Timeline for Lifting line bundles
Current License: CC BY-SA 3.0
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Feb 20, 2017 at 14:07 | vote | accept | George | ||
| Feb 19, 2017 at 21:15 | answer | added | inkspot | timeline score: 5 | |
| Feb 19, 2017 at 20:45 | comment | added | Jason Starr | @t3suji. I did not see your comment, or I would have held off on mine. | |
| Feb 19, 2017 at 20:44 | comment | added | Jason Starr | . . . So if $h^2(X,\mathcal{O}_X)$ is positive, then the obstruction to lifting modulo $p$, $p^2$, etc. might all vanish until we reach $p^d$. On the other hand, if $h^2(X,\mathcal{O}_X)$ equals $0$, then for every $d\geq 0$, for every ring homomorphism $W[[t_1,\dots,t_n]]\to W/p^dW$, this homomorphism lifts to $W[[t_1,\dots,t_n]]\to W$. | |
| Feb 19, 2017 at 20:43 | comment | added | t3suji | Another way to say this: if you managed to lift $L$ modulo $p^n$, then the obstruction to lifting modulo $p^{n+1}$ lie in $H^2(X,O_X)$. Just because you know that the obstruction is zero on a particular step, you don't know whether it will vanish on the next. On the other hand, if $H^2(X,O_X)=0$, all of the obstructions vanish. (Sorry Jason, did not mean to interrupt --- I did not realize you were about to add more.) | |
| Feb 19, 2017 at 20:38 | comment | added | Jason Starr | This is described very clearly in any reference on infinitesimal deformation theory, e.g., the beginning of Koll'ar's "Rational Curves on Algebraic Varieties", Mike Artin's TIFR book on "Deformations of Singularities", etc. Deeper are Illusie's books on the cotangent complex. The short answer is that the functor is pro-represented by a complete local ring $W[[t_1,\dots,t_n]]/\langle f_1,\dots,f_r \rangle$ where $n$ equals $h^1(X,\mathcal{O}_X)$ and $r$ equals $h^2(X,\mathcal{O}_X)$. If $r$ equals $1$, for instance, $f_1$ might be $p^d$. | |
| Feb 19, 2017 at 20:06 | comment | added | George | @DonuArapura Right! I forgot that. I'm still a bit confused though. Why is it then that $L$ will lift if $\mathrm{H}^2(X,\mathcal{O}_X)=0$? Your comment seems to suggest that the vanishing of this $H^2$ a priori only implies the liftability mod $p$. Am I missing something? | |
| Feb 19, 2017 at 19:58 | comment | added | Donu Arapura | There are separate obstructions for lifting mod $p^2$, $p^3$ etc. So even if you can do the first lift, you could get stuck at the $p^2$ step, where the obstruction space is only a $\mathbb{Z}/p^2$ module. | |
| Feb 19, 2017 at 19:50 | history | edited | George | CC BY-SA 3.0 |
added 4 characters in body
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| Feb 19, 2017 at 19:27 | history | asked | George | CC BY-SA 3.0 |