# Intuition behind differentials on smooth manifold

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).

## 1 Answer

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.

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