# Does $H^3\times I$ admit a Kähler metric?

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?

• It is parallelizable and every open parallelizable $n$-manifold immerses in $R^n$. Aug 1, 2023 at 17:00
• Following up on Moishe Kohan's superior explanation, if $M$ is an odd-dimensional stably parallelisable manifold without boundary, then $M\times(0,1)$ admits a complex structure and Kähler metric for the same reason. Aug 2, 2023 at 20:46

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 $$\mathbb{C}^2$$.