How To Assemble The Tangent Spaces Of A Manfold Nto A Coherent Whole

CHAPTER 7 VECTOR BUNDLES We next begn addressng the queston: how do we assemble the tangent spaces at varous ponts of a manfold nto a coherent whole? In order to gude the decson, consder the case of U R n an open subset. We reflect on two aspects. The frst aspect s that the total dervatve of a C functon should change n a C manner from pont to pont. Consder the C map f : U R m. Each pont x U gves a lnear map f x = Df(x) : R n R m the total dervatve whch s represented by an m n matrx, the Jacoban matrx of Df(x). The Jacoban matrx s a matrx of C functons n x. Whle for each x U there s a lnear map Df(x) : T U x = R n R m = T R m f(x) these ft together to gve a C map on the product U R n R m R m (x, v) (f(x), Df(x)(v)). The second aspect s that we wsh to defne vector felds. A vector feld s a choce of tangent vector at each pont. For an open subset U of R n, a vector feld s just gven by a functon g : U R n (as the reader probably learned n Advanced Calculus). In order to keep track of the tal, we wrte the vector feld as V : U U R n x (x, g(x)). Any C functon g gves a vector feld. The complcaton on a manfold M s that the vector wth tal at x M must be n the vector space T M x and these vector spaces change wth x. In ths chapter, we study the requred concepts to assemble the tangent spaces of a manfold nto a coherent whole and construct the tangent bundle. The tangent bundle s an example of an object called a vector bundle. Defnton 7.***. Suppose M n s a manfold. A real vector bundle over M conssts of a topologcal space E, a contnuous map π : E M and a real vector space V (called the fber) such that for each m M, π (m) s a vector space somorphc to V, and there exsts an open neghborhood U of m, and a homeomorphsm copyrght c 2002 µ U : π (U) U V Typeset by AMS-TEX

2 CHAPTER 7 VECTOR BUNDLES such that µ U (m, ) : {m} V π (m) s a lnear somorphsm. The bundle s smooth f E s a smooth manfold, π s smooth, and µ U s a dffeomorphsm. In these notes, all vector bundles wll be smooth. We may denote a vector bundle by π : E M (and suppress the vector space) or as E. If the dmenson of the vector space s m then the bundle s often called an m-plane bundle. A -plane bundle s also called a lne bundle. A bundle over a manfold s trval f t s smply the Cartesan product of the manfold and a vector space. The neghborhoods U over whch the vector bundle looks lke a product are called trvalzng neghborhoods. Note that µ W µ U : {m} V {m} V s a lnear somorphsm. Denote ths map h WU (m). Defnton 7.2***. If µ U : π (U) U V and µ W : π (W) W V are trval neghborhoods of a vector bundle then µ W µ U : (W U) V (W U) V (x, v) (x, h WU (x)v) where h WU : W U GL(V ). The h WU are assocated to each par of trval neghborhoods (U, µ U ) and (W, µ W ). They are called transton functons. Theorem 7.3***. Every smooth vector bundle has smooth transton functons,.e., h WU : W U GL(V ) s smooth. Proof. The map µ W µ U defnes h WU so the ssue s to see that h WU s smooth. Let h WU (x) be the matrx (h j (x)) j n a fxed bass for V. Then, µ W µ U (x, (r,, r n )) = (x, ( j h j(x)r j,, j h nj(x)r j )). To see that each h j (x) s smooth let r vary over e for =, n. Snce µ W µ U s smooth, so are ts coordnate functons. Example 7.4***. Lne Bundles Over S. We take the crcle to be S = {e θ θ R} the unt crcle n the complex plane {(cos θ, sn θ) θ R 2 }. One lne bundle over the crcle s ɛ S, the trval bundle π ɛ : S R S by π ɛ ((e θ, r)) = e θ. For the trvalzaton neghborhoods, only one s needed: take U = S. There s another, more nterestng lne bundle over S. Let E = {(e θ, re θ 2 ) r, θ R} and π γ : E S by π γ ((e θ, re θ 2 )) = e θ. Denote ths bundle γs Notce that π γ (e θ ) s a real lne n the complex plane. Two values of θ that dffer by 2π determne the same pont, so θ 2 s not well-defned. Nevertheless, the lne n the complex plane through e θ 2 s well defned snce e 2π 2 =. We now construct the trvalzng neghborhoods. Let U = S \ {} = {(e θ, re θ 2 ) θ (0, 2π)} and W = S \ { } = {(e θ, re θ 2 ) θ (π, 3π)}. Now, () µ U : πγ (U) U R (e θ, re θ 2 ) (e θ, r).

