Timeline for Converse of the relative canonical bundle $-K_{X/Z}$ of smooth morphism can not be ample?
Current License: CC BY-SA 4.0
14 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Aug 24 at 18:01 | comment | added | Will Sawin | No, the point is to deform $C$, not in $X$, but in $X$ relative to $Z$, i.e. to consider the space of deformations that fix the map to $Z$. Such deformations are parameterized by $H^0$ of the relative tangent bundle, so for such deformations to exist after Frobenius pullback it suffices for the relative tangent bundle to have positive degree. | |
| Nov 22, 2023 at 12:29 | comment | added | DVL-WakeUp | @JasonStarr Thank you for your patient reply. Now I know the main goal of doing Bend-and-Break. Hence finally we get a section curve $C'$ such that $K_X\cdot C'\geq0$. As $K_{X/Z}=K_X-f^*K_Z$, we have $$K_{X/Z}\cdot C'\geq-f^*K_Z\cdot C'\geq-\deg_ZK_Z.$$ Maybe at the first time we need to choose $C$ to be a rational curve? | |
| Nov 22, 2023 at 10:10 | comment | added | Jason Starr | Bend-and-break does not only apply to the Fano case. It applies whenever there is a curve whose anticanonical degree is positive. Consider the section curve. If the anticanonical degree is positive, use Bend-and-Break to "break" off a rational curve in a fiber so that the original section curve is algebraically equivalent to the broken-off rational curves union a "handle" that is also a section curve, but now with smaller anticanonical degree. If the anticanonical degree is still positive for this handle, repeat. | |
| Nov 22, 2023 at 3:26 | comment | added | DVL-WakeUp | @JasonStarr Thank you for your help. But I still not understand, could you give me more detail? Now $Z$ is a smooth projective curve with a (why rational?) section of $f$. Next I don't know how to use Bend-and-Break as we have no "Fano" condition here. So we can not use Frobenius to expand the dimension of the deformation space of maps from curves. By the way, I would appreciate it if you could include it in the answer below. | |
| Nov 21, 2023 at 16:21 | comment | added | Jason Starr | As KMM state, this follows from bend-and-break. Consider any curve $C$ that is not contained in a fiber, and perform base change by the morphism from $C$ to $Z$. This reduces to the case that $Z$ is itself a smooth, projective curve that admits a rational section of $f$. Now use Bend-and-Break to keep breaking off rational curves in fibers and driving "down" the curve class of the "handle" of the comb. Since there exists some ample divisor class, you cannot do this indefinitely. Once the handle becomes rigid, the relative anticanonical degree is nonpositive. | |
| Nov 21, 2023 at 16:15 | comment | added | DVL-WakeUp | @Sasha But $-K_{X/Z}$ need not be absolutely ample on $X$? | |
| Nov 21, 2023 at 14:54 | comment | added | Sasha | If $X \cong Y \times Z$ where $Y$ is a Fano variety and $f$ is the projection then $-K_{X/Z}$ is relatively ample. | |
| Nov 21, 2023 at 14:27 | history | edited | DVL-WakeUp | CC BY-SA 4.0 |
added 4 characters in body; edited title
|
| Nov 21, 2023 at 14:19 | history | edited | DVL-WakeUp | CC BY-SA 4.0 |
added 1 character in body
|
| Nov 21, 2023 at 13:42 | history | edited | DVL-WakeUp | CC BY-SA 4.0 |
added 172 characters in body
|
| Nov 21, 2023 at 13:35 | history | edited | DVL-WakeUp | CC BY-SA 4.0 |
edited body
|
| Nov 21, 2023 at 13:35 | comment | added | DVL-WakeUp | @MatthieuRomagny Oh I'm sorry. Now I have edited it. | |
| Nov 21, 2023 at 13:33 | comment | added | Matthieu Romagny | Is $Z$ equal to $Y$? | |
| Nov 21, 2023 at 13:21 | history | asked | DVL-WakeUp | CC BY-SA 4.0 |