Allen Hatcher - Cornell University

To motivate the definition of a vector bundle let us consider tangent vectors tothe unit 2 sphere S 2 in R3 . At each point x S 2 there is a tangent plane Px . Thisis a 2 dimensional vector space with the pointx as its zero vector 0x . Vectors vx Px arethought of as arrows with their tail at x . Ifwe regard a vector vx in Px as a vector in R3 ,then the standard convention in linear algebrawould be to identify vx with all its paralleltranslates, and in particular with the uniquetranslate τ(vx ) having its tail at the origin inR3 . The association vxfunction τ : T S2 R3 τ(vx )where T S2defines ais the set of all tangent vectors vx as x ranges2over S . This function τ is surjective but certainly not injective, as every nonzerovector in R3 occurs as τ(vx ) for infinitely many x , in fact for all x in a great circlein S 2 . Moreover τ(0x ) 0 for all x S 2 , so τ 1 (0) is a whole sphere. On the otherhand, the function T S 2 S 2 R3 , vxtopologize T S2 (x, τ(vx )) , is injective, and can be used to2as a subspace of S R3 , namely the subspace consisting of pairs(x, v) with v orthogonal to x .Thus T S 2 is first of all a topological space, and secondly it is the disjoint unionof all the vector spaces Px for x S 2 . One can think of T S 2 as a continuous familyof vector spaces parametrized by points of S 2 .The simplest continuous family of 2 dimensional vector spaces parametrized bypoints of S 2 is of course the product S 2 R2 . Is that what T S 2 really is? Moreprecisely we can ask whether there is a homeomorphism h : T S 2 S 2 R2 that takeseach plane Px to the plane {x} R2 by a vector space isomorphism. If we had suchan h , then for each fixed nonzero vector v R2 the family of vectors vx h 1 (x, v)would be a continuous field of nonzero tangent vectors to S 2 . It is a classical theoremin algebraic topology that no such vector field exists. (See §2.2 for a proof using

Basic Definitions and ConstructionsSection 1.15techniques from this book.) So T S 2 is genuinely twisted, and is not just a disguisedform of the product S 2 R2 .Dropping down a dimension, one could consider in similar fashion the space T S 1of tangent vectors to the unit circle S 1 in R2 . In this case there is a continuousfield vx of nonzero tangent vectors to S 1 , obtained by regardingpoints x S 1 as unit complex numbers and letting vx be thetranslation of the vector ix that has its tail at x . This leads to ahomeomorphism S 1 R T S 1 taking (x, t) to tvx , with {x} Rgoing to the tangent line at x by a linear isomorphism. Thus T S 1really is equivalent to the product S 1 R1 .Moving up to S 3 , the unit sphere in R4 , the space T S 3 of tangent vectors is againequivalent to the product S 3 R3 . Regarding R4 as the quaternions, an equivalence isthe homeomorphism S 3 R3 T S 3 sending (x, (t1 , t2 , t3 )) to the translation of thevector t1 ix t2 jx t3 kx having its tail at x . A similar construction using Cayleyoctonions shows that T S 7 is equivalent to S 7 R7 . It is a rather deep theorem, provedin §2.3, that S 1 , S 3 , and S 7 are the only spheres whose tangent bundle is equivalentto a product.Although T S n is not usually equivalent to the product S n Rn , there is a sensein which this is true locally. Take the case of the 2 sphere for example. For a pointx S 2 let P be the translate of the tangent plane Px that passes through the origin. For points y S 2 that are sufficiently close to x the map πy : Py P sendinga tangent vector vy to the orthogonal projection of τ(vy ) onto P is a linear isomorphism. This is true in fact for any y on the same side of P as x . Thus fory in a suitable neighborhood U of x in S 2 the map (y, vy )homeomorphism with domain the subspace of T S2 (y, πy (vy ))is aconsisting of tangent vectors atpoints of U and with range the product U P . Furthermore this homeomorphismhas the key property of restricting to a linear isomorphism from Py onto P for eachy U . A convenient way of rephrasing this situation, having the virtue of easilygeneralizing, is to let p : T S 2 S 2 be the map (x, vx ) x , and then we have a homeomorphism p 1 (U) U P that restricts to a linear isomorphism p 1 (y) {y} Pfor each y U . With this observation we are now ready to begin with the formaldefinition of a vector bundle.

