,n is the quotient of V k,n by the action of O(k). This gives identi-cations:

Gk,n = V k,n /O (k) = O(n)/ (O(k) O(n k)) .

4. Fibre Bundles and Vector Bundles

4.1. Fibre Bundles. A bundle means a triple ( E,p,B ), where p : E B is a (continuous) map.The space B is called the base space , the space E is called the total space , and the map p is calledthe projection of the bundle. For each bB , p 1(b) is called the bre of the bundle over bB .

Intuitively, a bundle can be thought as a union of bres f 1(b) for b B parameterized by Band glued together by the topology of the space E . Usually a Greek letter ( ,,,, etc) is used todenote a bundle; then E () denotes the total space of , and B () denotes the base space of .

A morphism of bundles ( , ) : is a pair of (continuous) maps : E () E ( ) and : B () B ( ) such that the diagram

E () - E ( )

B ()

p()

? - B ( )

p( )

?

commutes.The trivial bundle is the projection of the Cartesian product:

p: B F B, (x, y ) x.

Roughly speaking, a bre bundle p : E B is a locally trivial bundle with a xed bre F .More precisely, for any x B , there exists an open neighborhood U of x such that p 1(U ) is a trivial


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bundle, in other words, there is a homeomorphism U : p 1(U ) U F such that the diagram

U F x- p 1(U )

U

1

?========= U

p

?

commutes, that is, p((x , y)) = x for any x U and y F .Similar to manifolds, we can use chart to describe bre bundles. A chart (U, ) for a bundle

p: E B is (1) an open set U of B and (2) a homeomorphism : U F p 1(U ) such that p((x , y)) = x for any x U and y F . An atlas is a collection of charts {(U , )} such that{U } is an open covering of B .

Proposition 4.1. A bundle p: E B is a bre bundle with bre F if and only if it has an atlas.

Proof. Suppose that p: E B is a bre bundle. Then the collection {(U (x), x ) |x B } is an atlas.Conversely suppose that p: E B has an atlas. For any x B there exists such that x U and so U is an open neighborhood of x with the property that p| p 1 (U ) : p 1(U ) U is a trivialbundle. Thus p : E B is a bre bundle.

Let be a bre bundle with bre F and an atlas {(U , )}. The composite

1 : (U U ) F - p 1(U U )

1- (U U ) F has the property that

1 (x, y ) = ( x, g (x, y ))for any x U U and y F . Consider the continuous map g : U F F . Fixing any x,g (x, ) : F F , y g (x, y ) is a homeomorphism with inverse given by g (x, ). This givesa transition function

g : U U - Homeo(F, F ),where Homeo( F, F ) is the group of all homeomorphisms from F to F .

Exercise 4.1. Prove that the transition functions {g } satisfy the following equation (3) g (x) g (x) = g (x) x U U U .

By choosing = = , g (x) g (x) = g (x) and so(4) g (x) = x x U .

By choosing = , g (x) g (x) = g (x) = x and so

(5) g (x) = g (x) 1 x U U .We need to introduce a topology on Homeo( F, F ) such that the transition functions g are

continuous. The topology on Homeo( F, F ) is given by compact-open topology briey reviewed asfollows:

Let X and Y be topological spaces. Let Map( X, Y ) denote the set of all continuous maps fromX to Y . Given any compact set K of X and any open set U of Y , let

W K,U = {f Map( X, Y ) | f (K ) U }.


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24 JIE WU

Then the compact-open topology is generated by W K,U , that is, an open set in Map( X, Y ) is anarbitrary union of a nite intersection of subsets with the form W K,U .

Map( F, F ) be the set of all continuous maps from F to F with compact open topology. ThenHomeo(F, F ) is a subset of Map( F, F ) with subspace topology.Proposition 4.2. If Homeo(F, F ) has the compact-open topology, then the transition functionsg : U U Homeo(F, F ) are continuous.

