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Is there a $2$ - dimensional submanifold $S$ of $\mathbb{R}^3$ which can be parametrized with $x : U \subset \mathbb{R}^2 \to S$ such that :

$E=G=1$, $F=0$, $e=-g=1$ and $f=0$

Where $E,F,G$ and $e,f,g$ are the coefficients of the first and second fundamental form.

I think that such a submanifold exists. Indeed since the coefficients does not contradict the fact that the first fundamental form is symmetric non negative and the second fundamental form is symmetric I think such a submanifold exists.

But I am not able to prove it. I don't even know where to start!

Thanks in advance.

Arctic Char
  • 16,972
John Marjan
  • 195

1 Answers1

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Such a surfaces does not exist. Note that besides the symmetry of first, second fundamental form, they also need to satisfy the Gauss equation

$$2K = \det(I^{-1} II),$$

where $I, II$ are the first and second fundamental form respectively. However, from the first fundamental form we have $K = 0$, while $\det (g^{-1} A) = 1 \neq 0$. Thus such a surface does not exist.