Relation between symplectic manifolds and (almost) complex manifolds I'm a beginner in symplectic geometry, and I recently learned that every symplectic manifold has an almost complex structure.
I am curious about the converse. Does every almost complex manifold have a symplectic structure? If not, then does every complex manifold have a symplectic structure? I tried to search but I couldn't find relevant results. (If these are true) is there a reference for proofs of these results?
 A: Here is a counter-example for "almost complex": the sphere $\Bbb S^6$.
It has an almost-complex structure induced by multiplication of octonions. Namely, $\Bbb R^7$ is the only Euclidean space (apart from $\Bbb R^3$) carrying a binary cross product. If $\Sigma \subseteq \Bbb R^7$ is a orientable hypersurface with unit normal field $N\colon \Sigma \to \Bbb S^6$, define, for each $x \in \Sigma$, $J_x\colon T_x\Sigma \to T_x\Sigma$ by $J_x(v) = N(x)\times v$. In particular, one may take $\Sigma = \Bbb S^6$ itself.
But $\Bbb S^6$ does not carry any symplectic manifold, for topological reasons. Namely, all even order cohomologies of a compact symplectic manifold must be non-trivial as a consequence of Stokes's theorem, but $H^2_{\rm dR}(\Bbb S^6) = 0$.
It is an open problem whether $\Bbb S^6$ has integrable almost-complex structures. And if I recall correctly, the almost-complex structures defined on hypersurfaces of $\Bbb R^7$ via the above recipe are never integrable.
Complex examples are a bit more complicated, but apparently Calabi-Eckmann manifolds fit the bill. They're complex, but one can show (e.g. using the Künneth formula) that a product $\Bbb S^p \times \Bbb S^q$, with $p,q \geq 1$, carries a symplectic structure if and only if $p=q=1$ or $p=q=2$.
