# Doubly Ruled Surfaces

I get stuck in the following exercice:

Let $S$ be a doubly ruled surfaces by orthogonal lines.

1. Use the Gauss equations to prove that $K\equiv 0$;
2. Conclude that $S$ is a plane.

I tried to use a parametrization $X(u,v)=\alpha(u)+v\beta(u)$, usinge the fact that the lines are orthogonal and calculate the Christoffel symbols, but it did't worked.

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Could you remind us what a doubly ruled surface is? – Paul Dec 4 '12 at 1:12
A surface $S$ is doubly ruled if in each point $p\in S$ there is two straight lines passing through $p$ totally fully contained in $S$. – Jön Dec 4 '12 at 10:57

We don't even need Christoffel symbols or curvature:

We may assume that $S$ is given in the form $${\bf f}: \quad (x,y)\mapsto \bigl(x,y, f(x,y)\bigl)\qquad(*)$$ with $f(x,0)\equiv0$, $f(0,y)\equiv0$, i.e., the $x$-axis and the $y$-axis are lines on $S$.

By assumption through each point $(x,0,0)$ there is a line $g_x$ on $S$ which is orthogonal to the $x$-axis; therefore $g_x$ can be written in the form $${\bf g}_x:\quad s\mapsto\bigl(x, s, p(x) s\bigr)\ ,$$ where $p(x)$ denotes the slope of $g_x$ with respect to the $(x,y)$-plane.

Similarly, through each point $(0,y,0)$ there is a line $h_y$ on $S$ which is orthogonal to the $y$-axis; therefore $h_y$ can be written in the form $${\bf h}_y:\quad t\mapsto\bigl(t, y, q(y) t\bigr)\ ,$$ where again $q(y)$ denotes the slope of $h_y$ with respect to the $(x,y)$-plane.

The two lines $g_x$ and $h_y$ intersect at the point ${\bf r}_{xy}\in S$ characterized by ${\bf g}_x(y)={\bf h}_y(x)$. Looking at the third coordinate of ${\bf r}_{xy}$ we see that $$p(x)y=q(y) x\ .$$ As this is true for all $x$ and $y$ near $0$ there has to be a constant $c\in{\mathbb R}$ with $p(x)=cx$, $q(y)=cy$. This implies that the $f$ in $(*)$ is given by $f(x,y)=c\>xy$, or that $S$ is a parabolic hyperboloid, when $c\ne0$.

Since $g_x$ and $h_y$ should intersect orthogonally at ${\bf r}_{xy}$ we must have $$0={\bf g}_x'(y)\cdot {\bf h}_y'(x)=(0,1, cx)\cdot(1,0, cy)=c^2 x y\ ,$$ which implies $c=0$, so that in fact $f(x,y)\equiv0$, and $S$ is a plane.

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