# Compute the cohomology ring of $G=\langle u,v\mid uvuv=vuvu\rangle$.

I am trying to compute the integral cohomology ring of the following group: $$G=\langle u,v\mid uvuv=vuvu\rangle$$ First, I managed to compute the cohomology groups: $$H^0(G,\mathbb{Z})=\mathbb{Z}e_0,H^1(G,\mathbb{Z})=\mathbb{Z}e_1\oplus\mathbb{Z}e_2,H^2(G,\mathbb{Z})=\mathbb{Z}e_3$$ and $$H^i(G,\mathbb{Z})=0~\forall~i\geq 3$$. Hence, $$H^\ast(G,\mathbb{Z})=\mathbb{Z}e_0\oplus\mathbb{Z}e_1\oplus\mathbb{Z}e_2\oplus\mathbb{Z}e_3$$. To understand the ring structure we need to compute the cup product of the generators, which are: $$e_i\smile e_i=0~\forall~i=0,1,2,3$$ $$e_0\smile e_i=e_i~\forall~i=0,1,2,3$$ $$e_1\smile e_2=2e_3$$ and $$e_i\smile e_j=0$$ for every other $$i,j$$. I would like to know whether $$H^\ast(G,\mathbb{Z})$$ is isomorphic to a well known ring, such as a polynomial ring or some exterior algebra.

• I'm interested; how did you comute these $H^k$? Commented May 13 at 16:46
• Just a note: letting $x=uv$ gives an equivalent presentation $\langle x,u\mid [x^2,u]\rangle$. The normal closure $N=\langle x^2,u\rangle^G$ is such that $[G:N]=2$ and $N\cong\mathbb{Z}\times F_2$, with the $\mathbb{Z}$ factor coming from the central $x^2$ and the free $F_2$ factor generated by $u$ and $u^x$. Commented May 13 at 16:53
• @FShrike $G$ is an Artin group, and there is a canonical free resolution to compute the (co)homology of those groups. Commented May 13 at 17:06
• @SteveD Is this related with my problem? Because I dont see the relation of your claim with my problem. Commented May 13 at 17:08
• I was just finding a finite-index right-angled Artin group inside yours. This also follows from the fact your group is just $F_2\rtimes\mathbb{Z}$, with the generator of $\mathbb{Z}$ acting by swapping the generators of $F_2$. Commented May 13 at 18:27

If I understand you correctly, the ring you are looking at is a free $$\mathbb{Z}$$-module with basis $$e_0,e_1,e_2,e_3$$ where $$e_0$$ is the identity, $$e_3=\frac{1}{2}e_1\smile e_2$$, $$e_1^2=e_2^2=0$$, and the ring is graded commutative with $$e_1$$ and $$e_2$$ in degree $$1$$. Therefore, $$e_0,e_3$$ are even degree, and hence central, and
$$e_1\smile e_2=-e_2\smile e_1.$$
This ring is almost the exterior algebra $$\bigwedge \mathbb{Z}^2$$ with $$\mathbb{Z}^2$$ generated by $$e_1,e_2$$. Specifically, the subring generated by $$e_1,e_2$$ is the copy of $$\bigwedge \mathbb{Z}^2$$ spanned by $$e_0,e_1,e_2,2e_3$$ - an index 2 subgroup of the entire cohomology ring. Since $$e_3=\frac{1}{2}e_1\smile e_2$$, the desired ring $$R$$ is actually stricly between $$\bigwedge \mathbb{Z}^2$$ and $$\bigwedge \mathbb{Q}^2$$.
I doubt it has a name because there are a lot of rings $$R$$ with $$\bigwedge \mathbb{Z}^2\subset R\subset \bigwedge \mathbb{Q}^2$$.