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?

3$\begingroup$ It is parallelizable and every open parallelizable $n$manifold immerses in $R^n$. $\endgroup$– Moishe KohanAug 1, 2023 at 17:00

$\begingroup$ Following up on Moishe Kohan's superior explanation, if $M$ is an odddimensional stably parallelisable manifold without boundary, then $M\times(0,1)$ admits a complex structure and Kähler metric for the same reason. $\endgroup$– Michael AlbaneseAug 2, 2023 at 20:46
1 Answer
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 $S^4$ 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 $\mathbb{C}^2$.