# Ruled surface with negative/zero gaussian curvature.

So a surface $$S \in \mathbb{R}^3$$ is ruled if through each point $$p$$ there is a line in $$\mathbb{R}^3$$ entirely contained in $$S$$.

Show that the line through $$p$$ lies along an asymptotic direction.

Prove that if a surface is ruled, then $$K\leq0$$ at each point.

Can someone give a hint?

• For the second part, I would proceed by showing the following: If $K>0$ at some point $p \in S$, then for a small neighbourhood of $p$ in S, say $N_{p}$, $N_{p} \setminus \{p\}$ is contained in only one of the connected components of $\mathbb{R}^{3} \setminus T_{p}(S)$. This clearly rules out the existence of a line in $S$ through $p$. – Nick L Nov 5 '18 at 18:27

Approach 1: One can show that a ruled surface locally can be parametrised as $$x(u,v) = c(u) + v X(u),$$ where $$c$$ is a curve and $$X$$ a vector field along $$c$$. A calculation shows that the Gauss curvature is non-positive and that the lines $$v\mapsto x(u_0,v)$$ are asymptotic lines. This is probably not the quickest approach for this problem, but you can look up the calculations in, e.g. Kühnel - Differential Geometry. Curves-Surfaces-Manifolds (p. 85).
Approach 2: Let $$c\colon I \to S$$ be a unit speed parametrisation of a line in the surface $$S$$. Then $$T'=c'' = 0$$. You can easily show that $$S_N T$$ vanishes. ($$S_N$$ stands for the shape operator.)
Next you can use Euler's formula $$k_n = k_1 \cos \theta + k_2 \sin \theta$$. Here $$k_n$$ is the normal curvature in the direction $$\cos\theta e_1 + \sin\theta e_2$$, $$k_1$$, $$k_2$$ are the principal curvatures and $$e_1$$, $$e_2$$ are the principal directions. If there is one asymptotic direction, then $$k_1$$ and $$k_2$$ must have different signs and hence $$K$$ cannot be positive.