Let $\dfrac{xdy-ydx}{x^2+y^2}$ be a 1-form defined in $\mathbb{R}^2\backslash\{0\}$.
Where can I find a detailed proof that it is not exact? I would prefer a proof that doesn't use results about conservative vector fields (I know that way). I have searched Jefree Lee, Do Carmo & some others, but they all skip this part. Thanks a lot.
[Math] Classic example of a non exact form
differential-formsdifferential-geometry
Best Answer
I'm not sure whether this is acceptable or not, but the simplest thing to do (since the form is closed) is to attempt to integrate it to find supposed $\phi$ such that your form $\omega$ is $\mathrm d \phi$.
One can derive a contradiction most straightforwardly by going around a circle, say $x^2+y^2=1$. We use the path $x(t)=\cos t,y(t)=\sin t$. $$\int_0^{2\pi} \omega = \int_0^{2\pi} \frac{\cos^2 t +\sin^2 t}{1}\mathrm d t = 2\pi$$
But if $\omega=\mathrm d \phi$ then the expression on the left is $\phi((1,0))-\phi((1,0))=0$, a contradiction.
In response to your complaints about this method: The definition of exact is that $\omega=\mathrm d \phi$ for some $\phi$. The only result we've used is that integrals of derivatives are given by boundary values (a very mini version of Stokes's theorem), i.e. $$\int_{\mathbf x}^{\mathbf y} \mathrm d \phi = \left[\int_{t_x}^{t_y} \frac{\mathrm d \phi}{\mathrm d t} \mathrm d t = \right]\phi({\mathbf y})-\phi({\mathbf x})$$ which I don't think is really a mysterious result. If there's something you're not happy with do just say.