Metric tensor and Christoffel symbols of the hyperbolic n-space

Question

Let $\Bbb{H}^n:=\left\{(x_1,...,x_n)\in\Bbb{R}^n\mid x_n>0\right\}$ be the hyperbolic space and $g={d^2x_1+\dots+d^2x_n \over x_n^2}$ be the standard hyperbolic metric.

Looking at the $\left(\Bbb{H}^n,g\right)$ Riemannian manifold, I would like to calculate the metric matrix $[g_{ij}]$ at any point of the manifold, and moreover the $\Gamma^k_{ij}$ Christoffel symbols.

Can anyone help me with that? I'm not sure how to deal with the $x_n$ at the devider of the metric.

Answer 1

First, note that $\{\frac{\partial}{\partial x_i}\}_{i=1}^n$ is the basis of the tangent space at any point of $\mathbb{H}$. Hence, it follows immediately from the definition of the metric that $$g_{ij}=g(\frac{\partial}{\partial x_i},\frac{\partial}{\partial x_j})=\frac{\delta_{ij}}{x_n^2}.$$ From this, we can calculate $\frac{\partial}{\partial x_k}g_{ij}$. Many of them vanishes since $g_{ij}$ depends only on $x_n$. Finally, use the formula of the Christoffel symbol: (see here) $$\Gamma_{ij}^k=\frac{g^{kl}}{2}(\frac{\partial}{\partial x_j}g_{il}+\frac{\partial}{\partial x_i}g_{jl}-\frac{\partial}{\partial x_l}g_{ij}).$$

Answer 2

In order to compute the Christoffel symbols we need to analyse the summands $g_{ij,k}$ for $i,j,k \leq n$. Now as we already saw $(g_{ij})_{i,j} = \frac{1}{y_n^2} \mathbb{I}_n$ at a point $(y_1,...,y_n) \in \mathbb{H}^n$ which shows that $g_{ij,k} = \frac{d}{dy_k}g_{ij} $ is nonzero iff $i=j$ and $k=n$. Now using the Einstein convention we see that

$\Gamma_{ij}^k= \frac{1}{2}g^{kl}(g_{jl,i} - g_{ij,l}+ g_{li,j})=\frac{1}{2}g^{kk}(g_{jk,i} - g_{ij,k} + g_{ki,j}).$

Assume $i<n$ then we have

$ \Gamma_{ii}^n = \frac{1}{2}g^{nn}(g_{in,i}-g_{ii,n}+g_{ni,i}) = \frac{1}{2}g^{nn}(-g_{ii,n})= \frac{1}{2}y_n^2 \frac{d}{dy_n}(-\frac{1}{y_n^2}) = \frac{1}{y_n}. $

and for $k<n$ it is easy to see that $\Gamma^k_{ii} = 0$. On the other hand for arbitraty $k$ we find that

$\Gamma^k_{kn} = \frac{1}{2}g^{kk}(g_{nk,k} - g_{kn,k}+g_{kk,n}) =\frac{1}{2}g^{nn}g_{kk,n}= \frac{1}{2}y_n^2 \frac{d}{dy_n}(\frac{1}{y_n^2})=-\frac{1}{y_n} = \Gamma_{nk}^k$.

All other symbols $\Gamma_{ij}^k$ vanish. To see this observe that for $\Gamma_{ij}^k$ to not vanish one of the $i,j,k$ must be $n$. In case $k=n$ we find that $\Gamma_{ij}^n = \frac{1}{2}g^{nn}(g_{jn,i} - g_{ij,n}+ g_{ni,j})$ wich is nonzero iff $i=j$. This case was already discussed. In case $j=n$ we find that $\Gamma^k_{in} = \frac{1}{2}g^{kk}(g_{nk,i}-g_{in,k}+g_{ki,n})$. We can assume that $i<n$ because $\Gamma^k_{nn} = 0$ for all $k$ but $k=n$ wich was also discussed above. Now under the assumption that $i<n$ we have $\Gamma^k_{in} = \frac{1}{2}g^{kk}g_{ki,n}$ which is nonzero iff $i=k$ and this was also discussed above. We use symmetry of christoffel symbols to find that the case $i=n$ yields the same result as the case $j=n$.