6Chapter 1Vector Bundles1.1 Basic Definitions and ConstructionsThroughout the book we use the word “map" to mean a continuous function.An n dimensional vector bundle is a map p : E B together with a real vectorspace structure on p 1 (b) for each b B , such that the following local trivialitycondition is satisfied: There is a cover of B by open sets Uα for each of which thereexists a homeomorphism hα : p 1 (Uα ) Uα Rn taking p 1 (b) to {b} Rn by a vector space isomorphism for each b Uα . Such an hα is called a local trivialization ofthe vector bundle. The space B is called the base space, E is the total space, and thevector spaces p 1 (b) are the fibers. Often one abbreviates terminology by just callingthe vector bundle E , letting the rest of the data be implicit.We could equally well take C in place of R as the scalar field, obtaining the notionof a complex vector bundle. We will focus on real vector bundles in this chapter.Usually the complex case is entirely analogous. In the next chapter complex vectorbundles will play the larger role, however.Here are some examples of vector bundles:(1) The product or trivial bundle E B Rn with p the projection onto the firstfactor.(2) If we let E be the quotient space of I R under the identifications (0, t) (1, t) ,then the projection I R I induces a map p : E S 1 which is a 1 dimensional vectorbundle, or line bundle. Since E is homeomorphic to a Möbius band with its boundarycircle deleted, we call this bundle the Möbius bundle.(3) The tangent bundle of the unit sphere S n in Rn 1 , a vector bundle p : E S nwhere E { (x, v) S n Rn 1 x v } and we think of v as a tangent vector toS n by translating it so that its tail is at the head of x , on S n . The map p : E S nsends (x, v) to x . To construct local trivializations, choose any point x S n andlet Ux S n be the open hemisphere containing x and bounded by the hyperplanethrough the origin orthogonal to x . Define hx : p 1 (Ux ) Ux p 1 (x) Ux Rnby hx (y, v) (y, πx (v)) where πx is orthogonal projection onto the hyperplanep 1 (x) . Then hx is a local trivialization since πx restricts to an isomorphism ofp 1 (y) onto p 1 (x) for each y Ux .(4) The normal bundle to S n in Rn 1 , a line bundle p : E S n with E consisting ofpairs (x, v) S n Rn 1 such that v is perpendicular to the tangent plane to S n atx , or in other words, v tx for some t R . The map p : E S n is again given byp(x, v) x . As in the previous example, local trivializations hx : p 1 (Ux ) Ux Rcan be obtained by orthogonal projection of the fibers p 1 (y) onto p 1 (x) for y Ux .(5) Real projective n space RPn is the space of lines in Rn 1 through the origin.Since each such line intersects the unit sphere S n in a pair of antipodal points, we can