CHAPTER 7 VECTOR BUNDLES 3 Ths map s well defned snce θ (0, 2π), a restrcted doman whch allows us to determne θ from e θ 2. We smlarly defne (2) µ W : πγ (U) U R (e θ, re θ 2 ) (e θ, r) for θ (π, 3π). We next check the compatblty condton. The set U W s S \{, } = {(e θ, re θ 2 ) θ (0, π) (π, 2π)}. Suppose θ (0, π) then µ W µ U ((eθ, r)) = µ W ((e θ, re θ 2 )) = µ W ((e (θ+2π), re θ+2π 2 )) = (e (θ+2π), r) = (e θ, r) Notce that we had to change the expresson for the second coordnate because formulas () and (2) requre dfferent domans. We have that h UW (e θ )(r) = r for θ (0, π). Now, suppose θ (π, 2π), then µ W µ U ((eθ, r)) = µ W ((e θ, re θ 2 )) = (e θ, r) We have that h UW (e θ )(r) = r for θ (π, 2π). Therefore the transton functon h UW : U W Gl(, R) s { f Im(x) > 0 h UW (x) =. f Im(x) < 0 Example 7.5***. The Tautologcal Lne Bundle Over RP n Defne a Z 2 acton on S n R by ( ) (x, r) = ( x, r). We show that ths acton satsfes the hypotheses of Theorem 2.23***. Suppose (x, r) S n R. Take U an open neghborhood of x n S n that s entrely n one hemsphere. Then t follows that U U =, and U R and U R are dsjont neghborhoods of (x, r) and ( ) (x, r). Let E = S n R/Z 2. By 2.23***, E s a smooth manfold and the quotent map q : S n R E s a local dffeomorphsm. Let π E : E RP n by π E (x, r) = [(x, r)]. The followng dagram s a commutatve dagram of smooth maps, S n R π S n q E π E q RP n where π(x, r) = x and q s the quotent map from Example 2.25***, RP n. Let U be an open set n S n that s entrely n one hemsphere so that U U =. Then q U R :

4 CHAPTER 7 VECTOR BUNDLES U R π E (q(u)) s a dffeomorphsm and lnear on each fber. If V s another such open subset of S n then ( q V R ) q U R : U (V V ) R U (V V ) R (x, r) (x, h(x)r) where { f x U V h(x) = f x U V Hence, π E : E RP n s a vector bundle. Proposton 7.6***. Transton functons satsfy the followng property: h WU (x) = h WO (x) h OU (x) for x W O U. Proof. Ths property follows from the defnton of transton functons and s the equaton of the last coordnate of the equaton below. (µ W µ O ) (µ O µ U ) = µ W µ U The next theorem shows that a choce of transton functons consstent wth the propertes of the last proposton wll determne the vector bundle. A bundle can be defned by the glung (transton) functons. Frst a lemma. Lemma 7.7***. Suppose X s a set and {U I} s a collecton of subsets. If h j : U U j GL(n, R) for all (, j) I I satsfes h j (x) = h k (x) h kj (x) for x U U j U k, then they also satsfy () h (x) = I V for x U (2) h j (x) = (h j (x)) for x U U j Proof. Snce h (x) = h (x) h (x) and h (x) has an nverse, multply both sdes by h (x) to get (). Snce h j (x)h j (x) = h (x) = I V by the hypothess and (), h j (x) = h j (x). Theorem 7.8***. Suppose M s a manfold and {(U, ψ ) I} s a countable atlas for M. Suppose V = R m s a vector space and for all (, j) I I there s a C functon f j : U U j GL(V ) such that f kj (x) f j (x) = f k (x) for all x U U j U k. On {(u, v, ) I, u U, v V } let (u, v, ) (u, v, ) f and only f u = u, f (u)(v) = v. Then s an equvalence relaton. Furthermore, f E = {(u, v, ) I, u U, v V } and π([u, v, ]) = u, then π : E M s a smooth vector bundle wth fber V. Proof. We frst show that s an equvalence relaton, The functons f j also satsfy the hypotheses of Lemma 7.7***. The relaton s an equvalence relaton as reflexvty, symmetry and transtvty are guaranteed by condtons and 2 of Lemma 7.7***, and condton 3 s the hypothess on the f j s. We must verfy the varous requrements of the defnton of a vector bundle for π : E M. The longest part s done frst, showng that E s a manfold.

