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Mobius strips and similar manifolds

  1. Jan 8, 2012 #1
    If you have a strip and you bring it around so that the ends join, that is a manifold, call it X for convenience. If instead, you put a single twist in it before joining the ends, that is a Mobius strip, which is not homeomorphic to X. If you instead put two twists in it before you join the ends, that is another manifold, call it Y for convenience. Is Y homeomorphic to X?

    My intuition says that it probably is. Y can be obtained from X by cutting X, adding the two twists and then sticking the ends back together. It can be done so that in the neighbourhood of the cut, the transformation is the identity, so it doesn't alter the topology. And around the rest of the manifold, all you've done is warp it a bit which also doesn't alter the topology - it would appear that this transformation should indeed be a homeomorphism. The two manifolds both have two edges, they come around and join themselves and they don't have any holes or singularities or anything like that in them.

    On the other hand, Y can't be obtained from X just by stretching and bending, which usually means that the two should not be topologically equivalent - is this a counterexample that highlights a limitation of that understanding of topological equivalence? Or is the reasoning from the former paragraph just wrong? There is a clear qualitative difference between the two manifolds when they are viewed embedded in 3d space.
     
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  3. Jan 8, 2012 #2

    mathwonk

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    bending and stretching are not precise definitions, just intuitive metaphors. look up the precise definition of homeomorphism and try to apply it.
     
  4. Jan 9, 2012 #3

    lavinia

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    The double twist and the no twist are homeomorphic - as is easily seen because the gluing rules along the edges are the same. But they are different embeddings of the cylinder into 3 space. I do not know anything about equivalence of embeddings but there is probably more than one idea. E.g. there is a smooth homotopy of immersions that brings one embedding into the other. Or there is a conformal homeomorphism of 3 space that brings one into the other. My gut tells me that no definition of equivalence of embedding will work here but I really don't know.
     
  5. Jan 9, 2012 #4
    Yes, the caveat about viewing homeomorphisms as bending or stretching is that it's okay to cut things as long as you glue them back together in the same way.

    Bending and stretching would seem to be referring to isotopic embeddings, but you might want to try to think of it more loosely than that.

    Essentially, a strip with twists is what topologists would think of as a "framed" unknot. Another way you could think of it is as a loop with a vector field along it. The isotopy class is determined by the number of times the tip of that vector field winds around the loop. I don't know about all the different definitions of equivalent embeddings, but with the usual one, each different framing gives you a different isotopy class.

    I think if you use isotopy in S^3,though, then they are isotopic:

    http://en.wikipedia.org/wiki/Regular_isotopy
    http://en.wikipedia.org/wiki/Ambient_isotopy
     
  6. Jan 9, 2012 #5
    I think it's only necessary to go back to fundamental, totally precise definitions when more intuitive notions fail, and I don't think intuition has failed here.

    ---

    The thread was informative, thanks.
     
  7. Jan 9, 2012 #6

    lavinia

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    So are you saying that the meridian circle of the twisted torus and each if its edge circles form a non-trivial link? So moving the meridian circle onto the z-axis plus the point at infinity in the 3 sphere then projecting an edge circle onto the xy plane should give a loop that winds twice around the origin?
     
  8. Jan 9, 2012 #7

    lavinia

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    While bending and stretching are two pictures of homeomorphisms I do not believe that they cover the idea or the intuition completely. For instance a symmetry is neither a bend or a stretch. Also flows of fluids across manifolds are homeomorphisms can have incredible complexity and I do not see how to interpret them generally as stretches/ shrinks though maybe this can be done locally.
     
  9. Jan 10, 2012 #8
    Who is telling you that you have to do all this in 3-space? There is nothing in the topology of these spaces telling us that "we should view these things as embedded in 3-space". The double twist and no twist are homeomorphic despite the fact that we can't deform one to the other in 3-space because of the obvious homeomorphism, but also intuitively if we had another dimension to work in then we probably could remove the twist (correct me if I'm wrong).
     
  10. Jan 10, 2012 #9
    I believe it is indeed possible to remove the twist if it's embedded in R^4.

    Look at the two boundaries of the manifold, they're essentially two loops going through each other, suppose for convenience that they are both roughly in the xy plane, going around the z-axis. Take one of the loops and translate it perpendicularly to R^3, or perpendicularly to the xyz hyperplane or whatever the appropriate term is. Then translate it in the z direction. Then undo the first translation so that both boundaries again are entirely within R^3. Now the two boundaries are two separate loops in R^3. In other words, it's a cylinder. Also, throughout the transformation, let the non-boundary parts of the manifold just move in the most natural, smooth way.

    That doesn't require any boundaries to cross eachother or any other non-kosher stuff, so I think it's okay.

    As for why I thought initially of embedding them in R^3... the universe is 3 dimensional, so it is natural to consider such an embedding and interesting to examine how the objects behave in that setting.
     
  11. Jan 11, 2012 #10

    lavinia

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    Could you explain this in more detail?
     
  12. Jan 12, 2012 #11

    Bacle2

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    If I understood you well, there are no non-trivial S1-knots in ℝ4; the extra direction allows you to unknot them.
    direction allows you to undo
     
  13. Jan 13, 2012 #12

    lavinia

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    But a twisted cylinder is not a knot. So how do you unknot the knot when the two curves are part of the cylinder?
     
  14. Jan 13, 2012 #13

    Bacle2

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    Yes, my bad, I did not read carefully-enough. Let me think it through.
     
  15. Jan 14, 2012 #14
    Here's the way I view it:

    First, squeeze your twist into one segment of the strip and lay the rest on some plane, like a segment of a race track connecting the two ends of the twisted segment. The twisted segment looks a bit like a "loop the loop". Concentrate on this segment, this is what we want to "untwist" in ℝ4. If we cut this segment out, it is obvious how we can do this- just hold one end still and pull the other one round the loop in a circular motion.

    Of course, the problem is that when you connect the other ends to the loop, if you extended this motion to that part then you are going to get self intersection. However, it is obvious that you can do this motion with self intersection so that the extra segment only passes through the looped segment in a small area away from the looped segment.

    Fine, so we have our motion with self intersection in ℝ3 and the self intersection is only a problem in two disjoint areas of the loop. So do the whole motion again in ℝ4 except this time, continuously vary the fourth spatial coordinate in an area around the offending part of the unlooped segment so that this time when we get to the part of the motion with self intersection, the fourth spatial coordinates will now be different and we won't get self intersection. After the motion is done, put everything back into the ℝ3 subspace of ℝ4 to really see that what we've got is the unknotted version.

    Really, all we are doing is the "same thing we'd do to unknot in ℝ3 but lifting the loop up out into the extra dimension to avoid the self-intersection when it occurs".

    I hope that all makes sense!
     
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