Curvature of given metric space As my question 1 and 2, I still have many problems.
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 
$$\tag{1} g = ds^2 + \frac1M \sinh^2(\sqrt M s) d\Omega^2.$$
I want to compute the curvature of $(\mathbb R^n,g)$.I try to compute $\Gamma_{ij}^k$, but in different dimension,$d\Omega^2$ has different form , I can't get $g_{ij} $ for $i,j\ne0$.Then ,I don't know how to do it .
 A: (1) If $M>0$, define $$ \phi(s):=\frac{1}{\sqrt{M}} \sinh\
\sqrt{M}s,\ g_N:=dr^2 +\phi(r)^2 g_{S^{n-1}}$$
In this case we have a formula $$ K_{rad} = - \frac{\phi''}{\phi} $$
$$ K_{sph} = \frac{1-(\phi')^2}{\phi } $$
So in this case we have $$ K_{rad}=K_{sph}=-M $$
(2) Now we will prove that these formula : For $p=(r,\theta)\in
N={\bf R}^n$, consider $$ \exp_\theta : T_\theta S^{n-1}\rightarrow
S^{n-1} $$ Here $S^{n-1}$ has normal coordinate\
$$ F: {\bf R}^n={\bf R}\times T_\theta S^{n-1} \rightarrow N ,\ F(r, v) =
(r,\exp_\theta v ) $$
That is $$ (g_N)_{ij}=g_N(dF e_i, dF e_j) =\phi(r)^2 \delta_{ij} $$
(2.1) Here sectional curvature for radial direction is given by
$$K_{rad} =( D_{\partial_r}D_{\partial_{\theta }}
\partial_{\theta } - D_{\partial_{\theta} } D_{\partial_r}
\partial_{\theta } ,
\partial_r  )\frac{1}{|\partial_\theta |^2}$$ since $\partial_r\perp
\partial_{\theta} $ where $\partial_\theta = \partial_{\theta_i} =dF\ e_i $ for some $i$
Here
$$D_{\partial_r}\partial_r=0,\ \Gamma_{r\theta_j}^r=0
$$
If $(D_{\partial_\theta }\partial_r,\partial_{\theta_k})=0\ \ast$
for $i\neq k$ (cf. $(\nabla_{\ } \partial_r,\ )$ is symmetric so
that it can be diagonalized), then
$$\Gamma_{\theta\theta}^r=-\phi\phi',\ \Gamma_{\theta_jr}^{\theta_j}=\phi'/\phi $$
$$
K_{rad} =\frac{1}{|\partial_\theta |^2} \{ -\frac{1}{2}\partial_r^2
(
\partial_\theta,
\partial_{\theta } ) + |D_{
\partial_{\theta } } \partial_r |^2  \} = -\frac{\phi''}{\phi }$$
(2.2) $$K_{sph} =( D_{\partial_{\theta_j} }D_{\partial_{\theta }}
\partial_{\theta } - D_{\partial_{\theta} } D_{\partial_{\theta_j} }
\partial_{\theta } ,
\partial_{\theta_j}  )\frac{1}{\phi^4}$$ $$ = \frac{1}{\phi^2} \{ \underbrace{\partial_{\theta_j}
\Gamma_{\theta\theta}^{\theta_j} -\partial_\theta\Gamma_{\theta_j
\theta}^{\theta_j} +
\Gamma_{\theta\theta}^{\theta_k}\Gamma_{\theta_j\theta_k}^{\theta_j}
-
\Gamma_{\theta_j\theta}^{\theta_k}\Gamma_{\theta\theta_k}^{\theta_j}}_{=1}
\} + \frac{1}{\phi^2} \{
\Gamma_{\theta\theta}^r\Gamma_{\theta_jr}^{\theta_j} -
\underbrace{\Gamma_{\theta_j\theta}^r\Gamma_{\theta
r}^{\theta_j}}_{=0\ by\ \ast} \}
$$
( Here $\Gamma_{\theta\theta_k}^{\theta_j}$ is clearly $
\Gamma_{\theta\theta_k}^{\theta_j} (g_N)$ Since $\phi$ is a function
independent of $\theta_i$, then we have $
\Gamma_{\theta\theta_k}^{\theta_j}=\Gamma_{\theta\theta_k}^{\theta_j}
(g_{S^{n-1} })$ )
