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how can I show that the following map is a covering map of $T:=$ $S^1$ x $S^1$?

$\pi: T\rightarrow T$ with $(x,y)$ $\mapsto$ $(x^ay^b, x^cy^d)$, where $a,b,c,d \in \mathbb{Z}$ and $ad-bc=m\neq 0$.

Furthermore every $(x,y)$ has $|m|$ inverse images under the mapping $\pi$.

Many thanks, Alex

Alex
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  • Welcome to Math.StackExchange! It is sometimes helpful to include what you have tried in your question, so someone answering can help you understand what might be going wrong. – N. Owad May 14 '15 at 17:54
  • Hey, thanks. I tried to use the definition of covering map, which can be seen on this link: http://mathworld.wolfram.com/CoveringMap.html. But it is not so clear to me how I should construct the neigbourhood of a point in $T.$ – Alex May 14 '15 at 18:37

1 Answers1

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You're essentially having a matrix $(\begin{smallmatrix}a&b\\c&d\end{smallmatrix})$ act on $\Bbb R/\Bbb Z\times\Bbb R/\Bbb Z\cong\Bbb R^2/\Bbb Z^2$ in the obvious way.

If $A\in{\rm GL}_n(\Bbb R)\cap M_{n\times n}(\Bbb Z)$ then $A$ acts as an automorphism of $\Bbb R^n$ and maps $\Bbb Z^n$ into itself.

Consider the composition $\Bbb R^n/\Bbb Z^n\to A\Bbb R^n/A\Bbb Z^n\to \Bbb R^n/\Bbb Z^n$. The first part is an isomorphism, and the second part is a projection map since $A\Bbb R^n=\Bbb R^n$ and $A\Bbb Z^n\subseteq\Bbb Z^n$. The kernel of the latter is the finite group $\Bbb Z^n/A\Bbb Z^n$ which has cardinality $m=|\det A|$ (which can be seen by writing $A$ in Smith Normal Form so $A\Bbb Z^n$ simplifies nicely) and hence the map is $m$-to-$1$.

anon
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  • Thank you very much for the answer. If I understand it correctly you are showing why there has to be $|m|$ inverse images. I am also having trouble by showing that $\pi$ is a covering map. Or is it also in your answer? In this case I would have to ask for a little more details. – Alex May 14 '15 at 18:46
  • @user240592 Here's the visual: form a small enough disk around a point in $\Bbb R^2/\Bbb Z^2$ and its preimage will be $m$ disjoint ellipses. These ellipses $A$ maps homeomorphically onto said disk (which can be proved by lifting up to $\Bbb R^2\to\Bbb R^2$ presumably, I don't really do much bookkeeping to be honest). – anon May 14 '15 at 18:57