Let $f: M \rightarrow \mathbb{R}$ be a smooth function on a manifold $M$. Then $df:M\rightarrow TM^*$ denotes a cut on the cotangent bundle. Then $df$ with $p \in M$ is given by $$ df(p) = \sum_{j=1}^{n}df(p)\left(\frac{\partial}{\partial x_j}\bigg \vert_p\right)dx_j(p) = \sum_{j=1}^{n}\frac{\partial}{\partial x_j}\bigg\vert_p(f_p)dx_j(p) $$

I don't get the point where this $dx_j(p)$ comes from. Is this because of the chart of the manifold? Because for the map $f: M \rightarrow \mathbb{R}$, $[\gamma]_p$ beeing the germ in $p$, $df(p)([\gamma]_p) = (Id_M \circ f\circ\gamma)'(0)=(f\circ\gamma)'(0)$ is given for all $[\gamma]_p \in T^{geo}_pM$. Could anybody help me to derive the form in local coordinates? Maybe an illustrative explanation would help (as far as this seems possible).


If we fix local coordinates $(U, (x^i))$ on an $n$-manifold, then $(dx^a)$ is a local coframe of $TM$ and in particular for any $p \in U$, $(dx^a(p))$ is a basis of $T^*_p M$, so we have $$df(p) = \sum \lambda_a dx^a(p)$$ for some coefficients $\lambda_a$. Since $(dx^a(p))$ is dual to the basis $\left(\left.\frac{\partial}{\partial x^a}\right\vert_p\right)$, evaluating both sides at $\left.\frac{\partial}{\partial x^b}\right\vert_p$ gives $$\lambda_b = df(p)\left(\left.\frac{\partial}{\partial x^b}\right\vert_p\right) .$$ Substituting for $\lambda_b$ recovers the first equality.

  • $\begingroup$ Thank you, Travis! $\endgroup$ – John Smith Feb 18 at 20:03
  • $\begingroup$ You're welcome! I hope you found it useful. $\endgroup$ – Travis Willse Feb 18 at 20:09
  • $\begingroup$ Yes. I got the clue after work and you just confirmed my guess. :-) $\endgroup$ – John Smith Feb 18 at 20:12

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