A lower bound for $K^2_S$

Abstract

Let $(S,{\mathcal {L}})$ be a smooth, irreducible, projective, complex surface, polarized by a very ample line bundle ${\mathcal {L}}$ of degree $d > 35$. In this paper we prove that $K^2_S\ge -d(d-6)$. The bound is sharp, and $K^2_S=-d(d-6)$ if and only if d is even, the linear system $|H^0(S,{\mathcal {L}})|$ embeds S in a smooth rational normal scroll $T\subset {\mathbb {P}}^5$ of dimension 3, and here, as a divisor, S is linearly equivalent to $\frac{d}{2}Q$, where Q is a quadric on T.

Appendix

• (i) With the notation as in the proof of Proposition 2.2, consider the function

  $$\begin{aligned} \phi (a):= & {} -6a^3+a^2(-9m+5+3\epsilon )+a(2m(3\epsilon -4)-6\epsilon +10)\\&+3m^3+m^2(3\epsilon -13)+m(10-6\epsilon )+8. \end{aligned}$$

  We are going to prove that if $d\ge 18$ and $-m\le a \le \frac{1}{2}(m+\epsilon -1)$, then $\phi (a)\ge -d(d-6)$, and $\phi (a)=-d(d-6)$ if and only if $a=a^*$. To this purpose we derive with respect to a:

  $$\begin{aligned} \phi '(a)=-18a^2+2a(-9m+5+3\epsilon )+2m(3\epsilon -4)-6\epsilon +10. \end{aligned}$$

  This is a degree 2 polynomial in the variable a, whose discriminant is:

  $$\begin{aligned} \Delta =324m^2-936m+216m\epsilon +110+114\epsilon +36\epsilon ^2, \end{aligned}$$

  which is $>0$ when $m\ge 3$, hence when $d\ge 10$ (compare with (14)). Denote by $a_1$ and $a_2$ the real roots of the equation $\phi '(a)=0$, with $a_1<a_2$, and let I be the open interval $I=(a_1,a_2)$. Then $\phi '(a)>0$ if and only if $a\in I$. In particular $\phi (a)$ is strictly increasing for $a\in I$, and strictly decreasing for $a\not \in I$. Now observe that

  $$\begin{aligned} \phi (-m)=8, \quad \phi (-m+1)=-9m+17-3\epsilon , \quad \phi (-m+2)=0. \end{aligned}$$

  Notice that $-9m+17-3\epsilon \ge -9m+11$ because $0\le \epsilon \le 2$, $-9m+11\ge -9\frac{d-1}{3}+11$ since $\frac{d-1}{3}\le m\le \frac{d-3}{3}$, and $-3(d-1)+11>-d(d-6)$ if $d>7$. So for $a\in \{-m,-m+1,-m+2\}$ we have $K^2_S>-d(d-6)$. Moreover we have $\phi '(-m+2)=18m+6\epsilon -42>0$ if $d\ge 12$, and $\phi '(-1)=10m+6m\epsilon -12\epsilon -18>0$ if $d\ge 9$. Therefore $[-m+2,-1]\subset I$ and so $\phi (a)\ge \phi (-m+2)=0>-d(d-6)$ for $-m\le a\le -1$ and $d>12$. We also have:

  $$\begin{aligned} \phi (0)=(m-2)(3m^2-7m+3m\epsilon -4), \end{aligned}$$

  which is $\ge 0$ for $m\ge 3$, hence for $d\ge 12$. And

  $$\begin{aligned} \phi (1)=(m-1)(3m^2-10m+3m\epsilon +3\epsilon -17), \end{aligned}$$

  which is $\ge 0$ for $m\ge 5$, hence for $d\ge 18$. Moreover we have:

  $$\begin{aligned} \phi '(1)=2-26m+6m\epsilon <0. \end{aligned}$$

  Therefore $\phi (a)$ is strictly decreasing for $a\ge 1$. It follows that in the range $1\le a \le a^*:=\left[ \frac{m+\epsilon -1}{2}\right]$, the function $\phi$ takes its minimum exactly when $a=a^*$. Define p and q by dividing:

  $$\begin{aligned} m+\epsilon -1=2p+q, \quad 0\le q\le 1, \end{aligned}$$

  so that $p=a^*$. Notice that d is even if and only if $q=0$. We have:

  $$\begin{aligned} \phi (a^*)={\left\{ \begin{array}{ll} -d(d-6) &{} {{\mathrm{if}\; q=0}}\\ -\frac{1}{4}d^2+\frac{1}{2}d+\frac{35}{4} &{} {{\mathrm{if}\; q=1.}} \end{array}\right. } \end{aligned}$$

  Since when $d>5$ we have

  $$\begin{aligned} -\frac{1}{4}d^2+\frac{1}{2}d+\frac{35}{4}>-d(d-6), \end{aligned}$$

  by previous analysis it follows that, for any integer $-m\le a \le \frac{m+\epsilon -1}{2}$, one has $\phi (a)\ge -d(d-6)$, and $\phi (a)=-d(d-6)$ if and only if d is even and $a=\frac{m+\epsilon -1}{2}$.

