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?

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

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.

  • $\begingroup$ Wow, and I thought word problems were unsolvable! You must be some kind of wizard ;o) $\endgroup$ – Aaron Mazel-Gee May 18 '12 at 22:59
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    $\begingroup$ Not unsolvable, undecidable. Big difference. $\endgroup$ – JeremyKun Dec 24 '12 at 0:46

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