Frankel: 1-form chain rule with pullback

Question

Could someone explain me what is happening in the first step please? Also what is the method to prove similar chain rule for a vector?

Answer 1

So here we can assume $F:\mathbb R^k\to \mathbb R^n$ is differentiable. Let me denote the evaluation af a 1-form on a 1-vector as a pairing: $\langle \alpha, v\rangle=\alpha(v)$. With such notation the pullback definition for 1-form rewrites as: $$ \langle F^*\alpha, v\rangle=\langle \alpha, F_*v\rangle $$ Here $F_*$ is the linear map which in canonical basis is represented by the jacobian matrix of $F$.

Therefore the step you're asking about is:

$$ \langle F^* dx_i, \partial_r\rangle=\langle dx_i, F_*(\partial _r)\rangle $$ and since $x=F(u)$ or rather, in the author notations which in my opinion are here a bit confusing $x=x(u)$ the right hand side can be exactly rewritten as above.

Answer 2

He should write $$\frac{\partial x^j}{\partial u^r}\partial_j,$$ for the linear combination inside the argument of $dx^i$.