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Let $P^2 \mathrel{\tilde \times} \mathbb{R}^2$ be the $\mathbb{Z}_2$-quotient of $S^2 \times \mathbb{R}^2$, where the $\mathbb{Z}_2$ action on $S^2 \times \mathbb{R}^2$ is antipodal on $S^2$ and a reflection on $\mathbb{R}^2$. Similarly, $K^2 \mathrel{\tilde \times} \mathbb{R}^2$ is the $\mathbb{Z}_2$-quotient of $T^2 \times \mathbb{R}^2$.

Can we find a Kahler surface $M$ homeomorphic to $P^2 \mathrel{\tilde \times} \mathbb{R}^2$ or $K^2 \mathrel{\tilde \times} \mathbb{R}^2$?

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    $\begingroup$ Isn't $K^2 \tilde{\times} \mathbb{R}^2$ diffeomorphic to the cotangent bundle of $K^2$? According to mathoverflow.net/questions/26776/…, all cotangent bundles are Kähler. $\endgroup$
    – Marco Golla
    Commented Jul 16, 2021 at 11:21

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