# Flat Surfaces in $\mathbb{R}^3$ Can Be Bent Only Along Straight Lines

This is a problem out of Elementary Differential Geometry by Barrett O'Neill (Chapter 6 Section 3 Number 2).

Let $M$ be a flat surface in $\mathbb{R}^3$ with principal curvatures $k_1$ and $k_2$, where $k_1 = 0$ and $k_2$ is never zero. Want to show that the principal curves of $k_1$ are line segments in $\mathbb{R}^3$.

There is a hint: With $\{E_i\}$ principal and $\alpha''= \nabla_{E_1}E_1$ and use the connection equations.

So far I have concluded that $\alpha'= E_1$, $\alpha ''=E_2$. So, $$\frac{S(\alpha')\cdot \alpha'}{\alpha' \cdot \alpha'}=\frac{\alpha'' \cdot U}{\alpha' \cdot \alpha'}=\frac{E_2 \cdot E_3}{E_1 \cdot E_1}=0.$$ And $S(\alpha') \cdot \alpha' = 0$ implies $\alpha$ is asymptotic. Which means $\alpha$ is not bending away from the tangent plane. Which I suppose I already knew because $k_1$ is $0$. Also I know that the Gaussian curvature is $0$ so my surface is either a plane or trough-shaped. But $k_2$ is non-zero so $M$ cannot be a plane.

I'm not sure how any of this helps me, nor do I know how to use the hint. Also any pointers on why this is intuitively true. I have been looking at examples but cannot figure out why this has to be true.

• Hi, Can you show how do you get $\alpha'' = E_2$? If $\alpha$ is a line you should get $\alpha ''= 0$. – user99914 Apr 21 '14 at 7:45