I am trying to solve the following exercise of Barden's book:
You will need this other exercise to solve the first one.
I think I managed to do the first one ( (i) ) correctly. Now, about the $\theta^{-1}*d\theta(X)$ part of the second exercise( (ii) ): I managed to get the result that I know is correct - $\theta^{-1}*d\theta(X)=\nabla\times X$. However, I do not know how to justify some of the steps of my proof: I accepted them temporarily just so that I could get to the result. Here is the outline of my proof, along with the questions that arose:
$$\theta(X)=\theta(a^ie_i)\overset{1}{=}a^i\theta(e_i)\overset{2}{=}a^idx_i\,(\in \Omega^1(U))$$ Now $$d\theta(X)=d(a^idx_i)=...=da^idx_i$$ So $$*dv(X)=*(da^idx_i)=...=\Big( \frac{\partial a^3}{\partial x_1} - \frac{\partial a^2}{\partial x_3}\Big)dx_1 + ... + \Big( \frac{\partial a^2}{\partial x_1} - \frac{\partial a^1}{\partial x_2}\Big)dx_3$$ Hence $$\theta^{-1}(*dv(X))\overset{3}{=} \Big( \frac{\partial a^3}{\partial x_1} - \frac{\partial a^2}{\partial x_3}\Big)e_1 + ... + \Big( \frac{\partial a^2}{\partial x_1} - \frac{\partial a^1}{\partial x_2}\Big)e_3=\nabla\times X$$
Questions (referent to the numbered equality signs):
1) I know that $\theta$ is an isomorphism and hence linear. But what does it mean when the "vector space" it acts on is $\mathcal{X}(U)$? As far as I understand, $\mathcal{X}(U)$ is not really a vector space, but a module, the underlying ring being $\mathcal{C}^\infty(U)$... So does linearity of $\theta$ mean that $\theta(f\omega_1+g\omega_2)=f\theta(\omega_1)+g\theta(\omega_2)$? (I am not very familiar with modules).
2) Is this valid? I do not see why, though it seems natural.
3) Again, I use that $\theta(e_i)=dx_i$, so answering 2 probably answers this one.