# Why the following set is a cylinder?

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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What do you define as a cylinder? Because it's pretty intutitive; geometrically, looking at the plane with a loop in one direction and a straight line in another is pretty much the same thing as a cylinder. If you are having problems dealing with definitions now that's another question. – Patrick Da Silva Dec 23 '11 at 18:11
@Patrick, my understanding of the set $\mathbb{C}_a$ for some real number $a$ is the same as the set $\{x\in \mathbb{C} \mid 0 \leq Im(x) \leq 2\pi/a \}$. – LJR Dec 23 '11 at 18:22
I didn't ask for your definition of $\mathbb C_a$, I asked for your definition of a cylinder. Your definition is right there, I'm not blind (yet =P). If this is just a question of understanding in which sense this could be seen as a cylinder, then just think of the way that a cylinder has a loop in one direction (if you cut it in a way so that the cut leaves a circle, the loop is that circle precisely), and in the other direction, a straight line that goes to infinity. $\mathbb C_a$ has pretty much those properties, because $\{ x \in \mathbb C \, | \, 0 \le Im(x) \le 2\pi/a \}$ is that loop. – Patrick Da Silva Dec 23 '11 at 18:28
@user9791: Assuming $a\in\mathbb{R}$, the set you describe is almost in bijection with $\mathbb{C}_a$. The issue is that the bounding lines $\{Im(x)=0\}$ and $\{Im(x)=2\pi/a\}$ are glued together, giving a cylinder. (In the notation of the answer I posted below, where I didn't assume $a\in\mathbb R$, these lines are $L_0$ and $L_1$.) – Brad Dec 23 '11 at 18:30
Although if $a$ is an arbitrary complex number (I thought of it as a positive real number at first), you would still have a cylinder, but the loop would be $\{ x \in \mathbb C \, | \, x = \alpha \left( \frac{2\pi i}a \right), \alpha \in [0,1] \}$. – Patrick Da Silva Dec 23 '11 at 18:32

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.