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A Action of a vector on the pull-back of a function

  1. Oct 16, 2016 #1
    Consider how a vector ##\bf{v}##at ##x## of ##M##, as a differential operator, acts on the pull-back of a function:

    ##{\bf{v}}(F^{*}f) = (F_{*}{\bf{v}})(f) = df(F_{*}{\bf{v}})##

    I was wondering how to relate this to the following?

    ##\displaystyle{{\bf{v}}(F^{*}f)={\bf{v}}[f\{y(x)\}]=v^{i}\frac{\partial}{\partial x^{i}}[f\{y(x)\}]=v^{i}\bigg(\frac{\partial y^{j}}{\partial x^{i}}\bigg)\bigg(\frac{\partial f}{\partial y^{j}}\bigg)}##

    I ask this question because ##\displaystyle{F_{*}{\bf{v}}=\bigg(\frac{\partial y^{j}}{\partial x^{i}}\bigg)v^{i}}##, with the implicit Einstein summation convention,

    so that ##\displaystyle{(F_{*}{\bf{v}})(f)=v^{i}\bigg(\frac{\partial y^{j}}{\partial x^{i}}\bigg)f \neq v^{i}\bigg(\frac{\partial y^{j}}{\partial x^{i}}\bigg)\bigg(\frac{\partial f}{\partial y^{j}}\bigg)}##
     
    Last edited: Oct 16, 2016
  2. jcsd
  3. Oct 17, 2016 #2

    Orodruin

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    The first step of your final line is not how a vector acts on a function. In general, if ##V## is a vector in the manifold ##F## maps to, then ##V(f) = V^i\partial_i f = df(V)## by definition.
     
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