Let $H^3$ be the Heisenberg manifold. It is known that the first betti number of $H^3\times S^1$ is odd and therefore it does not support any Kähler metric. Now let $I=(0,1)$ or $I=[0,1]$, does it still hold that $H^3\times I$ admits no Kähler metric?



Does $H^3\times I$ admit a Kähler metric?
Note that $S^3$ embeds in $S^4$ and $S^3$ is the total space of the Euler class one circle bundle over $S^2$. It follows that the Euler class one circle bundle over $\Sigma\_g$ embeds in $S^4$ for all $g \geq 0$, see Proposition $7.2$ of [*Smoothly Embedding Seifert Fibered Spaces in $S^4$*](https://arxiv.org/abs/1810.04770) by Issa and McCoy for example. Since the Heisenberg manifold $H^3$ is the total space of the Euler class one circle bundle over $T^2$, we see that $H^3$ embeds in $S^4$.


As $H^3$ embeds in $S^4$, it also embeds in $\mathbb{R}^4 = \mathbb{C}^2$. Note that a tubular neighbourhood of $H^3$ in $\mathbb{C}^2$ is diffeomorphic to $H^3\times (0, 1)$ and inherits a complex structure and Kähler metric from $\mathbb{C}^2$.


