HainanZhang2014

TOPOLOGY OF FIBER BUNDLES by HAINAN ZHANG B.S., Kansas State University, 2012 A THESIS submitted in partial fulllment of the requirements for the degree MASTER OF SCIENCE Department of Mathematics College of Arts and Sciences KANSAS STATE UNIVERSITY Manhattan, Kansas 2014 Approved by: Major Professor David Auckly Copyright Hainan Zhang 2014 Abstract This report introduces the ber bundles. It includes the denitions of ber bundles such as vector bundles and principal bundles, with some interesting examples. Reduction of the structure groups, and covering homotopy theorem and some specic computation using obstruction classes, Cech cohomology, Stiefel-Whitney classes, and rst Chern classes are included. Table of Contents Table of Contents iv List of Figures v 1 Introduction to Fiber Bundles 1 1.1 Trivial Bundles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4 1.2 Vector Bundles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4 1.3 Principal Bundles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4 2 Example: The Bundle of Frames (;SO3;S2;SO2) 6 3 Trivialization and Section 14 4 Reduction of the structure group 17 5 Covering Homotopy Theorem 21 5.1 Hopf-bration: ( S1!S3!S2) . . . . . . . . . . . . . . . . . . . . . . . . . 24 6 Obstruction Theory 25 6.1 Stiefel-Whitney Classes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32 6.2 Chern Class . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33 7Cech Cohomology 35 Bibliography 40 iv List of Figures 6.1 Cell decomposition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27 6.2 Graph one of 1st Stiefel Class under section sigma . . . . . . . . . . . . . . . 28 6.3 Graph two of 1st Stiefel Class under section sigma . . . . . . . . . . . . . . . 29 6.4 Graph of 1st Chern class . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31 v Chapter 1 Introduction to Fiber Bundles A ber bundle is a space that is locally a product space, but globally may have a dierent topological structure. We can describe it as a continuous surjective map :E!Bthat in small neighborhood Ubehaves like the projection of UFtoU. Usually, we call the map the projection of the bundle. The space E,B,Fare known as total space, base space, and ber space respectively. In general, there are two kinds of ber bundles. The trivial bundle is BF, the mapis the canonical projection from the product space to the rst factor. In the non-trivial case, bundles will have totally dierent topological structures globally, such as the M obius strip and the Klein bottle, as well as non-trivial covering spaces. Denition 1. A ber bundle is a structure (;E;B;F ), whereE,B, andFare topological spaces and:E!Bis a continuous surjection satisfying a local triviality condition, which is called the bundle projection. The space Eis the total space, Bis called the base space of the bundle, and Fis the ber space. The locally triviality condition is that for any point x2B, there exists an open neighborhood 1 UxBsuch that1(Ux)is homeomorphic to the product space UxFsuch that the following diagram commutes: 1(Ux)UxF Ux//  proj (1.1) where proj :UxF!Uxis the natural projection. The set of all f(Uj;j)gis called a local trivialization of the bundle. Thus for any p in B, the preimage 1(p)is homeomorphic to Fand is called the ber over p. Every ber bundle :E!Bis an open map since the projection of products are open maps. Therefore Bcarries the quotient topology determined by the map . Transition Functions If letU=fUg2Ibe a open cover of the base space, then one can construct many dierent local homeomorphisms by taking dierent open sets. Then, the nonempty interscetion of two local homeomorphisms will give two dierent systems for the same point in the over- lap. Transition functions give a invertible transformation of the ber over the point in two dierent systems. Given a vector bundle E!B, and a pair of neighborhood UandUover which the bundle trivializes via two homeomorphisms, 2 h:UF!1(U) (1.2) h:UF!1(U) (1.3) the composite function is dened to be h1 h:UF!UF (1.4) whereUisU\U, which is well dened on the overlap by h1 h(x;v) := (x;v) = (x; (x)(v)) (1.5) where the map :U!Homeo(F) (1.6) sends x to (x) such that (x) :F!F (1.7) is the automorphism of the ber Fwithf! (x)(f) for allf2F. Lemma 1.   =  . Proof. Note is dened to be h1 h:UF!UF, therefore,   = h1 hh1 h =h1 h =  . Thus, by this fact, for all x2U\U\U andf2Fwe have   (x;f) =  (x;f)) (x;  (x)(f)) = (x;  (x)(f))) (x; (x)(  (x)(f))) = (x;  (x)(f)))(x;   (x)(f)) = (x;  (x)(f)))   =  3 1.1 Trivial Bundles LetE=BFand let:E!Bbe the projection onto the rst factor. Then Eis a ber bundle over B. The bundle Eis not just a locally product but globally one. Thus, a trivial bundle is homeomorphic to the Cartesian product. Lemma 2. Any ber bundle over a contractible CW-complex is trivial. Proof. LetBbe a contractible CW-complex. Since Bis contractible, there is a point x2B and a homotopy between idBand the constant map f:B!fxg. Then, we can dene thepullback withid B(E)wf(E). Therefore, E=Id Bandf=BEximply that EwBEx, which is a trivial bundle.1 1.2 Vector Bundles Denition 2. A vector bundle is a ber bundle in which every ber is a vector space. For every point in the base space B, there is an open neighborhood U and a homeomorphism h:URk!1(U)such that for all x 2U, 1.