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Defining the upper half-plane $H^{n} = \left\{ \ x = (x_{1}, \ldots, x_{n}) \in \mathbb{R}^{n} \ | \ x_{n} \geq 0 \ \right\}$.

In complex analysis there's this notion of conformally mapping the unit disk $D^{2}$ to the upper half plane $H^{2}$.

But is this true in general that $D^{n}$ (including its insides) can be conformally mapped to $H^{n}$? (this would have to be a Mobius transformation, via Liouville's theorem)

I feel like the stereographic projection should be able to do the trick somehow, but I'm not certain if this is indeed a conformal mapping. On top of this, the stereographic projection only maps the boundary of $\partial D^{n} = S^{n-1}$ to the boundary of $\partial H^{n}$...

EDIT: Just saw on wikipedia that the stereographic projection is indeed conformal. This resolves the first part of my question...what's left unclear to me is how do I map points on the inside of the disk to the parts of the upper half plane not on its boundary?

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  • $\begingroup$ The unit disk is usually denoted $D^2$ and $S^1$ is the unit circle. More generally, the $n-1$ sphere $S^{n-1}$ is the boundary of the $n$-disk $D^n$. $\endgroup$ – ziggurism Feb 14 '17 at 3:02
  • $\begingroup$ My mistake, I've edited the post now. $\endgroup$ – Greg.Paul Feb 14 '17 at 3:04
  • $\begingroup$ You may like to change the superscripts too. The unit disk $D^2$ is two dimensional, so usually gets a superscript 2. And $\partial D^n=S^{n-1}$ $\endgroup$ – ziggurism Feb 14 '17 at 3:04
  • $\begingroup$ Here I'll do it. $\endgroup$ – ziggurism Feb 14 '17 at 3:07
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Wikipedia gives the isometry between the Poincaré half plane model and the Poincaré disk model. Although the formulas given are for the two-dimensional case, they extend in the obvious way to the $n$-dimensional case, which are also isometric.

A point $(x,y)$ in the disk model maps to $$\left( \frac{2x }{x^2+ (1-y)^2} \ , \ \frac{1-x^2-y^2}{x^2+ (1-y)^2} \right) \,$$ in the halfplane model.

Since the boundary of the half-plane or disk is not included in these models, the formula is for the interior.

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  • $\begingroup$ Amazing. So $(x_{1},\ldots,x_{n-1},x_{n})$ maps to: $\left( \ \frac{2x_{1}}{x_{1}^{2}+\ldots+x_{n-1}^{2}+(1-x_{n})^{2}}, \ldots , \ \frac{2x_{n-1}}{x_{1}^{2}+\ldots+x_{n-1}^{2}+(1-x_{n})^{2}} , \frac{1 - x_{1}^{2} - \ldots - x_{n-1}^{2} - x_{n}^{2} }{x_{1}^{2}+\ldots+x_{n-1}^{2}+(1-x_{n})^{2}} \ \right)$ $\endgroup$ – Greg.Paul Feb 14 '17 at 3:24
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    $\begingroup$ Yep, exactly. Nice work. $\endgroup$ – ziggurism Feb 14 '17 at 3:26
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    $\begingroup$ And it's clear that this map reduces to stereographic projection on the boundaries. Again since the metrics are not defined on the boundaries, I guess we won't say that stereographic projection is an isometry, although it is still conformal wrt euclidean metric. $\endgroup$ – ziggurism Feb 14 '17 at 3:33

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