# Proving that a surface is isometric to the plane

A surface $S$ has first fundamental form $du^2 + G(u,v)dv^2$ and curvature $0$. Also the curve $u=0$ is a geodesic when parametrized by arclength.

Prove that $G(u,v) = 1$ i.e. that $S$ is isometric to the plane.

It seems like you should be able to do this using the formula for curvature $K=\frac{-\sqrt{G}_{uu}}{\sqrt{G}}$ along with the geodesic equations, but I can't get it to work.

• Are you sure that you're not told that all the $v$-curves ($u=\text{constant}$) are geodesics? Otherwise, all you have is that $\sqrt G_u=0$ when $u=0$. If you knew $\sqrt G_u=0$ on the entire coordinate patch, then you would be done. Why? [You don't even need to use Gaussian curvature for your argument.] – Ted Shifrin May 29 '14 at 17:46
• Trying it again, from the Gaussian curvature being $0$ we get that $\sqrt{G}_{u} = f(v)$ so if we have $\sqrt{G}_{u}=0$ for $u=0$ then we have it for the entire thing. But i don't see how this is enough to get that $G(u,v)=1$ – user153684 May 29 '14 at 20:30
• Ohhh ... I missed the hypothesis that $K=0$ the first time. My apologies. Yes, you're right. Well, then $G(u,v)=g(v)$, and you set $\tilde v = \int \sqrt{g(v)}dv$ ... It's not literally correct that $G=1$, but you can change coordinates to make it so. – Ted Shifrin May 29 '14 at 20:45

## 1 Answer

Know a curve $γ(s)=(u(s), v(s))$ on a surface parametrized by arc length is a geodesic if and only if

$$\dfrac{d}{ds}(Eu'+Fv')=\dfrac{1}{2}(E_uu'^2+2F_uu'v'+G_uv'^2)$$ o $$\dfrac{d}{ds}(Fu'+Gv')=\dfrac{1}{2}(E_vu'^2+2F_vu'v'+G_vv'^2).$$

Then, use the form

$$du^2+G(u,v)dv^2$$

If $K=0$ then $\sqrt{G}_{uu}=0$ so $$G(u, v)=A(v)u+B(v).$$ But at $u=0$, $x_u$ and $x_v$ are unit so $B(v)=1$.

Also, the curve $u=0$ is a geodesic with $v$ arc length. So the geodesic equation

$$\dfrac{d}{ds}(Eu'+Fv')=\dfrac{1}{2}(E_uu'^2+2F_uu'v'+G_uv'^2)$$

gives $0=G_u(0, v)$ and this means in our case $A(v)=0$. The first fundamental form is therefore $$ds^2=du^2+dv^2$$ also is isometric to the plane.