Worst Singularities of Plane Curves of Given Degree

Abstract

We prove that $\frac{2}{d},\frac{2d-3}{{(d-1)}^2},\frac{2d-1}{d(d-1)},\frac{2d-5}{d^2-3d+1}$ and $\frac{2d-3}{d(d-2)}$ are the smallest log canonical thresholds of reduced plane curves of degree $d⩾3$, and we describe reduced plane curves of degree d whose log canonical thresholds are these numbers. As an application, we prove that $\frac{2}{d},\frac{2d-3}{{(d-1)}^2},\frac{2d-1}{d(d-1)},\frac{2d-5}{d^2-3d+1}$ and $\frac{2d-3}{d(d-2)}$ are the smallest values of the $\mathit{\alpha }$-invariant of Tian of smooth surfaces in ${\mathbb{P}}^3$ of degree $d⩾3$. We also prove that every reduced plane curve of degree $d⩾4$ whose log canonical threshold is smaller than $\frac{5}{2d}$ is GIT-unstable for the action of the group ${\mathrm{PGL}}_3(\mathbb{C})$, and we describe GIT-semistable reduced plane curves with log canonical thresholds $\frac{5}{2d}$.

Introduction

Let $C_d$ be a reduced plane curve in ${\mathbb{P}}^2$ of degree $d⩾3$, and let P be a point in $C_d$. The curve $C_d$ can have any given plane curve singularity at P provided that its degree d is sufficiently big. Thus, it is natural to ask

Question 1.1

What is the worst singularity that $C_d$ can have at P?

Denote by $m_P$ the multiplicity of the curve $C_d$ at the point P, and denote by $\mathit{\mu }(P)$ the Milnor number of the point P. If we use $m_P$ to measure the singularity of $C_d$ at the point P, then a union of d lines passing through P is an answer to Question 1.1, since $m_P⩽d$, and $m_P=d$ if and only if $C_d$ is a union of d lines passing through P. If we use the Milnor number $\mathit{\mu }(P)$, then the answer would be the same, since $\mathit{\mu }(P)⩽{(d-1)}^2$, and $\mathit{\mu }(P)={(d-1)}^2$ if and only if $C_d$ is a union of d lines passing through P. Alternatively, we can use the number

$$\begin{array}{c}\hfill {\mathrm{lct}}_P({\mathbb{P}}^2,C_d)=\sup \{\mathit{\lambda }\in \mathbb{Q}|\text{the log pair}({\mathbb{P}}^2,\mathit{\lambda }{C}_d)\text{is log canonical at}P\},\end{array}$$

which is known as the log canonical threshold of the log pair $({\mathbb{P}}^2,C_d)$ at the point P or the log canonical threshold of the curve $C_d$ at the point P (see [4, Definition 6.34]). The smallest ${\mathrm{lct}}_P({\mathbb{P}}^2,C_d)$ when P runs through all points in $C_d$ is usually denoted by $\mathrm{lct}({\mathbb{P}}^2,C_d)$. Note that

$$\begin{array}{c}\hfill \frac{1}{m_P}⩽{\mathrm{lct}}_P({\mathbb{P}}^2,C_d)⩽\frac{2}{m_P}.\end{array}$$

This is well known (see, [4, Exercise 6.18] and [4, Lemma 6.35]). So, the smaller ${\mathrm{lct}}_P({\mathbb{P}}^2,C_d)$, the worse singularity of the curve $C_d$ at the point P is.

Example 1.2

Suppose that $C_d$ is given by $x_1^{n_1}{x}_2^{n_2}(x_1^{m_1}+x_2^{m_2})=0$ up to analytic change of local coordinates, where $m_1$ and $m_2$ are positive integers, and $n_1,n_2\in \{0,1\}$. Then

$$\begin{array}{c}\hfill {\mathrm{lct}}_P({\mathbb{P}}^2,C_d)=\min \{1,\frac{\frac{1}{m_1}+\frac{1}{m_2}}{1+\frac{n_1}{m_1}+\frac{n_2}{m_2}}\}\end{array}$$

by [8, Proposition 2.2].

