Wrapping a Rubber Band around a Cube

Question

Imagine a cube wrapped with string so that each of the six sides has two linked "L's" on it, like this image (or some reflection or rotation thereof).

You can get a lot of different knots and links depending on the particular choices made.

Are any of them the unknot or the unlink?

In other words, can any of them be achieved by wrapping real rubber bands around a real box?

Answer 1

Here's the geometric net of a cube:

I represented the cube as a graph $g$ of size $6$ vertices, each vertex of degree four (links to its neighbors).

Then I searched for an Eulerian path through the graph $g$ with the conditions that no three consecutive sides were collinear. (That is, there must be an L turn at each vertex.) Here is one such Eulerian path:

$\{a \to b \to f \to a \to d \to f \to e \to d \to c \to e \to b \to c \to a \}$

There are then $2^5$ distinguishable choices for "over-under." Some will be trivial, others (ummm.... "knot"!).

Answer 2

No, it's not possible. The bands must be linked together before being wrapped around the cube.

Fun question. I wrote code to (intelligently) manipulate and check the knotworks. None of them have the same invariant properties as an unlink—I checked the crossing number, the number of three-colorings, and the Jones polynomial evaluated at $t=9$. This was sufficient to rule out all 216 distinct configurations (1024 nominal configurations before accounting for rotational and reflective symmetry).

1. The wrapped cube face shown in the question has 180 degree rotational symmetry. This means there are two options for each cube face—unrotated, or rotated 90 degrees—plus two more options if you're allowed to reflect it in a mirror.

   Out of $4096=4^6$ nominal configurations, I found that there are actually only 216 distinct configurations once you account for rotational and reflectional symmetry. (I am not sure why the number is $216=6^3$—maybe someone can point out a combinatorial connection.)

2. Next, I computed the crossing numbers invariant for each knot. The crossing number of two strings is the (signed) number of times they wrap around each other. The crossing number of any pair of strings in the unlink is zero; hence the only candidates for unlinks are those with all zero crossing numbers.

   However, out of these 216 candidates, all of the multistring knots failed the test. The only ones with zero crossing numbers were the single-string knots which, of course, have no crossing numbers at all because there are no other strings. This made me suspect that none of these knots would end up being an unknot.

   There were 216 candidates, comprising 61 (one string), 90 (two strings), 54 (three strings), and 11 (four strings). Hence only 61 candidates remained.

3. I calculated the three-coloring invariant next. This invariant says: suppose you color each segment of a knot in one of three colors, where the underthread at a crossing counts as two separate before-and-after segments. How many ways are there to do this such that (1) at least two colors are used in the coloring, (2) at every crossing where three segments meet (an over crossing and two under crossing segments), they are all the same color or all different colors.

   The unknot cannot be three-colored this way, due to a technicality that it can only be a single color throughout. Hence our candidates are required to have this same three-colorable defect as well. Out of the 61 remaining single-threaded knots, only thirty knots qualify as being uncolorable.

4. Finally, I calculated the Jones polynomial for the remaining thirty knots. Technically, I calculated the value of the polynomial at $t=9$ because it made the math nice and meant I didn't have to develop a data structure for polynomials.

   Programming the Jones polynomial calculation was admittedly a bit hairy—at first, I worried that I made a mistake somewhere, but I hand checked the answers for a few knots and tried it out on some well-known knots like the trifoil and it got the correct answers. So I have good reason to believe the code is correct.

   The value of the Jones polynomial for the unknot is a constant $V_\bigcirc(x)=1$. However, none of the thirty knots has that Jones polynomial value. Hence none of these single-string cube formations are the unknot, QED. (For good measure, I also computed the Jones polynomial of all 216 original configurations, and found that none of them matched the value of the Jones polynomial for any unlink with one, two, three, or four strings.)

I am sure there are deeper explanations for why this particular configuration cannot be made into an unlink, but for now I hope this empirical check will serve as a useful source of information and exploration.