Basic Definitions and ConstructionsSection 1.17also regard RPn as the quotient space of S n in which antipodal pairs of points areidentified. The canonical line bundle p : E RPn has as its total space E the subspaceof RPn Rn 1 consisting of pairs (ℓ, v) with v ℓ , and p(ℓ, v) ℓ . Again localtrivializations can be defined by orthogonal projection.There is also an infinite-dimensional projective space RP which is the unionof the finite-dimensional projective spaces RPn under the inclusions RPn RPn 1coming from the natural inclusions Rn 1 Rn 2 . The topology we use on RP isthe weak or direct limit topology, for which a set in RP is open iff it intersects eachRPn in an open set. The inclusions RPn RPn 1 induce corresponding inclusionsof canonical line bundles, and the union of all these is a canonical line bundle overRP , again with the direct limit topology. Local trivializations work just as in thefinite-dimensional case.(6) The canonical line bundle over RPn has an orthogonal complement, the spaceE { (ℓ, v) RPn Rn 1 v ℓ } . The projection p : E RPn , p(ℓ, v) ℓ ,is a vector bundle with fibers the orthogonal subspaces ℓ , of dimension n . Localtrivializations can be obtained once more by orthogonal projection.A natural generalization of RPn is the so-called Grassmann manifold Gk (Rn ) ,the space of all k dimensional planes through the origin in Rn . The topology on thisspace will be defined precisely in §1.2, along with a canonical k dimensional vectorbundle over it consisting of pairs (ℓ, v) where ℓ is a point in Gk (Rn ) and v is avector in ℓ . This too has an orthogonal complement, an (n k) dimensional vectorbundle consisting of pairs (ℓ, v) with v orthogonal to ℓ .An isomorphism between vector bundles p1 : E1 B and p2 : E2 B over the samebase space B is a homeomorphism h : E1 E2 taking each fiber p1 1 (b) to the corresponding fiber p2 1 (b) by a linear isomorphism. Thus an isomorphism preservesall the structure of a vector bundle, so isomorphic bundles are often regarded as thesame. We use the notation E1 E2 to indicate that E1 and E2 are isomorphic.For example, the normal bundle of S n in Rn 1 is isomorphic to the product bundle S n R by the map (x, tx) (x, t) . The tangent bundle to S 1 is also isomorphicto the trivial bundle S 1 R , via (eiθ , iteiθ ) (eiθ , t) , for eiθ S 1 and t R .As a further example, the Möbius bundle in (2) above is isomorphic to the canonical line bundle over RP1 S 1 . Namely, RP1 is swept out by a line rotating throughan angle of π , so the vectors in these lines sweep out a rectangle [0, π ] R with thetwo ends {0} R and {π } R identified. The identification is (0, x) (π , x) sincerotating a vector through an angle of π produces its negative.SectionsA section of a vector bundle p : E B is a map s : B E assigning to each b B avector s(b) in the fiber p 1 (b) . The condition s(b) p 1 (b) can also be written as

8Chapter 1Vector Bundlesps 11, the identity map of B . Every vector bundle has a canonical section, the zerosection whose value is the zero vector in each fiber. We often identify the zero sectionwith its image, a subspace of E which projects homeomorphically onto B by p .One can sometimes distinguish nonisomorphic bundles by looking at the complement of the zero section since any vector bundle isomorphism h : E1 E2 musttake the zero section of E1 onto the zero section of E2 , so the complements of thezero sections in E1 and E2 must be homeomorphic. For example, we can see that theMöbius bundle is not isomorphic to the product bundle S 1 R since the complementof the zero section is connected for the Möbius bundle but not for the product bundle.At the other extreme from the zero section would be a section whose values areall nonzero. Not all vector bundles have such a section. Consider for example thetangent bundle to S n . Here a section is just a tangent vector field to S n . As we shallshow in §2.2, S n has a nonvanishing vector field iff n is odd. From this it followsthat the tangent bundle of S n is not isomorphic to the trivial bundle if n is evenand nonzero, since the trivial bundle obviously has a nonvanishing section, and anisomorphism between vector bundles takes nonvanishing sections to nonvanishingsections.In fact, an n dimensional bundle p : E B is isomorphic to the trivial bundle iffit has n sections s1 , · · · , sn such that the vectors s1 (b), · · · , sn (b) are linearly independent in each fiber p 1 (b) . In one direction this is evident since the trivial bundlecertainly has such sections and an isomorphism of vector bundles takes linearly independent sections to linearly independent sections. Conversely, if one has n linearlyPindependent sections si , the map h : B Rn E given by h(b, t1 , · · · , tn ) i ti si (b)is a linear isomorphism in each fiber, and is continuous since its composition witha local trivialization p 1 (U) U Rn is continuous. Hence h is an isomorphism bythe following useful technical result:Lemma 1.1.A continuous map h : E1 E2 between vector bundles over the samebase space B is an isomorphism if it takes each fiber p1 1 (b) to the correspondingfiber p2 1 (b) by a linear isomorphism.Proof:that hThe hypothesis implies that h is one-to-one and onto. What must be checked is 1is continuous. This is a local question, so we may restrict to an open set U Bover which E1 and E2 are trivial. Composing with local trivializations reduces to thecase that h is a continuous map U Rn U Rn of the form h(x, v) (x, gx (v)) .Here gx is an element of the group GLn (R) of invertible linear transformations ofRn , and gx depends continuously on x . This means that if gx is regarded as ann n matrix, its n2 entries depend continuously on x . The inverse matrix gx 1 alsodepends continuously on x since its entries can be expressed algebraically in termsof the entries of gx , namely, gx 1 is 1/(det gx ) times the classical adjoint matrix ofgx . Therefore h 1 (x, v) (x, gx 1 (v)) is continuous.

