# Is the Chow ring of a wonderful model for a hyperplane arrangement isomorphic to the singular cohomology ring?

In the article "Hodge theory for combinatorial geometries" by Adiprasito, Huh and Katz, it it claimed in the proof of theorem 5.12 that there is a Chow equivalence between the de Concini-Processi wonderful model $Y$ of an arrangement, and a certain toric variety $X$.

However, the source they cite only proves that

$$H^*(Y)\cong Ch(X)$$

It seems that to draw the conclusion that $Ch(Y)\cong Ch(X)$ it is used that $Ch(Y)\cong H^*(Y)$, but I do not know why that is the case?

Here is what I know: $Y\subset X$ as a closed subset, and the relevant maps are (should be) induced by this inclusion. We have a diagram $$\require{AMScd} \begin{CD} Ch(X) @>{ch}>> H^*(X)\\ @VVV @VVV \\ Ch(Y) @>{ch}>> H^*(Y) \end{CD}$$ and we know that the diagonal arrow is an isomorphism. To establish that $Ch(X)\cong Ch(Y)$ it is enough to show that $Ch(Y)\cong H^*(Y)$. Since the diagonal is an isomorphism, the map $Ch(Y)\to H^*(Y)$ is surjective. I do not see how to prove injectiveness though.

• I am not sure I remember well, but I think that the wonderful arrangement can be expressed as a tower of projective bundles over a toric variety. It is known that the Grothendieck standard conjectures are true for toric varieties, hence they are also true for towers of projective bundles over toric varieties. In particular, if $Y$ is a tower of projective bundles over a toric variety, then homological and algebraic equivalences do coincide on $Y$. Which proves the injectivity of $ch_Y$. – Libli Sep 2 '17 at 12:24

The isomorphism $H^\cdot(Y)\cong Ch^\cdot(X)$ is shown by
It's worth emphasizing that the inclusion $Y\subset X$ is not a homotopy equivalence, and generally $H^\cdot(X)\not\cong H^\cdot(Y)$.
The property that $H^\cdot(Y)\cong Ch^\cdot(Y)$ for iterated blowups of projective spaces I think is due to Sean Keel, Trans. Amer. Math. Soc. 330 (1992), 545-574, https://doi.org/10.1090/S0002-9947-1992-1034665-0, in line with Libli's comment above.
• Hi Graham, why do you say that $H^\ast(X)\neq H^\ast(Y)$? In general the cycle class map from the Chow ring to the cohomology ring is always an isomorphism for a smooth toric variety $X$. – Dan Petersen Feb 12 '18 at 5:20
• Hi, Dan. That's true when $X$ is complete. Here, $Y$ is projective (but not toric) while $X$ is toric (but not projective). – Graham Denham Feb 13 '18 at 14:17
• Here's a minimal example. Consider the arrangement of three distinct points in ${\mathbb P}^1$. Then $Y={\mathbb P}^1$. The fan for $X$ (the Bergman fan) consists of just the three coordinate rays in ${\mathbb R}^3/(1,1,1)$, so $X\cong {\mathbb P}^2-\{\text{3 pts.}\}$. Yet somehow $Ch^\cdot(X)\cong H^\cdot({\mathbb P}^1)$, so it has Poincaré duality (in particular) and it's nicer than we'd expect. – Graham Denham Feb 13 '18 at 14:28