2
$\begingroup$

This is a follow-up question to Can I wrap a suitcase with hair ties.

Now we know that it is possible to wrap a suitcase with hair ties without tying them together,

but can you do it with large rotational symmetry?

In other words, I am looking for a nontrivial link $L$ in a solid torus made by short circles that is trivial in the ambient euclidean space and such that $L$ is invariant with respect to a large cyclic group of rotations of the solid torus.

$\endgroup$
5
  • $\begingroup$ If you modify the question to ask for invariance under a cyclic group acting on $S^3$ without fixed points, then I think it will be possible. $\endgroup$
    – Ian Agol
    Commented Jun 16, 2023 at 2:29
  • 1
    $\begingroup$ @IanAgol I am interested. (Altho it has no applications for application to loom bracelet designs.) $\endgroup$ Commented Jun 16, 2023 at 11:12
  • 2
    $\begingroup$ @IanAgol - but this can be done by taking a disjoint union of many copies of the previous solution...... Suppose that U is the usual solid torus in S^3. Let V_i be n copies of the solid torus, each embedded in U, each meeting the meridian disk of U once, and no V_i linking any V_j (thought of inside of S^3). Then we can put the previous solution inside of each of the V_i and we are done. $\endgroup$
    – Sam Nead
    Commented Jun 16, 2023 at 13:34
  • 1
    $\begingroup$ @SamNead I did not formulate it precisely, but I wanted geometric symmetry. Your symmetry is only topological. $\endgroup$ Commented Jun 16, 2023 at 15:36
  • $\begingroup$ If you arrange the V_i symmetrically, and place the copies of the link inside the V_i symmetrically, then the result will have a geometric symmetry. (This is similar to the construction of lens spaces.) $\endgroup$
    – Sam Nead
    Commented Jun 16, 2023 at 16:35

1 Answer 1

5
$\begingroup$

Edit: my previous answer was incorrect

No, this one you cannot do. If one had such an arrangement of circles in the solid torus inside $R^3$, and quotiented by a rotation, then by the equivariant Dehn's lemma there would be a collection of essential disks in the complement of the split link that are invariant under the symmetry.

Since the link components get permuted, no disk can have a fixed point of the cyclic action. So they are disjoint from the axis of symmetry of the rotation. Hence the components are unknotted in the solid torus, a contradiction

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .