Chapter 4 Tangent Bundles, Vector Bundles, and Classification

CHAPTER

4 Tangent Bundles, Vector Bundles, and Classification

The tangent space to a manifold M" is built locally from copies of 0, x R", where 0,is a coordinate chart in M".If a point lies both in 0, x R" and 0, x R", a double representation is avoided (see Definition 2.5) by an identification or equivalence relation that is linear in the second variable. This situation arises in other, rather different situations. For example, we may consider a submanifold S P G R k . For each x E S P , we then let N i - P be the ( k - p)-dimensional space of vectors that begin at x and are perpendicular to the tangent space to S p at x, i.e., perpendicular to every vector in (TSP),. One may easily topologize the union

u

Nk-P,

XSSP

in a manner similar to the definition of the tangent space, and we write this space, often called the normal bundle, with the notation ( N S p ) k - POr, . for another example, if S P E N k , one may restrict attention, to the tangent space (TNk),,for x E S p . Over each point of the p-dimensional S p , one then has a k-dimensional vector space ( TNk),. The gist of all these constructions, whose geometric basis is really quite well motivated, is the local procedure of taking a bit of a manifold, forming a7

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a Cartesian product with some Euclidean space, and then performing some identifications that are linear in the second variable at each point. Such constructions are now called vector bundles, I presume because near any point they appear to be a bunch of vector spaces. They have played an increasingly important role in modern mathematics, culminating in the sophisticated subject of K-theory, which studies equivalence classes of vector bundles over a given space. Our goals in this chapter are the following: (i) analyze the structure of the tangent space to a differentiablemanifold from this somewhat deeper point of view; (ii) define vector bundles in general, and develop all the basic properties; and (iii) prove a classification theorem for vector bundles over a compact Hausdorff space. Hardly anything we do in this chapter cannot be done in some greater degree of generality-usually with much more work. We shall include at the end a brief survey of where one may find the more striking generatizations. It will be convenient at many places to discuss topological groups and their actions on spaces. This will be our starting point here. GROUPS ACTING ON SPACES

Definition 4.1 Let G be a group that is also endowed with the structure of a topological space. We call G a topological group if the map $:GxG+G

defined by $(x, y ) = xy-' is continuous. Remarks (1) It is clearly equivalent to require that the multiplication and inverse maps, respectively

(x,y)+xy

and

x+x-l,

be both continuous. (2) Naturally, any group that is endowed with the discrete topology is a topological group. (3) Gl(n; R), the group of invertible n x n real matrices, is clearly a topological group. (4) Products and subgroups afford easy ways of generating various examples of topological groups.

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Definition 4.2 Let G be a topological group and X a topological space. We say that G acts on X , or that there is an action of G on X , if we are given a continuous map p:GxX+X such that = p(g1g2,x ) for any gl, g2 E G and x (a) p(g,, p ( g 2 , 4) (b) if e E G is the identity element in G, and x E X ,

E

X ; and

p ( e , x ) = x.

Sometimes one calls G a transformation group (or a left transformation group acting on X ) .

Remarks (1) A right action, or right transformation group, would be defined by a map p : X x G -,X with the appropriate properties. (2) If G is a topological group, p : G x G + G, defined by p(x, y ) = xy defines an action of the topological group G on the topological space G (check this.) (3) If G E Gl(n; R) is any subgroup, there is an action of G on R", defined by left matrix multiplication. Specifically, if M E G, X E R", we set p(M,X ) = M X

E

R".

(4) If p(g, x) = y, p ( g - ' , y) = x (for p(e, x) = x and gg-' = g-'g = e, the identity element.) ( 5 ) A matrix M E Gl(n; R) is orthogonal if M - ' = M', where M i , the transpose of M , interchanges the rows and columns of M , If ( , ) denotes the inner product and X , Y E R", then

( M X ,M X )= (M'MX,X ) =( M - l M X , X ) =(X, X). Thus, if M is orthogonal llMXll = IIXII, so that left multiplication by an orthogonal matrix sends S"- to S"- . If the orthogonal matrices are denoted O(n),clearly O(n)is a closed subgroup of Gl(n; R) which acts, as a (left) transformation group, on S"- '. There are several kinds of transformation groups that are noteworthy. We discuss this more in Chapter 10, but two kinds of action are relevant to our needs here.

'

Definition 4.3 Given an action p : G x X

+

X and a set S E X , we write

Gs= { g E Glp(g, y) = y, for all y E S } . (a) We say that the action is eflectiue if Gx = { e } ,that is, the only element leaving all of X fixed is e, the identity.

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(b) We say that the action is transitiue if given x, y E X there is some

g E G such that p(g, x) = y .

Remarks (1) In Remark (2) before Definition 4.3., the action is visibly effective and transitive. (2) A trivial example of a transitive action that is not effective occurs when G is a nontrivial group but X is a single point. Similarly, if G = { e } ,the action of G on any X is effective. (3) The action of Gl(n; R) on R", as described in Remark (3) preceding Definition 4.3., is visibly both effective and transitive, omitting the origin. (4) Given an action p : G x X + X and a subgroup H c G, there is an action p n : H x X + X,

defined by p(g, x), when g E H. If p is effective, so is pH. On the other hand, if p,, is transitive, so is p. The following proposition clarifies the importance of transitive actions: Proposition 4.1 Let X be a Hausdorff space and p : G x X sitive action. Let xo E X and recall

-,X a tran-

G,, = (9 E G(PkI,xo) = xo>. Note that G,, is closed. Define G/G,, to be the quotient space of G by the equivalence relation g1 g2 if and only if g;'g2 E G,,. (G/G,, is called a coset space; in this case it is easily checked to be Hausdorff.) If G/G,, is compact, then there is a natural homeomorphism

-

4 : G/G,, -, X.

-

Proof Define 4({g}) = p(g, xo). Note that ifg g', so that (g')-'g E G,,, then 4%')) = cL((g')-'g, XO) = A g , XO) = 4({g)), so that 4 XO) = is well defined. Continuity is immediate. Because G acts transitively, 6 is immediately checked to be onto. Now if 4({Sd) = 4({92D,then

or acting by g;

' we get

cL(s1, xo) = pL(g2, xo),

xo = CL(g;lgl,

xo) = Cl(g;lgz,

xo).

But this implies, by definition, that g; 'g2 E G,, (equivalently; g; 'gl E G,,, because G,, is a subgroup). But this means that if 4({gl})= 4({g2}), then g1 g2. Therefore, 4 is 1-1.

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THE TANGENT BUNDLE

91

Naturally a 1-1 map from a compact Hausdorff space onto another is a homeomorphism. Remarks (1) A standard case to which this situation applies is the action of O(n),the group of n x n orthogonal matrices, on S"-' [see Remark ( 5 ) preceding Definition 4.33. The subgroup of O(n)that leaves a unit vector, such as x o = (1,0, . . . ,0) fixed, is easily checked to isomorphic to O(n - 1) in the form

with * an (n - 1) x ( n - 1) orthogonal matrix. In this case one occasionally sees written O(n)/o(n- 1) = S"-'. (2) In case G,, is a normal closed subgroup of G , G/G,, and hence X inherits the structure of a topological group. For example, let Go = 2, the group of integers, and G = R, the additive group of real numbers, in the case of R acting on the circle S' via the prescription that a acts on a point by a counterclockwise rotation of 2na. Then Proposition 4.1 tells us R/Z = S' . THE TANGENT BUNDLE

We next will utilize the action of Gl(n; R) on R" to exhibit some richer structure in the tangent space to a manifold. This will be a prototype for our vector bundles. Consider T , = 0, x R" with 4,: 0, + R" some coordinate chart or neighborhood in some manifold. T , is a building block in our construction of T M " . We note that we may define an action of Gl(n; R) on Taby AM, ( x , Y ) ) = ( x , M Y )

with M E Gl(n; R) and M y left matrix multiplication of y E R". More generally, the equivalencerelation that we used in the definition of T M "(Definition 2.5) places ( x , y) and ( x , J,,, ~(+,(x))y)in the same equivalence class whenever x E 0, n 0,. We now rewrite this in the language of actions as follows: there is a continuous map, J,, , I from +,(O, n 0,)to Gl(n; R) such that whenever x E 0, n O,,

,,

@ ,

(x, Y ) 1 ( x , J,,

0

4;

4+a(x))Y)-

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To put it differently, the equivalence relation in Definition 2.5 is obtained in terms of a continuous map J from 0, n 0, to Gl(n; R) that identifies two values of the second coordinate whenever the action J at the point takes the one second coordinate into the other. Using these ideas, we rewrite the definition of TM" is a way that remarkably close to the original definition of a manifold. Definition 4.4 (Rewording of TM" according to Definition 2.5) Let M" be a Cr-manifold, r 2 1, with a cover by coordinate charts { O,}, each with a given homeomorphism 4,: 0,sR". By the tangent space to M" we mean the space TM"endowed with a map n: TM" + M", specified by the following structure: Every point z E TM" has a neighborhood U , that is homeomorphic to 0, x R", where 0, is a coordinate chart for M". We write the homeomorphism $,: U , -,0, x R". If z E U , n U p , then in a neighborhood of $,(z) the map $, ;)I is 0

(x, Y )

+

(x, J+p 4; 0

l ( 4 d X ) )Y).

The map n: TM" + M" is defined in any U , by ~ ( xy), = x. The passage from tangent bundle to arbitrary vector bundle is clear. M" is generalized to a topological space covered by {O,}.The J's are no longer Jacobians in general, but (to assure that we have an equivalence relation) they are functions (continuous) to Gl(n; R) that satisfy J,. +, I = identity, as well as JbV +; 1J+@ +; 1 = JcPv +, 1, where defined. r)

(I

VECTOR BUNDLES

Definition 4.5 An n-dimensional real vector bundle over a topological spaceX, written@, X, p)consistsofaspaceEandacontinuousmapp: E + X such that for each x E X,p - ' ( x ) is homeomorphic to R", and for each x there is an open set 0,, x E 0,, and a homeomorphism

$,: p-'(o,)+ 0,x R" such that (a) if n,: 0, x R" + 0, is projection onto the first coordinate and Y E ~ - l ( O a ) , then na($a(y)) = P(Y). (b) Whenever 0, n 0, # 4, there is a continuous map gas: 0, n 0, + Gl(n; R) such that (i) g,, = 1, the identity, (ii) gapgsy= gay,where defined, and

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93

Remarks (1) The definition of a vector bundle has much in common with the definition of a manifold; it is specified in terms of local structure, with the overlap transformation being the identity in the first variable and a (continuous family of) linear transformations in the second variable. Note that if 0, is a smaller open set than 0, one may define $, and the gyBby restriction. (2) If M" is an n-manifold, TM" arises by choosing the 0, to be coordinate neighborhoods and setting gaB(x)= J,, ,;-I(@,(X)). If X E M" is a subspace, we set E = K '(X) and p = K ) E to get a vector bundle over X,( E , X,p). This is the restriction of TM" to X. For any space Y , a trivial n-dimensional vector bundle may be constructed by setting Y x R" = E. (3) A similar definition may be given over any topologized field (or even skew field). p- '(x) would then be an n-dimensional vector space over that field, and Gl(n; R) would be replaced by the general linear group over that field. For example, in the case of a complex vector bundle, we would have

$,-yo,) = 0,x C",

where C is the complex numbers. (4) Neither E nor X need be connected, but since the map p is onto (obviously) when E is connected, then so is X. But the converse is also true, because each set p - ' ( x ) is connected, so any two points in E lie in connected sets p - ' ( x ) and p - ' ( y ) , both of which would meet the connected set {(x,O) E Elx E X } (this latter being homeomorphic to X by p ( x ) = {(x,O)}). ( 5 ) If X = M k ,a k-dimensional manifold, and ( E , X,p ) is an n-dimensional (real) vector bundle over X, then E is an (n + k)-dimensional manifold. Depending on how much differentiabilityone assumes on X and the functions gab, one can prove differentiability on E. In particular, if X and all gmaare smooth, so is E . (6) The general philosophy will be to fix X and examine various possible vector bundles over X. For example, in the introduction to this chapter we looked at a submanifold M" G Rk and considered the tangent bundle to M", the normal bundle to M", namely (NM")k-",and (TIWk)lM",which is the restriction of T R k to the subspace M" as in (2) above. Over any point, the bundle (T R k )I M" clearly splits into perpendicular pieces, one in TM" and the other in (NM")k-".This is the operation of taking a direct sum of two vector subspaces that intersect only at the zero vector. In fact, we shall

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see that one can do almost anything with vector bundles that one could have done with vector spaces. First, however, it is important to look at the notion of a section, which is the obvious generalization of a tangent vector field. Definition 4.6 If (E, X , p ) is a vector bundle, a section of the bundle means a continuous map s:X+E

such that p s = lx, the identity on X (i-e.,p(s(x))= x for all x). 0

For example, one may always set s(x) = {(x,O)}, getting the so-called zero section. There may not necessarily exist a section s that is everywhere nonzero; this depends on certain topological invariants (see [25, 1021). But in a simple example, such as TS', we may put s(x) to be the unique unit vector in the tangent space to a point that points in, say, the counterclockwise direction. Because each p - '(x) is a vector space, we may form sums and scalar multiples of sections; the sections, therefore, form a vector space, which is usually infinite dimensional. CONSTRUCTIONS WITH VECTOR BUNDLES

We now look at the basic constructions that are possible with vector bundles. Definition 4.7 Given two vector bundles ( E l , X , p , ) and ( E 2 , X , p 2 ) ,we define their Whitney or direct sum as follows: For each x, choose neighborhoods U: and U: and suitable homeomorphisms

u,l x

R",

s=

u 0,

$::p;'(U+ u,' x R", where n and m are the dimensions of El and E 2 , respectively. Let Ox = U,l n U,Z. Consider the disjoint union $::p;l(U:)%

x R" x R".

xsx

Introduce the equivalence relation ( Z l , U l , u1)

( z 2 ,u 2 , v 2 )

if and only if

z , = z 2 , gXIx2(u1) (1) = u 2 , and g'$Jul)

= 02.