Log canonical thresholds of plane curves have been intensively studied (see, for example, [8]). Surprisingly, they give the same answer to Question 1.1 by

Theorem 1.3

([1, Theorem 4.1]) One has ${\mathrm{lct}}_P({\mathbb{P}}^2,C_d)⩾\frac{2}{d}$. Moreover, $\mathrm{lct}({\mathbb{P}}^2,C_d)=\frac{2}{d}$ if and only if $C_d$ is a union of d lines that pass through P.

In this paper we want to address

Question 1.4

What is the second worst singularity that $C_d$ can have at P?

To give a reasonable answer to this question, we have to disregard $m_P$ by obvious reasons. Thus, we will use the numbers $\mathit{\mu }(P)$ and ${\mathrm{lct}}_P({\mathbb{P}}^2,{\mathbb{C}}_d)$. For cubic curves, they give the same answer.

Example 1.5

Suppose that $d=3,m_P<3$ and P is a singular point of $C_3$. Then P is a singular point of type ${\mathbb{A}}_1,{\mathbb{A}}_2$ or ${\mathbb{A}}_3$. Moreover, if $C_3$ has singularity of type ${\mathbb{A}}_3$ at P, then $C_3=L+C_2$, where $C_2$ is a smooth conic, and L is a line tangent to $C_2$ at P. Furthermore, we have

$$\begin{array}{c}\hfill \mathit{\mu }(P)=\left\{\begin{array}{cc}& 1\text{if}{C}_3\text{has}{\mathbb{A}}_1\text{singularity at}P,\hfill \\ \hfill & 2\text{if}{C}_3\text{has}{\mathbb{A}}_2\text{singularity at}P,\hfill \\ \hfill & 3\text{if}{C}_3\text{has}{\mathbb{A}}_3\text{singularity at}P.\hfill \\ \hfill \end{array}\right.\end{array}$$

Similarly, we have

$$\begin{array}{c}\hfill {\mathrm{lct}}_P({\mathbb{P}}^2,C_3)=\left\{\begin{array}{cc}& 1\text{if}{C}_3\text{has}{\mathbb{A}}_1\text{singularity at}P,\hfill \\ \hfill & \frac{5}{6}\text{if}{C}_3\text{has}{\mathbb{A}}_2\text{singularity at}P,\hfill \\ \hfill & \frac{3}{4}\text{if}{C}_3\text{has}{\mathbb{A}}_3\text{singularity at}P.\hfill \\ \hfill \end{array}\right.\end{array}$$

For quartic curves, the numbers $\mathit{\mu }(P)$ and ${\mathrm{lct}}_P({\mathbb{P}}^2,{\mathbb{C}}_d)$ give different answers to Question 1.4.

Example 1.6

Suppose that $d=4,m_P<4$ and P is a singular point of $C_4$. Going through the list of all possible singularities that $C_P$ can have at P (see, for example, [6]), we obtain

$$\begin{array}{c}\hfill \mathit{\mu }(P)=\left\{\begin{array}{cc}& 6\text{if}{C}_4\text{has}{\mathbb{D}}_6\text{singularity at}P,\hfill \\ \hfill & 6\text{if}{C}_4\text{has}{\mathbb{A}}_6\text{singularity at}P,\hfill \\ \hfill & 6\text{if}{C}_4\text{has}{\mathbb{E}}_6\text{singularity at}P,\hfill \\ \hfill & 7\text{if}{C}_4\text{has}{\mathbb{A}}_7\text{singularity at}P,\hfill \\ \hfill & 7\text{if}{C}_4\text{has}{\mathbb{E}}_7\text{singularity at}P,\hfill \\ \hfill \end{array}\right.\end{array}$$

