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Is there a reason why there are no triply-ruled surfaces found in spatial geometry? Does it have to do with the fact that there are at most two dimensions/parameterizations for a surface? If that's so, then do 3D hyper-surfaces in a 4D+ space allow for triply ruled surfaces?

To be more clear, I am trying to find a proof of sorts that can demonstrate this fact, and whether or not the proof can be generalized to higher dimensional surfaces and spaces.

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The first Google result for "triply ruled surface" gives a book excerpt: "16.5 There are no non-planar triply ruled surfaces." books.google.com/… –  Qiaochu Yuan Jun 11 '11 at 22:52
    
"A treatise on analytic geometry in three dimensions" is a good source too: archive.org/details/cu31924001521065 –  JavaMan Jun 12 '11 at 1:18

2 Answers 2

Well, one can show that the only doubly rules surfaces are the plane, the hyperbolic paraboloid, and the single-sheeted hyperboloid, and none of these is non-planar and triply-ruled.

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One should probably add the hypothesis that the surface be connected and complete. –  Mariano Suárez-Alvarez Jun 11 '11 at 23:23
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I am asking how one can show this. –  Justin L. Jun 13 '11 at 6:02
    
@Justin: the classification of doubly ruled surfaces is given in many textbooks on differential geometry---for example, you can find it in the link provided by Qiaochu in a comment to your question. It is rather too elaborate to turn it into an answer here! –  Mariano Suárez-Alvarez Jun 13 '11 at 12:27

One way to think about the fact that no surface in space can be triply ruled as follows:

Firstly, one sees that a surface $S$ of degree $3$ or higher is not ruled at all; a cubic surface can contain at most $27$ lines in total, and higher degree surfaces typically contain no lines at all. (One way to prove this is via an argument with incidence varieties; see the sketch in this tricki entry.) So a non-planar ruled surface has to be a quadric (i.e. cut out by a degree $2$ equation).

On the other hand, if $\ell$ is a line passing lying on $S$, passing through a point $s \in S$, then $\ell$ lies in the tangent plane to $S$ at $s$. Since $S$ is a quadric, when you intersect it with a plane, the intersection is a (possibly degenerate) conic section, and so can contain at most two lines. Thus there are at most two lines on $S$ passing through any given point $s$, and hence a quadric is at most doubly ruled.

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Probably, one should somehow prove that a triply ruled surface is necessarily algebraic somewhere along the way, too. –  Mariano Suárez-Alvarez Jun 13 '11 at 12:28
    
@Mariano: Dear Mariano, You mean there are non-algebraic surfaces?! ... Cheers, –  Matt E Jun 13 '11 at 12:43

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