Proof. Given W K,U , we show that g 1 (W K,U ) is open in U U . Let x0 U U such thatg (x0) W (K,U ) . We need to show that there is a neighborhood V is x0 such that g (V ) W K,U ,or g (V K ) U . Since U is open and g : (U U ) F F is continuous, g 1(U ) is anopen set of (U U ) F with x0 K g 1 (U ). For each y K , there exist open neighborhoodsV (y) of x and N (y) of y such that V (x) N (y) g 1 (U ). Since {N (y) | y K } is an open cover

of the compact set K , there is a nite cover {N (y1), . . . , N (yn )} of K . Let V =n

j =1V (yj ). Then

V K g 1

(U ) and so g

(V ) W K,U

.

Proposition 4.3. If F regular and locally compact, then the composition and evaluation maps

Homeo(F, F ) Homeo(F, F ) - Homeo(F, F ) (g, f ) f gHomeo(F, F ) F - F (f, y ) f (y)

are continuous.

Proof. Suppose that f g W K,U . Then f (g(K )) U , or g(K ) f 1(U ), and the latter is open.Since F is regular and locally compact, there is an open set V such that

g(K ) V V f 1(U )

and the closure V is compact. If g W K,V and f W V ,U , then f g W K,U . Thus W K,V andW V ,U are neighborhoods of g and f whose composition product lies in W K,U . This implies that

Homeo(F, F ) Homeo(F, F ) Homeo(F, F ) is continuous.Let U be an open set of F and let f 0(y0) U or y0 f 10 (U ). Since F is regular an locally

compact, there is a neighborhood V of y0 such that V is compact and y0 V V g 10 (U ).If g W V ,U and y V , then g(y) U and so the evaluation map Homeo( F, F ) F F iscontinuous.

Proposition 4.4. If F is compact Hausdorff, then the inverse map

Homeo(F, F ) - Homeo(F, F ) f f 1

is continuous.

Proof. Suppose that g 10 W K,U . Then g 10 (K ) U or K g0(U ). It follows that

F K F g0(U ) = g0(F U )

because g0 is a homeomorphism. Note that F U is compact, F K is open and g0 W F U,F K .If g W F U,F K , then, from the above arguments, g 1 W K,U and hence the result.

Note. If F is regular and locally compact, then Homeo( F, F ) is a topological monoid, namelycompact-open topology only fails in the continuity of g 1 . A modication on compact-open topologyeliminates this defect [1].


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4.2. G-Spaces and Principal G-Bundles. Let G be a topological group and let X be a space.A right G-action on X means a(continuous) map : X G X, (x, g) x g such that x 1 = xand ( x g) h = x (gh). In this case, we call X a (right) G-space. Let X and Y be (right) G-spaces.A continuous map f : X Y is called a G-map if f (x g) = f (x) g for any x X and g G. LetX/G be the set of G-orbits xG , x X , with quotient topology.

Proposition 4.5. Let X be a G-space.1) For xing any g G, the map x x g is a homeomorphism.2) The projection : X X/G is an open map.

Proof. (1). The inverse is given by x x g 1 .(2) If U is an open set of X ,

1((U )) =gG

U g

is open because it is a union of open sets, and so (U ) is open by quotient topology. Thus is anopen map.

We are going to nd some conditions such that : X X/G has canonical bre bundle structurewith bre G. Given any point x X/G , choose x X such that (x) = x. Then

1(x) = {x g | g G} = G/H x ,

where H x = {g G | x g = x}.For having constant bre G, we need to assume that the G-action on X is free, namely

x g = x = g = 1

for any x X . This is equivalent to the property that

x g = x h = g = h

for any x X . In this case we call X a free G-space .Since a bre bundle is locally trivial (locally Cartesian product), there is always a local cross-

section from the base space to the total space. Our second condition is that the projection : X X/G has local cross-sections. More precisely, for any x X/G , there is an open neighborhood U (x)with a continuous map s x : U (x) X such that s x = id U ( x ) .

(Note. For every point x, we can always choose a pre-image of , the local cross-section meansthe pre-images can be chosen continuously in a neighborhood. This property depends on thetopology structure of X and X/G .)

Assume that X is a (right) free G-space with local cross-sections to : X X/G . Let x be anypoint in X/G . Let U (x) be a neighborhood of x with a (continuous) crosse-section s x : U (x) X .Dene

x : U (x) G - 1(U (x)) ( y, g) - s x (y) gfor any y U (x).

