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Thread: 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.

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

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

    $\displaystyle w=w_ie_i$

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

    What I don't understand is how can you use $\displaystyle \dfrac{\partial}{\partial x^i}$ as a basis, which is different than $\displaystyle \dfrac{\partial f}{\partial x^i}$
    Last edited by fobos3; Jul 6th 2010 at 01:24 PM.
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  4. #4
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    How are you defining the tangent space?

    If the tangent space at $\displaystyle p$ is as an (equivalence class of) curve(s) $\displaystyle \gamma$,$\displaystyle \gamma(0)=p$, which act on a function $\displaystyle f:M\to\mathbb{R}$ as $\displaystyle df(\gamma(t))\over dt$, then you can chuse a coordinate chart $\displaystyle \phi$ on $\displaystyle M$ and rewrite $\displaystyle {df(\gamma(t))\over dt} = {df(\phi^{-1}(\phi(\gamma(t))))\over dt} = {\partial f(\phi^{-1})\over \partial x^i}{d\phi^i(\gamma)\over dt}$ $\displaystyle = \left( {d\phi^i(\gamma)\over dt}{\partial\over \partial x^i} \right) f$ where$\displaystyle \partial\over\partial x^i$ acts on a function as $\displaystyle \partial f(\phi^{-1})\over\partial x^i$. Therefore the set of these is spanning and simple examples show it is linearly independant...
    Last edited by maddas; Jul 6th 2010 at 04:03 PM.
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