Fundamental group of Klein Bottle

Question

It is well know that the fundamental group of the Klein Bottle $G$ is defined by

$$G=BS(1,-1)=\langle a,b: bab^{-1}=a^{-1}\rangle.$$

I know, for example that $BS(1,2)$ can be defined as the group

$$BS(1,2)=\langle A,B\rangle $$

where

$$A=\left( \begin{array}{cc} 1 & 1 \\ 0 & 1 \\ \end{array} \right), B=\left( \begin{array}{cc} 2 & 0 \\ 0 & 1 \\ \end{array} \right).$$

These matrices satisfy the equation $BAB^{-1}=A^{2}$ and are free : there is not an integer $k$ such that $A^{k}=I$ or $B^{k}=I$. This implies that we obtain an "explicit description" of $BS(1,2)$ as the group generated by $A$ and $B$.

I know that the matrices

$$a=\left( \begin{array}{cc} 1 & 1 \\ 0 & 1 \\ \end{array} \right), b=\left( \begin{array}{cc} -1 & 0 \\ 0 & 1 \\ \end{array} \right)$$

satisfy the relation $bab^{-1}=a^{-1}$ but $b^{2}=I$. This implies that $BS(1,-1)$ is not generated by $a$ and $b$.

My question is: is there an "explicit description" for $G=BS(1,-1)$ with matrices or maybe another couple of objects?

Does this answer your question? Identifying $\langle a,b\mid a^2b^2\rangle$. — Shaun, Jul 23 '20 at 17:27
@PaulPlummer by deck transformation of $\mathbb{R}^{2}$ do you mean that $BS(1-1)$ can be defined as the group generated by the homeomorphism $f(x,y)=(x,y+1), g(x,y)=(x+1,1-y)$ ? — José Luis Camarillo Nava, Jul 23 '20 at 17:32
Yah, those are the "standard" maps. Seems pretty explicit to me (isometries of $\mathbb{R}^2$). — , Jul 23 '20 at 17:33
My motivation is give a linear action of $G$ over $M$ such that $H^{0}(G,M)=0$ but $H^{1}(G,M)\neq 0$ where $M=\mathbb{R}$ or $M=\mathbb{R}^{2}$ — José Luis Camarillo Nava, Jul 23 '20 at 17:39
@Shaun the group in your suggested duplicate is a different group from this one. — Steven Stadnicki, Jul 23 '20 at 19:02
While the groups are isomorphic, nonetheless I don't think the question is a duplicate. This one asks for a more "explicit description" with the suggestion of matrices, which is a good question. @Shaun — Lee Mosher, Jul 23 '20 at 19:31
That's a fair comment, @LeeMosher; I'll withdraw my vote to close. — Shaun, Jul 23 '20 at 19:33
@LeeMosher By the way how can we define a linear action of $G$ over $\mathbb{R}^{2}$? — José Luis Camarillo Nava, Jul 23 '20 at 19:34
That's an interesting question, but it would be better to ask it in a new post rather than as a comment to this post. — Lee Mosher, Jul 23 '20 at 19:39
Here is a meta-answer: $\operatorname{BS}(1,-1)$ contains $\mathbb{Z}^2$ as an index-$2$ subgroup. As $\mathbb{Z}^2$ is linear over $\mathbb{Z}$, we immediately have that $\operatorname{BS}(1,-1)$ is linear over $\mathbb{Z}$, as virtually-linear implies linear. So yes, there is an explicit description in terms of integer matrices, and it can be computed. — user1729, Jul 23 '20 at 21:30

Lee Mosher · Accepted Answer · 2020-07-23T18:56:19.727

Starting from the representation described by @PaulPlummer as a group of isometries of $\mathbb R^2$, you can obtain a representation as a group of linear transformations of $\mathbb R^3$.

To do this, one uses a standard embedding of the isometry group of $\mathbb R^2$ as a subgroup of $GL(3,\mathbb R)$. Each isometry of $\mathbb R^2$ can be written uniquely as $P \mapsto MP + Q$ for some $M \in O(2,\mathbb R)$ and some $Q \in \mathbb R^2$ (vectors are written in column format). The representing element of $GL(3,\mathbb R)$ is the matrix written in block form as $\pmatrix{M & Q \\ 0 & 1}$. If you then represent a column 2-vector $P$ as a column 3-vector $\pmatrix{P \\ 1}$ then matrix multiplication gives you $$\pmatrix{M & Q \\ 0 & 1}\pmatrix{P \\ 1} = \pmatrix{MP+Q \\ 1} $$

