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Math Help - Differential Geometry Bias Question

  1. #1
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    Differential Geometry Bias Question

    I started studying Riemannian geometry a day ago and in my book it says you form the following bias coordinates in the tangent space.

    e_i = \dfrac{\partial}{\partial x_i}

    And if you have a vector w(w_1,w_2,\ldots,w_n)

    w=w_ie_i

    What I don't understand is how can you make an operator in biasing coordinates. Does the above imply that w=w_ie_i=\dfrac{\partial w_i}{\partial x_i} i.e. v\dfrac{\partial}{\partial x}\equiv\dfrac{\partial v}{\partial x}
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  2. #2
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    I believe that is the Einstein's convention of writing sum. w=\sum_{i} {w^i * e_i }
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    Quote Originally Posted by xxp9 View Post
    I believe that is the Einstein's convention of writing sum. w=\sum_{i} {w^i * e_i }
    Yeah, I know that. What I don't understand is how an operator as a basis works. If you have a surface in \mathcal{R}^n the tangent space is \sum_i^n \dfrac{\partial f}{\partial x_i} and you can use that as a basis.

    What I don't understand is how can you use \dfrac{\partial}{\partial x^i} as a basis, which is different than \dfrac{\partial f}{\partial x^i}
    Last edited by fobos3; July 6th 2010 at 02:24 PM.
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  4. #4
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    How are you defining the tangent space?

    If the tangent space at p is as an (equivalence class of) curve(s) \gamma,  \gamma(0)=p, which act on a function f:M\to\mathbb{R} as df(\gamma(t))\over dt, then you can chuse a coordinate chart \phi on M and rewrite {df(\gamma(t))\over dt} = {df(\phi^{-1}(\phi(\gamma(t))))\over dt} = {\partial f(\phi^{-1})\over \partial x^i}{d\phi^i(\gamma)\over dt} = \left( {d\phi^i(\gamma)\over dt}{\partial\over \partial x^i} \right) f where  \partial\over\partial x^i acts on a function as \partial f(\phi^{-1})\over\partial x^i. Therefore the set of these is spanning and simple examples show it is linearly independant...
    Last edited by maddas; July 6th 2010 at 05:03 PM.
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