# Symplectic structure on the square of a 3-manifold

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.

• There is a complex structure on $S^3\times S^3$, see Calabi-Eckmann manifolds. I am not aware of any Kähler examples. – Michael Albanese Sep 30 at 20:14
• @MichaelAlbanese thank you – user164740 Sep 30 at 20:16
• Obviously, there is no symplectic structure (and hence no Kähler structure) on any smooth manifold homeomorphic to $S^3\times S^3$ because $H^2(S^3\times S^3) = 0$. – Robert Bryant Sep 30 at 20:23
• @RobertBryant: I meant no example where $M\times M$ is Kähler. Note that the question has been edited since my comment (previously it asked whether there was an example with $M\times M$ complex). – Michael Albanese Sep 30 at 20:28

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 (the converse also holds).
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, which implies that if $$M\times M$$ is Kähler, then either $$M$$ is a manifold admitting Nil geometry (the fundamental group is commensurable with the 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.
Here is a Kahler example. Consider a hyper-elliptic curve $$C$$ of positive genus with involution $$\sigma$$. Take $$C\times S^1$$ and quotient $$C\times S^1$$ by $$\mathbb Z_2$$ that rotates $$S^1$$ by a half-turn and acts by $$\sigma$$ on $$C$$. Now, to get a Kahler structure on $$M\times M$$ take $$C\times C\times E$$, where $$E$$ is an elliptic curve and quotient by an obvious action of $$(\mathbb Z_2)^2$$.