knot theory problems and solutions

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 Knot Theory Problems/Solutions Problem 1: Is the figure 8 knot equivalent to its mirror image? It turns out we can manipulate the one on the left using R-moves to get the one on the right. Problem 2: Construct a system of four links such that if any one link is r emoved the whole thing falls apart. (Basically a Borromean knot, but with four loops). In other words, show that: (or does not) R1 R3 R2 R2 = R2 =

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Knot Theory Problems and Solutions

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  • Knot Theory Problems/Solutions

    Problem 1: Is the figure 8 knot equivalent to its mirror image?

    It turns out we can manipulate the one on the left using R-moves to get the one on the right.

    Problem 2: Construct a system of four links such that if any one link is removed the whole thing falls

    apart. (Basically a Borromean knot, but with four loops).

    In other words, show that: (or does not)

    R1 R3

    R2

    R2 = R2

    =

  • There may be more ways to do this, but we found a way that just links all four loops together into one

    big loop.

    Problem 3: Show that a braid attached to its mirror image produces 0.

    This problem can be explained very easily intuitively. Lets call the braid A. In order to undo A (make it

    0) we need to go backwards through all the steps we did to make A (like all the crossings of the

    strings, etc). This means that we need to reverse all of our displacements. This can only be done by

    attaching a braid onto A that is perfectly symmetric to A about the center, which is the mirror image of

    A. Essentially, a mirror image just goes backwards through the steps that created the braid, which

    unravels the whole braid, producing straight strings.

    Problem 4: How many elements are there in the group of symmetries of a cube?

    First, we can put planes inside a cube to make symmetry.

    If any one loop is removed, the rest

    can slide apart. This concept can apply

    to any number of loops

    Basically, show that this works for any

    braid its mirror image.

  • There are also rotational symmetries. Think of every sides as being a different color (6 different colors).

    The cube can be positioned so that any of the six sides is facing up. With each of these 6 sides facing up,

    it can be turned horizontally 4 different ways. All together, there are 24 (6*4) rotational symmetries.

    In total, there are 33 symmetries (3+9+24).

    Problem 5: Show that the knot drawn below is an unknot.

    There are 3 planes that can be placed

    that are parallel to the sides of the

    cubes, like the one on the left.

    And there are 6 planes that can be

    drawn from the corners of the sides

    This should be solved using R-moves,

    but that takes a really long time, so

    Im just going to do it in a few moves

    so that everyone gets the general idea

    of how to solve it.