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Symplectic form on a 2-sphere

  1. Nov 9, 2010 #1

    The 2-sphere 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 one-form [tex]d \theta[/tex] well-defined? If yes, how? If not then how does one make [tex]\Omega[/tex] globally well defined?

    Thanks in advance :)

  2. jcsd
  3. Nov 9, 2010 #2
    The parametrization itself is not well defined at the north pole. You will need at least two charts to cover the 2-sphere 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.
  4. Nov 9, 2010 #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 one-form 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]?
  5. Nov 9, 2010 #4
    Yes, you can wind R onto the circle but you can't wind torus (product of two circles) onto the sphere.
  6. Nov 9, 2010 #5


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    In the case of S^1, people say that [itex]d\theta[/itex] is a 1-form. This is sloppy because [itex]\theta[/itex] is not defined at one point of S^1 (usually (-1,0)), but it is to be interpreted as such: "There is a globally defined 1-form [itex]\alpha[/itex] on S^1 such that with respect to the usual angle parametrization [itex]\theta[/itex], [itex]\alpha=d\theta[/itex] everywhere where [itex]\theta[/itex] is defined." And indeed, if you take the chart of S^1 that covers the whole of S^1 except (1,0) and associates to a point its angle [itex]\theta'[/itex], where the point (-1,0) is considered to have angle [itex]\theta'=0[/itex], you will find that [tex] d \theta = d\theta'[/tex] everywhere where both these 1-forms are defined. And in particular, there is only one way to patch the local 1-form [itex]d\theta[/itex] at (-1,0) to make it a global 1-form and that is to set it equal to [itex]d\theta'[/itex] at that point.

    In the case of S^2, the area form [itex]\sin{\varphi} d \varphi \wedge d \theta[/itex] is not defined on a whole "half-slice" of S^2. Show that it can be patched in a unique way to give a globally defined 2-form 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.
  7. Nov 10, 2010 #6
    Thank you.
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