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I have been trying to find the FG of the Klein bottle, and I was wondering if someone could verify that this process is correct.

After triangulating it, I then found a maximal tree (shown in yellow) as follows:

Tree

I then shaded in purple to get the maximal contractable subspace. Then I considered the generators on the remaining 1-simplices as follows:

FG

Using the relation implied by the bottom left corner, I get: $bab=a$

So is it correct that $\pi(Klein Bottle)=\{a, b | bab=a \}$ ?

Thank you so much!

thinker
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1 Answers1

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Yes, that is right.

If you forget about the internal simplices then it's easy to see that what you've got is correct: you start off with two loops $a$ and $b$ and then attach a 2-dimensional disk along the path given by $baba^{-1}$ (this is the outside of the square that you've drawn). This means that that loop is now contractible, so you need to add $baba^{-1}$ as a relator, which is the same as saying $bab = a$.

Josh Hunt
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  • Thank you so much! I tried to do a similar thing with a sphere just now here http://math.stackexchange.com/questions/1778421/fundamental-group-of-the-sphere-via-triangulation is it possible for you to see if it is correct? thank you! – thinker May 09 '16 at 18:36