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Constructing a chart with coord. basis equal to given basis at one pt.

  1. May 19, 2014 #1
    Suppose we have a manifold ##M## and at ##p \in M## we have a basis for the tangent space of vectors ##X_i##. Since ##M## is a manifold, there exists a local chart ##(U,\phi)## about ##p##. Now the question is, given such a chart , how can we construct a new chart in a such that ##X_i = \left. \frac{\partial}{\partial x^\mu} \right|_{p}##.

    I know there is a theorem that says given commuting vector fields we can find a chart such that these vector fields locally is the coordinate basis of that chart.

    However, I want to prove in a transparent manner the less general statement above; that we can construct a coordinate system such that the vector ##X_i## _in the tangent space at p_ is the coordinate basis ##\left. \tfrac{\partial}{\partial x^\mu} \right|_{p}## just at p.
     
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  3. May 19, 2014 #2

    Ben Niehoff

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    Use the standard construction for Riemann normal coordinates, but forget the fact that the basis is orthonormal.
     
  4. May 19, 2014 #3

    pasmith

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    Let [itex](U,\phi)[/itex] be a chart on [itex]M[/itex] whose domain contains [itex]p[/itex], and consider the chart [itex](U, A \circ \phi)[/itex] where [itex]A : \mathbb{R}^n \to \mathbb{R}^n[/itex] is linear and invertible. This chart is smoothly compatible with the original chart [itex](U,\phi)[/itex].

    The transition function [itex]A[/itex] then pushes forward to a linear map [itex]A_{*} : T_pM \to T_pM[/itex], giving [tex]
    \left.\frac{\partial}{\partial x^i}\right|_p = \frac{\partial \tilde x^j}{\partial x^i}
    \left.\frac{\partial}{\partial \tilde x^j}\right|_p =
    (A_{*})_i{}^j \left.\frac{\partial}{\partial \tilde x^j}\right|_p[/tex] where [itex](x^i)[/itex] are the coordinate functions of the chart [itex](U,\phi)[/itex] and [itex](\tilde x^j)[/itex] are those of the chart [itex](U, A \circ \phi)[/itex]. Since [itex]\tilde x^j = A^j{}_i x^i[/itex] we have that [tex]\frac{\partial \tilde x^j}{\partial x^i} = (A_{*})_i{}^j = A^j{}_i.[/tex] If [itex]{X_j}[/itex] is a basis for [itex]T_pM[/itex] we may then set [tex]
    \left.\frac{\partial}{\partial \tilde x^j}\right|_p = X_j[/tex] in the above to obtain [tex]
    \left.\frac{\partial}{\partial x^i}\right|_p = A^j{}_i X_j.[/tex]
     
    Last edited: May 19, 2014
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