# Prove $\omega=\lambda d\theta + dg$ for a 1-form $\omega$ on $\mathbb{R}^2\setminus0$ with $d\omega=0$.

I've recently been attempting to self-study Spivak's Calculus on Manifolds and have hit a snag in the section discussing integration on chains. My confusion centers around the nuance between $$f^{*}$$ and $$f_{*}$$. Spivak defines their relationship as $$f^{*}\omega\left( p \right)\left( v_{1}, \dots, v_{n} \right) = \omega \left( f \left( p \right) \right) \left( f_{*}\left( v_{1} \right), \dots, f_{*}\left( v_{n} \right) \right)$$. Here, $$f_{*}(p)(v) =\left(Df(p)(v)\right)_{f(p)}$$.

(Problem 4-30) If $$\omega$$ is a 1-form on $$\mathbb{R}^2 \setminus \left\{ 0 \right\}$$ s.t. $$d\omega = 0$$, prove $$\omega = \lambda d\theta + dg$$ for a $$\lambda \in \mathbb{R}$$ and $$g: \mathbb{R}^2 \to \mathbb{R}^2 \setminus \left\{ 0 \right\}$$.

Spivak provides a hint to equate $$c_{R,1}^*\left( w \right) = \lambda_{R}dx + d \left( g_{R} \right)$$ and show all $$\lambda_{R} = \lambda$$ for all possible $$\lambda_{R}$$.

I was considering an application of Stoke's Theorem $$\int_{\partial c} \omega = \int_{c} d\omega$$ as follows. I define $$c'=\{(x, y) \in \mathbb{R}^2:|(x, y)| < R\}$$

$$\int_{c_{R,1}} \omega = \int_{[0, 1]} c^*_{R,1} \left( \omega \right) = \int_{[0, 1]} \lambda_R dx + d \left( g_R \right) = \int_{c'}dw = 0$$

Do you all have suggestions for materials that could elaborate on k-forms, exterior calculus, dual spaces, $$f^*$$, and volume elements? I have purchased a copy of Munkres's Analysis on Manifolds, and are there any web resources defining the terms present in each text.

Best Regards.

• You might check out my YouTube lectures (linked in my profile). They — and my text — are far, far more concrete than Spivak's book, and I do not recommend dealing with the chains unless you're more advanced. ... So far as your question is concerned, you cannot apply Stokes's Theorem as you tried to because the origin is inside your $c'$. However, you can apply Stokes's Theorem to show that the integrals of $\omega$ over circles of different radii centered at the origin are all equal. Dec 23, 2020 at 6:46