Exercise 4.2. Let X be a (right) free G-space with local cross-sections to : X X/G . Then thecontinuous map x : U (x) G 1(U (x)) is one-to-one and onto.

We need to nd the third condition such that x is a homeomorphism. Let

X = {(x, x g) | x X, g G} X X.


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A function : X - G

such that x (x, x ) = x for all (x, x ) X is called a translation function . (Note. If X is a free G-space, then translation function is uniquebecause, for any ( x, x ) X , there is a unique g G such that x = x g, and so, by denition, (x, x ) = g.)

Proposition 4.6. Let X be a (right) free G-space with local cross-sections to : X X/G . Then the following statements are equivalent each other:

1) The translation function : X G is continuous.2) For any x X/G , the map x : U (x) G 1(U (x)) is a homeomorphism.3) There is an atlas {(U , } of X/G such that the homeomorphisms

: U G - 1(U )satisfy the condition (y,gh ) = (y, g) h, that is is a homeomorphism of G-spaces.

Proof. (1) = (2) . Consider the (continuous) map

: 1(U (x)) - U (x) G z ((z), (s x ((z)) , z)) .Then

x (y, g) = (s x (y) g) = ( y, (s x (y), s x (y) g)) = ( y, g),

x (z) = x ((z), (s x ((z)) , z)) = s x ((z)) (s x ((z)) , z) = z.Thus x is a homeomorphism.

(2) = (3) is obvious.(3) = (1). Note that the translation function is unique for free G-spaces. It suffices to show

that the restriction (X ) : X ( 1(U ) 1(U )) = 1(U )

- Gis continuous. Consider the commutative diagram

(U G)=- 1(U )

G

(U G)

?=========== G.

(X )

?

Since

(U G)(( y, g), (y, h )) = g 1his continuous, the translation function restricted to 1(U )

(X ) = (U G) (( )) 1

is continuous for each and so (X ) is continuous.


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Now we give the denition. A principal G-bundle is a free G-space X such that

: X X/G

has local cross-sections and one of the (equivalent) conditions in Proposition 4.6 holds.Example. Let be a topological group and let G be a closed subgroup. Then the action of Gon given by (a, g ) ag for a and g G is free. Then translation function is given by (a, b) = a 1b, which is continuous. Thus /G is principal G-bundle if and only if it has localcross-sections.

4.3. The Associated Principal G-Bundles of Fibre Bundles. We come back to look at brebundles given by p : E B with bre F . Let {(U , )} be an atlas and let

g : U U - Homeo(F, F )be the transition functions. A topological group G is called a group of the bundle if

1) There is a group homomorphism

: G - Homeo(F, F ).2) There exists an atlas of such that the transition functions g lift to G via , that is,

there is commutative diagram

U U g- Homeo(F, F )

U U g - G,

6

(where we use the same notation g .)3) The transition functions

g : U U - Gare continuous.

4) The G-action on F via is continuous, that is, the composite

G F id F - Homeo(F, F ) F evaluation- F

is continuous.We write = {(U , g )} for the set of transition functions to the atlas {(U , )}.

Note. In Steenrods denition [11, p.7], is assume to be a monomorphism (equivalently, theG-action on F is effective, that is, if y g = y for all y F , then g = 1.).

We are going to construct a principal G-bundle : E G B . Then prove that the total space

E = F G E G

and p : E B can be obtained canonically from : E G

B . In other words,all bre bundles can obtained through principal G-bundles through this way. Also the topologicalgroup G plays an important role for bre bundles. Namely, by choosing different topological groupsG, we may get different properties for the bre bundle . For instance, if we can choose G to betrivial (that is, g lifts to the trivial group), then bre bundle is trivial. We will see that the bundlegroup G for n-dimensional vector bundles can be chosen as the general linear group GL n (R ). The


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vector bundle is orientable if and only if the transition functions can left to the subgroup of GL n (R )consisting of n n matrices whose determinant is positive. If n = 2 m, then GL m (C ) GL2m (R ).The vector bundle admits (almost) complex structure if and only if the transition functions can leftto GL m (C ). (For manifolds, one can consider the structure on the tangent bundles. For instance,an oriented manifold means its tangent bundle is oriented.)

