MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

You and I decide to play a game:

To start off with, I provide you with a frictionless, perfectly spherical sphere, along with a frictionless, unstretchable, infinitely thin magical rope. This rope has the magical property that if you ever touch its ends to each other, they will stick together and never come apart for all eternity. You only get one such rope, but you are allowed to specify its length.

Next, I close my eyes and plug my ears as you do something to the rope and the sphere. When you are done with whatever you have decided to do, you give me back the sphere and rope. Then I try my best to remove the rope from the sphere (i.e., make the smallest distance from a point on the rope to a point on the sphere at least 1 meter). Of course, since the rope is not stretchable, the total length of the rope cannot increase while I am trying to remove it from the sphere.

If I succeed in removing the rope from the sphere, I win. Otherwise, you win. Who has the winning strategy?

EDIT: To clarify, Zeb is looking for an answer with a finite length, piecewise smooth rope, and the sphere should be rigid.

share|cite|improve this question
Should the rope length be finite? And the embedding (piecewise) smooth? I don't know if there's a silly way to do it using a non-smooth embedding (Peano curve?) or infinitely long rope, but the title suggests that that isn't what you're after. Can you confirm that? – Loop Space Dec 7 '09 at 12:51
It's a fun question, but see Ian Agol's answer below. It looks like there are fake solutions that use friction or possibly rope with non-zero thickness. With infinitely thin rope which is either slippery or does not touch itself, there is strong evidence that the answer is no. – Greg Kuperberg Dec 8 '09 at 18:38
Is it true that any convex body can be removed? – Anton Petrunin Dec 23 '09 at 21:55
@Anton, you should post it as another question! – Ilya Nikokoshev Dec 27 '09 at 19:51
See also the related question, "Hanging a ball with string." – Joseph O'Rourke Sep 30 '15 at 22:54

11 Answers 11

up vote 50 down vote accepted

Without loss of generality we can assume that rope is everywhere tangent to the sphere. Then some infinitesimal Möbius tranform of its surface will shorten the wrapping length while preserving the crossing pattern (and therefore will loosen the rope without causing it to pass through itself). Once it is proved, moving in this direction will eventually allow the sphere to escape.

Proof. let $u$ be conformal factor. Since Möbius tranform preservs total area $\oint u^2=1$ . Thus, $\oint u<1$. It follows that for a sutable rotation of $S^2$, we get length decreasing family of Möbius tranforms.


  • The same proof works for link made out of 3 circles.

  • It is easy capture sphere in a link from 4 circles. (4 strings go around 4 faces of regular tetrahedron on $S^2$ and link an the vertexes as in the answer of Anton Geraschenko, the first picture)

  • BTW, Can one capture a convex body in a knot?

share|cite|improve this answer
Let $u$ be conformal factor. Since Möbius tranform preservs total area $\oint_{S^2}u^2=1$. Thus $\oint_{S^2}u<1$. It follows that for sutable rotation of $S^2$, we get length decreasing family of Möbius tranforms. QED – Anton Petrunin Dec 10 '09 at 5:43
Could somebody expand this solution for me? I'd really like to understand it, but it's a bit too terse for me now. Here's what (I think) I understand so far. Pick the standard Riemannian metric on $S^2$ so that the total surface area is 1. Then the pullback of this metric along any Möbius transform is obtained by scaling the original metric by a function $u$. Preservation of total area implies $\int_{S^2}u^2=1$, so $\int_{S^2}u\le 1$ (strict for non-trivial transforms), so any Möbius transform decreases length somewhere, but how do you get one that decreases the total length of the knot? – Anton Geraschenko Dec 16 '09 at 1:04
If you choose a random orientation of the sphere before applying the mobius transform, the expectation of the length of the image of the knot is smaller than the original length of the knot (by linearity of expectation). Thus, some orientation of the sphere has to work. – zeb Dec 16 '09 at 2:59
Can someone explain how this proof generalizes to two or three links and why it fails for four? Thanks! – aorq Feb 5 '10 at 19:57
Isn't an equilateral triangle an obvious positive, if degenerate, answer to the question "Can one capture a convex body in a knot"? It then seems that one could also thicken the triangle a bit to get a 3D answer. (No idea if one could also round the edges to get a strictly convex shape.) The equilateral triangle case I'm talking about has on one side 3 strings running parallel to edges and meeting at the center (with appropriate knot) and on the other side the strings continuing to separate the 3 tips from central hexagon. – Yaakov Baruch Feb 23 '10 at 21:10

