Hamiltonian reduction in symplectic geometry

Question

If $V$ is symplectic and $W^\perp \subseteq W\subseteq V$, then why is $W$ a pre-symplectic vector space? Why is $W/W^\perp$ symplectic?

Answer 1

If $V$ is symplectic, then it comes equipped with a symplectic form $\omega$. Then $\omega|_W$ is a skew-symmetric bilinear form on $W$, so that $(W, \omega|_W)$ is a pre-symplectic vector space.

To see that $W/W^\perp$ is symplectic, we need to check that $\omega|_W$ descends to a non-degenerate $2$-form on $W/W^\perp$. The natural pre-symplectic form on $W/W^\perp$ is $$\bar{\omega}: W/W^\perp \otimes W/W^\perp \longrightarrow \Bbb R,$$ $$\bar{\omega}([w], [v]) = \omega(w, v),$$ where $w$ (respectively $v$) is any choice of representative for $[w]$ (respectively $[v]$). It is trivial to check that $\bar{\omega}$ is well-defined since any two representatives of an element of $W/W^\perp$ differ by an element of $W^\perp$. To see that $\bar{\omega}$ is non-degenerate, suppose that for some fixed $[w] \in W/W^\perp$, $$\bar{\omega}([w],[v]) = 0 \text{ for all } [v] \in W/W^\perp.$$ This is equivalent to saying $$\omega(w, v) = 0 \text{ for any representative } w \text{ of } [w] \text{ and all } v \in W.$$ This implies that any representative $w$ of $[w]$ is an element of $W^\perp$, and hence $[w] = 0$. Therefore $\bar{\omega}$ is non-degenerate and $(W/W^\perp, \bar{\omega})$ is a symplectic vector space.