Proposition 4.7. If is the set of transition functions for the space B and topological group G,then there is a principal G-bundle G given by

: E G - B

and an atlas {(U , )} such that is the set of transition functions to this atlas.

Proof. The proof is given by construction. Let

E =

U G ,

that is E is the disjoint union of U G. Now dene a relation on E by

(b,g,) (b , g , ) b = b , g = g (b)g .

This is an equivalence relation by Equations (3)-(5). Let E G = E/ with quotient topology andlet {b,g,} for the class of (b,g,) in E G . Dene : E G B by

{b,g,} = b,

then is clearly well-dened (and so continuous). The right G-action on E G is dened by

{b,g,} h = {b,gh, }.

This is well-dened (and so continuous) because if ( b , g , ) (b,g,), then

(b , g h, ) = ( b, (g (b)g)h, ) = ( b, g (b)(gh), ) (b,gh, ).

Dene : U G 1(U ) by setting

(b, g) = {b,g,},

then is continuous and satises (b, g) = b and

(b, g) = {b,1 g, } = {b,1, } g

for bU and g G. The map is a homeomorphism because, for xing , the map

(U U ) G - U G (b, g , ) (b, g (b)g )

induces a map 1(U ) U G which the inverse of . Moreover,

(b, g (b)g) = {b, g (b)g, } = {b,g, } = (b, g)

for b U U and g G. Thus the {(U , g )} is the set of transition function to the atlas{(U , )}.


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Let X be a right G-space and let Y be a left G-space. The product over G is dened by

X G Y = X Y/(xg,y) (x,gy)

with quotient topology. Note that the composite

X Y X - X - X/G

(x, y ) x xfactors through X G Y . Let p : X G Y X/G be the resulting map. For any x X/G , choosex 1(x) X , then

p 1(x) = 1(x) G Y = x Y/H x ,where H x = {g G | xg = x}. Thus if X is a free right G-space, then the projection p: X G Y X/G has the constant bre Y .

Proposition 4.8. Let : X X/G be a (right) principal G-bundle and let Y be any left G-space.Then

p: X G Y - X/Gis a bre bundle with bre Y .

Proof. Consider a chart ( U , ) for : X X/G . Since the homeomorphism : U G 1(U ) is a G-map, there is a commutative diagram

U Y === ( U G) G Y =- 1(U ) G Y === p 1(U )

U

U ?

=========== U

U ?

============= U

p

?========== U

p

?

and hence the result.

Let be a (right) principal G-bundle given by : X X/G . Let Y be any left G-space. Thenbre bundle

p: X G Y - X/Gis called induced bre bundle of , denoted by [Y ].

Now let p : E B is a bre bundle with bre F and bundle group G. Observe that the actionof Homeo(F, F ) on F is a left action because ( f g)(x) = f (g(x)). Thus G acts by left on F via : G Homeo(F, F ).

A bundle morphism

E () -

E ( )

B ()

p()

? - B ( )

p( )

?


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30 JIE WU

is call an isomorphism if both and are homeomorphisms. ( Note. this means that ( 1 , () 1)are continuous.) In this case, we write = .

Theorem 4.9. Let be a bre bundle given by p: E B with bre F and bundle group G. Let G

be the principal G-bundle constructed in Proposition 4.7 according to a set of transitions functionsto Then G [F ]= .

Proof. Let {(U , )} be an atlas for . We write for in the proof of Proposition 4.7. Considerthe map given by the composite:

1(U ) G F idF =

(U G )G F === U F =- p 1(U ).

From the commutative diagram

((U U ) G ) G F (b, g , y) (b, g , g (b)(y))

((b, g , ), y) ((b, g (b)g , ), y)- ((U U ) G ) G F

(U U ) F (b, y) (b, g (b, y)) - (U U ) F

p 1(U U )

= ?

=========================================== p 1(U U ),

= ?

the map induces a bundle map

E G

G F -

E ()

B () ?

====== B (). ?

This is an bundle isomorphism because is one-to-one and onto, and is a local homeomorphic byrestricting each chart. The assertion follows.

