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Cutting a hole in the projective plane

  1. Feb 16, 2012 #1
    According to this book I'm reading, if you cut out a closed disc in the projective plane, then the complement of the interior of this disc is topologically a Mobius strip with boundary.

    Looking at the constructions of real projective space and the mobius strip from squares though, I can't see how this works?
     

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  2. jcsd
  3. Feb 16, 2012 #2

    morphism

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    Have you tried doing the edge identifications with a physical piece of paper? To make things easier, instead of cutting out a small disk, cut out a large square.

    Alternatively, if you want to visualize it using drawings like the ones you've attached, what you could do is cut the square from the midpoint of the left edge to the midpoint of the right edge.This'll give you two rectangles with three edges and part of the fourth edge. The edge identifications on the three edges are aba and aba (resp., and same a and b). Mark down the endpoints of the partial edge as x and y (on both rectangles). Now do the identifications on each piece separately. What you ought to get is two 'incomplete' mobius strips, which when you glue together give you a mobius strip with boundary.
     
    Last edited: Feb 16, 2012
  4. Feb 16, 2012 #3

    lavinia

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    Projective space can be constructed as the quotient of the 2 sphere by the atnipodal map, the map that sends each point to its opposite point across the middle of the sphere.

    A disc around the north pole is identified to the opposite disc around the south pole and becomes a single disc in projective space. The complement of these two discs is an equatorial cylinder. What is its image in projective space?

    You should be able to see this from the square identifications.
     
  5. Feb 17, 2012 #4

    mathwonk

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    imagine a sphere from which a disc and its antipode are removed. then remove a hemisphere defined by a plane containing the diameter joining the centers of both discs.

    then identify the antipodes on the remaining set, and note this gives a mobius strip. i.e. the only remaining antipodes are on the vertical edges of the remaining strip.
     
  6. Feb 19, 2012 #5
    Thanks, it's a lot easier to see now :)
     
  7. Dec 18, 2012 #6
    "According to this book I'm reading, if you cut out a closed disc in the projective plane, then the complement of the interior of this disc is topologically a Mobius strip with boundary."
    Are you trying to say that removing an open disc in the projective plane renders the Mobius strip?
    The punctured projective plane is homeomorphic to the mobius strip and this doesn't seem to agree with the fact that the disc has to be open....
     
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