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A ruled surface $S$ shall be defined as surface consisting of straight line segments. It is commonly known (cf. [BER, p.362] or [STR, p.93] - bibliography at the end) that a ruled surface allows for a parameterisation $$\phi_R:(u,v)\mapsto \gamma(u) + vd(u),$$ where $\gamma$ is a curve and $d$ a vector field. However, most of the basic literature does not care about smoothness. Therefore, I asked myself: If $S$ posesses a regular (linear independent partial derivatives) parametrisation $\phi_\circ\in\mathcal{C}^k$, what can I say about the smoothness of the ruled surface parametrisation $\phi_R$ and its ingredients $\gamma$ and $d$?

I tried to answer this question by proving that also $\gamma\in\mathcal{C}^k$ and that $d\in\mathcal{C}^{k-1}$ is possible. However, this topic contains the reference to a paper [USH] which claims that in general $d\in\mathcal{C}^0$ is the best you can get. This is in accordance with [HAR], who give another counter example. Now again, for both counter examples there were parts of the proves that did not make sense to me.

I would be most happy if you could explain my mistakes in (a) understanding the proves in literature and/or (b) in my own proof?


[USH, p. 415]

Therefore, the change of variables (3) [$(u,v)\mapsto(x,y)$] is not $\mathcal{C}^1$ smooth. And since the radius vector $r(x,y)$ is $\mathcal{C}^\infty$, the composition $r(u,v) \mapsto r(x(u,v),y(u,v))$ is not $\mathcal{C}^1$.

Is this correct reasoning? As a counter example, the composition $f(g(x))$ of $f:x\mapsto x^2$ and $g:x\mapsto \vert x\vert$ is continuously differentiable, although $g$ is not.


[HAR, p. 917]

The last statement follows from the example $S: z = (y)^4/(2-x)^3$. [...] $S$ has the parametrization $x=v, y=(u/4)^{1/3}(2-v), z=(u/4)^{4/3}(2-v)$, which is linear in $v$ [...]. This parametrization is continuous but not of class $\mathcal{C}^1$. An argument [...] shows that $S$ has no $\mathcal{C}^1$ parametrization of the desired type.

Obviously, choosing $x=v, y=u(2-v), z=u^4(2-v)$ is a simple reparametrisation which is still ruled, but continuously differentiable.


My proof

Finally, coming to my attempt of proving the following (probably wrong) hypothesis: If $\phi_\circ\in\mathcal{C}^k$ is a regular parametrisation of a ruled surface, then for the ruled surface parametrisation it is possible to choose $\gamma\in\mathcal{C}^{k}$ and $d\in\mathcal{C}^{k-1}$.

As $\phi_\circ$ is a regular parametrisation and therewith locally injective, I may locally apply the inverse function theorem. In this way, I project the "rulings" (these are the straight lines on the surface) to their original images in the parameter space. There I construct a regular $\mathcal{C}^k$ curve $\xi$ which is transversal to the original images of the rulings (for existence, cf. this topic). Then $\gamma := \phi_\circ \circ \xi$ is also $\mathcal{C}^k$ and regular.

Now for the director: As the surface $S$ consists of straight line segments, I may find a mapping $u\in I \subset\mathbb{R}\mapsto L(u)\subset S$, where $L(u)$ is a straight line segment. Every such line segment has an up to orientation unique unit vector $d(u)$ which is collinear with $x-y$ for all $x,y\in L(u)$. This means the straight lines provide existence of the vector field $d$, such that the ruled surface parametrisation $\phi_R$ exists.

If I additionally assume that $S$ is compact, then it is trivial that the parametrisations $u\mapsto \xi(u)$ and $u\mapsto d(u)$ may share the same parameter interval. I believe this is also true for non-compact $S$ as then I may choose $u\in\mathbb{R}$, however, I'm a bit vague here as I only consider compact $S$.

The only thing left is a statement about the smoothness of $d$. The director $d$ is the partial derivative of $\phi_R$ with respect to $v$, which again is the directional derivative of $\phi_\circ$ with respect to the direction $\phi_\circ^{-1}(d)$. By assumption, this directional derivative and therewith the director is $\mathcal{C}^{k-1}$.

Edit: I think I got the point where my proof goes wrong. When I specify that the partial derivative $\partial \phi_R / \partial v$ is equal to the directional derivative $\vec{\partial}_d \phi_\circ$, there is a change of parameters involved. If this change of parameters is not smooth enough, I cannot extract smoothness from the original parametrisation $\phi_\circ$.