I would try to "gift wrap" the sphere. Take the 1-skeleton of the octahedron, and take an Eulerian path. Then make each vertex into a twist, and pull tight, so that it looks like the union of three orthogonal great circles. I have no idea though whether this has any length decreasing deformations. Here's a crude stereographic projection, teased apart so you can see the crossings:

alt text

See lower left Japanese glass float:

alt text

There's some variations one could make to this construction, if this doesn't work. Something like this:

glass floats

Edit: Actually, I just realized that this won't work. One may deform the 1-skeleton of the octahedron by rotating both great circles toward the 3rd without changing length. So you can rotate the whole thing onto a great circle, then slip it off. The Japanese get around this issue by tying little knots where the strands clasp each other, which prevents the verices from moving. But of course this isn't allowed in the formulation of the problem.

Edit: I think the answer is "no". I found this abstract at Mathscinet, although I haven't looked up the article. Pronin proves that a (locally) minimal 1-complex (network) on a sphere is unstable. Since a piecewise linear "knot" on the surface of a sphere has underlying space a (multi) graph, one can deform the graph to decrease length, and the length of the knot decreases. The only thing I'm not quite sure about is whether Pronin allows multiple edges, but I'd be willing to bet that the same argument (whatever it is) works in this case.

share|cite|improve this answer
+1 for the beautiful pictures! What's the last one? – Scott Morrison Dec 8 '09 at 6:42
What does "locally minimal" mean? Apparently not the literal meaning... – Reid Barton Dec 8 '09 at 17:31
meaning that it is a critical point for length (like a geodesic is locally minimal). For a graph, this is equivalent to saying that the lengths of edges are geodesics, and the sum of the unit tangent vectors at a vertex = 0. – Ian Agol Dec 8 '09 at 17:54
@Scott: Type "glass float" into Google image search. – S. Carnahan Dec 10 '09 at 20:49

Edit: This solution is incorrect, but I'm leaving it here because I think it's still interesting.

Here's a solution Scott Morrison and I came up with.

Choose a trivalent graph on the sphere with the vertices connected by segments of great circles such that any slight pertubation of the vertices would result in an a greater total edge length. The really symmetric tetradodecahedron on the sphere is such a graph. Now replace each vertex with a "clove hitch vertex" and each edge by a pair of strands:

enter image description here

Introduce some twists along the edges in order to make the whole link into a single knot. Two strands running between a pair of vertices can't be separated without making them longer, so the only way to deform the knot-around-the-sphere is to deform the "underlying graph", and we've chosen the graph so that any deformation would result in strictly larger total edge length.

The end result (without the twists to make a single component) would look like this if you used a tetrahedron instead:

enter image description here

If you don't believe that the symmetric tetrahedron is minimal in this way, either prove it in another answer, or ask another question! Edit: As some of you have pointed out, the tetrahedron is not such a minimal graph, and we were being too greedy asking for a minimal graph where the vertices are so far apart. So you have to find some graph which actually is minimal. I think a dodecahedron should do the trick (edit: it doesn't), but I don't know how to prove it.

share|cite|improve this answer
Awesome! I'm voting you both up for this one :) – Andrew Critch Dec 8 '09 at 0:34
I think the regular spherical tetrahedron is maximal, not minimal, in terms of total edge length. Consider the 1-parameter family of tetrahedra with one vertex at the north pole and three rotationally symmetric vertices at angle x from the north pole. The total length (of the edges of the tetrahedron) for this family is 0 for x=0 and 3 \pi for x=\pi. The lengthiest member of this family is the regular tetrahedron. – Kevin Walker Dec 8 '09 at 2:13
@j.c.: Thanks. I added some new pictures. – Anton Geraschenko Sep 30 '15 at 18:47

The regular dodecahedron is not a local minimum for total edge length either. Consider the five vertices Vi, i = 1, ..., 5 of a face together with the "center" C of the face. The spherical triangle CV1V2 has angles 72°, 60°, 60°, so V1V2 > CV1. Therefore if we move all five of the Vi to C, the total edge length decreases. (This part would be true even if the edges joining the Vi to the rest of the dodecahedron did not extend to pass through C, by the triangle inequality.)