()(x;v) =xfor all vectors v in Rk 2.v!(x;v)is an isomorphism between the vector space Rkand1(x), and the vector bundle over the elds are dened similarly. 1.3 Principal Bundles A principal bundle Pis a mathematical object which formalizes some essential features of the Cartesian product XGof a topological space with a structure group Gequipped with: 1. An action of GonP, analogous to ( x;g)h= (x;gh ) for a product space 4 2. A natural projection onto X, this is just the projection onto the rst factor, ( x;g)!x Denition 3. A principle G-bundleP, whereGdenotes any topological group, is a ber bundle:P!Xtogether with a continuous right action PG!Psuch thatGpreserves the bers of Pand acts freely and transitively on them i.e. let Pxbe a ber over any point x, then8x0andx12Px,9someg2Gsuch thatx0g=x1 The denition implies that each ber of the bundle is homeomorphic to the group Gitself. Unlike the product space, principal bundles Plack a preferred choice of identity section. Likewise, there is no general projection onto Ggeneralizing the projection onto the second factor,XG!Gwhich exists for the Cartesian product. 5 Chapter 2 Example: The Bundle of Frames (;SO3;S2;SO2) Every rotation maps an orthonormal basis of R3to another orthonormal basis. Like any lin- ear transformation of a nite-dimensional vector space, a rotation can always be represented by a matrix. The orthonormality condition can be expressed as RTR=I, whereRis any ro- tation matrix in R3andIis the 33 identity matrix. Matrices for which this property holds are called orthogonal matrices. The group of the orthogonal matrices is called orthogonal group, and SO3:=fA2Mat 33jATA=I;det(A) = 1gis the subgroup of the orthogo- nal group with determinant +1. It is called the special orthogonal group, and the det(A) = 1 Similiarly, SO2is the orientation preserving orthogonal group in R2. In order to carry out a rotation using the point (x, y) to be rotated is written as a vector, then multiplied by a matrix from the rotating angle : 2 64x0 y03 75=2 64cossin sincos3 752 64x y3 75 (2.1) 6 where (x0,y0) are the coordinates of the point after rotation, and the formulae for x0andy0 can be seen to be x0=xcosysin y0=xsin+ycos: Note that unlike higher dimension, the group of vector rotations in R2is commutative. Before showing the bundle structure of ( ;SO3;S2;SO2), let us consider the map e: SO3=SO2!S2dened by e([A]) =A
e3whereA2SO3ande3= [0;0;1]T. Here, the equivalence class of [A] is dened via the equivalence AA
2 64B0 0 13 75 whereB2SO2. We are trying to show that this map is a bijective continuous closed map, therefore, a homeomorphism. Proof. We rst show that this map is well-dened. Let [ A1] = [A2], soA1=A2
2 64B0 0 13 75. Thene[A1] = [A1]
e3= [A2]
2 64B0 0 13 75
e3= [A2]
e3=e[A2]. Assume e[A1] =e[A2] so thatA1
e3=A2
e3, whereA1;A22SO3. NowAT 2A1
e3= AT 2A2
e3=I3
e3=e3, thus, there exists one element B2SO2such thatAT 2A1=2 64B0 0 13 75, which implies that A1=A2
2 64B0 0 13 75. Therefore, [ A1] = [A2] implies that eis injective. 7 For each vector !u=2 66664x y z3 777752S2, there exists an A2SO3=SO2with A=2 66664xzp 1z2yp 1z2x yzp 1z2xp 1z2y p 1z2 0z3 77775 such that e([A]) =A
e3=2 66664xzp 1z2yp 1z2x yzp 1z2xp 1z2y p 1z2 0z3 777752 666640 0 13 77775=2 66664x y z3 77775 where the way to get A is applying the spherical system by composing the rotation of and the rotation of . Since rotation of is on the xy-plane, then R() =2 66664cos()sin() 0 sin() cos() 0 0 0 13 77775(2.2) and the rotation of can be xed on the xz-plane without lost generality, then R() =2 66664cos() 0 sin() 0 1 0 sin() 0 cos()3 77775(2.3) therefore, their composition is 8 R()R() =2 66664cos()sin() 0 sin() cos() 0 0 0 13 777752 66664cos() 0 sin() 0 1 0 sin() 0 cos()3 77775(2.4) =2 66664cos() cos()sin() cos() sin() sin() cos() cos() sin() sin() sin() 0 cos( )3 77775(2.5) Thus, set cos() sin() =x;sin() sin() =y;cos() =z; we can solve each entry of the matrix which is exactly where we get A. Therefore, eis surjective. By the construction of e, the map actually is the projection of the third column by mul- tiplyingAand the third base element. If we just consider the map :SO3!S2dened by (2 66664a11a12a13 a21a22a23 a31a32a333 77775) =2 66664a13 a23 a333 77775 Note that SO3Mat 33(R) is a subspace of R9, and R9is homeomorphic to R6R3 which is the product topological space. Thus, is just the projection from R9toR3. Then, applying the result that the projection from the product topology is continuous will lead thateis also continuous. 9 SO3is closed as it is the inverse image of a closed set S2, and bounded by computing its norm, therefore, it is compact. It is clear that S2is a Hausdor Space. By the closed map lemma, every continuous map from a compact space to a Hausdor space is closed and proper, we conclude that the continuous map eis closed. Hence, we just proved that eis a homeomorphism by the fact that a bijective continuous map is homeomorphism if and only if it is a closed map. In fact, when His a closed subgroup of a Lie group G, the projection G!G=H will be a principalH-bundle. The example of Q2Rshows that one must make some assumption on H. We will now show directly the SO3!S2is a principal SO2-bundle by replacing HtoSO2 andGtoSO3. Dene a map :SO3!S2by(A) =A
