Consider the Lobachevsky plane defined as
$$\mathbb{R}^2_+ = \left\{{(x,y)\in{\mathbb{R}^2};y>0}\right\}$$
with the metric given by $$g_{11} = g_{22} = \frac{1}{y^2}, g_{12} = 0$$
Calculate the curvature at each point.
The only thing I think I could do is to apply this two formulas:
$$K = \frac{R_{1212}}{g_{11}g_{22}-g^2_{12}}$$
where $$R_{klij} = \frac{1}{2}(\frac{d^2}{dx^jdx^k}g_{il} + \frac{d^2}{dx^idx^l}g_{jk} - \frac{d^2}{dx^jdx^l}g_{ik} - \frac{d^2}{dx^idx^l}g_{jl} + g_{rs}Γ^r_{jk}Γ^s_{il} - g_{rs}Γ^r_{jl}Γ^s_{ik} )$$
But I am not sure, anyone could help me? Thakyou very much!
It pays off to show in general that the curvature of $$\frac{{\rm d}x^2+{\rm d}y^2}{h(x,y)^2}$$is $K = h(\triangle h) - \|\nabla h\|^2$. For $h(x,y)=y$ this gives $K=-1$ as expected.