positions of regular cubes in Euclidean space with all its vertices without distinction

Question

Let $P$ be the standard regular cube, centered at the origin of $\mathbb{R}^3$ with all its vertices on the unit sphere.

If the vertices are labelled by $1,2,3,4,5,6,7,8$, then the collection of all such positions of the cube in $\mathbb{R}^3$ is $O(3)$.

Now we regard all its vertices to be the same without distinction. As a result, the collection of all such positions of the cube (with vertices the same without distinction) in $\mathbb{R}^3$ is $O(3)/Sym(P)$, where $Sym(P)$ is the symmetric group of the regular cube.

Question: any reference for $O(3)/Sym(P)$? What is the mod 2 cohomology ring $$ H^*(O(3)/Sym(P);\mathbb{Z}/2)? $$

Answer 1

If the polyhedron $P$ in ${\mathbb R}^3$ has orientation-reversing symmetries then the orbit space $O(3)/Sym({P})$ is equal to $SO(3)/G$ for $G=G{(P})$ the orientation-preserving elements of $Sym{(P})$. If $P$ has no orientation-reversing symmetries then $O(3)/Sym{(P})$ has two components each homeomorphic to $SO(3)/G$ so we may as well restrict attention to $SO(3)/G$ in all cases.

When $P$ is a cube the group $G$ is the octahedral group of order 24 and $SO(3)/G=S^3/\tilde G$ for $\tilde G$ the binary octahedral group of order 48. This has a presentation $\langle a,b \ |\ a^2=b^3=(ab)^4\rangle $ from which one can easily compute the abelianization $H_1(SO(3)/G)$ to be $\mathbb Z_2$ (=${\mathbb Z}/2$). Thus by the universal coefficient theorem and Poincaré duality $H^*(SO(3)/G;{\mathbb Z}_2)$ is isomorphic to $H^*({\mathbb R}P^3;{\mathbb Z}_2)$ additively. This is also an isomorphism multiplicatively for the cup product structure. To see this we can use the general fact that the Bockstein $\beta:H^1(X;{\mathbb Z}_2)\to H^2(X;{\mathbb Z}_2)$ is the same as the cup-product square, $x\mapsto x^2$. (It suffices to check this in the universal example ${\mathbb R}P^\infty = K({\mathbb Z}_2,1)$.) Thus if $x$ is a generator of $H^1(SO(3)/G;{\mathbb Z}_2)$ then $x^2$ is a generator of $H^2(SO(3)/G;{\mathbb Z}_2)$ since $\beta$ is nonzero on classes coming from elements of order 2 in $H_1(X;{\mathbb Z})$. Since $x^2$ generates $H^2(SO(3)/G;{\mathbb Z}_2)$ it follows from Poincaré duality that $x^3$ generates $H^3(SO(3)/G;{\mathbb Z}_2)$.

For each symmetry group $G$ the manifold $SO(3)/G$ is Seifert fibered over $S^2$ with at most three singular fibers. Cup products in these manifolds have been computed with ${\mathbb Z}_2$ coefficients in a paper by J. Bryden, C. Hayat-Legrand, H. Zieschang, and P. Zvengrowski called "The cohomology ring of a class of Seifert manifolds" in Top. Appl. 105 (2000), 123-156. There is a later paper by Bryden and Zvengrowski generalizing this to all orientable Seifert manifolds and ${\mathbb Z}_p$ coefficients for $p$ any prime.

An interesting case is when $P$ is a "brick", a rectangular parallelepiped with three different edge lengths. The symmetry group $G$ in this case is ${\mathbb Z}_2 \times {\mathbb Z}_2$ and $\pi_1(SO(3)/G)$ is the quaternion group of order 8, so $H_1(SO(3)/G;{\mathbb Z})= {\mathbb Z}_2 \times {\mathbb Z}_2$. The ring $H^*(SO(3)/G;{\mathbb Z}_2)$ is the quotient $$ {\mathbb Z}_2[x,y]/(x^3,y^3,x^2+y^2+xy) $$ Thus $x^2$ and $y^2$ generate $H^2$, while $H^3$ is generated by $x^2y=xy^2$, with $x^3=y^3=0$. Additively this cohomology ring is the same as $$ H^*({\mathbb R}P^3\#{\mathbb R}P^3;{\mathbb Z}_2)={\mathbb Z}_2[x,y]/(xy,x^3+y^3,x^4,y^4) $$ but the ring structures differ by whether there are nonzero cubes or not.