Fundamental Group of Klein Bottle acts on $\mathbb{R}$ It is well know that the Fundamental Group of the Klein Bottle can be defined (up to isomorphism) as the group with two generator and one relation
$$BS(1,-1)=\langle a,b: bab^{-1}=a^{-1}\rangle $$
In Algebraic Topology this fundamental group is defined as the group $G$ of homeomorphisms of $\mathbb{R}^{2}$ generate by $f,g:\mathbb{R}^{2}\rightarrow\mathbb{R}^{2}$ where:
$$f(x,y)=(x,y+1), g(x,y)=(x+1,1-y)$$
I want to find a linear action of $G$ on $\mathbb{R}$, e.g, an operation $"\cdot":G\times \mathbb{R}\rightarrow \mathbb{R}$ such that:
1.- $g\cdot(x_{1}+x_{2})=g\cdot x_{1}+g\cdot x_{2}$
2.- $g_{1}\cdot (g_{2}\cdot x)=(g_{1}g_{2})\cdot x$
3.- $1_{G}\cdot x=x$,
for all $g,g_{1},g_{2}\in G$ and $x, x_{1}, x_{2}\in \mathbb{R}$.
My motivation: find a solvable notnilpotent group $G$ acting linearly on a finite dimensional vector space $M$ such that
$H^{0}(G,M)=0$ but $H^{k}(G,M)\neq 0$ for some $k$.

Any one-dimensional representation of this group, will, as for any group, factor through its abelianization, which is $\mathbb{Z} \oplus \mathbb{Z}/2\mathbb{Z}$.
Concretely, we can have $b$ act by multiplication by any non-zero scalar, and $a$ act by multiplication by $-1$ (we could also have $a$ acting trivially), and these are all the representations.
