Nef divisors with few global sections 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.
 A: If $A$ is a principally polarized abelian variety, and $L$ the line bundle coming form the polarization, then $H^0(A,L)$ is one dimensional.
A: How about the trivial divisor $\mathcal{O}_X$. It is nef because the intersection with any curve is 0. But $H^0(X, \mathcal{O}_X)=\mathbb{C}$.
A: Here is an example in a slightly different vein. There is a surface with a nef line bundle on it so that every power of the line bundle has global sections of dimension one. This example is possibly due to Cutkosky? I got it out of Rob Lazarsfeld's book PAGI.
Let $C$ be a curve of genus $g \geq 1$ and let $P$ be a line bundle of degree zero that is not torsion, that is, $P^{\otimes m} \neq O_C$ for any $m \geq 1$. There are plenty of them since $Pic^0(C)$ is a non-trivial abelian variety and therefore torsion is countable. Consider the vector bundle $V = O_C \oplus P$ and let $X = \mathbb{P}(V)$ be the associated projective bundle. This is a ruled surface. Consider the line bundle $O_X(1)$.
Let us compute the sections of $O_X(m)$. Recall that $H^0(X, O_X(m)) = H^0(C, Sym^m V)$. Since $P$ is degree zero and not torsion, $H^0(C, O_C(kP)) = 0$ for all $k \geq 1$. But $Sym^m V = O_C \oplus V'$ where $V'$ is a direct sum of line bundles of the form $O_C(kP)$ for $k \geq 1$. It follows that $h^0(X, O_X(m)) = 1$ for all $m \geq 1$.
But I claim that $O_X(1)$ is nef. Indeed, $X$ is a surface and if $0 \neq D \in |O_X(1)|$ (we just showed it exists) then recall that, since $V = O_C \oplus P$ with $P$ degree zero, $(D^2) = deg(V) = deg(P) = 0$. So $D$ is nef.
There is a similar but scarier example, due to Mumford, of a ruled surface so that $h^0(X, O_X(m)) = 0$ for all $m \geq 1$ and yet $O_X(m)$ is nef.
A: Consider a fibration $f \colon S \to C$, where $S$ is a smooth surface and $C$ a smooth curve of genus strictly greater then $0$. If $F$ is any fibre of $f$, then $F$ is nef and $h^0(S, \mathcal{O}_S(F))=1$. 
A: Another extremely simple example:
Let $S = \mathrm{Bl}_p \mathbb{P}^2$ be the blowup of $\mathbb{P}^2$ at a point, and look at the divisor $H-E$ on $S$, where $E$ is the exceptional divisor and $H$ is the pullback of a hyperplane class on $\mathbb{P}^2$.  Notice $h^0(H-E)=2 = \dim S$, since sections correspond to lines in $\mathbb{P}^2$ passing through $p$ (up to scalars).  But $(H-E)^2=0$, so $H-E$ is nef.
A: 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.
A: Very simple example: let X be the blowup of P^2 at 9 general points ("general" meaning "not the base locus of a pencil of cubics"). Then there is a unique cubic curve C in P^2 passing through the 9 points. 
If C' is the proper transform of C on the surface X, it defines a line bundle L on X with a 1-d space of global sections: h^0(X,L)=1. Here's why: any global section of L must vanish exactly along the curve C', because otherwise we'd get a curve D' linearly equivalent to C' but intersecting it properly. Projecting to P^2, D' would give a cubic curve passing through the 9 blown-up points but intersecting C properly, which is impossible by the generality assumption.
Also, the self-intersection number drops by 1 each time you blow up, so L^2 = C^2 - 9 = 0. Therefore L is nef. 