CHAPTER 7 VECTOR BUNDLES 5 We frst ntroduce the map µ. Let µ : π (U ) U V by µ ([u, v, ]) = (u, v). Ths map s a bjecton: f u U then the class [u, v, ] has (u, v, ) as the unque representatve wth the last coordnate by the defnton of as f (u) = I V. We verfy that E s a smooth manfold by usng Theorem 2.5*** Let φ = (ψ I V ) µ so φ : π (U ) ψ (U ) V R n V = R n+m. We clam that E s a smooth manfold wth atlas A = {(π (U ), φ ) I}. We verfy the condtons of Theorem 2.5*** wth X = E and A. The frst two condtons are true, π (U ) E and I π (U ) = E by constructon. Next, the map φ s a bjecton snce both µ : π (U ) U V and ψ I V : U V ψ (U ) V are bjectons. The fourth condton s also mmedate: φ : π (U ) ψ (U ) V R n V and φ : π (U ) π (U j ) ψ (U U j ) V R n V are both open. The last condton s that φ j φ : φ (π (U ) π (U j )) φ j (π (U ) π (U j )) s smooth. We now show that the map s smooth. Suppose x ψ (U U j ) and v V. Then, φ j φ (x, v) = (ψ j I V ) µ j µ (ψ I V )(x, v) = (ψ j I V ) µ j µ (ψ (x), v) = (ψ j I V ) µ j ([ψ (x), v, ]) = ψ j I V µ j ([ψ (x), f j (ψ (x))(v), j]) by the equvalence relaton = ψ j I V (ψ (x), f j (ψ (x))(v)) = (ψ j ψ (x), f j (ψ (x))(v)) We have shown that φ j φ (x, v) = (ψ j ψ (x), f j (ψ (x))(v)). The frst coordnate s smooth snce M s a manfold. The second coordnate map s smooth snce t s a composton of the followng smooth maps: ψ(u U j ) R m f (U U j ) R m f 2 GL(n, R) R m f 3 R m where f (y, v) = (ψ (y), v) s smooth snce ψ s smooth, f 2 (x, v) = (f j (x), v) s smooth by Theorem 7.3***, and f 3 s smooth snce t s evaluaton of a lnear functon at a vector,.e., matrx multplcaton. The hypotheses of Theorem 2.5*** are verfed. We show n the next two paragraphs that E s second countable and Hausdorff, and hence a manfold. Snce I s countable, {π (U ) I} s countable and so E s a second countable space. Suppose [u, v, ], [u, v, j] E. If u u, then there are open sets O and O 2 that separate u and u n M. Hence π (O ) and π (O ) separate [u, v, ] and [u, v, j] n E. If u = u, then [u, v, ], [u, v, j] π (U ) whch s Hausdorff. Therefore E s Hausdorff and E s a manfold. We now show that π : E M s a vector bundle. The vector space structure on π (m) s defned by the structure on V: a[m, v, ] + b[m, v, ] = [m, av + bv, ]. Ths s well defned snce f [m, v, ] = [m, w, j] and [m, v, ] = [m, w, j] then f j (m)(w) = v and

6 CHAPTER 7 VECTOR BUNDLES f j (m)(w ) = v so a[m, w, j] + b[m, w, j] = a[m, f j (m)(w), ] + b[m, f j (m)(w ), ] = [m, af j (m)(w) + bf j (m)(w ), ] = [m, f j (m)(aw + bw ), ] = [m, av + bv, ] The map µ : π (U ) U V s a dffeomorphsm. Ths was shown above as these maps are part of the manfold structure for E (Theorem 2.2***). µ (m, ) : V π (m) s an somorphsm as µ (m, )(v) = µ (m, v) = [m, v, ]. π s smooth snce f m s n the coordnate neghborhood gven by (U, ψ ), then [m, v, ] s n the coordnate neghborhood gven by (π (U ), φ ) and ψ π φ (x, v) = x for (x, v) φ (π (U )) R n V. π Defnton 7.9***. A bundle map between two vector bundles E π M and E 2 2 M2 s a par of smooth maps f : E E 2 and g : M M 2 such that π 2 f = g π and f x : π (x) π 2 (g(x)) s lnear. π Defnton 7.0***. Suppose that E π M and 2 E2 M are bundles over a manfold M. A bundle equvalence between these bundles s a a bundle map (f, I M ) over the dentty s an somorphsm on each fber. Please note the result Exercse 2***: f E an E 2 are vector bundles over the manfold M and f E s bundle equvalent to E 2 by a bundle equvalence f, then E 2 s bundle equvalent to E by f whch s also a bundle equvalence. Defnton 7.***. A secton of a vector bundle, E π M s a smooth map s : M E such that π s = I M. Notaton 7.2***. Suppose E s E π M a vector bundle wth fber V. Let Γ(E) or Γ(E) denote the sectons of the vector bundle E. Example 7.3***. The sectons of the trval bundle. The sectons of ɛ n M are smooth maps s : M M Rn wth π(s(x)) = x for all x M. Hence, the frst coordnate of s(x) s x and s(x) = (x, f(x)) for any smooth functon f : M R n. Therefore, C (M, R n ) = Γ(ɛ n M ) by the correspondence: f f C (M, R n ) then s Γ(ɛ n M ) wth s(x) = (x, f(x)). Example 7.4***. The sectons of lne bundles over S We frst note that the functon p : R S by p(θ) = (cos θ, sn θ) R 2 or e θ C s a local dffeomorphsm. An nverse of p : (θ 0 π 2, θ 0 + π 2 ) p(θ 0 π 2, θ 0 + π 2 ) s p (x, y) = arcsn(y cos θ 0 x sn θ 0 ) + θ 0. Of course, p sn t globally well defned as p (cos θ, sn θ) = {θ + 2kπ k Z} In Example 7.3*** t was shown that the sectons of ɛ S were gven by the smooth functons f : S R. We also note that these sectons can be gven by the functons {g : R R g(θ + 2π) = g(θ)},.e., perodc functons of the angle rather than of a pont

