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Let $\text{CH}_4$ be the molecule of Methane: enter image description here

The four hydrogen atoms form vertices of a regular tetrahedron with the carbon atom in the center of the regular tetrahedron. Here we regard all atoms to be points in $\mathbb{R}^3$ with volume zero (or we can consider the centers of these atoms, which are points with volume zero). We want to embed the $\text{CH}_4$ molecule into $\mathbb{R}^3$ such that the $\text{C}$-atom is at the origin $0$.

enter image description here

The collection of all such embeddings of $\text{CH}_4$ molecule into $\mathbb{R}^3$ form a manifold $G$. In fact, $$ G=O(3)/{\frak S}_4=SO(3)/{\frak A}_4, $$ where ${\frak S}_4$ is the symmetric group and ${\frak A}_4$ the alternating group. Here we regard $O(3)$ as the isometric embedding Lie group of the $\text{CH}_4$ molecule with the four hydrogen atoms distinct, labelled by $1,2,3,4$. Then we quotient the permutation action of $S_4$ on the four hydrogen atoms, regarding all the four hydrogen atoms the same and do not distinguish them.

Question: as a manifold, what is the cohomology ring (with cup product) $$ H^*(G;\mathbb{Z}_2) $$ and the Steenrod square $Sq$'s acting on the cohomology ring?

A contradiction with the following answer by W. Schlieper:

We consider a vector bundle $$ \xi: \mathbb{R}^4\to O(3)\times_{S_4}\mathbb{R}^4\to O(3)/S_4 $$ where $S_4$ permutes the order of coordinates in $\mathbb{R}^4$. The action of $S_4$ on $\mathbb{R}^4$ may have determinant $-1$, hence it is not contained in $SO(4)$, but only in $O(4)$. Therefore, the first Stiefel-Whitney class $w_1(\xi)\neq 0$. However, $w_1(\xi)\in H^1(O(3)/S_4;\mathbb{Z}/2)$. This contradicts your result $$ H^1(O(3)/S_4,\mathbb{Z}/3) \simeq H^2(O(3)/S_4,\mathbb{Z}/3) \simeq \mathbb{Z}/3. $$ Why we have such a contradiction?

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    $\begingroup$ I don't see that your $G$ is a Lie group, because I don't think $\mathfrak S_4$ is a normal subgroup of $O(3)$. $\endgroup$
    – Andreas Blass
    Commented Oct 3, 2015 at 6:31
  • $\begingroup$ In response to the apparent contradiction: could you also write the vector bundle as $\mathbb{R}^4\to SO(3)\times_{A_4}\mathbb{R}^4\to SO(3)/A_{4}$? Then it is clear that the bundle is orientable. The point is that $\pi_1(O(3)/S_4)\neq S_4$. $\endgroup$
    – Mark Grant
    Commented Oct 6, 2015 at 6:37
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    $\begingroup$ In the question, $S_4$ acts on $O(3)$, which is not a connected group. Equivalently, one can divide $SO(3)$, which is connected, by an action of $A_4$; $A_4$ itself however will not be the fundamental group because $SO(3)$ is not simply connected, so $SO(3) \simeq \mathbb{RP}^3$ is a cover, but not the universal cover. $\endgroup$
    – W. Cadegan-Schlieper
    Commented Oct 6, 2015 at 17:02
  • $\begingroup$ Just being curious, why is the methane molecule mentioned in the beginning? It seems to me a tetrahedron works equally well. And I didn't realize it's a cohomology question until I scrolled down almost half of the page. $\endgroup$
    – Honglu
    Commented Nov 22, 2015 at 22:04

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Over $\mathbb{Z}/2$ it turns out to be pretty boring. Your space $G$ (which you really shouldn't call that, as it's merely a manifold, not a group), which you define as a quotient of $O(3)$ or equivalently as one of $SO(3)$, can be thought of as the quotient of this group's double cover $SU(2) \cong S^3$ by the binary tetrahedral group $2T$, which turns out to be your fundamental group as $S^3$ is simply connected.

The first homology group over $\mathbb{Z}$ will be $2T/[2T,2T]$, but $[2T,2T]$ is the quaternion group $Q_8$, and $2T/Q_8 \simeq \mathbb{Z}/3$, so $H_1(G,\mathbb{Z}/2)=H^1(G,\mathbb{Z}/2)=0$ and by Poincaré duality, $H^2(G,\mathbb{Z}/2)=0$ will have the same values as that of the $3$-sphere.

Over $\mathbb{Z}/3$ you'll get $H^1(G,\mathbb{Z}/3) \simeq H^2(G,\mathbb{Z}/3) \simeq \mathbb{Z}/3$ and a nontrivial Bockstein, though.

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  • $\begingroup$ Thanks, Schlieper! Why we can use Poincar\'{e} duality? The Poincar\'{e} duality requires the manifold to be orientable. However, $G$ may not be orientable. $\endgroup$
    – QSR
    Commented Oct 8, 2015 at 8:27
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    $\begingroup$ Over $\mathbb{Z}/2$, orientability is actually not required as orientation mod $2$ is trivial. But for more general coefficient groups, $G$ will in fact turn out to be oriented because $S^3$ is oriented and the action of $2T$ is entirely of rotations of $S^3$, which all preserve the orientation. $\endgroup$
    – W. Cadegan-Schlieper
    Commented Oct 8, 2015 at 17:10

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