Is there a surface X(u,v) such that $E=1+u^2$, $F=0$ and $G= 1+v^2$

Question

I'm doing an exercise of my mathematics degree university that asks if there exists such a surface, I have tried to find the explicit parametrization but it's too difficult, so I decided to use other results.

So far, since the parametrization X(u,v) is orthogonal ($F=0$), the Gaussian curvature is $K=\frac{1}{-2\sqrt{EG}}\left(\frac{\delta}{\delta u}\frac{G_u}{\sqrt{GE}}+\frac{\delta}{\delta v}\frac{E_v}{\sqrt{GE}}\right)$ and knowing $E=1+u^2$, $F=0$ and $G= 1+v^2$ we got \begin{equation} K=0 \end{equation} How do I know if there exists such a surface if I don't have L,M,N?

Answer 1

Use that formula for Guassian curvature in terms of Christoffel symbols: $$K=\frac{-1}{E}(\frac{\partial}{\partial u}\Gamma_{12}^{2}-\frac{\partial}{\partial v}\Gamma_{11}^{2}+\Gamma_{12}^{1}\Gamma_{11}^{2}-\Gamma_{11}^{1}\Gamma_{12}^{2}+\Gamma_{12}^{2}\Gamma_{12}^{2}-\Gamma_{11}^{2}\Gamma_{22}^{2})$$

knowing the relationships with the first fundamental form:

Answer 2

HINT: Reparametrize. Define $\tilde u,\tilde v$ so that $\tilde E=\tilde G = 1$ and $\tilde F = 0$. The big clue is that $E=E(u)$ and $G=G(v)$. Undoing this, you can reparametrize the (flat) plane to get your desired $E,F,G$.