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The Klein bottle using the Euler caracteristic + orientability caracterisation

  1. Apr 6, 2007 #1

    quasar987

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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

    [tex]\chi(M\#N)=\chi(M)+\chi(N)-2[/tex]

    then I proved by induction that

    [tex]\chi(M^{\#_k})=(k+1)\chi(M)-2k[/tex]

    which can be solved for k:

    [tex]k=\frac{\chi(M^{\#_k})-\chi(M)}{\chi(M)-2}[/tex]

    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

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

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

    [tex]k=\frac{\chi(K)-1}{1-2}=1-\chi(K)[/tex]

    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:

    http://en.wikipedia.org/wiki/Classification_theorems_of_surfaces#Construction_from_polygons

    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

    [tex]\chi(K)=1-3+2=0[/tex]

    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

    mathwonk

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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

    quasar987

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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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