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

1. Oct 16, 2016

### spaghetti3451

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. Oct 17, 2016

### Orodruin

Staff Emeritus
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.