FALL 2012 MATH 8230 (VECTOR BUNDLES) LECTURE …

1 FALL 2012 math 8230 ( vector BUNDLES) LECTURE NOTES. 1. DEFINITIONS: vector BUNDLES AND STRUCTURE GROUPS. A vector bundle over a topological space M (or with base space M ) is, essentially, family of vector spaces continuously parametrized by M . (I'm using the letter M to denote the base space of the vector bundle as a concession to the fact that in most of the applications we'll be interested in the base space will be a smooth manifold; however for basic definitions and results it there is no need to restrict to this case.) One way of making this precise is as follows: Definition A (real, rank-k) vector bundle over a topological space M is a continuous map : E M where E is a topological space such that, for all m M : (i) the fiber Em := 1 ({m}) is equipped with the structure of a vector space over R. (ii) There is an neighborhood U M of m and a local trivialization : 1 (U) U Rk which, for each x U, maps the fiber E x to {x} Rk by a linear isomorphism.

2 Remark In practice one says things like, Let E be a vector bundle over M instead of Let : E M be a vector bundle over M . But one should keep in mind that, even so, the vector bundle E refers not just to the topological space but also to the projection map and the local trivializations. Example There is one very obvious example (for any M and any k): Simply take E = M Rk and let be the projection. For item (ii) in Definition one can just set U = M (for any m M ), so that 1 (M ) = M Rk , and let : 1 (M ) M Rk be the identity map. Appropriately, this bundle is called the trivial bundle of rank k over M .. Definition Let E : E M and F : F M be two vector bundles over M . An isomorphism from E to F is a homeomorphism : E F such that for each m . M restricts to Em as a linear map (and hence a linear isomorphism, since is a homeomorphism) from Em to Fm.

3 The vector bundle : E M is called trivial it is isomorphic to the trivial rank-k bundle over M for some k. Exercise Suppose that E : E M and F : F M are two vector bundles and : E F. is a continuous map that maps each fiber Em to Fm by a linear isomorphism. Prove that is an isomorphism. (The nontrivial part of this is showing that 1 is continuous. Hint: By restricting to appropriate local trivializations, you can reduce to the case where is a continuous map of the trivial bundle U Rk to itself which restricts to each {x} Rk as a linear automorphism, and you should be able to show then that the fact that is continuous implies that 1 is also continuous.). At first glance one might not be sure whether there exist any nontrivial bundles. A first example of one is given by the following. Example Let E denote the topological space which can be expressed as a quotient in either of the following two equivalent ways: [0, 1] R R R.

4 E= = . (1, t) (0, t) (s + 1, t) (s, t). 1. 2 FALL 2012 math 8230 ( vector BUNDLES) LECTURE NOTES. You should recognize this topological space as an (open, with no boundary) M bius strip. Where we identify the circle S 1 with R/Z, there is a continuous map : E S 1 defined by ([s, t]) = [s]. It is not hard to see that this is a rank-1 vector bundle over S 1 . Indeed, using the second expression for E above, for any [s0 ] S 1 (where s0 R), we have 1 ([s0 ]) =. {[s0 , t]|t R}, which is naturally identified with R and so inherits the vector space structure of R. To construct a local trivialization around s0 , let U = {[s]|s0 1/2 < s < s0 + 1/2}, and then define : 1 (U) U R by ([(s, t)]) = ([s], t) for s0 1/2 < s < s0 + 1/2. This is easily seen to be a homeomorphism (it is well-defined and injective because s is taken from an open interval of length only one) which satisfies the requirements of a local trivialization.

5 There are various ways of seeing that this M bius bundle is not trivial. For one thing, if it were trivial then the open M bius strip would be homeomorphic to the open cylinder, which it is not, though showing this is not exactly straightforward (but perhaps already familiar to you). There is also a somewhat easier argument (which is best phrased in terms of the language of the upcoming Definition ) in which one starts from a system of local trivializations for the M bius bundle and derives a contradiction from the way that a hypothetical isomorphism to the trivial bundle would have to act with respect to the transition functions for the local trivializations. Example As one learns in an introductory smooth manifolds course, for any smooth man- ifold M there is a naturally associated vector bundle with rank equal to dim M , namely the tangent bundle : T M M , whose fiber at a point m M is the tangent space Tm M.