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CONSTRUCTIONS WITH VECTOR BUNDLES

Here, we write g!&, for the function from the bundle Eiconcerning the overlap of the sets Ox, and Ox,, containing the points z I and z,. The quotient space S ' / - is defined to be El 0 E , , the Whitney sum. Define p1 0 p 2 : El 0 E , -, X by (pl 0 p2) ( ( z ,u, u)} = z. It is trivial to check that E l 0 E , is a vector bundle of dimension (n + m),over X . We observe that the coordinate change functions in the second variables, namely the gap, for the Whitney sum bundle El 0 E , , are the following matrices:

When we clarify the notion of isomorphism of bundle, we shall show that if M" c_ Rk is a submanifold (7-Rk)lM"= ( T M " ) @ ( N M " ) k - " .

Formalizing our earlier remarks about restriction, we have

Definition 4.8 Let (E, X , p ) be a vector bundle and A c X a subspace; one defines the vector bundle restricted to A to be ( E A , A , p A ) ,where E , = p - ' ( A ) and p , = p I E,, that is, p applied to elements of E,. There is no difficultyin showing that when A is nonempty ( E , , A, p,) is a vector bundle of the same dimension as E. On occasion, we write EIA for E,. Definition 4.9 An n-dimensional vector bundle ( E , X , p ) is called trivial if there is a homeomorphism 4 : E 5 X x R",such that R 4 = p , i.e., for any e E E, n($(e))= p(e), where R is the projection of X x R" onto the first factor and where $ is linear on each p - ' ( x ) . 0

Notice that for any X and n there is always a trivial bundle E = X x R" with exactly one set U , = E and the obvious homeomorphism. One quickly verifies that a sum of trivial bundles is trivial, and a restriction of a trivial bundle to a subset is still trivial. Since TRk is trivial (because Rkhas a single coordinate chart), TRkIM" is trivial for any submanifold M" c Rk. Since

TRkI M" = T M " 0 ( N M " ) k - n if TM" is nontrivial, one can see that a nontrivial bundle may be a summand of a trivial bundle. Note that a trivial bundle has a wealth of nonzero sections. For any v E R", v # 0, set s(x) = (i-l(x,v).

Then s is clearly a nonzero section (check!) But, it is well known from algebraic topology (see [25]) that the tangent bundle to an even-dimensional

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sphere is nontrivial; in fact, there is no nowhere vanishing tangent vector field. Hence, TS'' is a nontrivial vector bundle. I believe it will be helpful to explore in full detail an elementary example. Let S c R3 be a surface, that is, a 2-dimensional submanifold. For each x E S, let N , be the line through x perpendicular to S. Let N be the (disjoint) union of all the lines N , . If z E N , c N , we define p(z) = x. N is topologized so that locally it looks like 0 x R, where 0 is a (small) open set in S. Then it is easy to check that N has the structure of a 1-dimensional vector bundle over S whose functions gaptake values in Gl(1,R), which is the multiplicative group of nonzero real numbers. Such a 1-dimensionalvector bundle ( N , S, p ) is often called a line bundle. If S is an open band (S = S' x (0,l)), then this normal line bundle ( N , S , p ) is trivial because S has two sides, say an inside and an outside, and we define 6:N + S x R by Q(z) = (p(x),r(x)),where r(x) is the positive distance from z to S along N , in case z lies on outside, and otherwise the negative of the distance from z to S along N , . Such a Q easily meets all the conditions of Definition 4.9. But if S is the (open) Mobius band, that is, the interior of the quotient space obtained by gluing the two vertical sides AB of a closed box (Fig. 4.1) in reverse direction, then N is not trivial. To see this, observe that a trivial bundle would have many nonzero sections s. But if you follow a continuous family of nonzero vectors all normal to the Mobius band once around the circle in the center of the band, you will find that the vector at the original point points in the opposite direction to that when you started. Hence, a supposed nonzero normal vector field cannot be well defined. We conclude that a normal line bundle to a Mobius band embedded in R3 is nontrivial. A

8

Figure 4.1 A

One may now easily verify that if S is a surface in R3, there is a homeomorphism ( T S ) @ ( N S ) = TR3IS.

The right-hand side is always trivial, being the restriction of a trivial bundle to a subspace. We have just showed that N may be nontrivial. For completeness, consider finally a compact, orientable surface. It is well known that such surfaces are homeomorphic to S2, the torus, or to a generalized torus with rn holes, rn > 1. It is shown in algebraic topology (see [25, 1031) that only the torus has a nonvanishing tangent vector field, or section of the

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97

tangent bundle. In all other cases the tangent bundle must be nontrivial. On the other hand, because each such surface has a well-defined inside and outside (see [ZS]), N clearly has a nowhere vanishing section, consisting of all unit vectors in each N , that point, for example, outside. Then just as we had remarked for the open band above, one sees that N is a trivial line bundle. Lastly, we note that the compact 2-dimensional manifolds that are not orientable, namely the Klein bottle and the real projective plane, do not embed in R3, so that we cannot ask similar basic questions about line bundles. We must now focus on the proper notion of a map between vector bundles. This concept combines the ideas of continuous maps of spaces, with linear transformations of vector spaces.

Definition 4.10 A map or homomorphism of vector bundles f:(El,Xl,Pl)

+

(Ez,Xz,P1)

consists of a pair of continuous maps fElE1

such that p 2 0 fE and such that

= fx

0

+

fx:X,

E,,

+

x,

p1 (which implies that fE maps p ; '(x) to p ; '(fX(x)))

f E IP;

'(x) :P;

' ( 4 P ; '(fx(x)) +

is a linear transformation of vector spaces. We may rephrase this latter condition by selecting z E p ; '(Ox) 0,x R", f(z) E p ; '( V,) U , x R", with dim El = n, dim E2 = m, and specifying that is in a neighborhood of $(z) the map x f 0 0

(U?4

-+

(fxb),Y(U) . 4

for some y(u) E L(R"; R"), the set of linear transformations from R" to R". Remarks (1) The function y(u) from an open neighborhood of $(z) to L(R",R"), the set of m x n real-valued matrices, is easily checked to be

continuous because the maps fx and fE are continuous. (2) Trivial examples of maps of vector bundles are the identity map and the inclusion of ( E A ,A , p A ) in ( E , X, p ) when A E X is a subspace. Slightly less trivial examples come from choosing fx to be the identity, El = E , , but the maps y(u) all to be the negative of the identity matrix. The obvious inclusions El + El0 E , or projections El0 E , + E , form further examples.

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Definition 4.11 (a) The most frequently encountered case of a map of vector bundles is the case where Xl = X, and f x is the identity. The set of all bundle maps from (El ,X,pl) to (E, ,X,p,), for which f , = l,, the identity, will be denoted

Hom(E1, E2). (b) The subset of the set Hom(E,,E,) consisting of all maps f having the property that for each x E X the restriction f Ip ; '(x) is an isomorphism of vector spaces is denoted WE,,Ed Note that Iso(El,Ez) is empty unless dimE, = dimE,. If there is an element in Iso(El, E,), we occasionally will write El NN E, . Remarks (1) The above definitions may be somewhat clarified by the following elegant construction. Suppose we are given two vector bundles over the same space X. We shall form a vector bundle that at each point consists of the vector space of linear transformations from p ; '(x) to p ; '(x). Choose an open neighborhood 0, of x small enough that p;'(O,) and p;'(O,) are homeomorphic to 0, x R" and 0, x R" by homeomorphisms i,h, and xx, respectively. Our new bundle is formed locally from the spaces

0, x L(R";R"), where L(R";Wm)is the vector space of linear transformations from R" to R". If x E 0, for a similar 0,, then ( x , f ) is defined to be equivalent to ( x , f ' ) , where f ' is xy O

x i ' ( f ) i , h x $; O

'

all over the point x. We define p ( { ( x , f ) } )= x. A tedious but elementary verification shows that this is a rn x n-dimensional vector bundle over X,which we denote Lin(E, ,E,). (2) In the language of (l), Hom(E,,E,) consists of the sections of Lin(E,, E,), because a section will be a continuous assignment of a linear map from p ; '(x) to p ; '(x), from which a suitable fE is easily fabricated. (3) Hom(E,,E,) may be topologized by either by (2) above, in which case we define an open neighborhood of a section of Lin(El,E,) in terms of a metric on E,, or else, in a more elementary fashion, by choosing a positive continuous function E ( X ) on X and then saying that f ,g E Hom(El, E,) belongs to an E ( X ) open neighborhood, provided that for each y E p ; '(x) =

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99

x x Iw" whose distance from s x 0 is less than l,f(y) and g ( y ) are closer than E(x). (4) For m = n, Iso(El,El)is nonempty, for it contains the identity. For any El and E , , Hom(E, ,E,) is nonempty, for it contains the zero map. Iso(E,,E , ) E Hom(El,E 2 ) is clearly open, because any map sufficiently near a linear transformation whose determinant is nonzero, is a linear transformation whose determinant is nonzero. ( 5 ) Iff: (El, X,pl) + ( E , , X , p , ) is an isomorphism, then one may easily define f- by setting

fi'lP;l(x)

=( f l P 3 4 ) - ' .

The expression on the right is merely the inverse of the linear transformation of vector spaces flp;'(x). It is clear that fi' is continuous, and to check that it is a map of vector bundles, note simply that if y:O -P Gl(n;R) is continuous, then T(x) = (y(x))-' is also continuous. (This may be checked using a standard formula for expressing the inverse of an invertible matrix in terms of determinants.) (6) Given a fixed f E Hom(E,, E,)

{XI flp; '(x)

is an isomorphism of vector spaces}

is visibly an open subset of X. There are many other important constructions on vector bundles, such as the dual and the tensor product (consult [4, 511). For our immediate purposes the next notion, the induced bundle, is basic. This will give a way to build many new vector bundles out of a fixed bundle. It is a special case of the abstract notion, from category theory, of a pull-back (see [79,97]). Definition 4.12 Let (E, X , p ) be an n-dimensional vector bundle over a space X,and let f : Y + X be a continuous map. Define

f-'E

= { ( y , e )E Y x

E(f(y) = ~ ( e ) } .

This set is given the relative topology of a subset of Y x E. We define p ' : . f - l E -, Y by p ' ( y , e )= y. Now, notice that if we have 0, E X and a homeomorphism $,:p-'(O,) 0, x R", then we may define another homeomorphism

$L:(P')-'(~-

'(0,))f- '(0,) x Rn +

by setting

$XX, e) = (x, xz$a(e)h

+

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where n2:Oax 08” --t R” is projection onto the second factor. The map is evidently continuous and onto. Because x is the first factor in the definition of $:, the only way $&could fail to be 1-1 would be if $&(x,el)= $l(x,e,) for different el and e,. But p(el) = p(e,), since (x, el) and (x, e,) both belong to f-’E,so that el,e, E p-’(x). Of course, K, 0 $a is 1-1 on each vector space p-’(x). Finally, the reader may easily check that $1 takes a small basis open set in (p’)- ‘(f-‘(Oa))into an open set, showing that $&is a homeomorphism. The fact that the “interchange maps” $; $&-’ come from continuous maps into the general linear group follows in a direct and formal way from the similar condition on t,bfl $; Hence f - ’ E is an n-dimensional vector bundle over X , which we write (f-’E,X,p’) and call the induced bundle. Generally speaking, this terminology poses no danger of confusion with the inverse image of a set under a mapping. 0

0

’.

