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Consider the following configuration space of triples of points. $$\begin{align}C &= \left\lbrace (z_1,z_2,z_3) \in (\mathbb C^*)^3, z_1 \ne z_2, z_1 \ne z_3, z_2 \ne z_3\right\rbrace \\&\phantom{abcde}\setminus \left\lbrace (z_1,z_2,z_3) \in (\mathbb C^*)^3, |z_1| = |z_2| = |z_3|\right\rbrace \,,\end{align}$$ i.e., it is a complement of $\mathbb C^3$ to a union of $6$ hyperplanes and $\left(S^1\right)^3 \times \mathbb R$.

There are some evident nontrivial cohomology classes, like the ones represented by forms $d\log(z_i), d\log (z_i - z_j)$, and the one represented by the pull-back of a closed $1$-form under a moment map. But are these the only generators of the cohomology ring?

How to compute its cohomology ring? Is it possible to write down explicitly the closed forms generating the ring and list the complete set of relations?

Khashayar Baghizadeh
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user79456
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    I think you mean $(S^1)^3 \times \mathbb R^+$. The cohomology of the complement of a union of hyperplanes is well-understood; it would be generated as an algebra by the 6 corresponding classes you mention. But I don't know what to do about the weird four-dimensional piece you're removing. – Hugh Thomas Apr 09 '16 at 16:51
  • @Hugh Thomas, well, $\mathbb R^+$ is homeomorphic to $\mathbb R$... That is precisely my problem. I know the description of the cohomology ring of a hyperplane complement but have no idea what to do with this additional piece either. – user79456 Apr 09 '16 at 20:52
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    to compute (co)homology groups you may use Alexander duality, but it says nothing about a ring structure – Andrey Ryabichev Apr 19 '16 at 10:45
  • @AndreyRyabichev I am mainly concerned about the ring structure. My hope was that one is able to use an idea, similar to Arnol'd treatment of the pure braid group classifying space. But I have no clue how one can produce an analogous argument. – user79456 Apr 20 '16 at 06:54

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This paper by Burt Totaro, supplies a spectral sequence converging to the cohomology of $C$. The $E_2$ page is the bigraded algebra $$\mathbb{Z}[a_1,a_2,a_3,G_{12},G_{23},G_{31}]$$ with $a$, $b$, $c$ the three generators from $H^*((\mathbb{C}^*)^3)$ of bidegree $(1,0)$ and $G_{ij}$ the class of $d\log(z_i - z_j)$, of bidegree $(0,1)$, and relations $$a_i^2, a_i G_{ij} - a_jG_{ij}, G_{12}G_{23} + G_{23}G_{31} + G_{31}G_{12}, (G_{ij})^2 .$$ The differential is given by the diagonal class in $H^*((\mathbb{C}^*)^2)$, which is $0$, and so $H^*(C)$ is isomorphic to the same algebra, with the total grading.

I believe the same result can be obtained from the Goresky-Macpherson formula, since $C$ is a hyperplane complement in $\mathbb{C}^3$.

ronno
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