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Symplectic form on a 2sphere 
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#1
Nov910, 04:12 AM

P: 3

Hi,
The 2sphere is given as example of symplectic manifolds, with a symplectic form [tex]\Omega = \sin{\varphi} d \varphi \wedge d \theta[/tex]. Here the parametrization is given by [tex](x,y,z) = (\cos{\theta}\sin{\varphi}, \sin{\theta}\sin{\varphi}, \cos{\varphi})[/tex] with [tex] \varphi \in [0,\pi],\ \theta \in [0, 2\pi) [/tex]. Now my question is, at the points [tex]\varphi = 0, \pi[/tex], which are the north and the south pole, is the oneform [tex]d \theta[/tex] welldefined? If yes, how? If not then how does one make [tex]\Omega[/tex] globally well defined? Thanks in advance :) Ram. 


#2
Nov910, 05:44 AM

P: 1,411

The parametrization itself is not well defined at the north pole. You will need at least two charts to cover the 2sphere unambigously. Thus it is simpler to consider [tex]S^2[/tex] as embedded in [tex]\mathbf{R}^3[/tex] and define
[tex]\omega_u(v,w)=\langle u,v\times w\rangle[/tex] whre [tex]u\in S^2[/tex] and [tex]v,w\in T_u S^2.[/tex] Then you can show that this expression, in coordinates, is identical to the one you are given. 


#3
Nov910, 07:25 AM

P: 3

Thanks a lot for the reply. Now I understand what makes [tex] S^2 [/tex] a symplectic manifold.
However, the parametrization not being well defined does not necessarily lead to the oneform not being well defined, does it? For example the usual parametrization [tex] \theta [/tex] on [tex] S^1 [/tex] is not well defined globally, however [tex] d \theta [/tex] is. Something else happening with [tex] S^2 [/tex]? 


#4
Nov910, 07:49 AM

P: 1,411

Symplectic form on a 2sphere
Yes, you can wind R onto the circle but you can't wind torus (product of two circles) onto the sphere.



#5
Nov910, 09:40 AM

Sci Advisor
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P: 4,771

In the case of S^2, the area form [itex]\sin{\varphi} d \varphi \wedge d \theta[/itex] is not defined on a whole "halfslice" of S^2. Show that it can be patched in a unique way to give a globally defined 2form on S^2, so that talking about "the area form [itex]\sin{\varphi} d \varphi \wedge d \theta[/itex] on S^2" is not ambiguous. 


#6
Nov1010, 05:38 AM

P: 3

Thank you.



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