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I am currently learning intersection theory of smooth algebraic varieties and I have the following question.

Let $X$ be a smooth projective variety and $\mathcal{F}$ a vector bundle on $X$. Then the $i$th Chern class of $\mathcal{F}$ is an element of the $i$th Chow group $A^i(X)$ of $X$. What about the converse? Can every class in $A^i(X)$ be realized as the $i$th Chern class if a vector bundle? Clearly, this is true for $A^1(X)$ and it is true for all $i$ if $X$ is the projective space. Is it true in general? If not, is it true for some nice varieties, for example for Grassmannians?

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It is not true in general. The counterexample that I know is $X$ a general hyperplane section of $LGr(3,6)$ and the class of a line on $X$.

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  • $\begingroup$ Thanks! Do you know if it is true for Grassmannians? $\endgroup$ – Hans Sep 30 '18 at 10:01
  • $\begingroup$ @Hans: I don' know. What is true for Grassmannians (and not true in the counterexample) is that Chern classes of bundles generate the Chow ring. So, for Grassmannians any class can be represented as a polynomial of Chern classes of vector bundles. $\endgroup$ – Sasha Sep 30 '18 at 10:07
  • $\begingroup$ That I knew, I think. I guess you take the special Schubert cycles which generate the Chow ring and can be realized as the Chern classes of the universal sub and quotient bundles respectivley. thanks anyways! $\endgroup$ – Hans Sep 30 '18 at 10:19

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