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.