and $\mathit{\mu }(P)<6$ in all remaining cases. Similarly, we get

$$\begin{array}{c}\hfill {\mathrm{lct}}_P({\mathbb{P}}^2,C_4)=\left\{\begin{array}{cc}& \frac{5}{8}\text{if}{C}_4\text{has}{\mathbb{A}}_7\text{singularity at}P,\hfill \\ \hfill & \frac{5}{8}\text{if}{C}_4\text{has}{\mathbb{D}}_5\text{singularity at}P,\hfill \\ \hfill & \frac{3}{5}\text{if}{C}_4\text{has}{\mathbb{D}}_6\text{singularity at}P,\hfill \\ \hfill & \frac{7}{12}\text{if}{C}_4\text{has}{\mathbb{E}}_6\text{singularity at}P,\hfill \\ \hfill & \frac{5}{9}\text{if}{C}_4\text{has}{\mathbb{E}}_7\text{singularity at}P,\hfill \\ \hfill \end{array}\right.\end{array}$$

and ${\mathrm{lct}}_P({\mathbb{P}}^2,C_4)>\frac{5}{8}$ in all remaining cases.

Recently, Arkadiusz Płoski proved that $\mathit{\mu }(P)⩽{(d-1)}^2-\lfloor \frac{d}{2}\rfloor$ provided that $m_P<d$. Moreover, he described $C_d$ in the case when $\mathit{\mu }(P)={(d-1)}^2-\lfloor \frac{d}{2}\rfloor$. To present his description, we need

Definition 1.7

The curve $C_d$ is an even Płoski curve if d is even, the curve $C_d$ has $\frac{d}{2}⩾2$ irreducible components that are smooth conics passing through P, and all irreducible components of $C_d$ intersect each other pairwise at P with multiplicity 4. The curve $C_d$ is an odd Płoski curve if d is odd, the curve $C_d$ has $\frac{d+1}{2}⩾2$ irreducible components that all pass through $P,\frac{d-1}{2}$ irreducible component of the curve $C_d$ are smooth conics that intersect each other pairwise at P with multiplicity 4, and the remaining irreducible component is a line in ${\mathbb{P}}^2$ that is tangent at P to all other irreducible components. We say that $C_d$ is Płoski curve if it is either an even Płoski curve or an odd Płoski curve.

Each Płoski curve has unique singular point. If $d=4$, then $C_4$ is a Płoski curve if and only if it has a singular point of type ${\mathbb{A}}_7$. Thus, if $d=4$, then $\mathit{\mu }(P)={(d-1)}^2-\lfloor \frac{d}{2}\rfloor =7$ if and only if either $C_4$ is a Płoski curve and P is its singular point or $C_4$ has singularity ${\mathbb{E}}_7$ at the point P (see Example 1.6). For $d⩾5$, Płoski proved

Theorem 1.8

([10, Theorem 1.4]) If $d⩾5$, then $\mathit{\mu }(P)={(d-1)}^2-\lfloor \frac{d}{2}\rfloor$ if and only if $C_d$ is a Płoski curve and P is its singular point.

This result gives a very good answer to Question 1.4. The main goal of this paper is to give an answer to Question 1.4. using log canonical thresholds. Namely, we will prove that

$$\begin{array}{c}\hfill {\mathrm{lct}}_P({\mathbb{P}}^2,C_d)⩾\frac{2d-3}{{(d-1)}^2}\end{array}$$

provided that $m_P<d$, and we will describe $C_d$ in the case when ${\mathrm{lct}}_P({\mathbb{P}}^2,C_d)=\frac{2d-3}{{(d-1)}^2}$. To present this description, we need

Definition 1.9

The curve $C_d$ has singularity of type ${\mathbb{T}}_r$ (resp., ${\mathbb{K}}_r,{\tilde{\mathbb{T}}}_r,{\tilde{\mathbb{K}}}_r$) at the point P if the curve $C_d$ can be given by $x_1^r=x_1{x}_2^r$ (resp., $x_1^r=x_2^{r+1},x_2{x}_1^{r-1}=x_1{x}_2^r,x_2{x}_1^{r-1}=x_2^{r+1}$) up to analytic change of coordinates at the point P.

