0

I'm having a little trouble computing the inverse of the Möbius transform in $\mathbb{R^n}$, as outlined here in "higher dimensions". I assume it exists because it goes on to say that it forms a group.

$$f(x) = b + \frac{\alpha A(x-a)}{|x-a|^2}, \quad x, a, b \in \mathbb{R}^n, \alpha \in \mathbb{R}$$

Where $A$ is an orthogonal matrix. It seems like it would be elementary, but not quite sure how to deal with the norm, and scouring the literature no one seems to bother computing it.

Anton Vrdoljak
  • 2,717
jjyyss
  • 33

1 Answers1

0

Figured it out, inspired by the line of reasoning here

First, I started with $f(x) = x/|x|^2$. The inverse of this $f^{-1}(y) = y/|y|^2$. It is straightforward from there.

jjyyss
  • 33