Negative curves of small genus on surfaces

Abstract

Let X be a smooth geometrically irreducible projective surface over a field. In this paper we give an effective upper bound in terms of the Néron–Severi rank of X for the number of irreducible curves C on X with negative self-intersection and geometric genus less than $b_1(X)/4$, where $b_1(X)$ is the first étale Betti number of X. The proof involves a hyperbolic analog of the theory of spherical codes. More specifically, we relate these curves to the hyperbolic kissing number, and then prove upper and lower bounds for the hyperbolic kissing number in terms of the classical Euclidean kissing number.

Appendix: A calculus exercise

In this appendix we complete the proof of Lemma 6.1, whose notation we now assume. As in that proof, we begin by fixing $0 < d_1 \le d_2 \le d_3 = 1$. Let $\underline{0} = (0,0)$ be the origin in $\mathbb {R}^2 = \mathbb {R}^{n-1}$, and let $S(d_1,d_2)$ be the set of triples $(z_1,z_2,z_3) = (\underline{0},z_2, z_3)$ with $z_2, z_3 \in \mathbb {R}^2$ that satisfy (6.1). Then (6.1) implies that $|z_1 - z_2| = |z_2|$, $|z_1 - z_3| = |z_3|$, and $|z_2 - z_3|$ are bounded above and below by positive constants. It follows that $S(d_1,d_2)$ is compact. The law of cosines gives

$$\begin{aligned} \mathrm {cos}(\theta _1) = \frac{|z_1 - z_2|^2 + |z_1 - z_3|^2 - |z_2 - z_3|^2}{2\cdot |z_1 - z_2| \cdot |z_1 - z_3|}, \end{aligned}$$

(9.1)

where the denominator on the right is bounded away from 0. Thus $\mathrm {cos}(\theta _1)$ is a continuous function on $S(d_1,d_2)$, so it attains its maximum. We now assume this maximum occurs at $(z_1,z_2,z_3) = (\underline{0},z_2,z_3)$. As noted in Sect. 6, to prove Lemma 6.1 it will suffice to show that $\theta _1 \ge \phi _0$.

Let $\theta _2$ and $\theta _3$ be the angles at $z_2$ and at $z_3$ between the sides of the triangle with vertices at $z_1, z_2, z_3$, respectively. Since we observed in Sect. 6 that (6.1) implies that $z_1, z_2$, and $z_3$ are not collinear, all of $\theta _1, \theta _2$ and $\theta _3$ lie in the open interval $(0,\pi )$. Suppose $\theta _3 \ge \pi /2$. Then (6.1) gives

$$\begin{aligned} (d_1 + d_2)^2\ge & {} |z_1 - z_2|^2 \nonumber \\\ge & {} |z_1 - z_3|^2 + |z_2 - z_3|^2 \quad \quad (\mathrm {since}\ \theta _3 \ge \pi /2) \nonumber \\\ge & {} d_1^2 + d_3^2 + d_2^2 + d_3^2. \end{aligned}$$

(9.2)

This gives

$$\begin{aligned} 2 d_1 d_2 \ge 2 d_3^2 = 2. \end{aligned}$$

However, $d_1 \le d_2 \le d_3 = 1$, so this is only possible if $d_1 = d_2 = d_3 = 1$ and if all of the inequalities in (9.2) are equalities. Hence $(z_1, z_2, z_3) = (\underline{0}, z_2,z_3)$ is a right triangle with side lengths $|z_1 - z_2| = |z_2| = d_1 + d_2 = 2$, $|z_1 - z_3| = \sqrt{d_1^2 + d_3^2} = \sqrt{2}$, and $|z_2 - z_3| = \sqrt{d_2^2 + d_3^2} = \sqrt{2}$. This means that $\theta _1 = \pi /4 \ge \phi _0$ in this case, as claimed.

As noted in Sect. 6, the main fact we can now apply is that since $(z_1, z_2, z_3) \in S(d_1,d_2)$ minimizes $\theta _1$, we cannot move $z_1$, $z_2$ and $z_3$ in $\mathbb {R}^2$ and then translate $z_1$ back to $\underline{0}$ in such a way that the inequalities (6.1) still hold with the same $d_1, d_2$ and $d_3 = 1$ but a smaller value for $\theta _1$.

