Homology groups of spherical $3$-manifolds Let $G$ be a finite subgroup of $SO(4)$, acting freely on $\mathbb{S}^3$. How can we compute $H_2(\mathbb{S}^3/G; \mathbb{Z})$? Is it zero?
 A: As Steve D suggests in the comments, this follows from Poincaré duality. The quotient $S^3/G$ is orientable because $G$ is a subgroup of $SO(4)$, and therefore
$$H_2(S^3/G; \mathbb{Z}) \cong H^1(S^3/G; \mathbb{Z}) \cong \operatorname{Hom}(\pi_1(M/G), \mathbb{Z}) = \operatorname{Hom}(G, \mathbb{Z}) = 0.$$
More generally, for any finite group $G$ which acts freely on a rational homology sphere $X$ with $\dim X = 2n + 1 \geq 3$, we have $H_{2n}(X/G; \mathbb{Z}) = 0$. To see this, first note that $X/G$ must be orientable. There are two ways to do this:

*

*As $ \pi : X \to X/G$ is a finite covering, $\pi^* : H^*(X/G; \mathbb{Q}) \to H^*(X; \mathbb{Q})$ is injective, so $b_i(X/G) \leq b_i(X)$ for all $i$. As $X$ is a $(2n+1)$-dimensional rational homology sphere, its only non-zero Betti numbers are $b_0(X) = b_{2n+1}(X) = 1$. Note that $b_0(X/G) = 1$, while $b_{2n+1}(X/G) = 1$ if and only if $X/G$ is orientable, and all other Betti numbers are zero. As $\chi(X/G) = 0$, we must have $b_{2n+1}(X/G) = 1$ and hence $X/G$ is orientable.


*Let $f : X \to X$ be a fixed-point free homeomorphism. By the Lefschetz fixed point theorem, the alternating sum of traces of $f_*$ on rational homology is zero. As $X$ is a rational homology sphere, it only has non-zero rational homology in degrees $0$ and $2n+1$ in which case it is one dimensional. On degree zero homology, $f_*$ is always the identity, while on degree $2n + 1$ homology, it acts by the identity if and only if $f$ is orientation-preserving. As the alternating sum of traces is zero, we see that $f_*$ must be the identity on $H_{2n+1}(X; \mathbb{Q})$ and hence $f$ is orientation-preserving. In particular, $G$ must be an orientation-preserving action, and hence $X/G$ is orientable.
Now that we know $X/G$ is orientable, we see that $H_{2n}(X/G; \mathbb{Z}) \cong H^1(X/G; \mathbb{Z})$ by Poincaré duality. As mentioned in point 1 above, we have $b_1(X/G) \leq b_1(X) = 0$ so $H^1(X/G; \mathbb{Z})$ has rank zero. As the first cohomology group is always torsion-free, we conclude that $H^1(X/G; \mathbb{Z}) = 0$ and hence $H_{2n}(X/G; \mathbb{Z}) = 0$.