Remarks (1) There are cases of the induced bundle, which we have already considered. For example, given (E,X , p ) and a subspace A E X with inclusion map i: A + X,then the restriction ( E A ,A,p,), from Definition 4.8, is identical to the induced bundle (i- ‘ E , A,$) (check this!). (2) If f : Y -,X is a constant map (that is, f(y) = xo for all y E Y ) then (f-’E, Y , p ) is always a trivial bundle. To sse this, define

h: Y x p - ’ ( x o ) + f-’E by h( y , e) = ( y, e). (Check details here.) (3) Clearly, if one uses the identity map for f; the induced bundle is precisely the original bundle. In addition, one checks that if we are given two maps Z.1:Y%XX,

and a bundle ( E , X , p ) , then there is a natural isomorphism of bundles (g 0 h)-’

E x h-’(g-’E).

(4) Given two bundles over X , one may construct a product of these bundles over X x X in an obvious fashion. The Whitney sum of the two bundles then appears as the induced bundle from the product bundle by the diagonal map A: X + X x X,defined by A(x) = (x, x). (5) The basic classification theorem (in the compact case, our Theorem 4.2) shows how to recover any bundle in terms of bundles that are induced from a “universal” bundle. But before we can get to that stage, we need some further preparatory material. Just like with vector spaces, there are natural notions of sub and quotient bundles and 1-1 and onto maps.

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Definition 4.13 (a) A map or homomorphism of vector bundles over X (Definition 4.1 l), say f :(El, X , p l ) + ( E , , X , p , ) , is called a monomorphism if each linear transformation f E I p ; ' ( x ) is a 1-1 map (often called a monomorphism) of vector spaces for every x E X. It is called an epimorphism if each j E I p ; I ( x ) is onto. It is called an isomorphism if it is both a monomorphism and a epimorphism (compare with Definition 4.1lb). (b) A bundle ( E l , X , p , ) is a subbundle of ( E 2 , X , p , ) if each p ; ' ( x ) is a 3 vector subspace of p ; '(x) and there are maps (for suitable 0,)I),: p ; '(0,) 0 RdirnEZ defining the bundle E , that, when restricted to p ; '(Om), map 1 p;'(O,) homeomorphically to 0, x RdirnE1, where RdirnE1is a fixed, chosen subspace of RdimE2, thus defining the structure of the bundle E l .

For example, both Eland E , would be subbundles of the vector bundle El @ E2, in a natural way. Now, in the category of vector spaces, a vector subspace of a vector space F c E is the image of a 1-1 map, the inclusion, and conversely, the image of a 1-1 map is a vector subspace. This sort of characterization is also valid for vector bundles, but as we now see it is a bit more difficult to prove. We shall now assume that our bundles are over X , and for maps f,f x is the identity. Proposition 4.2 (El,X,pl) is a subbundle of ( E , , X , p , ) if and only if (El,X , p , ) is the image of a 1-1 map or monomorphism of a vector bundle to (E,, X , p , ) (that is, there is a 1-1 map of bundles, f , such that the image of f is precisely El). Proof The necessity is easy. A subbundle of a vector bundle is clearly a vector bundle, and it is the image of the inclusion map i : El 5 E , . On the other hand, suppose we are given a monomorphism of vector bundles

f:(E,,X,p,)

+

(E,,X,p,)

with our convention that f, = 1,. It follows that p 2 o f E = p l , or in other words, the following diagram is commutative :

X

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in the sense that if one follows the maps from El to X around all possible routes in the diagram, then the maps (two in this case) with common domain and range are equal. Now, for x E 0, E X,suitable 0,, we have homeomorphisms $,:p;l(O,) 3 0 , x R",

$,:p;l(Od 5 0 , x R",

with n = dim E2 and rn = dim El. The map $, sends f ( p ; ' ( x ) ) onto an rn-dimensional linear subspace of x x R",which we may write x x R", for some R" E R". It follows that in some small open set about x, IcI, maps every subset p;'(y) (often called the fiber over y) to some rn-dimensional subset of y x R" whose projection to y x R", chosen above, is onto. (This is the usual remark that if one has a continuous family of linear maps and one is onto, then all others sufficiently near to it are onto.) It follows that near x one may construct a continuous family of linear transformations 1(x) E Gl(n; R) that send f ( p ; '( y)) by projection onto y x R" and map the orthogonal complement of R" identically to itself. Define $&(z) = 1 x ~ ( P , ( z0) $)a *

This is visibly a homeomorphism near any point in p ; ' ( x ) ,and because 1 is a continuous homomorphism in Gl(n; R), this new coordinate neighborhood is trivially checked to be compatible with the structure of the bundle ( E , ,X , p2). But our construction is precisely what is needed to exhibit (El ,X,pl) as a subbundle of the vector bundle. Remark The reader should compare this material with the definition of a submanifold (Definition 1.5). There in order to describe a submanifold one might need to take more charts for the larger manifold than in the original presentation; an extreme case would be submanifolds of R",since this latter needs only one chart. In the case of a vector subbundle constructed from the image of a monomorphism of vector bundles, in the Proposition 4.2, we see that we may need to pass to smaller neighborhoods, and hence more neighborhoods, to exhibit the subbundle structure. This is why in Definition 4.13b we speak of suitable neighborhoods defining the bundle structure and in Proposition 4.2 we pass to a smaller neighborhood and take new coordinate homeomorphisms in terms of I(x), which are thus compatible with the original bundle structure. Naturally, with either the original or the new neighborhoods the identity map from E 2 to itself is an isomorphism of vector bundles (Definition 4.3a). One often hears this situation described in terms of the different neighborhoods yielding the same vector bundle structure on E , .

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Finally, we observe that in the extreme case where ( E l , X , pl) is a nontrivial bundle that is a subbundle of the trivial bundle ( E , , X , p , ) , one needs only one neighborhood, all of X , to describe the bundle structure on E,, but more than one neighborhood if one wishes to describe the subbundle structure of ( E l , X , p , ) . A specific example of this interesting phenomenon is TS2as a subbundle of TR3 I S2, where S 2 is, of course, a submanifold of R3. We are now ready to define a quotient bundle. Definition4.14 Let ( E , , X , p , ) be a subbundle of the bundle ( E , , X , p , ) . The quotient bundle ( E , / E , , X , p 3 ) is then defined as follows: (a) For each x E X , set p ; ' ( x ) = p i ' ( x ) / p ; ' ( x ) , simply the quotient of one vector space by a vector subspace. Let Ez/Ei =

u P; '(XI-

xex

(b) Topologize E , / E , by choosing for each x E X an open set 0,, x E 0,, with a homeomorphism

$,:piyo,)5 0,x

R",

n = dim&, such that +,Ip;'(O,) is a suitable homeomorphism for E l , from p ; ' ( O x )to 0, x R". Then we set

$,:p;I(o,)+ 0,x R"-", m = dim E l , by selecting a copy of R"-" that is the orthogonal complement to R" c R" with projection map R : R" + R"-" and putting $a({YI)

= (1 x

0

$,(Y).

This map is clearly well defined, because the addition of any element in p;'(O,) is not detected after one applies the map 1 x R . It is visibly also 1-1 and onto. The topology on E J E , is then defined by giving p;'(O,) the open sets that result from applying to open sets in 0, x R"-", and extending this to topologize E 2 / E . We now note (c) If { y) E p i '(0,) n p ; '(OD), $p $ ; is clearly evaluated on a pair ( x ,4 as 0

'

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4. TANGENT BUNDLES, VECTOR BUNDLES, AND CLASSIFICATION

where x(z') = z. But because the action of gmPsends p;'(x) to itself, this is well defined and therefore creates a bundle structure on E J E , . Finally, (d) Define p , ( { y } ) = p 2 ( y ) . Then the final verifications that we have a bundle are immediate. Of course, there is always a bundle map

(E2 x,P A 9

+

@,/El

9

x,P A ,

that is, for every x, a linear transformation of p ; '(x) onto p ; '(x). It is defined by sending each point to its equivalence class. Remarks (1) For our purposes the most important examples will be furnished by manifolds and their tangent bundles. Let M" c Rk be a submanifold. Write TM" for the tangent bundle, thought of as the collection of tangent spaces to M" in R k ;write NM" for the ( k - +dimensional normal bundle and TRkI M" for the restriction of the (trivial) tangent bundle to M". Then we have the following isomorphisms: TRkI M" = T M " @ NM", T M " = TRkI M"/NM", N M " = TRkI M"1TM".

( 2 ) Let s: M" + TM" be a nonzero section. Then s defines a I-dimensional vector bundle, which over each point x consists ofthe 1-dimensionalsubspace of (TM"), containing the vector s(x). As before, this line bundle is trivial, the map to M" x R being that map which associates to such z E ( T M " ) , the pair (x, a), where z = as(x). If for a tangent bundle there is a continuous inner product defined on the tangent space at each point, then there will be a well-defined notion of orthogonal complement.If we have a r-dimensional subbundle of TM",say E, then there will be an (n - r)-dimensional subbundle that consists at each point of the orthogonal complement. The orthogonal complement to a line bundle will therefore be, for example, an (n - 1)-dimensional subbundle. The continuous inner product will exist under very general circumstances; see Chapter V or [103]. In the compact case ( X being compact), one may also infer the existence of an orthogonal complement from Lemma 4.6. In other terms, for a given subbundle (El, X , pl) of the bundle ( E , , X , p,), one seeks a bundle E:, where E , = El @ E:. One may then easily find suitable E: locally, and the full bundle E: can also be fabricated from the local pieces, using a partition of unity. The question of how to find an orthogonal complement for a subbundle is often called the problem of splitting, and it is known to be solvable for any vector bundle over a paracompact base space.

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THE CLASSIFICATION OF VECTOR BUNDLES

Our next goal is to classify the n-dimensional vector bundles over a fixed space X. Naturally, we really want the isomorphism classes of such bundles, for which we write v"(X). We first observe that the correspondence that passes from X to v"(X) gives us something more than a mere set, for if f : Y + X is a continuous map, the induced bundle construction (Definition 4.12) gives a function, called v(f),

v(f):v " ( X )+ v"( Y). This construction has the following already verified properties : V(1X) = L ( X ) ,

(1)

that is, it transforms identity maps to identity maps; v(g il .f)= v(f) "

w

(2)

(this was Remark (3) after Definition 4.12). In the language of category theory, our construction is a contravariant functor from the category of topological spaces and continuous maps to the category of sets and maps. All this not withstanding, it is of great importance to gain an understanding of v"(X) in terms of the space X. Such an understanding should subsume in one result the basic facts alluded to above, such as that every bundle over Rk is trivial, i.e., v"(rWk) consists of a single element, whereas v'(S') has two elements, the ordinary and Mobius bands, etc. We should also be able to build up v"(X)in terms of v" applied to special, simple subspaces of X . The answer that we shall get for the structure of v"(X)(see Theorem 4.1) will not always be easy to calculate, but it will have a simple and beautiful format. In rough terms, we shall construct a space B, so that, for each X , v"(X)will be a sort of natural quotient of the set of continuous maps from X to B,. One will be able to infer, in particular, that for many X,o"(X) is finitely computable. For reasons of space and convenience, we shall present our main theorem when X is a compact Hausdorff space. An interested reader might pursue the most general theorem in the beautiful paper of Dold [26]. Our basic theorem will pass from maps to bundles, using the induced bundle construction. For that reason the initial part of our presentation will treat properties of maps. Then, we shall need to take a close look at the way in which two bundles defined over two subspaces of X may be glued together to form one bundle over their union.

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Definition 4.15 Let f,g: X + Y be two continuous maps from X to Y . We say that f and g are homotopic, written f z g, if there is a continuous map

where I = [0, 11,

F:X x I +

such that for any x E X F(x, 1) = g(x).

F(x, 0) = f(x),

F is often referred to as the homotopy between f and g . Remarks (1) It is easy to check that this is an equivalence relation. Use F(x, 1 - t ) to check symmetry, and change scale from I to [0, 1/21 and glue together two maps of X x [0,1/2] to show that the relation is transitive. (2) One denotes by [X,Y] the set of homotopy classes of maps, that is, equivalence classes of maps under the relation "homotopic," from X to Y . (3) If we consider maps from X to Iw", then any two are homotopic. Simply define, for each x E X , the map F(x, t ) by the formula F(x, t ) = t f ( x )

+ (1 - t)g(x).

