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By Werner Hildbert Greub

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Extra resources for Connections, Curvature, and Cohomology: Cohomology of principal bundles and homogeneous spaces

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We can even limit our search to tangent bundles. Thus we ask the following question. Is it true that for every smooth manifold M the tangent bundle T M is trivial (as a vector bundle)? Let us look at some positive examples. 45. T S 1 ∼ = RS 1 Let θ denote the angular coordinate on the circle. Then ∂ is a globally defined, nowhere vanishing vector field on S 1 . We thus get a map ∂θ RS 1 → T S 1 , (s, θ) → (s ∂ , θ) ∈ Tθ S 1 ∂θ which is easily seen to be a bundle isomorphism. Let us carefully analyze this example.

A bundle isomorr phism E → KM is called a trivialization of E, while an isomorphism Kr → E is called a framing of E. A pair (trivial vector bundle, trivialization) is called a trivialized, or framed bundle. 44. Let us explain why we refer to a bundle isomorphism ϕ : KrM → E as a framing. Denote by (e1 , . . , er ) the canonical basis of Kr . , as (special) sections of KrM . The isomorphism ϕ determines sections fi = ϕ(ei ) of E with the property that for every x ∈ M the collection (f1 (x), . .

Often, when the field of scalars is clear from the context, we will use the simpler notation E ⊗ F . The tensor product has the following universality property. 1. For any bilinear map φ : E × F → G there exists a unique linear map Φ : E ⊗ F → G such that the diagram below is commutative. ι '' w E ⊗ F '' Φ ') u φ E×F . G The proof of this result is left to the reader as an exercise. Note that if (ei ) is a basis of E, and (fj ) is a basis of F , then (ei ⊗ fj ) is a basis of E ⊗ F , and therefore dimK E ⊗K F = (dimK E) · (dimK F ).

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