Timeline for Why do the subgroups of $Hdg^{2k}(X)$ generated by the cycle classes and Chern classes of vector bundles coincide in algebraic variety $X$?
Current License: CC BY-SA 3.0
7 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jul 18, 2011 at 14:23 | answer | added | Benoît | timeline score: 2 | |
| Jul 15, 2011 at 15:55 | vote | accept | Jun Lu | ||
| Jul 12, 2011 at 13:44 | comment | added | Minhyong Kim | Given the standard comparison isomorphism between $K$-theory and $K'$-theory, I guess the argument above must work on regular schemes (say Noetherian). | |
| Jul 12, 2011 at 13:14 | answer | added | Donu Arapura | timeline score: 10 | |
| Jul 12, 2011 at 11:07 | comment | added | Minhyong Kim | If $X$ is a smooth projective variety, then for any subvariety $V$ of codimension $r$, we can resolve $O_V$ by a finite complex of vector bundles $E_.$. But then $c_r(E_.)=[V]$ as an algebraic cycle. This kind of thing is discussed extensively in Fulton's book on intersection theory. I forget when this works in the non-projective case. In the projective case, the construction of this resolution is definitely very algebraic, and unlikely to carry over to complex manifolds. | |
| Jul 12, 2011 at 7:50 | history | edited | Jun Lu | CC BY-SA 3.0 |
added 1 characters in body; edited title
|
| Jul 12, 2011 at 7:43 | history | asked | Jun Lu | CC BY-SA 3.0 |