I read from an article that the space $X=S^5/Z_q$ is not a Lens space because the orbifold action is not compatible with the action of the Hopf fibration $S^1\longrightarrow S^5\longrightarrow CP^2$. However, an article tells me that the Leray spectral sequence gives the following torsion terms
$H_1(S^5/Z_q)=Z_q$
$H_3(S^5/Z_q)=Z_q$
The question is: are there ways to compute the homologies for these spaces for all $q$'s? How in general an homology is computed, or maybe how can I see (more or less easily) that some homology of these such spaces have a torsion term?
I don't know how the Leray spectral sequence works but still I'm interested in ways of computing homologies for spaces that are modded by some orbifold action (let's say also abelian orbifolds). Are there any practical references or some way to compute such things?
Best Answer
If the action is free, then you have the Cartan-Leray spectral sequence that has second page $E^2_{p,q} = H_p(G, H_q(X)) \implies H^{p+q}(X/G)$. So if $H_q(X)$ is nice enough (this is the case here : $X$ is $S^5$ so it only has two nontrivial homology groups), and if you know the homology of $G$ well enough (this is the case here : $G=\mathbb{Z/q}$ is a cyclic group, whose homology is well-known to be $\mathbb{Z}$ in degree $0$, $0$ in positive even degrees, and $\mathbb{Z/q}$ in odd degrees), then you can get access to the homology of $X/G$ by this spectral sequence.
Here, the $E^2$ page is particularly nice since it has only two nonzero rows (when $H_q(S^5) \neq 0$, that is $q=0$ or $5$)
The action being free tells us that $g$ is homotopic to $-id$ for all $g\in G$ , but then it must induce $id$ on homology because $5$ is odd, therefore $H_p(G,H_q(X))$ is with trivial action of $G$ on $H_q(X)$.
So we have two nonzero rows $q=0,5$ and so the differentials don't do anything until page $6$, but then the bidegree is $(-6, 5)$ and so the $5$ first groups of the bottom row don't change even then, and they don't change afterwards : therefore $E^\infty_{p,0} = E^2_{p,0}$ for $p\leq 5$.
Moreover $E^\infty_{p,q}=0$ for $p+q\leq 4, p<4$, so that actually $H_n(S^5/(\mathbb{Z/q}))=E^\infty_{n,0}$ for $n\leq 4$, therefore for $n\leq 4$, $H_n(S^5/(\mathbb{Z/q}))= E^2_{n,0} = H_n(G,H_0(X)) = H_n(G,\mathbb{Z})$, and this together with the knowledge of the group homology of cyclic groups is enough to conclude.
Note that this works (with appropriate modifications) for any free action of a group $G$ on an odd dimensional sphere, so it's a very useful tool in determining for instance which groups can act freely on spheres; but as you see here it also allows us to get from the algebraic computation of the homology of $G$ to the computation of the homology of this quotient $S^n/G$.