CHAPTER 7 VECTOR BUNDLES 7 on the crcle. Gven a secton s Γ(ɛ S ) wth s(cos θ, sn θ) = ((cos θ, sn θ), f(cos θ, sn θ), the functon g s g(θ) = f(cos θ, sn θ). Conversely, gven a perodc functon g let the secton s be s(cos θ, sn θ) = ((cos θ, sn θ), g(θ)). The secton s well-defned snce g s perodc and t s smooth snce p has a smooth local nverse. We now descrbe the sectons of γs. The bundle was descrbed n Example 7.4***. We show that Γ(γS ) correspond to the functons {g : R R g(θ + 2π) = g(θ)}. Gven a functon g take s to be the secton s(e θ ) = (e θ, g(θ)e θ 2 ). Conversely, gven a secton s, t can be wrtten as s(e θ ) = (e θ, h(e θ )), where by the defnton of γs, h(e θ ) s a real multple of e θ 2. Then let g(θ) = h(e θ )e θ 2. Remark 7.5***. A secton of a vector bundle s a way of choosng an element n each fber that vares n a smooth manner. One speaks of an element of a vector space and the approprate generalzaton to a vector bundle usually s a secton of a vector bundle. Example 7.6***. The zero secton Every vector space has a dstngushed element: zero, the addtve dentty. If E π M s a vector bundle, let 0 x π (x) be the zero. Let z : M E be defned by z(x) = 0 x for all x M. The secton z s called the zero secton. We check that ths map s a smooth secton. If x M take an open neghborhood that s both part of a coordnate neghborhood (U, φ) and a trvalzng neghborhood for E, µ : U U R m. The map z s smooth on U f and only f Z = (φ I V ) µ z φ : φ(u) φ(u) R m s smooth. The map Z s Z(x) = (x, 0) whch s a smooth map of an open subset of R n to R n R m. Therefore, z s a smooth secton. Proposton 7.7***. A bundle π : E M wth an n-dmensonal fber s a trval bundle f and only f t has n sectons {s,, s n } such that s (x),, s n (x) are a bass for π (x) for each x M. Proof. If f : M R n E s a bundle equvalence over M, then let s be defned by s (x) = f(x, e ). The sectons {s,, s n } are the requred set of sectons. Note that s (x) = (x, e ) are the requred sectons for the trval bundle. If π : E M s a bundle wth sectons {s,, s n } such that s (x),, s n (x) are a bass for π (x) for each x M, then f : M R n E defned by f(x, (a,, a n )) = n = a s (x) s a smooth map and somorphsm on each fber. Therefore f s a bundle equvalence. Theorem 7.8***. If N s an n-manfold, then there s a vector bundle π N : T N N such that π N (x) = T M x. Furthermore, f M s an m-manfold, and f : N M s a smooth map, then there s a bundle map f : T N T M defne by f T Nx = f x for each x N,.e., the bundle map on T N x s the dervatve map. The frst part of the theorem asserts the exstence of a vector bundle. The substance, the justfcaton to call the bundle the tangent bundle s n the second paragraph. The dervatve map at each pont combnes to form a smooth map. It ndcates that the