Note that ${\mathbb{T}}_2={\mathbb{A}}_3,{\mathbb{K}}_2={\mathbb{A}}_2,{\tilde{\mathbb{T}}}_2={\tilde{\mathbb{K}}}_2={\mathbb{A}}_1,{\tilde{\mathbb{K}}}_3={\mathbb{D}}_5,{\tilde{\mathbb{T}}}_3={\mathbb{D}}_6,{\mathbb{K}}_3={\mathbb{E}}_6$ and ${\mathbb{T}}_3={\mathbb{E}}_7$. Furthermore, since we assume that $d⩾3$, the formula in Example 1.2 gives

$$\begin{array}{c}\hfill {\mathrm{lct}}_P({\mathbb{P}}^2,C_d)=\left\{\begin{array}{cc}& \frac{2d-3}{{(d-1)}^2}\text{if}{C}_d\text{has}{\mathbb{T}}_{d-1}\text{singularity at}P,\hfill \\ \hfill & \frac{2d-1}{d(d-1)}\text{if}{C}_d\text{has}{\mathbb{K}}_{d-1}\text{singularity at}P,\hfill \\ \hfill & \frac{2d-5}{d^2-3d+1}\text{if}{C}_d\text{has}{\tilde{\mathbb{T}}}_{d-1}\text{singularity at}P,\hfill \\ \hfill & \frac{2d-3}{d(d-2)}\text{if}C\text{has}{\tilde{\mathbb{K}}}_{d-1}\text{singularity at}P,\hfill \\ \hfill \end{array}\right.\end{array}$$

where $\frac{2}{d}<\frac{2d-3}{{(d-1)}^2}<\frac{2d-1}{d(d-1)}<\frac{2d-5}{d^2-3d+1}⩽\frac{2d-3}{d(d-2)}$. In this paper we will prove

Theorem 1.10

Suppose that $d⩾4$ and ${\mathrm{lct}}_P({\mathbb{P}}^2,C_d)⩽\frac{2d-3}{d(d-2)}$. Then one of the following holds:

1. $m_P=d$,

2. the curve $C_d$ has singularity of type ${\mathbb{T}}_{d-1},{\mathbb{K}}_{d-1},{\tilde{\mathbb{T}}}_{d-1}$ or ${\tilde{\mathbb{K}}}_{d-1}$ at the point P,

3. $d=4$ and $C_d$ is a Płoski quartic curve (in this case ${\mathrm{lct}}_P({\mathbb{P}}^2,C_d)=\frac{5}{8})$.

This result describes the five worst singularities that $C_d$ can have at the point P. In particular, Theorem 1.10 answers Question 1.4. This answer is very different from the answer given by Theorem 1.8. Indeed, if $C_d$ is a Płoski curve, $d>3$ and P is its singular point, then

$$\begin{array}{c}\hfill {\mathrm{lct}}_P({\mathbb{P}}^2,C_d)=\frac{5}{2d}>\frac{2d-3}{{(d-1)}^2}.\end{array}$$

The proof of Theorem 1.10 implies one result that is interesting on its own. To describe it, let us identify the curve $C_d$ with a point in the space $|{\mathcal{O}}_{{\mathbb{P}}^2}(d)|$ that parameterizes all (not necessarily reduced) plane curves of degree d. Since the group ${\mathrm{PGL}}_3(\mathbb{C})$ acts on $|{\mathcal{O}}_{{\mathbb{P}}^2}(d)|$, it is natural to ask whether $C_d$ is GIT-stable (resp., GIT-semistable) for this action or not. For small d, its answer is classical and immediately follows from the Hilbert–Mumford criterion (see [9, Chapter 2.1]).

