Skip to main content
9 events
when toggle format what by license comment
Feb 28, 2013 at 4:03 comment added Sándor Kovács @rick: If both $L$ and $L^∗$ have non-zero global sections, then they are both trivial. Or in other words a global section of $L^*$ would correspond to an effective divisor $F$, but then $F\sim −E$ and hence the effective divisor $E+F\sim 0$, but that is impossible on a projective variety.
Feb 27, 2013 at 21:38 comment added rick I'm in the case $L.L=2$. From what you say if $kL\sim E$ then $h^0(X,kL)=h^0(X,E)=1$. From Riemann Roch and Serre duality i get $h^0(X,L)+h^0(X,L^*)\ge 2+\frac{kL.kL}{2}$. You say i have a contradiction because $h^0(X,L)\ge 2+k^2$. But why is $h^0(X,L^*)=0$?
Feb 27, 2013 at 19:54 comment added Sándor Kovács Exactly. So you use Riemann-Roch and perhaps Kodaira vanishing for the higher cohomology and of course the fact that the Picard number is $1$. :) –
Feb 27, 2013 at 13:27 comment added J.C. Ottem Yes, let's see. If $k=1$, then $H^0(X,L)=k$, but for an ample $L$, one always has $H^0(X,L)=L^2/2+2\ge 3$ since the higher cohomology vanishes.
Feb 27, 2013 at 6:57 comment added Sándor Kovács I meant "@JC", not "@LC"... sorry.
Feb 27, 2013 at 1:54 comment added Sándor Kovács @L.C.: That was my first thought, too, but why is it obvious that that cannot happen with, say, $k=1$? There exist ample divisors that are fixed, for instance a theta divisor on an abelian variety. You are right that this is impossible on a $K3$, but I think you need to prove that. Essentially this is what I did in my answer below.... Cheers!
Feb 27, 2013 at 1:07 comment added J.C. Ottem If $|L|$ had a fixed component $E$, then in $Pic(X)$ we must have $E\sim kL$ for some $k>0$, which is clearly impossible..
Feb 27, 2013 at 0:58 answer added Sándor Kovács timeline score: 4
Feb 27, 2013 at 0:47 history asked rick CC BY-SA 3.0