# Looking for a biholomorphism [closed]

Is there a biholomorphism between $\mathbb{H}_2 \times S^1$ and a half-space of $\mathbb{R}^3$?

## closed as unclear what you're asking by Andrew D. Hwang, user147263, Michael Galuza, user91500, Claude LeiboviciAug 26 '15 at 7:42

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• What do you mean by a biholomorphism in this context? – Michael Albanese Aug 21 '15 at 19:05
• A conformal map from one space to the other whose inverse is also conformal? Possible? – user6818 Aug 21 '15 at 19:21
• No, these spaces aren't even homeomorphic! – Kyle Aug 21 '15 at 19:21
• Ah! Right! So is there even a one-way conformal map? – user6818 Aug 21 '15 at 19:25
• If you want "conformal" you'll have to specify metrics on the domain and target.... Are you willing to take a hyperbolic plane crossed with a Euclidean line as your half-space of $\mathbf{R}^{3}$? – Andrew D. Hwang Aug 21 '15 at 19:40

No, because these spaces aren't even homeomorphic. Any half-space of $\mathbb{R}^3$ is contractible, but $\mathbb{H}_2 \times S^1$ deform retracts to $S^1$ and thus has nontrivial $\pi_1$.
• I don't think so. As suggested in Andrew D. Hwang's comment, you can get a cover $\mathbb{H}_2 \times \mathbb{R} \to \mathbb{H}_2 \times S^1$, so it's possible if you view $\mathbb{R}^3$ as $\mathbb{H}_2 \times \mathbb{R}$. But I doubt this is what you're looking for. On the other hand, I don't think it's possible if you have the standard conformal structure on $\mathbb{R}^3$. It seems like there should be an obstruction coming from curvature or something, using the fact that $\mathbb{R}^3$ is flat. That's just a guess, since I don't know much about this stuff. – Kyle Aug 21 '15 at 20:12
• You think there is a covering map from $\mathbb{H}_2 \times (\mathbb{R}^+ \cup \{0\}) \rightarrow \mathbb{H}_2 \times S^1$ which is conformal? – user6818 Aug 21 '15 at 20:19