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Fundamental polygon of a Mobius strip

  1. Mar 1, 2013 #1
    Hey I am having a little bit of difficulty.

    The classification theorem for 2 - manifolds tells me that every 2 -manifold has the following representation:

    1) connect sum of n-tori
    2) connect sum of n-projective planes
    3) a sphere

    Now, using Massey's book there is a very algorithmic way to take a polygon given its edge representation (say aa*bc*d*bcd*, something like that) and get to one of those forms.

    But such a simple example, the Mobius strip, I get lost on and I'm not sure how to realize it.

    When I think of the Mobius strip as a CW complex, I envision the 1 skeleton as a square with 3 1-cells connected to 2 0-cells, and a 2 cell as the middle of the "square", so the following edge orientation: aba*d. How do I realize this as a connect sum of tori, or projective planes? Massey's algorithm only seems to work when edges are paired up, so just having the one "b" and the one "d" means it doesn't work
  2. jcsd
  3. Mar 2, 2013 #2
    The Mobius strip is not a 2-manifold since it has a boundary.
    If you remove the boundary, then it's still not a 2-manifold since it's not compact.

    So the classification theorem does not apply.
  4. Mar 2, 2013 #3
    oh... oh wow. thank you so much. Just out of curiosity, I know that S1 is homotopic to the Mobius strip, so should that have told me right away that the M-S is not a 2-manifold? (because S1 is not a 2-manifold)
  5. Mar 2, 2013 #4
    Not really, since the punctured plane [itex]\mathbb{R}^2\setminus \{0\}[/itex] is also homotopy equivalent to [itex]S^1[/itex]. But [itex]\mathbb{R}^2\setminus \{0\}[/itex] is a 2-manifold.
  6. Mar 3, 2013 #5
    But how is R^2 - {0} a 2-manifold? It is homeomorphic to the punctured sphere with N and S poles missing via stereographic projection (my thinking here is if you only left out the north pole, then you'd still map to the origin because of the projection through the S pole, but if you take out the S pole as well, you'd stereographically project to R^2 - {0}, and the punctured sphere isn't a 2-manifold because it's not closed as its complement would be the two singleton points N and S and singleton points aren't open.

    EDIT: Nevermind, there is no condition that the space has to be closed, not sure what I was thinking.

    To be honest, I'm really confused by the fact that you can have spaces where one is a 2 manifold and the other isn't but they are homotopy equivalent. I'm going to have to think about that. I get that homotopy equivalence is not as "strong" as homeomorphism, but I don't know I still have to wrap my head around this because I still have this notion of sameness when thinking of homotopy and I guess I'm thinking of it too strongly.
  7. Mar 3, 2013 #6
    If you want to express that two spaces are the exact same thing, then you need homeomorphisms. Homeomorphisms preserve all the topological properties.

    Homotopy equivalence does not preserve all topological properties. Being a manifold, being compact, etc. are all not preserved by homotopy equivalences. So a homotopy equivalence does not means that two spaces are the same.

    However, homotopy equivalence do preserve some interesting properties. Most of these properties have to do with algebraic topology. For example, they preserve the fundamental groups, the homology groups, being path connected, etc. In particular, a homotopy equivalence can't change how many "holes" your space has (since holes can be measures by homology).
    It might be a bit inaccurate, but I'd say that notions which can be defined in algebraic topology tend to be preserved by homotopy equivalence. But other notions are usually not.
  8. Mar 4, 2013 #7


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    You can always consider the interior of the Mobius Strip, where x is in [0,1], y is in
    (0,1) . Then you identify the left side with the right side, with opposite orientations.
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