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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 ?

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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 skew-symmetric 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. – Spiro Karigiannis Jun 24 '11 at 11:12
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 non-degenerate... – Stefan Waldmann Sep 29 '11 at 14:40
up vote 9 down vote accepted

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.

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This was too long for a comment---A 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.

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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. – Giuseppe Jun 25 '11 at 17:56
that's what I was trying to say perhaps not in the clearest way... that's why $S^4$ is not a symplectic manifold. – Daniel Pomerleano Jun 25 '11 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 even-dimensional 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 $n>1$.

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Don't you mean $H^2_{dR}(S^{2n}) = 0$? – Qiaochu Yuan Jun 25 '11 at 18:21
@Qiaochu Yuan: Sure, you are right, I need to edit lightly the answer. Thank you very much. – Giuseppe Jun 25 '11 at 18:32

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