Timeline for if f is birational, is the pushforward map on the numerical groups surjective?
Current License: CC BY-SA 3.0
13 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Oct 14, 2011 at 23:44 | vote | accept | Yosemite Sam | ||
| Oct 12, 2011 at 7:45 | comment | added | Yosemite Sam | @qing liu: thanks, that was very helpful. | |
| Oct 11, 2011 at 7:03 | comment | added | Qing Liu | @konb: yes, if $C$ has dimension at least 1, its generic points live in the open subset where $f$ is an isomorphism, so $Z_i(X)\to Z_i(Y)$ is surjective for $i=1,2$. As you work over an algebraic closed base field, $f_∗$ is also surjective on $Z_0$. | |
| Oct 10, 2011 at 22:33 | comment | added | Yosemite Sam | @liu: what if $f$ is contracting a curve. can I at least say that $f$ is surjective on $Z_1$? | |
| Oct 10, 2011 at 22:05 | comment | added | Qing Liu | @konb: If $C\in Z(Y)$ (say irreducible) is not in the image of the exceptional locus of $f$, then yes, taking the strict transform and pushing-down you get $C$. But otherwise you just get $0$ when pushing-down the pre-image of $C$. Now if you suppose that $Y$ is regular, by Chow's moving lemma, you can move $C$ out of the image of the exceptional locus, so $CH(X)\to CH(Y)$ is surjective. | |
| Oct 10, 2011 at 20:48 | comment | added | Yosemite Sam | after all surjectivity on the level of cycles $Z_1(X) \to Z_1(Y)$ should imply surjectivity on the numerical groups! | |
| Oct 10, 2011 at 20:47 | comment | added | Yosemite Sam | I think it should be possible to still use the numerical groups in the setting of the question (although I lack the technical knowledeg to define them). But even with the smoothness assumption (but without $f$ flat) I don't see why you would have surjectivity. Perhaps given a curve $C \in Z_1(Y)$ one could take the preimage and push it down again. Is it that simple? | |
| Oct 10, 2011 at 20:09 | answer | added | Sándor Kovács | timeline score: 7 | |
| Oct 10, 2011 at 19:43 | answer | added | Student | timeline score: 1 | |
| Oct 10, 2011 at 15:21 | comment | added | Damian Rössler | Suppose $X$ and $Y$ are smooth over a field (otherwise, it is not quite clear how to define numerical equivalence). Then the equations in your post are valid if you work with the Chow theory groups ${\rm CH}^\bullet(X)_{\bf Q}$ and ${\rm CH}^\bullet(Y)_{\bf Q}$ instead of the groups $N_*(X)$ and $N_*(Y)$ (see Fulton's book on intersection theory). In particular the morphism $f_*:{\rm CH}^\bullet(X)_{\bf Q}\to {\rm CH}^\bullet(Y)_{\bf Q}$ is surjective. By definition of Chow theory, this implies that $f_*:N_*(X)\to N_*(Y)$ is also surjective. But maybe your issue lies somewhere else ? | |
| Oct 10, 2011 at 13:10 | comment | added | Yosemite Sam | It's the group of cycles modulo numerical equivalence. When your variety is smooth one cane use the intersection pairing to define it. If I'm not mistaken there should be a way to define it over more general schemes by using intersections with cartier divisors. (or perhaps over C one could use the homological equivalence and pretend they are equal). Makes sense? | |
| Oct 10, 2011 at 10:53 | comment | added | Damian Rössler | What is the definition of $N_*(X)$ ? | |
| Oct 10, 2011 at 10:45 | history | asked | Yosemite Sam | CC BY-SA 3.0 |