Let $X=F_{\mathbb{Z_7}}(S^3,2)$ be the orbit configuration space, given by pairs of points $(x,y)$ of $S^3$ which dont lie on the same orbit, i.e., $x\neq g(y)$ for any given $g \in \mathbb Z_7$. The projection on the first coordinate gives $F_{\mathbb{Z_7}}(S^3,2)$ the structure of bundle over $S^3$ with fibre $S^3\backslash\mathbb Z_7 \simeq \bigvee_6 S^2$ that admits a section.
Question: How do I compute its homology groups via the serre spectral sequence ${\displaystyle E_{p,q}^{2}=H_{p}(S^3,H_{q}(\vee_6 S^2))\Rightarrow H_{p+q}(X)}$ and what exactly went wrong on my attempt.
My incorrect attempt:
Using that $H_p(S^3;\mathbb Z^6)$ is $\mathbb Z^6$, if $p=0,3$ and $0$ otherwise and $H_q(\bigvee_6 S^2) $ is $\mathbb Z^6$, if $q=0,2$ and $0$ otherwise, we get for the $E^2$ terms of the spectral sequence: $E_{p,q}^{2}=\mathbb Z^6$ if $p=0,3 \wedge q=0,2$ and otherwise $E_{p,q}^{2}=0$.
Therefore the differential map $d_2:E_{p,q}^{2} \to E_{p-2,q+1}^{2}$ is trivial for every $p,q$ (either $p\neq 0,3$ or $p-2 \neq 0,3)$. This implies $E_{p,q}^{2} = E_{p,q}^{\infty}$. Furthermore we get the terms $E_{p,q}^{\infty}$ isomorphic to the successive quotients $F^p_{p+q}/F^{p-1}_{p+q}$ in a filtration $0 \subset F^0_{p+q} \subset \dots \subset F^{p+q}_{p+q} = H_{p+q}(X)$ of $H_{p+q}(X)$.
Hence we have
- $F^0_{0}/ 0=E_{0,0}^{\infty}=\mathbb Z^6$
- $F^3_{3}/F^2_{3} =E_{3,0}^{\infty}=\mathbb Z^6$
- $F^0_{2}/ 0=E_{0,2}^{\infty}=\mathbb Z^6$
- $F^3_{5}/F^2_{5} =E_{3,2}^{\infty}=\mathbb Z^6$
and all the other quotients are $0$, since the other $E_{p,q}^{\infty}$ are zero.
We conclude:
$F^2_{3} = F^2_{5}=0$
$H_{0}(X) = F^0_{0}= \mathbb Z^6$
$H_{2}(X) = F^0_{2}= \mathbb Z^6$
$H_{3}(X) =F^3_{3}= \mathbb Z^6$
$H_{5}(X) = F^3_{5}= \mathbb Z^6 $
$H_n(X)=0$ for $n\neq 0,2,3,5$.
THe formula you wrote for $H^0(\bigwedge^6 S^2)$ is not correct; this wedge of spheres is connected and so its zeroth cohomology has rank one.