Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

A gömböc is a $3-$dimensional convex body (having uniform density) which has exactly one stable and one instable equilbrium position (see http://en.wikipedia.org/wiki/G%C3%B6mb%C3%B6c). Such a convex body keeps its properties under tiny perturbations. There exists thus a gömböc given by algebraic equations. What is the simplest (small degree and small coefficients) polynomial $P(x,y,z)\in\mathbb R^3$ such that $P(x,y,z)\leq 0$ defines a gömböc?

Added after Stopple's remarks: Existence of an algebraic gömböc is not obvious. I should have asked first for existence.

Two other questions concerning these objects are:

Is the set of all (algebraic) gömböcs connected?

Are there any gömböcs in dimension $>3$? (There are none in dimension $2$, every $2-$dimensional convex set has at least $4$ equilibrium points.)

share|improve this question
Where you claim "Such a convex body keeps its properties under tiny perturbations", the wikipedia page actually seems to say the opposite: "The shape of those bodies is very sensitive to small variation." So why should there be any algebraic solutions? –  Stopple Apr 6 '11 at 18:56
Thank you, Anton. Concerning Stopple's comment, I do not see any contradiction: tiny can be very tiny! –  Roland Bacher Apr 6 '11 at 19:06
I don't mean to sound argumentative, but I don't see why there are algebraic solutions. Neither wikipedia, nor the Math Intelligencer article wikipedia references, refer to any. –  Stopple Apr 6 '11 at 23:11
I agree that this is not obvious. The boundary of a gömböc can of course be approximated arbitrarily well by an algebraic surface but it is not obvious that such a surface defines a body which is convex. Moreover, small perturbation might create additional equilibria as shown in the paper of Domokos and Varkonyi. I should definitively have asked for existence! –  Roland Bacher Apr 7 '11 at 6:22
Semi-algebraic sets, are, however, doable, as I remark in my answer. –  Daniel Litt Apr 7 '11 at 6:28
show 1 more comment

1 Answer

Aside from the last question, about high-dimensional gömböcs, these questions are certain to all be open, and are also totally inaccessible via the methods of Domokos and Varkonyi (who invented the gömböc and proved that it satisfied the constraints of Arnold's conjecture). For reference, here is their paper. To address Stopple's comment the paper uses throughout the stability of the desired properties (being a gömböc) under small perturbations, so there are semi-algebraic gömböcs (albeit likely very complicated ones, whose geometry is unlikely to be in any way revealing).

I went to a talk a couple years ago by Domokos and Varkonyi, where they described their methods and listed some of their conjectures. As I understood it at the time, they found the gömböc (and certain other shapes with the same equilibrium properties, albeit with much less deviation from a sphere) essentially by guessing a form for the solution, and then doing a search by computer within the limited parameter space this guess gave them. Certainly the first two questions you ask (about algebraicity of gömböcs and connectivity of the space of algebraic gömböcs) are inaccessible via this method. Indeed, Domokos and Varkonyi do not even have reasonable conjectures as to the minimal number of faces in a polyhedral gömböc, as I recall; their best candidate has many thousands of faces.

I will sketch what I believe to be an answer to your last question, which I'd say is the most interesting of the three you ask (why are algebraic gömböcs better than, say, piecewise smooth gömböcs?).

High-Dimensional Gömböcs Exist

The idea is to take proceed by induction; take an $n$-dimensional gömböc and revolve it around the axis connecting its two equilibria. It's not hard to convince yourself that this is a gömböc; I'll be a bit more explicit, albeit still a bit sketchy.

Let $G$ be a gömböc and $L$ the line passing through the two equilibria of $G$---say $G$ is good if for $x\in L$, the cross-section of $G$ perpendicular to $L$ at $x$ contains $x$, or is empty; note that Domokos and Varkonyi's gömböc is good (in fact it is easy to see that any gömböc is good, but I don't feel like proving this). I claim that if a good gömböc exists in dimension $n$, a good gömböc exists in dimension $n+1$, which suffices by induction. Indeed, place an $n$-dimensional gömböc $G^n$ in $\mathbb{R}^{n+1}$, and let $L$ be the line passing through the two equilibria (called $S$ and $U$, for stable and unstable). Let $G^{n+1}$ be the body formed by revolving $G^n$ about $L$. Then $G^{n+1}$ clearly is convex (by goodness of $G^n$) and has a stable equilibrium at $S$ and an unstable equilibrium at $U$. We must check that it has no other equilibria. But this is clear; indeed, any point on the surface of $G^{n+1}$ other than $S$ or $U$ belongs to some rotated copy of $G^n$, in which it is not an equilibrium point; thus it is not an equilibrium point of $G^{n+1}$. Goodness of $G^{n+1}$ is clear.

Note that there is an interesting question asked in the Domokos/Varkonyi paper, which they answer in dimension 3; namely which configurations of stable and unstable equilibria are permissible among convex homogenous solids? They show that given some convex homogenous solid, they can make small local modifications to add points of equilibria, without affecting other equilibria. It seems to me that their methods generalize to $k$ dimensions straightforwardly, so by producing a high-dimensional gömböc we have also shown that we may have as many stable and unstable equilibrium points as we desire, as long as there is one of each.

Unfortunately Domokos and Varkonyi have not addressed a slightly more subtle question; namely, there are various types of unstable equilibria (e.g. saddles) which may be classified as follows: consider the vector field on a (smooth) gömböc whose value at a point is the force exerted on the gömböc when it is resting on that point; equilibria are points where the vector field vanishes. Then the signature of the derivative of this vector field gives a reasonable classification of equilibria, assuming they are isolated. It would be interesting to know what are the permissible configurations of these slightly more refined equilibria.

share|improve this answer
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.