Timeline for $H$ very ample, $f$ finite, is there uniform $C=C(\mathrm{deg}(f))$ for $C f^* H$ very ample?
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9 events
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| Jul 16, 2018 at 19:17 | comment | added | Hacon | Since $f^*H$ is ample and gg, then $K_X+(dim X+1)f^*H $ is gg (and something similar will be very ample)...But to answer your question you would need to control when $cf^*H-K_X$ is nef (in abx's example, if $f_*O_X=O_Y\oplus (-tH)$, and say $K_Y=0$, then $c\geq t$ can be arbitrarily big). Maybe one can show something similar with $K_{X/Y}$ instead of $K_X$, but I doubt you can do any better. | |
| Jul 16, 2018 at 16:27 | comment | added | Stefano | @abx I see. Thank you for all the comments. I wil try to think about the last one | |
| Jul 16, 2018 at 16:24 | comment | added | abx | Well, some kind of bound on the ramification would eliminate the counter-examples that I have in mind. But I have no idea of what would be a reasonable formulation. | |
| Jul 16, 2018 at 16:15 | comment | added | Stefano | @abx Thank you for the reply. Are there conditions on the ramification that one could put to get such a statement? | |
| Jul 16, 2018 at 16:13 | comment | added | abx | No, it is not enough — already with degree 2 and dimension 1. | |
| Jul 16, 2018 at 16:12 | history | edited | Stefano | CC BY-SA 4.0 |
added 58 characters in body
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| Jul 16, 2018 at 16:12 | comment | added | Stefano | @abx I am fixing the degree and the dimension. I forgot to include the dimension in the statement. Is it still not enough? | |
| Jul 16, 2018 at 16:09 | comment | added | abx | What do you fix?? If you want a universal $C$ depending only on $\deg(f)$, this doesn't exist already for $\deg(f)=2$. | |
| Jul 16, 2018 at 15:25 | history | asked | Stefano | CC BY-SA 4.0 |