cohomology ring of a quotient space

Question

what is the cohomology ring $$ H^*((S^3\times S^3\setminus \{e\})/(a,b)\sim (ab,b^{-1});\mathbb{Z}_2)? $$ Here the unit $e$ and the product $ab$ is of the Lie group $S^3=Sp(1)$.

Answer 1

Let $\sigma:S^3\times S^3\to S^3\times S^3$ such that $\sigma(a,b)=(ab,b^{-1})$. As $\sigma\circ\sigma$ is the identity map, $\sigma$ is a homeo of order $2$.

The subset $X=S^3\times(S^3\setminus\{e\})$ is invariant under $\sigma$ and, if we call $G$ the group generated by $\sigma$ (which is cyclic of order two), you are asking for the cohomology of the quotient $Y=X/G$. It is easy to see that $\sigma$ does not have fixed points on $X$, so the action of $G$ on $X$ is properly discontinuous and, in particular, the quotient map $X\to Y$ is a two-sheeted covering.

There is a convergent spectral sequence of algebras with second page $E_2^{p,q}=H^p(G,H^q(X,\mathbb Z_2))$ converging to $H^\bullet(Y,\mathbb Z_2)$; this was constructed, for example, by Grothendieck in his famous Tôhoku paper.

Now $H^\bullet(X)=H^\bullet(S^3\times(S^3\setminus\{e\}))$ is easy to compute, as the inclusion $S^3\times\{u\}\to S^3\times(S^3\setminus\{e\})$ is a homotopy equivalent for all $u\in S^3\setminus\{e\}$.

The shape of the spectral sequence implies that $E_3=E_2$ and $E_\infty=E_4$. Moreover, looking at the corner we see at once that $H^0(Y,\mathbb Z_2)=H^1(Y,\mathbb Z_2)=\mathbb Z_2$. Since the spectral sequence is of algebras, and using the standard calculation of the cohomology algebra of the cyclic group of order $2$ with coefficients in $\mathbb Z_2$, and the fact that the space $Y$ is a $3$-manifold so that it cohomology vanishes in dimensions larger than $3$, we see that the differential $d_3^{0,2}$ is an isomorphism (otherwise the limit would have infinitely many non-zero group), and from that we conclude that $H^p(Y,\mathbb Z_2)=0$ for all $p\geq2$.

The ring structure is therefore trivial, up to mistakes here. But, in any case, looking at this spectral sequence should be quite enough.