Example 1.11

([9, Chapter 4.2]) If $d=3$, then $C_3$ is GIT-stable (resp., GIT-semistable) if and only if $C_3$ is smooth (resp., $C_3$ has at most ${\mathbb{A}}_1$ singularities). If $d=4$, then $C_4$ is GIT-stable (resp., GIT-semistable) if and only if $C_4$ has at most ${\mathbb{A}}_1$ and ${\mathbb{A}}_2$ singularities (resp., $C_4$ has at most singular double points and $C_4$ is not a union of a cubic with an inflectional tangent line).

Paul Hacking, Hosung Kim and Yongnam Lee noticed that the log canonical threshold $\mathrm{lct}({\mathbb{P}}^2,C_d)$ and GIT-stability of the curve $C_d$ are closely related. In particular, they proved

Theorem 1.12

([5, Propositions 10.2 and 10.4], [7, Theorem 2.3]) If $\mathrm{lct}({\mathbb{P}}^2,C_d)⩾\frac{3}{d}$, then the curve $C_d$ is GIT-semistable. If $d⩾4$ and $\mathrm{lct}({\mathbb{P}}^2,C_d)>\frac{3}{d}$, then the curve $C_d$ is GIT-stable.

This gives a sufficient condition for the curve $C_d$ to be GIT-stable (resp, GIT-semistable). However, this condition is not a necessary condition. Let us give two examples that illustrate this.

Example 1.13

([13, p. 268], [5, Example 10.5]) Suppose that $d=5$, the quintic curve $C_5$ is given by

$$\begin{array}{c}\hfill x^5+(y^2-xz{)}^2(\frac{x}{4}+y+z)=x^2(y^2-xz)(x+2y),\end{array}$$

and $P=[0:0:1]$. Then $C_5$ is irreducible and has singularity ${\mathbb{A}}_{12}$ at the point P. In particular, it is rational. Furthermore, the curve $C_5$ is GIT-stable (see, for example, [9, Chapter 4.2]). On the other hand, it follows from Example 1.2 that

$$\begin{array}{c}\hfill \mathrm{lct}({\mathbb{P}}^2,C_5)={\mathrm{lct}}_P({\mathbb{P}}^2,C_5)=\frac{1}{2}+\frac{1}{13}=\frac{15}{26}<\frac{3}{5}.\end{array}$$

Example 1.14

Suppose that $C_d$ is a Płoski curve. Let P be its singular point, and let L be a general line in ${\mathbb{P}}^2$. Then

$$\begin{array}{c}\hfill \mathrm{lct}({\mathbb{P}}^2,C_d+L)=\mathrm{lct}({\mathbb{P}}^2,C_d)={\mathrm{lct}}_P({\mathbb{P}}^2,C_d)=\frac{5}{2d}<\frac{3}{d}.\end{array}$$

On the other hand, if d is even, then $C_d$ is GIT-semistable, and $C_d+L$ is GIT-stable. This follows from the Hilbert–Mumford criterion. Similarly, if d is odd, then $C_d$ is GIT-unstable, and $C_d+L$ is GIT-semistable.

In this paper we will prove the following result that complements Theorem 1.12.

Theorem 1.15

If $\mathrm{lct}({\mathbb{P}}^2,C_d)<\frac{5}{2d}$, then $C_d$ is GIT-unstable. Moreover, if $\mathrm{lct}({\mathbb{P}}^2,C_d)⩽\frac{5}{2d}$, then $C_d$ is not GIT-stable. Furthermore, if $\mathrm{lct}({\mathbb{P}}^2,C_d)=\frac{5}{2d}$, then $C_d$ is GIT-semistable if and only if $C_d$ is an even Płoski curve.

Example 1.14 shows that this result is sharp. Surprisingly, its proof is very similar to the proof of Theorem 1.10. In fact, we will give a combined proof of both these theorems in Section 3.

