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Let $\mathbb{C}$ be the complex plane. Define a set: $\mathbb{C}_{a} = \mathbb{C}/((2\pi i/a)\mathbb{Z})$ such that $2\pi i/a \not\in \mathbb{Q}$. Here $\mathbb{Z}$ is the set of integers and $\mathbb{Q}$ is the set of rational numbers. Why $\mathbb{C}_a$ is a cylinder? Thank you very much.

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    @Brad, thank you very much.2011-12-23

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Let's simplify the notation by setting $w = \frac{2\pi i}{a}$, so that $\mathbb{C}_a = \mathbb{C}/w\mathbb{Z}$.

For each integer $n$, consider the line $L_n = \{nw + tiw: t\in\mathbb R\}$. That is, $L_n$ is the line through $nw$ which is orthogonal to the line $\mathbb{R}w$.

Next, for each $n$, let $R_n$ denote the rectangular region bounded by the lines $L_n$ and $L_{n+1}$, but not including the line $L_{n+1}$ itself. Let $q:\mathbb{C}\to \mathbb{C}_a$ be the quotient map, sending $z$ to the coset $z+w\mathbb{Z}$. Then for any given $n$, the map $q$ restricts to a bijection from $R_n$ to $\mathbb{C}_a$, and the lines $L_n$ and $L_{n+1}$ are identified under $q$.

So what happens is that $\mathbb{C}_a$ is realized by taking the closure of the rectangular region $R_n$ (for any choice of $n$) and gluing together the bounding lines $L_n$ and $L_{n+1}$. This gives a cylinder.