• (ii) We are going to prove that inequality (12) implies inequality (13), and that (13) is impossible for $d>24$. First we recall that

  $$\begin{aligned} d-1=4p+q, \quad 0\le q\le 3, \end{aligned}$$

  and that

  $$\begin{aligned} {{t=0\, \mathrm{for}\, 0\le q \le 2, \quad \mathrm{and} \quad t=1\; \mathrm{for}\; q=3.}} \end{aligned}$$

  (15)

  Next we recall that from the definition (10) of the function k(i) we have (for $d\ge 9$):

  $$\begin{aligned} \sum _{i=1}^{d-4}(i-1)(d-k(i))= & {} \sum _{i=1}^{p}(i-1)(d-4i)+p(d-k(p+1))\nonumber \\= & {} {\left( {\begin{array}{c}p\\ 2\end{array}}\right) }d-8{\left( {\begin{array}{c}p+1\\ 3\end{array}}\right) }+tp. \end{aligned}$$

  (16)

  Taking into account that

  $$\begin{aligned} p=\frac{d-1-q}{4}, \end{aligned}$$

  and therefore that

  $$\begin{aligned} {\left( {\begin{array}{c}p\\ 2\end{array}}\right) }=\frac{1}{32}\left( d^2-6d-2dq+5+6q+q^2\right) \end{aligned}$$

  and that

  $$\begin{aligned} 8{\left( {\begin{array}{c}p+1\\ 3\end{array}}\right) }=\frac{1}{48}\left( d^3-3d^2-3d^2q-13d+6dq+3dq^2+15+13q-3q^2-q^3\right) , \end{aligned}$$

  from (16) we get:

  $$\begin{aligned}&\sum _{i=1}^{d-4}(i-1)(d-k(i))=\frac{d}{32}\left( d^2-6d-2dq+5+6q+q^2\right) \\&\qquad -\frac{1}{48}\left( d^3-3d^2-3d^2q-13d+6dq+3dq^2+15+13q-3q^2-q^3\right) \\&\qquad +t\left( \frac{d-1-q}{4}\right) \\&\quad =\frac{1}{96}\left( d^3-12d^2+d(41+6q-3q^2+24t)-30-26q+6q^2+2q^3-24t-24tq\right) . \end{aligned}$$

  Inserting this equality and (11) into (12), we get:

  $$\begin{aligned}&(d-3)\left( \frac{1}{8}d^2-\frac{3}{4}d+1\right) \\&\quad \le -\frac{1}{96}\left( d^3-12d^2+d(41+6q-3q^2+24t)-30\right. \\&\quad \left. -26q+6q^2+2q^3-24t-24tq\right) \\&\qquad +(d-4)\left( \frac{1}{8}d^2-\frac{1}{2}d+\frac{3}{8}+\frac{1}{4}q-\frac{1}{8}q^2+t\right) . \end{aligned}$$

  Multiplying by 96 and simplifying we obtain (13), i.e. that

  $$\begin{aligned}&-d^3+24d^2+(-9q^2+18q-125+72t)d-2q^3+42q^2\\&\quad -70q+174-360t+24tq \ge 0. \end{aligned}$$

  Finally denote by C(d, q, t) the left member of (13), i.e. set

  $$\begin{aligned}&C(d,q,t):=-d^3+24d^2+(-9q^2+18q-125+72t)d-2q^3\\&\quad +42q^2-70q+174-360t+24tq. \end{aligned}$$

  Then we have (compare with (15)):

  $$\begin{aligned} C(d,q,t)={\left\{ \begin{array}{ll} d^2(24-d)-125d+174 &{}\quad {{\mathrm{if}\; (q,t)=(0,0)}}\\ d^2(24-d)-116d+144 &{}\quad {{\mathrm{if}\; (q,t)=(1,0)}}\\ d^2(24-d)-125d+186 &{}\quad {{\mathrm{if}\; (q,t)=(2,0)}}\\ d^2(24-d)-80d &{}\quad {{\mathrm{if}\; (q,t)=(3,1)}}, \end{array}\right. } \end{aligned}$$

  from which it is evident that (13) is impossible for $d>24$.