For the Klein bottle group, the two generators are:

A translation $1$ unit to the right: $$P \mapsto M_1 P + Q_1, \qquad M_1 = \pmatrix{1 & 0 \\ 0 & 1}, \qquad Q_1 = \pmatrix{1 \\ 0} $$
A glide reflector, gliding up the $y$-axis by $1$ unit: $$P \mapsto M_2 P + Q_2, \qquad M_2 = \pmatrix{-1 & 0 \\ 0 & 1}, \qquad Q_2 = \pmatrix{0 \\ 1} $$

score 2 · Answer 2 · edited Jul 24 '20 at 09:06

2

Finally , $G$ can be described as the group of $2\times 2$ matrices generated by two matrix $A,B$ such that $BAB^{-1}=A^{-1}$ and $A^{k}\neq I$, $B^{k}\neq I$ for all $k$. Let $A=\left( \begin{array}{cc} a & b \\ c & d\\ \end{array} \right)$ and $B=\left( \begin{array}{cc} \lambda & 0 \\ 0 & \mu \\ \end{array} \right)$ (to simplify computations). Working whith the equation $BAB^{-1}=A^{-1}$ and assuming that $ad-bc=1$ and $b,c\neq 0$ we have that

$A=\left( \begin{array}{cc} a & b \\ c & a\\ \end{array} \right)$ and $B=\left( \begin{array}{cc} \lambda & 0 \\ 0 & -\lambda \\ \end{array} \right)$

This family of matrices satisfy the relation $BAB^{-1}=A^{-1}$.

If $A=\left( \begin{array}{cc} 2 & 3 \\ 1 & 2\\ \end{array} \right)$ and $B=\left( \begin{array}{cc} 2 & 0 \\ 0 & -2 \\ \end{array} \right)$, then $BAB^{-1}=A^{-1}$ and $A^{k}\neq I$, $B^{k}\neq I$, for all $k$

Hence $G\equiv \langle A,B\rangle$.

edited Jul 24 '20 at 09:06

user1729

32,369

answered Jul 23 '20 at 23:31

José Luis Camarillo Nava

1,023
6
12

There could be other relations in your matrix group so you haven't shown it is $G$ yet – Jul 24 '20 at 00:33
@PaulPlummer In that case a new question comes to me: is there two $2\times 2$ matrices $A,B$ such that $G\equiv <A,B>$ ? – José Luis Camarillo Nava Jul 24 '20 at 01:35
I think it is likely that this will do it but you just haven't fully shown it(shouldn't be too bad it) – Jul 24 '20 at 01:59
Building on JCAA's deleted answer: $G=\langle A, B\rangle$ is in fact $\operatorname{BS}(1, -1)$. The idea is to prove that $G=\mathbb{Z\rtimes Z}$ with the appropriate action, and the only thing needed to prove this is to show that $\langle A\rangle\cap\langle B\rangle=1$. To see this, note that $A$ contains only positive integers, and so any positive power of $A$ also contains only positive integers (as components in such a power are obtained by adding components in $A$). In particular, $A^k$ is non-diagonal for all $k>0$. cont. – user1729 Jul 24 '20 at 08:46
In contrast, every power of $B$ (positive power or negative power) is diagonal. Hence, $\langle A\rangle\cap\langle B\rangle$ is trivial, as required. – user1729 Jul 24 '20 at 08:51
I am rather surprised that you found this example in $2\times2$ matrices, so good job! In my original comment to the question, I had in mind starting with a copy of $\mathbb{Z}^2$ in $\operatorname{GL}(3, \mathbb{Z})$, and then the rank would increase to construct a copy of $\operatorname{BS}(1, -1)$. To find an example over integer matrices the rank would have to be at least $3$, as $\operatorname{GL}(2, \mathbb{Z})$ is virtually free (your group $G$ does not lie in $\operatorname{GL}(2, \mathbb{Z})$, as $B^{-1}$ is not integer-valued). – user1729 Jul 24 '20 at 09:05
@user1729 So... are you agree whit my answer? Does A and B generate the Group G? – José Luis Camarillo Nava Jul 24 '20 at 13:58
@JoseCamarillo Lets be precise: The matrices $A$ and $B$ generate a group isomorphic to the fundamental group of the Klein bottle. (But they do not generate the fundamental group of the Klein bottle. In my above comments, I was writing $G$ as shorthand for $\langle A, B\rangle$, so my $G$ is different from your $G$.) – user1729 Jul 24 '20 at 14:17

Fundamental group of Klein Bottle

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