8 CHAPTER 7 VECTOR BUNDLES constructon of the tangent bundle s the phlosophcally correct manner to combne the tangent spaces nto a coherent whole. Proof of Theorem 7.8***. For U an open set n R n, the tangent bundle s a smple cross n product U R n. A bass n the fber over a pont p s a p x = where x represents p the equvalence class of the path t te + p. Observe that T U = U R n agrees wth the usual notons from Calculus as dscussed n the begnnng of ths chapter. We now construct the tangent bundle for a smooth n-manfold M. Let {(U, φ ) I} be a countable atlas, and let f j : U U j GL(R n ) be defned by f j (u) = (φ j φ ) φ (u) = D(φ j φ )(φ (u)). Frst note that snce φ j φ s a C functon on an open set of R n nto R n, each matrx entry of the Jacoban of φ j φ s a C functon. Therefore, D(φ j φ )(φ (u)) = f j (u) s a smooth functon. We next show that these maps satsfy the condtons of Theorem 7.8*** and so defne a vector bundle, T M M. The above functons f j (u) = (φ j φ ) φ (u) satsfy the condton for transton functons. Suppose u U U j U k, then (φ j φ k ) φ k (u)(φ k φ ) φ (u) = (φ j φ ) φ (u) by the chan rule. Hence f jk (u) f k (u) = f j (u). The hyposthess of Theorem 7.8*** are satsfed and so gve a well-defned bundle, the tangent bundle of M. The next task s to show that f f : N M s a smooth map, then there s a bundle map f : T N T M defne by f T Nx = f x for each x N. The map f s well defned and lnear on each tangent space T N x by the gven formula. It must be shown that t s a smooth map from T N to T M. Let π N : T N N and π M : T M M be the projectons for the tangent bundles. Take any pont n T N. It s n the fber over some pont n N, say x N. Let (W, ψ) be a chart on M wth f(x) W. By constructon of the tangent bundle, (W ) s open n T M and π M Ψ : π M (W ) ψ(w ) Rm v (ψ(π M (v)), ψ πm (v)(v)) s a coordnate chart. Let (U, φ) be a chart on N wth x U and f(u) W. Agan, by constructon of the tangent bundle, π N (U) s open n T N and Φ : π N (U) φ(u) Rn v (φ(π N (v)), φ πn (v)(v)) s a coordnate chart. We check smoothness by usng Proposton 2.8***. Suppose (x, v) Φ(π (U) R n R n. Then compute Ψ f Φ ((x, v)) = (ψ f φ (x), ψ f(φ (x) f φ (x) φ x (v)) = (ψ f φ (x), (ψ f φ ) x (v)).

CHAPTER 7 VECTOR BUNDLES 9 The last lne follows by the chan rule, Theorem 5.3*** part 2. Ths map s C n (x, v) snce t the dervatve map for ψ f φ a C functon between open subsets of R n and R m. Defnton 7.9***. A secton of the tangent bundle of a manfold M s called a vector feld on M. Example 7.20***. The tangent bundle to the 2-sphere T S 2. Let S 2 = {(x, y, z) R 3 x 2 + y 2 + z 2 = }. Recall from Calculus that the tangent plane to S 2 translated to (0, 0, 0) s P (x,y,z) = {(v, v 2, v 3 ) R 3 (x, y, z) (v, v 2, v 3 ) = 0}. Let E = {( x, v) R 3 R 3 x = and v P x } and π : E S 2 by π(( x, v)) = x s a vector bundle whch s bundle equvalent to the tangent bundle. We frst examne the tangent bundle of S 2 usng the atlas constructed n Example 2.9A*** for S 2. Let U = S 2 \ {(0, 0, )} and U 2 = S 2 \ {(0, 0, )}. Let φ : U R 2 for = and 2 be φ (x, y, z) = ( 2x z, 2y z ) and φ 2(x, y, z) = ( 2x +z, 2y +z ). Then φ (x, y) = 4x ( x 2 +y 2 +4, 4y x 2 +y 2 +4, x2 +y 2 4 x 2 +y 2 +4 ) and φ 2 φ (x, y) = ( 4x s then g 2 : U U 2 Gl(2, R) defned by x 2 +y 2, 4y x 2 +y 2 ). The transton functon (Eq ***) g 2 ( x) = D(φ 2 φ )(φ ( x)) and D(φ 2 φ ( 4(y 2 x 2 ) )((x, y)) = 8xy (x 2 +y 2 ) 2 (x 2 +y 2 ) 2 8xy 4(x 2 y 2 ) (x 2 +y 2 ) 2 (x 2 +y 2 ) 2 It s C snce φ and D(φ 2 φ ) are C. We now turn to π : E S 2. For each x S 2, we note that π ( x) = P x s a vector space, a two dmensonal subspace of R 3. We defne homeomorphsms h : U R 2 π (U ) for = and 2. These maps wll be used to show E s a 4-manfold. They wll commute wth the projectons and be lnear on each fber. Ther nverses wll be the trvalzaton maps for the bundle. The compatblty condton for the charts s the smoothness of the transton functons. Let h : U R n π (U ) ( x, v) ( x, Dφ (φ ( x)) v) and h 2 : U 2 R n π (U 2 ) ( x, v) ( x, Dφ 2 (φ 2( x)) v) The nverses are ) µ : π (U ) U R n ( x, v) ( x, Dφ ( x) v) and µ 2 : π (U 2 ) U 2 R n ( x, v) ( x, Dφ 2 ( x) v) These maps are homeomorphsms as Dφ (φ ( x))dφ ( x) = I P x and Dφ ( x)dφ (φ ( x)) = I R 2

