Are there nef divisors D on a complex projective manifold X such that $h^0(X,D)$ is less than or equal to $\dim X$?


Edit: In fact I'm interested in nef line bundles D, not just divisors.



Nef divisors with few global sections
Here's an interesting example from complex algebraic surfaces: the so called Godeaux surface. This is a surface, $S$ on which the canonical bundle is $ample$ and yet $h^{0}(S,K\_{S})=0$.


To construct such an $S$ we start with a quintic, $S'$ in $\mathbb{P}^{3}$ defined by the Fermat form:


$X\_{0}^{5}+X\_{1}^{5}+X\_{2}^{5}+X\_{3}^{5}=0$


The group $\mathbb{Z}\_{5}$ acts on $S'$ by:


$[X\_{0},X\_{1},X\_{2},X\_{3}]\rightarrow [X\_{0},\zeta X\_{1},\zeta^{2}X\_{2},\zeta^{3}X\_{3}]$


Where in the above $\zeta$ is a primitive fifth root of unity. One readily checks that this action is fixed point free so we get a smooth quotient $S=S'/\mathbb{Z}\_{5}$.


We will compute the invariants of $S$ by first computing those of $S'$ and then taking the quotient. The Lefschetz theorem on hyperplane sections tells us that $h^{1}(S')=0$. Next by adjunction, $K\_{S'}=H$ is a hyperplane section of $S'$. Applying the exact sequence of sheaves:


$ 0\rightarrow O\_{\mathbb{P}^{3}}(H-S')\rightarrow O\_{\mathbb{P}^{3}}(H)\rightarrow O\_{S'}(H)\rightarrow 0 $


we obtain the exact sequence of cohomology groups:


$ 0=H^{0}(\mathbb{P}^{3},-4H)\rightarrow H^{0}(\mathbb{P}^{3},H)\rightarrow H^{0}(S',K\_{S'})\rightarrow H^{1}(\mathbb{P}^{3},-4H)=0$


It follows that $h^{0}(S',K\_{S'})=h^{0}(\mathbb{P}^{3},H)=4$ and thus the holomorphic Euler characteristic of $S'$ is:


$\chi(S')=1-h^{1}(S')+h^{0}(S',K\_{S'})=5$


Now, since $S$ is a quotient of $S'$ by a free action of a finite group, the holomorphic forms on $S$ are just the $\mathbb{Z}\_{5}$ invariant forms on $S'$. In particular $h^{1}(S)=0$. Further, $\chi(S)=\frac{1}{5}\chi(S')=1$. Thus we have:


$h^{0}(S,K\_{S})=\chi(S)+h^{1}(S)-1=0$


So the canonical bundle of $S$ has no sections.


On the other hand, the canonical bundle of $S$ is manifestly ample. Indeed, $\pi\_{1}(S)=\mathbb{Z}\_{5}$ so $S$ is irrational. And:


$K\_{S}^{2}=\frac{1}{5}K\_{S'}^{2}=\frac{1}{5}\mathrm{deg(S')}=1 >0$


So a sufficiently high power of $K\_{S}$ maps $S$ birationally onto its image. It is not hard to see that for sufficiently high powers this birational map must be an isomorphism so $K\_{S}$ is ample.


