A developable surface is a smooth surface whose Gaussian curvature vanishes everywhere. A ruled surface is a surface where for each point there must be a line passing through the point lying on the surface. See https://en.wikipedia.org/wiki/Developable_surface and https://en.wikipedia.org/wiki/Ruled_surface for both definitions. I know that a ruled surface need not be a developable surface, say a hyperboloid for example. However, I'm not able to show that all developable surfaces must be ruled.

It is a folklore that any developable surface must be ruled, and there are two references so far I know. One is Pogorelov's text book "exterior geometry of convex surfaces", but unfortunately I cannot get access to it. The other one is Hartman and Nirenberg's 1959 paper https://www.jstor.org/stable/2372995?seq=1#metadata_info_tab_contents, but in that paper they proved much more. I personally care about an answer which gives a simple understanding of the folklore without going into too many technical details.

I'm not sure if my intuition is correct. The following is how I visualize in my mind: If a surface $z=u(x,y)$ has vanishing Gaussian curvature everywhere, then the Hessian of $u$ is degenerate and has nontrivial kernels. In other words, the second fundamental form must be the squared of a $1$-form, say $II=\left(f(x,y)dx+g(x,y)dy\right)^2$. Hence along the curve defined by $dy/dx=-f(x,y)/g(x,y)$ the second fundamental form vanishes and this means the normal to the surface is constant along the lifting of the above curve in the surface. Then the only possibility is that the curve must be a line.

I don't have much geometry background. I wonder is the above intuition correct? Or can anyone give me a readable reference? Any comments or ideas would be really appreciated.

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    $\begingroup$ It's clear from your expression for $II$ that you mean developable surface in $\Bbb E^3$, but NB the claim is already false in $\Bbb E^4$: the flat torus is compact; ruled surfaces are not. $\endgroup$ Oct 6, 2019 at 3:20

1 Answer 1


I am sure it is written somewhere, but I do not know where.

First note that one can bend a plane disc from three sides leaving a flat triangle in it. The obtained surface is developable, but formally speaking it is not ruled. So you should change the definition of ruled surface to allow flat regions in it.

Now assume that second fundamental form does not vanish at a point $p$. Consider the tangent plane at $p$ and look at its intersection with the surface. Note/prove two things: (1) the surface (locally) lies on one side of the plane, (2) it is locally convex around the intersectio, in particular it intersect the plane along a line segment or a single point (3) the intersection is a segment with ends on the boundary (or unbounded in one or two directions) --- otherwise one could perturb the plane in a strict local supporting position in this case the curvature is positive --- a contradiction.

Assume that the second fundamental at $p$ vanishes. If $p$ is a limit of points $p_n$ with nonvanishing second fundamental form, then passing to the limit we get that $p$ (as well as each $p_n$) lie on a line segment from boundary to boundary. Otherwise $p$ lies in interior of a plane region. As a result your surface is a collection of line segments from boundary to boundary plus an open flat region.

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    $\begingroup$ Very nice. Is it also easy to see what happens when the second fundamental form vanishes at a point but not in any neighborhood of it? $\endgroup$
    – Deane Yang
    Oct 6, 2019 at 17:02
  • $\begingroup$ @Anton Petrunin, very nice argument! Thank you very much. $\endgroup$
    – student
    Oct 6, 2019 at 18:00
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    $\begingroup$ @DeaneYang if it is a limit point of points with nonvanishing second fundamental form, then you get it as a limit case; otherwise it lies in interior of a plane region. $\endgroup$ Oct 6, 2019 at 18:01
  • $\begingroup$ @AntonPetrunin, thanks. I did manage to see that right after I posted my question. $\endgroup$
    – Deane Yang
    Oct 6, 2019 at 18:11

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