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.

  • $\begingroup$ 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.] $\endgroup$ – Ted Shifrin May 29 '14 at 17:46
  • $\begingroup$ 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 $G(u,v)=1$ $\endgroup$ – user153684 May 29 '14 at 20:30
  • 1
    $\begingroup$ 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. $\endgroup$ – Ted Shifrin May 29 '14 at 20:45

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


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


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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.