Problem : Equip the unit disk $D^n \subset \mathbb{R} $ with the metric g defined in the identity chart $(x^1,...,x^n) :D^n \rightarrow \mathbb{R} $ by
$g_{ij} = \frac {\delta_{ij}}{1 - ||x||^2 } + \frac {x^i x^j}{({1 - ||x||^2})^2 } $
where $||x|| = \sqrt {(x^1)^2+...+(x^n)^2}$ is the euclidean norm. I am expected to compute the components of curvature tensor of $(D^n,g)$ in the canonical coordinates and deduce that $(D^n,g)$ has sectional curvature $K\equiv -1 $
My attempt : I am proceeding in following way -
The Christoffel's symbol is given by formula $\Gamma_{ij}^k = \frac{1}{2} g^{kl} [\partial_i g_{jl} + \partial_i g_{il} - \partial_i g_{ij} ] $
The term $\partial_i g_{jl} $ will be
$\partial_i g_{jl} = \partial_i \left [ \frac {\delta_{jl}}{1 - ||x||^2 } + \frac {x^j x^l}{({1 - ||x||^2})^2 } \right ] = \frac{2x^i\delta_{jl} + \delta_{ij}x^l + \delta_{il}x^j}{(1-||x||^2)^2} + \frac{4x^jx^lx^i}{(1-||x||^2)^3} $
Also, the inverse of metric $g_{kl}$ is given by
$ g^{kl} = \frac{\delta^{kl}(1-||x||^2)^2}{\delta_{kl}(1-||x||^2) + x^kx^l} $ (I am not sure about this)
And the formula for the curvature tensor is -
$ R_{kij}^l = \partial_i \Gamma_{kj}^l - \partial_k \Gamma_{ij}^l + \Gamma_{kj}^p \Gamma_{ip}^l - \Gamma_{ij}^p \Gamma_{kp}^l $
If we want to come up with some simplified final expression for the curvature tensor, overall computations look too tedious and lengthy. Is there any possibility to simplify the computations to get simplified final expression for curvature tensor and deduce that sectional curvature $K\equiv -1 $