why normal coordinate is good guy ? [closed]

We know on $\nabla$ in Riemannian geometry by changing the coordinate system we get the different results. In Riemannian geometry for simplifying local calculations, we use of normal coordinates because Christoffel symbols of the connection vanishes. My question is where we can use of normal coordinates in Riemannian manifolds. Moreover, why normal coordinate is good guy ?

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This is erring on the side of incomprehensible due to your grammar and diction. – Chris Gerig Jun 24 at 18:44
If you can, please try to rephrase your question. If French is better for you, you could probably ask your question in French on the meta and someone (e.g. me) could translate? – David Corwin Jun 24 at 18:57
Thanks, for the comments, I try to revise my question. – Hassan Jolany Jun 24 at 20:13
I agree it is incomprehensible, but it also hints at the OP's confusion. You certainly do not get different results. A clever / convenient choice of coordinate system may vastly simplify some proofs, but the results are always the same, that's the whole point of coordinate invariance. – Spiro Karigiannis Jun 24 at 20:48

1 Answer

Any tensor can be written in a basis of vector fields and covector fields on a small neighborhood. The two most convenient such frames are an orthonormal basis or a coordinate basis $\partial/\partial x^i.$

The first has the advantage that inner product terms vanish, but it is not necessarily the case that mixed derivatives agree in reversed order, that is $\nabla_{e_i} e_j \neq \nabla_{e_j} e_i$ because $\nabla_{e_i} e_j - \nabla_{e_j} e_i = [e_i, e_j].$

The second has the advantage that the mixed derivatives work, but inner products are not so good, $\langle \partial/\partial x^i, \partial/\partial x^j \rangle = g_{ij}.$

If you have some very complicated tensor relationship to work out, the strength of tensors is that they occur in linear spaces that can be evaluated pointwise. That is, we can demand normal coordinates around some point $p$ of the manifold. At the point $p,$ both $g_{ij}$ is the identity matrix and all $[e_i, e_j] = 0.$ Everything gets easier at that point.

About to be closed...tensors probably began as objects in physics requiring a number of subscripts, and something with a bunch of subscripts/superscripts was called a tensor if it changed in the proper way when switching to a new coordinate system. And one ought to go through a number of such calculations to have a ground level feel for the concept. Now we also have these vector bundles where a good deal of derivative information is encoded, and these take pointwise values. Probably enough.

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