How to compute the homology groups of spherical space forms?

Question

Let $S^n /\Gamma$ be a spherical space form where $\Gamma<O(n+1)$ is a finite subgroup. How can we find the homology groups $H_*(S^n /\Gamma,\mathbb Z)$?

Answer 1

This is a nice question with a nice answer. Since $\Gamma$ acts freely, either $n$ is even and $\Gamma$ is $\Bbb Z/2$ acting antipodally or $n$ is odd and $\Gamma \subset SO(n+1)$. I will only bother with the second case.

Because $\Gamma \subset SO(n+1)$, your action arises from the restriction to a unit sphere of $\Gamma \curvearrowright \Bbb R^{n+1}$. I will call this representation $V$. Then your sphere is the unit sphere $S(V)$, and the space you care about is $S(V)/\Gamma$.

First, give $S(V)$ a $\Gamma$-invariant cell decomposition: this arises by choosing a cell decomposition of $S(V)/\Gamma$ and taking the naturally induced cell decomposition on the universal cover. In fact, we may choose a cell decomposition of $S(V)/\Gamma$ which has one $0$-cell and one $n$-cell (by Morse theory, if you like).

Write $C_*(S(V))$ for the associated cellular chain complex; this is a chain complex of free $\Bbb Z[\Gamma]$-modules, with $H_*(S(V)) = \Bbb Z$ in degrees $0$ and $n$.

Write $$C = \bigoplus_{k \geq 0} C_*(S(V))[k(n+1)].$$ The $[j]$ means that the chain complex is shifted up by $j$ in degree. We make $C$ into a $\Bbb Z[\Gamma]$-chain complex as follows. The differential on $C_j(S(V))[k(n+1)]$ is the same as on $C_j(S(V))$ for $k = 0$ or for $j > 0$. The only difference is the map $$d: C_0(S(V))[k(n+1)] \to C_n(S(V))[(k-1)(n+1)].$$

To describe this map, first, we use what we know about the homology groups. Because $H_0(S(V)) \cong \Bbb Z$, we find that there is an isomorphism $$C_0(S(V)) \cong B_0(S(V)) \oplus \Bbb Z$$ as $\Bbb Z[\Gamma]$-modules. (The second term is the invariant chain given by taking the sum over the $\Gamma$ many $0$-cells.) Similarly, we may find a $\Gamma$-equivariant isomorphism $$C_n(S(V)) \cong \Bbb Z \oplus Z_n(S(V)).$$

For $k>0$, we define the map $$d: C_0(S(V))[k(n+1)] \to C_n(S(V))[(k-1)(n+1)]$$ to send $B_0(S(V))$ to zero and to send $\Bbb Z$ to $\Bbb Z$. This is $\Gamma$-equivariant, so $C$ is a chain complex of free $\Bbb Z[\Gamma]$-modules. Furthermore, $C$ is a free resolution of $\Bbb Z$ (given by the augmentation $C_0(S(V)) \to \Bbb Z$, where we just sum over the weights on our $0$-cells). Therefore, the group homology $H_*(\Gamma;\Bbb Z)$ may be computed as $H_*(C \otimes_{\Bbb Z[\Gamma]} \Bbb Z)$.

What is this quotient? It is the chain complex given by assembling together shifted copies of $C_*(S(V)/\Gamma)$, with the map $C_0(S(V))/\Gamma)[k(n+1)] \to C_n(S(V)/\Gamma)[(k-1)(n+1)]$ just the identity map $\Bbb Z \to \Bbb Z$. Therefore, its homology groups are shifted copies of the reduced homology groups of $S(V)/\Gamma$, except that we have killed off the top-degree element of homology.

In particular, for $j < n$, we find $$H_j(S(V)/\Gamma;\Bbb Z) \cong H_j(\Gamma;\Bbb Z),$$ and in fact $H_j(\Gamma;\Bbb Z)$ enjoys a periodicity: for $j > 0$, we have $$H_j(\Gamma;\Bbb Z) \cong H_{j+n+1}(\Gamma;\Bbb Z)!$$

OK, I'll admit this is a bit backwards: how do you compute group homology in this case? You probably start by writing down exactly the manifold $S(V)/\Gamma$, and compute its homology (by understanding the geometry well enough to write down a cell decomposition), and exploit the periodicity we just found! But the connection was worth stating. I doubt you will find a more satisfying answer - you will likely have to compute by hand. And at least this gives meaningful answers for eg $H_1, H_2$. And group (co)homology can be computed algorithmically.

Answer 2

I like Mike Miller's answer because it provides a clear and direct link with group homology. Here's another way of seeing this link, which I share because it's always interesting to have different points of view (though I'm not 100% sure of how different the two are).

The point is that when you have a free $G$-complex $X$ for a group $G$, then you have the Cartan-Leray spectral sequence whose abutment is $H_*(X/G)$, and whose second page is $E^2_{p,q} = H_p(G,H_q(X))$ where $H_q(X)$ is a $G$-module in the obvious way. In particular if you know the homology of $G$ well enough and if $H_q(X)$ is often zero (or at least often has $H_p(G,H_q(X)) = 0$) then this spectral sequence can help you compute $H_*(X/G)$.

