This is the set up of my problem:
Let $M$ be a manifold with local coordinates $x^1,\dots, x^n$. Let $x^1,\dots,x^n,\xi_1,\dots,\xi_n$ denote the induced coordinates on $T^\ast M$. Let $f:M\to M$ be a diffeomorphism and consider the induced map
$$F:T^\ast M \to T^\ast M \ , \ (p,\xi)\mapsto (f(p),(f^\ast)^{-1}(\xi)).$$ This is the inverse of the pullback map. We get a new local coordinate system on $T^\ast M$ given by $\widetilde x^i=f(x^i)$ and $\widetilde\xi_i=(f^\ast)^{-1}_x(\xi_i)$
I know the chain rule gives $$\frac{\partial}{\partial x_i}=\frac{\partial \widetilde x^j}{\partial x^i}\frac{\partial}{\partial\widetilde x^j}.$$For simplicity let $x=(x^1,\dots,x^n)$ and $\xi=(\xi_1,\dots,\xi_n)$. I know that if the $\xi$'s were coordinates on the tangent bundle we would have $$\xi_i=\frac{\partial\widetilde x^j}{\partial x^i}\widetilde\xi_j$$ but I am not sure if this is true as coordinates on the cotangent bundle.
Anyway, I was told that since $(f^\ast)^{-1}(x,\xi)=\widetilde\xi$ it follows that $$\widetilde\xi_i=\frac{\partial x^j}{\partial \widetilde x^i}\xi_j$$I am hoping that someone can explain it to me. I know it has something to do with the inverse transpose of the change of basis matrix. That is, working with induced coordinates $x^1,\dots,x^n,v^1,\dots,v^n$ on the tangent bundle we get the formula $$v^j=\frac{\partial\widetilde x^i}{\partial x^j}\widetilde v^i$$so that $$\frac{\partial \widetilde v^j}{\partial v^i}=\frac{\partial \widetilde x^j}{\partial x^i}$$and we are flipping the roles of $i,j$ and the tildes.
Best Answer
Let $f:\mathbb{R}^n\to\mathbb{R}^n$ be a smooth function. Let $\xi\in T_{f(0)}^*\mathbb{R}^n$ a cotangent vector, and say $\xi=(\xi_1,\ldots,\xi_n)$, when the coordinates are with respect to the dual of the standard basis i.e. $\xi=\sum \xi_idx^i$. By definition of the pullback, $$f^*\xi=\xi\circ df,$$ and if we put $f^*\xi=u\in T_0^*\mathbb{R}^n$ we have $$\xi\circ df=u,$$ and so $$\xi=u\circ df^{-1}.$$ If we want to express the coordinates $\xi_1,\ldots,\xi_n$, we let $J$ denote the derivative matrix of $f$ and get $$(\xi_1,\ldots,\xi_n)=(u_1,\ldots,u_n)J^{-1},$$ or rather $$\left(\begin{array}{c}\xi_1\\ \vdots\\ \xi_n\end{array}\right)={J^{-1}}^T\left(\begin{array}{c}u_1\\\vdots\\u_n\end{array}\right).$$