As we move the Vi symmetrically towards C along the edges CVi at constant speed, the total edge length is some smooth function which is smaller at the end of the motion than at the beginning. I want to prove the beginning is not a local minimum. If it were, there would be a second critical point in the interior of the motion. But that contradicts Torricelli's theorem—the angle CV1V2, which is 180° minus the angle between two of the edges incident on V1, is decreasing during the motion, and thus only once equal to 60°.

share|cite|improve this answer
nice argument! (slicker than my argument for the cube, too) What do you think about the graph you get by shrinking the top and bottom faces of the dodecahedron to points? The only motion I can think of for that one that might decrease length is moving the remaining ten vertices all upwards simultaneously. – zeb Dec 8 '09 at 20:22
I didn't follow all the details, but I'll ask anyway: Can this be adapted to show that any trivalent graph with a pentagonal face can't be a local minimum? Because every trivalent planar graph has a face with <= 5 sides. – David Speyer Dec 8 '09 at 22:17
I did try briefly but was unsuccessful. One first needs some point to play the role of C. I suppose it could even be the case that any point inside the face will do; my intuition for spherical geometry is very bad. – Reid Barton Dec 9 '09 at 17:30

Some "climber's intuition" suggested that it might be possible to do this with an ellipsoid. The idea is to tie two clove hitches near the pointy ends:

   alt text

This doesn't seem to quite work -- either clove hitch can move towards the centre and simultaneously loosen. Maybe it's possible to get the ellipsoid just right so that it's "pointy enough the clove hitches can't fall off the end", and "round enough that the clove hitches can't loosen fast enough".

Even better, after looking at this, Anton and I realised that you can build "trivalent vertices" essentially using a clove hitch. He's about to post a followup answer explaining that geodesic trivalent graphs can be realised by "tight" knots.

share|cite|improve this answer

Well, here's a proof that the symmetric tetrahedron is not a local minimum - in fact, I claim that it's basically a local maximum!

Fix the positions of three of the vertices $v_1$, $v_2$, $v_3$ of the symmetric tetrahedron. First, let's find the point $p$ on the sphere for which the sum of the lengths of geodesics connecting $p$ to $v_1$, $v_2$, $v_3$ is minimal. By Torricelli's Theorem, this point must either be one of $v_1$, $v_2$, $v_3$, or a point where all edges leaving it meet at $120$ degrees. Thus, $p$ is one of $v_1$, $v_2$, $v_3$, the fourth vertex of the symmetric tetrahedron, or the antipodal point to one of the four points I already mentioned. By direct calculation, we see that $p$ is the antipodal point to the fourth vertex of the symmetric tetrahedron.

Now if we move the fourth vertex to any point $q$ nearby, and draw three great semicircles connecting it to the antipodal point $q'$ and passing through the other three vertices of the symmetric tetrahedron, we see that the sum of the lengths from $q$ is three pi minus the sum of the lengths from $q'$. Since the sum of the lengths from $q'$ is at least the sum of the lengths from $p$, the sum of the lengths from $q$ is at most the sum of the lengths from the top vertex of the symmetric tetrahedron. Thus, we can basically move the top vertex anywhere we like without increasing the total length of the string.

The next thing you will probably be tempted to try is the symmetric cube (using the same trick to handle the vertices of degree three). In this case, each vertex actually is at a strict local minimum if you hold the other vertices fixed. However, I'm pretty certain that it's possible to move all four vertices on the top face simultaneously either to the top of the sphere or to the equator of the sphere without increasing the total length during the process.

Edit: Here's a proof that we can move the vertices on the top face of the cube all upwards or downwards without increasing the total length. Let $x$ be the angle that the line connecting a vertex on the top face to the center of the sphere makes to the plane through the equator of the sphere. We are going to calculate the total length of all string above the equator as a function of $x$ - it's going to be $4x + 4(\mbox{angle between adjacent vertices})$. The dot product of the vectors corresponding to adjacent vertices is $\sin^2(x) = \frac{1-\cos(2x)}{2}$. Letting $a = \cos(2x)$, we see that the total length above the equator is $2\arccos(a) + 4\arccos(\frac{1-a}{2})$. Thus, since $\arccos$ is a concave function for $a$ between $0$ and $1$, that total length achieves its maximum when $a = \frac{1-a}{2}$, i.e. $cos(2x) = a = \frac{1}{3}$, which is the initial angle we started out with on the cube.

