Let $X$ be a scheme. Let $\mathcal E$ be a locally free sheaf of rank $r$ on $X$ and let $s$ be a section of $\mathcal E$. Then the zero scheme of $s$ is defined as follows: Consider the homomorphism $\mathcal O_X \to \mathcal E$ induced by $s$, taking duals, we obtain $\mathcal E^\vee \to \mathcal O_X$. Then $Z(s)$ is defined to be the scheme associated to the sheaf of ideals $\mathop{\mathrm{im}}(\mathcal E^\vee \to \mathcal O_X)$.

First question: is there some natural conditions of regularity defined on $s$. For example, when $\mathcal E = \mathcal L$ is an invertible $\mathcal O_X$-module, then $s$ is said to be regular if and only if the induced homomorphism $\mathcal O_X \to \mathcal L$ is injective, and in this case $Z(s)$ is an effective Cartier divisor on $X$. So I think the condition of regularity should satisfy that if $s$ is regular, then any generic point of irreducible components of $Z(s)$ has codimension $r$ in $X$.

Now suppose we have already defined some conditions of regularity. Assume now that $X$ is a smooth projective variety and let $s$ be a "regular" section. Consider the $r$-cycle associated to $Z(s)$. Prove the following statement: The class of the $r$-cycle associated to $Z(s)$ in the $r$-th Chow group $\mathop{\mathrm{CH}}^r(X)$ equals the $r$-th Chern class $c_r(\mathcal E)$ of $\mathcal E$.

Hence the linear equivalence class of $Z(s)$ is independent of the choice of $s$ and we get a well-defined map $$\{\text{locally free sheaves of rank } r \text{ on } X\} / \{\text{isomorphisms}\} \to \mathop{\mathrm{CH}}\nolimits^r(X),$$ is this map bijective? (When $r = 1$, we obtain an isomorphism $\mathop{\mathrm{Pic}}(X) \to \mathop{\mathrm{Cl}}(X)$.)

If the map above is bijective, then the group structure on $\mathop{\mathrm{CH}}\nolimits^r(X)$ should induces a group structure on $\{\text{locally free sheaves of rank } r \text{ on } X\} / \{\text{isomorphisms}\}$, and what is it? (When $r = 1$, it is tensor products of invertible $\mathcal O_X$-modules. However, we cannot simply take the tensor product of two locally free sheaves of rank $r$).

  • 3
    $\begingroup$ Locally $s$ is given a collection of functions $f_1,f_2\ldots $. $s$ would be regular if the sequence is regular in the usual sense of commutative algebra. Concerning your 3rd paragraph, it's a bit more complicated. See p 151 of Grothendieck, La theorie des classes de Chern. The last paragraph seems pretty far off. The closest correct statement involves the relation between $K_0(X)$ and $CH^*(X)$. $\endgroup$ – Donu Arapura Nov 22 '20 at 17:11
  • 4
    $\begingroup$ Of course the local description globalises to a definition in terms of the Koszul complex: a section $\mathscr E^\vee \to \mathcal O$ is regular if and only if the Koszul complex $0 \to \wedge^r \mathscr E^\vee \to \ldots \to \mathscr E^\vee \to \mathcal O$ is exact. $\endgroup$ – R. van Dobben de Bruyn Nov 22 '20 at 21:13

The question of what regularity means was addressed in the comments, and I don't have any more to add. In particular, if $s$ is a regular section of a rank $r$ bundle $\mathcal{E}$, the class of $Z(s)$ in the Grothendieck group $K_0(X)$ is $\sum (-1)^i [\wedge^i \mathcal{E}^\vee]$. When $X$ is a nonsingular variety, $r$th Chern class of this is $(-1)^r(r-1)![Z(s)]$, where $[Z(s)]$ is the class in $CH^r(X)$. See, for example, page 151 of Grothendieck, La théorie des classes de Chern.

Finally, let me point out that the (normalized) $r$th Chern class generally won't give a bijection between the pointed set of isomorphism classes of rank $r$ bundles and $CH^r(X)$. To see this, let $X=\mathbb{P}^2$ and $r=2$. Then I'll leave you to check that $$c_2(T_X)= c_2(O(1)\oplus O(3))\in CH^2(X)\cong \mathbb{Z}$$ but the bundles are not isomorphic. Probably the best you can hope for along the lines you are suggesting is the isomorphism $K_0(X)\otimes \mathbb{Q}\cong CH^*(X)\otimes\mathbb{Q}$, when $X$ is a nonsingular variety.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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