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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
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