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 $N$...


1 Answer 1


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 and here), 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. $\square$


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