[BER] Berger, Marcel; Gostiaux, Bernard, Differential geometry: manifolds, curves, and surfaces., Graduate Texts in Mathematics, 115. New York etc.: Springer-Verlag. XII, 474 p.; DM 98.00 (1988). ZBL0629.53001.

[STR] Struik, D. J., Lectures on classical differential geometry, Cambridge, Mass.: Addison-Wesley Press. VIII, 221 p. (1950). ZBL0041.48603.

[USH] Ushakov, Vitaly, Parametrisation of developable surfaces by asymptotic lines, Bull. Aust. Math. Soc. 54, No. 3, 411-421 (1996). ZBL0890.53004.

[HAR] Hartman, Philip; Nirenberg, Louis, On spherical image maps whose Jacobians do not change sign, Am. J. Math. 81, 901-920 (1959). ZBL0094.16303.

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$\newcommand\C{\mathcal C}\newcommand\ga{\gamma}\newcommand\sgn{\operatorname{sgn}}$The main question in your post appears to be as follows: "If $\phi\in\mathcal{C}^k$ is regular (linear independent partial derivatives), what can I say about the smoothness of $\gamma$ and $d$?"

At least as far as $d$ is concerned, the answer to this question is actually trivial: Since $d(u)$ is the partial derivative of $\ga(u,v)$ with respect to $v$, the condition
$\phi\in\mathcal{C}^k$ implies the desired condition $b\in\C^{k-1}$.

Apparently, you wanted to ask the following, much less trivial question: If the ruled surface is $\C^k$-smooth and regular, does it follow that $d$ is $\C^{k-1}$-smooth? This is the kind of question considered in your cited paper [USH].

The answer to this question is no: in general, the best you can get is $b\in\C^0$, even if the ruled surface is $\C^\infty$-smooth and regular.

E.g., let \begin{equation} g(u):=e^{-1/u^2} \end{equation} for real $u\ne0$, with $g(0):=0$. Let then \begin{equation} \ga(u):=\tfrac12\, \big(u,2-g(u),2\big) \end{equation} and \begin{equation} d(u):=\tfrac12\, \big(u\sgn u,-g(u)\sgn u,2\big) \end{equation} for $u\in(-2,2)$.

Then the image-set \begin{equation} S:=\{\ga(u)+v\,d(u)\colon u\in(-2,2),v\in(-1,1)\} \end{equation} is $\C^\infty$-smooth and regular (see details on this right after the picture below), and $d\in\C^0$. However, $d\notin\C^1$, since $d'(0-)=(-1/2,0,0)\ne(1/2,0,0)=d'(0+)$.

The only way to eliminate this discontinuity of $d'$ by a re-parametrization of the ruled surface $S$ is to replace the parameter $u$ by a parameter $t$ such that $\dfrac{du}{dt}\Big|_{u=0}=0$; however, such a re-parametrization of $S$ will not be regular. So, there is no $\C^\infty$-smooth regular parametrization of the ruled surface $S$ such that $d\in\C^1$.


The ruled surface $S$ is obtained by gluing together two ruled surfaces, $S^-$ and $S^+$, where (i) $S^-$ is the union of the (relatively) open intervals $\{(1-t)(0,1,0)+t(u,1-g(u),2)\colon t\in(0,1)\}$ with one endpoint at $(0,1,0)$ and the other endpoint on the curve $\{(u,1-g(u),2)\colon u\in(0,2)\}$ and (ii) $S^+$ is the union of the open intervals $\{(1-t)(0,1,2)+t(u,1-g(u),0)\colon t\in(0,1)\}$ with one endpoint at $(0,1,2)$ and the other endpoint on the curve $\{(u,1-g(u),0)\colon u\in(-2,0)\}$. The gluing is done along the open interval $\{(1-t)(0,1,0)+t(0,1,2)\colon t\in(0,1)\}$ with endpoints at $(0,1,0)$ and $(0,1,2)$.

Here is the corresponding picture:

enter image description here

enter image description here


Details of why $S$ is $\C^\infty$-smooth and regular: Informally, this follows because (i) $S$ is obviously $\C^\infty$-smooth at all points of $S$ with $u\ne0$ and (ii) $S$ is $\C^\infty$-smooth at all points of $S$ with $u=0$, since the degree of flatness of $S$ at such points is $\infty$.

