Let $X$ be a smooth quasi-projective variety over $\mathbb{C}$ and $Z$ be a closed subvariety of codimension $k$.

Q1. How to define a cycle class $[Z]\in H^k(X,\Omega_X^{k})$ ?

Q2. More general, which are necessary conditions to have a "good" cycle class map of this type? The "good" means it is coincide with the usual cycle class map when $X$ is smooth projectve complex variety. I mean that if $X$ can be define over arbitrary field or in the case $X$ is a variety but maybe not smooth quasi-projective.

  • $\begingroup$ This looks rather strange; one usually defines cycle classes only for those cohomology theories that satisfy some sort of the homotopy invariance property (that fails for the cohomology you consider unless $X$ is also proper). $\endgroup$ – Mikhail Bondarko May 19 '15 at 20:39
  • $\begingroup$ I find this in subsection 1.1 of Variations de structure de Hodge et zéro-cycles sur les surfaces générales, where the author give a cycle class in $H^2(X,\Omega_X^2)$ $\endgroup$ – mwZhang May 20 '15 at 1:13
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    $\begingroup$ In that paper $X$ is proper, so this just the standard definition. $\endgroup$ – abx May 20 '15 at 5:13
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    $\begingroup$ There does indeed exist a cycle class map in Hodge cohomology for any smooth variety. See, for example, the article by El Zein "Complexe dualisant et applications à la classe fondamentale d'un cycle". $\endgroup$ – naf May 20 '15 at 9:08

Here's one way to obtain a cycle class, at least over a field $k$ of characteristic 0 where resolutions of singularities are available:

Let $Z \subset X$ be a subvariety of codimension $c$ (possibly singular), and let $\pi: \tilde{Z} \to Z$ be a resolution of singularities. This means $\pi$ is proper and birational, and $\tilde{Z}$ is smooth. Let $f = \iota \circ \pi : \tilde{Z} \to X$ where $\iota: Z \to X$ is the inclusion.

The (dual of the) differential $d f^\vee: f^*\Omega_X^* \to \Omega_{\tilde{Z}}^*$ induces homomorphisms of Hodge cohomology groups $$ f^*: H^q(X, \Omega_X^p) \to H^q(\tilde{Z}, \Omega_{\tilde{Z}}^p) \text{ for all } p, q $$ Now, $H^{\dim \tilde{Z}}(\tilde{Z}, \Omega_\tilde{Z}^{\dim \tilde{Z}}) \xrightarrow{\mathrm{Tr}, \simeq} k$ via the trace map, and the composition $$ H^{\dim \tilde{Z}}(X, \Omega_X^{\dim \tilde{Z}}) \xrightarrow{f^*} H^{\dim \tilde{Z}}(\tilde{Z}, \Omega_{\tilde{Z}}^{\dim \tilde{Z}}) \xrightarrow{\mathrm{Tr}, \simeq} k $$ is an element of $$ H^{\dim \tilde{Z}}(X, \Omega_X^{\dim \tilde{Z}})^{\vee} \simeq H^{\dim X -\dim \tilde{Z}}(X, \Omega_X^{\dim X -\dim \tilde{Z}}) $$ This isomorphism comes from Poincare duality -- since $\Omega_X^{\dim \tilde{Z}} \otimes \Omega_X^{\dim X - \dim \tilde{Z}} \xrightarrow{\wedge} \omega_X$ is a perfect pairing it induces an isomorphism $\Omega_X^{\dim X - \dim \tilde{Z} } \simeq \mathrm{Hom}(\Omega_X^{\dim \tilde{Z}}, \omega_X)$. Hence we've assigned a cohomology class, say $\eta(Z) \in H^c(X, \Omega_X^c)$ to $Z$. Well, strictly speaking we've assigned it to the morphism $f : \tilde{Z} \to X$; it'd take more work to prove $\eta(Z)$ is idependent of the resolution $ \pi : \tilde{Z} \to Z$ (idea: any 2 resolutions of $Z$ can be dominated by a third).


An alternative approach without using resolution of singularities is given in Lipman's "Blue Book":

Dualizing sheaves, differentials and residues on algebraic varieties. Astérisque No. 117 (1984)

See Chap. 3, Remark (ii) on page 39.

In short, for a $d$-dimensional variety $V$ there is a canonical map $c_V \colon \Omega^n_V \to \omega_V$ that provides an element in $$ H^{N-d}_V(X,\Omega^{N-d}_X) $$ where $V \hookrightarrow X$ is the embedding of $V$ into a regular variety $X$ of dimension $N$. Take $Z := V$ in your notation. Of course there is map from cohomology with supports to usual cohomology.


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