# Complex-analytic isometry of hyperbolic ball and half-space

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.

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 $$\mathbb B^n_R$$?