# Why are the integers appearing in lens spaces coprime?

I have a past paper question for a first course in algebraic topology, which asks one to calculate the first three homology and homotopy groups for the space $L_n$, defined as follows:

Let $G=\{z\in\mathbb C|z^n=1\}\cong\mathbb Z_n$ act on $S^3=\{(z_1,z_2)| |z_1|^2+|z_2|^2=1\}$ by $z(z_1,z_2)=(zz_1,zz_2)$ and define $L_n$ to be the quotient space. (The question claims that the action of $G$ is properly discontinuous.)

This space $L_n$ looks suspiciously like a lens space $L(p,q)$ with $p=q=n$, with the exception that $p,q$ are not coprime, as usually required in the definition.

I think I can manage to calculate homology and homotopy groups for the usual lens space, but I am a bit lost as to what $L_n$ is.

My question is, why are $p,q$ in the definition of a lens space $L(p,q)$ required to be coprime?

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Wikipedia says coprimality is required for the action to be free. –  Qiaochu Yuan Jun 16 '13 at 1:20
@QiaochuYuan In that case the quotient is a manifold if and only if $p,q$ are coprime. I guess that doesn't matter for homology and homotopy, so for all pure topoloical properties, the space as given in the question is a lens space, then, correct? –  Earthliŋ Jun 16 '13 at 1:23
Recall that $L(p,q)$ is the quotient of $S^3$ by the action of $\mathbb{Z}/p\mathbb{Z}$ (thought of as the $p$th roots of $1$) given by $z(z_1,z_2) = (zz_1, z^q z_2)$. Hence, the space $L_n$ is actually the space $L(n,1)$ in the $(p,q)$ notation (and note that $\gcd(n,1) = 1$, as is required for the $(p,q)$ notation.)
As Qiaochu (and Wikipedia) point out, requiring $\gcd(p,q)=1$ is equivalent to asking that the action be free. Free actions are nice for two reasons. First, as you pointed out, freeness guarantees the quotient is a manifold. Second, it also guarantees the quotient map $S^3\rightarrow L(p,q)$ is a covering, which reduces much (but not all) of the algebraic topology considerations of $L(p,q)$ to those of $S^3$.
Thank you very much, I seem to have missed the notation $L(p,q)=L(p;1,q)$. Is a free action really necessary for getting covering map, though? I thought that a covering is a purely topological phenomenon, not depending on the manifold structure. Even in the non-free case I expect the alg. top. considerations still to reduce to those of $S^3$ (and $p$, of course). Am I expecting too much? –  Earthliŋ Jun 16 '13 at 3:12
Well, consider the action of $G=\mathbb{Z}/2\mathbb{Z}$ on $S^1$ given by reflection in the line $y=0$. This is almost a free action covering map (has two bad points), yet the orbit space $S^1/G$ is homeomorphic to $[0,1]$. In particular, it is simply connected so has no nontrivial covering. Reflection $S^n$ in the hyperplane $x_n = 0$ gives a similar example. –  Jason DeVito Jun 16 '13 at 3:44