This argument immediately generalizes to maps from X to any convex subset of R". (4) Iff 'Y g and Y has several connected components, then for any x, f(x) and g(x) belong to the same connected component. (5) If n > 0, then the identity map 1:S" -, S" is never homotopic to the constant map g, defined by g(x) = a, for some fixed a and x E S". For if f N g, we have

F:S" x I

+

S"

with F(x, 1) = a for all x E S". But then F defines a map on the quotient space of S" x I , with the subset {(x, 1)[x E S"} shrunk to a point, that is,

F:D"+'+ S", because the quotient space is identified with either the upper hemisphere of

'

'

Snf or, equivalently,D"' (check !). This map is still the identity on all pairs {(x, 0)}, that is, the boundary of D"". But we proved [Corollary 3.3 (lemma

to the Brouwer fixed point theorem)] that there is no such map.

The interest, at this time, in homotopic maps, is the following remarkable proposition. Proposition 4.3 Let X be a compact Hausdorff space and f,g :X + Y be two homotopic maps, or in symbols, f N g. Let (E, Y , p ) be an n-dimensional vector bundle over Y .

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Then the two induced bundles f-'E and g - ' E (Definition 4.12) are isomorphic (Definition 4.11b). Proof We need two preliminary steps. (a) Given a vector bundle ( F , X , q) and a closed subset B 5 X . Let s: B -+ F B be a section over B, i.e., s: B + F is a continuous map such that whenever b E B, q(s(b))= b. Then there is a section so:X -, F of the bundle such that sol B = s; i.e., if b E B, s(b) = so(b). Of course, locally F is a product 0, x R" for suitable 0, G X.Then, the

1

usual Tietze extension theorem, applied to maps to R", shows that there is a section SO,: 0,+ F

such that

s0,IBnO,= s l B n O , .

This resolves the question in the small. We remark that, by compactness, we need only consider a finite number of sets 0,, say O,,, . . . , Oak. Let . . . , cPk be a partition of unity (Theorem 1.4) subordinate to the That is, open cover O,,, . . . , 0,".

+,,

(i) 0 I + i I 1, (ii) + i ( x ) # 0 implies x E Oai,and (iii) + i ( x ) = 1.

xi

We define si:x + F

by setting i f y E Oai otherwise.

{+i(y)soajy) si(y)=

0

si is trivially checked to be a section, with continuity coming from the fact

that must vanish at the boundary of Oa8. Finally, we define s O ( Y ) = C si(y)* i

The addition makes sense, because each s i ( y ) lies in the vector space q - ' ( y ) , and continuity is clear. Now, if y E B, we may calculate

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4. TANGENT

BUNDLES,VECTOR BUNDLES, AND CLASSIFICATION

because when y E B n Oai,so,. and s agree on y. But this last sum is just I

because the 4;s are a partition of unity. Our second preliminary step is now the following: (b) Given (El, X , pl) and ( E 2 ,X , p 2 ) as two n-dimensional vector bundles over a compact Hausdorff space X , let A c X be a nonempty closed subset. Let Ei 1 A denote the restriction of Ei to A (Definition 4.8). Suppose we have an isomorphism f : E , ( A% E 2 ( A . Then there is an open set U , A E U E X , and an isomorphism f l : E l ( U 9 E21U

such that fl and f agree on all points of p;'(A). In other words, the isomorphism over A extends to an isomorphism over U . The easy way to see this is to use the remark following Definition 4.11. f is merely a section of the bundle Lin(El A, E 2 I A), which is an isomorphism on each fiber p ; ' ( x ) ,x E A. By step (a) above the section extends, so that we have fl:El + E 2 , which is a map of bundles so that over A, fl(x) = f ( x ) . But fl is a map of bundles, and points y near enough points where fl / p i ' ( x ) has nonzero determinant are points for which fl( p ; ' ( y )has nonzero determinant [see Remark (6) following Definition 4.111. Thus, fl is an isomorphism over the desired open set. If one wishes to avoid the bundle Lin(E, I A, E z I A ) and extensions to Lin(E,, Ez),a local argument, much as in (a) above, may also be given.

I

To complete the proof, let F:XxI-+Y

describe the homotopy between our given maps f and g, with F(x, 0) = f ( x ) . Let n;: X x I -+ X be projection on the first factor. Define then, for any fixed t E I, F,(x) = F(x, t).

We consider the two bundles F-'E and a-'F; ' E . One may check at once from the definitions that they are isomorphic on the subset X x { t } c X x I. Note that bundles induced by n; are essentially constant in the second coordinate.

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109

We may use (b)above to conclude that F - ' E and n-' F; ' E are isomorphic on an open set U containing X x { t } . Because X is compact we see at once that there is some E > 0, depending on t, such that X x ( t - E, t

+ E) E U ,

and consequently F - ' E and z-'F; ' E are isomorphic on X x (t - E, t + E). On the set X x {tl}, F - ' E is the same as FL'E, identifying X x {tl} with X.On the set X x {tl}, z-'F;'E is the same as z-'F;'E on X x {t}, because of the definitions of z and of an induced bundle. Thus, for all t l , It - t,l < E, the bundles F , ' E are isomorphic. Let v"(X)be the set of classes of isomorphic, n-dimensional vector bundles over X,given the discrete topology. Define T:I +

V"(X)

by z(tl) = { F ; IE}. We have just proved that T is locally constant. Therefore, is continuous. Since I is connected and v"(X)is discrete, z must be constant. Therefore FL'E are isomorphic for all choices of t l E [0,1]. Select t , to be 0 and 1, and the proof of the proposition is complete. T

Remarks (1) Proposition 4.3 can be used to show that in various cases v"(X)contains one element, or since there is always the trivial bundle X x R", every bundle is isomorphic to the trivial bundle; i.e., every bundle is trivial. For example, let X s Rk be a convex subset. Suppose (E, X , p ) is an ndimensional vector bundle, which we write E for short. If 1: X + X is the identity map, then by Remark (3) after Definition 4.12, 1-'E is isomorphic to E. But by Remark (2) after Definition 4.15, 1 is homotopic to a constant map g : X + X , where g ( x ) = b for some fixed b E X . According to Remark (2) after Definition 4.12, g - ' E is a trivial bundle, and thus so is E. More generally, if A is a subspace of R", form the cone on A, denoted C ( x , A), which consists of all straight line segments from x , where x 4 A is fixed, to a point in A. Then it is easily shown that the identity map from C ( x , A ) to itself is homotopic to the constant map that sends everything to x. Consequently v"(C(x,A ) ) consists of a single element in this case as well. (2) If n > 0 and X is homeomorphic to Y , then there is a 1-1 correspondence (isomorphism of sets) between the sets v"(X)and v"(Y ) ,for if the spaces are homeomorphic, there are continuous maps J':X+ Y

and

g:Y+X

such that g 6 f = l x , the identity on X , and J' g = 1,. We define maps 0

4:vn(Y)-+v"(X)

and

+:v"(X)+v"(Y)

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4. TANGENT BUNDLES, VECTOR BUNDLES, AND CLASSIFICATION

by the induced bundle construction, that is,

4 ( { J q=) {f-'Jq

and

$ ( { F ) ) = {g-'F}.

The obvious fact that the induced bundles, via a given map, of two isomorphic bundles, yield isomorphic bundles, means that these maps are well defined. But then it is clear that the compositions $ 0 4 and 4 0 $ are the identity maps, so that 4 (or +) is a 1-1 correspondence. (Compare Remark (3) following Definition 4.12.) (3) Our Proposition 4.3 now suggests a very important generalization of Remark (2)above. We say that the spaces X and Y have the same homotopy type if there are continuous maps f :X + Y and g: Y + X whose respective compositions are homotopic to (rather than merely equal to) the identity; specifically, we require g 0 f N lx and f g N 1,. For example, R and a single point have the same homotopy type because we set 0

f:pt

-+

R

by f(pt) = 0

and g:R+pt as the unique such map. Then g 0 f = lpt,so trivially g 0 f z lpt,and the homotopy between f 0 g and 1, is given by F(x,t) = xt.

If X has the homotopy type of a point, we shall say that X is contractible. Similarly, S2 has the homotopy type of R3 - (O,O,O).Both the Mobius and the ordinary band have the same homotopy type as the equatorial circle that runs around them. Clearly, we see from these examples that nonhomeomorphic spaces may have the same homotopy type, while trivially homeomorphic spaces always have the same homotopy type. Because (Proposition 4.3) the passage to induced bundle does not see a difference between homotopic maps, one may directly adopt the proof in (2) above to prove that, if X and Y have the same homotopy type, then v"(X) and v"(Y)are in 1-1 correspondence. If X is contractible, v"(X) is a single element. For example, one may check without much difficulty that v'(S') has two elements, one represented by the ordinary band B and one represented by the Mobius band M ; one thinks in terms of R identified with (- 1, l), and B = S' x (- 1,l).Then one has v'(S') = v'(R2 - (0,O))= v'(B) = v'(M), etc. In our drive towards the main theorem, we shall want to form quotient spaces of spaces over which we have bundles, and we shall want to glue together spaces over which we have bundles and get nice, well-defined

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111

bundles as an end product. The following set of lemmas systematically treats these basic technical questions and forms the foundation of our main results. Lemma4.1 Let ( E , X , p ) be an n-dimensional vector bundle over a compact Hausdroffspace X . Let A be a closed subset over which the restricted bundle EA or E 1 A (Definition 4.8)is trivial, i.e., there is an isomorphism of vector bundles f from p - ' ( A )to A x R". Then there is an n-dimensional vector bundle E/p-'(A), over the space X / A , defined as follows:

-

(a) E / p - ' ( A ) = E / - , where is the equivalence relation that identifies two points of p-'(A), if their images under the isomorphism map f have the same second coordinates. (b) p": E / p - ' ( A )+ X / A is defined by ' ( { e l ) = {:)point

if P ( 4 4 A A / A in X / A

if p(e) E A .

(Here X / A means the quotient space of X under the equivalence relation that identifies all of A with a single point, and A / A simply refers to that point.)

Proof It is clear from the definitions that E / p - ' ( A ) is a space, p" is a continuous map onto X / A , and p-'(pt) is homeomorphic to R" for any point, which again we simply abbreviate pt. It is also obvious that if 0, is an open set, from the definition of ( E , X , p ) , that does not meet A, then p-'(O,) = p - l ( O , ) = 0 , x R". We must check that p is locally a Cartesian product in a neighborhood of A / A ; i.e., we must find an open 0 c X / A , A/A E 0, with a suitable homeomorphism p- '(0) 5 0 x R". Now, according to step (b) in the proof of Proposition 4.3, the isomorphism f :p - ' ( A )+ A x R" extends to an isomorphism fl of bundles over some open set U , with A c U . Let 0 be the image of U in X / A ; it follows immediately, from the definition of the quotient topology, that 0 is open. We then see at once that fl yields the desired homeomorphism

p"-'(o)+ox

R"

It is routine to check that the coordinate transformation maps gaa for this new bundle are linear and have the properties required by Definition 4.5. Definition 4.16 Let ( E , X , p ) be a bundle and A E X a closed subset. (a) An isomorphism of bundles f :E A + A x R" (just as in the previous lemma) is called a triuiulizution of E over A .

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(b) Two trivializations of E over A, f,g:EA+A

X

R"

are called homotopic if there is a trivialization of the n-dimensional vector bundle (E x I, X x I, p x 1,), I = [0,1], over the subspace A x I, written F : ( E x I)Axl + ( A x I) x R", such that ifeEp-'(A), F(e,O) = f(e)

and

F(e, 1) = g(e).

Clearly, the concept is inspired by the notion of homotopic maps and offers a relevant way to compare trivializations. Note also that for any continuous map I: X -,Gl(n; R), the map

f;:xx R " + X

x

R"

defined by &(x, u) = (x, l(x)u) is a trivialization of the vector bundle that is its domain; consequently, there is a great wealth of possible trivializations of a vector bundle, and hence a need to look at homotopic ones. The next lemma shows that if we replace one trivilazation by a homotopic one then the construction of Lemma 4.1 does not suffer.