0 CHAPTER 7 VECTOR BUNDLES by the chan rule. Note that the transton functon for the bundle E s g 2 ( x) = (Dφ 2 ( x)dφ (φ ( x)) = D(φ 2 φ )(φ ( x)) whch s the same as we computed for the tangent bundle. We further pursue Example 7.20*** and prove that T S 2 s not trval. We actually show more. We prove that S 2 doesn t have any nowhere zero secton. The proof of ths theorem requres that the reader knows some of the bascs of homotopy. We wll requre () The defnton of a homotopy class. (2) The fact that the map f : S R 2 \ {(0, 0)} defned by f(e θ ) = e 2θ or f((cos θ, sn θ)) = (cos 2θ, sn 2θ) s not homotopc to a constant map. We state the specfc elements we wll use n the followng lemma. Lemma 7.2***. Suppose g : S GL(2, R) s a contnuous functon. For any contnuous functon f : S R 2 \ {(0, 0)}, let G(f) : S R 2 \ {(0, 0)} by G(f)(p) = g(p)(f(p)). We then have () If f : S R 2 \ {(0, 0)} be any contnuous map, then the map f s homotopc to f. (2) If f, h : S R 2 \ {(0, 0)} and f s homotopc to h, then G(f) s homotopc to G(h). (3) Let S S 2 be the equator and g the transton functon that from T S 2 defned n (Eq I***) of Example 7.20***. Furthermore, let ι : S R 2 \ {(0, 0)} be the constant map ι(p) = (, 0). Then the map G(ι) s not homotopc to a constant map. Proof. To show tem we gve a homotopy from f to f. Let H : S [0, ] R 2 \{(0, 0)} by ( ) cos πt sn πt H(p, t) = f(p). sn πt cos πt Item 2 s also easy to see by the approprate homotopy. Suppose that H : S [0, ] R 2 \ {(0, 0) s the homotopy between f and h, then (p, t) g(p)(h(p, t)) s the homotopy between G(f) and G(h). To show tem 3, we frst wrte p = (x, y, 0) for p on the equator of S 2 and recall that φ (p) = (2x, 2y) for the map φ, stereographc projecton (see Example 7.20***). Snce the transton functon g(p) s the matrx gven n (Eq I***) of Example 7.20***, G(ι)(p) = D(φ 2 φ )(φ (p))(ι(p)) = ( 4((2y) 2 (2x) 2 ) 8(2x2y) ((2x) 2 +(2y) 2 ) 2 ((2x) 2 +(2y) 2 ) 2 8(2x2y) 4((2x) 2 (2y) 2 ) ((2x) 2 +(2y) 2 ) 2 ((2x) 2 +(2y) 2 ) 2 ( ) y = 2 x 2 2xy ) ( ) 0 The last lne follows snce x 2 + y 2 =. Wrtng x = cos θ and y = sn θ, G(ι)(cos θ, sn θ) = (cos 2θ, sn 2θ) usng the double angle denttes. By tem, G(ι) s homotopc to the map (cos θ, sn θ) (cos 2θ, sn 2θ), whch s not homotopc to a constant map.

CHAPTER 7 VECTOR BUNDLES Theorem 7.22***. There s no nonzero vector feld on S 2. As a consequence of ths theorem and Proposton 7.7***, T S 2 s not equvalent to a trval vector bundle. Proof. We use proof by contradton. Suppose V s a nowhere zero secton of T S 2,.e., V s a nowhere zero vector feld. We examne the two maps that characterze the vector feld n the trvalzng neghborhoods for T S 2 as gven n Example 7.20***. We use the maps and notaton from Example 7.20***. For each p S 2 \ {(0, 0, )}, φ (V p ) = (φ (p), f(p)) R 2 R 2 \ {(0, 0)} and for each p S 2 \ {(0, 0, )}, φ 2 (V p ) = (φ 2 (p), h(p)) R 2 R 2 \ {(0, 0)}. The mages of the maps f and g avod (0, 0) precsely because the vector feld s nowhere zero. It s easy to see that f and g are homotopc to constant maps. The null homotopy for f s H : S [0, ] R 2 \ {(0, 0)} by H(p, t) = f(φ (tφ (p))) and for h s K : S [0, ] R 2 \ {(0, 0)} by K(p, t) = h(φ (tφ (p))). The maps f and h are also related by the transton functon on the sphere by h(p) = g 2 (p)f(p). In other words, f f s null homotopc then G(f) s also null homotopc. Ths concluson contradcts tem 3 of the lemma. Therefore T S 2 doesn t have a nonzero secton. Dualty and Dual Bundles. We revew the noton of dualty from lnear algebra. Suppose V s a vector space. Let V be the dual vector space Hom(V, R), the vector space of lnear functons from V to R. The relatonshp between V and V s usually expressed n the evalutaton parng: V V R (f, v) f(v) We gve the basc propertes of V n the next two theorems. The frst theorem s usually covered n a lnear algebra course. The second sn t dffcult, but has a flavor dfferent from a frst course n lnear algebra. Theorem 7.23. For any fnte dmensonal vector space V, let V be the dual space of lnear functonals. () V s a vector space of the same dmenson as V. (2) Suppose that e,, e n s a bass for V. Let e ( n j= a je j ) = a. Then e,, e n s a bass for V called the dual bass. (3) Let V and W be vector spaces and f : V W be a lnear map. Let f : W V by f (g) = g f. Proof. We can add lnear functonals and multply be a constant. If fact V = Hom(V, R) s somorphc to the vector space of n by matrces. The somorphsm s gven by a choce of bass. If the choce of bass s e,, e n, then e s the matrx wth a n the th row and zeros elsewhere. Ths shows tems and 2. Item 3 s a short computaton. Suppose g, g 2 V. Then, f (ag + bg 2 ) = (ag + bg 2 ) f = ag f + bg 2 f = f (ag ) = af (g ) + bf (ag 2 )

