# Short proof that a surface has constant Gaussian curvature

Let $$M \subset \Bbb R^3$$ be a regular surface with a parametrization where the induced metric is $$ds^2=\frac{du^2+dv^2}{(u^2+v^2+c)^2}$$

Where $$c>0$$ is constant. Prove that the Gaussian curvature of $$M$$ is constant.

I succeeded to solve it by computing the Riemannian curvature by the definition, but it was very long and not elegant. Is there a simpler way to calculate it, by using the direct definition of Gaussian curvature (i.e using a parametrization)?

The curvature of $$\frac{du^2+dv^2}{h(u,v)^2}$$ is given by $$K = h\,\triangle h - \|\nabla h\|^2.$$Here $$h(u,v)=u^2+v^2+c$$ gives that $$K = 4(u^2+v^2+c) -(4u^2+4v^2) = 4c.$$The proof of this formula is a direct application of Gauss' formula $$K = \frac{-1}{\sqrt{EG}}\left(\left(\frac{(\sqrt{G})_u}{\sqrt{E}}\right)_u+\left(\frac{(\sqrt{E})_v}{\sqrt{G}}\right)_v\right)$$for orthogonal parametrizations.

• Thank you, but this is actually how I originally solved it. I didn't learn the above formula, so I had to calculate $K$ by definition. Is there another way to do it, maybe by finding a parametrization? – user401516 Feb 23 at 19:32
• When you write the metric as a combination of $du$ and $dv$, a parametrization is already implied. The given metric expression is already local. We are already working in coordinates, and the fact that $M$ is embedded in $\Bbb R^3$ is irrelevant. Also, unless you want to try using big guns, it can't get much simpler than using the formula for the curvature of a conformal metric to the flat one. – Ivo Terek Feb 23 at 19:35

I will give an answer using solely differential forms which I believe gives a more elegant solution as it rewrites the above formulae using less variables; hopefully you are familiar with them. In my opinion (not necessarily right or more instructive), it is easier to remember this flow of ideas than the formula-based approach.

The induced metric (first fundamental form) can be written as $$I=\theta_1^2+\theta_2^2,$$ where in your case you have $$\theta_1=\dfrac{du}{u^2+v^2+c}$$ and $$\theta_2=\dfrac{dv}{u^2+v^2+c}$$.

Also, we know that the Gaussian curvature appears in the Gauss equation $$d\omega_{12} = K\theta_1\wedge\theta_2,\tag{1}\label{Gauss}$$ where $$\omega_{12}$$ is the $$\textit{connection form}$$ satisfying the structure equations $$d\theta_1 + \omega_{12}\wedge\theta_2=0$$ $$d\theta_2 - \omega_{12}\wedge\theta_1=0$$

Computing the exterior derivatives you get $$d\theta_1=\dfrac{2v\;du\wedge dv}{(u^2+v^2+c)^2}$$ and $$d\theta_2=-\dfrac{2u\;du\wedge dv}{(u^2+v^2+c)^2}$$. Using these along with the two equations above you get $$\omega_{12}=\dfrac{-2v\;du+2u\;dv}{u^2+v^2+c},$$ which replaced in (1) gives you $$\dfrac{4c\;du\wedge dv}{(u^2+v^2+c)^2}=K\theta_1\wedge\theta_2$$ from which you can conclude $$K=4c$$.

Hope it helps and shows a less computational way of finding out the Gauss curvature.