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Let $E$ be a rank $n$ locally free sheaf on a smooth $n$ dimensional variety $X$, and $s\in H^0(X,E)$. If $\dim Z(s)=0$ (which is the expected dimension), then we can understand the cohomology class of $Z(s)$ as a chern class of $E$. Can anything be said about this class if $\dim Z(s)>0$?

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  • $\begingroup$ I don't know what $Z(s)$ is but if it is the zero locus of $Z$, your identification with Chern class seems to be incorrect if the intersection with the zero section is not transversal due to lack of information of multiplicity. $\endgroup$
    – Z. M
    Aug 3, 2021 at 15:20
  • $\begingroup$ @Z.M I meant for $Z(s)$ to denote the zero locus with its natural scheme structure, so including multiplicity information. $\endgroup$ Aug 3, 2021 at 15:56
  • $\begingroup$ But in that case, how do you define the cohomology class of $Z(s)$ in that generality (i.e. for arbitrary subscheme)? Note that the usual intersection theory uses Chow's moving lemma to transform an arbitrary intersection into a proper intersection, in which case $\dim Z(s)=0$. $\endgroup$
    – Z. M
    Aug 3, 2021 at 21:47
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    $\begingroup$ Let $z:Z(s)\to X$ be the inclusion of the zero-scheme. Assume that $Z(s)$ is smooth. Let $N$ be the normal bundle of $Z(s)$ in $X$. There is a natural monomorphism $N\to z^*(E)$, whose quotient $V$ (the so-called excess normal bundle) is locally free. We then have the formula $$ c^{\rm top}(E)=z_*(c^{\rm top}(V)). $$ This is a consequence of the "excess intersection formula" (see eg Fulton's book). Eg, if $s$ is the zero-section, the formula is tautological and if $Z(s)$ is discrete then you recover the fact that $ c^{\rm top}(E)$ is the cycle class of the zero-scheme. $\endgroup$ Aug 5, 2021 at 14:52
  • $\begingroup$ @DamianRössler Thanks, I did not know that! Should read up on my interestion theory I guess... $\endgroup$ Aug 5, 2021 at 16:34

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Not much can be said. For instance, let $X = \mathbb{P}^n$ and $E = \mathcal{O}(1)^{\oplus n}$. Then the zero locus of a section might be any linear subspace in $X$. In particular, its cohomology class is $H^k$, where $H$ is the hyperplane class and $k$ may be any integer in the range $0 \le k \le n$.

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