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
$$BS(1,2)=langle A,Brangle $$
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?

José Luis Camarillo Nava · Answer

Finally , $G$ can be described as the group of $2times 2$ 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 $b,cneq 0$ 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 $Gequiv langle A,Brangle$.

Lee Mosher · Answer

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:
$$P mapsto M_2 P + Q_2, qquad M_2 = pmatrix{-1 & 0 \ 0 & 1}, qquad Q_2 = pmatrix{0 \ 1}
$$

Fundamental group of Klein Bottle

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