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?


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.

  • $\begingroup$ 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." $\endgroup$ – Justin Campbell Jul 12 '12 at 20:22
  • $\begingroup$ 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. $\endgroup$ – Justin Campbell Jul 12 '12 at 20:32
  • $\begingroup$ Yes, it is straightforward, do Carmo book. It is also eight pages. $\endgroup$ – 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.


$\def\RR{\mathbb{R}}$I had trouble extracting this from Struik, probably because I don't know enough classical facts about envelopes of families of planes, so I'll write up my solution. I'll say more about how Struik confuses me below. (On rereading your question, I was confused at an earlier point than you: I didn't get why everywhere flat implied $X$ contains line segments. Note that Gauss map = constant on a curve doesn't imply that curve is a line; consider the circle $\{ x^2+y^2=1,\ z=0 \}$ in $z=(x^2+y^2-1)^2$. So my answer addresses the thing which you say is clear.)

Here is the set up. We have a smooth surface $X \subset \RR^3$. The Gauss map is denoted $N : X \to S^2$. The area form on $S^2$ is denoted $\omega$, and the pullback $N^* \omega$ is the Gaussian curvature. There are three related concepts discussed in Struik:

(1) $N^{\ast} \omega=0$.

(2) There is a family of planes $H(u) = \{ \vec{x} : \vec{x} \cdot \vec{p}(u) + a(u)=0 \}$ dependent on a parameter $u$, such that $X$ is locally the envelope of the $H(u)$. (See p. 66.) From what I can learn from Wikipedia, the envelope is defined as the union of the characteristic lines $$\ell(u) = \{ \vec{x} : \vec{x} \cdot \vec{p}(u) + a(u)=\vec{x} \cdot \vec{p}'(u) + a'(u)=0 \}.$$

(3) There is a family of lines $\ell(u) = \{ \vec{b}(u) + v \vec{z}(u) \}$ such that $X$ is locally the union of the $\ell(t)$. Moreover, $\det(\vec{b}', \vec{z}, \vec{z}')=0$. (See page 90, but note that I have changed the variable names so that the $u$'s in (2) and (3) will be related and no unrelated quantities have the same name.)

Condition (3) without the determinant condition is the condition for $X$ to be ruled.

Now, at some point, Struik must show that $\ell(u) \subset X$. I can't figure out where he does this! I see where he build the family of places (p. 91), and I can do it in a cleaner way. I see where he writes down the formula for $\ell(u)$ (p. 66, equation 4-1). But I can't see where he shows that $\ell(u) \subset X$. If he is using Wikipedia's definition of envelope, he needs to do this to show $(1) \implies (2)$. If he is using some other definition, then his discussion about the family of planes $H(u)$ may establish $(1) \implies (2)$, but the fact that $\ell(u) \subset X$ is undoubtedly part of $(3)$, so it has to be proved somewhere.

Since $N^{\ast} \omega=0$, and $\omega$ is everywhere nonzero, we see that the derivative $DN$ everywhere has rank $<2$. Let $A$ be the locus in $X$ where $\mathrm{rank} \ DN=0$ and $U$ the locus where $\mathrm{rank} \ DN=1$. So $A$ is closed and $U$ is its open complement. To make life simpler, I'll work in the interior of $A$ or of $U$, and not at the boundary. See this paper for more about the boundary.

In the interior of $A$, life is very easy. $DN$ is everywhere zero, so $N$ is locally constant. Let $B$ be a connected component of $A$ and let $\vec{n}$ be the constant value of $N$ on $B$. Set $\lambda(\vec{x}) = \vec{n} \cdot \vec{x}$, for $\vec{x} \in B$. Then $d \lambda=0$ on $B$, so $\lambda$ is constant, and $B$ is contained in a plane. (2) and (3) both clearly hold.

In the interior of $U$, life is more interesting. By the constant rank theorem, $N$ locally factors through a curve $C$, with $q: X \to C$ a submersion and $\vec{p}: C \to S^2$ an immersion. Choosing a parameter $u$ on $c$, this is the function $\vec{p}$ from (2). Shrinking $X$ further, I believe I can assume that $q$ has connected fibers.

Fix $u \in C$ and consider the fiber $q^{-1}(u)$. Since $q$ is a submersion, this fiber is a curve. At every point $x$, the tangent vector to this curve is in $T_x X$, and thus perpendicular to $\vec{p}(u)$. So the function $x \mapsto \vec{p}(u) \cdot x$ has derivative $0$ on $q^{-1}(u)$, and is thus a constant. Let $-a(u)$ be the constant value of this function.

With these definitions of $\vec{p}(u)$ and $a(u)$, the plane $H(u)$ is tangent to $X$ along the entire curve $q^{-1}(u)$. So $q^{-1}(u)$ and $\ell(u)$ are a curve and a line in the plane $H(u)$. We know that $q^{-1}(u) \subset X$, and in fact $H$ is tangent to $X$ along $q^{-1}(u)$. We don't yet know that $\ell(u) \subset X$. We will establish this by (locally) showing that $\ell(u) = q^{-1}(u)$.

Okay, so let's do it. Let $x \in q^{-1}(u)$. Choose local orthogonal coordinates on $T_x X$ such that $\mathbb{R} (1\ 0)^T$ is the kernel of $(DN)_x$. We use the same coordinates on $T_{N(x)} S^2$. So $DN$ looks like $\left( \begin{smallmatrix} 0 & \ast \\ 0 & \ast \end{smallmatrix} \right)$. But now we use a crucial fact about the Gauss map -- its differential is self adoint! In other words, this matrix is symmetric and must look like $\left( \begin{smallmatrix} 0 & 0 \\ 0 & \ast \end{smallmatrix} \right)$. So the image of $(DN)_x$ is $\mathbb{R} (0\ 1)^T$. Now, the image of $(DN)_x$ is $p'(u)$, and the kernel of $(DN)_x$ is $T_x q^{-1}(u)$. So we deduce that $T_x q^{-1}(u) \perp p'(u)$.

Also, $T_x q^{-1}(u) \subset T_x X = p(u)^{\perp}$, so $T_x q^{-1}(u) \perp p(u)$. So the tangent line $T_x q^{-1}(u)$ points in direction orthogonal to $p(u)$ and $p'(u)$. In particular, this tangent line has direction independent of the choice of $x$, so $q^{-1}(u)$ is a straight line. And its direction is parallel to $\ell(u)$.

It remains to check that $q^{-1}(u)$ is not only parallel to $\ell(u)$, but equal. In other words, we know that $p(u) \cdot$ and $p'(u) \cdot$ are constant along $q^{-1}(u)$, but we need to know that the constant values are $-a(u)$ and $-a'(u)$. For the first, $q^{-1}(u) \subset H(u)$, where $p(u) \cdot x +a(u)=0$. For the second, choose a section $x : C \to X$ of the map $q$. (Implicit function theorem!) So $p(u) \cdot x(u) + a(u)=0$ for all $u$. Differentiating, $p'(u) \cdot x(u) + p(u) \cdot x'(u) + a'(u)=0$. But $x'(u) \in T_x X = p(u)^{\perp}$, so the second term is $0$, and $p'(u) \cdot x(u) + a'(u)=0$ as desired. $\square$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.