We may assume that $z_3$ is a point on the positive real axis by rotating both $z_2$ and $z_3$ around $z_1 = \underline{0}$. Since $0< \theta _1< \phi _0 < \pi /2$, the point $z_2$ now lies in the upper right quadrant. Let $C_1$ be the circle of radius $|z_1 - z_2| = |z_2|$ around $z_2$ in $\mathbb {R}^2$. Suppose that

$$\begin{aligned} |z_1 - z_3| < d_1 + d_3. \end{aligned}$$

(9.3)

Let $z'_2 = z_2$ and $z'_3 = z_3$. We now move $z_1 = \underline{0}$ to a point $z'_1$ on $C_1$ that lies in the upper left quadrant and is very close to $z_1$. The only side length which changes is then $|z_1 - z_3|$, which becomes $|z'_1 - z'_3| > |z_1 - z_3|$ since $z'_3 = z_3$ lies on the positive part of the real line. Since $|z_1 - z_3| < d_1 + d_3$, all of the inequalities in (6.1) will hold with the same $d_1, d_2, d_3$ when we replace $(z_1,z_2,z_3)$ by $(z'_1, z'_2, z'_3) = (z'_1,z_2,z_3)$ if $z'_1$ is a point on $C_1$ that lies in the upper left quadrant and is close enough to $z_1$. We now show that the angle $\theta '_1$ between the sides meeting at $z'_1$ of the triangle $(z'_1,z'_2,z'_3)$ satisfies $\theta '_1 < \theta _1$. This will contradict the minimality of $\theta _1$ and show that (9.3) cannot hold.

Let $\theta '_3$ be the angle between the sides of $(z'_1,z'_2,z'_3)$ meeting at $z'_3 = z_3$. Since $z'_1$ lies in the upper left quadrant and on the same side of the line between $z'_2 = z_2$ and $z'_3 = z_3$ as the origin $z_1 = (0,0)$, we have $0< \theta '_3< \theta _3 < \pi /2$. Thus $0< \mathrm {sin}(\theta '_3) < \mathrm {sin}(\theta _3)$. The law of sines now gives

$$\begin{aligned} \frac{\mathrm {sin}(\theta '_1)}{\mathrm {sin}(\theta '_3)}= & {} \frac{|z'_2 - z'_3|}{|z'_1 - z'_2|}\nonumber \\= & {} \frac{|z_2 - z_3|}{|z_1 - z_2|}\nonumber \\= & {} \frac{\mathrm {sin}(\theta _1)}{\mathrm {sin}(\theta _3)}. \end{aligned}$$

(9.4)

Since $\mathrm {sin}(\theta '_3) < \mathrm {sin}(\theta _3)$ we conclude that $\mathrm {sin}(\theta '_1) < \mathrm {sin}(\theta _1)$. Since we took $z'_1$ to be close to $z_1 = (0,0)$ on $C_1$, we can ensure that that $\theta '_1$ is close to $\theta _1$. Since $0< \theta _1< \phi _0 < \pi /2$, we conclude that $\theta '_1 < \theta _1$, contradicting the minimality of $\theta _1$. Thus (9.3) is false, so

$$\begin{aligned} z_1 = \underline{0} = (0,0) \quad \mathrm {and} \quad z_3 = (d_1+ d_3, 0) \end{aligned}$$

(9.5)

after rotating $z_3$ as above so that it lies on the positive real line.

Now suppose that

$$\begin{aligned} d_1^2 + d_2^2< |z_1 - z_2|^2 < (d_1 + d_2)^2. \end{aligned}$$

(9.6)

Recall that $z_2$ is a point in the upper right quadrant, and that we have reduced to the case in which (9.5) holds. We let $z'_2 = z_2$ and $z'_3 = z_3$. Define $C_2$ to be the circle with center $z_3 = (d_1 + d_3, 0)$ and radius $d_1 + d_2$, so that $C_2$ contains $z_1 = \underline{0}$ by (9.5). Consider points $z'_1$ very close to $z_1$ on $C_2$. The only edge distance that can change on replacing $(z_1,z_2,z_3)$ by $(z'_1,z'_2,z'_3)$ is $|z_1 - z_2|$. Since $|z'_1 - z'_2| = |z'_1 - z_2|$ will be close to $|z_1 - z_2| = |z_2|$ if $z'_1$ is close to $z_1$, we conclude from (9.6) that all the inequalities in (6.1) will hold if $(z_1,z_2,z_3)$ is replaced by $(z'_1,z'_2,z'_3) = (z'_1,z_2,z_3)$ and $z'_1$ is any point on $C_2$ sufficiently close to $z_1$.