Basic Definitions and ConstructionsSection 1.19As an example, the tangent bundle to S 1 is trivial because it has the section(x1 , x2 ) ( x2 , x1 )for (x1 , x2 ) S 1 . In terms of complex numbers, if we setz x1 ix2 then this section is z iz since iz x2 ix1 .There is an analogous construction using quaternions instead of complex numbers. Quaternions have the form z x1 ix2 jx3 kx4 , and form a division algebraH via the multiplication rules i2 j 2 k2 1 , ij k , jk i , ki j , ji k ,kj i , and ik j . If we identify H with R4 via the coordinates (x1 , x2 , x3 , x4 ) ,then the unit sphere is S 3 and we can define three sections of its tangent bundle bythe formulasz izor(x1 , x2 , x3 , x4 ) ( x2 , x1 , x4 , x3 )z jzor(x1 , x2 , x3 , x4 ) ( x3 , x4 , x1 , x2 )z kzor(x1 , x2 , x3 , x4 ) ( x4 , x3 , x2 , x1 )It is easy to check that the three vectors in the last column are orthogonal to each otherand to (x1 , x2 , x3 , x4 ) , so we have three linearly independent nonvanishing tangentvector fields on S 3 , and hence the tangent bundle to S 3 is trivial.The underlying reason why this works is that quaternion multiplication satisfies zw z w , where · is the usual norm of vectors in R4 . Thus multiplication by aquaternion in the unit sphere S 3 is an isometry of H . The quaternions 1, i, j, k formthe standard orthonormal basis for R4 , so when we multiply them by an arbitrary unitquaternion z S 3 we get a new orthonormal basis z, iz, jz, kz .The same constructions work for the Cayley octonions, a division algebra structure on R8 . Thinking of R8 as H H , multiplication of octonions is defined by(z1 , z2 )(w1 , w2 ) (z1 w1 w 2 z2 , z2 w 1 w2 z1 ) and satisfies the key property zw z w . This leads to the construction of seven orthogonal tangent vector fields onthe unit sphere S 7 , so the tangent bundle to S 7 is also trivial. As we shall show in§2.3, the only spheres with trivial tangent bundle are S 1 , S 3 , and S 7 .Another way of characterizing the trivial bundle E B Rn is to say that there is acontinuous projection map E Rn which is a linear isomorphism on each fiber, sincesuch a projection together with the bundle projection E B gives an isomorphismE B Rn , by Lemma 1.1.Direct SumsGiven two vector bundles p1 : E1 B and p2 : E2 B over the same base space B ,we would like to create a third vector bundle over B whose fiber over each point of Bis the direct sum of the fibers of E1 and E2 over this point. This leads us to definethe direct sum of E1 and E2 as the spaceE1 E2 { (v1 , v2 ) E1 E2 p1 (v1 ) p2 (v2 ) }There is then a projection E1 E2 B sending (v1 , v2 ) to the point p1 (v1 ) p2 (v2 ) .The fibers of this projection are the direct sums of the fibers of E1 and E2 , as we

