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Let $E$ be a nonzero, non effective divisor on a smooth quartic surface $X \subset \mathbb P^3$ (i.e. $H^0(X, E) =0$). Then is it possible that $H^1(X, E) \neq 0$?

Euler Characteristic computation suggests that it is possible iff $E^2 \neq -8$. Are there any concrete examples to this $E$?

Thanks in advance.

Proj
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1 Answers1

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Let $L_1$ and $L_2$ be a pair of intersecting lines on $X$ (using Bertini Theorem it is easy to check that a smooth quartic $X$ containing such a pair of lines exists). Set $$ E = L_1 - L_2. $$ Then obviously $H^0(X,E) = 0$, using Serre duality $H^2(X,E) = 0$, but $$ E^2 = (L_1 - L_2)^2 = L_1^2 - 2L_1\cdot L_2 + L_2^2 = - 2 - 2 - 2 = -6, $$ hence by Riemann--Roch $$ \chi(E) = \frac12E^2 + 2 = -3 + 2 = -1, $$ hence $H^1(X,E) \ne 0$.

Sasha
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  • Here both $E$ and its dual is non effective? In my context $E=cH-2D$ where $c$ is a negative number, $H$ is the hyperplane class and $D$ is an effective divisor. Can we say anything in such situation? Also my argument $E^2=-8$ is wrong as in that case $H^0(X, E^*)$ is zero or not is unknown. – Proj May 30 '23 at 08:17
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    Yes, sure, you can take $c = 1$ and $D = L$ to be a line, $X$ sufficiently general (e.g., such that all planes passing through $L$ intersect $X$ along $L$ and an irreducible cubic curve). – Sasha May 30 '23 at 08:52
  • In my case $c <-3$ is a negative number. Can we put some restriction on $c$ and $D$ such that $h^1(X, E)$ does vanish? – Proj May 30 '23 at 09:04