Surface analog of clothoid: curvatures covering $\mathbb{R}$

Question

The clothoid $C$, a.k.a. the Euler spiral, is one among many curves with the property that its curvatures cover $\mathbb{R}$ in the sense that, for every $x \in \mathbb{R}$, there is a point $p \in C$ such that the curvature at $p$ is $x$:
         
I am seeking surface analogs:

Are there examples of surfaces $S$ embedded in $\mathbb{R}^3$ with the property that for every $x \in \mathbb{R}$, there is a point $p \in S$ such that the Gaussian curvature at $p$ is $x$?

Although my main interest is in surfaces in 3D, one could ask the same question for Riemannian manifolds whose sectional curvatures cover $\mathbb{R}$.

Answer 1

Constructing such a surface is not too hard.

Note that for a smooth surface, Gaussian curvature is a continuous function. Hence if you have a connected surface it suffices to have points of arbitrarily large (in absolute value) Gaussian curvatures of either sign.

So consider the graph of the function

$$ z = f(x,y) = \cos(2\pi x) + x y^2 $$

We have that

$$ \mathrm{d}f = (-2\pi \sin (2\pi x) + y^2)\mathrm{d}x + 2xy \mathrm{d}y $$

and in particular

$$ \mathrm{d}f(n,0) = 0 $$

for $n \in \mathbb{Z}$. At those points the Gaussian curvature is simply the determinant of the Hessian

$$ K(n,0) = \det\begin{pmatrix} - 4\pi^2\cos(2\pi n) & 2\cdot 0 \\ 2\cdot 0 & 2n \end{pmatrix} = - 8 \pi^2 n$$

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Do you perhaps intend to add other criteria to your surface? The clothoid has the property that every curvature value is realised by exactly one point. This is of course not possible for a surface, but maybe you want a surface where every the sets $K^{-1}(k)$ are all homeomorphic or something like that?

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If you want a surface that is contained in a compact set in $\mathbb{R}^3$, consider the following map:

Let $D = \{ (\theta,s)\in \mathbb{R}^2 : \theta\in (-\pi,\pi), |s| < 1, |s^2 \tan \theta| < \frac12\}$

Let $\phi:D\to \mathbb{R}\times\mathbb{R}_+ \times\mathbb{S}^1$, the cylindrical coordinate representation of $\mathbb{R}^3$, be given by

$$\phi(\theta,s) = (s,\frac12 \tan\theta s^2,\theta) $$

The principle curvatures at point $(0,s)$ are $\{ 1, -\tan\theta\}$ and so the Gauss curvataure is $-\tan\theta$.

Our choice of domain guarantees that $\phi$ is an embedding and that $\phi(D)$ is contained in a ball of sufficiently large radius.

Answer 2

Here is Willie Wong's function $f(x,y) = \cos(2\pi x) + x y^2 $ at two different scales:
     

Answer 3

when reading about the problem, I almost immediately had the idea to define a surface via the combination of two clothoids in a similar fashion as two circles are combined to define a torus:

The first clothoid is defined via the $u$ parameter in the $xy$-plane as usual and will be traced out by a second clothoid that is defined via the $v$ parameter and, as the origin of the second clothoid moves along the first clothoid, $x(u)$ corresponds to $z(v)$ and, $y(u)$ corresponds to the orthogonal distance to the $xy$-plane's clothoid after a point's projection into the $xy$-plane.

I apologize for this rather coarse description; hopefully the example helps despite.