# Thread: Differential Geometry Bias Question

1. ## 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. $v\dfrac{\partial}{\partial x}\equiv\dfrac{\partial v}{\partial x}$

2. I believe that is the Einstein's convention of writing sum. $w=\sum_{i} {w^i * e_i }$

3. Originally Posted by xxp9
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 $\dfrac{\partial f}{\partial x^i}$

4. 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...