Symplectic orthogonal I don't understand a detail in theses lectures notes about symplectic geometry (page 142). Let $(M,G,\mu, \omega)$ an Hamiltonian $G$-space and assume $0$ is a regular value of $\mu$. Then, we can see that $(T_p \mathcal O_p)^{\omega} = \ker \text{d}\mu_p = T_p{\mu^{-1}(0)}.$ Which means that $T_p \mathcal O_p$ and $T_p\mu^{-1}(0)$ are orthocomplement. In particular if the action of $G$ on the fiber is free then for $p \in \mu^{-1}(0)$, $T_p \mathcal O_p$ is isotropic.
I wanted to have some pictures in mind, so the only example I have is $S^2$, with action by rotation along $z$-axis and with symplectic form $\text{d}h \wedge \text{d}\theta$ which is not really good because in this case $T_p \mathcal O_p = T_p \mu^{-1}(0)$. I would like a case where the equality is strict for remember it better (and maybe understand a bit better what's going on : this formalism is a bit abstract for me).
 A: You wrote " In particular if the action of $G$ on the fiber is free then for $p∈μ^{−1}(0)$, $T_pO_p$ is isotropic ". I would like to point out that this isotropy of the orbit follows more from '$G$ acts on the fiber' than from 'the action is free', which is more closely related (although not quite) to $0$ being a regular value of $\mu$. This later fact nevertheless implicitly enters the proof of the former statement, through the equality $ker \, d\mu_p = T_p \mu^{-1}(0)$. I am not sure if this equality is necessary for the proof, but it certainly simplifies the matter.
Indeed, for a Hamiltonian action, the moment map $\mu$ is $G$-equivariant. It follows from this that the elements of $(T_pO_p) \cap (T_p \mu^{-1}(0))$ correspond to elements in (the Lie algebra of) the stabilizer subgroup of $\mu(p) = 0$, which is here the (Lie algebra of the) whole group $G$ as the $G$-action on the moment space $\mathfrak{g}^{\ast}$ is linear. (In particular, $G$ acts on $\mu^{-1}(0)$.) Hence $(T_pO_p) \cap (T_p \mu^{-1}(0))$ is all of $T_pO_p$, hence the isotropy (since $(T_pO_p)^{\omega} = T_p\mu^{-1}(0)$).
Note that an orbit has dimension at most the dimension of $G$, whereas the dimension of $\mu^{-1}(0)$ (assuming that $0$ is a regular value) is $\dim M - \dim \mathfrak{g}^{\ast} = \dim M - \dim G$. In the example you have, $\dim G = \dim M - \dim G =1$. In order to get something less trivial, you only have to consider a manifold $M$ of higher dimension. For instance, $S^2 \times S^2$ with the product symplectic form, and consider the $S^1$-action which acts as the usual rotations on the first factor and which acts trivially on the second factor. In this case, $\mu^{-1}(0) = [equator] \times S^2$.