e3. The similarly way as we did foreshows that is a surjective projection. Now we check the local triviality condition. By doing the stereographic projection, let U=S2[0;0;1]T, dene a map :R2!S2. Pick a point ( x;y;0)2R2, then the line equation !r(t) =t(x;y;0) + (1t)(0;0;1) = (tx;ty;t1): Since this point is on the unit sphere S2, we get t=2 x2+y2+ 1 by solvingt2x2+t2y2+t22t+ 1 = 1. So the corresponding point on the sphere is 10 (2x x2+y2+ 1;2y x2+y2+ 1;1x2y2 x2+y2+ 1): Thus, we can nd the new basis for every element of R2by the dierentiation respect to x andy. Denote the new basis by ( e1;e2;e3), e1=@x jj@xjj= (22x2+ 2y2 (x2+y2+ 1)2;4xy (x2+y2+ 1)2;4x (x2+y2+ 1)2) e2=@y jj@yjj= (4xy (x2+y2+ 1)2;2x22y2+ 2 (x2+y2+ 1)2;4y (x2+y2+ 1)2) e3= (2x x2+y2+ 1;2y x2+y2+ 1;1x2y2 x2+y2+ 1) Now dene h:USO2!1(U) by h((x;y);2 64cossin sincos3 75) = (cose1(x;y) + sine2(x;y);sine1(x;y) + cose2(x;y);e3(x;y)) (2.6) = (e1(x;y);e2(x;y);e3(x;y))2 66664cossin0 sincos0 0 0 13 777752SO3:(2.7) Then, dene g:1(U)!USO2byg(A) = (1(A
e3);(e1(x;y);e2(x;y);e3(x;y)1A) as a well-dened continuous map. It is easy to check that h1 g=id1(U)and g1 h=idUSO 2. Therefore, his a homeomorphism. 11 Similarly, we can constrcut a map h+:U+SO2!1(U+) by h+((x;y);2 64cossin sincos3 75) = (cosf1(x;y) + sinf2(x;y);sinf1(x;y) + cosf2(x;y);f3(x;y)) (2.8) = (f1(x;y);f2(x;y);f3(x;y))2 66664cossin0 sincos0 0 0 13 777752SO3(2.9) forU+=S2[0;0;1]T, where the basis ( f1;f2;f3) is f1=@x jj@xjj= (22x2+ 2y2 (x2+y2+ 1)2;4xy (x2+y2+ 1)2;4x (x2+y2+ 1)2) f2=@y jj@yjj= (4xy (x2+y2+ 1)2;2x22y2+ 2 (x2+y2+ 1)2;4y (x2+y2+ 1)2) f3= (2x x2+y2+ 1;2y x2+y2+ 1;x2+y21 x2+y2+ 1) and deneg+:1(U+)!U+SO2byg+(A) = (1(A
e3);(f1(x;y);f2(x;y);f3(x;y)1A) as a well-dened continuous map. It is easy to check that h1 +g+=id1(U+)and g1 +h+=idU+SO 2. Therefore, h+is a homeomorphism. Hence, combining the fact of the closed subgroup of the Lie group, we claim that this structure is a princiapl SO2-bundle. It can also be veried that SO3is not homeomorphic toS2SO2by computing fundamental groups. The fundamental group 1(SO2S2) is equal to1(SO2)1(S2), which is homeomorphic to Zf1g=Z. While the fundamental 12 group1(SO3) is homeomorphic to Z2, so (;SO3;S2;SO2) is not a trivial bundle. 13 Chapter 3 Trivialization and Section Denition 4. A section of a ber bundle is a continuous right inverse of the projection . If:E!Bis a ber bundle, then a section is a continuous map s:B!Esuch that (s(x)) =xfor allx2B The most common question regarding any ber bundle is whether or not it is trivial. There is a nice lemma to answer it, but it is not true for the other ber bundles. Lemma 3. A principal bundle is trivial if and only if it admits a global section. Proof. If a principal G-bundlePis trivial (where Gdenotes any topological group), then, P=BG, whereGandBare ber space and base space, is just the Cartesian product. Dene:P!Bas a ber projection, and proj :BG!Bwithproj (b;g) =bas a Cartesian projection, and we will be able to nd a section s:B!Psuch thatsis just the inclusion map i:B!BGwithi(b) = (b;1). Thus,(s(b)) =proj (i(b)) =proj ((b;1)) = b8b2B, which implies that the inclusion map iis a global section. Now if a principal G-bundlePadmits a global section. A ber bundle :P!Bwith a continuous right action :PG!Pdened by(p;g) =pgfor allg2G, which actually 14 denes an equivalence class of any element pinPsuch thatppg. Now letBGbe a Cartesian product and dene a map :BG!Pby(b;g) =(s(b);g) =s(b)g, we want to show that is an homeomorphism, thus, Pis a trivial bundle. Check injective: Let (b1;g1) =(b2;g2), so(s(b1);g1) =(s(b2);g2), which implies s(b1)g1 =s(b2)g2. Then, by the denition of section, we have b1=(s(b1)) andb2=(s(b2)), wheres(b1);s(b2)2P. Sincedenes the equivalence class such that ppg8g2G, s(b1)s(b1)g1ands(b2)s(b2)g2for someg1;g22G. Thus,b1=(s(b1)) =([s(b1)g1]) = (s(b1)g1) andb2=(s(b2)) =([s(b2)g2]) =(s(b2)g2). So,b1=b2andg1=g2since s(b1)g1=s(b2)g2=s(b2)g1. Hence, (b1;g1) = (b2;g2). Check surjective: Given any p2P, set(p) =bfor someb2Bby the projection map : P!B. By the map s:B!P, we know that s(b)2P. Then, there exists some g2Gsuch thatp=s(b)g. Then,p=s(b)g=(s(b);g) =(b;g) =((p);g) =(s((p));g) =s(p)g. Note that [p][s((p))] since(p) =(s((p))). Therefore, there exists b=(p)2Band g2Gwithp=s(b)gsuch thatp=(b;g). Check continuous: By the denition of the map :BG!Pdened by (b;g) = (s(b);g) =s(b)g. We know that depends on two continuous map ands, thus,is a continuous map by the fact that the composition of the continuous maps is still continuous. Check open: Now we pick any open set W2BGto show that it maps to an open set inP. PickUb, whereUbis the open neighborhood of the arbitrary points binB. Consider the local triviality hb:UbG!1(Ub) which is a homeomorphism, therefore, h1 b:UbG!UbGis dened by h1 b(x;g) =h1 b(s(x)g) = (x;Pr((h1 b(s(x)g)))) = (x;Pr((h1 b(s(x))))g), where Pris just the natural projection from UbGtoG, therefore, 15 continuous. h1 bis continuous since the composition of two continuous map is continuous. Let :UbG!UbGdened by ( x;g) = (x;Pr((h1 b(s(x))))1g) and  :UbG! UbGdened by ( x;g) = (x;Pr((h1 b(s(x))))g), where and  are both continuous by denition. Now,  = 1 and  = 1 . Thus,  is a homeomorphism and jUbG:UbG!1(Ub) is a homeomorphism as =hb. Thus,jUbGis open and (Ub\W) is open. Therefore, (W) =([(Ub\W)) =[((Ub\W)) is open. Hence, is open. 16 Chapter 4 Reduction of the