# Symplectic Form Preserved by Orthogonal Transformation

I'm trying to prove that the symplectic form

$$\omega = d(\cos\theta) \wedge d\phi$$

is preserved by the action of $SO(3)$ on $S^2$ where $\phi$ and $\theta$ are spherical polars. Now $SO(3)$ simply acts by

$$\theta \mapsto \theta + \epsilon, \ \phi \mapsto \phi + \delta$$

and writing this diffeomorphism as $F:S^2 \to S^2$ I compute

$$F^*(\omega) = d(\cos(\theta + \epsilon))\wedge d(\phi + \delta) = \cos(\epsilon) d(\cos\theta)\wedge d\phi - \sin(\epsilon) d(\sin \theta)\wedge d \phi$$

Is this correct? It doesn't seem like the form is invariant under $SO(3)$. Perhaps it's only meant to be a local symplectomorphism though.

Am I allowed to claim that it is a local symplectomorphism because it gives the right result in the limit as $\epsilon \to 0$? I think that would be right, because it would mean that the Lie derivative vanishes.

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How many Euler angles are there? Can you compute new $\theta$ and $\phi$ in terms of them? Can you represent these transformations as $\theta\mapsto\theta+\epsilon$, $\phi\mapsto\phi+\delta$?
Ah right - so I need another transformation because SO(3) is 3 dimensional. I can't work out what it is though. Is there some nice formula that you know? I know that I could solve this problem by finding the relevant vector fields and showing that they Lie derive $\omega$, but I'd prefer to do a direct solution if possible! Many thanks! –  Edward Hughes May 16 '13 at 21:54
For a direct solution, one should really express new $\theta$, $\phi$ in terms of old ones and three Euler angles. I guess the formulas will be rather cumbersome. The action you have written is not of $SO(3)$ but rather of $S^1\times S^1$, and it does not act nicely on $S^2$. –  O.L. May 16 '13 at 22:53
Here's a hint: Can you see that the standard area 2-form on $S^2$ is $SO(3)$-invariant?