# [SOLVED] Algebraic Topology: Fundamental group of Klein bottle

• April 25th 2010, 02:05 PM
JJMC89
[SOLVED] Algebraic Topology: Fundamental group of Klein bottle
Prove that the fundamental group of a Klein bottle is $G = \{ a^{m}b^{2n+\epsilon} \ ;\ m \in \mathbb{Z}, \ n \in \mathbb{Z} \ \epsilon = 0 \ or\ 1, \ ba=a^{-1}b\}$, i.e. G is the group on two generators $a ,\ b$ with one relation $ba=a^{-1}b$
• April 25th 2010, 03:18 PM
Hey bro.

Consider the Klein bottle as the unit square with opposite edges identified (with twists). FIx as basepoint b the origin. There is a natural covering map $p:(\mathbb{R}^2,0)\to(\mathrm{Klein\ bottle},0)$ defined by tiling the plane with squares and turning every one into a Klein bottle. If you look at the points that are identified, you will see a diamond pattern. Start with the unit square. The corners (0,0) and (1,1) are identified. Then look at the square directly above this one. The corners (1,1) and (0,2) are identified. Moving up again, the corners (0,2) and (1,3) are identified, and moving right, the corners (1,3) and (2,2) are identified.

Draw a picture!

A moment's thought will convince you that in fact, the fibre $p^{-1}(0)$ consists of the diamond pattern $\Phi := \{ (x,x+2k) : x\in \mathbb{Z}, k\in\mathbb{Z}\}$.

For every path in the Klein bottle, we can lift that path to the covering space, the plane, and then find a unique homotopic path which transverses only the edges of squares. In this way, paths in the Klein bottle are in one to one corospondence with "taxicab" paths in the plane that start at the origin and end at a point in $\Phi$; call these paths good.

Let a mean "move right by 1 unit" and b mean "move up by 1 unit". If you think of starting at the origin, then a word $a^mb^{2n+\epsilon}$ defines a good path, ie a path that ends up at a point of $\Phi$. In addition, the identification we used above (a corner of a sqaure is identified with the corner opposite it) means that if we move up by 1 and then move right by 1, it is the same as if we move left by 1 and then up by 1. This is the relation $ba = a^{-1}b$. Its easy to see that this gives all the proper identifications ie it identifies all the points in $\Phi$. (Some other relations that work are $ab^{-1} = a^{-1}b$ and $a^{-1}b=a^{-1}b^{-1}$; this is easy to see algebraically, what are their interpretation geometrically?).

Hopefully you can tighten everything up into a rigorous argument, or at least get started :] The main point will be to show that the map from good path to words in a and b is a isomorphism.
• April 25th 2010, 05:57 PM
aliceinwonderland
Quote:

Originally Posted by JJMC89
Prove that the fundamental group of a Klein bottle is $G = \{ a^{m}b^{2n+\epsilon} \ ;\ m \in \mathbb{Z}, \ n \in \mathbb{Z} \ \epsilon = 0 \ or\ 1, \ ba=a^{-1}b\}$, i.e. G is the group on two generators $a ,\ b$ with one relation $ba=a^{-1}b$

Use Van Kampen's theorem.

Let a Klein bottle be K such that $K = U \cup V$. I'll omit the base point for clarity. You may need to include base points and their transforms for the more rigourous proof.

The choice for U and V for K for Van Kampen can be:

U: K-{y}, where the point y is the center point of the square.
V: the image of the interior of the square under identification.

Since V is simply connected, we apply the following theorem.

Theorem. Assume V is simply connected. Then, $\psi_1:\pi_1(U) \rightarrow \pi_1(K)$ is an epimorphism, and its kernel is the smallest normal subgroup of $\pi_1(U)$ containing the image of $\phi_1(\pi_1(U \cap V))$, where $\phi_1$ is a group homomorphism $\phi_1:\pi_1(U \cap V) \rightarrow \pi_1(U)$.

It is easily seen that $\psi_1$ is an epimorphism. We know that $\pi_1(U)$ is a free group with two generators, let's say a, b (deformation retracts to figure 8 space). Since $\pi_1(U \cap V)$ is the infinite cyclic group generated by, let's say r, then $\phi_1(r)=aba^{-1}b$. Then the kernel of $\psi_1$ is the the group generated by $aba^{-1}b$ by the above theorem.

By the first isomorphism theorem, we see that $\pi_1(K)= $.
• April 25th 2010, 06:04 PM