6 (Thus, loosely speaking, Tm M consists of all possible velocity vectors of curves passing through m. For a review of this see any introductory text on smooth manifolds, [Lee, Chapter 3], or [U1, Section ]). Exercise Show that T S 1 S 1 is trivial, but that the M bius bundle over S 1 is not. Remark In fact, every rank-1 vector bundle over S 1 is either trivial or isomorphic to the M bius bundle , as we should be able to prove within the next few weeks. In general, a smooth manifold M such that T M is trivial is called parallelizable. The example of S 1 should not be considered representative, as most manifolds are not parallelizable the only spheres with this property are S 0 , S 1 , S 3 , and S 7 , as was proven by Kervaire and Bott- Milnor in the late 1950s. I'd now like to give an equivalent definition of a vector bundle , which will have some useful generalizations when we wish to speak of vector bundles carrying some additional structure.

7 Definition A (real, rank-k) vector bundle over a topological space M is a continuous map : E M where E is a topological space such that there is an open cover {U } A of M and a collection of local trivializations : 1 (U ) U Rk such that (i) Each is a homeomorphism, which for every x U maps 1 ({x}) to {x} Rk . (ii) For each , A, the map 1 k k : (U U ) R (U U ) R has the form 1. (x, v) = (x, g (x)v). where for every x U U , g (x): Rk Rk is a linear map. This definition is clearly closely related to Definition ; in particular it should be clear that a vector bundle in the sense of Definition gives a vector bundle in the sense of Definition (just take the open cover to consist of all the sets U in Definition obtained for various values of m M ). Conversely, given a vector bundle in the sense of Definition and m M , one can choose so that x U and then is an obvious candidate for the local trivialization around m FALL 2012 math 8230 ( vector BUNDLES) LECTURE NOTES 3.

8 Required in Definition (ii). But perhaps at this point one notices a difference between the two definitions, namely that in Definition I assumed that the fibers Em were all vector spaces and that the local trivializations were fiberwise linear, whereas I didn't say anything about a vector space structure on the fibers in Definition This discrepancy is resolved by the following proposition, whose proof generalizes in important ways and so should be paid close attention. Proposition Given a vector bundle in the sense of Definition , for each m M there is a unique vector space structure on the fiber Em = 1 ({m}) such that for every A with m U . the local trivialization restricts as a linear isomorphism | Em : Em {m} Rk . Proof. For any v, w Em and c R, our job is to define v + w, cv Em in such a way that the vector space axioms are satisfied and such that each of the restrict to Em as a linear isomorphism to Rk.

9 If we fix an such that m U , there is one and only one way of trying to do this: If v Em we have (v) = (m, v ) for some unique v Rk , and if is to be a linear isomorphism we must define addition + and scalar multiplication by v + w = 1. (m, v + w ) c v = 1. (m, cv ). These are well-defined algebraic operations on Em which inherit the vector space axioms from {m} Rk because we have defined the operations by forcing the bijection | Em : Em {m} Rk to intertwine them with the standard operations on {m} Rk . What remains to be checked is that these operations are independent of the choice of with m U ; here we use a fact that we haven't used before, namely that the transition functions . g (m) from Definition (ii) are linear. So choose , A such that m U U ; we are to show that for any v, w Em and c R we have v + w = v + w and c v = c v. Note first that, where as before v Rk is defined by the property that (v) = (m, v ) and likewise for v , we have.

10 (m, v ) = (v) = ( 1. ) (v). = ( 1. )(x, v ) = (x, g (m)v ). Thus v = g (m)v , and likewise w = g (m)w . Therefore, v + w = 1 1. (m, v + w ) = (m, g (m)v + g (m)w ).. = 1 1. (m, g (m)(v + w )) = 1 1. (m, v + w ) = (m, v + w ). = v + w, where in the third equality we have used the fact that g (m) is linear. In exactly the same way one shows that c v = c v. Thus the operations + and are in fact independent of the local trivialization used to define them, and so give well-defined vector space operations on Em satisfying the required properties.. This result gets at an important idea which I want to introduce now we'll come back to it in more detail later. Note first of all that one could imagine a version of Definition which is exactly the same except that in the final line one eliminates the requirement that the g (m). are linear and just requires them to be homeomorphisms from Rk to itself.