A smooth manifold $M$ admits a symplectic structure if there is an alternating
non degenerate $2$-form $\omega \in \Lambda^2(M)$ that is also closed i.e. $d\omega = 0.$
Usually we can express obstructions to the existence of certain tensors in terms of vanishing of some cohomology classes.
On the other hand, due to the "integrability" condition $d\omega = 0$ one should expect that a set of necessary and sufficient condition for $M$ to admit a symplectic structure cannot be expressed just in homological terms.
In order to support this guess I am thus looking for a concrete example:
Find $M_1, M_2 $ (possibly) compact smooth manifolds (of the same dimension) such that
$M_1$ is symplectic
$M_2$ does not admit a symplectic structure
$M_1$ is homotopically equivalent to $M_2$
Best Answer
There are many examples in four dimensions, in fact there are infinitely many examples where $M_1$ and $M_2$ are actually homeomorphic.
Let $M_1$ be a simply connected Kähler surface which is not spin. Then $M_1$ is symplectic and is homeomorphic to the smooth manifold $M_2 = b^+\mathbb{CP}^2\# b^-\overline{\mathbb{CP}^2}$ by Freedman's Theorem. However, if $b^+ > 1$, if follows from Seiberg-Witten theory that $M_2$ does not admit a symplectic form; in particular, $M_1$ and $M_2$ are not diffeomorphic. Taubes proved that a symplectic manifold with $b^+ > 1$ has a non-zero Seiberg-Witten invariant. However, all the Seiberg-Witten invariants of $M_2$ vanish because $M_2$ is diffeomorphic to the connected sum of two manifolds with $b^+ \geq 1$.
An explicit example of such a pair is $M_1 = \operatorname{Bl}_p(K3) = K3\#\overline{\mathbb{CP}^2}$, the blowup of $K3$ at a point, and $M_2 = 3\mathbb{CP}^2\# 20\overline{\mathbb{CP}^2}$.