In this paper we will also prove one application of Theorem 1.10. To describe it, we need

Definition 1.16

([12, Appendix A], [3, Definition 1.20]) For a given smooth variety V equipped with an ample $\mathbb{Q}$-divisor $H_V$, let ${\mathit{\alpha }}_V^{H_V}:V\to {\mathbb{R}}_{⩾0}$ be a function defined as

$$\begin{array}{c}\hfill {\mathit{\alpha }}_V^{H_V}(O)=\sup \left\{\mathit{\lambda }\in \mathbb{Q}\left|\begin{array}{cc}& \text{the pair}\left(V,\mathit{\lambda }{D}_V\right)\text{is log canonical at}O\hfill \\ \hfill & \text{for every effective}\mathbb{Q}\text{-divisor}{D}_V{\sim }_{\mathbb{Q}}{H}_V\hfill \\ \hfill \end{array}\right.\right\}.\end{array}$$

Denote its infimum by $\mathit{\alpha }(V,H_V)$.

Let $S_d$ be a smooth surface in ${\mathbb{P}}^3$ of degree $d⩾3$, let $H_{S_d}$ be its hyperplane section, let O be a point in $S_d$, and let $T_O$ be the hyperplane section of $S_d$ that is singular at O. Similar to ${\mathrm{lct}}_P({\mathbb{P}}^2,C_d)$, we can define

$$\begin{array}{c}\hfill {\mathrm{lct}}_O(S_d,T_O)=\sup \{\mathit{\lambda }\in \mathbb{Q}|\text{the log pair}(S_d,\mathit{\lambda }{T}_O)\text{is log canonical at}O\}.\end{array}$$

Then ${\mathit{\alpha }}_{S_d}^{H_{S_d}}(O)⩽{\mathrm{lct}}_O(S_d,T_O)$ by Definition 1.16. Note that $T_O$ is reduced, since the surface $S_d$ is smooth. In this paper we prove

Theorem 1.17

If ${\mathit{\alpha }}_{S_d}^{H_{S_d}}(O)<\frac{2d-3}{d(d-2)}$, then

$$\begin{array}{c}\hfill {\mathit{\alpha }}_{S_d}^{H_{S_d}}(O)={\mathrm{lct}}_O(S_d,T_O)\in \{\frac{2}{d},\frac{2d-3}{{(d-1)}^2},\frac{2d-1}{d(d-1)},\frac{2d-5}{d^2-3d+1}\}.\end{array}$$

Similarly, if $\mathit{\alpha }(S_d,H_{S_d})<\frac{2d-3}{d(d-2)}$, then

$$\begin{array}{c}\hfill \mathit{\alpha }(S_d,H_{S_d})=\underset{O\in S_d}{inf}\{{\mathrm{lct}}_O(S_d,T_O)\}\in \{\frac{2}{d},\frac{2d-3}{{(d-1)}^2},\frac{2d-1}{d(d-1)},\frac{2d-5}{d^2-3d+1}\}.\end{array}$$

If $d=3$, then we can drop the condition ${\mathit{\alpha }}_{S_d}^{H_{S_d}}(O)<\frac{2d-3}{d(d-2)}$ in Theorem 1.17, since $\frac{2d-3}{d(d-2)}=1$ in this case. Thus, Theorem 1.17 implies

Corollary 1.18

([3, Corollary 1.24]) Suppose that $d=3$. Then ${\mathit{\alpha }}_{S_3}^{H_{S_3}}(O)={\mathrm{lct}}_O(S_3,T_O)$.

If $d⩾4$, we cannot drop the condition ${\mathit{\alpha }}_{S_d}^{H_{S_d}}(O)<\frac{2d-3}{d(d-2)}$ in Theorem 1.17 in general. Let us give two examples that illustrate this.

