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My apologies if this question is too naive.

Let $X$ be a smooth projective complex variety. There is a natural map $A^{\bullet}(X) \to H^{2\bullet}(X)$ of graded rings from the Chow ring of $X$ to the integral cohomology of $X$ given by taking Poincaré duals of fundamental classes. (Is there a convenient name for it?) For some particularly nice varieties, e.g. projective spaces and more generally Grassmannians, this map is an isomorphism.

What are some more general $X$ for which $A^{\bullet}(X) \to H^{2\bullet}(X)$ is an isomorphism?

I think it suffices that $X$ have a stratification by affine spaces. (Is there a convenient name for such spaces?) In this case there is apparently a difficult theorem of Totaro asserting that $A(X)$ is free abelian on the strata, which I think is also the case for the integral cohomology via cellular cohomology, and degrees and intersections ought to match as well. Are there interesting families of examples where $X$ doesn't admit such a stratification?

Edit: A somewhat more general sufficient condition, if I've understood my reading correctly, is that the Chow motive of $X$ is a polynomial in the Lefschetz motive. Guletskii and Pedrini showed that this is true (edit: rationally) of the Godeaux surface so this is more general than admitting a stratification by affine spaces. Are there interesting families of examples where this doesn't hold either?

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Maybe you have seen this (or it is of no help - or both) but perhaps relevant is and some of its references (e.g. Fulton's book on Young Tableaux). –  Benjamin Dickman Dec 10 '13 at 6:50
It follows from Proposition 2.1(1) here: that this is the case for any variety with a full exceptional collection. –  Daniel Litt Dec 10 '13 at 7:01
Your question is (at least conjecturally) equivalent to the Chow motive of X being a polynomial in the Lefschetz motive (with rational coefficients). This is not true integrally since the map can be an isomorphism even if there is torsion in the cohomology (e.g., Enriques, Godeaux,...). –  ulrich Dec 11 '13 at 7:34

1 Answer 1

up vote 14 down vote accepted

[I've incorporated or addressed comments of Dan Petersen and Daniel Litt into this. My thanks to them.]

One sometimes says that $X$ admits a cellular decomposition if it admits a stratification by affine spaces. The isomorphism of the type you mention was known before Totaro, cf Fulton's Intersection Theory 19.1.11 (in my edition). This applies to flag varieties. Also such an isomorphism holds for toric varieties for similar reasons.

A surface with $p_g=q=0$ i.e. $\dim H^0(X,\Omega^2)=\dim H^0(X,\Omega^1)=0$, and $A_0(X)\cong \mathbb{Z}$ satisfies the condition of $A(X)\cong H^*(X)$ by the Lefschetz $(1,1)$ theorem. Note that Bloch has conjectured that the second condition follows from the first. One has examples of surfaces of Kodaira dimension zero $\ge 0$, such as Enriques surfaces (Bloch-Kas-Lieberman) and Godeaux surfaces (Voisin) where these conditions hold. These do not have cellular decompositions.

Finally, assuming the Hodge and Bloch-Beilinson conjectures [thanks Dan], an isomorphism $A(X)\otimes \mathbb{Q}\cong H^{2*}(X,\mathbb{Q})$ holds if and only if the Hodge numbers $h^{pq}=0$ for $p\not= q$.

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Thanks for the response! My reference for the claim that the cellular result is due to Totaro is Eisenbud and Harris' 3264 and All That, in a comment below Proposition 1.19, although admittedly this is still a draft. –  Qiaochu Yuan Dec 10 '13 at 7:36
About the necessity of tensoring with $\mathbb{Q}$ in the last statement - Kollar's example: on a very general degree $48$ hypersurface in $\mathbb{P}^4$, every curve has even degree, and $h^{p, q} = 0$ for $p\neq q$, $p+q \neq 3$. –  Piotr Achinger Dec 10 '13 at 7:40
I don't understand your last statement. Wouldn't the Hodge conjecture just imply surjectivity of the cycle class map? –  Dan Petersen Dec 10 '13 at 7:46
Not to pile on, but I also don't see why injectivity holds for $0$-cycles in your surface example. This seems to imply Bloch's conjecture for surfaces with $h^1(X)=0$, which is open to my knowledge. –  Daniel Litt Dec 10 '13 at 8:38
@DonuArapura Actually I think your original statement was OK, you just need a further conjecture! If you assume both the Hodge conjecture and the Bloch-Beilinson conjectures on filtrations of Chow groups, then the cycle class map from Chow to cohomology is an isomorphism (with $\mathbf Q$ coefficients) if and only if all off-diagonal Hodge numbers vanish. –  Dan Petersen Dec 10 '13 at 13:54

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