Let $X$ be a K3 surface. I know that $Pic(X)\simeq H^{1,1}(X,\mathbb{Z}):=H^{1,1}(X)\cap H^2(X,\mathbb{Z})$. Let's suppose that $H^{1,1}(X,\mathbb{Z})$ is one dimentional and generated by $\omega=c_1(L)$. Why is that the linear system $|L|$ can not have fixed components?
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$\begingroup$ 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.. $\endgroup$– J.C. OttemCommented Feb 27, 2013 at 1:07
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1$\begingroup$ @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! $\endgroup$– Sándor KovácsCommented Feb 27, 2013 at 1:54
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$\begingroup$ I meant "@JC", not "@LC"... sorry. $\endgroup$– Sándor KovácsCommented Feb 27, 2013 at 6:57
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1$\begingroup$ 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. $\endgroup$– J.C. OttemCommented Feb 27, 2013 at 13:27
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1$\begingroup$ Exactly. So you use Riemann-Roch and perhaps Kodaira vanishing for the higher cohomology and of course the fact that the Picard number is $1$. :) – $\endgroup$– Sándor KovácsCommented Feb 27, 2013 at 19:54
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Let's say $D$ is a fixed irreducible component. If $D^2\geq 0$, then Riemann-Roch shows that $D$ moves, so it can't be a fixed component. Then since $D$ is irreducible, $D^2=-2$ (again by Riemann-Roch) and hence $D$ is a $(-2)$-curve (use adjunction for this). However, if there is a $(-2)$-curve on your $K3$ surface, then its Picard number has to be at least $2$ which contradicts your assumption. (If the Picard number is $1$, then every effective curve is ample and clearly a $(-2)$-curve is not ample.)
answered Feb 27, 2013 at 0:58