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[edit] Vector bundles and covariant derivatives
I don't understand this section. I'm afraid that there is some mismatch in the usage of e. If e ∈ E, then it makes no sense to multiply it or translate by it. In other hand, if e is an element of the fiber, then the multipication and addition is defined on the fiber only and not on E. More precisely, these operations on Ex (as a subset of E) depend on the local trivialization of the bundle. —Preceding unsigned comment added by 86.101.197.154 (talk) 07:03, 20 February 2008 (UTC)
- The fibers in a vector bundle can be given the structure of a vector space in natural fashion. This does not depend on the local trivializations as you suggest. This follows from the fact that the transition functions are linear isomorphisms (or that the structure group is GL(n)). Compare definitions 1 and 2 in the vector bundle article. It's a good exercise to show that these are equivalent. -- Fropuff (talk) 00:22, 21 February 2008 (UTC)
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- Ok, Take two local trivializations with transfer function L. Suppose that in the first trivialization e is mapped to (x,f) and in the second trivialization to (x,L(f)). Multiplying e by λ in the trivializations yield (x,λ f) and (x, λ L(f)) = (x, L(λ f)). These are really the images of the same element in the given two trivializations.
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- Take another point p in this fiber. The image of it in the trivializations are (x,g) and (x,L(g)) respectively. Shifing p by e in the first trivialization yields (x,g+f) while in the second one (x,L(g)+L(f)) = (x,L(g+f)). The result is the same again. You are right, thank you. —Preceding unsigned comment added by 86.101.197.154 (talk) 06:25, 21 February 2008 (UTC)
