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Knots in 4D

  1. Sep 15, 2008 #1
    I've seen the statement repeated in many places that a stable knot comprising 1D curves cannot be created in 4D, since it's always possible to untie the knot by moving in the 4th dimension (whereas it is possible to create a stable knot in 4D using 2D surfaces). Can anyone point me to an authorative source for the proof of this statement (ideally available online)?

    Also, is anyone aware whether this statement remains true if the 4th dimension allows movement in only one direction (e.g. it represents time)? Intuitively I would guess that it wouldn't.
  2. jcsd
  3. Sep 15, 2008 #2


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    Here's the key paragraph from the WikiP article
    "In four dimensions, any closed loop of one-dimensional string is equivalent to an unknot. We can achieve the necessary deformation in two steps. The first step is to 'push' the loop into a three-dimensional subspace, which is always possible, though technical to explain. The second step is changing crossings. Suppose one strand is behind another as seen from a chosen point. Lift it into the fourth dimension, so there is no obstacle (the front strand having no component there); then slide it forward, and drop it back, now in front. An analogy for the plane would be lifting a string up off the surface."

    I wouldn't say WikiP is reliably authoritative, various articles I've looked at had misleading stuff. But this particular article seems OK, and what they say here is actually a short mathematical proof. You can see how it works. So in some sense we don't need to rely on authority in this case. For context, if you need help understanding the proof, you might look at the rest of the article.
    Last edited: Sep 15, 2008
  4. Sep 15, 2008 #3


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    saltlick if you can only move in one direction in that dimension it doesn't really make sense to speak of untying a 4D knot, right?

    The prove you look for is elementary so I don't know where to find it. If you look at the two dimensional projection of the knot distinguishing over and under crossings you get a link diagram. The three reidemeister moves take any diagram of one knot into that of the same know in 3D (and thus in 4D) but in 4D you can change undercrossings for overcrossing by the process you described. But now it is clear that all knots are equivalent to the unknot by the following procedure: Take a loop at the perimeter of the diagram. Remove all internal edges except for one using Reidemeister move III then remove the loop (first changing an undercrossing to an overcrossing if neccessary) by a type I move, this will remove one or two loops from the diagram. Rinse repeat until you are either left with one loop, that is, the unknot.
  5. Sep 15, 2008 #4


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    Hi Saltlick! :smile:

    Here's another way of looking at it:

    Look at two strings along the x and y axes, crossing each other at O, with the x-axis string on top.

    Make a sphere around O, and deform the x-axis string in the x-z plane so that it goes round the top hemisphere of the sphere.

    In 3D, you can't slide it round onto the mirror-image position on the bottom hemisphere because the positions through which it slides form a simply connected (2D) surface whose boundary is the vertical great circle in the x-z plane, which means it must be one vertical hemisphere or the other, which means that it includes one of the two points where the y-axis string crosses the sphere.

    But in 4D, you can slide it, because the slidey surface is still connected and 2D, and it only has two points it must miss, which is easy! :smile:
  6. Sep 15, 2008 #5
    Codimensions add to -1

    The heuristic rule is that two manifolds can be knotted when their codimensions add to -1. An n-dim manifold embedded in d-dim ambient space has codim d-n. Generically, the intersection between manifolds of dim m and n has codim (d-m) + (d-n) = 2d-m-n, and is thus a manifold with dimension k = m+n-d. If k = -1 the manifolds do not intersect (generically) but are knotted. In particular, if m = n = 1, k = -1 if d = 3.
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