Assume $M$ is a $2n-$dimensional differentiable manifold. Let $(U\_{i})$ be a open covering of $M$. With respect to this covering let $\rho\_{i}$ be a partition of unity. Assume that on each $U\_{i}$ we have a symplectic form $\omega\_{i}$. Is then $\omega := \sum\_{i} \rho\_{i} \omega\_{i}$ a symplectic form ? If, not what condition on the functions $\rho\_{i}$ should be, so that $\omega$ is symplectic ?



symplectic form with partition on unity
Suppose you are in $\mathbb{R}^{2n}$, and endow it with a symplectic form $\omega$. Let $U\_1 = \{x\_1>-\varepsilon\}$ and $U\_2=\{x\_1<\varepsilon\}$, with two symplectic forms $\omega\_1 = \omega|\_{U\_1}$, $\omega\_2=-\omega|\_{U\_2}$. Notice that if $n$ is even, the $\omega\_i$ induce the same orientation on the overlap of $U\_1$ and $U\_2$.


Now, by the intermediate value theorem, for any partition of unity $\{\rho\_1, \rho\_2\}$ subordinated to our cover, there's a point where the form $\rho\_1\omega\_1 + \rho\_2\omega\_2$ vanishes.


So, in general, you need to have some compatibility conditions for the $\omega\_i$'s on the intersections of the $U\_i$'s, otherwise there's no hope.


