# Divisors with positive Iitaka dimension

Let $X$ be a non-singular projective variety, and $D$ a divisor on $X$.

Saying that $D$ has positive (meaning non-zero) Iitaka dimension is equivalent to the function $n \mapsto h^0(\cal{O}(D))$ being strictly increasing for sufficiently large $n$?

Does every effective divisor have positive Iitaka dimension? If not, what are counterexamples?

• Any irreducible curve with negative square on a surface has Iitaka dimension 0. – abx Mar 20 '14 at 11:26
• To supplement abx's comment, whenever you blow up (say a subvariety in a smooth variety) you will get an exceptional divisor with Iitaka dimension 0. – user5117 Mar 20 '14 at 11:42
• Thanks for the comments and examples! Do you know if the property of $n \to h^0(\cal{O}(nD)$ being strictly increasing has a special name in the literature? – Marcos Jardim Mar 25 '14 at 10:09

• Thanks for the comments and examples! Do you know if the property of $n \to h^0(\cal{O}(nD))$ being strictly increasing has a special name in the literature? – Marcos Jardim Mar 25 '14 at 10:12
• There is the following lemma which might be useful (again this is in Lazarsfeld, Positivity, 2.1.38) Let $L$ a line bundle on a smooth variety, set $\kappa = \kappa(X,L)$. Then there are constants $a,A>0$ such that $a m^\kappa \le h^0(X,L^m) \le A m^\kappa$ for all sufficiently large $m\in \mathbb{N}(X,L)$. So if the exponent of your line bundle is $1$, for example if $L$ is big, then this might happen. – aegbert Mar 25 '14 at 19:19
(edited) Indeed Lazarsfeld's example mentioned by user132885 answers your first question: take $X=Y\times Z$ and $\mathcal{O}_X(D)=L\boxtimes M$, with $L$ ample and $M$ torsion, say of order $p$. Then $h^0(mD)=h^0(L^m)$ if $m$ is a multiple of $p$, and $0$ otherwise. The Iitaka dimension of $D$ is $\dim Y$.