2
$\begingroup$

For a smooth projective complex variety $X$ over $\mathbb{C}$, there is a natural map from its Chow ring $\mathbb{A}^*(X)$ into even integral cohomology $H^{2*}(X)$ of its (often implicitly identified) analytification. The question of when this is an isomorphism is subtle, but I have two (possibly) simpler questions, neither of which is hopefully ridiculous.

First one is pretty quick: isn't it true that while it's obviously difficult for this to be surjective, it is always surjective into the subring of even cohomology representable by closed submanifolds? My understanding is that this is one implication of GAGA; every closed analytic subspace is also a closed algebraic subvariety.

My main question, though, is whether the map is always injective - or, at least, if there's nice sufficient (+ necessary would be even better!) conditions for it to be injective. That is, whether a linear combination of closed submanifolds representing 0 in homology implies that the associated algebraic cycle is rationally equivalent to zero. The reverse (proving the map is well-defined) follows nicely from the homotopy-esque definition of rational equivalence given in Fulton, but I can't see any obvious approach here.

$\endgroup$
  • 1
    $\begingroup$ This condition probably won't be of much help to you, but there is a factorization $CH^*\to \mathbb{Z}\otimes_{MU^*(*)} MU^*\to H^*$, and the latter map is known to not be a injection or surjection in even the case of complex analytic manifolds. This is proven here, and in section 7 there is given an example of a smooth complex projective variety where the map is not an injection, so the conditions would have to be pretty strong. $\endgroup$ – Pax Kivimae Jul 21 '15 at 6:16
1
$\begingroup$

The cycle map $CH^*\rightarrow H^{2*}$ is almost never injective, in fact the group $CH^*$ is really huge. Most of the time, it is not finitely generated.

Take the example of a smooth projective curve $C$. Then $CH^1(C)$ is the Picard group and $H^2(C)=\mathbb{Z}$. The kernel of the cycle map is an abelian variety of dimension $g$, where $g$ is the genus of the curve. Hence the map is injective if and only if the curve is $\mathbb{P}^1$.

In general, a sufficient condition would be that the variety admit a cellular decomposition, that is a stratification by affine spaces. This is the case for instance of projective spaces and grassmanian varieties. But it is a very strong condition.

It is probably not a necessary condition, though I do not have counter examples...

$\endgroup$
  • $\begingroup$ Oh wow, that obvious counterexample should have occurred to me. Thank you! $\endgroup$ – Peter Xu Jul 21 '15 at 18:37

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.