Two conditions on divisors on surfaces

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.

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.

• thank you very much for the answer. your answer shows that in general these two are not related. Is it known that for any speciific surface these two are closely related? (e.g surfaces where $K_X$ is ample and moreover $h^1(K_X)=h^1(\mathcal O_X)=0$? and in general for any surface is it natural to expect that one could find examples of divisors satisfying the second condition? Commented May 6, 2022 at 7:35
• I don't think that your two conditions are closely related — I don't see what extra properties one should impose to get them related. As for condition $(ii)$, it is of course satisfied if $D$ is sufficiently ample, but you can find many other cases.
– abx
Commented May 6, 2022 at 8:26