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?
[Math] Complete developable surface in $\mathbb{R}^3$ is ruled
curvaturedifferential-geometrysurfaces
Related Solutions
Short answer: the curvature is a property of the second derviative, and it's perfectly possible for a $C^1$ function to have a badly behaved second derivative.
In one variable, think about the function $f(t) = |t|^{3/2}$; it is $C^1$ but $f''(0) = +\infty$.
To turn this into a surface with infinite curvature, try $$X(u,v) = (u, |u|^{3/2}, v)$$ i.e. just extending this curve in the $z$ direction, forming a sheet. You'll find the normal curvature at the origin in the direction normal to the $xy$-plane is infinite.
Almost every sentence requires a correction. To begin with:
- "A sphere can be uniquely defined as a closed two dimensional manifold with constant positive Gaussian curvature which can be determined by measuring length and angles on the surface alone without taking reference to some ambient space."
No. The issue is that in math there are several objects, all called "a sphere." In the context of your question, you can say
"A round 2-dimensional sphere is a 2-dimensional Riemannian manifold which is closed, simply-connected and of constant Gaussian curvature."
Instead of calling it "round" you can also say "homogeneous." But some adjective is needed since in RG the word "sphere" is normally reserved for a (topological or smooth) 2-dimensional manifold, without any chosen Riemannian metric.
- The next sentence is almost right:
The total surface area, $A$, and the Gaussian curvature, $K$, of the sphere are related by the Gauss-Bonnet theorem as $$ A\cdot K~=4\pi \tag{1}. $$ The adjective "local" you are using is meaningless: By the very definition, Gaussian curvature is not just local, but an infinitesimal notion, depending on the 2nd derivative of the metric tensor. Saying "local curvature" is no more meaningful than saying "local derivative" or "local Taylor series."
- The next line is almost entirely meaningless:
"This [what does "this" refer to here?] is an equivalence relation [do you know what an equivalence relation is?] that allows to define static spherically symmetric spacetime as a fiber bundle with base $\mathbb{R}^{+}$ (surface area) and fiber $\mathbb{S}^2$ (two-sphere)."
There is no equivalence relation here at all. A fiber bundle over a line is always trivial, so you are really talking about the manifold $M=S^2\times {\mathbb R}^+$. You did not even attempt to specify a semi-Riemannian metric on this manifold, so you are not getting any spacetime here. In what sense is ${\mathbb R}^+$ "surface area"? (It is like saying "${\mathbb N}$ (age).") Maybe you are talking about a 1-parameter family of round spheres $(S^2, ag_1), a\in {\mathbb R}^+$, where $g_1$ is a Riemannian metric of curvature $+1$ on $S^2$ (unique up to isometry). The parameter $a$ equals the area of the Riemannian manifold $(S^2, ag_1)$. You can then equip $M$ with the semi-Riemannian metric $$ h(x,t)=tg_1(x) - dt^2, x\in S^2, t\in {\mathbb R}^+. $$ One can describe such a semi-Riemannian metric abstractly, without writing an explicit formula. For instance, you can say that $(M,h)$ is a semi-Riemannian manifold admitting a surjective semi-Riemannian submersion $f: (M,h)\to ({\mathbb R}^+, -dt^2)$, whose fibers $f^{-1}(t)$ (equipped with metrics induced from $(M,h)$) are isometric to round spheres of area $t$.
I stop here.
One reference that I like is the classical textbook on semi-Riemannian geometry,
B. O'Neill, Semi-Riemannian Geometry with Applications to Relativity, Academic Press, New York, (1983).
Best Answer
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.