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?

  • $\begingroup$ Where do stereographic coordinates come in? To be more specific: Are you saying that you are required to use stereographic coordinates to prove that $\omega$ is indeed symplectic? $\endgroup$ – Kyle Jun 15 '15 at 20:01
  • $\begingroup$ No, I was just asking for the expression of $\omega$ in stereographic coordinates but I also made this. Thanks a lot for your answer and patience. $\endgroup$ – L.F. Cavenaghi Jun 25 '15 at 0:45

You certainly don't need stereographic coordinates. By definition, you need to check that

  1. $\omega$ is a smooth 2-form,
  2. $d \omega=0$, and
  3. $\omega$ 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 $c \in \mathbb{R}$.


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