Lemma 4.2 Iff and g are homotopic trivializations of (E, X,p ) over the closed subset A of the compact Hausdorff space X,then the bundles E / p - '(A) constructed by Lemma 4.1 from either f or g are isomorphic. Proof

Define p : ( X / A )x

z-bx

x I/A x

I

by the formula p ( { x } , t ) = {x,t>. It is the obvious map induced from the projection X x I + X x I / A x I. If we use Lemma 4.1 on the trivialization of E x I over A x I, guaranteed by the assumption that f and g are homotopic (Definition 4.16b), then we may define a bundle (E x I ) / ( p x 11)-'(,4 x I) over X x I / A x I. We may then form the induced bundle p - w x

I)l(P

x W ( A x 1))

over the space ( X / A ) x I. We note that the restriction of this bundle to ( X / A ) x (0) is precisely the construction of Lemma 4.1 according to the map f, while the restriction to ( X / A ) x (1) is that construction according to g (direct verification). We might also recall [Remark (1) after Definition 4.121 that the restriction ofa bundle to a subspace is precisely the same bundle as the induced bundle formed by the inclusion map from the subspace to the whole space.

THE CLASSIFICATION OF VECTOR BUNDLES

113

Therefore, because of Proposition 4.3, we shall only need to check that the inclusion maps and

i,: X / A --+ ( X / A ) x I

i , :X / A -+ ( X / A ) x I

defined by ij({e}) = ({e},j ) are homotopic. But that is completely trivial, for if we define G : ( X / A )x I

+( X / A )x

I

to be the identity, G({e}, t ) = ({e}, t), then G({e},O) = io({e}) and G ( { e } ,1) = i,({e}), completing the proof.

As an application, we may now prove the following result: Lemma 4.3 Let X be a compact Hausdorffspace and A a closed subspace. We suppose that A is contractible; that is, A has the homotopy type of a single point. Let f:X X / A be the projection map. Then the map -+

f : v"( X / A )

-+

u"( X )

defined by f ( { E } ) = {f-'E}, the induced bundle construction, is a 1-1 correspondence of sets. Proof We need to construct an inverse for f. Let ( F , X , q ) be an ndimensional vector bundle over X . Then the restriction, FA is automatically trivial by Remark (3) following Proposition 4.3. Let two trivializations

h : F , + A x R",

g : F A - + Ax R"

be given. Then g h-' is an isomorphism of trival bundles, so that (by Definition 4.10) there is a (necessarily continuous) map 0

y: A + Gl(n; R)

such that 9 h-

l ( 4 Y ) = (a, y ( 4 y ) .

Because A is contractible, l A is homotopic to a constant map. But then y = y 0 1, must also be homotopic to a constant map. Now Gl(n;R) has two pathwise connected components. If we require h and g to preserve orientation on q-'(pt) for any fixed point of A, g 0 h-' will lie in the component of the identity matrix. (This is no loss of generality because h : F , +. A x R" can be followed by l A x r, when r ( x l , . . . ,x,) = ( - xl, x2, . . . , xn), in case of reversed orientation.) But then y is homotopic to a constant map in the path component of the identity matrix. Any two

114

4. TANGENT BUNDLES, VECTOR BUNDLES, AND CLASSIFICATION

such maps into a path connected space are homotopic (reparametrize the path to have length 1 and move down the path), so y is homotopic to the constant map that takes all A to the identity matrix. We conclude at once that g o h-' is homotopic to the identity 1. But ifg o h- 1: 1, g N 1 h = h, or g 'v h. Thus, we may use Lemma 4.2 to conclude that-up to isomorphism-there is a unique bundle F/q- '(A) over X / A . We must verify that this construction is the inverse of the map f so as to get our 1-1 correspondence or isomorphism of sets. If we are given ( F , X , q ) ,then f - ' ( F / q - ' ( A ) ) is precisely

'

0

x,

{(x,{ e ) ) l x E { e l E F/q-'(A), f ( x ) = 4"({e))). But since q ( { e } )= q(e),the condition so expressed simply means q(e) = f ( x ) . It is clear that ~:(F,X,q)-tf-'(F/q-'(A))

defined by the formula +(e) = (q(e),{ e } ) is a map of bundles. Now, if x 4 A, 4 restricted to q - ' ( x )is trivially an isomorphism of vector spaces. But when x E A, it is also an isomorphism, because F/q-'(A) is defined by an equivalence relation that keeps the second coordinates in A x R"fixed. Therefore, 4 is an isomorphism of vector bundles. This shows that f applied to the construction F/q- '(A) yields the identical bundle, up to isomorphism, to that which we started with. To complete the proof that we have an inverse map to f,we must show that f(E)followed by our construction from Lemma 4.1, that is, T(E)/p-'(A), is isomorphic to E for any bundle E over X / A . If we take (E, X / A , p) and form f- 'E, this then consists of { ( x ,e)Ix

x,

p(e) = f ( 4 ) . If we pass to f- E / p - '(A), we may form a map

'

E

e E E,

$:f-'E/p-'(A)

+E

by $ ( { x , e ) } = e. This is well defined because different but equivalent pairs (x,e), with x E A, have the same second coordinate after applying the trivialization. Once again, one may check that $ is an isomorphism on every fiber; for example, if we apply $ to say jF-'(A/A), our formula for $ is the identity map. Hence f - ' E / p - ' ( A ) is isomorphic to E, as desired. Remark For a large class of spaces, A contractible implies that X and X / A have the same homotopy type. In such cases, Lemma 4.3 would follow at once from Remark (2) after Proposition 4.3. (This is a well known, but infrequently published fact in algebraic topology. The reader may consult the general material in [113].)

THE CLASSIFICATION OF VECTOR BUNDLES

115

As an easy example, take the plane R2, let the closed interval A = [ - 2,2] live in the x-axis, and let B be the union of two closed line segments, from ( - L O ) to (0,l) and from (+1,0) to (0,l). Let X = A u B . Then X / A is homeomorphic to S ' , so that v " ( X )= v"(S') for all n 2 0. The construction of Lemma 4.1 is often referred to as shrinking or collapsing. We must now study in a similar fashion another basic construction, often called gluing or clutching. Lemma 4.4 Let X be a compact Hausdorff space with X I and X , being closed subsets. Suppose X I n X, is not empty; i.e., X , n X, # We shall assume that we are given two n-dimensional vector bundles ( E l ,X I ,p l ) and (E2,X , , p , ) and an isomorphism of the bundles restricted to X I n X , , say

0.

(EAxI0 x 2 .

~ : ( E I ) nxz x,

Then there is an n-dimensional vector bundle El

Uf

E2

over X Iu X , whose restriction to X I (resp. X , ) is isomorphic to E l (resp. E,). We shall write the projection map as P1.2:EI UfE2

+

XI u x,.

-

Proof This is quite straightforward. We introduce an equivalence relation in the disjoint union E l u E2 by asking that e, e,, if either (a) e l = e2, or (b) el E W J X , ,,X2, e2 €(EAX, XI' and f(eJ = e2, or ( 4 precisely the same as (b) except that the subscripts are interchanged and f is replaced with f - I . Set El u FE2 to be the quotient space of the disjoint union by this equivalence relation. Because f is a homomorphism (indeed isomorphism) of vector bundles, one may define the obvious projection map pl,,: El uf E, + X I u X , , by pl,,(e) = pl(e) if e E E,, otherwise p2(e). One should verify local triviality, i.e., the existence of suitable open sets 0,, with p-'(O,) homeomorphic to 0, x R". First, if x E ( X , u X , ) ( X I n X , ) , we choose an open neighborhood 0;of x that lies in one of the sets X , or X , , but not both. If necessary, choose a smaller open set 0, over which the relevant bundle, E, or E , , is trivial; i.e., say pY1(O,) = 0, x R". Then, it is clear from our definition that El uf Ef is also trivial over 0,. On the other hand, if x E XI n X , , one begins with a closed set 0,whose interior contains x over which, say, El is trivial. Using the isomorphismf and part (b) of the proof of Proposition 4.3, one easily sees that this trivialization n X I n X , to a trimay be extended from a trivialization of El over vialization of E, over (0, n X I n X , ) u (0, n X I n X , ) , for some

116

4.

TANGENT BUNDLES, VECTOR BUNDLES, AND CLASSIFICATION

0,c X,,x E 0,, over which E , is trivial, i.e., isomorphic to the bundle 0, x R". But then it is clear that we have a trivialization of Eluf E , over 0,u 0,. The final assertion, about the restrictions to X I and X , , is immediate. Remark If g l : E l + El is an isomorphism of vector bundles over X1, 9,: E 2 --t E 2 is an isomorphism of vector bundles over X,,and we have two

isomorphisms of the restrictions f:(El)XlnX1

+

(E2).~1nXz, f':(E;)X,nx,

+

(EJx~~x,

that satisfy 9, o f = f' g1I X , n X,, then g1 and g, define an isomorphism 0

El uf E z 4 El vr E;.

Definition4.17 With the above data, suppose that we are given two isomorphisms fo,fi:(El)x,nx,

+

(Ez)x~~x~*

We say that these isomorphisms are homotopic if, when we write IL: X x I + X for the projection, we have an isomorphism H:(n-'El)(Xl

nX I ) x I

(n-'E&,

-+

n x z )x I

that restricts to fo and fl on the endpoints, that is, the values 0 and 1 in I = [O, 13. We need one more basic lemma. Lemma 4.5 Two homotopic isomorphisms fo,fi:(E1)X1

nX2

+

(E2)Xl n X 2

determine glued bundles that are isomorphic, that is,

El

UJ0 E2

= El U f , E2.

Proof Define i,:X + X x I by setting i,(x) = (x,t), so that io and il are the inclusions of the faces of the cylinder. Because foand fl are homotopic, we have an isomorphism

H : ( ~ - ~ E l ) n[ xx 2l) x I 4 (n-lE&l n X 2 ) x 1 At any specific value o f t E I , H describes an isomorphism Ht:(El)xl n X 2

--t ( E A x I

n~

2

9

THE CLASSIFICATION OF VECTOR BUNDLES

117

which is the same as the isomorphism obtained after applying i;' to each bundle (because the restriction of a bundle to a subspace is the same as the induced bundle obtained from the inclusion map of the subspace). I claim that for any r E Z El u H 1

E , x i;'((n-'E,)

UH

(TC-IEZ)).

For given r E El uHI E , , if pl,,(e) # X , n X , , we may map e to the same point, in the restriction of n-'E, uHn -'Ez to n - l E 1 or n-lE, at time t. But ifp,,,(e) E X ln X , , since the identification map H in n-lEI uHa - ' E , is precisely H,for this value t, the desired map from El uH1 E , to i; '( (n-'El) uH(n-'E,)) may therefore be chosen to be the identity. The resulting map is visibly an isomorphism of each fiber, as desired. Returning to i , : X --$ X x I , clearly all these maps it are homotopic (by simple reparametrization). By Proposition 4.3, all the bundles i; ((n-'E1)uH(n-lE,)) are isomorphic for any choice of t . Taking t = 0 or 1 and using the isomorphism from the previous paragraph gives our desired result. We now have completed our analysis of how to build bundles over unions, and we have a good understanding on how to cope with contractible subspaces, or subspaces over which a given bundle is trivial. The most logical problem to tackle next, indeed an important case of the classification theorem, is the problem of classifying bundles over a space that is the union of two contractible subspaces. The simplest example is the sphere S", which is the union of the closed upper and lower hemispheres D: and D! ,each of which is homeomorphic to the disk

D" = { X

E

I

R" llXll I 1).

Of course D" is a convex subset of R" and hence is contractible. The classification of bundles over spheres was achieved in the 1930s, and with little more difficulty we may extend it to bundles over suspensions, defined as follows: Definition 4.18 If X is a topological space, we define the suspension of X , written S X , to be the quotient of X x I by the equivalence relation (x, 1)

(u,O)

-

(y, 1) (u.0)

for all x, y E X , for all u,u E X .

In other words, we collapse (separately) the upper and lower faces of X x Z to single points. We observe that the subspaces of SX defined by C + X = {{(x,r)j E S X I t > & )

and

C-X={{(~,~)}ESXI~I$}

118

4. TANGENT BUNDLES, VECTOR BUNDLES, AND CLASSIFICATION

are both cones over the set of all { (x, i)}, which is obviously homeomorphic to X because no distinct pairs (x, t ) with t # 0, t # 1 are equivalent. In terms of symbols, we then may write SX = C + X u C - X

and

X =C+X nC-X.

Remark If X is nonempty, SX is always pathwise connected. (Check!) If X is a compact, Hausdorff space, then so is S X . Naturally, S(S") = S"". Proposition4.4 If X is a compact Hausdorff space and xo E X , then there is a natural isomorphism of sets

v"(SX) x [X;Gl(n;R)]o/-,

-

where[X,Gl(n; R)],is thesetofthose homotopyclassesofmapsX + Gl(n; R) that send xo into the identity matrix, and identifies { g } with the class of x --* ag(x)a- for any a E Gl(n; R) not in the path component of the identity. Since any two such a may be joined by a path, yielding a homotopy, the choice of a is immaterial.