10Chapter 1Vector Bundleswanted. For a relatively painless verification of the local triviality condition we maketwo preliminary observations:(a) Given a vector bundle p : E B and a subspace A B , then p : p 1 (A) A isclearly a vector bundle. We call this the restriction of E over A .(b) Given vector bundles p1 : E1 B1 and p2 : E2 B2 , then p1 p2 : E1 E2 B1 B2is also a vector bundle, with fibers the products p1 1 (b1 ) p2 1 (b2 ) . For if we havelocal trivializations hα : p1 1 (Uα ) Uα Rn and hβ : p2 1 (Uβ ) Uβ Rm for E1 andE2 , then hα hβ is a local trivialization for E1 E2 .Then if E1 and E2 both have the same base space B , the restriction of the productE1 E2 over the diagonal B {(b, b) B B} is exactly E1 E2 .The direct sum of two trivial bundles is again a trivial bundle, clearly, but thedirect sum of nontrivial bundles can also be trivial. For example, the direct sum ofthe tangent and normal bundles to S n in Rn 1 is the trivial bundle S n Rn 1 sinceelements of the direct sum are triples (x, v, tx) S n Rn 1 Rn 1 with x v , andthe map (x, v, tx) (x, v tx) gives an isomorphism of the direct sum bundle withS n Rn 1 . So the tangent bundle to S n is stably trivial: it becomes trivial after takingthe direct sum with a trivial bundle.As another example, the direct sum E E of the canonical line bundle E RPnwith its orthogonal complement, defined in example (6) above, is isomorphic to thetrivial bundle RPn Rn 1 via the map (ℓ, v, w) (ℓ, v w) for v ℓ and w ℓ .Specializing to the case n 1 , the bundle E is isomorphic to E itself by the map thatrotates each vector in the plane by 90 degrees. We noted earlier that E is isomorphicto the Möbius bundle over S 1 RP1 , so it follows that the direct sum of the Möbiusbundle with itself is the trivial bundle. To see this geometrically, embed the Möbiusbundle in the product bundle S 1 R2 by taking the line in the fiber {θ} R2 that makesan angle of θ/2 with the x axis, and then the orthogonal lines in the fibers form asecond copy of the Möbius bundle, giving a decomposition of the product S 1 R2 asthe direct sum of two Möbius bundles.Example:The tangent bundle of real projective space. Starting with the isomor-nphism S Rn 1 T S n NS n , where NS n is the normal bundle of S n in Rn 1 ,suppose we factor out by the identifications (x, v) ( x, v) on both sides of thisisomorphism. Applied to T S n this identification yields T RPn , the tangent bundle toRPn . This is saying that a tangent vector to RPn is equivalent to a pair of antipodaltangent vectors to S n . A moment’s reflection shows this to be entirely reasonable,although a formal proof would require a significant digression on what precisely tangent vectors to a smooth manifold are, a digression we shall skip here. What we willshow is that even though the direct sum of T RPn with a trivial line bundle may notbe trivial as it is for a sphere, it does split in an interesting way as a direct sum ofnontrivial line bundles.

Basic Definitions and ConstructionsSection 1.111In the normal bundle NS n the identification (x, v) ( x, v) can be written as(x, tx) ( x, t( x)) . This identification yields the product bundle RPn R sincethe section x ( x, x)is well-defined in the quotient. Now let us consider

To motivate the definition of a vector bundle let us consider tangent vectors to the unit 2 sphere S2 in R3. At each point x S2 there is a tangent plane P x. This is a 2 dimensional vector space with the point xas its zero vector 0x. Vectors vx Px are thought of as arrows with their tail at x. If we regard a vector vxin Pxas a vector in R 3,