A Topological Menagerie

Some of the party games that young teenagers play have surprisingly rich mathematical easy in mind "Entanglement" is one such game in which braces are linked with strings tied tightly to their wrists, as shown in Figure 1a, and then challenged to disentangle themselves, ofttimes leading to Figure 1b. In fact there is a quick solution (needles to say missing the point of the game): simply push a bit of the girl's string below the boy's at his wrist, pass the resulting noose over his hand, and then pluck it free on the other side, as shown in Figure 2

One way to formulate this mathematically is to ask for an isotopy to dislodge the meridian loop m from a rigid wire w embedded in the 3-sphere, as shown in Figure 3a. (The wire exhibits the boy with his attached string, where the upper noose of the wire is his right hand, while the meridian shows the girl with her string.) Of course this is easy: just slide m along w to a point just below the upper bight of w, and then strain it out and pull it not on the top. Or put differently, mark that w is isotopic to the trivial wire *____* and in like manner the meridian slips right not on

A similar puzzle attributed to Stewart Coffin [2] was considered by dint of Inta Bertuccioni in a novel issue of this MONTHLY [1] Here the wire w is configured slightly differently (as in Figure 3b) and there is in fact no solution, that is, m cannot be isotoped not upon w. Bertuccioni proves this by means of an explicit calculation showing that m shows a nontrivial element in the fundamental cluster of the complement of if.

The end of this note is to give a more conceptual examination of the impossibility of solving Coffin's gravel We then generalize the trial using an elementary but nontrivial come in knot theory to present to view that all but one of the vast "menagerie" of possible embarrasss suggested by entanglement and Coffin's confuse are also unsolvable.

First consider Coffin's pose The wire w consists of sum of two units unknots ?« joined by an arc ?± as illustrated in Figure 4a. From this single obtains a knot k by the agency of banding the components of ? together along a, shown in Figure 4b Actually there are infinitely many knots that can be formed in this way, for there may be twists in the band. The individual pictured is readily seen to be the square knot.

Now note that if m could be isotoped not upon w, then it certainly could be isotoped not upon k, since the complement of k contains the full tale of w (with a thickened a bit). on the contrary then the lift k of k to the infinite cyclic overlay of the complement of m would consist of an infinite number of copies of k (Note that the full quantity of m is a solid torus, and the overlay is an infinite solid cylinder.) However, it is easily seen that the composings of k are unknotted, contradicting the fact that k is a nontrivial knot (the square knot). Indeed, viewing k as the closure of a tangle with axis m the link k is obtained through composing infinitely many copies of this tangle, as illustrated in Figure 5 (see for example, [4]) Thus each composing of k looks like a lengthy worm that can be contracted by pushing from its "free" extreme point

For a general gravel in the menagerie we allow any embedding of the wire for which ? is an unlink, while the arc ?± can be arbitrary. individual such puzzle is shown in Figure 6a. The foregoing argument demonstrates that there is no faith for a solution unless the associated knot k is trivial, because each composing of the lift k is unknotted (as seen in Figure 6b; in general slide m shut to an endpoint of ?± before taking the cover) However, by the agency of a theorem of Marty Scharlemann [5] k is trivial if and single if the arc ?± is isotopic (fixing ?) to a trivial arc. Thus "most" of the poses in the menagerie, indeed all on the other hand those equivalent to entanglement, have no solution.

From a practical point of view individual might ask how to recognize when a is trivial. This is actually a difficult question. However there is a simple proof for the triviality of the homotopy class of ?±: The fundamental cluster of the complement of ? is unrestrained of rank two, with generators x and y corresponding to the sum of two units components of ?. The arc ?± oriented from the x-loop to the y-loop of ? determines a word in x and y also denoted ?± Since the arc ?± can be made to spiral around ? at its endpoints, its homotopy class corresponds to an equivalence class of words: ?± ~ ?? if and solitary if there exist integers m and n for which ?± = x^sup m^??^sup n^ Thus the arc a is homotopically trivial if and single if the word ?± ~ 1 For example, the entanglement stagger has ?± = xy ~ 1 as wait fored whereas Coffin's puzzle has ?± - yx * 1 and the bewilder in Figure 6a has a = yxyx * 1

Of course there are many homotopically trivial arcs ?± that are isotopically nontrivial, or equivalently, whose associated knots are nontrivial. Consider, for example, the confound in Figure 7, which has a = xy ~ 1 The associated knot has Jone polynomial t^sup -5^ - t^sup -4^ - t^sup -1^ + 2 - t + t^sup 2^ + t^sup 5^ - t^sup 6^ (see [3]; its Alexander polynomial is trivial), in like manner this puzzle has no solution.