5
$\begingroup$

I have a few questions about nonplanar "stick circuits" (or hexagons and higher $n$-gons) that you might be able to help with:

(I know that $n=6$ is the minimum number of points to form a stick knot.)

  1. Given $n=6$ points in $\mathbb{R}^3$ in general position connected by a specific "stick circuit" (nonplanar hexagon), what test can be done to see if it forms a stick knot vs. an unknot?

  2. Given $n=6$ points in $\mathbb{R}^3$ in general position, there are 60 different stick circuits connecting them. True or false, at least one forms a knot?

  3. Given $n=6$ points in $\mathbb{R}^3$ in general position, does the minimum-length stick circuit on these $n$ points ever form a knot? ("knotted 6-point traveling salesman problem with return)

All these can be generalized to $n > 6$.

These questions occurred to me over the last few days. I suspect (1) has a known answer but I have no idea about (2) or (3).

$\endgroup$
5
  • 1
    $\begingroup$ For (2), no. Take the six points to be approximately the vertices of a regular octahedron. $\endgroup$ Jun 10, 2012 at 2:57
  • 3
    $\begingroup$ Actually, a regular octahedron (or close to it) is a configuration where I would always expect a stick knot. A pentagonal pyramid (or close to it) would be more challenging. Gerhard "Ask Me About System Design" Paseman, 2012.06.09 $\endgroup$ Jun 10, 2012 at 3:13
  • $\begingroup$ For (1) I know no simple test. However, all my mental images of 6-stick nots involve a stick with two acute angles at either end. I imagine this is a necessary condition. Gerhard "Or My Imagination Is Limited" Paseman, 2012.06.09 $\endgroup$ Jun 10, 2012 at 3:22
  • $\begingroup$ If the knot consists entirely of edges of the octahedron (or any other convex polytope) then it is a Jordan curve in that convex surface, therefore unknotted. The other possibilities can easily be handled case by case. $\endgroup$ Jun 10, 2012 at 4:04
  • $\begingroup$ (3) was asked a few days ago on m.se, math.stackexchange.com/questions/154385/… $\endgroup$ Jun 10, 2012 at 6:19

4 Answers 4

10
$\begingroup$

The answer to question (2) is

  • no for $n=6$,
  • yes for $n=7$.

For $n=6$, take, for example, the following six points as vertices of a straight-line (stick) embedding of $K_6$:

$A = (-2,-2,1)$, $B= (2,-2,0)$, $C= (0,2,0)$, $D= (-1,-1,0)$, $E= (1,-1,1)$, $F= (0,1,2)$

The projection onto the $xy$ plane has crossing number $3$.

a projection of K_6 with crossing number 3

Moreover, the crossings are between disjoint pairs of edges. Therefore, since every nontrivial knot has at least three crossings, there is at most one possible cycle that could form form a nontrivial knot; that is, the cycle $AECDBF$ formed by the six edges participating in the crossings. But by the above-below relations at the crossings, this cycle clearly forms an unknot.


For $n=7$, Conway and Gordon proved that every embedding of $K_7$ in $\mathbb{R}^3$ contains a Hamiltonian cycle forming a nontrivial knot, using the parity of the sum of the quadratic terms of the Conway polynomials of the Hamiltonian cycles as an invariant.

Edit: See also J. L. Ramírez Alfonsín, Spatial Graphs and Oriented Matroids: the Trefoil, Discrete and Computational Geometry 22:149--158 (1999) for the following stronger result:

Every stick embedding of $K_7$ in $\mathbb{R}^3$ contains a Hamiltonian cycle forming a (left-handed or a right-handed) trefoil.

$\endgroup$
1
$\begingroup$

A partial test for (1) is provided by the Fary-Milnor theorem. See also this question.

$\endgroup$
0
$\begingroup$

There is a problem about matching red and blue dots in the plane in pairs by straight line segments, with a length minimal matching involving no crossings. I imagine (3) could be answered negatively by similar reasoning.

Gerhard "Ask Me About System Design" Paseman, 2012.06.09

$\endgroup$
0
$\begingroup$

A paper by Gerard Venema and Tom Clark classified stick knots with 6 segments (using the lengths of the segments); they are using chains for their knots.

$\endgroup$
1
  • $\begingroup$ I realize now that you're looking specifically for circuits - my apologies. $\endgroup$ Jun 12, 2012 at 1:10

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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