Conformal diffeomorphism of higher dimensional unit ball

Question

I have read one lecture note regarding the conformal diffeomorphism of hypersphere, they construct following conformal map: $$f=\phi_p^{-1}\circ D_t\circ \phi_p$$ where $\phi_p$ is the steregraphic projection w.r.t point $p\in \mathbb{S}^{n}$, $D_t$ is the dilation map on $\mathbb{R}^n$. I think I understand this map, it maps a domain of sphere to Euclidean space, then dilate it, finally pull it back. It's pushing all the points of sphere to point $p$ expect $-p$. And it is conformal map because it is composition of conformal maps. Now I am wondering what is the conformal diffeomorphism group of unit ball? Or is there any specific conformal diffeomorphism of unit ball except those trivial ones. I know it has something to do with liouville's theorem, but I couldn't find an example.

Answer 1

Let $B^{n}$ be the unit ball in $\mathbb{R}^{n}$ and $S^{n-1}$ be the unit sphere in $\mathbb{R}^{n}$, so that $S^{n-1} = \partial B^{n}$.

The conformal automorphisms of $S^{n-1}$ are the Möbius transformations of $S^{n-1}$, i.e., finite compositions of sphere inversions in $\overline{\mathbb{R}^{n-1}}$ if you identify $S^{n-1} \approx \overline{\mathbb{R}^{n-1}}$ by stereographic projection. The conformal map you mention in your question is an example of Möbius transformation.

On the other hand, you can consider Möbius transformations of $\overline{\mathbb{R}^{n}}$ which preserves $B^n$, i.e. finite composition of inversions with respect to spheres that are orthogonal to $\partial B^n$. It turns out that they are the same as the previous ones: they also preserve $\partial B^n = S^{n-1}$ so induce a conformal automorphism of $S^{n-1}$, and all of them are obtained this way. Moreover, they are also all the conformal automorphisms of $B^{n}$.

Note: If you equip $B^{n}$ with the Poincaré metric $g$, i.e. the unique complete conformal Riemannian metric in $B^{n}$ of constant curvature $-1$, you obtain the Poincaré ball model of hyperbolic space $\mathbb{H}^n$. It turns out that $G$ is also the group of isometries of $\mathbb{H}^n = (B^n, g)$.

In conclusion: $$G = \mathrm{Aut}(B^{n}) = \mathrm{Möb}(B^{n}) = \mathrm{Möb}(S^{n-1}) = \mathrm{Aut}(S^{n-1}) = \mathrm{Isom}(\mathbb{H}^{n})$$

As a Lie group, $G$ is identified to the Lorentz group $O^+(n,1)$ via the hyperboloid model of $\mathbb{H}^n$ (obtained via another stereographic projection).

I haven't proven anything, so you would need to work or do some research if you want the proofs. But none of it is actually hard to prove.