Formally, for $(u,v)\in(-2,2)\times(-1,1)$,
\begin{equation} \ga(u)+v\,d(u)=(x,y,z):=\Big(\frac{u(1+v\sgn u)}2,1-\frac{1+v\sgn u}2\,g(u),1+v\Big). \end{equation} The correspondence \begin{equation} \begin{aligned} (u&,v)=\Big(\frac{2x}{1+(z-1)\sgn x},z-1\Big)\in(-2,2)\times(-1,1) \\ &\!\updownarrow \\ (x&,z)=\Big(\frac{u(1+v\sgn u)}2,1+v\Big) \in\{(\xi,\zeta)\colon0<\zeta<2,\zeta-2<\xi<\zeta\} \ \end{aligned} \end{equation} is a homeomorphism. It also follows that \begin{equation} S=\big\{\big(x,Y(x,z),z\big)\colon 0<z<2,z-2<x<z\big\}, \end{equation} where \begin{equation} \begin{aligned} Y(x,z)&:= 1-\frac{1+(z-1)\sgn x}2\,g\Big(\frac{2x}{1+(z-1)\sgn x}\Big) \\ &:= 1-x\, h\Big(\frac{2x}{1+(z-1)\sgn x}\Big), \end{aligned} \end{equation} $h(u):=g(u)/u$ for real $u\ne0$, with $h(0):=0$, so that $h\in\C^\infty$, with $h^{(k)}(0)=0$ for all $k=0,1,\dots$. It follows that the function $Y$ is $\C^\infty$-smooth on the set $\big\{\big(x,z\big)\colon 0<z<2,z-2<x<z\big\}$. Thus, $S$ is indeed $\C^\infty$-smooth. Also, it follows that $S$ is regular, since the vectors $\partial_x \big(x,Y(x,z),z\big)=\big(1,\partial_x Y(x,z),0\big)$ and $\partial_z \big(x,Y(x,z),z\big)=\big(0,\partial_z Y(x,z),1\big)$ are linearly independent.


Finally, concerning the various proofs you discuss/offer.

(i) Your proof I do not understand at all. In particular, I do not know what $\phi_\circ$ is or what you meant by "project the rulings (= straight lines) to the parameter space". Anyhow, as my example shows, your proof cannot be correct.

(ii) Concerning the example in the paper [HAR], cited by you: Here you are indeed right, in that the [HAR] example does not work. Concerning the [HAR] example, [USH] notes on p. 419: "Unfortunately, this is an unsuccessful example, since the surface even has an analytic standard parametrisation."

(iii) A large part of [USH] is devoted to clarification of previously given Klingenberg's example. Your objection to the reasoning on p. 415 of [USH] is somewhat valid. However, in your counterexample in this regard, the derivative of the map $x\mapsto x^2$ at $x=0$ is $0$. Apparently, [USH] forgot/neglected to mention that the corresponding map has a nonzero Jacobian determinant.

Anyhow, the example given above in this answer seems much simpler than Klingenberg's. The main idea of this example was borrowed from [USH], though.

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  • $\begingroup$ Thank you very much for putting that much work in my question! You definitely helped me with the two papers I cited. I'll edit my proof such that it gets easier to understand. I'm somewhat convinced that there has to be some problem in it, as I get the much stronger result than [USH]. However, I need some thinking over your example as it primarly shows that there is a ruled surface parametrisation with $d\in\mathcal{C}^0$ - where to rigorously contradict my attempted hypothesis, it needs to show that there is no possible parametrisation with $d\in\mathcal{C}^1$. $\endgroup$ Apr 20 at 6:14
  • $\begingroup$ To be more precise, it needs to show that there is no possible parametrisation with $d\in\mathcal{C}^1$ without loosing regularity of $\gamma$. For example, when I plugin $u=t^3$ the director $d$ becomes continuously differentiable with respect to $t$, but at the same time $\gamma$ becomes singular in $0$ as its derivative vanishes. $\endgroup$ Apr 20 at 7:18
  • $\begingroup$ I think I got it: If I specify some parameter change $t\mapsto u$, then the director derivative is $dd/dt = (du/dt sgn(u), -dg/du du/dt, 0)$, which is only continuous if $du/dt$ vanishes in $0$. But then the curve derivative $d\gamma/dt = 1/2 (du/dt , -2 dg/du du/dt, 0)$ automatically vanishes, as well. Maybe you could add this to your answer to make it a complete counter-example for my hypothesis? $\endgroup$ Apr 20 at 7:32
  • $\begingroup$ @BenjaminBauer : Thank you for your appreciation of this work. As you requested, I have added a paragraph on why it is impossible to eliminate the discontinuity of $d'$ without losing the regularity. $\endgroup$ Apr 20 at 12:42
  • $\begingroup$ I have also added an explanation on how this ruled surface was obtained. $\endgroup$ Apr 20 at 12:43

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