# Symplectic form on a 2-sphere

1. Nov 9, 2010

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?

Ram.

2. Nov 9, 2010

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.

3. Nov 9, 2010

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 $$S^2$$?

4. Nov 9, 2010

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

5. Nov 9, 2010

### quasar987

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.

6. Nov 10, 2010