This theorem tells that any bre bundle with a bundle group G is an induced bre bundle of a principal G-bundle. Thus, for classifying bre bundles over a xed base space B , it suffices toclassify the principal G-bundles over B . The latter is actually done by the homotopy classes from

B to the classifying space BG of G. (There are few assumptions on the topology on B such as B isparacompact.) The theory for classifying bre bundles is also called (unstable) K -theory , which isone of important applications of homotopy theory to geometry. Rough introduction to this theoryis as follows:

There exists a universal G-bundle G as : EG BG . Given any principal G-bundle over B ,there exists a (continuous) map f : B BG such that , as a principal G-bundle, is isomorphic to


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the pull-back bundle f G given by

E (f G ) = {(x, y ) B EG | f (x) = (y)} - EG

B

(x, y ) x

? f - BG.

?

Moreover, for continuous maps f, g : B BG , f G = gG if and only if f g, that is, there is acontinuous map (called homotopy ) F : B [0, 1] BG such that F (x, 0) = f (x) and F (x, 1) = g(x).In other words, the set of homotopy classes [ B,BG ] is one-to-one correspondent to the set of isomorphic classes of principal G-bundles over G.

Seminar Topic: The classication of principal G-bundles and bre bundles. (References: forinstance [8, pp.48-58] Or [9, 10].)

4.4. Vector Bundles. Let F denote R , C or H -the real, complex or quaternion numbers. Ann-dimensional F -vector bundle is a bre bundle given by p: E B with bre F n and an atlas{(U , )} in which each bre p 1(b), b B , has the structure of vector space over F such thateach homeomorphism : U F n p 1(U ) has the property that

|{b} F n : {b} F n - p 1(b)

is a vector space isomorphism for each bU .Let be a vector bundle. From the composite

(U U ) F n - p 1(U U )

1- (U U ) F n ,

the transition functions

g : U U -

Homeo(Fn

, Fn

)have that property that, for each x U ,

g (x) : F n - F n

is a linear isomorphism . It follows that the bundle group for a vector bundle can be chosen asthe general linear group GL n (F ). By Theorem 4.9, we have the following.

Proposition 4.10. Let be an n-dimensional F -vector space over B . Then there exists a principal GLn (F )-bundle GL n ( F ) over B such that = GL n ( F ) [F n ]. Conversely, for any principal GLn (F )-bundle over B , GL n ( F ) [F n ] is an n-dimensional F -vector bundle over B .

In other words, the total spaces of all vector bundles are just given by E (GL n (F ) ) GL n ( F ) F n .

4.5. The Construction of Gauss Maps. The Grassmann manifold Gn,m (F ) is the set of n-dimensional F -subspaces of F m , that is, all n-F -planes through the origin, with the topology de-

scribed as in the topic on the examples of Manifolds. (If m = , F =

j =1F .) Let

E ( mn ) = {(V, x) Gn,m (F ) F m | x V }.


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Exercise 4.3. Show that p: E ( mn ) Gn,m (F ) (V, x) V

is an n-dimensionalF

-vector bundle, denoted by m

n . [Hint: By reading the topic on the examples of manifolds, check that V n,m (F ) Gn,m (F ) is a principal O(n, F ), where O(n, R ) = O(n), O(n, C ) =U (n) and O(n, H ) = Sp( n). Then check that E ( mn ) = V n,m (F ) O (n, F ) F n .]

A Gauss map of an n-dimensional F -vector bundle in F m (n m ) is a (continuous) mapg : E () F m such that g restricted to each bre is a linear monomorphism.Example. The map

q: E ( mn ) F m (V, x) xis a Gauss map.

Proposition 4.11. Let be an n-dimensional F -vector bundle.1) If there is a vector bundle morphism

E ()u-

E ( mn )

B ()

p()

? f - Gn,m (F )

p( mn )

?

that is an isomorphism when restricted to any bre of , then q u : E () F m is a Gaussmap.

2) If there is a Gauss map g : E () F m , then there is a vector bundle morphism (u, f ) : mn such that qu = g.