Recall that $\theta _2$ is the angle at $z_2$ between the sides of the triangle $(z_1,z_2,z_3)$ adjoining $z_2$. Let $\theta '_2$ be the corresponding angle for the triangle $(z'_1,z'_2,z'_3)$. If $z'_1$ lies in the upper half plane and is sufficiently close to $z_1$, it is on the other side of the line between $z_2$ and $z_1 = \underline{0}$ from $z_3$. It follows that $\theta '_2 > \theta _2$ in this case. We find similarly that $\theta '_2 < \theta _2$ in case $z'_1$ is a point of $C_2$ that lies in the lower half plane and is sufficiently close to $z_1$. Thus we can in either case choose a $z'_1$ on $C_2$ arbitrarily close to $z_1$ for which

$$\begin{aligned} 0< \mathrm {sin}(\theta '_2) < \mathrm {sin}(\theta _2). \end{aligned}$$

(9.7)

Since $|z_1 - z_3| = d_1 + d_2 = |z'_1 - z_3|$ and $|z_2 - z_3| = |z'_2 - z'_3|$, the law of sines gives

$$\begin{aligned} \frac{\mathrm {sin}(\theta '_1)}{\mathrm {sin}(\theta '_2)}= & {} \frac{|z'_2 - z'_3|}{|z'_1 - z'_3|}\nonumber \\= & {} \frac{|z_2 - z_3|}{|z_1 - z_3|}\nonumber \\= & {} \frac{\mathrm {sin}(\theta _1)}{\mathrm {sin}(\theta _2)}. \end{aligned}$$

(9.8)

Now (9.7) shows $\mathrm {sin}(\theta '_1) < \mathrm {sin}(\theta _1)$. Since $\theta '_1$ will be close to $\theta _1< \phi _0 < \pi /2$ for $z'_1$ close to $z_1$, we conclude that $\theta '_1 < \theta _1$, which contradicts the minimality of $\theta _1$. Thus the hypothesis (9.6) must be false, and so

$$\begin{aligned} d_1^2 + d_2^2 = |z_1 - z_2|^2 \quad \mathrm {or} \quad |z_1 - z_2|^2 = (d_1 + d_2)^2. \end{aligned}$$

(9.9)

We now apply the law of cosines, together with (9.5) and $d_2^3 + d_3^2 \le |z_2 - z_3|^2$ from (6.1). This gives

$$\begin{aligned} \mathrm {cos}(\theta _1)= & {} \frac{|z_1 - z_2|^2 + |z_1 - z_3|^2 - |z_2 - z_3|^2}{2 \cdot |z_1 - z_2| \cdot |z_1 - z_3|} \nonumber \\\le & {} \frac{|z_1 - z_2|^2 + (d_1 + d_3)^2 - d_2^2 - d_3^2}{2 \cdot |z_1 - z_2| \cdot (d_1 + d_3)} \end{aligned}$$

(9.10)

where $d_1 \le d_2 \le d_3 = 1$.

Suppose first that $d_1^2 + d_2^2 = |z_1 - z_2|$ in (9.9). Then (9.10) becomes

$$\begin{aligned} \mathrm {cos}(\theta _1) \le \frac{d_1^2 + d_2^2 + (d_1+1)^2 - d_2^2 - 1}{2 \cdot \sqrt{d_1^2 + d_2^2} \cdot (d_1 + 1)} = \frac{d_1}{ \sqrt{d_1^2 + d_2^2}} = \frac{1}{\sqrt{1 + (d_2/d_1)^2}} \le \frac{1}{\sqrt{2}} \end{aligned}$$

(9.11)

since $0 < d_1 \le d_2$. This forces $\theta _1 \ge \pi /4$, contradicting $\theta _1< \phi _0 < \pi /4$.

The remaining possibility in (9.9) is that $|z_1 - z_2| = d_1 + d_2$. Then (9.10) gives

$$\begin{aligned} \mathrm {cos}(\theta _1)\le & {} \frac{(d_1+d_2)^2 + (d_1 + 1)^2 - d_2^2 - 1^2}{2 \cdot (d_1+d_2) \cdot (d_1 + 1)}\nonumber \\= & {} \frac{(d_1 + d_2 + 1)d_1 }{(d_1+d_2) \cdot (d_1 + 1)}\nonumber \\\le & {} \left( 1 + \frac{1}{d_1 + d_2}\right) \cdot \left( \frac{1}{1 + 1/d_1}\right) \nonumber \\\le & {} \frac{3}{4} \cdot \end{aligned}$$

(9.12)

since $0 < d_1 \le d_2 \le d_3 = 1$. This gives $\theta _1 \ge \phi _0 = \mathrm {arccos}(3/4)$, which completes the proof of Lemma 6.1.