Symplectic form on 2-sphere? I am trying to put the symplectic form of the 2-sphere defined by
$\omega_u(v,w) := \langle u,v\times w\rangle,$ where $u \in \mathbb{S}^2$ and $v, w \in T_u\mathbb{S}^2$. I just can't do this in stereographic coordinates.
Would you help me?
 A: You certainly don't need stereographic coordinates.  By definition, you need to check that


*

*$\omega$ is a smooth 2-form,

*$d \omega=0$, and

*$\omega$ is non-degenerate, i.e. for all $v \in T_u S^2$, there exists $w \in T_u S^2$ such that $\omega_u(v,w)\neq 0$.


You could try to verify (1) using stereographic coordinates, but that seems unnecessary. The function $\mathbb{R}^3 \times \mathbb{R}^3 \times \mathbb{R}^3 \to\mathbb{R}$ given by $(u,v,w)\mapsto \langle u,v \times w\rangle$ is smooth. This restricts to $\omega$ on  $$TS^2 \times_{S^2} TS^2 =\{(u,v,w) \mid u \in S^2 \text{ and } v,w \in T_uS^2\} \subset \mathbb{R}^3 \times \mathbb{R}^3 \times \mathbb{R}^3,$$
so $\omega$ is smooth. Since there are no nonzero 3-forms on a 2-dimensional manifold, we have $d\omega=0$. Finally, consider a fixed $v \in T_u S^2$. Then for any other $w \in T_u S^2$ that is linearly independent from $v$, the vector $v \times w \in \mathbb{R}^3$ is perpendicular to $T_u S^2$ and nonzero. This means that $v \times w$ is a nonzero multiple of $u$, so $\omega_u(v,w)=\langle u, c u\rangle = c$ for some nonzero $c \in \mathbb{R}$.
