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Let $S_3$ be the symmetric group of order $3$. What is the cohomology ring

$$ H^*(S_3;\mathbb{Z})?$$

My attempt: I want to use mathematical induction on $n$ for $S_n$. For $n=1$, $S_1$ is trivial. Hence $$ H^*(S_1;\mathbb{Z})=\mathbb{Z}.$$ For $n=2$, $BS_2=\mathbb{R}P^\infty$. Hence $$ H^*(S_2;\mathbb{Z})=H^*(\mathbb{R}P^\infty;\mathbb{Z})=\mathbb{Z}_2[\alpha],$$ where $|\alpha|=2$. But I do not know how to continue.

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  • $\begingroup$ Dear Prof. Chris, how did you compute the cohomology? I have not found ways to compute the cohomology groups and the ring structure $\endgroup$ – QSH Feb 13 '15 at 5:09
  • $\begingroup$ What are the dimensions of $x,y$? $\endgroup$ – QSH Feb 13 '15 at 5:25
  • $\begingroup$ Maybe it's worthwhile having a look at the book "Cohomology of finite groups" by Adem and Milgram. It covers all sorts of computational methods, and Chapter VI is completely devoted to computations of cohomology of symmetric groups. In the specific case at hand - $S_3$ - it is probably more helpful to view it as dihedral group rather than symmetric group. There are general computations for cohomology of dihedral groups available. $\endgroup$ – Matthias Wendt Feb 13 '15 at 15:05
  • $\begingroup$ Identical math.SE question. $\endgroup$ – Najib Idrissi Feb 16 '15 at 20:49
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Your ring for $S_2=\mathbb{Z}_2$ is wrong, because $H^0=\mathbb{Z}$. It should be $\mathbb{Z}[\alpha]/(2\alpha)$.

The cohomology groups of $S_3=\mathbb{Z}_3\rtimes \mathbb{Z}_2$ are computed using the $p$-primary decomposition, $H^i(S_3)=H^i(S^3)_{(2)}\oplus H^i(S^3)_{(3)}$. I provided this long computation as an answer on Math StackExchange, here. It is $\mathbb{Z}_2$ in degrees 2 mod 4, and $\mathbb{Z}_6$ in nonzero degrees 0 mod 4, and 0 in odd degrees, and $\mathbb{Z}$ in degree 0.

This group has periodic cohomology of period 4, which means the generator $x\in H^4(S_3)=\mathbb{Z}_6$ induces isomorphisms $H^i\cong H^{i+4}$ via cup product. And the generator $y\in H^2(S_3)=\mathbb{Z}_2$ satisfies $y^2=3x$ because $H^4(S^3)_{(3)}=\mathbb{Z}_3$ while $H^2(S^3)_{(3)}=0$.

Thus $H^\ast(S_3)\cong\mathbb{Z}[x,y]/(6x,2y,y^2-3x)$ where $|x|=4$ and $|y|=2$.


Alternative: A group $G$ with periodic cohomology has its cohomology ring isomorphic to the associated graded ring of the representation ring $R(G)$. In general, $H^\ast(G)$ and $R(G)$ are related by a spectral sequence, and this was shown by Atiyah in "Characters and Cohomology of Finite Groups". Section 13 computes the generators explicitly for $S_3$, but he didn't completely write down the ring structure: Here $R_{2k-1}(S_3)=R_{2k}(S_3)$ and $R_2(S_3)=\lbrace \alpha,\beta\rbrace$ and $R_4(S_3)=\lbrace 2\alpha,\alpha+\beta\rbrace$, such that $\alpha^2=\alpha\beta=2\alpha$ and $\beta^2=3\beta-\alpha$. Thus $R_6(S_3)=R_2(S_3)R_4(S_3)=\lbrace 4\alpha,2\alpha+\beta^2\rbrace$, so that $\alpha\,\text{mod}\,R_4(S_3)$ generates $H^2(S_3)$ and $\alpha+\beta\,\text{mod}\,R_6(S_3)$ generates $H^4(S_3)$. These generators are related by $3(\alpha+\beta)=4\alpha+\beta^2\equiv 2\alpha=\alpha^2\,\text{mod}\,R_6(S_3)$, so we get agreement with the ring that I gave.

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