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.

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 $a$ and $b$ are two loops connected by a point.

Also you have the boundary map that goes $abab^{-1}=1$, so I did some calculations I got basically this $\dfrac{\mathbb{Z}\langle ab,b\rangle}{\langle(ab)^2=b^2\rangle}$.

But, I don't think that is correct. But, yeah how you calculate the fundamental group of the Klein bottle?

share|improve this question
    
I changed $<\cdots>$ to $\langle\cdots\rangle$. –  Michael Hardy May 18 '12 at 16:25
    
@MichaelHardy thanks I just didn't know what the notation for the brackets are. –  simplicity May 18 '12 at 16:42
2  
I don't remember the answer, but note that you can simplify this presentation to $\langle a,b | aba=b \rangle$. –  Aaron Mazel-Gee May 18 '12 at 18:54
add comment

1 Answer

up vote 6 down vote accepted

The short answer is that your presentation is isomorphic to the (slightly more) "standard" presentation - so you computed a correct answer.

More detail -

According to Wolfram and using $K$ to denote the Klein bottle, we have $\pi_1(K) \cong \langle c,d \rangle/cdc^{-1}d$ whereas you have (according to Aaron Mazel-Gee's comment) $\pi_1(K)\cong \langle a,b\rangle /abab^{-1}$.

The question is, then, are these isomorphic?

Well, we have $cdc^{-1}d = e$ so, taking the inverse of both sides gives $d^{-1}cd^{-1}c^{-1} = e$. But this has the same form as the relation between $a$ and $b$, so now the isomorphism is clear: we map $a$ to $d^{-1}$ and $b$ to $c$.

That is, define $f:\langle a,b \rangle /abab^{-1}\rightarrow \langle c,d\rangle/cdc^{-1}d$ by $f(a) = d^{-1}$ and $f(b) = c$.

I claim this is well defined, for \begin{align*} f(abab^{-1}) &= f(a)f(b)f(a)f(b)^{-1} \\\ &= d^{-1}cd^{-1}c^{-1} \\\ &= (cdc^{-1}d)^{-1} \end{align*} as it should. (Technically, I'm defining $f$ on $\langle a,b\rangle$ and proving it descends to the quotient.)

Finally, rather than show this is 1-1 and onto, instead, show that $g:\langle c,d\rangle/ cdc^{-1}d\rightarrow \langle a,b\rangle/ abab^{-1}$ defined by $g(c) = b$ and $g(d) = a^{-1}$ is well defined, and the inverse to $f$, so $f$ is the desired isomorphism.

share|improve this answer
    
Wow, and I thought word problems were unsolvable! You must be some kind of wizard ;o) –  Aaron Mazel-Gee May 18 '12 at 22:59
2  
Not unsolvable, undecidable. Big difference. –  JeremyKun Dec 24 '12 at 0:46
add comment

Your Answer

 
discard

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.