Symplectic manifold

Remarks

Example: $\R^{2n}$ with coordinates $(x_1, \ldots, x_n, y_1, \ldots, y_n)$ and $\omega_0 = \sum_{k=1}^n dx_k \wedge dy_k$. It has constant coefficients, so $d\omega_0 = 0$. If $v \ne 0$ has a nonzero component $v_{x_k}$, then $\omega_0(v, \partial_{y_k}) = v_{x_k} \ne 0$, and similarly if $v_{y_k} \ne 0$ one pairs with $\partial_{x_k}$; so $\omega_0$ is nondegenerate.

Non-example: the sphere $S^4$ has no symplectic form. On a closed manifold of dimension $2n$, $\omega^n$ is a volume form, so the class $[\omega]^n$ is nonzero and hence $[\omega] \ne 0$ in $H^2_{dR}(M)$; but $H^2_{dR}(S^4) = 0$.

Clause 2 says that each $(T_p M, \omega_p)$ is a symplectic vector space, so $\dim M$ is even. Other standard examples are oriented surfaces with an area form and cotangent bundles $T^* Q$ with the form $-d\theta$ built from the tautological one-form $\theta$. Darboux's theorem says that near each point there are coordinates in which $\omega = \omega_0$.