share|cite|improve this answer
I checked the statement about the cube too, but your argument is a lot slicker than mine was. – Reid Barton Dec 8 '09 at 5:59

Adding to Zeb's proof that the tetrahedron can be deformed, one should notice that any tassellation of the sphere containing at least one hexagon (fullerene type) won't be rigid either. In fact already in the plane the regular hexagon can be inflated (at the expense of the 6 outgoing rays) with no change in total length, much more so on a positively curved surface. Like Zeb, I'm also skeptical about the cube. Perhaps the dodecahedron has a chance, since its pentagons are quite a bit flatter than the faces of either the cube or tetrahedron (but then the inflation procedure is also cheaper than for the triangle or the square).

share|cite|improve this answer

Since Anton's beautiful solution makes use of the symmetry of the sphere, I wonder how similar results could be proven, or counterexamples given, for any other convex shape, including 2-dimensional ones - i.e. infinitely thin 3-dimensional objects. I can't figure a way to tie a cube or a square, but it seems that an equilateral triangle could be tied by 3 connected loops starting from some knot at the center and going over each vertex (forming 3 equilateral triangles 1/3 the size of the original).

(I would have liked to just leave this post as a comment to Anton's proof, but I'm not allowed to do that. Should it perhaps be a new question?)

share|cite|improve this answer
It would make an excellent new question. – j.c. Dec 29 '09 at 2:16

I think the sphere cannot be removed. I think that in physical knot theory that is knot theory with constraints on quantities as length and thickness there are models where a quantity called entanglement is related to the curvature of the containing sphere. So in this case any transformation would have to be within a certain distance of the sphere. I suspect that this is enough to keep the sphere locked in. Here is a paper on knot theory that deals with entanglement:

share|cite|improve this answer

It seems that both the 2-agon and the octahedron (which after all is a collection of 3 somewhat constrained 2-agons) can be shrunk off the sphere, but with 0 derivative at the start, which means that they come close. Perhaps the icosahedron could work (as for tying 5 edges into one vertex, a tangle of simple knots should do). But notice that inflating a triangle (even though it's fairly flat on an icosahedron) will trade off with shrinking not 3 (tetrahedron) or 6 (octahedron) other edges, but 9 of them. Such trade off would increase the total length on a plane, but on the sphere it depends on the curvature, i.e. the size of the triangles (similarly to what I pointed out re. the dodecahedron in an earlier post).

I suspect the just like no hexagon at all can be allowed on a graph tessellating the sphere, so no square or pentagon can be allowed to belong to any vertex other than a 3-edged one. If that's the case, probably the cube (which no one seems to have ruled out yet), icosahedron and dodecahedron are the only candidates for graphs with strictly locally minimal total length.

Other tessellations of the sphere with mixed polygons probably shouldn't work either. For example, considering a spherical triangular prism (2 squares and a triangle meeting at each of the 3 vertices): while inflating a triangle is harder than on a spherical tetrahedron, inflating a square will be easier than on a spherical cube (whose square is less curved). Similarly for a pentagonal prism: advantageous from the point of view of its 5 squares, but not from that of its 2 pentagons (when compared with the dodecahedron).

Very nice problem!

share|cite|improve this answer
A single great circle also has 0 derivative at the start. I'm not convinced that this is a sign of coming close. – David Eppstein Dec 8 '09 at 14:58

Reid, excellent proof. It works for the cube too and even kills the icosahedron. In this last case I don't know what the angles of a triangle are exactly, so let's just say 60+ each. Then joining each vertex to the center gives smaller triangles with 120, 30+ and 30+ degree angles and it's then clear that shrinking a triangle towards its center will again reduce the total length. More generally, your idea seems to kill the possibility any small enough face with 3, 4 or 5 edges, and one of those is certainly needed somewhere! My bet is now confidently on the sphere always escaping.

share|cite|improve this answer
Generally, this kind of answer belongs in a comment. – Harry Gindi Dec 8 '09 at 18:32
Though, of course, brand new users can't comment. – Noah Snyder Dec 8 '09 at 18:43
Touch\'e, sir. (I don't know how to actually write the correct letter there, and jsmath doesn't interpret that, so I'm leaving it like that.) – Harry Gindi Dec 9 '09 at 2:41

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.