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 ?

3$\begingroup$ Essentially, the reason it works for Riemannian metrics is because a convex linear combination of positive definite inner products is still a positive definite inner product. This is not true for skewsymmetric bilinear forms (symplectic forms) nor is it true for inner products with indefinite signature. The existence of such structures on manifolds is a topological question. $\endgroup$– Spiro KarigiannisCommented Jun 24, 2011 at 11:12

$\begingroup$ It fails for two reasons: first there is no guarantee that your $\omega$ is still closed. Second, there is no guarantee that yor $\omega$ is still nondegenerate... $\endgroup$– Stefan WaldmannCommented Sep 29, 2011 at 14:40
3 Answers
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.
This was too long for a commentA conceptual way to understand this is to say for a vector bundle defined using patching, the transition functions live in GL(n,R). GL(n,R) is homotopy equivalent to O(n) by Polar decomposition so one can always define a Riemmanian metric(From this point of you can imagine a vector bundle defined on $X\times I$ and deforming your transition functions by that homotopy equivalence above). The restriction to each end must be isomorphic by a well known theorem.
The symplectic group on the other hand is not homotopy equivalent to GL(n,R) and thus there are obstructions to giving your tangent bundle the structure of a symplectic vector bundle or equivalently giving your manifold a symplectic form. $S^4$ for example is not a symplectic manifold for a simple reason. If $\omega$ were your symplectic form it would necessarily be exact. But for any hypothetical symplectic form on $S^4$, $\omega\wedge\omega$ would be a volume form, so no dice.

$\begingroup$ Dear Daniel Pomerleano, excuse me, but, if $\omega$ is a symplectic form on a compact connected manifold $M$, should not its cohomology class be necessarily nonzero? infact $[\omega]=0$ imply $[\omega^n]=[\omega]^n=0$ and this last contradicts $\int_M\omega\neq 0$. Bye. $\endgroup$– agtCommented Jun 25, 2011 at 17:56

$\begingroup$ that's what I was trying to say perhaps not in the clearest way... that's why $S^4$ is not a symplectic manifold. $\endgroup$ Commented Jun 25, 2011 at 18:36
I would develop a little the remark of Spiro Karigiannis on the topological obstruction for a manifold to have a symplectic structure.
Two first rough necessary conditions for the existence are: the manifold has to be evendimensional and orientable.
Apart from this, we have also the cohomology ring condition:
If a connected compact $2n$dimensional manifold $M$ has a symplectic structure, then there exists $u\in H_{dR}^2(M)$ such that $u^k\neq 0\in H_{dR}^{2k}(M)$ for $k=1,\ldots,n$, and in particular $H_{dR}^{2k}(M)\neq 0$,for $k=1,\ldots,n$.
Proof. It is sufficient to establish that, for any symplectic form $\omega$ on $M$, we have $[\omega]^n=[\omega^n]\neq 0$. Because of the nondegeneracy of $\omega$, we have that $\omega^n$ is a volume form, so its integral over $M$ is not zero and hence $[\omega^n]\neq 0$.
This condition and the computation $H_{dR}^{2}(S^{2n})=0$ for any $n>1$ imply the nonexistence of symplectic structures on $S^{2n}$ for all $n>1$.

2$\begingroup$ Don't you mean $H^2_{dR}(S^{2n}) = 0$? $\endgroup$ Commented Jun 25, 2011 at 18:21

$\begingroup$ @Qiaochu Yuan: Sure, you are right, I need to edit lightly the answer. Thank you very much. $\endgroup$– agtCommented Jun 25, 2011 at 18:32