Is the One-Form d \theta Well-Defined at the North and South Pole on a 2-Sphere?

[rbayadi]

Hi,

The 2-sphere is given as example of symplectic manifolds, with a symplectic form \Omega = \sin{\varphi} d \varphi \wedge d \theta. Here the parametrization is given by (x,y,z) = (\cos{\theta}\sin{\varphi}, \sin{\theta}\sin{\varphi}, \cos{\varphi}) with \varphi \in [0,\pi],\ \theta \in [0, 2\pi).

Now my question is, at the points \varphi = 0, \pi, which are the north and the south pole, is the one-form d \theta well-defined? If yes, how? If not then how does one make \Omega globally well defined?

Thanks in advance :)

Ram.

 

[arkajad]

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 S^2 as embedded in \mathbf{R}^3 and define

\omega_u(v,w)=\langle u,v\times w\rangle

whre u\in S^2 and v,w\in T_u S^2.

Then you can show that this expression, in coordinates, is identical to the one you are given.

 

[rbayadi]

Thanks a lot for the reply. Now I understand what makes S^2 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 \theta on S^1 is not well defined globally, however d \theta is. Something else happening with S^2?

 

[arkajad]

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

 

[quasar987]

Thanks a lot for the reply. Now I understand what makes S^2 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 \theta on S^1 is not well defined globally, however d \theta is. Something else happening with S^2?

In the case of S^1, people say that d\theta is a 1-form. This is sloppy because \theta 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 \alpha on S^1 such that with respect to the usual angle parametrization \theta, \alpha=d\theta everywhere where \theta 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 \theta', where the point (-1,0) is considered to have angle \theta'=0, you will find that d \theta = d\theta' everywhere where both these 1-forms are defined. And in particular, there is only one way to patch the local 1-form d\theta at (-1,0) to make it a global 1-form and that is to set it equal to d\theta' at that point.

In the case of S^2, the area form \sin{\varphi} d \varphi \wedge d \theta 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 \sin{\varphi} d \varphi \wedge d \theta on S^2" is not ambiguous.

 

[rbayadi]

Thank you.