First and third homology of $S^5/Z_q$ and Leray spectral sequence

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?

• What action of $\mathbb{Z}_q$ do you have in mind? – Qiaochu Yuan Mar 7 at 19:53
• I know that a Lens space relate a five-sphere with a 3-dimensional complex space. Taking three coordinates (z1,z2,z3) of C3 I have in mind an action of Zq as (\lambda z1, \lambda z2, \lambda^(-2) z3) where \lambda is a q-root of the unity. Is it what you're asking? – Alessandro Mininno Mar 8 at 7:30

Edit: the spectral sequence argument to prove my theorem is given more concisely in Max's answer

The geometric intuition here is that if we're considering free actions of a topological group $$G$$ on a sphere $$S^k$$, then the quotient $$S^k/G$$ is "homotopically the same" as the classifying space $$BG$$ in degrees that are small relative to $$k$$ (in particular they have the same homotopy and homology groups in these degrees), essentially since $$BG$$ is a quotient of a contractible space by a free action, and $$S^k$$ is $$(k-1)$$-connected so it "looks contractible" below degree $$k$$. Computing the homology of $$BG$$ is usually done via spectral sequence. In our case when $$G=\mathbb{Z}/n$$ for $$n> 1$$ then the homology of $$BG$$ is isomorphic to $$\mathbb{Z}/n$$ in every odd degree and is $$0$$ otherwise (see the bottom of my post), so this is one way of seeing where the torsion is coming from in $$H_1$$ and $$H_3$$ of $$S^5/(\mathbb{Z}/n)$$.

If you're interested in seeing how the spectral sequence argument goes, I think I figured out what they were aiming for. I just noticed in one of your comments that you're very new to proofs in general so this might be a bit advanced, but believe it or not it's relatively nice as far as spectral sequences go.

My spectral sequence argument actually establishes the following:

Theorem: If $$G$$ is a topological group with an action on $$S^k$$ such that the quotient map is a fibre bundle (for finite groups it's enough that the action is free), and $$BG$$ is the classifying space of principal $$G$$-bundles, then

$$H_p(S^k/G) \cong H_p(BG) \text{ for }p\leq k.$$

You can also get this result via proving the analogue for homotopy groups first and then showing that implies the homology result. In our particular case of $$\mathbb{Z}/n$$ acting on $$S^5$$, this result combined with the spectral sequence computation in the Aside tells us that $$H_1(X)\cong H_3(X) \cong \mathbb{Z}/n$$, regardless of the free action we started with.

Proof via spectral sequence: Suppose $$G$$ is a topological group (for our case it will be $$\mathbb{Z}/n$$) and suppose it admits a free action on $$S^k$$ such that the quotient map $$S^k \to X= S^k/G$$ is a fibre bundle. This is in particular a fibration so we could try study its Leray-Serre spectral sequence, but it turns out a slightly different fibration is more convenient here. Instead, use the fact that this principal $$G$$ bundle is classified by a map to the classifying space $$BG$$, such that

$$S^k \to X \to BG$$ is equivalent to a fibration. If $$G$$ is a discrete group the spectral sequence for this fibration is usually called the spectral sequence for a covering space.

The $$E^2$$ page has groups

$$E^2_{p,q} \cong H_p(BG ; H_q(S^k))$$

and it converges to $$H_{p+q}(X)$$. These groups can only be non-zero when $$q =0$$ or $$k$$, and the $$q=0$$ row is just a copy of $$H_*(BG)$$. If $$\pi_1 BG\cong \pi_0 G \neq 0$$ we need to use twisted coefficients when $$q = k$$ where the action of $$\pi_0 G$$ on $$H_k(S^k)$$ is induced by the original action on the space $$S^k$$, and so the $$k$$-th row of the $$E^2$$ page (along with the differentials) will depend on the particular action. What doesn't depend on the action is that the only non-zero differentials are $$d_{k+1}$$ and so the groups $$\{E^r_{p,0}\}_{p\leq k}$$ are never hit by any differentials for $$r >1$$, so must survive as $$\{H_p(X)\}_{p\leq k}$$. The result follows since $$E^2_{p,0}\cong H_p(BG)$$.

