-1
$\begingroup$

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

$\endgroup$

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

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

  • 1
    $\begingroup$ What do you mean by a biholomorphism in this context? $\endgroup$ – Michael Albanese Aug 21 '15 at 19:05
  • $\begingroup$ A conformal map from one space to the other whose inverse is also conformal? Possible? $\endgroup$ – user6818 Aug 21 '15 at 19:21
  • $\begingroup$ No, these spaces aren't even homeomorphic! $\endgroup$ – Kyle Aug 21 '15 at 19:21
  • $\begingroup$ Ah! Right! So is there even a one-way conformal map? $\endgroup$ – user6818 Aug 21 '15 at 19:25
  • 1
    $\begingroup$ 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 $\mathbf{R}^{3}$? $\endgroup$ – Andrew D. Hwang Aug 21 '15 at 19:40
1
$\begingroup$

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 $\pi_1$.

$\endgroup$
  • $\begingroup$ Yes. I realized that after typing the question. But is there even a one-way conformal map? $\endgroup$ – user6818 Aug 21 '15 at 19:27
  • $\begingroup$ 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. $\endgroup$ – Kyle Aug 21 '15 at 20:12
  • $\begingroup$ 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? $\endgroup$ – user6818 Aug 21 '15 at 20:19

Not the answer you're looking for? Browse other questions tagged or ask your own question.