2 CHAPTER 7 VECTOR BUNDLES Theorem 7.24***. The followng are propertes of. Suppose V, W and X are fnte dmensonal vector spaces. () If h : V W and f : W X are lnear, then (f h) = h f (2) I V = I V (3) If f : W V s lnear and one-to-one, then f s onto. (4) If f : W V s lnear and onto, then f s one-to-one. (5) If f : W V s an somorphsm, then f s an somorphsm and (f ) = (f ). (6) The map F : GL(V ) GL(V ) by F (f) = f s a C map. Proof. We check each tem. Proof of tem : (f h) (α) = α f h = h (α f) = f h (α) Proof of tem 2: IV (α) = α I V = α = I V Proof of tem 3: The functon f s one-to-one f and only f there s a functon h such that h f = I V. Hence (h f) = IV. By and 2, f h = I V. Snce f has a rght nverse, f s onto. Proof of tem 4: The functon f s onto f and only f there s a functon h such that f h = I V. Hence (f h) = IV. By and 2, h f = I V. Snce f has a left nverse, f s one-to-one. Proof of tem 5: Snce f s one-to-one and onto, so s f by 3 and 4. Now, f f = I V. Items and 2 mply, (f ) f = I V, therefore (f ) = (f ) Proof of tem 6: We pck charts on GL(V ) and GL(V ) by pckng bases for V and V. Suppose {e = n} s a bass for V. Let e {e = n} as a bass for V. If f s represented by the matrx M n the e bass then f s represented by M T n the e bass. We check ths fact. n n f ( a e )( n n b j e j ) = ( a e ) = j= = = ( = n,j= n,j= V, where e ( n j= b je j ) = b. Take a m j b j j= k= m j a e j )( n m kj b j e k n b k e k ) So, f ( n = a e ) = n,j= m ja e j and M T represents f n the e bass. If we consder GL(V ) and GL(V ) as open sets n R n2 va matrx coordnates, then F s M M T. Snce transposng s just a permutaton of the entres, the map s C and n fact, lnear. If π : E M s a vector bundle E wth fber the vector space V, then we can then construct E, a dual bundle ρ : E M. The fber should be V, but how should the fbers be assembled? Consder the followng: each pont n V gves a map from V to R, so k=

CHAPTER 7 VECTOR BUNDLES 3 each pont n ρ (x) should gve a map from π (x) to R. A secton of ρ : E M pcks out a pont n each fber and vares n a smooth manner. Therefore, we want a secton of ρ : E M to gve a map to R whch s lnear on each fber. Another way to thnk of the above descrpton s that a secton of the dual bundle should gve a bundle map of E to ɛ M, the trval lne bundle over M. Before proceedng to the constructon of the cotangent bundle, we frst undertake a dscusson of the meanng of a dual of a bundle,.e., the co n cotangent. For a vector space the meanng s expressed n the evaluaton parng, V V R (f, v) f(v). For vector bundles, there should be a parng for each x M and ths parng should vary smoothly n x. Therefore there should be an evaluaton parng, (2) Γ(E ) Γ(E) Γ(ɛ M ) wth (f, s)(x) = (x, f(x)(s(x))). Note that ths condton s guded by Remark 7.5***. Defnton 7.25***. Suppose E s a vector bundle π : E M. The vector bundle ρ : E M s the dual bundle to E f ρ (x) = (π (x)) for all x M and f E satsfes the followng property: There s an evaluaton parng (2) F : Γ(E ) Γ(E) Γ(ɛ M ) defned by F (f, s) = f(x)(s(x)) for all x M. The parng s the usual parng on each fber, but the fbers ft together smoothly so the evaluaton vares smoothly over M. The reader should notce that the dual bundle to the trval bundle s agan a trval bundle snce the dual to the bundle π : M V M s the bundle π : M V M. Theorem 7.26***. Suppose E s a vector bundle π : E M wth fber V, then there s a dual vector bundle E, π : E M wth fber V. Furthermore, f U = {(U, ϕ ) I} s a countable atlas for M wth each U also a trvalzng neghborhood for E and g j : U U j GL(V ) are the transton functons for E, then f j : U U j GL(V ) defned by f j (x) = (g j (x)) for each x U U j forms a set of transton functons for E. Proof. The cover U, the vector space V and the functons f j satsfy Theorem 7.8*** and so defne a vector bundle E. To see ths fact, we must check the condton the f j must satsfy. f k (x)f kj (x) = (g k (x)) (g jk (x)) = (g jk (x)g k (x)) by propertes of *, Theorem 7.24*** = (g j (x)) by propertes of transton functons, Proposton 7.6*** = f j (x)