structure group Denition 5. Given a principal G-bundlePwith:P!Band a monomorphism H!G, whereGis the structure group, then a reduction of the structure group to His an principal H-bundleQsuch thatQHG=P. IfWis a rightG-space and Xis a leftG-space, the balanced product WGXis the quotient space WX=, where (wg;x )(w;gx ). Equivalently, we can simply convert Xto a rightG-space, and take the orbit space of the diagonal action ( w;x)g= (wg;g1x); thusWGX= (WX)=G. The following special cases should be noted: 1. IfX=is a point,WG=W=G 2. IfX=Gwith the left translation action, the right action of Gon itself makes WGGinto a right G-space, and the action map WG!Winduces aG-equivalent homeomorphism WGG=W. LetGandHbe topological groups. A ( G;H )-space is a space Y equipped with a left G-action and right H-action, such that the two actions commute: ( gy)h=g(yh).Note that ifYis a (G;H )-space and Xis a rightG-space,XGYhas a right H-action dened by 17 [x;y]h= [x;yh ]; similarly YHZhas a leftG-action dened by g[y;z] = [gy;z]ifZis a left H-space.2 Proposition 1.The balanced product is associative up to a natural isomorphism: Let X be a right G-space,Ya (G;H )-space, and Za leftH-space. Then there is a natural homeomorphism (XGY)HZ=XG(YHZ): Proof. By the denition, the balanced product XGYis the quotient group XY= where (xg;y)(x;gy). Thus, (XGY)HZis the quotient group (( XGY)Z)= where ([x;y]h;z)([x;y];hz) and [x;y] is the equivalence class dened as above. Therefore, ((xg;y)h;z)((x;gy)h;z)((x;gy);hz)(x;gy;hz ). Similarly, the balanced product YHZis the quotient group YZ=where (yh;z)(y;hz). Thus,XG(YHZ) is the quotient group ( X(YHZ))=where (xg;[y;z])(x;g[y;z]) and [y;z] is the equivalence class. Therefore, ( xg;(yh;z))(x;g(yh;z))(x;g(y;hz))(x;(gy;hz )) (x;gy;hz ). Proposition 2.The bundle E=PG(G=H ) associated with P with standard ber G=H can be identied with P=H . An element ( p;gH )2PG(G=H ) maps into the element pg2P=H . Consequently, P(E;H ) is a principal bundle over the base E=P=H with structure group H. The projection P!Emapsp2Pintopg2E. Proof. The proof is straightforward by the denition of the ber bundle, except the local triviality of the bundle P(E;H ), which follows from local triviality of E(B;G=H;G;P ) and G(G=H;H ). LetUbe an open set of Bsuch that1 E(U)=UG=H and letVbe an open set ofG=H such thatp1(V)=VH, wherep:G!G=H is the projection. Let Wbe the open set of 1 E(U) which corresponds to UVunder the identication 1 E(U)=UG=H . If:P!E=P=H is the projection, then 1(W)=WH.3 18 Theorem 1. The structure group of a principal G-bundlePcan be reduced to Hif and only ifE=PG(G=H )!Badmits a section, where His the closed subgroup of the Lie group G. Proof. Suppose that the structure group of Pcan be reduced to a principal H-bundle Q such thatQHG=P. Then,E=PG(G=H ) =QHGG(G=H ) =QHGGGH= QHGH=QH(G=H ) by the proposition of the balanced product, where corresponds to any point in H. Since the identity coset in G=H is anH-xed point, then we can dene a maps:!G=H . Applying the functor QH() to the map s, there is a morphism, B!QH(G=H ) which is equivalent to B!PG(G=H ), a section.2 Suppose that PG(G=H )!Badmits a section s:B!PG(G=H ). By the proposition above,E=PG(G=H )=P=H . Dene:P!P=H to be the projection, and let Qbe the set of points x2Psuch that(x) =s((x)). In other words, Qis the inverse image ofs(B) by the projection . For every b2B, there isx2Qsuch that(x) =bsince 1(s(b)) is non-empty. Given xandyin the same ber of P, ifx2Qtheny2Qif and only ifx=yhfor someh2H. Thus, the restriction jQ:Q!Q=H is induced by , soQ is a closed sub-manifold of P.3By the fact that G=H is a smooth manifold and G!G=H is a submersion since His the closed subgroup of the Lie group G, thusQis a principal bundle imbedded in PwithQHG=P.4 The balanced product let one considers many ber bundles from one principal bundle. Consequently, given any ber bundle E!B, one constructs a principal Homeo(F)-bundle such that P =a UHomeo(F)=; where the equivalence class is ( x;f) = (x; (x)(f)), and is the transition function. Example 1.As a concrete example, every even-dimension real vector space is the underlying 19 real space of a complex vector space: it admits a linear complex structure. A real vector bundle admits an complex structure if and only if it is the underlying real bundle of a complex vector bundle. This is a reduction along the inclusion GL(n;C)!GL(2n;R). Example 2.LetG=GL(n) andH=GL+(n). A reduction to GL+(n)-bundle is a ber orientation, and we can dene a map GL(n)=GL+(n)!Z2by sending the equivalence class [A] to the sign of det(A). Example 3.LetG=GL(n) andH=O(n). ThenG=H is homeomorphic to the group of upper triangular matrices with positive diagonal entries, and so is contractible. This implies that the associated bundle PGL(n)GL(n)=O(n) admits a section. Hence, any GL(n)-bundle is induced from an O(n)-bundle. Example 4.LetG=GL(n;R) andH=SL(n;R), whereHG. ThenG=H =GL(n;R)=SL(n;R)= R, therefore, a reduction to SL(n) cooresponds to an oriented ber volume form. These examples demonstrate the fact that the structure group of a real vector bundle of rank n can always be reduced to O(n), which can be reduced to SO(n) if and only if the vector bundle is orientable. And the vector bundle is orientable if its structure group may be reduced to GL+(n). 