Timeline for if f is birational, is the pushforward map on the numerical groups surjective?
Current License: CC BY-SA 3.0
7 events
| when toggle format | what | by | license | comment | |
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| Oct 14, 2011 at 23:44 | vote | accept | Yosemite Sam | ||
| Oct 11, 2011 at 6:55 | comment | added | Sándor Kovács | Jason, I didn't mean to suggest that any such map $X\to Y$ may be embedded as the contraction of the exceptional locus, but I think there should be an example such that. My feeling is that if you just take a sufficiently general contraction, there is no reason that the exceptional locus over its image should have a section. You probably know more about this than I actually. Anyway, given the additional requirement of konb given in his comment above that it should be a flopping contraction, I am not sure that it can happen, but I am also not sure that it can't. | |
| Oct 11, 2011 at 0:12 | comment | added | Jason Starr | S'andor, everything you write is correct. And one can even write down explicit examples where the "index" gcd(d) is not 1 (your advisor's "Trento examples" are quite close to this, and I have another paper with similar families of 1-parameter families of Calabi-Yau hypersurfaces). But these are not examples of birational morphisms, as the OP asked. Are you suggesting there is a way to embed X --> Y as the contraction of the exceptional locus in a larger ambient variety? | |
| Oct 10, 2011 at 22:29 | history | edited | Sándor Kovács | CC BY-SA 3.0 |
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| Oct 10, 2011 at 22:27 | comment | added | Yosemite Sam | all right, I'll lay down all the cards. The case I really only care about $N_1$ and a flopping contraction between threefolds (here codimension of exceptional locus is two). But, I would really like avoid tensoring with Q. (I guess I should have said that from the start right?) | |
| Oct 10, 2011 at 22:14 | comment | added | Qing Liu | Yes if $f : X\to Y$ a surjective morphism of algebraic varieties, then $f_*: Z(X)\to Z(Y)$ is surjective when tensored by $\mathbb Q$: take an irreducible cycle $C$ on $Y$ and its generic point $\xi$, take any closed point in $f^{-1}(\xi)$ and its Zariski closure $D$ in $X$. Then $f_*D$ is a multiple of $C$. In general one can't remove the multiplicity (consider integral $Y$ and $X\to Y$ with generic fiber a conic without rational point). | |
| Oct 10, 2011 at 20:09 | history | answered | Sándor Kovács | CC BY-SA 3.0 |