4 CHAPTER 7 VECTOR BUNDLES whch s the requred property. It remans to check the property of the parng (2). Ths condton s a local condton and so we check t n the trvalzatons. Suppose that µ : π (U ) U V by µ(e) = (π(e), h (e)) and γ : π (U ) U V by γ(e) = (π(e), k (e)). If s and s 2 are sectons then for x U the parng (2) s (s 2 (x), s (x)) (x, k (s 2 (x))(h (s (x)))). Snce s, s 2, h, and k are smooth, the parng s also smooth, but we must check that the parng s well-defned. If x U j, then f j (x)(k j (s 2 (x))) = g j (x)(k j(s 2 (x))) = k (s 2 (x)) and g j (x)(h j (s (x))) = h (s (x)). Hence, So the parng s well-defned. k (s 2 (x))(h (s (x)) = f j (x)(k j (s 2 (x)))(g j (x)(h j (s (x)))) Example 7.27***. The Cotangent Bundle = g j (x)(k j(s 2 (x)))(g j (x)(h j (s (x)))) = (k j (s 2 (x)))(g j (x)g j (x)(h j (s (x)))) = k j (s 2 (x))(h j (s (x))) Suppose M s a smooth n-manfold. The dual bundle to the tangent bundle s called the cotangent bundle. The cotangent space at a pont x s (T M x ) whch we denote T M x. The bundle dual to the tangent bundle s (T M) n the dualty notaton, but s usually denoted T M. ) ϕ j (m). Note that f g j are the transton functons from whch we constructed the tangent bundle above, then f j (x) = (g j (x)). Ths s the constructon specfed by Theorem 7.25*** to construct the bundle dual to the tangent bundle. Call t the cotangent bundle and denote t T M. If φ : N M s a smooth map between smooth manfolds, then there s an nduced map φ φ(x) : T M φ(x) T N x defne by φ φ(x) (γ) = γ φ x. Let f j (m) = (ϕ j ϕ Example 7.28***. A secton of the cotangent bundle, df. Suppose f : M R, then let df x = π f where π : T R = R R R s projecton n the fber drecton,.e., π(a p x ) = a. Suppose that φ : M N s a smooth map and an onto map. Then, φ (df x ) = df x φ = π f φ = d(f φ). Suppose f : M R then we wll show that df defnes a secton of T M. If x U, then x d(f ϕ ) ϕ (x) s a smooth map snce x (f ϕ ) ϕ (x) s smooth. In the notaton of Theorem 7.8***, df x s represented by [x, d(f ϕ ) ϕ (x), ]. We have that f x U U j, then [x, d(f ϕ ) ϕ (x), ] = [x, d(f ϕ j ) ϕj (x), j] as f j (x)((d(f ϕ j d(f ϕ ) ϕj (x)) = (ϕ j ϕ ) d(f ϕ j ) ϕj (x) = d(f ϕ j ϕ j ϕ ) ϕ (x) usng the equvalence relaton of Theorem 7.8***. ) ϕ (x) =

CHAPTER 7 VECTOR BUNDLES 5 Remark 7.29. Sectons of the tangent and cotangent bundle are both used to generalze types of ntegraton. Sectons of the tangent bundle are used to ntegrate, solve a dfferental equaton. Sectons of the cotangent bundle are used to compute defnte ntegrals. Exercses Exercse ***. Let π : E M be a vector bundle. Show that t has sectons that are not dentcally zero. Ths next exercse justfes the termnology bundle equvalence. Exercse 2***. If E an E 2 are vector bundles over the manfold M and f E s bundle equvalent to E 2 by a bundle equvalence f, then E 2 s bundle equvalent to E by f whch s also a bundle equvalence. Show that bundle equvalence s an equvalence relaton. Exercse 3***. Show that every vector bundle over an nterval s trval. Exercse 4***. Show that up to bundle equvalence, there are exactly two dstnct lne bundles over the crcle. Ths next exercse ntroduces the noton of a Remannan metrc. Exercse 5***. Suppose that E s a vector bundle π : E M. A Remannan metrc on E s a choce of nner product <, > x for each fber π (x) such that there s an nduced map on sectons <, >: Γ(E) Γ(E) Γ(ɛ M ) defned by < s, s 2 > (x) =< s (x), s 2 (x) > x. a. Show that every vector bundle has a Remannan metrc. Ths argument wll requre a partton of unty. b. Show that E s bundle equvalent to E.