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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 \mathbb{R}P^2\#\mathbb{R}P^2 as I should.
I started with the result that for two surfaces M and N, we have
\chi(M\#N)=\chi(M)+\chi(N)-2
then I proved by induction that
\chi(M^{\#_k})=(k+1)\chi(M)-2k
which can be solved for k:
k=\frac{\chi(M^{\#_k})-\chi(M)}{\chi(M)-2}
Now, the Klein bottle is not orientable, so it must be homeomorphic to (\mathbb{R}P^2)^{\#_k} for some k. And since the Euler caracteristic is a topological invariant, it must be that
\chi((\mathbb{R}P^2)^{\#_k})=\chi(K)
And so, according to the above formula for k, with \chi(M)=\chi(\mathbb{R}P^2)=1,
k=\frac{\chi(K)-1}{1-2}=1-\chi(K)
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
\chi(K)=1-3+2=0
But this means that k=1, which is incorrect. It should be 2.
Where did I go wrong?
I started with the result that for two surfaces M and N, we have
\chi(M\#N)=\chi(M)+\chi(N)-2
then I proved by induction that
\chi(M^{\#_k})=(k+1)\chi(M)-2k
which can be solved for k:
k=\frac{\chi(M^{\#_k})-\chi(M)}{\chi(M)-2}
Now, the Klein bottle is not orientable, so it must be homeomorphic to (\mathbb{R}P^2)^{\#_k} for some k. And since the Euler caracteristic is a topological invariant, it must be that
\chi((\mathbb{R}P^2)^{\#_k})=\chi(K)
And so, according to the above formula for k, with \chi(M)=\chi(\mathbb{R}P^2)=1,
k=\frac{\chi(K)-1}{1-2}=1-\chi(K)
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
\chi(K)=1-3+2=0
But this means that k=1, which is incorrect. It should be 2.
Where did I go wrong?
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