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

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Michael Albanese
Michael Albanese
  • 19.4k
  • 9
  • 87
  • 161

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 $I=[0,1]$, isdoes it still hold that $H^3\times I$ admits no 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 $I=[0,1]$, is it still hold that $H^3\times I$ admits no 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?

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Does $H^3\times I$ admit 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 $I=[0,1]$, is it still hold that $H^3\times I$ admits no Kähler metric?