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Download Lectures on fibre bundles and differential geometry by J.L. Koszul, S. Ramanan PDF

By J.L. Koszul, S. Ramanan

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We shall denote the space of projectable vector field by ℘. It is easy to see that if X, Y ∈ ℘, then X + Y ∈ ℘ and p(X + Y) = pX + pY. Moreover ℘ is a submodule of C (P) regarded as an U (V)-module (but not an U (P)-submodule). For f ∈ U (V) and X ∈ ℘, we have p((p∗ f )X) = f (pX). Thus p : ℘ → C (V) is an U (V)homomorphism and the kernel is just the module N of vector fields on P tangential to the fibre. Furthermore, for every X, Y ∈ ℘, we have [X, Y] ∈ ℘ and p[X, Y] = [pX, pY]. 44 Proposition 7.

We may assume that V is simply connected in which case Φ(ξ) = Φr (ξ), the general case being an easy consequence. Moreover, since it is enough to take d1 η, d2 η to be horizontal, we may consider the values of K on the manifold M(ξ). We may therefore restrict ourselves to the case when Φr (ξ) = G, and M(ξ) = P. Let I be the subalgebra of G generated by 4. Holonomy Groups 58 75 the values of K. Let L be the set of vector fields X on P such that γ(X) is a function with values in I . This is an U (P)- submodule.

H is defined by h(y, ξ) = ξ. By proposition (3). 3, the principal bundle Pq is trivial if and only if there exists a differentiable cross section for Pq over Y. , the diagram is commutative. P ??    p   Y q // X λ 33 We now assume that q is surjective and everywhere of rank = dim X. If Pq is trivial we shall say that P is trivialised by the map q. Let q be a differentiable map Y → X which trivialises P. Consider the subset Yq of Y × Y consisting of points (y, y′ ) such that q(y) = q(y′ ).

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