Let $X$ be a smooth projective surface and $D$ be an effective Cartier divisor (not necessarily ample) on $X$. Is there a connection between these two conditions?


$(i)$ for a large enough $n$, the linear system $|nD|$ is base point free (semiample divisor)


$(ii)$ $h^1(\mathcal O\_X(D)^{\otimes t})=0$ for all $t >0$. (I couldn't find any specific name for divisors satisfying this condition)


Is the second one much more stronger and therefore one can rarely find divisor satisfying these conditions?


Is there any specific instance when these two become equivalent?


Any remark from anyone is welcome.



Two conditions on divisors on surfaces
Neither condition implies the other.


$(i)\, \not\!\Rightarrow\, (ii)$: Take for $X$ a surface with $K$ ample, but $h^1(K)=h^1(\mathscr{O}\_X)>0$ (e.g. a product of 2 curves of genus $>1$). Then take $D=K$.


$(ii)\, \not\!\Rightarrow\, (i)$: Consider a smooth cubic curve $C\subset \mathbb{P}^2$, take $9$ general points on $C$, and take $X=\mathbb{P}^2$ blown up at these 9 points, $D=$ the proper transform of $C$. The normal bundle $N$ of $D$ in $X$ has degree $0$, and since the points are general it is not torsion. Therefore $H^0(N^{t})=H^1(N^t)=0$ for all $t>0$. The cohomology exact sequence of


$$0\rightarrow \mathscr{O}\_X((t-1)D)\rightarrow \mathscr{O}\_X(tD)\rightarrow N^{t}\rightarrow 0$$


gives $h^{0}(tD)=1$ and $h^{1}(tD)=0$ for all $t>0$. So $(ii)$ holds, but $\lvert tD\rvert$ contains only one curve.


