I'm currently reading Nakahara's Geometry, Topology and Physic and I didn't understand the exemple 6.7 about the $S^2$ sphere.
Exemle 6.7 Geometry, Topology and Physics
I didn't understand the last sentence, why $\Omega$ isn't an exact form sinc it's write as $\Omega = -d(\cos (\theta) d\phi)$ ?
My question is maybe stupid but I think I really miss something...
The problem here is that $\cos (\theta) d\phi$ is not a $1$-form on all of $S^2$.
Imagine the simpler example of the circle $S^1$ with coordinate $\theta$. We often write the standard 1-form as $d\theta$, however this it is not exact since $\theta$ is not a function. Every point $x\in S^1$ is "mapped" by $\theta$ to countable many values $x+2\pi n$. The notation, although standard, is misleading in this case.
What happends with $\cos \theta d\phi$ is that it is not defined at the poles. We can see this from writing it in cartesian coordinates as $$\cos \theta d\phi = z \frac{xdy-ydx}{x^2+y^2}$$.
Note however that the original 2-form is since both $d\theta$ and $$\sin \theta d\phi = \sqrt{x^2+y^2}\frac{xdy-ydx}{x^2+y^2} =\frac{xdy-ydx}{\sqrt{x^2+y^2}} $$ are defined everywhere.