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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.

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    $\begingroup$ There is a complex structure on $S^3\times S^3$, see Calabi-Eckmann manifolds. I am not aware of any Kähler examples. $\endgroup$ – Michael Albanese Sep 30 at 20:14
  • $\begingroup$ @MichaelAlbanese thank you $\endgroup$ – user164740 Sep 30 at 20:16
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    $\begingroup$ Obviously, there is no symplectic structure (and hence no Kähler structure) on any smooth manifold homeomorphic to $S^3\times S^3$ because $H^2(S^3\times S^3) = 0$. $\endgroup$ – Robert Bryant Sep 30 at 20:23
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    $\begingroup$ @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). $\endgroup$ – Michael Albanese Sep 30 at 20:28
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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.

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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 $(\mathbb Z_2)^2$.

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