I have met this in diff. geometry class which states:
We are to determine if there exists a regular surface in $ R^3 $, $ S = f(u,v) $ with fundamental forms as follows:
$ I = \begin{bmatrix} cos^2(u) & 0 \\ 0 & 1 \\ \end{bmatrix} $
$ II = \begin{bmatrix} 1 & 0 \\ 0 & cos^2(u) \\ \end{bmatrix} $
I know that if S is a compact surface this cannot happen because the Gaussian curvature satisfies $ K=-1 $ and for a compact surface at least one point has positive K but can a general regular surface in $ R^3 $ have these fundamental forms? Also obviously $ cos^2(u) \neq 0 $ as first fundamental form should be invertible. Thanks for the help
There are at least two ways to calculate the Gaussian Curvature, when $M$ is a regular surfaces in $\mathbb R^3$.
By the second fundamental form: $$ K = \frac{\det II}{\det I} = \frac{\cos^2 u}{\cos^2 u} = 1.$$
By the first fundamental form alone (this equation holds when $F = 0$, there's another formula when $F\neq 0$):
$$ K = -\frac{1}{2\sqrt{EG}} \left(\frac{\partial }{\partial u} \frac{G_u}{\sqrt{EG}} + \frac{\partial}{\partial v} \frac{E_v}{\sqrt{EG}}\right) = 0$$
Thus such a surface do not exist.