1. Introducing Vector Bundles Further Reading: [Hat, Chapter 1], [MS74, Chapter 2]

Home , Bundle (mathematics)

1. Introducing vector bundles Further reading: [Hat, Chapter 1], [MS74, Chapter 2]. Vector bundles (or at least, tangent bundles) appear quite naturally when one tries to work with differential manifolds, since in order to define derivatives we must define what a “tangent vector” to a manifold is. Given ann-manifoldM embedded inR N , we can define the tangent spaceTM ofM to be the set of points (x, v) withx M andv R N such thatv is tangent toM atx. The set of points withfirst coordinatex is∈ ann-dimensional∈ vector space. Since we generally think of manifolds as existing independently of the embeddings we want an independent definition of “a family of vector spaces over a space”; this is exactly the notion of a vector bundle. Definition 1.1 ([MS74, Chapter 2]).A vector bundle on a spaceB (generally called the base space) is a spaceE (generally called the total space) together with a mapp:E B and the structure of a vector space on eachfiberp 1(x) forx B, satisfying the extra condition: − ∈ (VB) There exists an integerk (the rank ofE) such that for every pointx B there exists a ∈ neighborhoodx U B and a homeomorphismϕ :U R k p 1(U). This homeomor- ∈ ⊆ x × − phism must satisfy the condition thatp ϕ x =π B (the projection ontoB) and that for k◦ 1 everyy U, the restrictionϕ x y Rk :R p− (y) is a linear homeomorphism. ∈ | × We often dropp from our notation. Whenn = 1 we will call such a vector bundle a line bundle. For any pointx B we callp 1(x) thefiber overx. ∈ − This definition thus says that a vector bundle is a continuous family of vector spaces overB. (In some formulations, the integerk only has to exist locally, so that ifB is not connected it can have different ranks over different connected components. We do not care about this in the current discussion, so we will stick to vector bundles of constant rank.) Some important examples of vector bundles: Example 1.2. The trivial bundle is the bundleB R k B where the map is just projection onto thefirst coordinate. × Example 1.3. As mentioned before, we can define the tangent bundleTM to a manifold embedded inR n by taking the set of points (x, v) withx M andv tangent toM atx. However, there is also an intrinsic definition. ∈ For any smoothn-manifoldM,TM is defined to be the set x M TxM. (Recall thatT xM, the tangent space atx, is defined as the vector space of derivations.)∈ This comes with a natural mapp:TM M which projects onto thefirst coordinte. To de�fine the topology onTM, let n (Uα,ϕ α:U α R ) be a smooth atlas onM. The local coordinates (x 1, . . . , xn) onU α give local { } 1 2n coordinates (∂/∂x 1,...,∂/∂x n) onT xM. Thus we can define a map ϕα:p − (Uα) R by

(a1x1 + +a nxn, b1∂/∂x 1 + +b n∂/∂x n) (a1, . . . , bn). ··· ··· � We define the topology onTM via these maps: a subsetA M is open exactly when ϕ α(A U α) is open inR 2n for allα. ⊆ ∩ Note that this is a linear mapfiberwise by definition. Also note that this construction� proves not only thatTM is a vector bundle overM but also that it is a 2n-manifold. Example 1.4. Suppose that we have a manifoldM embedded inR N . The normal bundle ofM is the set of points (x, v) withx M andv orthogonal toM atx. As above, this is naturally a bundle by projecting onto theM∈-coordinate. This does not exist independently of the embedding. Example 1.5. Suppose that we have a vector bundlep:E B. For everyα we have a homeomor- phismϕ :p 1(U ) U R n. Thus for everyα,β we have a composite homeomorphism α − α α × 1 (ϕα U U Rn )− ϕβ U U Rn n | α∩ β × 1 | α∩ β × n (Uα U β) R p− (Uα U β) (Uα U β) R ∩ × 2 ∩ ∩ × which is the identity after projection to thefirst coordinate and a linear homeomorphism on each fiber. Thus this map gives a smooth map g :U U GL (R). αβ α ∩ β n This satisfies:

(1)g αα is uniformly the identity. 1 (2)g αβ(x) =g βα(x)− for allx U α U β. (3)g (x)g (x)g (x) = 1 for∈ allx U∩ U U . αβ βγ γα ∈ α ∩ β ∩ γ Now suppose that we have a collection of suchg’s which satisfy these conditions. Then we can assemble a bundleE onB by taking E= U R n/ , α × ∼ α � where for anyx U α U β we say that (x, v) (x, g αβ(x) v). The conditions above exactly state that is an equivalence∈ ∩ relation, and the smoothness∼ conditions· on theϕ are enforced because ∼ α eachg αβ is smooth. For example, we can use this to construct the Mobius bundle. This is a bundle overS 1. We deﬁne 1 1 1 it using the atlasU 1 =S north andU 2 =S south . We deﬁne the functiong 12:S poles by letting it be 1 on the\{ part of}S 1 with negative\{ x-coordinate} and 1 on the part ofS\{1 with} positivex-coordinate.− This last example is an example of a procedure that is often done in mathematics. We take an object that we understand (Rn,C n, trivial bundle, ring) glue a whole bunch of them together in a nice way, and produce a new object ((real/complex) manifold, vector bundle, scheme) which is more general and interesting, while still retaining many of the properties of the simpler object. As always, now that we have a bunch of examples of vector bundles we want to know when two vector bundles are isomorphic.

