$\newcommand{\cyclic}[1]{\langle #1\rangle}$ This is a seemingly straightforward question from Galot, Hulin, Lafontaine: Riemannian Geometry, Third Edition.
We have a model of hyperbolic space embedded in the Minkowski space $\mathbb{R}^{n+1}$ with the quadratic form $\langle x,x\rangle=-x_0^2+\sum_{i=1}^nx_i^2$. $H^n$ is the submanifold $\{x\in \mathbb{R}^{n+}|\langle x,x\rangle=-1, x_0>0\}$. The quadratic form induces a metric $g=-dx_0^2+\sum_{i=1}^ndx_i^2$ on $H^n$. I am asked to show the following
Let $f$ be the "pseudo-inversion" with pole $s=(-1,0,...,0)$ defined by $$f(x)=s-\frac{2(x-s)}{\langle x-s,x-s\rangle}.$$ Show that $f$ is a diffeomorphism from $H^n$ onto the unit disk $D:=\{x\in \mathbb{R}^n| |x|<1\}$ and that $(f^{-1})^*g = 4\sum_{i=1}^n\frac{dx_i^2}{(1-|x|^2)^2}$ where $|x|^2$ is the usual Euclidean norm squared in $\mathbb{R}^n$.
I am unable to see how this is a diffeomorphism onto $D$ or compute the desired pullback.
What I have done (edited) is compute
$$\cyclic{f(x)-s,f(x)-s} = \left\langle{-\frac{2(x-s)}{\cyclic{x-s,x-s}}, -\frac{2(x-s)}{\cyclic{x-s,x-s}}}\right\rangle = \frac{4}{\cyclic{x-s,x-s}}$$ $$f(f(x)) = s-\frac{2(f(x)-s)}{\cyclic{f(x)-s,f(x)-s}} = s-\frac{2\left(-\frac{2(x-s)}{\cyclic{x-s,x-s}}\right)}{\frac{4}{\cyclic{x-s,x-s}}} = s-(-(x-s))=x$$
So $f$ is in some sense its own inverse. Then in coordinates, since $-x_0^2 + x_1^2+...+x_n^2 = -1$, we have $-(x_0+1)^2+x_1^2+...+x_n^2 =-2x_0-2$ so
\begin{align*}f(x_0,x_1,...x_n) &= (-1,0,...,0)-\frac{1}{-2x_0-2}(2x_0+2,2x_1,...,2x_n) \\&= (-1,0,...,0)-(-1,\frac{2x_1}{-2x_0-2},..., \frac{2x_n}{-2x_0-2})\\ &= (0,\frac{x_1}{x_0+1},...,\frac{x_n}{x_0+1}) \end{align*}
Here are my problems:
- It is not clear to me that $f(x)$ lies in $D$, or even that it can be identified with a point in $\mathbb{R}^n$. There is a remark in the book about the hyperplane $x_0=0$, and that would make this clearly a map onto $\mathbb{R}^n$ but not onto $D$. Furthermore, none of the points where $x_0=0$ are in $H^n$ as we have written it. I am clearly misunderstanding something.
Edit: Now, I think I can see that $f(x)\in \{x\in \mathbb{R}^{n+1}| |x|<1, x_0=0\}$ since if we start with an $x\in H^n$, $x_1^2+...+x_n^2=x_0^2-1$
$$|f(x)|^2 = \frac{x_1^2+...+x_n^2}{(x_0+1)^2} = \frac{x_0^2-1}{(x_0+1)^2} = \frac{x_0-1}{x_0+1}$$
When $x_0>0$, $-1<|f(x)|^2<1$.
- I think I can see that this function is a diffeomorphism by noting it's a composition of the smooth invertible maps $x\mapsto \frac{x}{\langle x,x\rangle}=-x$, $x\mapsto x+s$, and scaling. But I don't know how to compute its inverse function...
Edit: If $f$ is its "own inverse", then does it makes sense to compute $f^*g$? It appears that this will give the same result as pulling back the map in the comments?