Fremdenverkehr

Eintrag vom: 28.06.2013.

Background I am trying to compute the first integral homology group of the Klein bottle and refer to Ted Shifrin's answer to the following question: Homology groups of the Klein bottle. The Klein bottle (K) admits a simplicial decomposition according to the following fundamental polygon representation: Now the relevant chain ...

 

I want to calculate the Klein bottle. So I did it by Van Kampen Theorem. However when I'm stuck at this bit. So I remove a point from the Klein bottle to get $\\mathbb{Z}\\langle a b\\rangle$ where...

 

I'm trying to see the fundamental group of the Klein bottle minus a point without success. I know how to solve the torus minus a point giving a deformation retraction to the wedge sum of two circle...

 

I'm trying to find the universal covering space of the Klein bottle. I know that R2 R 2 ${\mathbb{R}}^{2}$ covers the Klein bottle but I don't know how to prove I found this proof on internet: Someone knows why this quotient map is a covering map or have an alternative solution? I found this solution a little bit weird due my lack of experience on this subject if fact I'm a really ...

 

The Klein bottle is the quotient space of the torus by the action of an orientation reversing involution. So it is not orientable. One can choose this involution to be rotation by 180 degrees along one generating circle followed by reflection along the second. The top Z homology of the Klein bottle is zero so it is not orientable.

 

The Klein bottle is the boundary of the twisted $2$ -disk bundle (the solid Klein bottle). Gluing two solid Klein bottles along their boundaries yields a twisted $2$ -sphere bundle.

 

2. ???? ?? aba^ {-1}=b^ {-1} ??? ?? a ??????? b ????????? ?????????????? ??????? \pi_1 (T^2)\cong \langle a b\mid aba^ {-1}b^ {-1}=1\rangle \cong \mathbb Z^2. ????????? ab=ba. ???????? ab=b^ {-1}a. ????????????????

 

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I am using Do Carmo's Riemannian Geometry and struggling to solve a problem. The problem is: Show that the mapping $F:\\mathbb{R}^2\\to\\mathbb{R}^4$ given by $$F(x ...

 

Think of the bottle as two Möbius strips glued along their edges. A closed loop that circles one of the bands (along its center say) is not a boundary but if you follow it twice it becomes one: you can think of it as a loop that follows precisely the edge of one of the original bands so it's just the boundary of that band. This gives you the "torsion" piece of the homology and it shows ...