TITLE: Cohomology ring of a configuration space
QUESTION [7 upvotes]: 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?

REPLY [2 votes]: 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 $\mathbb{C}^3$.