'

Proof We shall construct two maps between these sets that are inverses of one another. We begin with the easier. Given a homotopy class {g} in [X;Gl(N;R)],, we consider the trivial bundles C + X x R" and C-X x R" over C + X and C - X . We define an isomorphism f over the restrictions to C+X n C - X = X , by (x,

4 * (x, g(x)u).

Using Lemma 4.4,we see that we have an n-dimensional vector bundle over SX, namely (C+X x R") U f ( C - X x 02"). Observe that Lemma 4.5 assures us that the construction does not depend on the choice of g, in its homotopy class (check).If we replace g by an equivalent map, in the sense of -, then we change the orientation on each piece of the bundle, C + X x R" and C-X x R", but the resulting bundle is of course isomorphic to the original. We write this class of bundles, coming from (91, as ml). To define a second map, going in the other direction, we let (E, S X , p ) be a bundle over SX. The subspaces C + X and C - X being contractible, it and ( E ) c - x are trivial; follows at once that the restricted bundles (QCtx we choose two trivializations f':(&

tx +C + X

x R"

f - :(E)c-x + C - X x R".

and

THE CLASSIFICATION OF VECTOR BUNDLES

119

We assume that f- is chosen such that 'f and f- are the same map of p-'({(xO,4)))to R". Write f,' = ~ + ~ P - ~ ( C n+ CX - X ) and f; = , f - l p - ' (C+X n C - X ) . Now, the map

(fi) (f;)-l:X 0

x R" + X x R"

is clearly a bundle map of trivial bundles. By Definition 4.10, we must have a (continuous) map g : X -,Gl(n; R), so that

(f,')o(fo)-'(x, 4 = (x,g(x)u). Because of the way in which we chose'f and f-,it follows that g(x,) is the identity matrix. It is now necessary to see that this construction is independant of our choices of trivialization. Suppose therefore that 'f was replaced by another trivialization g+:(E),+, + C+X x R" preserving the original map on the inverse image of (xO,4).Then as before

'f

0

( g + ) - l : C + x x R" -+ C'X x R"

defines a map y: C + X -+ Gl(n; R), which (because C + X is contractible) must be homotopic to a constant map. Our conventions to the effect that 'f and 9' agree on one fiber (inverse image of (x,, i))show that this map may be assumed constant at the identity matrix. Restricting attention to C + X n C - X = X , the mapf;(g,')-', which is precisely the map (x, 4 + (x, y(x)u),

must be homotopic as an isomorphism of bundles (Definition 4.17), to the identity map. We may also easily do this in such a way that y(xo) remains at the identity matrix during the homotopy. It follows at once that the two bundle isomorphisms 9,'

o(fo)-'

and

( f ; ) o ( g W "(S;)"(fo)-l

=(fiWfo)-'

are homotopic isomorphisms, and therefore the homotopy class of g is unchanged i f f ' is replaced by g+. A similar argument may be given for f-. It is easy to see that choosing the reverse orientations on p - ' ( C + X ) and p-l(C-X) produces a map equivalent to y, in that it has the form x -,a g(x) u-' (for we can let a be a fixed orientation reversing the linear isomorphism of R" to itself and proceed as above). Therefore, the equivalence class of g: X + Gl(n; R) is well defined, starting from a bundle (E,SX,p). Using the remark after Lemma 4.4, one checks without difficulty that an isomorphic bundle yields the same element of [X, Gl(n; R)IO/-. Thus, we have our map 0

0

0:

vn(SX)+ [X,Gl(n; R)],/-.

120

4. TANGENT BUNDLES, VECTOR BUNDLES, AM) CLASSIFICATION

We must check that the compositions of z and CT are the respectiveidentities. Given g : X --t Gl(n; R), construct f :X x R" + X x R" by f ( x , u) = ( x , g(x)u), and a bundle over C + X u C - X = SX built from the isomorphism f, i.e., E = ( C + X x R")

( C - X x R").

If we choose f' and f- to be the identity maps in each case, we clearly get o ( { E } )= { g } , proving that CT z is the identity. Equivalent g's yield isomorphic bundles by reversing orientation, again getting equivalent 9's. On the other hand, given a bundle (E, S X , p), choose suitable trivializations 'f and f- and form g by ( f : ) ( f ; ) - ' ( x , u ) = (x,g(x)u). To pass back to a bundle from g, one forms 0

0

(C'X x R") Uf (C-xx R"),

where f = ( f i ) &)-I. An isomorphism from ( E , S X , p ) to this latter bundle is formed by taking f' on ( E ) c + xand f - on (E)c-x, compatibility on the overlap being immediate. This shows that z CT is also the identity, completing our proof. 0

0

Remark The statement and proof of this proposition are complicated by the equivalence N , and automorphisms a 0 g ( x ) o a - ' . A simpler result may be had if one restricts attention to oriented vector bundles, whereby an orientation is chosen over some point and all coordinate transformations gmaare required to lie in the connected component of the identity matrix (compare [103], as well as our problems and projects at the end of this chapter).

While Proposition 4.4 is a satisfactory classification theorem for understanding vector bundles over suspensions, we note that few spaces are suspensions, and therefore one needs a modified approach to get the general result. The next two lemmas deal with the splitting of bundles into separate pieces; they are important technical tools. The following lemma is actually valid for paracompact X , although we shall need it only when X is a compact Hausdorff space. Lemma4.6 Assume we are given two vector bundles over X , say ( E , , X , p , ) and ( E 3 ,X , p3), and a map of bundles f : E2

E3

(naturally, we assume fx = lX) that maps every p ; ' ( x ) onto p ; ' ( x ) , x E X . Then there is a subbundle E l of E , and an isomorphism E l 0 E3" E,.

THE CLASSIFICATION OF VECTOR BUNDLES

121

El is the subset of elements in each p ; '(x) that are mapped to 0 E p ; '(x) (clearly a subbundle). Suppose that we are able to find a map of vector bundles

Proof

p : E 3 + E2

such that f p = lE3,where lE3means the identity map on E 3 . Let then i : El + E2 be the inclusion of the subbundle El, defined in the statement of the lemma, into E , . We form the map 0

i

+p : E l0E3+

E,,

where (i + p)(a,b)= i(a) + p(b) and addition takes place in the fiber over each point p ; '(x). There is no difficulty in checking that i + p is a map of bundles. If we temporarily restrict attention to points lying over a given xo E X,we note the following:

+

+ +

+

(a) i pis 1-1, because if(i p)(e,,e3) = 0, i ( e l ) p(e3) = 0, and hence f ( i ( e l ) p ( e 3 ) ) = 0. But since f 0 i ( e l ) = 0, by definition of El, and .f 0 p(e3) = e 3 , we must have e3 = 0. But then, because i ( e l ) p(e,) = 0, we have i ( e l ) = 0, or el = 0. Thus, ( i p)(el,e3) = 0 implies el = 0, e3 = 0,

+

+

proving the claim. (b) Given e, E p ; '(xo) c E , . we observe that f ( p ( e 3 ) ) = e 3 , so that f ( e 2 - p .f(e2)) = f(e2) - f(e,) = 0. Then e, - p f ( e 2 ) belongs to E l , so we have 0

0

(i + d ( e 2 - P " f ( e z ) , f ( e , ) ) = e2 - P

Thus, i + p is onto.

O

f e z ) +P

O

f ( e 2 ) = e2.

We conclude that for any xo E X i + p is an isomorphism of fibers over + p is an isomorphism of vector bundles. Thus, we have to construct our map p, often called a right inverse forf: We shall use a partition of unity argument, passing from the local to the global. For every x E X , select an open neighborhood 0, such that both bundles E , and E , are trivial over Ox, i.e., xo, and hence i

+,:p;'(o,)5 0, x Naturally, k I n. The map + 3 ' j

[w",

f 0&':Ox

43:p;1(0,) % 0, x Rk. x R"-+O,

x

Rk

is onto. be the subspace of R",annihilated by 43 f 0 & for first coLet ordinate x; let n be the projection of R" onto R"-k.Form the sum map 0

h = ( 4 , ~ ~ f 0 & ' + 1x ~ ) : O , x R"+(O,x

',

R k ) @ ( O x xR n - k ) .

122

4. TANGENT BUNDLES, VECTOR BUNDLES, AND CLASSIFICATION

Of course at x the map h is visibly onto, and for this fixed x the domain and range are vector spaces of the same dimension. Thus, h is an isomorphism of vector spaces at x. By step (b) in the proof of Proposition 4.3, h is an isomorphism of vector bundles in a neighborhood of x. Therefore, in this possibly smaller neighborhood a right inverse to 43 f 4; may be found by taking the composition of the inclusion of 0, x R"into the Whitney sum (0,x Rk)@(O, x R"-") and then applying h-'. (1 x n annihilates such an element, and h applied to it is the same as 43 f 4; applied to it.) Using the homeomorphisms 42 and $ 3 , we easily convert this to a right inverse for 0

0

f:pil(0)

+

0

0

p ; W

for some suitably small neighborhood of x, say 0. Finally, choose a finite cover by such O's, a partition of unity, and add up the products ofthe functions in the partition of unity and the right inverses for f in the respective sets. Lemma 4.7 Let (E, X,p ) be a vector bundle over a compact Hausdorff space. Then there is an integer N > 0 and an onto map of bundles f : X x RN + E. That is, f is a map of bundles, which is a linear transformation that maps the points over x in the domain onto those over x in the range, for all x E X . We continue to assume f x = lX. Proof This, too, is a rather standard localization argument. Let S ( E ) be the vector space ofsection of E, i.e., continuous maps s: X E with p s = l X .Addition and multiplication are carried out in each set $,(p-'(x)) just as they would be in a vector space and then carried back to E by $;.' Of course S ( E ) is almost always an infinite-dimensional vector space over the real numbers. But note that there is a natural map --f

0

f l : X x S(E) + E

defined by fl(x, s) = s(x); such a map is often called an evaluation map. If we can find a finite-dimensional subspace of S(E), say F, such that the restriction of fi to X .x F maps X x F onto E, and if we can show that fl is linear on each fiber, i.e., linear over each point x E X , we will be done. But the linearity on each fiber is immediate, because if $ p : p - l ( O p )5 0, x R"

THE CLASSIFICATION OF VECTOR BUNDLES

123

is a homeomorphism coming from the definition of E and x E O,, then the composition X x S(E) 2 E

3 0,x

R"

is defined for given x E 0, and s E S(E), and is given by ($,

fl)(X,

4 = bqs(x));

therefore,

w,

O

fl)(X,S,

+ s2) = $,(SAX) + SZ(X)) = $&l(X)) + $,(SZ(X))

because of the definition of addition. Similarly, (+, 0 fl)(x, us) = a($, o fl)(x,s), for any real number a. (Compare Definition 4.10 with the identity map on X . ) Thus, we are reduced to locating our finite-dimensional subspace F. For each point x E X , choose a suitable OBj and define sections s j , . . . ,s! as follows : sj(x) = $Pj'(x,(O, . . . ,1, . . . , O ) ) ,

where the 1 occurs in the ith place. Notice that the . . . ,.'.form a basis for p - '(x) for any point x E O P j and , that if c( is any positive number, us{, . . ,as! would also be a basis for p - ' ( x ) . Select a finite cover by 0,,,. . . , OPmand a partition of unity ( $ j } subordinate to this cover. Then the collection of sections obtained by multiplying all the sections obtained for the O,,, . . . , Obmby the respective $1, . ,$,,, gives a finite family of functions, so that for any y E X the family contains a positive multiple of a basis for p - ' ( y ) . If we choose F to be the subspace spanned by these sections, $j(x)d(x), . . . , $j(x).'.(x), for j = 1, . . . ,m, then fl clearly maps X x F onto E , completing our proof.

4,

.

.

Corollary 4.1 If E is a vector bundle over a compact Hausdorff space X , then there is another vector bundle over X , say El, such that E @ E l is trivial. Proof

Apply Lemma 4.6 to the conclusion to Lemma 4.7.

Remarks Corollary 4.1 suggests an interesting algebraic interpretation for vector bundles. We think of all trivial vector bundles as a sort of zero element, and the Whitney sum of bundles as an addition. Then the previous corollary shows that we have an additive inverse operation. With proper definitions this approach gives rise to the important K-theory (see [4,88]), which asociates an abelian group with every X.Unfortunately, this is largely beyond the scope of this book.

