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Suppose I am given a $2n$ dimensional manifold $M$ and I want to put a symplectic form on it. How could I determine whether or not it admits one and if I do know that it admits one what are my options for (explicitly) constructing some?

My current knowledge of techniques is the following:

(1) If we are in $\mathbb R^{2n}$ or $\mathbb C^n$ then we can take global coordinates and construct the standard ones.

(2) If we are on a tangent bundle we can construct the canonical one using the tautological 1-form.

(3) If the Riemannian holonomy (assuming we put some metric on $M$ -- although I'm confused about why the metric we choose matters...) group is contained in the symplectic group then we can transfer symplectic forms on $\mathbb R^{2n}$ to $M$. My understanding here though is that we can say the form looks a certain way in certain coordinates at a point on the manifold but not really what it looks like locally.

(4) Brute force- define locally in coordinates and then show well-defined.

(5) And I am aware of this cohomology existence result.

So are there any other techniques? In particular, ways like method $(2)$, where one can construct the form globally but still know what it looks like locally.

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    $\begingroup$ There is a theorem of Marsden-Weinstein-Meyer which gives you a symplectic structure on an Hamiltonian G-space : example include moduli space of planar polygons and more. Roughly speaking, you have an action of a compact Lie group $G$ on a symplectic manifold $X$, a map $\mu : X \to \mathfrak g^*$ with some conditions. Then, you will get a symplectic structure on $\mu^{-1}(0)/G$. $\endgroup$ – user171326 Jan 7 '17 at 18:06
  • $\begingroup$ One more precision : the canonical symplectic structure is on the cotangent bundle, not the tangent bundle. If you have a riemannian metric you can identify both of course. $\endgroup$ – user171326 Jan 7 '17 at 19:10
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In your list of constructions, you are certainly missing coadjoint orbits: If $G$ is any Lie group with Lie algebra $\mathfrak g$, let $\mathfrak g^*$ be the dual vector space to $\mathfrak g$. This carries a natural representation of $G$, defined by $(g\cdot\lambda)(X):=\lambda(Ad(g^{-1})(X))$ with $Ad$ denoting the usual adjoint action. Any orbit of $G$ in this representation is a smooth submanifold (since it is a homogeneous space of $G$) which inherits a canonical symplectic structure, usually called the Kirillov-Kostant-Souriau form.

However, I think that overall your question is rather misleading: By Darboux's theorem, all symplectic forms look the same locally, i.e. locally around any point, you can find coordinates in which a given symplectic form looks like the standard form on $\mathbb R^{2n}$. Global existence of symplectic forms is a very difficult question, and apart from the obvious topological obstructions that occur in your question (and which basically only concern compact manifolds) non-existence results depend on subtle invariants.

By the way, the condition in (3) is a typical example for why I consider your question as misleading. The condition on holonomy says that the given metric $g$ is a Kähler metric. This implies that its so-called fundamental two-form is a symplectic form. Hoever, Kähler metrics are much more restrictive objects than symplectic forms, and even if a manifold admits one Kähler metric, then there is an infinite dimensional space of metrics on $M$ which are not Kähler. So this is a method which only works if you have guessed the right result from the beginning. This does not mean that the condition is not interesting, but you get the existence of a symplectic form from existence of a much stronger and much more rigid structure.

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In real dimension 4, there is a complete topological characterization of which smooth closed 4-manifolds admit symplectic forms. By a theorem of Donaldson all symplectic 4-manifolds admit the structure of a Lefschetz pencil compatible with the symplectic form, and by a theorem of Gompf admiting a Lefschetz pencil implies the existence of a compatible symplectic form.

As for obstructions, the most basic is $H^2_{dr}\ne 0$. There is a canonical way to place an almost complex structure on the tangent bundle of any symplectic manifold, and therefore there are algebraic relations in terms of the characteristic classes of the tangent bundle that need to be solved. For instance in dimension 4 the Wu relations need to be satisfied (c.f. https://mathoverflow.net/questions/63439/how-can-we-detect-the-existence-of-almost-complex-structures). For open manifolds this turns out to be the only obstruction due a theorem of Gromov (any open almost-complex manifold admits a symplectic structure).

In dimension 4 there are finer invariants related to the gauge-theoretic invariants and Heegard Floer homology. A famous result of Taubes in this direction is that if $X,Y$ are closed 4-manifolds with $b_2^+>0$ then $X\#Y$ does not admit a symplectic structure.

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  • $\begingroup$ Thanks for the nice answer. I've accepted the other one because it was first, but both answers are very helpful. $\endgroup$ – Ashley Jan 8 '17 at 20:47

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