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The Klein bottle using the Euler caracteristic + orientability caracterisation

  1. Apr 6, 2007 #1


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    I went off on my own to study the Euler caracteristic + orentability caracterisation of closed surface and I must have gotten lost somewhere, because I do not find that the Klein bottle is homeomorphic to [itex]\mathbb{R}P^2\#\mathbb{R}P^2[/itex] as I should.

    I started with the result that for two surfaces M and N, we have


    then I proved by induction that


    which can be solved for k:


    Now, the Klein bottle is not orientable, so it must be homeomorphic to [itex](\mathbb{R}P^2)^{\#_k}[/itex] for some k. And since the Euler caracteristic is a topological invariant, it must be that


    And so, according to the above formula for k, with [itex]\chi(M)=\chi(\mathbb{R}P^2)=1[/itex],


    So all that remains to do is to triangulate K and find its Euler caracteristic. A fundamental polygon for the Klein bottle is provided here:


    I believe it can be triangulated simply by drawing a side across one of the diagonal. All the vertices are identified, so V=1. There are 3 distinct sides, so E=3. My triangulation has 2 triangles so F=2, thus


    But this means that k=1, which is incorrect. It should be 2.

    Where did I go wrong?
    Last edited: Apr 6, 2007
  2. jcsd
  3. Apr 6, 2007 #2


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    well your euler characteristic calculation seems correct, since just looking at a picture of a klein bottle (the picture in my head) shows it obviously has the same edges faces and vertices as a torus, hence has the same euler chracteristic, namely zero. so the error is likely somewhere else.
  4. Apr 6, 2007 #3


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    Whoah! There's no error at all !

    [tex]\mathbb{R}P^2\#\mathbb{R}P^2 = (\mathbb{R}P^2)^{\#_1}[/tex]

    k=1 is correct! :grumpy:

    Edit: No! k should be 2 but I made an error in the very begining by misinterpreting the #_k notation. I thought k refered to the number of # symbol, while it refers to the number of spaces you glue together.
    Last edited: Apr 6, 2007
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