Showing that two quotient spaces are homeomorphic

Question

Let $M\in GL_n(\mathbb{Z})$

I would like to show that $\mathbb{R}^n/\mathbb{Z}^n \times [0,1]/\sim,$ quotiented by the equivalence relation $(x,0) \sim (Mx,1)$ is homeomorphic to $\mathbb{R^n \times \mathbb{R}}/\sim'$ quotiented by the equivalence relation $(x,y) \sim'(k+M^qx,y+q)$ for all $k\in\mathbb{Z}^n$ and $q\in\mathbb{Z}$.

I have a difficult time finding homeomorphism for quotient spaces and I don't know if there are any general tricks that help. So what I tried is the most naive way : guessing.

I did it by trial and error, the first thing to test out would be the obvious:

Let $$\varphi: (\mathbb{R^n \times \mathbb{R}}/\sim') \rightarrow (\mathbb{R}^n/\mathbb{Z}^n \times [0,1]/\sim)$$ $$(x,y)\rightarrow(\bar{x},\dot{y})$$ where the bar denotes reduction$\pmod {\mathbb{Z}^n}$ and the dot reduction$\pmod 1$, i.e. identifying a real with its fractional part.

But then a problem arises: I want $\varphi((x,y))=\varphi((k+M^qx,y+q))$, but instead I get $$(\bar{x},\dot{y})=(\overline{M^qx},\dot{y})$$

This is almost what I'm looking for, except for this $M^q$ which I would like to do away with. In other words, I would like to add to the "bar" modulo relation the additional identification: $M \equiv Id$.

But I don't know how to proceed from there. Should I just "do it", i.e. pretend that it is possible to define the "bar" reduction as a reduction $\pmod {\mathbb{Z}^n}$ and at the same time be such that $\overline{M^qx}=\bar{x}$ for all $q\in\mathbb{Z}$ and $x\in\mathbb{R^n}$? I want to make sure that it makes sense to do that because it doesn't seem as natural as "casting out" $\mathbb{Z}^n$, which can be interpreted naturally as identifying vectors of $\mathbb{R}^n$ with the vector formed by their fractional parts. Here, it would be casting out all vectors of the form $(M^q-Id)x$, but then this depends on $x$. Am I on the right track or is it completely illicit?

Answer 1

The first space is clearly $X=\mathbb{R^n\times R / Z^n\rtimes Z}$; the action being given by $(v,q)\cdot (x,y) = (v+M^qx, y+q)$. This is indeed an action : clearly $(0,0)$ acts trivially, and $$(v,q)(v',q')(x,y) = (v,q)(v'+M^{q'}x,y+q') = (v+M^q(v'+M^{q'}x), y+q'+q) = (v+M^qv' + M^{q+q'}x, y+q+q')=(v+M^qv', q+q')\cdot (x,y)$$, and this is the mutliplication in the semi-direct product.

How about the first one ? Well let's take a map $f:(\mathbb{R/Z})^n\times [0,1]\to X$ defined by $f(x,t) = [(x,t)]$ where $[]$ denotes the class modulo the action. Note that it is well-defined because if $x=y\mod \mathbb{Z}^n$, then $(x,s) = (y,s) \mod \mathbb{Z}^n\rtimes \mathbb{Z}$

Now $f(x,0) = [(x,0)]$ and $f(Mx,1) = [(Mx,1)]$ but $(Mx,1) = (0,1)\cdot (x,0)$ so $f(x,0) =f(Mx,1)$, so $f$ factors through $\overline{f}: (\mathbb{R/Z})^n\times [0,1]/\sim\to X$.

Moreover, $f(x,t) = f(y,s)\implies \exists v,q \mid (v,q)\cdot (x,t) = (y,s)$. But this implues $q=0$ or $1$ (have a look at the second coordinate). If $q=0$, then this implies $x=y \mod \mathbb{Z}^n$, but we started from $x,y\in (\mathbb{R/Z})^n$, so this implies $x=y$. Otherwise, $q=1$ so it implies $(x,t) = (x,0)$ and $(y,s) = (y,1)$. And now the same argument shows that $y=Mx$, so that actually our above factorization was just right : $\overline{f}$ is injective.

Moreover, it's quite clear that $\overline{f}$ is surjective (if you have $[(x,y)]$, then $y=s \mod\mathbb{Z}$ for some $s\in [0,1]$ (maybe two if $y$ is an integer), so $[(x,y)]=[(M^{-q}x,s)]$ for some $q$)

So $\overline{f}$ is a continuous bijection. But wait, $\mathbb{R/Z}^n$ and $[0,1]$ are compact, therefore so is there product, and so is any of its quotient. Moreover, $X$ is $T_2$ as the action is properly discontinuous (or you could prove it by hand, it's not that hard) : you have a continuous bijection from a compact space to a $T_2$ space: it must be a homeomorphism.