Proof. (1) is obvious. (2). For each b B (), g( p() 1(b)) is an n-dimensional F -subspace of F mand so a point in Gn,m (F ). Dene the functions

f : B () Gn,m (F ) f (b) = g( p() 1(b)) ,

u : E () E ( mn ) u(z) = ( f ( p(z)) , g(z)) .The functions f and u are well-dened. For checking the continuity of f and u, one can look at alocal coordinate of and so we may assume that is a trivial bundle, namely, g : B () F n F mrestricted to each bre is a linear monomorphism. Let {e1 , . . . , e n } be the standard F -bases for F n .Then the map

h : B - F m F m b (g(b, e1), g(b, e2), . . . , g (b, en ))is continuous. Since g restricted to each bre is a monomorphism, the vectors

{g(b, e1), g(b, e2), . . . , g (b, en )}

are linearly independent and so(g(b, e1), g(b, e2), . . . , g (b, en )) V n,m (F )

for each b, where V n,m (F ) is the open Stiefel manifold over F . Thus

h : B - V n,m (F ) b (g(b, e1), g(b, e2), . . . , g (b, en ))


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is continuous and so the composite

f : Bh- V n,m (F )

quotient-- Gn,m (F )

is continuous. The function u is continuous because the composite

E ()u - E ( mn )

E () E ()

? (f p()) g- Gn,m (F ) F m ?

is continuous. This nishes the proof.

Let be a vector bundle and let f : X B () be a (continuous) map. Then the induced vector f is the pull-back

E (f

) = {(x, y ) X E () |f (x) = p(y)}-

E ()

X ? f - B ().

p

?

Proposition 4.12. There exists a Gauss map g : E () F m ( n m ) if and only if = f ( mn )

over B () for some map f : B () Gn,m (F ).

Proof. = is obvious.= Assume that has a Gauss map g. From Part (2) of Proposition 4.11, there is a commutative

diagramE ()

u- E ( mn )

B ()

p()

? f - Gn,m (F ).

p( mn )

?

Since E (f mn ) is dened to be the pull-back, there is commutative diagram

E ()u- E (f mn )

B ()

p() ?

====== B (),

p ?

where u restricted to each bre is a linear isomorphism because both vector-bundle has the samedimension and the Gauss map g restricted to each bre is a monomorphism. It follows that


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34 JIE WU

u : E () E (f mn ) is one-to-one and onto. Moreover u is a homeomorphism by considering alocal coordinate.

We are going to construct a Gauss map for each vector bundle over a paracompact space. First,we need some preliminary results for bundles over paracompact spaces. (For further information onparacompact spaces, one can see [3, 162-169].

A family of C = {C | J } of subsets of a space X is called locally nite if each x X admits aneighborhood W x such that W x C = for only nitely many indices J . Let U = {U } andV = {V } be two open covers of X . V is called a renement of U if for each , V U for some .

A Hausdorff space X is called paracompact if it is regular and if every open cover of X admits alocally nite renement.

Let U = {U | J } be an open cover of a space X . A partition of unity , subordinate to U , isa collection { | J } of continuous functions : X [0, 1] such that

1) The support supp( ) U for each , where the support

supp( ) = {x X | (x) = 0 }is the closure of the subset of X on which = 0;

2) for each x X , there is a neighborhood W x of x such that |W x 0 for only nitelymany indices J . (In other words, the supports of s are locally nite.)

3) The equation

J

(x) = 1

for all x X , where the summation is well-dened for each given x because there are onlynitely many non-zeros.

We give the following well-known theorem without proof. One may read a proof in [2, pp.17-20].

Theorem 4.13. If X is a paracompact space and U = {U } is an open cover of X , then thereexists a partition of unity subordinate to U .

Lemma 4.14. Let be a bre bundle over a paracompact space B . Then admits an atlas with countable charts.

Proof. Let {(U , | J } be an atlas for . We are going to nd another atlas with countablecharts.

By Theorem 4.13, there is a partition of unity { | J } subordinate to {U | J }. LetV = 1 (0, 1] = {bB | (b) > 0}.

Then, by the denition of partition of unity, V V

U

. For each bB , let

S (b) = { J | (b) > 0}.Then, by the denition of partition of unity, S (b) is a nite subset of J .

Now for each nite subset S of J dene

W (S ) = {bB | (b) > (b) for each S and S }


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= S S

( ) 1(0, 1].

Then W (S ) is open because for each b W (S ), by denition of partition of unity, there exists aneighborhood W b of b such that there are nitely many supports intersect with W b; and so the above(possibly innite) intersection of open sets restricted to W b is only a nite intersection of open sets.

