Compute of curvature In the answer of this question,for the given metric 
$$g_M = ds^2 + \frac1M \sinh^2(\sqrt M s) d\Omega^2,$$ 
how to compute the curvature? Whether the hyperbolic space means $M=\{x\in R^n:x_n>0\}$?
Besides, I also don't know how to compute $\nabla^2s_M $.
I think my foundation of Riemannian geometry is very weak, hope for detail answer, thanks very much.
 A: First, the hyperbolic manifold is the manifold $(\mathbb R^n , g)$ given by one chart $\mathbb R^n$, where in spherical coordinates $(\theta^0= s, \theta^1, \cdots, \theta^{n-1})$, the metric is given by 
$$\tag{1} g = ds^2 + \frac1M \sinh^2(\sqrt M s) d\Omega^2.$$
Note that this is fairly explicit. The manifold you have in mind
$$\{ x\in \mathbb R^n : x_n >0\}$$
is just another realization of the hyperbolic manifold. However in that coordinates $(x_1, \cdots, x_{n-1}, x_n)$, the metric is of different form: It's 
$$g = \frac{1}{x_n^2} \left( dx_1^2 + \cdots dx_n^2\right).$$
But in our situation $(1)$ is more suitable as the Hessian $\nabla^2 s$ you want to calculate are already given by the spherical coordinates $s = \theta^0$. 
By definition, $\nabla^2 s$ is a $(0,2)$-tensor given by 
$$\nabla^2_{ij} s = \frac{ \partial^2 s}{\partial x^i \partial x^j} -\Gamma_{ij}^k \frac{\partial s}{\partial x^k}.$$
If we use the polar coordinates $(\theta^0= s, \theta^1, \cdots, \theta^{n-1})$, then as $\frac{\partial \theta^0}{\partial \theta^i} = \delta_{0i}$,
$$\nabla^2 _{ij} s = - \Gamma_{ij}^0 .$$
The Christoffel symbols is given by 
$$\begin{split} \Gamma_{ij}^0 &= \frac{1}{2} g^{k0} \left(\frac{\partial g_{jk}}{\partial \theta^i} + \frac{\partial g_{ki}}{\partial \theta^j}- \frac{\partial g_{ij}}{\partial \theta^k}\right)\\
&= \frac{1}{2} \left(\frac{\partial g_{j0}}{\partial \theta^i} + \frac{\partial g_{0i}}{\partial \theta^j}- \frac{\partial g_{ij}}{\partial \theta^0}\right)\\
&= -\frac 12 \frac{\partial g_{ij}}{\partial \theta^0}
\end{split}$$
as $g_{i0}$ is always constant (either $1$ or $0$). So 
$$\nabla^2_{ij}s = \begin{cases} 0 &\text{if either }i= 0 \text{ or }j= 0 \\\frac{1}{\sqrt M} \sinh(\sqrt M s)\cosh(\sqrt M s) \left\langle \frac{\partial }{\partial \theta^i},  \frac{\partial }{\partial \theta^i} \right\rangle & \text{otherwise.}\end{cases}$$
(Note here $\langle \cdot, \cdot\rangle $ is the standard metric on the unit sphere). Thus 
$$\nabla^2 s = \frac{1}{\sqrt M}\sinh(\sqrt M s)\cosh(\sqrt M s) d\Omega ^2,$$
as claimed in Anthony's answer. 
