Borromean strings

We’ll begin with an intriguing set of three interlinked rings, the Borromean rings. They can’t be realised as solid circular tori, but ellipses work just fine. Here is a visualisation:

If you remove any one of the three rings, the other two rings are unlinked. However, it is impossible to separate the rings without breaking one of them. Borromean rings have appeared in the structure of molecules, snakes and in heraldry. It turns out that its structure is equivalent to the standard three-strand braid, which likewise has the property that removing any one strand unlinks the remaining two. Here’s a picture from Wikipedia:

The Borromean string problem

On this topic, I propose the following group-theoretic puzzle:

Consider strings over the alphabet {a, b, c, a’, b’, c’}, where a’, b’ and c’ are the inverses of a, b and c, respectively. Formally, these strings are elements of the free group on three generators. Does there exist a string S with the following properties?

  1. In the free group, S is not equivalent to the identity;
  2. When all instances of any symbol and its inverse are removed (or, equivalently, by replacing that symbol with the identity), the string S is equivalent to the identity.

For example, a b a’ c a b’ a’ c’ is almost a ‘Borromean string’, since it works when b is removed or when c is removed:

  • Removing b, we have a a’ c a a’ c’ = a a’ c c’ = c c’ = identity;
  • Removing c, we have a b a’ a b’ a’ = a b b’ a’ = a a’ = identity;
  • However, removing a gives b c b’ c’, which is not the identity, so it is not a Borromean string.

Feel free to discuss any progress/solutions you have in the ‘comments’ section below:

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14 Responses to Borromean strings

  1. Kim Whittlesey needed to solve this problem for her Ph.D. dissertation. The answer is yes, they exist for arbitrary numbers of letters: just take commutators.

    Define [a,b] = a b a’ b’. This is nontrivial in the free group, but becomes trivial if either a or b is.

    Then [[[[[a,b],c], d], e], f], for example, is a nontrivial word in the free group, but becomes trivial if you mod out by any of a/b/c/d/e/f.

  2. ggendler says:

    I think I have a solution, linked here:

    http://pastebin.com/ZNjMh3GV

  3. Some time ago I solved an equivalent problem: How to hang a picture at the wall, using a single cord and N nails, such that the image falls down no matter which one of the nail breaks. (It’s some kind of “optimized Murphy’s Law”)

    If you look at the cords using fundamental groups, you have exactly the same group structure with the same conditions.

    My solution was to use an arbitraty combinations of commutators [x,y] where each nail (group element) appears exactly once. For example, with 3 nails you’ll get solutions like:

    [a, [b, c]] (= a b c b’ c’ a’ c b c’ b’)

    [[a, b], c]

    [[c, a], b]

    where [x,y] := x y x’ y’.

    This is easily extended to more nails. For example, here are two of the many solutions for 4 nails:

    [a, [b, [c, d]]]

    [[a, b], [c, d]]

  4. apgoucher says:

    Interesting work, folks! I particularly like your demonstration that it generalises to arbitrary numbers of generators, similar to the Borromean rings themselves. Is there a solution avoiding the use of commutators?

    • ggendler says:

      If you mean “does there exist a solution that is not built up entirely of commutators?” then yes, I can construct one:

      abcb’c’a’bcb’c’acbc’b’cbc’b’a’

      And in general, for n letters including p, let x be a solution to the same problem in all n letters except p. then a non commutator solution is:

      pxpxp

      • ggendler says:

        That should read pxp’xpx’x’p’

        • I’m not sure if that should really count as non-commutator solution, as it heavily depends on the commutator [p’,x] (= p’xpx’) in the center of that pattern. More generally:

          If q is a working pattern, so is q^y for any y. (where the superscript “^” denotes the conjugate, i.e. q^y := y’qy)

          This is simply because the conjugation q^y vanishes if and only if q vanishes. In this particular case, setting q := [p’,x] and y := x’p’ leads to:

          [p’,x] ^ (x’p’) = pxp’xpx’x’p’

          More generally, can “blow up” any working pattern (for which you still need commutators!) by injecting arbitrary conjugations into the formula. For example,

          [[a,b], c]

          can be blown up to

          [[a^(cab),b]^c, c^(ab)]^(aabca)

          So I’d like to propose a more difficult question: Is there any working pattern that cannot be described as “commutator solution decorated with conjugations”?

          • apgoucher says:

            Yes, I’ve found a Borromean string which is very non-trivial, containing no balanced proper substrings (therefore no commutators):

            (c c a c’ c’ c’ a’ b’ b’ a a a b b c b’ b’ b’ c’ a’ a’ c c c a a b a’ a’ a’ b’ c’ c’ b b b)

            I haven’t checked its minimality, but I strongly suspect that there are no shorter examples.

            EDIT: Actually, I’ve found a shorter example (b c c b’ c’ a’ a’ c a b b a’ b’ c’ c’ b c a a c’ a’ b’ b’ a).

      • apgoucher says:

        Sorry, I asked that question very vaguely. The question I intended to ask is:

        Does there exist a Borromean string such that no proper substring is balanced (i.e. each letter appears precisely as many times as its inverse)? The entire string must be balanced, of course, for it to reduce to the identity when you remove all instances of a particular symbol.

  5. Erik Demaine says:

    This paper explores the picture-hanging problem from a more efficient/algorithmic perspective, as well as covering the history of the problem:
    http://erikdemaine.org/papers/PictureHanging_FUN2012/

  6. Pingback: GMT | Complex Projective 4-Space

  7. Marc LeBrun says:

    If someone were inclined to do some counting, the OEIS would value sequences with definitions like “number of [insert restriction here] Borromean strings of length n on k symbols”

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