In our example $X=S^n$; and I won't copy Mike Miller's argument to explain why $X$ can be seen as a free $\Gamma$-complex (oh and I'll also only be considering the odd $n$ case, and $\Gamma \subset SO(n+1)$), so $H_q(X)$ is often $0$ (it's always, except when $q=0,n$)

So what happens in our spectral sequence (you should definitely try to draw it a bit to see what's happening) is that it's stable up to the $r=n+1$th page (the differentials have bidegree $(-r,r-1)$), so we might as well move straight to that page, let $r=n+1$; then $E^r_{p,q} = H_p(\Gamma,H_q(S^n))$, and the only interesting differentials are $d^r : E^r_{p,0}\to E^r_{p-(n+1), n}$.

Now there are two things to notice :

1) As noted in the comments below Mike's answer, we only care about $H_k(S^n/\Gamma)$ for $k<n$ : indeed for $k=n$, $S^n/\Gamma$ is an orientable closed $n$-manifold, so its $n$th homology is $\mathbb{Z}$ (or the coefficient ring, under mild assumptions, e.g. a PID), and for $k>n$, the $k$th homology is $0$ (because it's an $n$-manifold, or an $n$-dimension CW-complex, whichever you're most comfortable with)

2) If $g\in \Gamma$ acts on $S^n$, then since $g$ has no fixed point it is homotopic to $-id : S^n\to S^n$, hence induces the same map on $0$th and $n$th homology, which is known to be $id$ in the odd case; therefore $H_p(\Gamma, H_q(S^n)) = H_p(\Gamma, \mathbb{Z})$ with the trivial action on $\mathbb{Z}$ for $q=0,n$.

With these things in mind, for $p<n$, $d^r_{p,0}: H_p(\Gamma, \mathbb{Z})\to 0$, hence $E^\infty_{p,0} = E^r_{p,0} = H_p(\Gamma, \mathbb{Z})$

Now let $F_\bullet$ be a filtration on $H_*(S^n/\Gamma)$ such that $F_pH_{p+q}(S^n/\Gamma)/F_{p-1}H_{p+q}(S^n/\Gamma) = E^\infty_{p,q}$. Notice that since $E^\infty_{p,q} = 0$ if $q\neq 0,n$ and we're only interested in $p+q<n$, it only matters what $E^\infty_{p,0}$ is, so we are done trying to figure out more terms in the $\infty$th page of the spectral sequence.

So let $k<n$, then $F_pH_k(S^n/\Gamma)/F_{p-1}H_k(S^n/\Gamma) = E^\infty_{p, k-p}$, which is $0$ as long as $p\neq k$ (because $k-p < n$), so by induction you can prove that $F_pH_k(S^n/\Gamma) = 0$ for $p<k$, and $F_kH_k(S^n/\Gamma) = F_kH_k(S^n/\Gamma)/F_{k-1}(S^n/\Gamma) = E^\infty_{k,0} = H_k(\Gamma, \mathbb{Z})$; but now it's also clear that the higher $F_l/F_{l-1}$'s are $0$.

To conclude, we need to know that the filtration we got on $H_k(S^n/\Gamma)$ is finite; but that's just a consequence of how it was built. This then tells us that $H_k(S^n/\Gamma) = F_kH_k(S^n/\Gamma) = H_k(\Gamma, \mathbb{Z}), k<n$, so we find the same result as Mike gave.

Note that my answer seems long because I took my time with the details of the computation after we got the spectral sequence, but this is mostly because I'm new to spectral sequences and I wanted to make sure everything was clear to me; I assume an expert in spectral sequence could have written this in a much more concise manner.

Oh and a last thing : Mike says that it's a bit tautological because to compute $H_k(\Gamma, \mathbb{Z})$, we'd usually try to compute the homology of $S^n/\Gamma$, but there's also a purely algebraic construction of group homology, and this can be used for some computations; the example I gave in the comments is the computation of the homology of the lens space. Indeed, the homology of a finite cyclic group is easy to compute thanks to an easy free $\mathbb{Z}[G]$-resolution of $\mathbb{Z}$: if $P=\displaystyle\sum_{x\in G}x$ and $G=\langle g\rangle$, then there's a resolution of the form $\dots \to \mathbb{Z}[G]\to^{d_2} \mathbb{Z}[G]\to^{d_1} \mathbb{Z}[G]\to^{d_2}\mathbb{Z}[G]\to^{d_1} \mathbb{Z}[G]\to^\epsilon \mathbb{Z}$; where $d_1(1) = g-1$, $d_2(1) = P$; which makes $H_k(\mathbb{Z/nZ}, \mathbb{Z})$ really easy to compute.

Of course, as Mike pointed out, there are also algorithms to compute group homology.