Let's consider the following abstract surface on the unit disk $\mathbb{D}=\left\{(u,v) \in \mathbb{R}^{2} \ | \ u^2+v^2 < 1\right\}$: $$S = \left\{\mathbb{D}, \left\{E=G=\frac{4}{(1-u^2-v^2)^2}, \ F = 0\right\}\right\}$$ ($E,G,F$ denote the coefficients of the first fundamental form.)
Is $S$ necessarily complete? I have tried using arguments related to Gaussian curvatures to show that $S$ must be complete (Using the Brioschi Formula, we can get that the Gaussian curvature $K$ is always equal to $-1$)....but I got stuck on it. Any help will be greatly appreciated!
You are missing the square in the denominator of $E=G=\dfrac4{(1-u^2-v^2)^2}$.
By Hopf-Rinow theorem, we just need to show there is a point $p\in\mathbb{D}$ for which $\mathbb{D}$ is geodesically complete at $p$, i.e., you can extend geodesics infinitely starting at $p$ in every direction. The point $p=0$ (i.e., $(u,v)=(0,0)$) is a convenient choice here. Then noting Euclidean reflections across diameters of $\mathbb{D}$ are isometries, we know the geodesics through $0$ are the diameters. So it suffices to show the length of a Euclidean radius is infinite in this metric: $$ \int_0^1\frac{2\,\mathrm{d}r}{1-r^2}=\infty $$