TITLE: Sphere curvature as calculated from Liouville's equation
QUESTION [6 upvotes]: Liouville's equation for Gauss curvature tells us, that when Riemannian metric has the form $f^2(du^2+dv^2)$, then its Gauss curvature $K$ is expressed by the following equation:
$$-Kf^2=\Delta_{0}log(f)$$
where $\Delta_{0}=\frac{\partial^2 }{\partial u^2}+\frac{\partial^2 }{\partial v^2}$.
Consider a sphere with metric (due to stereographic projection):
$$\frac{du^2+dv^2}{(1+u^2+v^2)^2}$$
Therefore we can apply Liouville's equation to find the curvature of the sphere with $f=\frac{1}{1+u^2+v^2}$.However, calculating the curvature of the sphere in this way I get $K=4$, which is not true. And I cann't spot my mistake. 
Here are some calculations:
$$log(f)=-log(1+u^2+v^2)$$
$$\frac{\partial^2 }{\partial u^2}log(f)=-\frac{2(1-u^2+v^2)}{(1+u^2+v^2)^2}$$
$$\frac{\partial^2 }{\partial u^2}log(f)=-\frac{2(1+u^2-v^2)}{(1+u^2+v^2)^2}$$
$$\Delta_{0}log(f)=-\frac{2(1-u^2+v^2)}{(1+u^2+v^2)^2}-\frac{2(1+u^2-v^2)}{(1+u^2+v^2)^2}=\frac{-4}{(1+u^2+v^2)^2}$$
Therefore:
$$-K\frac{1}{(1+u^2+v^2)^2}=\frac{-4}{(1+u^2+v^2)^2}$$
Hence Gauss curvature of the sphere is (not):
$$K=4$$
My question: where is that mistake?

REPLY [4 votes]: Your formula 
$$\frac{du^2+dv^2}{(1+u^2+v^2)^2}$$
is incorect. It should be 
$$\frac{4(du^2+dv^2)}{(1+u^2+v^2)^2}$$