20 Chapter 5 Covering Homotopy Theorem Denition 6. Given a map :E!B, and any topological space X. Then, (X;)has the homotopy lifting property if: 1. For any homotopy f:XI!B 2. For any ef:X!Eliftingf0=fjXf0gsuch thatf0=ef, there exists a homotopy ef:XI!Eliftingfsuch thatf=efwithef0=efjXf0gas the following diagram. X E XI B//ef0 i0?? ef  //f(5.1) Denition 7. A Hurewicz-bration is a continuous map :E!Bsatises he homotopy lifting property with respect to any space,and a Serre-bration is a continuous map which has the homotopy lifting property with respect to any nite CW-complex. 21 Example 5.The projection map BF!Bis a trivial bration, and the ber over every point is homeomorphic to F. Indeed, given fandef0, dene ef:XI!BFby ef(x;t) = (f(x;t);Pr(ef0(x))). Example 6.Any covering space E!Bis bration. And, in fact, for a covering space the liftingefis uniquely determined. Example 7.Any ber bundle E!Bover paracompact base space B is a bration, e:g:any CW-complex. Example 8.In general,any ber bundle in which the base space is para-compact is a bration. A map:E!Bis a ber bundle with ber FifBhas a coverUandh:UF!1(U) is a homeomorphism for each U2U. But, without the para-compact hypothesis on the base, any ber bundle is at least a Serre-bration. Theorem 2. Any ber bundle :E!Bis a Serre-bration. Proof. Let:E!Bbe a ber bundle, and suppose given a lifting diagram InE InI B//f i0  //h LetUbe a covering for Bon which we have the local trivialization. I can divide Ininto sub-cubesCandIinto subintervals J,such that each CJis in a single h1(U). For each C, I can build a lift on each CJ, starting with the Jcontaining 0, so I have a lift along the initial point of the interval. Moreover, the cube Cborders other cubes, so I suppose that some union D of faces of @Chas a lift along DI. For the xed CandJ, the ber bundle is the trivial ber bundle UF!U. A lift in the diagram 22 C UF CI U//f i0?? 9eh  //h must be given in the rst coordinate by the map h, so it remains to describe the second coordinate of the lift. By assumption, we already have a lift C[DDJ!UF!F. Since the space C[DDJis a retract of CJ, so I can compose the lift with a choice of a retraction CJ!C[DDJ.5 Corollary 1.IfF!E!Bis a ber bundle and Bis paracompact and contractible, then Eis trivial. Proof. LetP!Bbe the associated principal bundle, we will show that it has a section. Since the base Bis contractible, it is homotopy equivalent to a single point . Leth: BI!Bbe a contraction such that h(b;0) =b0andh(b;1) =b. Now pushforward of 1(b0) to get the commutative square diagram, then there is a map eh:BI!Psuch that B P BI B//eh0 i0?? eh  //h So we dene :B!Pby(b) =eh(b;1). SincePhas a section, Pis trivial, then P=BHomeo(F). Thus, E=PHomeo(F)F=BF. 23 5.1 Hopf-bration: (S1!S3!S2) We can identify R4withC2andR3withCRby letting 1. (x1;x2;x3;x4)as(z0=x1+ix2;z1=x3+ix4) 2. (x1;x2;x3) as (z=x1+ix2;x=x3). ThusS3is identied with the subset of all ( z0;z1)2C2such thatjz0j2+jz1j2= 1, and S2is identied with the subset of all ( z;x)2CRsuch thatjzj2+x2= 1. Then the Hopf-bration is dened by (z0;z1) = (2z0z1;jz0j2jz1j2): The rst component is a complex number and the second one is real. Then, we need to check that the image of maps into S2, which can be veried by squaring two parts of the image (2z0z1)2+ (jz0j2jz1j2)2= 4jz0j2jz1j2+jz0j42jz0j2jz1j2+jz1j4(5.2) = (jz0j2+jz1j2)2= 1 (5.3) It is easy to see that this is a surjective map since S2!S3is an inclusion map. Then, if two points on S3map to the same point on the S2such that(z0;z1) =(w0;w1), then (w0;w1) must equal to ( z0;z1) for some complex number withjj2= 1. Since the set of the complex numbers withjj2= 1 forms the unit circle in the complex plane, it follows that for each point x2S2, the inverse image 1(x) is isomorphic to S1. This structure admits a local trivialization which implies that it is a ber bundle, hence, a bration since the base space S2is a paracompact CW-complex. 24 Chapter 6 Obstruction Theory Suppose we want to construct a section from a CW-complex Xinto a bundle Ewith ber F. We do this by induction: given a section :X(k)!Eon the kth-skeleton X(k)and a (k+1)-celli:Dk+1!X, we want to extend over thei. The obstruction to extend over a (k+1)-cell is an element of k(F), the k-th homotopy group of the ber.6 Dene the pullback i(E) :=f(p;q)2Dk+1Eji(p) =(q)g, whereq:=(i(p)) and the ber ofi(E) overx2Xis the ber of Eoveri(p)2X. Therefore, it admits the bundle structure i(E) SkDk+1?? // Thus,i(E) admits a trivialization such that :i(E)!Dk+1F 25 then, there exists a continuous projection P2maps to the ber space F. Now we can dene the obstruction class O(i) = [P2(b;i)]2k(F), whereO2Ck+1 CW(X;k(F)). These obstructions t together to give a cellular cochain OonXwith coecients in this k. In fact, this cochain is a cocycle, so it denes an obstruction class O(E) inHk+1(X;k(E)). Then there exists a cross-section over the (k+1)-skeleton if and only if a certain well dened obstruction class is zero. If the cochain is 0, then there exists a map :Dk+1!F. Then the section extending to (k+1)-skeleton e:Xk+1!Eis dened to be P2(1(v;(v))), wherev2Dk+1. Example 9.(1st Stiefel-Whitney class) LetE!Xbe a real vector bundle with structure group GL(n;R) as example 2 in section 2, whereXis a CW-complex space. Then, Eis orientable if and only if its structure group can be reduced to the subgroup GL+(n;R). Therefore, we have the associated bundle Z(E) =GL(n;E)GL(n;R)(GL(n;R=GL+(n:R))) with ber Z2as the following diagram, Z2 Z(E) X X//i ??  //Idx(6.1) where the orientation is a section . So, we build kinductively on the k-th skeleton X(k). Dene0:X(0)!Eas a section such thatk=IdsinceX(0),!Xis an inclusion map. Note that the ber Z2is a group so that its 0th-homotopy group 0(Z2) =Z2. In this case, the obstruction cocycle is the 1st Stiefel-Whitney class w 1(E)2C1 CW(X;Z2). Hence, a section dened on 0-cells is 26 extendable if and only if the 1st Stiefel-Whitney class w 1(E)2H1(X;Z2) is 0. (1). It is a cocycle since its coboundary is 0. Indeed, let i:D2!Xbe 1-cells and (w0 1(E))(i) =w0 1(E)(@i) = (0(1;e2)0(1;e1))+(0(1;e3)0(1;e2))+:::+(0(1;e1) 0(1;en)) = 0 where eiis the 0-cells of D2andenneed not be distinct. Figure 6.1 :cell decomposition (2). The 1st Stiefel-Whitney class is independent of the choice of sections. Let's pick two distinct sections and. Then, dene4;2C1 CW(X; 0(Z2)) by4;(x) =(x)(x)2 0(Z2). Thus, 27 (4;)(e) =4;(@e) (6.2) =4;(1;e)4;(1;e) (6.3) = ((1;e)(1;e))((1;e)(1;e)) (6.4) =w 1w 1= 0 (6.5) Figure 6.2 :1st Stiefel Class of total space E Exercise 1.Computew 1(E) andw 1(E), where;:S(0)!E. Let us rst consider the functionwhere the chosen two 0-cells are mapping into dierent sides of the total space E, and we can dene an orientation of S1as the graph shows by separating S1into two arcs denoted D+andD. By drawing two dierent arcs of the projections from Eto S1, there is no continuous mapping from the boundary of Dto the bundle space since the two points on Eis arc-wise disconnected, so does D+. Therefore, I conclude that w 1(E)(D) =w 1(E)(D+) = 1, which implies that w 1(E)(S1) = 1 + 1 = 2 = 0 in Z2. Let another function maps two 0-cells into the same side of the total space as the graph shows. Then, w 1(E)(D) =w 1(E)(D+) = 0, sow 1(E)(S1) = 0. It actually shows that 28 the obstruction class is independent with the choice of the functions. Therefore, the zero obstruction class implies that we can extend the function from 0-cell to 1-cell. Figure 6.3 :1st Stiefel Class of total space E0 Exercise 2.Computew 1(E0) andw 1(E0) of the dierent bundles E0, whereandare the same as part B. By separating the base space S1into two parts D+andDas we did above. Then,w 1(E0)(D) = 1 andw 1(E0)(D+) = 0, which implies that w 1(E0)(S1) = 1 + 0 = 1 inZ2. We know that the obstruction class is independent with the choice of the functions, thus,w 1(E0)(D) = 0 and w 1(E0)(D+) = 1 which can also be proved by the graph. So, w 1(E0)(S1) = 1 + 0 = 1. Hence, we can not extend any function on 0-cells over 1-celsl on the vector bundle E0. Example 10.(1st Chern class) Recall the denition of the complex vector bundle: a complex vector bundle Eof complex dimension n over Band projection map :E!B, together with the structure of a complex vector space in each ber 1(b) with local triviality such thath:UCn!1(U) is a homeomorphism which maps each ber 1(b) complex linearly onto bCn Just as the structure group of a real vector bundle can be reduced to the orthogonal group 29 O(n), the structure group of a rank n complex vector bundle can be reduced to the unitary groupU(n). Every complex vector bundle Eof rank n has an underlying real vector bundle ERof rank 2n, obtained by discarding the complex structure on each ber. Construct the composition GL(n;C)!C!S1, where the map from the complex linear groupGL(n:C) to the multiplicative group of complex number Cis a determinant func- tion, and the map from the total space to the base space S1is an argument function. Then, there exists a reduced bundle structure of the complex vector bundle Pwith the base space Xdenoted as E:=PGL(n;C)(GL(n;C)=Ker(argdet)), whereGL(n;C)=Ker(argdet) is homeomorphic to S1. Since0(S1) = 0 and 1(S1) =Z, the obstruction class O(E) is an element in C2 CW(X;Z). In this case, the 1st Chern class c1(E) is dened to be the obstruction classO(E) , an element in 2nd cohomology group H2(X;Z). Exercise 3.Consider the complex vector bundle :E!S2with ber C, where the total bundleEis the tangent space of S2. To compute the rst Chern class c1(E)(S2). Cut the 2-sphere into two halves with one labeled D2 +and another one D2 as the graph shows, which are homeomorphic to D2, a one dimensional disk over C. By pointing out the vectors on the equator on both halves, I take the projection of the vectors from the equator to the boundary of the disk as graphs. Dene :D1 +!R3by (x;y) = (x;y;p 1x2y2). Then, taking the patrial derivative to x, @x (x;y) = (1;0;x(1x2y2)1 2); (6.6) normalizing this vector, we get @x (x;y) jj@x (x;y)jj=((1x2y2)1 2;0;x) (1y2)1 2(6.7) 30 Now, plugging in some values for ( x;y), we can tell that the orientation of the vectors on D1 +does not change, while the orientation of the vectors on D1 changes twice by dening the similar parametrization function. Therefore, the rst Chern class is c1(E)(S2) =c1(E)(D1 +) +c1(E)(D1 +) = 0 + 2 = 2, which can be veried by our denition related to the Euler class of S2,c1(E)(S2) = e(ER2)(S2) =P(1)k(]of the k-cell) = (1)0(1) + (1)1(0) + (1)2(1) = 1 + 0 + 1 = 2 Figure 6.4 :Cell decomposition of torus Exercise 4. Let us see another example denoted as C!T(T2)!T2, where the total space is the tangent space of the 2-torus. By denition, T2=C=(ZiZ). Thus, we