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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...

2PiOmega
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1 Answers1

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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.

Bjorn
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  • I'm not sure to understand : $\theta$ map a point $p$ of $S^1$ to multiple value of x in $\mathbb{R}$ so it's not a 1-form. But $cos(\theta)$ is the same for all the $\theta+2\pi$ and we can associate one unique real $\cos(\theta)$ to each point of $S^1$? – 2PiOmega Apr 08 '21 at 12:59
  • What is the value of $\theta$ at the poles? – Zippy Apr 08 '21 at 17:01
  • @Zippy 0 and $\pi$ I think? – 2PiOmega Apr 08 '21 at 20:07
  • @2PiOmega You are correct, and the reason I gave in my answer was incorrect. I have since edited my answer. Let me know if you still find mistakes in it. – Bjorn Apr 08 '21 at 22:38