Let S and S be two subsets of J such that S = S and |S | = |S | = m > 0, where |S | is thenumber of elements in S . Then there exist S S and S S because S = S but S andS has the same number elements. We claim that

W (S ) W (S ) = .

Otherwise there exists bW (S ) W (S ). By denition W (S ), (b) > (b) because S and S . On the other hand, (b) > (b) because bW (S ), S and S .

Now dene

W m =bB

|S | = m

W (S (b))

for each m 1. We prove that (1) {W m | m = 1 , . . .} is an open cover of B ; and (2) restricted toW m is a trivial bundle for each m. (Then {W m } induces an atlas for .)

To check {W m } is an open cover, note that each W m is open. For each bB , S (b) is a nite setand bW (S (b)) because (b) = 0 for S (b) and (b) > 0 for S (b). Let m = |S (b) |, thenbW m and so {W m } is an open cover of B .

Now check that restricted to W m is trivial. From the above, W m is a disjoint union of W (S (b)).It suffices to check that restricted to each W (S (b)) is trivial. Fixing S (b), for any x W (S (b)),then

(x) > (x)for any S . In particular, (x) > 0 for any x W (S (b)). It follows that W (S (b)) V U .Since restricted to U is trivial, restricted to W (S (b)) is trivial. This nishes the proof.

Note. From the proof, if for each bB there are at most k sets U with bU , then B admitsan atlas of nite (at most k) charts. [In this case, check that W j = for j > k .]

Theorem 4.15. Any n-dimensional F -vector bundle over a paracompact space B has a Gauss mapg : E () F . Moreover, if has an atlas of k charts, then has a Gauss map g : E () F kn .

Proof. Let {(U i , i )}1 i k be an atlas of with countable or nite charts, where k is nite or innite.Let { i } be the partition of unity subordinate to {U i }. For each i, dene the map gi : E () F nas follows: gi restricted to p() 1(U i ) is given by

gi (z) = i (z)( p2 1i (z)) ,

where p2 1i is the composite

p() 1(U i ) 1i- U i F n

projection p2

- F n ;


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and gi restricted to the outside of p() 1(U i ) is 0. Since the closure of 1i (0, 1] is contained in U i ,gi is a well-dened (continuous) map. Now dene

g : E () k

i =1

F n = F kn g(z) =k

i =1

gi (z).

This a well-dened (continuous) map because for each z, there is a neighborhood of z such thatthere are only nitely many gi are not identically zero on it.

Since each gi : E () F n is a monomorphism (actually isomorphism) on the bres of E () overb with i (b) > 0, and since the images of gi are in complementary subspaces of F kn , the map g is aGauss map.

This gives the following classication theorem:

Corollary 4.16. Every vector bundle over a paracompact space B is isomorphic to an induced vector bundle f ( n ) for some map f : B Gn, (F ). Moreover every vector bundle over a paracompact

space B with an atlas of nite charts is isomorphic to an induced vector bundle f

( mn ) for some mand some map f : B Gn,m (F ).

Remarks: It can be proved that f mn = g mn if and only if f g : B Gn,m (F ). From this,one get that the set of isomorphism classes of n-dimensional F -vector bundles over a paracompactspace B is isomorphic to the set of homotopy classes [ B, G n, (F )].

For instance, if n = 1 and F = R , G1, (R ) BO (1) B Z / 2 RP , where BG is so-calledthe classifying space of the (topological) group G, and [B, RP ] = H 1(B, Z / 2), which states thatall line bundles are by the rst cohomology with coefficients in Z / 2Z . In particular, any real linebundles over a simply connected space is always trivial.

If n = 1 and F = C , then G1, (C ) BU (1) BS 1 CP , and [B, C P ] = H 2(B, Z ), whichstates that all complex line bundles are by the second integral cohomology.

If n = 1 and F = H , then G1, (H ) B Sp(1) BS 3

HP

, and so [B, G 1, (H )] = [B, H P

].However, the determination of [ B, H P ] is very hard problem even when B are spheres. If B = S n ,then [B, H P ] = n 1(S 3) that is only known for n less than 66 or so, by a lot of computationsthrough many papers. Some people even believe that it is impossible to compute the generalhomotopy groups n (S 3).

