# Complete developable surface in $\mathbb{R}^3$ is ruled

Let $X \subset \mathbb{R}^3$ be a complete smooth surface which is developable in the sense that its Gaussian curvature is identically zero. Wikipedia claims that such a surface is necessarily ruled, which makes perfect geometrical sense, but how does one rigorously prove this? Since one of the principal curvatures vanishes at each $x \in X$, certainly $X$ contains a short line segment through $x$, but how does the completeness hypothesis ensure that $X$ contains the entire line?

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Took a bit of digging. You want to look at some older books, in this case Dirk J. Struik, Lectures on Classical Differential Geometry. A surface in $\mathbb R^3$ is indeed developable if and only if the Gauss curvature is identically zero. This is on page 91 of the Dover reprint. What the word developable means needs work: it means there is a one-parameter family of planes of which the surface is an envelope.

The concept of envelope is explicit in the relationship between Pascal's Theorem and Brianchon's Theorem in projective geometry: http://en.wikipedia.org/wiki/Pascal%27s_theorem and http://en.wikipedia.org/wiki/Brianchon%27s_theorem as a conic is either thought of as generated by a family (set) of points or by a family of tangent lines.

See Struik, table on page 72.

Alright, now, there are ruled surfaces such as the hyperboloid of revolution $x^2 + y^2 - z^2 = 1$ that are ruled. They are not developable, which is a stronger condition...in M. do Carmo, Differential Geometry of Curves and Surfaces, he goes so far as to say the a ruled surface with an extra condition is called developable, this is formula (9) on page 194... the ruled surface is given by $$\vec{x}(t,v) = \alpha(t) + v \, w(t)$$ which is formula (8). The condition to have a developable surface is that $(w, \dot{w}, \dot{\alpha})$ be always linearly dependent, which he writes as the determinant of the evident three by three matrix being identically zero. Note that this includes cones of revolution, with a singular point. It is not until page 408 that do Carmo proves that a complete surface with vanishing Gaussian curvature is a cylinder or a plane.

So, I feel that you are mixing two issues. Vanishing Gauss curvature shows ruled, but the result could have singularities. If complete, simply meaning no self intersections or singularities, the surface is a cylinder, well, or a plane. The hyperboloid is not developable.

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Is there a straightforward proof of the last claim (that a surface satisfying my hypotheses is a cylinder)? By the way, I don't use "complete" to mean "no self intersections or singularities": this is subsumed in the word "smooth." Rather, I meant "complete as a metric space." –  Justin Campbell Jul 12 '12 at 20:22
Since the ambient space $\mathbb{R}^n$ is complete, in this case complete = (topologically) closed. But I wanted to avoid using the word "closed," since in this setting that sometimes means "compact," and of course ruled surfaces are never compact. –  Justin Campbell Jul 12 '12 at 20:32
Yes, it is straightforward, do Carmo book. It is also eight pages. –  Will Jagy Jul 12 '12 at 20:32

@Will Jagy One parameter family of planes is understood containing ruled generators in case of cone /cylinder, but in case of tangential developable surfaces what is the generating plane family?.. not intuitively obvious (to me). The line of stiction is said to play this role, but its source from any " plane" is cloudy. Regards.

OK, it is stated on pages 71 and Table on page 72 Struik,as envelope of osculating planes of curve which is itself the edge of regression...settles my question.

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