# Smoothness of ruled surface (asymptotic) parameterisations

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.

$$\newcommand\C{\mathcal C}\newcommand\ga{\gamma}\newcommand\sgn{\operatorname{sgn}}$$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:  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 $$\begin{equation} S=\big\{\big(x,Y(x,z),z\big)\colon 0 where \begin{equation} \begin{aligned} Y(x,z)&:= 1-\frac{1+(z-1)\sgn x}2\,g\Big(\frac{2x}{1+(z-1)\sgn x}\Big) \\ &:= 1-x\, h\Big(\frac{2x}{1+(z-1)\sgn x}\Big), \end{aligned} \end{equation} $$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. 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.

• 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 $d\in\mathcal{C}^1$. Apr 20 at 6:14
• 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. Apr 20 at 7:18
• 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? Apr 20 at 7:32
• @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. Apr 20 at 12:42
• I have also added an explanation on how this ruled surface was obtained. Apr 20 at 12:43