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The question was originally asked here Doubly Ruled Surfaces and I am following the hint provided by the OP. That is, first show that $K\equiv0$ and then deduce that the surface is a plane.

Let the surface be $x(u,v)$. If we suppose w.l.o.g. that locally the orthogonal lines are $u$- and $v$-curves, then we have $F\equiv 0$ and we know that $K=-\frac{1}{2\sqrt{EG}}\Big(\Big(\frac{E_v}{\sqrt{EG}}\Big)_v+\Big(\frac{G_u}{\sqrt{EG}}\Big)_u\Big)$. I am confused about how to use the fact that the parameter curves are straight lines. Do we have $x_{uu}\cdot x_v=0$ which would imply $E_v=0$ and similarly, $x_{vv}\cdot x_u=0$ which would imply $G_u=0$?

If we know that $K\equiv 0$, how do we deduce that the surface is a plane?

Community
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  • When K=0 , we have several possibilities. Ruled surfaces can be shown to have K<0, due to generator twist. Only the plane has K=0. Developable surfaces like cones,cylinders and developable helicoids are among the K=0 surfaces. – Narasimham Feb 29 '16 at 16:52
  • Is there a easy way to see that a developable surface with with orthogonal rulings must be a plane? –  Feb 29 '16 at 17:08

1 Answers1

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This question looks familiar. Yes, your approach to show $K=0$ is correct. Now remember that a plane is characterized by having $0$ shape operator (or $0$ second fundamental form), so see what you can figure out about the second fundamental form.

Ted Shifrin
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  • This is what confuses me actually. I think because we already assume a local parametrization by parameter curves, we cannot also assume they are simultaneously parametrized by arclength (because a reparametrization might render them non-orthogonal)? So $x_{uu}$ is not necessarily zero. But if we write $x_{uu}$ as a linear combination of the basis ${x_u,x_v,n}$ which is orthogonal now, does the fact that the $u$-curve is a line implies that $x_{uu}$ is a multiple of $x_u$ and that $x_{uu}\cdot n=e=0$? But then we do not even need to show that $K\equiv0$. –  Feb 29 '16 at 18:25
  • Yes, this is perfect. So you'll get $e=g=0$. You'll soon see why the first part was relevant. – Ted Shifrin Feb 29 '16 at 19:18
  • Okay. I suppose after we show that $e=f=g=0$ for all points, the shape operator is always zero which can also be seen as zero times identity and then we use the fact that $K\equiv0$ to rule out the sphere case. –  Feb 29 '16 at 21:17
  • Well, the issue is: How are you going to show $f=0$? But then, from a more basic standpoint, you should have seen an easy proof (before you got to the theorem you're referring to about the sphere) that if the shape operator is always $0$, then the surface is a (subset of a) plane. After all, the unit normal vector never changes! – Ted Shifrin Feb 29 '16 at 23:40
  • I finally realized what I missed. I confused orthogonal parametrization which only guarantees $F=0$ with the parametrization where both parameter curves are line of curvatures which gives $F=f=0$, as we use both parametrizations often. If $K=0$ and $e=g=0$, $f$ has to be zero everywhere. And if the second fundamental form is zero, $n_u$ and $n_v$ are both zero, which implies $n$ constant and the surface is a plane. –  Mar 01 '16 at 00:26
  • Correct, @user269715. No lines of curvature at all in this problem. So how do you get $f=0$ from your hypotheses? – Ted Shifrin Mar 01 '16 at 00:30
  • We know that $K=\frac{eg-f^2}{EG-F^2}=0$ (which is a formula that works for any regular surfaces without additional assumptions) and that $e=0$ for all points. This implies $f=0$ for all points. –  Mar 01 '16 at 00:44
  • Good job. So now you know what the problem had you show $K=0$ first. (I'm actually curious to know where you got the problem. I had not seen it anywhere else when I put it in my differential geometry text.) – Ted Shifrin Mar 01 '16 at 00:47
  • We are using a mixture of the text you wrote (and put online, I suppose) and do Carmo's Differential Geometry of Curves and Surfaces. This problem is from your text. –  Mar 01 '16 at 01:33
  • Thought so ... :) – Ted Shifrin Mar 01 '16 at 01:42