It is well known, and not difficult to prove that a vector bundle $E$ over a (smooth) manifold $M$ together with a metric gives rise to orthonormal frames (by Gram-Schmidt). An consequnece is that the transition maps $g_{ab}:U_a\cap U_b\rightarrow O(n)$ gives orthogonal matrices since it maps an orthonormal basis to an orthonormal basis.
But now i want to prove the converse: given an cover $(U_a)$ of $M$ and cocycles $g_{ab}:U_a\cap U_b\rightarrow O(n)$ then the vector bundle obtained by the cocycle construction gives a natural metric.
I know how the construction of the vectorbundle works, but how to bring a metric into this? Someone a solution for this problem? Thank you.
With Each $U_a$ you associate the trivial vector bundle $E_a=U_a\times \mathbb R^n$, with the standard metric on its fibers. Then you define $E$ as the disjoint union of these $E_a$'s, modulo de equivalence relation that identifies $(x,v)\in E_b$ with $(x,g_{ab}(x)v)\in E_a$. Since the $g_{ab}$ take values in $O_n$, the metrics on the $E_a$'s pass to the quotient, thus defining a global metric on $E$.