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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 $\mathbb{R}$.

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It's got to be named after Clotho... –  Will Jagy Jul 11 '13 at 18:59
@Will: Right! Or rather, Clotho is named after the root "to spin." clothoid: "From Gk. kloth, from klothein "to spin" + epenthetic vowel -o- + eides "form," → -oid; because the curve is reminiscent of the thread that winds around a weaving loom." –  Joseph O'Rourke Jul 11 '13 at 19:11
Alright, now we need pictures of thread winding around a weaving loom. Or paintings or.. what did that Albrecht Durer do? –  Will Jagy Jul 11 '13 at 19:29
Dürer---etchings. –  Joseph O'Rourke Jul 11 '13 at 20:16

3 Answers 3

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

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

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?

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.

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As a side remark: the graph of the function $y = x\sin x$ also attains, as curvature, all real numbers. –  Willie Wong Jul 11 '13 at 10:34
Considering that the Gauss curvature is a scalar function and that a surface is parametrised by two, perhaps a more interesting question to ask is for an example of a surface where the ordered pair (mean curvature, Gauss curvature) (or similarly, the pair of principle curvatures) cover $\mathbb{R}^2$. –  Willie Wong Jul 11 '13 at 10:51
I like these questions, Willie! The pair of principle curvatures is especially appealing. –  Joseph O'Rourke Jul 11 '13 at 11:27
@Willie Wong: For a surface in $\mathbb{R}^3$, one always has $H^2\ge K$, so the image of the map $(H,K)$ will never be onto $\mathbb{R}^2$. Maybe you want the image to be onto the set allowed by this inequality. –  Robert Bryant Jul 11 '13 at 19:28
@RobertBryant: right. AM-GM of course. There's also the small complication that the pair of principle curvatures are not uniquely ordered for the other version. Perhaps for pair of principle curvatures version a better target will be $\{(x,y)\in\mathbb{R}^2: x \geq y\}$. –  Willie Wong Jul 12 '13 at 7:37

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

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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.

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I just noticed that I have misinterpreted the problem; what I tried to do, is to find a surface such that for each pair of principal curvatures there is a point of S with exactly that pair of principal curvatures. –  Manfred Weis Jul 12 '13 at 18:50
You are answering the followup question, "Surface in 3D that realizes all pairs of principal curvatures"! –  Joseph O'Rourke Jul 12 '13 at 19:01

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