Let $X=[-1,1]\times\mathbb{R}$.
- Find all numbers $\lambda,a,b\in\mathbb{R}$ with the property that $$n\cdot (x,y):= (\lambda^nx,a+by+\lambda n)$$ defines an action of the additive group $(\mathbb{Z},+)$.
- For the falues of $\lambda,a,b$ that you found compute the resulting quotients $X/\Gamma$.
Can somebody help me? I know the definition of a group action, etc. But how to find these numbers is a mystery...
To check that you have a group action of $X$ you need to show that $$m\cdot (n \cdot (x,y) ) = (m+n) \cdot (x,y) \in X. \tag{1}$$
Since $\lambda^n x$ has to be in $[-1,1]$, $\lambda$ has to be in $[-1,1]$ and because we usually do not define $0^0$ we want $\lambda \ne 0$.
So if you write out both sides of $1$ it becomes
$$ m\cdot (n \cdot (x,y) ) = (\lambda^{m+n}x, a+ b(a+by+\lambda n)+ \lambda m) $$
and
$$ (m+n) \cdot (x,y) = (\lambda^{m+n}x, a+ by + \lambda (m+n)).$$
Checking the axiom for the group action now reduces to show that $$ ab +(b^2-b)y +\lambda n (b-1) =0.$$
Thus $b=0$ or $b=1$. Since $b=0$ implies $\lambda = 0$ we can exclude this case. So we have $b=1$, $a=0$ and $\lambda \in [-1,1]\setminus \{0\}$.
To compute the it is maybe convenient to get rid of the $\lambda$ in the second component. To do so, define $$ n * (x,y)= (\lambda^n x ,y+ n) $$ and $f \colon X \to X, \ (x,y) \mapsto (x,\lambda y)$, then $f\circ (n* (x,y) )= n \cdot f((x,y) )$.
For $\lambda = 1$ the quotient is just $[-1,1] \times S^1$. I havent figured out the other cases ... for $\lambda \in (0,1)$ maybe something like $[-1,1]^{cd}\times S^1$ where $[-1,1]^{cd}$ denotes $[-1,1]$ with the codiscrete toplogy. For $\lambda = -1$ you get $[0,1]\times S^1$ with say some strange neighborhoods around $\{0\} \times S^1$ since you glue $[-1,0] \times S^1$ and $[0,1]\times S^1$ with a rotation. And the last case $\lambda \in (-1,0)$ is probably some mixture of the last ones. Disclaimer: It is just a guess what the quotients are, maybe it is just nonsense.