have the graph of 31 the square representing the 2-torus, which is homeomorphic to the unit circle. Since A1and A2are equivalent, then we can dene a continuous function a:A1!A2bya(z) =z+i, where z is a complex number. So this map preserves the sign of the vector, therefore, the orientation of the vectors on A1andA2is the same. We can dene another continuous functionb:B1!B2byb(z) = 1 +z, which also preserves the orientation of the vectors on B1andB2. Note that the four vertices are identied by gluing together, so the orientation of the vectors on four sides preserves. Hence, the orientation of the vectors on the tangent bundle preserves, which implies that the rst Chern class c1(TT2)(T2) = 0. It can also be veried by computing the Euler class of T2,e(T2) = (1)0(1) + (1)1(2) + (1)2(1) = 0 6.1 Stiefel-Whitney Classes The Stiefel-Whitney classes are a set of topological invariants of a real vector bundle that describe the obstructions to constructing independent sections. Let Hi(B;Z2) denote the i-th singular cohomology group of B with coecients in Z2, here are four axioms which characterize the Stiefel-Whitney cohomology.7 Axiom 1.For each vector bundle Ethere corresponds a sequence of cohomology classes wi(E)2Hi(B(E);Z2);i= 0;1;2;:::; called the Stiefel-Whitney classes of E. The class w0(E) is the unit element 12H0(B(E);Z2); andwi(E) = 08inifEis an n-dimensional bundle. 32 Axiom 2.Iff:Y!Xis covered by a bundle map from EtoE0, then wi(E) =fwi(E0); wherefis the pullback. Axiom 3.IfEandE0are vector bundles over the same base space, then wn(EE0) =nX i=0wi(E)[wni(E0); where[denotes the cap product. For example, w1(EE0) =w1(E) +w1(E0), andw2(E E0) =w2(E) +w1(E)w1(E0) +w2(E0). Axiom 4.For the line bundle E1 1over the circle P1(real projective plane), the Stiefel-Whitney classw1(E1 1) is non-zero. Proposition 3.IfEiis isomorphic to E0thenwi(E) =wi(E0). Proposition 4.IfEis a trivial vector bundle then wi(E) = 0 fori>0. Proposition 5.IfEis trivial then wi(EE0) =wi(E0) 6.2 Chern Class The Chern classes are a set of topological invariants of a complex vector bundle that describe the obstructions to constructing independent sections. Let H2i(B;Z) denote the 2i-th sin- gular cohomology group of B with coecients in Z, here are four axioms which characterize the Chern cohomology.7 Axiom 5.For each complex vector bundle Ethere corresponds a sequence of cohomology classes ci(E)2H2i(B(E);Z);i= 0;1;2;:::; 33 called the Chern classes of E. The class c0(E) is the unit element 12H0(B(E);Z); andci(E) = 08inifEis an n-dimensional bundle. Axiom 6.Iff:Y!Xis covered by a bundle map from EtoE0, then ci(E) =fci(E0); wherefis the pullback. Axiom 7.IfEandE0are vector bundles over the same base space, then cn(EE0) =nX i=0ci(E)[cni(E0); where[denotes the cap product. For example, c1(EE0) =c1(E)+c1(E0), andc2(EE0) = c2(E) +c1(E)c1(E0) +c2(E0). Axiom 8.For the line bundle E1 1over the circle CP1(complex projective plane), the Chern classc1(E1 1) is1. Proposition 6.IfEiis isomorphic to E0thenci(E) =ci(E0). Proposition 7.IfEis a trivial vector bundle then ci(E) = 0 fori>0. Proposition 8.IfEis trivial then ci(EE0) =ci(E0) 34 Chapter 7 Cech Cohomology In general, to dene homotopy groups, one must pick a base point. We did not need the base point in our discussion of the rst Stiefel-Whitney classes as one does not need a base point to dene an element of 0(Z2). Similarly, one does not need a base point to discuss an element of 1(S1) =Z=H1(S1). In order to work with dierent base points, one needs to use the cohomology group with twisted coecients. Then, Cech cohomology is a tool applies abelian sheaf cohomology by using coverings and systems of coecients on the covering and all non-empty nite inter- sections. More generally, it applies to non-abelian cohomology, therefore, can be used to compute classes of ber bundles. Cech cohomology is obtained using an open cover of a topological space and it arise using purely combinatorial data. The idea being that if one has information about the open sets that make up a space as well as how those sets are glued together one can deduce global properties of the space from the local data. 35 LetU=fU:2Agbe an open cover of a connected manifold M. For0;


;n2A, we denote U0


k=U0\


\Uk or, equivalently, in multi-index notation, if a=f0;


;kg Ua=\ i2aUi Let X be a topological space, and let Ube an open cover of X. Dene a simplicial complex N(U), called the nerve of the covering as follows: 1.There is one vertex for each element of U2.There is one edge for each pair U1,U22U such thatU1\U26=?3.There is one k-simplex for each k+1-element subset fU0;


;Ukg ofUfor whichU0\


\Uk6=? The ideal of Cech cohomology is that, if we choose a nice cover Ucontaining of suciently small open sets, the resulting simplicial complex N(U) should be a good conmbinatorial model for the space X. For such a cover, the Cech cohomology of X is dened to be the simplicial cohomology of the nerve. Now letXbe a topological space, and Let Fbe the abelian group of coecients. Let Ube an open cover of X. An q-simplex ofN(U) is an ordered collection of q+1 sets chosen fromUsuch that the intersection of all these sets is non-empty. This intersection is called thesupport of. Now let= (Ui)i2f0;:::;qgbe such a q-simplex. The j-th partial boundary of is dened to be the (q-1)-simplex obtained by removing the j-th set from , that is 36 @j= (Ui)i2f0;:::^j;:::;qg the boundary of is dened as the alternating sum of the partial boundaries @=qX j=0(1)j@j A q-cochain ofUwith coecient in Fis a map which associates to each q-simplex and we denote the set of all q-cochains of Uwith coecients in FbyCq(U;F). In fact, all one needs is a way to associate an abelian group to an open say F(U)(which functionsU!F ) and a homeomorphism iU V:F(U)!F (V) for the subset V ,!Usuch that 1.iU U=idandiU ViV W=iU W. 2. LetU=[U, ifiU U