124

4.

TANGENT BUNDLES, VECTOR BUNDLES, AND CLASSIFICATION

We are now ready to examine the universal bundles and to tackle the basic classification theorem. Definition4.19 We recall (from Definition 2.4) that Gn,mis the Grassman manifold of n-dimensional vector subspaces of Rn+m. (a) If f:X x Rk + E is an onto mapping of vector bundles over X (compact Hausdorff here) and dim E = n, we define a map

f ‘: x

+ Gk-n,n

that assigns to each point x E X the null space or kernel of the linear transformation f a s a map of the set x x Rk to p-’(x). (b) Set F c Gk-n,nx Rk to be subspace consisting of all ( u , x ) with u containing the point x; this makes sense because u is a (k - n)-dimensional subspace of Rk. F is trivially checked to be a subbundle of the trivial vector bundle (Gk-n,nx Rk,Gk-n,n,p)with p having projection on the first factor. We form the quotient bundle Ek-,,,,,

= Gk-n,n

X

Rk/F:

It is an n-dimensional vector bundle over Gk-n,n,which we write

It is called the universal or classifying bundle for n-dimensional vector bundles. Remarks (1) The map in Definition 4.19a has some remarkable properties. For example, if F is any n-dimensional bundle over X and the bundle map f : X x Rk -+ F is onto, then

f ’: X

+

Gk-n,n

has the property that

f’-‘ ( E k - n , n )

= F,

where Ek-n,nis the universal bundle defined in Definition 4.19b. This is immediate because the left-hand side is by definition

{ .Y, t } I Y E X , (u,t ) E G k - n . k x Rk, f’(y) = u } , that is, the quotient in which are identified to a point those pairs (y, t) E X x Rksuch that all t E f’(y), i.e., f ( t ) = 0. But these pairs, with this identification, yield precisely the quotient of X x Rkwith the subbundle, which is the (

~

1

THE CLASSIFICATION OF VECTOR BUNDLES

125

null space off, identified to a point; since f is onto, this is isomorphic to F (check). (2) Let M" s Rk be a smooth submanifold, TM" and NM" be the tangent and normal bundles. As usual, let TRk M" be the restriction of the tangent bundle to Rk (which is trivial) to the submanifold M". Then there is a natural projection

I

f:T R k1 M" -+ T M "

equivalent to taking the quotient bundle of the domain by NM", to which we may apply Definition 4.19a. The resulting map f ' assigns to each x E M" the translation of the normal space at x down to the origin. The induced bundle from the universal bundle must then be, according to Remark (1) above, isomorphic to the tangent bundle to M". There remains one small hurdle between us and the statement of the main classification theorem. That is, we must consider the limit of a family of Grassman manifolds. For any k > 0, let p: R n + k

-+

Rntk-1

be the projection map that ignores the last coordinate. If 1 is an (n - 1)dimensional subspace of R n + k - l , its inverse image p - ' ( I ) is an n-dimensional subspace of R" ". This procedure clearly defines a continuous inclusion map jn-l:Gn-l,k

+

Gn,k*

It is easy to check that if (Ek-n,", GkPn,", p,) is the universal or classifying bundle (Definition 4.19b) then the bundle induced from this by jk-n-l is precisely ( E k - n - 1 ,n G k - n - 1,a P n ) * Now j,- defines a map of sets of homotopy classes by ( j n - J,({f}) = { j n - l o f ) , i.e., t

9

(jn-l)*:[X9Gn-l,kl-+

[X,Gn,kI

is defined by composition with j,- 1 .

Definition 4.20 We define the limit (direct limit or injective limit) lim[X, G i , m ]

--*

to be the quotient of the disjoint union

U [ X , GimI i

by the equivalence relation that identifies

if;,1 E [ X , Gi1,mI

with

{A*)E [X, G i 2 , m I

126

4. TANGENT BUNDLES, VECTOR BUNDLES, AND CLASSIFICATION

when there is an i3 2 max(i,, iz) such that (jiJ*

.

* *

(ji,+i)*(jiJ*({fi})=

(A,)*

* *

.(j~~+l)*(ji~)*({j~}).

In other words, two elements represent the same element, in the limit, if after a suitable number of (js)*'s they have a common image in some [X,Gs,m]. We note that there is a natural map

[XiGi,m]

+

@[X, Giml

that associates to every class of maps its equivalence class in the limit. We may now state our main theorem. Theorem 4.1 If X is a compact Hausdorff space, there is an isomorphism of sets (i.e., a 1- 1 correspondence)

l%[x;Gk-,,"] X

V"(x).

(Here k is the variable for the purpose of taking the limit. We write k - n because that form occurs naturally in Definition 4.19.) Proof

An element in the limit is represented as a homotopy class

{f}E LX, G k - n , n l for some k, perhaps very large. The induced bundle j - ' ( E k - , , J is an ndimensional vector bundle over X.By Proposition 4.3 and the fact that the maps jk-,,- carry, by the induced bundle construction, universal bundles into universal bundles, we see that this construction yields a well-defined map

@:lh[X,Gk-,,J + v"(X). We shall define a map Y: v"(X)+ lim[X, Gk-,J

that will be the inverse for @. Given an n-dimensional vector bundle E, we may use Lemma 4.7 to find a map g:x x Rk -+ E

that is onto. Using Definition 4.19a, we obtain a map 9':

x

-+

Gk-,,".

We shall show that for sufficiently large i the homotopy class of Oi)* . . . (jk-.)*({g'))

THE CLASSIFICATION OF VECTOR BUNDLES

127

depends only on the original bundle, and not on the choice of g or k. For brevity write (ji,k-A*

= (ji)*.

. .(jk-n)*.

Suppose that we are given two onto mappings of vector bundles

gl:Xx Rkl + E For each t, 0 I t I 1, form

and

gz:Xx Rk2 + E.

G,:X x Rkl x Rk2 + E, defined by G,(x, y,, y,) = (1 - r)gl(x,yl) + tq2(x,y3).This makes sense because the operations take place in p - ' ( x ) c E, which is a vector space, for each x. It is completely trivial to check that this a bundle map, or homoonto E . By definition 4.19, we arrive at a homomorphism from X x Rk1+k2 topy of maps or

G':X X I

+ Gk,+k2-",,,

defined by G(x, t ) = G;(x). It is clear that Gb is the same as ( j k l + k 2 - , , k l - , , ) * applied to g;. But unfortunately C; differs from the map ( j k l + k 2 - n , k 2 - n ) * applied to g; in that G , constitutes projection onto the last, rather than the first, coordinates, acting on g;. The proof will be complete if we can find a homotopy of the identity map of G k ,+ k z - n , n to a map that permutes the coordinates in such a way that the last k, coordinates become the first k2 coordinates, etc. To do this, observe , n itself the that for a map of R k l + k 2 to itself to induce a map of G k , + k 2 - n to map must take linear subspaces, such as the ( k , + k , - n)-dimensional subspace, to themselves. Any orthogonal transformation will have this property. Select an orthogonal transformation of Rk1+k2that takes the last kz coordinates to the first k , coordinates. Assuming that k l > 1, which is no loss of generality here, we may further arrange the matter so that the determinant of the orthogonal transformation is + 1. Because the orthogonal matrices of determinant + 1 are pathwise connected, we may connect this orthogonal transformation (viewed as a matrix) to the identity by a path of orthogonal matrices of determinant + 1. This gives a homotopy of our original orthogonal transformation, which takes the last k2 coordinates to the first, to the identity transformation, and correspondingly a suitable homotopy of maps of Gk, + k 2 - n , n to itself.

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4. TANGENT BUNDLES, VECTOR BUNDLES, AND CLASSIFICATION

We conclude that (jkl+kZ-n,kl - n ) * ( d l )

(jkl+k2-n,~~-n)*(g~),

showing that Y is well defined. We must finally check that Q, and Y are inverses of one another. Consider a map f:X + Gk-n,n;then Q,({f}) is represented by the induced bundle f-'(Ek-n,n),where we recall that Ek-n,n is the quotient of the trivial bundle Gk-n,nx Rk by the subbundle made up of pairs (x,y) with x E Gk-n,nand y E x . Of course, f-'(Ek-,,,,,) is then a quotient bundle of the bundle induced by the map f from the trivial bundle Gk-n,nx Rk. Consider the projection map from f-'(Gk-,,,,, x Rk) to f - ' ( ~ ~ - , , defined ~) by ( x , e ) + ( x , {el), and call it g . Because f-'(Ek-n,n) = {(x, { e } ) l x E x,

{e}

E Ek-n,n,

f ( x ) = Pn(e))

and f-*(Gk-",,,x Rk) = { ( x , e ) l xE X ,

e E R ~ , f ( x ) = n(e)},

being projection on first factor, we see that for each x E X the null space or kernel of g consists of all o E Rk such that o E f ( x ) . But this means (by Definition 4.19a) that g'(x) = f ( x ) , and thus, by checking the definitions,

7c

YJ O @({f)) = {f}.

On the other hand, given the bundle (E, X,p), we select a map g : X x Rk

E

that is onto (lemma 4.7). Form g': X + Gk-n,nand consider (9')- l ( E k - n , n ) ,

with E k - n , n the universal bundle as defined above. We then define h:(g')-'(Ek-n,n)

+

E

x,

by h(x,(g'(x), { y})) = g(& y), wfiere x E (g'(x),{ y}) E E k - n , n , with g'(x) = p,(y). Because the map g annihilates its own kernel or nullspace, this is then trivially checked to be a well-defined bundle map. Now, given z E E, we select (x, w ) E X x Rk such that g(x, w)= z. Then immediately h(x, g(x), {w})= z, proving that h is onto. But the domain and range of h are both ndimensional vector bundles over X,so that, over each point x , h is an onto map from a finite-dimensional vector space to one of the same dimension. Since such a linear map is always an isomorphism, by basic linear algebra,

EXAMPLES OF CLASSIFICATION

129

we see that h is an isomorphism. This means that 0 '-' Y({E}) = (E},

completing our proof. Note that the last part of the proof points to the need to use the limit in the proof, for we cannot control how large a k is needed in the map g: X x Rk + E without some knowledge of X, nor can we say for what sort of k is a given class in the limit represented by a specific map. EXAMPLES OF CLASSIFICATION

To achieve applications of the main theorem, one must naturally know facts about [X, Gk-,,J, or more exactly

If X is in some sense a finite-dimensional space, then there will be a large

K such that

Furthermore, Gk-n.nis well known in many respects, so that one may often say a great deal about v"(X). I propose to list here, with precise references, many of the basic facts about Gk -n,n. (i) Gk-n,nis a homogeneous space (see [47,103]). The orthogonal group O(k),as in Remark ( 5 )following Definition 4.2, acts on Rk as a transformation group in such a way that a linear space of a given dimension is mapped into a linear space of the same dimension. Hence a transformation from O(k) maps Gk-n,ninto itself. I claim that the action is transitive, for given two ( k - n)-dimensional subspaces of Rk, let us say y, and y2, we choose bases for y, and y,, say u : , . . . ,ui-" and u i , . . . ,u:-", as well as bases for the orthogonal complements, say w : , . . . ,w; and wi,. . . ,wz.Without loss of generality, these bases may be assumed to be orthonormal (i.e., mutually perpendicular and of length 1). Define an orthogonal transformation 4 : R k + Rk

by setting 4(0\)= ui and $ ( w j ) = w i and extending by linearity. It is then clear that when this orthogonal transformation acts on Gk-n,nit takes y, to y , , as we wished. We write this action as p:O(k)x Gk-n,n+ Gk-,,+.

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4. TANGENT BUNDLES, VECTOR BUNDLES, AND CLASSIFICATION

Since O(k)is compact, Gk-,,,,, is homeomorphic to O(k)/GO, where Go = { g E O(k)lp(g,x,) = xo> for some fixed x o E Gk-n,n(see Proposition 4.1). We suppose, as a matter of convenience, that XO={tO,.

..,tk)ltk-n+l

=tk-n+~="*=tk=O}.

Naturally, an orthogonal transformation will take the (k - n)-dimensional subspace to itself if and only if it takes the orthogonal complement x i to itself. The restriction of an orthogonal transformation to a subspace that it carries into itself is still an orthogonal transformation. It follows that Go in our basis must be the group of orthogonal matrices of the form

where A, E O(k - n) and A, E O(n).In other words Go = O(k - n) x O(n),

and identifying these groups as subgroups of O(k)as we have done, we may write Gk-n,n= O(k)/O(k- n)

X

O(n).

