Appendix 1 Fiber Bundles Let E and B be two sets endowed with topologies. (These are also called topological spaces.) Let E be “larger” than B in some sense. Moreover, let there exist a continuous projection mapping Π : E → B. The triple (E,B,Π) is called a bundle. The topological space B is called the base space. (See the reference [5].) Example (A1.1). Let M and M # be two C0 -manifolds. The Cartesian product manifold M × M # can be defined. The bundle (M × M #,M,Π) is a product bundle or trivial bundle with the projection mapping Π(p, p#):=p for all p in M and all p# in M #. (See fig. A1.1). 2 M (p, p#) M x M# M# Π p M Figure A1.1: A product or trivial bundle. Example (A1.2). A circular cylinder can be constructed by glueing the opposite edges of a rectangular sheet of paper. Topologically speaking, the 257 258 Appendix 1. Fiber Bundles cylinder is the product manifold S1 ×I, where S1 is the closed circle and I is an open linear segment. The product bundle (S1 ×I,S1, Π) is obtained by defining the projection mapping Π(p, q):=p for p ∈ S1 and q ∈I. The usual chart on the circle is furnished by θ ≡ x = χ(p),θ∈ [−π,π]. The points θ = −π and θ = π must be identified. (See fig. A1.2) (Distinguish between the projection Π and the angle π !) Identify a a a S1x I (p,q) (p,q) Topologically b b b Π Π Identify p p BS1 χ −π π −π π x θ x Identify Identify Figure A1.2: The product bundle (S1 ×I,S1, Π). −1 Consider the special bundles such that the subset Fp := Π (p) (for some p ∈ B ) can be mapped in a continuous and one-to-one manner into a typical set F. (Such a mapping is called a homeomorphism.) In that case, Fp is called the fiber over p,andF is the typical fiber. It may happen that there exists a group G of homeomorphisms taking F into itself. Moreover, let the base space allow a covering of countable open subset Uj ’s. (See fig. A1.3). A fiber bundle (E,B,Π,G) is a bundle (E,B,Π) with a typical fiber F and the group of homeomorphisms G of F into itself and a countable open covering of B by Uj ’s. Moreover, the following assumptions must hold. I. Locally, the bundle is a product manifold. II. There exist homeomorphisms Hj and hj such that −1 Hj :Π(Uj) −→ Uj × F for every j. −1 hj :Π(Uj) −→ F for every j. Appendix 1. Fiber Bundles 259 F h1, p E ξ -1 h1, p h2 , p h2, p Π p B U1 U2 Figure A1.3: A fiber bundle (E,B,Π,G). Thus, Hj(ξ):=(Π(ξ),hj (ξ)) = (p, hj(ξ)) ∈ Uj × F. (Note that the re- striction hj,p := hj|F (p) is a homeomorphism from the fiber Fp into F.) −1 III. Let p ∈ Uj ∩ Uk. The mapping hj,p ◦ [hk,p] is a homeomorphism of F −1 into itself. Thus, hj,p ◦ [hk,p] constitutes an element of the structure group G. Remark: In the case of a trivial or product bundle, the group G contains only the identity element. Example (A1.3). A circular cylinder can be constructed by glueing (or identifying) the opposite edges of a rectangular sheet of paper. (See example A1.2.) Mathematically speaking, the circular cylinder is the product manifold S1 ×I, where S1 is a closed circle and I is an open linear segment. The product bundle (S1 ×I,S1, Π) is obtained by defining the projection mapping Π(ξ) ≡ Π(p, y):=p for p ∈ S1 and y ∈I. Let us choose the circle S1 of unit radius and the interval I := (−1, 1) ⊂ R. A possible coordinate chart for the unit circle S1 is furnished by the usual angle θ ≡ x = χ(p),x∈ [−π,π]. (Caution: Distinguish between the number π and the projection Π from the context!) An atlas for S1 is provided by the union U ∪ U, where D = χ(U):= {x : −π<x<π/2} and D = χ(U):={x :0< x< 3π/2}. In the intersection U ∩U, the corresponding coordinate transformation is furnished by x = x. (See fig. A1.4.) The intersection U∩U is the intersection of subsets corresponding to D∩D = V := {x :0<x<π/2}. Moreover, W := {x : −π<x<−π/2}⊂D and W4 := {x : π<x< 3π/2}⊂D. 260 Appendix 1. Fiber Bundles Identify Identify a c a c ξ ξ ζ b d Π d ΠΠb p q p χ U U χ −π π/ −π/ π/ π 3 2 2 o x = x 2 D D h( ξ ) ξ ζ ξ h( ξ ) I I H H H H (p,h( ξ ) ) (p,h( ξ ) ) (q, h( ζ ) ) (q, h( ζ ) ) Figure A1.4: Fiber bundle (S1 ×I,S1, Π,G) and various mappings. (Note that subsets W and W4 are topologically identified.) The areas corre- sponding to χ−1(V ) ×I are shaded. Moreover, the areas corresponding to χ−1(W ) ×I and χ−1(W4) ×I, respectively, are dotted. (See fig. A1.4.) By the definition of various mappings, we obtain that H(ξ)=(Π(ξ),h(ξ)) = (p, h(ξ)) , H(ξ)=(Π(ξ), h(ξ)) = (p, h(ξ)), for all p ∈ U and all p ∈ U. Similarly, we get H(ζ)=(Π(ζ),h(ζ)) = (q,h(ζ)) , H(ζ)=(Π(ζ), h(ζ)) = (q,h(ζ)), for all q ∈ U ∩ U. Since two coordinate charts coincide in U ∩ U, the “vertical coordinates” satisfy h(ζ)=h(ζ). For the other part, χ−1(W ) ×I and χ−1(W4) ×I inside U ×I and U ×I are identified. Therefore, the “vertical coordinates” again Appendix 1. Fiber Bundles 261 Identify Identify a c b d ξ h( ) ξ ζ I I ξ h( ξ ) c b d a H H H H (p,h( ξ ) ) ζ (p, h( ξ ) ) (q, h( ) ) (q, h( ζ ) ) U x I U x I Figure A1.5: The Moebius strip as a fiber bundle. satisfy h(ξ)=h(ξ). Thus, in the union U ∪ U, we always have h ◦ (h)−1 = h ◦ h−1 = e =: identity . Therefore, the structural group is given by the singleton set G = {e}. 2 Example (A1.4). Now we shall discuss the Moebius strip as a fiber bun- dle. It is obtained by twisting and glueing edges of a strip of paper. It cannot be described globally as a product manifold. However, locally it is a product of U ×I, where U ⊂ S1 and I is a line segment. We need some mechanism for twisting to occur. Let us choose the line interval I := (−1, 1) ⊂ R and S1 as the unit circle. We choose exactly the same chart as in the preceding example. (See figs A1.3, A1.4, and A1.5.) The intersection U ∩ U is the intersection of subsets corresponding to D ∩ D =: V = {x :0<x<π/2}. Moreover, W := {x : −π<x<−π/2}⊂D and W4 := {x : π<x< 3π/2}⊂D (as in the preceding example). From the definition of various mappings, we conclude that H(ζ)=(q,h(ζ)) for all ζ ∈ Π−1 ◦ χ−1(V )=Π−1 ◦ χ−1(V ). Since two charts match exactly in V , we must have identical “vertical” coordinates. Therefore, h(ζ)=h(ζ)andh ◦ h−1 = h ◦ h−1 = e = identity for all ζ ∈ Π−1 ◦ χ−1(V ). However, in the other parts, χ−1(W ) ×I in U ×I and χ−1(W4) ×I in U ×I have opposite orientations incorporating the twist. Therefore, the “vertical” coordinates satisfy h(ξ)=−h(ξ) 262 Appendix 1. Fiber Bundles for all ξ ∈ Π−1 ◦ χ−1(W ). Thus, in such cases h ◦ h−1 = h ◦ h−1 = g, g2 = e. Therefore, the structural group is furnished by G = {e, g}. (This group is isomorphic to the permutation or symmetric group S2.) 2 A bundle (E,M,Π,G) is said to be a Ck -differentiable fiber bundle provided EandM are Ck -differentiable manifolds, G is a Lie group, and a covering of M by a Ck -differentiable atlas exists. 1T Let us define the tangent bundle 0 (M) over the base manifold M. It is the space of ordered pairs (p, vp) for all p in M and all tangent vectors ∈ 1T vp Tp(M). It constitutes the fiber bundle ( 0 (M),M,Π,G) in the following manner. The fiber at p is Tp(M) and is isomorphic to the typical fiber F = V, the N -dimensional (real) vector space. Projection Π is defined by Π(p, vp)=p. Covering of M is by a typical atlas. The Lie group G = G(N,R) is the group of invertible mappings of N -dimensional real vector space V into itself. A cross section of a general bundle (E,B,Π) is a mapping σ : B → E such that Π ◦ σ = identity. A tangent (or contravariant) vector field V 1T on M is a cross section of the tangent bundle 0 (M). A tangent vector field V maps each point p in M into a tangent vector V (p) ∈ Tp(M)bytherule V : p −→ (p, V (p)) = (χ−1(x)), V (χ−1(x)). (See fig. A1.6.) rT In a similar manner, we can construct a tensor bundle ( s (M),M,Π) r −→ r −1 r −1 and a tensor field sT : p (p, sT(p)) = (χ (x)), sT(χ (x)).