The problem:
Demonstrate that If $\xi^n=(E,\Pi,M)$ is a trival vector bundle with a riemannian metric, then there exist an isomorphism $\psi:E\to$ M x $R^n$ Such that $\psi $ is an isometry on each fiber.
What I have got:
Since $\xi^n$ is a trivial bundle I have an isomorphism $\psi:E\to$ M x $R^n$.
The Riemannian metric gives me an inner product on each fiber. So i can define a norm and find with Gram-Schmidt an ortonormal basis for each fibre, And construct an isometry with $\psi$ and a coordinate transformation. But i dont underestand why it would result in a vector bundle isomorphism.
I would appreciate a lot any help!
The point is that the Gram-Schmidt process is continuous (it is even smooth if this makes sense). Your vector bundle isomorphism will give you $n$ sections that form a basis of $E_x$ for each $x\in M$. If you apply the Gram-Schmidt process to this you will get $n$ smooth sections that are orthonormal at each point $x\in M$. This gives an isometry with the trivial bundle with the standard inner product.