Why is the tangent space to the orbit through $p\in\mu^{-1}(0)$ an isotropic subspace of $T_pM$? I'm reading symplectic geometry notes by Ana Cannas da Silva. 
The set up is a Hamiltonian action $G\curvearrowright(M,\omega)$ of a Lie group $G$ on the symplectic manifold $(M,\omega)$, with moment map $\mu:M\to \mathfrak{g}^*$. Denote by $\mathfrak{g}_p$ the Lie algebra of the stabilizer of $p\in M$. Then we have
$$\ker d_p\mu=(T_p\mathcal{O}_p)^{\omega_p}$$
$$\text{image} (d_p\mu)=\mathfrak{g}_p^o$$
Where $\mathcal{O}_p$ is the $G$-orbit through $p$, and $\mathfrak{g}_p^0$ the annihilator of $\mathfrak{g}_p$.
Now knowing this we proceed to note that: 


*

*$G$ acts freely on $\mu^{-1}(0)$

*$\Rightarrow$ $T_p\mu^{-1}(0)=\ker d_p\mu$ for $p\in \mu^{-1}(0)$

*$\Rightarrow$ $T_p\mu^{-1}(0)=(T_p\mathcal{O}_p)^{\omega_p}$ for $p\in \mu^{-1}(0)$


This is both clear, with the first implication being obvious, and the second following immediately from the result above.
However, now it is stated that in particular, $T_p\mathcal{O}_p$ is an isotropic subspace of $T_pM$ for $p\in\mu^{-1}(0)$. But for some reason I don't see how this follows. Why is is obvious that
$$T_p\mathcal{O}_p\subset (T_p\mathcal{O}_p)^{\omega_p}=T_p\mu^{-1}(0)$$
This would be obvious if $\mathcal{O}_p\subset \mu^{-1}(0)$ but I don't see why this would have to be the case. Why would the orbit of $p$ under the action of $G$ by completely contained in some level set of the moment map?
 A: For $p \in \mu^{-1}(0)$, $\mathcal{O}_p \subset \mu^{-1}(0)$ because the moment map $\mu$ is $G$-equivariant, i.e. it intertwines the hamiltonian $G$-action on $(M, \omega)$ with the co-adjoint action of $G$ on $\mathfrak{g}^{\ast}$ (which has $\{0\}$ as an orbit). As such, $\mu^{-1}(0)$ contains the orbits it intersects.
This is true regardless of the $G$-action on $M$ being free or not, since we have in all generality
$$ T_p \mathcal{O}_p \subseteq T_p \mu^{-1}(0) \subseteq \mathrm{ker}(d_p\mu) = (T_p \mathcal{O}_p)^{\omega_p} \, .$$

Let us discuss when the equality $T_p \mu^{-1}(0) = \mathrm{ker}(d_p\mu)$ holds for all $p \in \mu^{-1}(0)$. The aforementioned string of inclusions implies (with $\mathrm{dim} \, M = 2n$)
$$ \mathrm{dim} \, T_p \mathcal{O}_p \le \mathrm{dim} \, T_p \mu^{-1}(0) \le \mathrm{dim} \, \mathrm{ker}(d_p\mu) = \mathrm{dim} \, (T_p \mathcal{O}_p)^{\omega_p} = 2n - \mathrm{dim} \, T_p \mathcal{O}_p \, . $$
The rank-nullity theorem states that $\mathrm{dim} \, \mathrm{ker}(d_p\mu) = 2n - \mathrm{dim} \, \mathrm{im}(d_p\mu)$. Applied to our situation, we conclude that $\mathrm{dim} \, \mathrm{im}(d_p\mu) = \mathrm{dim} \, T_p \mathcal{O}_p$. Since $\mathcal{O}_p$ is a $G$-homogeneous space, its dimension is at most $g := \mathrm{dim} \, G$. (Remark : Note that an orbit also has dimension at most $n$.)
It happens to be precisely $g$ if and only if the action is semi-free (at least in a neighbordhood of $\mu^{-1}(0)$) i.e. if the stabilizer of a point is a discrete subgroup. In particular, this is the case if the action is free, in which case any orbit is diffeomorphic to $G$.
In such a case, we have $\mathrm{dim} \, \mathrm{im}(d_p\mu) = g = \mathrm{dim} \, \mathfrak{g}^{\ast}$. In other words, $\mu$ is a submersion in a neighbordhood of $\mu^{-1}(0)$. Hence $0$ is a regular value of $\mu$. As such, $\mu^{-1}(0)$ is an embedded submanifold of $M$ (in particular, it makes sense to speak of $T_p \mu^{-1}(0)$, something we implicitly assumed until now) whose tangent space at a point $p$ is precisely the kernel of $d_p\mu$. This proves the equality under the assumption of (semi-)freeness.

If you drop the freeness assumption, you can find situations where $T_p\mu^{-1}(0) \subsetneq \mathrm{ker}(d_p\mu)$. For instance, consider the hamiltonian $\mathbb{R}$-action on $(\mathbb{R}^2, \omega_{st})$ induced by the hamiltonian function $H(x,y) = x^2+y^2$ ; $\mu^{-1}(0) = \{(0,0)\}$ which is 0-dimensional, but $\mathrm{ker}(d_{(0,0)}\mu) = T_{(0,0)}\mathbb{R}^2$ which is 2-dimensional. This example can be modified to apply to an $S^1$-action on $S^2$, proving that even compactness assumptions (which usually yield much more constrained examples) do not suffice to force the equality $T_p\mu^{-1}(0) = \mathrm{ker}(d_p\mu)$.
