Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

A problem asks me to find all the covering spaces of a Klein bottle. This needs to calculate all the subgroups of the fundamental group of the Klein bottle. But I don't have any idea how to do it.

I googled it and an article says

The subgroups of the fundamental group of the Klein bottle are either trivial, free of rank one, free Abelian of rank two, or non-Abelian of rank two.

I don't know how to get the result and what is the concrete form of the subgroups (which is needed to calculate the covering spaces.)

Can you please help? Thank you.

share|improve this question
1  
As I recall, the best way to solve this particular problem is to figure out what the covering spaces of the Klein bottle are geometrically, and then use that to deduce what the subgroups of the fundamental group are. –  Charles Staats Apr 21 '11 at 16:08
    
I agree with Charles. In particular, every covering space is a quotient of the universal cover, and there aren't too many possibilities for what this can be. –  Qiaochu Yuan Apr 21 '11 at 17:31
1  
The Klein bottle group is a semi-direct product $\mathbb Z \rtimes \mathbb Z$ where $\mathbb Z$ acts on $\mathbb Z$ by its sole non-trivial involution. So any subgroup of the Klein bottle group is a semi-direct product $A \rtimes B$ where $A, B \subset \mathbb Z$, and there's basically just three possibilities for each of $A$ and $B$, up to isomorphism and the action of $B$ on $A$, etc.. –  Ryan Budney Apr 21 '11 at 17:57
    
@Charles @Qiaochu : But the problem is to find all covering spaces. If I don't know all the subgroups, how can I know I haven't miss something? –  Roun Apr 22 '11 at 0:56
1  
@Ryan: Well, I'm not quite familiar with semi-direct product. But your statement is not true when it is direct product. Not all subgroups of $\mathbb{Z}\times\mathbb{Z}$ are in the form $A\times B$ where $A,B\subset\mathbb{Z}$. –  Roun Apr 22 '11 at 1:06

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.