I'm trying to prove that the $n$-dimensional Poincaré ball and half-space models of hyperbolic space are isometric. Here the Poincaré ball $\mathbb B^n_R$ ($n$-ball of radius $R$) has the Riemannian metric $$ h^1_R = 4R^4 \frac{(du^1)^2 + \cdots + (du^{n})^2}{\left(R^2 - |u|^2\right)} $$ and the half-space $\mathbb H^n_R = \left\{\left(x^1, \ldots, x^{n-1}, y \right) : y > 0 \right\}$ has the Riemannian metric $$ h^2_R = R^2 \frac{\left(dx^1\right)^2 + \cdots + \left(dx^{n-1}\right)^2 + dy^2}{y^2}. $$ Strategy: Rewriting the coordinates in $\mathbb B^n_R$ as $(u, v)$, with $u \in \mathbb R^{n-1}$ and $v \in \mathbb R$, let $\kappa : \mathbb B^n_R \to \mathbb H^n_R$ be the generalized Cayley transform $$ \kappa(u,v) = (x,y) = \left( \frac{2R^2 u}{|u|^2 + (v-R)^2}, \: R\frac{R^2 - |u|^2 - v^2}{|u|^2+ (v-R)^2}\right) $$ This is a diffeomorphism, so we just need to prove it's an isometry, i.e. that $\kappa^* h^2_R = h^1_R$.

What I've tried: My reference text (Lee's "Riemannian Manifolds: An Introduction to Curvature") suggests first proving this for $n=2$ dimensions using the complex differential forms: $$ h_R^2 = R^2 \frac{dz\, d\overline z}{(\mathrm{Im}\,z)^2} \quad \textrm{and} \quad h_R^1 = 4R^4 \frac{dw \, d\overline w}{\left(R^2 - |w|^2\right)^2} $$ and using the fact that $F^*\left(dz\,d\overline z\right) = |F'(w)|^2dw\,d\overline w$ for any holomorphic diffeomorphism $F(w) = z$, and then proving it in higher dimensions by conjugating $\kappa$ with a "suitable orthogonal transformation" to reduce it to the $2$-dimensional case. I've succeeded in proving the $n=2$ case, but I'm having trouble with the general case.

EDIT: I've tried two different strategies, both leading to some roadblocks.

Strategy 1: I've managed to show that the differential of $\kappa$ is $$ d\kappa_{(u,v)} = \left( \begin{array}{cc} A(u,v) & \mathbf b(u,v) \\ -\mathbf b(u,v)^T & c(u,v) \end{array}\right) $$ where $A = A(u,v)$ is a symmetric $(n-1) \times (n-1)$ matrix, $\mathbf b = \mathbf b(u,v)$ is a vector in $\mathbb R^{n-1}$, and $c$ is a function from $\mathbb B^n_R$ to $\mathbb R$. If the matrices $A(u,v)$ are simultaneously diagonalizable, then there's an orthogonal matrix $Q$ that diagonalizes $A$, and I think $Q$ as a map $\mathbb C^n \to \mathbb C^n$ will do what we need to do, although some of the computational details in this case are still a little lost on me. Am I on the right track?

Strategy 2: For each $u \in \mathbb S^{n-1}$ we choose an orthogonal transformation $\tilde Q = \tilde Q_u \in SO(n-1)$ that sends $u$ to $(1,0, \ldots, 0)$. We further assume the mapping $u \mapsto \tilde Q_u : \mathbb S^{n-1} \to SO(n-1)$ is smooth. Then $Q_u := \tilde Q_u \times \mathrm{Id} : \mathbb R^n \to \mathbb R^n$ leaves the $v$-axis invariant and maps the plane $P_u$ containing the origin, $(0,v)$, and $(u,0)$ to the $(u^1, v)$-plane, which we denote $P_0$. (Note $u$ without indices represents a point in $\mathbb R^{n-1}$, whereas its components are $u^i$.) As in the $2$-dimensional case, $\kappa|_{P_0}$ maps $P_0 \cap \mathbb B_R^n$ to $P_0 \cap \mathbb H^n_R$ isometrically, and $\kappa$ maps the plane $P_u$ to itself (since the image of every point in $P_u$ under $\kappa$ is a linear combination of $(u,0) \in \mathbb S^{n-1} \times \{0\}$ and $(0,1)$). In particular, it's straightforward to show that $\kappa = Q_u^{-1} \circ \kappa \circ Q_u$ for every $u \in \mathbb S^{n-1}$. So, since $\kappa$ is a composition of isometries, $\kappa$ itself should be an isometry. But this only shows $\kappa$ is an isometry in each plane through the $v$-axis. How can we improve this to show $\kappa$ is an isometry on $\mathbb B^n_R$?


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.