Deﬁnition 1.6. Two vectors bundlesp 1:E 1 B andp 2:E 2 B are isomorphic if there exists a smooth homeomorphismf:E 1 E2 such that f E1 E2

p1 p2 B

commutes and such that its restriction to the preimage of any pointx B is a linear isomorphism. ∈ Note that if two vector bundles overB are isomorphic they must have the same dimension. Remark 1.7. Note that we have not used any property ofR when defining vector bundles. Thus we could define complex vector bundles in exactly the same way as we defined real vector bundles, but using the structure of complex vector spaces instead of real ones. We could go further. LetF be any space. Afiber bundle withfiberFp:E B is a map of topological spaces such that for every pointx B there exists a neighborhoodx U B and a ∈ ∈ ⊆ mapι:U F E such thatpι=1 U andι is a homeomorphism onto its image. Now× suppose that we want to remember some extra structure onF , such that it is a real/complex vector space, that it has an action by a group, or something else. We can write down exactly the same information, but impose the extra condition that for everyy U, the mapι y F :y 1 ∈ | × × F p− (y) is an isomorphism respecting this structure. Let’s look at some examples of when vector bundles are isomorphic. 3 Example 1.8. LetB=S 1 and considerTS 1. A point inTS 1 is a point inS 1 together with a vector tangent toS 1 at that point. In other words, we can write a point ofTS 1 as a point (cosθ, sinθ) and a vector ( λ sinθ,λ cosθ). This gives us a condinuous mapTS 1 R3 given by − ((cosθ, sinθ),( λ sinθ,λ cosθ)) (cosθ, sinθ,λ). − The image of this map is exactlyS 1 R, so we see thatTS 1 is isomorphic to the trivial bundle on S1. × What we did here is, in effect, define a nice basis for eachfiber in a way which is continuous overS 1. This let us “straighten out” the structure ofTS 1 and show that it is trivial. It is actually possible to do this in general. Definition 1.9.A section of a vector bundlep:E B is a maps:B E such thatp s=1 . ◦ B Every vector bundle has at least one section: the section which sends everything to 0. (This is called the zero section.) We can use sections to prove in very simple cases that vector bundles are not isomorphic. Note that every isomorphism of vector bundles must preserve the zero sections 0, so wheneverE 1 andE 2 are isomorphic vector bundles, the topological spacesE s (B) andE s (B) are homeomorphic. 1\ 0 1\ 0 Lemma 1.10.TS 1 is not isomorphic to the Mobius bundle. 1 1 1 1 1 Proof. ConsiderTS s0(S ) and Mobius s0(S ). SinceTS is trivial, this is isomorphic toS (R1 0 ), which is not\ connected. However,\ Mobius s (S1) is connected (as we all know from× \{ } \ 0 cutting a Mobius band down the middle). Thus these are not homeomorphic.� We can also use sections to try and “straighten out” a bundle to show that it is trivial. Proposition 1.11. Letp:E B be ann-dimensional vector bundle. There existn sections s1, . . . , sn:B E such that for allx B,s 1(x), . . . , sn(x) are linearly independent if and only if E is isomorphic to a trivial bundle. ∈ n Proof. IfE is isomorphic toB R then we can defines i:B E bys i(b) = (b, ei) for afixed n × basise 1, . . . , en ofR . Thens 1, . . . , sn are linearly independent at each point. Conversely, suppose that the sectionss 1, . . . , sn exist. Then we define a bundle mapf:E B Rn in the following manner. For a point (b, v) E, writev= n a s (b); this is always possible× ∈ i=1 i i sinces 1(b), . . . , sn(b) are linearly independent. Then map (b, v) to (b,(a 1, . . . , an)). This map is � continuous since thes i are, and afiberwise isomorphism by definition. Thus it is an isomorphism of vector bundles. � Here we implicitly used the following lemma:

Lemma 1.12 ([Hat, Lemma 1.1]). A continuous mapf:E 1 E2 between vector bundles is an isomorphism if it is a linear isomorphism on eachfiber. 1 Proof sketch. It suffices to check thath − is continuous. Since this is a local question it suffices to do it for trivial bundles, which we can do directly. � We get an interesting result as a corollary. This result will end up being very important later when we consider Euler classes. Corollary 1.13.TS 1 has no everywhere-nonzero section. Proof. Suppose thatTS 1 had an everywhere-nonzero section. Then by Proposition 1.11 it would be trivial. However, we just proved that it is not. � Remark 1.14. The question of the existence of nonzero sections has a natural higher-dimensional analog: given a vector bundleE B, how many everywhere-linearly-independent sections do there exist? 4 Wefinish up this section with an important result about homotopy groups which we will often use later: Theorem 1.15. LetE B be afiber bundle withfiberF . Then there is a long exact sequence of homotopy groups

πnF πnE πnB πn 1F π1F π1E πnB. ··· − ··· Note that for a vector bundle, since theﬁbers are contractible this says thatπ nE ∼= πnB. This makes sense since we can write down a general contraction sending each point (b, v) to (b, tv) which will be the identity att = 1 and the zero section att = 0.