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.(adsbygoogle = window.adsbygoogle || []).push({});

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?

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# The Klein bottle using the Euler caracteristic + orientability caracterisation

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