Let $M$ be a connected closed orientable 3-manifold. Assume $M$ is not the direct product of a surface and the circle.


Can there be a symplectic or Kähler manifold homeomorphic to $M\times M$? I think this might work if $M$ is a non-trivial circle bundle over the torus.



Symplectic structure on the square of a 3-manifold
Let $M$ be a 3-manifold fibering over $S^1$, so there exists a fibration $\Sigma \to M \to S^1$.
Then $M\times M$ will admit a symplectic structure.


There is a symplectic structure on $M\times S^1$, associated to the fibration $\Sigma \to M\times S^1 \to S^1\times S^1=T^2$ which is trivial in the second factor, by a [result of Thurston](https://www.jstor.org/stable/2041749) (the [converse also holds](http://doi.org/10.4007/annals.2011.173.3.8)).


Similarly, there is a fibration $\Sigma \to M\times M \to S^1\times M$ which is trivial in the second factor. Hence by the result in the fourth paragraph of Thurston's paper, $M\times M$ admits a symplectic structure.


These manifolds cannot be Kähler usually. See Theorem 1.2 of [Biswas-Mj-Seshadri](https://arxiv.org/abs/1101.1162), which implies that if $M\times M$ is Kähler, then either $M$ is a manifold admitting [Nil geometry](https://en.wikipedia.org/wiki/Geometrization_conjecture#Nil_geometry) (the fundamental group is commensurable with the [Heisenberg group](https://en.wikipedia.org/wiki/Heisenberg_group) $H$), or $M$ is finitely covered by $\Sigma \times S^1$ for some surface $\Sigma$ of positive genus. See Dmitri Panov's answer for a non-trivial example. Question 4.4 in the paper leaves open the possibility of whether $H\times H$ can be Kähler.