Seminar Topic: Gauss Maps and the Classication of Vector Bundles. (References: for instance [8,pp.26-29,31-33].)

A vector bundle is called of nite type if it has an atlas with nite charts. Given two vectorbundles and over B , the Whitney sum is dened to be the pull-back:

E () - E () E ()

B ?

diagonal- B B.

p() p() ?

Intuitively, is just the brewise direct sum.


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Proposition 4.17. For a vector bundle over a paracompact space B , the following statement areequivalent:

1) The bundle is of nite type.2) There exists a map f : B Gn,m (F ) such that is isomorphic to f mn .3) There exists a vector bundle such that the Whitney sum is trivial.

Proof. (1) = (2) follows from Corollary 4.16. (2) = (1) It is an exercise to check that mn isof nite type by using the property that the Grassmann manifold Gn,m (F ) = O(m, F )/O (n, F ) O(m n, F ) is compact, where O(n, R ) = O(n), O(n, C ) = U (n) and O(n, H ) = Sp( n). It followsthat = f mn is of nite type.

(2) = (3). Let ( mn ) be the vector bundle given by

E (( mn )) = {(V, v) Gn,m (F ) F m | vV }

with canonical projection E (( mn )) Gn,m (F ). Then mn ( mn ) is an m-dimensional trivialF -vector bundle. It follows that

f ( mn ( mn )) = f (

mn ) f ((

mn ))

is trivial. Let = f (( mn )). Then is trivial.(3) = (2). The composite

E () - E () = B F m - F m

is a Gauss map into nite dimensional vector space, where m = dim( ). By Proposition 4.12,there is a map f : B Gn,m such that = f mn .

Corollary 4.18. Let be a F -vector bundle over a compact (Hausdorff) space B . Then there is a F -vector bundle such that is trivial.

In the view of (stable) K -theory, the Whitney sum is an operation on vector bundles over a (xed)base-space, where the trivial bundles (of different dimensions) are all regarded as 0. In this sense,the Whitney sum plays as an addition (that is associative and commutative with 0). The bundle with property that is trivial for some means that is invertible. Those who are interested

in algebra can push notions in algebra to vector bundles by doing constructions brewisely. Moregeneral situation possibly is the sheaf theory (by removing the locally trivial condition) that is prettyuseful in algebraic geometry. In algebraic topology, people also study the category whose objects are just continuous maps f : E B with xed space B , or even more general category whose objectsare diagrams over spaces. In the terminology of bre bundles, a map f : E B is called a bundle(without assuming locally trivial).


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38 JIE WU

5. Tangent Bundles and Vector Fields

6. Cotangent Bundles and Tensor Fields

7. Orientation of Manifolds

8. Tensor Algebras and Exterior Algebras

9. DeRham Cohomology

10. Integration on Manifolds

11. Stokes Theorem

References

[1] R. Arens, Topologies for homeomorphism groups , Amer. Jour. Math. 68 (1946), 593-610.[2] Lawrence Conlon, Differentiable Manifolds, second Edition , Birkh auser Boston [2001].[3] J. Dugundji, Topology , Allyn and Bacon, Boston, MA [1966].[4] M. A. Kervaire and J. W. Milnor, Groups of homotopy spheres I , Ann. Math. 77 504-537, (1963).

[5] Serge Lang, Differential Manifolds , Addison-Wesley Publishing Company, Inc. [1972].[6] Victor Guillemin and Alan Pollack, Differential Topology , Prentice Hall, Inc, Englewood Cliffs, New Jersey

[1974].[7] Dominic G. B. Edelen, Applied Exterior Calculus , John Wiley & Sons, Inc. [1985].[8] Dale Husemoller, Fibre Bundles, second edition , Graduate Texts in Mathematics Springer-Verlag [1966].[9] J. Milnor, Construction of universal bundles I , Ann. Math. 63 (1956), 272-284.

[10] J. Milnor, Construction of universal bundles II , Ann. Math. 63 (1956), 430-436.[11] Norman Steenrod, The Topology of Fibre Bundles , Princeton University Press, Princeton, New Jersey, Ninth

Printing, [1974].

Intro to Differ en Ti Able Manifolds

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