Suppose $X/\mathbb C$ is a projective $\mathbb Q$-factorial variety with wild singularities. Let $N$ be a nef **$\mathbb R$-Cartier** divisor. Then is it possible that there are infinitely many curves $C\_i \subset X$, such that $C\_i \cdot N>0$ are infinitely close to $0$?


Notice that if $N$ is a $\mathbb Q$-Cartier divisor, then if $C \cdot N>0$, it must satisfy $C \cdot N > \frac 1 m$ for some $m$ such that $mN$ is Cartier. Also, if $N$ is an **ample** $\mathbb R$-Cartier divisor, then it can be written (in the ample cone) as $\sum r\_i A\_i$ with $r\_i \in \mathbb R\_{>0}$ and $A\_i$ $\mathbb Q$-Cartier ample divisor, hence the statement does not hold either.


I am trying to work with Kodaira/Nagata example of a surface with infinitely $(-1)$-curves [Hartshorne's book V Ex 4.15(e)], but I have no idea of finding such $N$...



Infinitely small intersections with nef $\mathbb R$-Cartier divisors
This is possible. Basically, if $N$ is a point on the boundary of the nef cone that is not in the span of the rational points on the boundary, then we can approximate $N$ arbitrarily closely by rational points in the interior. For this, we need a variety whose nef cone is not a rational polyhedral cone:


**Example.** Let $X = E \times E$, where $E$ is a non-CM elliptic curve. Then $\operatorname{NS}(X) = \mathbb Z f\_1 \oplus \mathbb Z f\_2 \oplus \mathbb Z \delta$, where $f\_1 = [p \times E]$, $f\_2 = [E \times p]$, and $\delta = [\Delta]$. Moreover, the intersection matrix with respect to the basis $(f\_1,f\_2,\delta)$ is given by
$$\begin{pmatrix} 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0 \end{pmatrix}.$$
Letting $H = f\_1 + f\_2 + \delta$, the nef cone is given by the inequalities $D^2 \geq 0$, $D \cdot H \geq 0$, i.e.
\begin{align\*}
xy + yz + zx \geq 0,\\
x + y + z \geq 0.
\end{align\*}
See for example Lazarsfeld's *Positivity in algebraic geometry I*, **Lemma 1.5.4**.


The key observation is that the boundary of the nef cone contains rays that are not defined over $\mathbb Q$. For example, it contains the point $N = (1,\sqrt{2},\sqrt{2}-2)$, which satisfies $N^2 = 0$.


We will use the following lemma.


**Lemma.** If $\alpha \in \mathbb R/\mathbb Z$ has infinite order, then the set $\mathbb N \alpha \subseteq \mathbb R/\mathbb Z$ is dense.


In other words, if $\alpha \in \mathbb R$ is irrational, then the fractional parts of $n\alpha$ for $n \in \mathbb N$ are dense in $[0,1]$. This is a strengthening of the Dirichlet approximation theorem.


*Proof.* This is well-known. It's sometimes stated for $\mathbb Z \alpha$ instead of $\mathbb N \alpha$ (e.g. [here](https://mathoverflow.net/a/127728/82179) and [here](https://math.stackexchange.com/a/272713)), but the proof can be easily modified. $\square$


**Example** (continued). Let $\varepsilon > 0$. Applying the lemma to the interval $(-\varepsilon,0) \pmod{\mathbb Z}$, we conclude that there exist $m,n \in \mathbb N$ such that $n - \varepsilon < m \sqrt{2} < n$, i.e. 
$$0 < n - m\sqrt{2} < \varepsilon.$$
Consider the element $D = (m,n,n-2m)$; this is an approximation of $mN$. We compute
\begin{align\*}
D^2 &= 2\bigg(mn + n(n-2m) + (n-2m)m\bigg) \\
&= 2\big(n^2 - 2m^2\big) > 0,
\end{align\*}
and $D \cdot H = 2(2n - m) > 2(n - m\sqrt{2}) > 0$, so $D$ is in the interior of the nef cone, hence ample. Since $K = 0$ and $\chi(\mathcal O\_X) = 0$ on an abelian surface, Riemann–Roch gives
$$h^0(D) - h^1(D) + h^0(-D) = \frac{1}{2} D^2.$$
This forces $h^0(D) > 0$, since $D^2$ is positive and $h^0(-D) = 0$. Thus, there exists an effective curve $C \in |D|$. Finally, we compute
\begin{align\*}
C \cdot N &= \bigg(m\sqrt{2} + n\bigg) + \bigg(n(\sqrt{2}-2) + (n-2m)\sqrt{2}\bigg) + \bigg(m(\sqrt{2}-2)+(n-2m) \bigg) \\
&= 2n\sqrt{2} - 4m = 2\sqrt{2}\bigg(n-m\sqrt{2} \bigg) < 2\sqrt{2} \varepsilon.
\end{align\*}
Since $\varepsilon$ is arbitrary, we conclude that there exist effective curves $C$ such that the product $C \cdot N$ is arbitrarily small. $\square$