f=iU U
g8, thenf=g. 3. For any f2F(U) such that iU U
f=iU U
f. Then,f2Usuch thatf=iU U
f. Such a structure is called a Sheaf. The cochain groups can be made into a cochain complex ( Ck(U;F);) by dening the coboundary operator q:Cq(U;F)!Cq+1(U;F) by (q!) =Pq j=0(1)jresj@jj jj!(@j), whereresj@jj jj!(@j) is the restriction morphism on the intersection. It also satises the composition that q+1q= 0. A q-cochain is called a q-cocycle if it is in the kernel of , henceZq(U;F) := ker(q: Cq(U;F)!Cq+1(U;F)) is the set of all q-cocycles. Thus a (q-1)-cochain f is a cocycle if 37 for all q-simplices the cocycle conditionPq j=0(1)jresj@jj jjf(@j) = 0 holds. In particular, a 1-cochain f is a 1-cocycle if f(B\C)jUf(A\C)jU+f(A\B)jU= 08U=A\B\C, wherefA;B;Cg2U. A q-cochain is called a q-coboundary if it is in the image of andBq(U;F): =im(q1: Cq1(U;F)!Cq(U;F)) is the set of all q-coboundaries. For instance, a 1-cochain f is a 1-coboundary if there exists a 0-cochain h such that f(U) = (h)(U) =h(A)jU-h(B)jU8 U=A\B, wherefA;B2Ug . Then, the Cech cohomology of Uwith values inFis dened to be the cohomology of the cochain complex ( Ck(U;F);). Thus the q-th Cech cohomology is given by Hq(U;F) =Hq((Cq(U;F);)) =Zq(U;F)=Bq(U;F) The Cech cohomology of X is dened by considering renements of open covers. If Vis a renement ofUthen there is a map in cohomology H(U;F)!H(V;F). The open covers ofXform a directed set under renement, so the above map leads to a direct system of abelian groups. The Cech cohomology of X with values in F is dened as the direct limit H(X;F) = lim!UH(U;F). Related to other cohomology IfXis homotopy equivalent to a CW-complex, then the Cech cohomology H(U;F) is naturally isomorphic to the singular cohomology H(U;F). IfXis a dierential manifold, then H(U;F) is naturally isomorphic to the de Rham cohomology of X. For some less well-behaved spaces that fails for the closed topologist's sine curve, its Cech cohomology H1(X;Z) =Z, whereas H1(X;Z) = 0. 38 Example 11.Compute the H(S1;Z) if the open cover U=U1[U2=f(x;y)jx2+y2= 1;y< 1 2g[f(x;y)jx2+y2= 1;y>1 2g.U1\U2=f(x;y)jx2+y2= 1;1 2<y<1 2g, which is the dis- connected two components. Note, for any continuous map f2C0(U1!Z), its image f(U1) is connected since U1is connected, but the only connected subspaces of Zare?and singleton points which implies that f(U1) = singleton point, therefore, isomorphic to Zsince theU1 is not empty. Also, the disconnection of U1\U2will giveC0(U12!Z) two distinct compo- nents. Then, C0(N(U)) =C0(U1!Z)C0(U2!Z)=ZZ, andC1(N(U)) =C0(U12! Z) =C0(U+ 12!Z)C0(U 12!Z)=ZZ. Dene1:C1(N(U))!C0(N(U)), 0:C0(N(U))!C1(N(U)), and1:C1(N(U))!C2(N(U)). The maps qwill all be 0 map since Cq(N(U)) = 08q2. Note, for any continuous map f2C0(N(U)) has two distinct components with f= (f1;f2). Thus,0f=0(f1;f2) =f1jU12f2jU12, so Ker0=f(f1;f2)2C0(N(U))j0(f1;f2) =f1jU12f2jU12= 0g=f(f1;f2)2C0(N(U))jf1jU12= f2jU12g=f(f1;f2)2C0(N(U))jf1=f2g=(ZZ)=Z=Zsincef1andf2are both entirely constant map. And, Im1= 0 sinceC1(N(U)) = 0. Thus, H0(S1;Z) =Ker0=Im1= ff1;f2jf1=f2g=ZZ=Z=Z.Ker1=fg= (g1;g2)2C1(N(U))j1(g1;g2) = 0g= 0 sinceC2(N(U)) is 0. So, Ker1=C1(N(U)) which generates by ( g1;g2), thus, iso- morphic to ZZ, whereg1generatesC0(U+ 12andg2generatesC0(U 12. And Im(0) = fg2C1(N(U))jg=0f= (f1;f2) =f1jU12f2jU12=g1+g2g=Z. Thus, H1(S1;Z) = Ker1=Im0=fg1;g2g=(g1+g2)=ZZ=Z=Z. Therefore, Hq(S1;Z) =8 >< >:Zforq= 0;1 0 forq2. As the 1-dimensional sphere can be constructed using CW-complex, its Cech cohomology is the same as its singular cohomology. 39 Bibliography [1] Michael Hutchings. Introduction to higher homotopy groups and obstruction theory, 2011. URL http://math.berkeley.edu/ ~hutching/teach/215b-2011/homotopy. pdf. [2] Stephen Mitchell. Notes on principal bundles and classifying spaces, 2011. URL http: //www.math.washington.edu/ ~mitchell/Notes/prin.pdf . [3] Shoshichi Kobayashi and Katsumi Nomizu. Foundations of dierential geometry. Vol. I . Wiley Classics Library. John Wiley & Sons, Inc., New York, 1996. ISBN 0-471-15733-3. Reprint of the 1963 original, A Wiley-Interscience Publication. [4] Ralph Abraham and Jerrold E. Marsden. Foundations of mechanics . Ben- jamin/Cummings Publishing Co., Inc., Advanced Book Program, Reading, Mass., 1978. ISBN 0-8053-0102-X. Second edition, revised and enlarged, With the assistance of Tudor Rat iu and Richard Cushman. [5] Bertrand Guillou. Class notes math 527 (spring 2011) week 6, 2011. URL http://www. ms.uky.edu/ ~guillou/S11/527-Week6.pdf . [6] Norman Steenrod. The Topology of Fibre Bundles . Princeton Mathematical Series, vol. 14. Princeton University Press, Princeton, N. J., 1951. [7] John W. Milnor and James Stashe. Kharakteristicheskie klassy . \Mir", Moscow, 1979. Translated from the English by M. A. Stanko. 40