(ii) For certain nice spaces X, one may say a great deal about [X, Gk-n,n]. These X are built up out of closed disks D" = {X E R" I llXl[ I 1>.The details are as follows (see [25, 1131). A finite cell complex is a finite union of spaces X o , X', . . . , Xp. X o is a finite disjoint union of points. Given X k , Xk+' may be formed as follows. Take a finite collection of maps

a l , . . . ,a,:Sk + X k ,

where Sk is the k-sphere. Form the disjoint union Z = xk u ei+1 u

u

where the ej"" are disjoint copies of D k + ' .X k + ' is the quotient space of 2 by the equivalence relation that identifies any point y lying in the boundary of some e f + (which is precisely Sk) with the point ai(y) E X k . In other words, points interior to any ef" are identified only to themselves, while points on the boundary of ef+' are attached to their images, under the maps ai, in Xk. For example, we may write the circle S' as a cell complex in several ways: S' = eo u el, as in Fig. 4.2a, or S' = ep u e i u eb u e t , as in Fig. 4.2b. There are naturally many other possibilities using more cells.

'

8 . 1

EXAMPLES OF CLASSIFICATION

Figure4.2

131

(b)

(a)

For the torus, S' x S ' , we may describe the structure of a cell complex as shown in Fig. 4.3. Here the torus is X 2 ; X o consists of two points. X' is three circles, and may be displayed as a cell complex as shown in Fig. 4.4. There are two cells of dimension 2 in this description of S' x S'. Of course, there are many other ways to write S' x S' as a cell complex.

Figure 4.3

Figure 4.4

Observe that if X is a cell complex, then each subspace Xi(called the iskeleton) is also a cell complex. Also, any quotient space X / X ' becomes a cell complex when the single point, which is the equivalence class of Xi,is treated as a single zero cell. The Grassman manifolds Gn,kmay be given the structure of a cell complex in such a way that the maps jn-i:Gn-i.k

+

Gn,k

take individual cells into individual cells of no greater dimension. This procedure, due to Ehresmann ([31,78]), is as follows.

132

4.

TANGENT BUNDLES, VECTOR BUNDLES, AND CLASSIFICATION

Fix as filtration [wo E [w'

c

9

.

.

=

Rk.

Select a sequence of positive integers oi,1 5 i 5 k 1

o1 5

- n, so that

. . . 5 b k - n 5 k.

Let e(ol, . , . ,ok-,,) denote the set of all (k - n)-dimensional subspaces of X in Rk, so that X meets Rut in an i-dimensional subspace and meets RUi-l in ( i - 1)-dimensional subspace for all i. The sets e(oi, . . . ,0k-n) may be shown to give a cell decomposition of Gk-,,". The number ofj-cells in Gk-n,n is easily expressed in terms of partitions. (iii) Note that if X and Yare cell complexes andf: X -+ Y is a continuous map, then f is homotopic to a map that takes i cells into a union of cells of no higher dimension (see [102,1131for this cellular approximation theorem). It is not difficult to see from (ii) above that the cells of Gk-n,nof dimension ~ pfor, p c k - n, are independent of k or n. We may then infer (iv) If X is a finite complex all of whose cells have dimension less that k - n, then (jn)*:

Cx,

Gk-n,nl

LX, G k - n +

1,111

is an isomorphism of sets. Referring to the definition of limit, we see that in this case [ X , Gk-n,n]

lim[X, Gi-.,,].

This fact may now be exploited to make a full analysis of the situation in low-dimensional cases. (v) (a) If X is a cell complex of dimension less than or equal to 2 (i-e., no cells of dimension 3 or greater), then line bundles over X are given by + lim[X, Gk-1.11. By (iv) above this is isomorphic as a set to G1,3] [x, RP3] [x, G3.11 [x, (see the remarks after Definition 2.4). In particular, we can check that there are two isomorphism classes of line bundles over the circle by noting that there are two homotopy classes of maps S' -+ RP3 (in fact, 7cI(RP3)has two elements (see [59, 1021). (b) To get at u2(X),consider [X, G3,2] = [X,G2,3]. Some easy calculations are now possible. Suppose, for example, that X is a suspension such as S2 = S(S'). u2(S2)= [S2,G3.J = [S', Gl(2; R)lO/-

by Proposition 4.4.Gl(2;R) is homeomorphic to O(2) x

[w3,

so it has the

EXAMPLES OF CLASSIFICATION

133

homotopy type of 0(2), which is homeomorphic to two disjoint copies of S' (see [20]). Then U Z ( S 2 ) = [Sl, S'l0/h.. [S',S'],

is the integers, and

-

identifies at most two elements together, so

uz(Sz)is in 1-1 correspondence with the integers.

In this correspondence, the tangent bundle to Sz corresponds to + 2 (this may be checked by examining the identification of two disks along the common boundary and examining the overlap map, as in Proposition 4.4). For further calculations and applications, consult [103,531. Lastly, a few remarks about orientable bundles and the case of complex vector bundles are in order here. By our basic Definition 4.5., a vector bundle (E, X , p ) has a local structure given by homeomorphisms $a:P-'(Ua)

+

U a x R".

The coordinate overlap maps $@ 0 $-, are completely determined functions gap:U , n U , -,Gl(n; R).

If the structure of the vector bundle may be determined in such a way 1 iat every g,@(x)has positive determinant, then a choice of an orientation in any p - ' ( x ) may be carried over to a choice of an orientation in p - ' ( y ) for any y in the same connected component of X as the point x. Hence, if X is connected and every gap(x)is of positive determinant, then every p - ' ( x ) may be given a specific, well-defined orientation. In this case the bundle is called orientable, and if in addition an orientation is specified on some p - ' ( x ) ,say by giving a definite homeomorphism h : p - '(x) 5 R", then we call the bundle oriented. For example, every bundle over Sz is orientable. We simply cover S2 with two overlapping open sets, each slightly larger than a hemisphere, and observe the overlap may be assumed to be homeomorphic to S' x (- E, E ) for some small E > 0. In particular, the overlap of the two open sets is connected. Therefore, it we choose the homeomorphisms $, and $@ so that at one point 0 $,-' preserves orientation, it will preserve orientation at every other point in S' x ( - E , E ) . Equivalently, the determinant of g,@(x) will be positive for all .K in the overlap. (Since any homeomorphism $, may may be composed with $@

1"- x h: u, x R"

-+

u, x

R"

for any linear homeomorphism h: R" + R", consistent with the definition of the bundle, we may choose any orientation we wish for $, or $@ restricted to a given point.)

134

4. TANGENT BUNDLES, VECTOR BUNDLES, AND CLASSIFICATION

On the other hand, the two orientations on TS2 give equivalent vector bundles (see problem 2), although inequivalent oriented bundles. For oriented bundles, one may develop a theory of classification similar to Theorem 4.1.One works with oriented Grassman manifolds, defined as homogeneous spaces [see Remark (l), after Theorem 4.11 in terms of the special orthogonal groups SO@),consisting of orthogonal matrices of determinant + 1. The resultant classifying spaces are two-sheeted covering spaces of our Grassman manifolds, because for each ( k - n)-dimensional subspace of Rk there are two oriented (k - +dimensional subspaces (consult [103]). The theory extends to vector bundles defined over any topological field or skew field, but the case of complex vector bundles defined over the field of complex numbers is especially important and interesting. All the basic definitions carry over immediately. The classification is, in fact, slightly simpler, because there are no difficulties of orientation, as we had in Propositions 4.4,in the case of complex vector bundles. This results from the fact that the general linear group over the complex numbers, Gl(n; C), is connected (see problem 3). In this case, the classifying space is

where the unitary group U(k)is the group of k x k complex matrices whose inverses equal the conjugates of their transposes. U(k)is connected, so that all bundles and homomorphisms are automatically oriented. For further developments of this beautiful and interesting subject, consult [4, 1031. The Classification of bundles is an important and interesting topic, and it has been studied in depth by many authors. Generalizations of Theorem 4.1 include the classification of a general principal bundle (the group being any topological group), as well as extending the results to cover more general X than compact Hausdorff spaces. For the early history, see Steenrod‘s book [103]. The most beautiful and general result in the area of classification is due to Dold [26]. PROBLEMS AND PROJECTS

1. (a) Use our results to show that there is no nontrivial line bundle over the sphere S2. One may show that [S’, 0(1)], is trivial or that [S’, RP3] is trivial (the latter because [S2, RP3]= [S2,S’], since S3 is a covering space of RP3and, from “cellular approximation,” any mapf: S2 + S3is homotopic to one that misses a point and thus factors through S3 - pt = D3, which is contractible).

PROBLEMS AND PROJECTS

135

(b) Describe the line bundles over S' u S', consisting of two copies of a circle attached at a single point. (c) Let X and Y be spaces, and xo E X . yo E Y be single points. Define i,:X + X x Y and i 2 : Y + X x Y by i l ( x )= (x, yo) and i2(y) = (xo, y). Let n 1 and 7c2 be the projections such that n1 i l = 1, and n2 0 i2 = l y . Deduce from Theorem 4.1 that if E is a nontrivial vector bundle over X (resp. Y), then n; '(E) (resp.n; ' E )is a nontrivial vector bundle over X x Y. As an example, exhibit nontrivial line bundles over the torus S' x S'. (d) In remark (lb) following Theorem 4.1, we describe a cell complex structure on S' x S'. If we identify to a single point all cells except one 2dimensional cell, we have a map g : S ' x S' + S2. Examine the induced bundle g-'(TS2).Does it have a nonzero section? (Is it trivial or nontrivial?) What is the bundle i;'(g-'TS2), with il:S1 -, S' x S' as in (c) above? 2. (a) The maps 0

9 ' : s' -, O(2)

by

cos8 -sin8 cos 8 )

Q ~ ( o=)(sin 8

and

are not homotopic. Following each map by the inclusion of O(2) c Gl(2; R) we get two bundles in v2(S2),which are equivalent because g1 and g2 are related by gl(8) = Ag2(8)A-',

where

A=

(y i).

Conclude as a consequence of this sort of analysis that the bundles TS2, with different orientations, are equivalent vector bundles (identical bundles) but inequivalent oriented bundles, and that the maps S' + G1(2;R), from Proposition 4.4, for the two orientations are not homotopic but are equivalent under -. (b) If n is odd and I E Gl(n; R) is the identity matrix, then det( - I ) = - 1. Nevertheless, - I, commutes with every matrix in Gl(n; R). Conclude that for n odd the equivalence relation in Proposition 4.4 is trivial, and that v"(SX) = [X,Gl(n; rW)IO.

(c) If X = S and n is odd, [X, Gl(n; R)], = n,(Gl(n; R)) the mth homotopy group. Gl(3; R) has the homotopy type of 0(3), which is two disjoint copies of SO(3).SO(3) is homeomorphic to RP3,and S3 is a twofold covering

136

4.

TANGENT BUNDLES, VECTOR BUNDLES, AND CLASSIFICATION

space of RP3.Thus conclude

u3(s3) = n2(S3), which by "cellular approximation" must be trivial, since any cellular map misses a point (and thus lies in the contractible D3 = S 3 - pt). Hence u3(S3) has only one (trivial) element. Consult [1031 for details here. (d) Show that every map S2 --t S' is homotopic to a constant (because R + S' is a covering space) and, thus that v2(S3)and v'(S3)must also consist of a single (trivial) element. 3. The unitary group U(n) is the group of isomorphisms of the n-dimensional complex vector space @" that preserve the inner product

( Y , Z )= y171

+ . . + y&, *

where Y = ( y l , . . . ,yn), etc, and Ti means the complex conjugate of zi. But the set (YEcnI,/pj=

I}

is clearly the sphere SZn-', because @" is homeomorphic to R2". Fixing Y E S2"-', verify that the action of U(n) on S2"-l defined by A+AY

is transitive and effective. Conclude that s2n-1

- U(n)/U(n- 1)

(choose suitable Y using Proposition 4.1). Verify that if G,, and GIG,, are path connected, so is G, in this case. Calculate U(l) and conclude that U(n) is path connected for all n. State the correct version of Proposition 4.4 for complex vector bundles over a suspension (compare [4]). 4. Show that an n-dimensional vector bundle (E, X, p) is trivial if and only if there are n sections sI,. . . ,s, and for each x the vectors sl(x),. . . ,s,(x) are linearly independent. Show that if the bundle is orientable, then it is trivial if and only if there are n - 1 such sections, linearly independent at a point. (The last section may be determined from orientability.) Find 2-dimensional vector bundles with 0, 1, or 2 sections that are linearly independent at every point (0 such sections means simply that there is no everywhere nonvanishing section of the bundle). (These examples can be built with tangent bundles.)