Suppose that $M$ is a smooth manifold, $\alpha:I\to \mathbb{R}^n$ a representation of a curve on $M$ and $\phi:M\to \mathbb{R}^n=(x_1,...,x_n)$ a coordinate chart. So, $\gamma(t)=\phi^{-1}\circ\alpha(t)$ is the $\alpha$-th coordinate curve iff $x_\alpha(\gamma(t))=t+c_\alpha$ and $x_i(\gamma(t))=c_i$ where $c_i$'s are suitable constants.

NOW why it holds that for our $\gamma$ with $\dot{x}_\alpha=\frac{d\gamma}{dt}$ that

$\dot{x_\alpha}(\frac{\partial}{\partial x^i})=\delta^\alpha_i$

? I.e. why the vector $\frac{\partial}{\partial x^\alpha}$ is the velocity vector to the $\alpha$-th coordinate curve?

  • $\begingroup$ What is $\dot{x_{\alpha}}$ ? $\endgroup$ – Sou Nov 26 '17 at 14:22
  • $\begingroup$ @Sou燈馬想: The derivative of the $x_{\alpha}$ coordinate variable. $\endgroup$ – Faraad Armwood Nov 26 '17 at 14:22
  • $\begingroup$ How can $\dot{x}_{\alpha}$ act on $\partial_i$ ? I think it should act on functions. $\endgroup$ – Sou Nov 26 '17 at 14:26

As the user in the comments pointed out, we have only defined the action of tangent vectors on functions $f: M \to \mathbb{R}$. Observe how this doesn't fall into the category for evaluation,

$$(x^{\alpha})' \frac{\partial}{\partial x^i} = x^{\alpha}_* \underbrace{\left( \frac{d}{dt} \right) \frac{\partial}{\partial x^i}}_{\textbf{doesn't make sense}}$$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.