Aside: The integral homology of $$BG$$ where $$G=\mathbb{Z}/n$$ is

$$H_p(B\mathbb{Z}/n) \cong \begin{cases} \mathbb{Z} & \text{if}\, p=0 \\ \mathbb{Z}/n & \text{if}\, p \text{ is odd} \\ 0 &\text{otherwise} \end{cases}$$

This can be seen via another spectral sequence argument. First you do the usual computation of the spectral sequence for $$S^1 \to S^\infty \to \mathbb{CP}^\infty$$ and find out that the $$E^2$$ page is concentrated in the first two rows, and at the same time compute the cohomology of $$\mathbb{CP}^\infty$$ and show that every non-trivial differential must be an isomorphism. Now restrict the $$S^1$$ action on $$S^\infty$$ to the subgroup $$\mathbb{Z}/n$$ and construct $$BG$$ as the quotient $$S^\infty/G$$, and notice that you get a map from $$BG \to \mathbb{CP}^\infty$$ whose fibre is $$S^1/n\mathbb{Z}$$. In fact this gives a fibration

$$S^1 \to B\mathbb{Z}/n \to \mathbb{CP}^\infty$$

and a map of fibrations from $$S^1 \to S^\infty \to \mathbb{CP}^\infty$$ which is a degree $$n$$ map on the fibres. By considering the induced map on spectral sequences you can deduce the result.

• Thank you! You gave to me what I was looking for! – Alessandro Mininno Mar 11 at 8:27

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 $$S^n/G$$.

• Both the answers you gave to me were wonderful! Thank you very much! I chose the other because he made some.example for me that allowed me to explain and extend what I asked also for other cases. Thank you anyway! – Alessandro Mininno Mar 11 at 8:26
• Both the answers you gave to me were wonderful! Thank you very much! I chose the other because he made some.example for me that allowed me to explain and extend what I asked also for other cases. Thank you anyway! – Alessandro Mininno Mar 11 at 8:26

For a Lens space $$S^{n-1}/G$$ is true that

$$H_j(S^{n-1}/G;Z)=\begin{cases} Z & if\, j=0 \\ |G|-torsion & if \, 1\leq j \leq n-2 \\ |G|-torsion \, or \, Z & if \, j=n-1 \end{cases}$$

In particular, if $$G-$$action preserves the orientation of $$S^{n-1}$$, then $$H_{n-1}(S^{n-1}/G;Z)\cong Z$$.

This is a proporsition that can be proved using the following isomorphism

$$H^*(X/G;k)\cong H^*(X;k)^G$$

where $$X$$ is a locally compact Hausdorff space, $$G$$ is a finite group and $$k$$ is a field of characteristic zero or coprime to |G| and then using the universal coefficient theorem. I am satisfied with this proof. Fonts and proof can be found in many books but very recently in arxiv.org/abs/1711.01748

• When you wrote "orbifold action" above, I was thinking you meant "action which may or may not be free". For a free action, I agree with this answer. But if the answer is not free, then it's less clear to me. – Jason DeVito Mar 8 at 14:57
• Is the action that I wrote in the comments to my question free? I'm not very familiar with the language so I'm actually asking for confirmation if this answer and with my definition of Lens space on a five-sphere, I solve my problem :) – Alessandro Mininno Mar 8 at 15:03
• Free means that the only element of $G$ which fixes any point of $M$ is the identity (which fixes all of $M$). For your action, $\lambda$ fixes a point $(z_1,z_2,z_3)\in S^5\subseteq \mathbb{C}^3$ iff $\lambda z_1 = z_1, \lambda z_2 = z_2, and \lambda^{-2} z_3 = z_3.$ Now, because $(z_1,z_2,z_3)\in S^5$, at least one of the $z_i$ is non-zero. If it is $z_1$, the equation $\lambda z_1 = z_1$ forces $\lambda = 1$, so we are good there. Likewise if $z_2\neq 0$. However, if $z_1 = z_2 = 0$, then we only have the third equation $\lambda^{-2} z_3 = z_3$.... – Jason DeVito Mar 8 at 15:24
• As $z_3\neq 0$, this gives $\lambda^{-2} = 1$. If $q$ is odd, then $\lambda^{-2} = 1$ implies $\lambda = 1$, so we have a free action when $q$ is odd. On the other hand, when $q$ is even, $\lambda = -1$ is a $q$-th root of unity, and then we find that $-1\in G$ fixes the point $(0,0,\pm 1)\in S^3$, so the action is not free in this case. Unrelated, but I don't see how to use the listed isomorphism to conclude anything about $H_j(S^{n-1}/G;\mathbb{Z})$ since $\mathbb{Z}$ is not a field... – Jason DeVito Mar 8 at 15:25
• If I understood you correctly you are making two fairly good points: 1. If q is even the proposition doesn't seem to work, although in the article they don't say that G needs to act freely on the sphere. 2. The proof uses an isomorphism defined with fields and Z is not a field. I cannot answer this second point but I think that that isomorphism is defined as consequence of Leray spectral sequence and it seems that using that sequence on this spaces it can be proved the torsion part, despite Z not being a field – Alessandro Mininno Mar 8 at 16:01