Show that a surface $\Phi$ with first fundamental form $ds^2 = udu^2 + vdv^2$ cannot
be locally isometric to a right circular cylinder.
Attempt:
The coefficients of the first fundamental form are:
$$
E = u , \quad F = 0 \quad \text{and} \quad G = v
$$
Now for the cylinder $C: \rho(u,v) = (r\cos u, r \sin u, v)$ we have:
$$
E _{cyl} = \rho_u^2 = r^2, F_{cyl} = 0 \quad \text{and} \quad G_{cyl}= \rho_v^2 = 1
$$
At this point, I cannot think of a way to proceed, to prove that these two surfaces cannot be locally isometric.
Obviously, $E, G$ are not equal to $E_{cyl}, G_{cyl}$ everywhere. Is this fact enough to prove the statement?
Best Answer
Suppose there exists a (local) isometry $F\colon U\subset\Phi \to C$ from (a part of) $\Phi$ to the circular cylinder. Then $F$ must preserve the Gauss curvature, i.e. $$ K(p) = K(F(p)) $$ for all $p\in U$. We know that the Gauss curvature $K=0$ everywhere on the circular cylinder. However, there is no open set on $\Phi$ where the Gauss curvature vanishes. One can easily verify this. We can use the following formula for $K$ for orthogonal parametrizations (i.e. parametrizations with $F=0$) $$ K = -\frac{1}{2\sqrt{EG}}\left(\frac{\partial}{\partial u}\frac{E_u}{\sqrt{EG}} +\frac{\partial}{\partial v}\frac{G_v}{\sqrt{EG}} \right) $$ and just substitute the values $E=u$ and $G=v$.
This is a contradiction and we conclude that there does not exist a local isometry from the surface to the cylinder.