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Consider the sphere $S^2\subset\mathbb{R}^3$ in cylindrical coordinates $(\theta, z)$ (away from poles $z=\pm 1$) with symplectic structure $\omega=d\theta\wedge dz$. I want to show that the vector field $X=\partial_\theta$ is hamiltonian. That is, the contraction of $\omega$ by $X$, denoted $i_X\omega$ is exact.

One way is straightforward: since the first de Rham group of the sphere is trivial the 1-form $i_X\omega$, being closed, has to be exact.

Question 1 -- why is $i_X\omega$ closed?

Question 2 -- Suppose I don't want to use my knowledge about de Rham groups and I want to show directly ("by hands") that $i_X\omega$ is exact. How to do this properly? Here is my work:

Since $X=\partial_\theta$ we get $$ i_X\omega(\partial_\theta)=\omega(X,\partial_\theta)=0 \\ i_X\omega(\partial_z)=\omega(X,\partial_z)=1 $$ Hence $i_X\omega=dz$ and so it is exact, at least locally on $U=S^2\setminus\{z=\pm 1\}$.

Now, to show the global exactness I would have to find a suitable change of charts and check that the transformed $\omega$ is exact around z=1 and z=-1.

If all I said is correct, then what would be a suitable change of charts? Or, is there an easier way?

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up vote 3 down vote accepted

Away from the poles $z = \pm 1$, we have $$\iota_X \omega = \iota_{\partial_\theta} (d\theta \wedge dz) = (d\theta \wedge dz)(\partial_\theta, -) = dz$$ by the definition of the wedge product and $d\theta$, $dz$. Therefore $X$ has Hamiltonian function $$H(\theta, z) = z$$ away from $z = \pm 1$, as $$\iota_X \omega = dH.$$ Note that $\partial_\theta = 0$ at the poles $z = \pm 1$ and the poles are also critical points of the height function $H = z$, so we also have $$\iota_X \omega = dH$$ at $z = \pm 1$ (simply because $0 = 0$). Therefore $\partial_\theta$ is a Hamiltonian vector field on $S^2$ with Hamiltonian function $H = z$.

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Thanks for the great reply. I don't get why can we say that $\partial_\theta=0$ at the poles? –  Heitor Fontana Feb 2 '13 at 10:10
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@boib: $\partial_\theta$ is a smooth global vector field tangent to each circle $z = \text{const}$. At $z = \pm 1$ these circles actually degenerate to single points, so you can see that $\partial_\theta$ has to be zero at these points. If you actually compute $\partial_\theta$, you will see that it has constant magnitude on each circle $z = \text{const}$, and the magnitude becomes smaller as we go to circles with larger $|z|$, until the magnitude becomes zero at $z = \pm 1$. –  Henry T. Horton Feb 2 '13 at 19:24
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