Example 1.19

Suppose that $d=4$. Let $S_4$ be a quartic surface in ${\mathbb{P}}^3$ that is given by

$$\begin{array}{c}\hfill t^{3}x+t^{2}yz+xyz(y+z)=0,\end{array}$$

and let O be the point [0 : 0 : 0 : 1]. Then $S_4$ is smooth, and $T_O$ has singularity ${\mathbb{A}}_1$ at O, which implies that ${\mathrm{lct}}_O(S_4,T_O)=1$. Let $L_y$ be the line $x=y=0$, let $L_z$ be the line $x=z=0$, and let $C_2$ be the conic $y+z=xt+yz=0$. Then $L_y,L_z$ and $C_2$ are contained in $S_4$, and $O=L_y\cap L_z\cap C_2$. Moreover,

$$\begin{array}{c}\hfill L_y+L_z+\frac{1}{2}{C}_2\sim 2H_{S_4},\end{array}$$

because the divisor $2L_y+2L_z+C_2$ is cut out on $S_4$ by $tx+yz=0$. Furthermore, the log pair $(S_4,L_y+L_z+\frac{1}{2}{C}_2)$ is not log canonical at O, so that ${\mathit{\alpha }}_{S_4}^{H_{S_4}}(O)<1$ by Definition 1.16.

Example 1.20

Suppose that $d⩾5$ and $T_O$ has ${\mathbb{A}}_1$ singularity at O. Then ${\mathrm{lct}}_O(S_d,T_O)=1$. Let $f:{\tilde{S}}_d\to S_d$ be a blow up of the point O. Denote by E its exceptional curve. Then

$$\begin{array}{c}\hfill (f^{\ast }(H_{S_d})-\frac{11}{5}E{)}^2=5-\frac{121}{25}>0.\end{array}$$

Hence, it follows from Riemann–Roch theorem there is an integer $n⩾1$ such that the linear system $|f^{\ast }(5n{H}_{S_d})-11nE|$ is not empty. Pick a divisor $\tilde{D}$ in this linear system, and denote by D its image on $S_d$. Then $(S_d,\frac{1}{5n}D)$ is not log canonical at P, since ${\mathrm{mult}}_P(D)⩾11n$. On the other hand, $\frac{1}{5n}D{\sim }_{\mathbb{Q}}{H}_{S_d}$ by construction, so that ${\mathit{\alpha }}_{S_d}^{H_{{}_d}}(O)<1$ by Definition 1.16.

This work was carried out during the author’s stay at the Max Planck Institute for Mathematics in Bonn in 2014. We would like to thank the institute for the hospitality and very good working condition. We would like to thank Michael Wemyss for checking the singularities of the curve $C_5$ in Example 1.13. We would like to thank Alexandru Dimca, Yongnam Lee, Jihun Park, Hendrick Süß and Mikhail Zaidenberg for very useful comments.

Preliminaries

In this section, we present results that will be used in the proof of Theorems 1.10, 1.15, 1.17. Let S be a smooth surface, let D be an effective non-zero $\mathbb{Q}$-divisor on the surface S, and let P be a point in the surface S. Write

$$\begin{array}{c}\hfill D=\sum _{i=1}^r{a}_i{C}_i,\end{array}$$

where each $C_i$ is an irreducible curve on the surface S, and each $a_i$ is a non-negative rational number. Let us recall

Definition 2.1

([4, § 6]) Let $\mathit{\pi }:\tilde{S}\to S$ be a birational morphism such that $\tilde{S}$ is smooth. Then $\mathit{\pi }$ is a composition of blow ups of smooth points. For each $C_i$, denote by ${\tilde{C}}_i$ its proper transform on the surface $\tilde{S}$. Let $F_1,\dots ,F_n$ be $\mathit{\pi }$-exceptional curves. Then

$$\begin{array}{c}\hfill K_{\tilde{S}}+\sum _{i=1}^r{a}_i{\tilde{C}}_i+\sum _{j=1}^n{b}_j{F}_j{\sim }_{\mathbb{Q}}{\mathit{\pi }}^{\ast }(K_S+D)\end{array}$$

for some rational numbers $b_1,\dots ,b_n$. Suppose, in addition, that ${\sum }_{i=1}^r{\tilde{C}}_i+{\sum }_{j=1}^n{F}_j$ is a divisor with simple normal crossings. Then the log pair (S, D) is said to be log canonical at P if and only if the following two conditions are satisfied:

• $a_i⩽1$ for every $C_i$ such that $P\in C_i$,

• $b_j⩽1$ for every $F_j$ such that $\mathit{\pi }(F_j)=P$.

Similarly, the log pair (S, D) is said to be Kawamata log terminal at P if and only if $a_i<1$ for every $C_i$ such that $P\in C_i$, and $b_j<1$ for every $F_j$ such that $\mathit{\pi }(F_j)=P$.

Using just this definition, one can easily prove

Lemma 2.2

Suppose that $r=3,P\in C_1\cap C_2\cap C_3$, the curves $C_1,C_2$ and $C_3$ are smooth at $P,a_1<1,a_2<1$ and $a_3<1$. Moreover, suppose that both curves $C_1$ and $C_2$ intersect the curve $C_3$ transversally at P. Furthermore, suppose that (S, D) is not Kawamata log terminal at P. Put $k={\mathrm{mult}}_P(C_1·C_2)$. Then $k(a_1+a_2)+a_3⩾k+1$.

Proof

Put $S_0=S$ and consider a sequence of blow ups

where each ${\mathit{\pi }}_j$ is the blow up of the intersection point of the proper transforms of the curves $C_1$ and $C_2$ on the surface $S_{j-1}$ that dominates P (such point exists, since $k={\mathrm{mult}}_P(C_1·C_2)$). For each ${\mathit{\pi }}_j$, denote by $E_j^k$ the proper transform of its exceptional curve on $S_k$. For each $C_i$, denote by $C_i^k$ its proper transform on the surface $S_k$. Then

$$\begin{array}{c}\hfill K_{S_k}+\sum _{i=1}^n{a}_i{C}_i^k+\sum _{j=1}^k(j(a_1+a_2)+a_3-j)E_j^k{\sim }_{\mathbb{Q}}{({\mathit{\pi }}_1\circ {\mathit{\pi }}_2\circ \cdots \circ {\mathit{\pi }}_k)}^{\ast }(K_S+D),\end{array}$$

and ${\sum }_{i=1}^n{C}_i^k+{\sum }_{j=1}^k{E}_j$ is a simple normal crossing divisor in every point of ${\cup }_{j=1}^k{E}_j$. Thus, it follows from Definition 2.1 that there exists $l\in \{1,\dots ,k\}$ such that $l(a_1+a_2)+a_3⩾l+1$, because (S, D) is not Kawamata log terminal at P. If $l=k$, then we are done. So, we may assume that $l<k$. If $k(a_1+a_2)+a_3<k+1$, then $a_1+a_2<1+\frac{1}{k}-a_3\frac{1}{k}$, which implies that

$$\begin{array}{cc}\hfill l+1⩽& l(a_1+a_2)+a_3<(l+\frac{l}{k}-a_3\frac{l}{k})+a_3=l+\frac{l}{k}+a_3(1-\frac{l}{k})\hfill \\ \hfill ⩽& l+\frac{l}{k}+(1-\frac{l}{k})=l+1,\hfill \end{array}$$

because $a_3<1$. Thus, the obtained contradiction shows that $k(a_1+a_2)+a_3⩾k+1$.

$\square$

Corollary 2.3

Suppose that $r=2,P\in C_1\cap C_2$, the curves $C_1$ and $C_2$ are smooth at $P,a_1<1$ and $a_2<1$. Put $k={\mathrm{mult}}_P(C_1·C_2)$. If (S, D) is not Kawamata log terminal at P, then $k(a_1+a_2)⩾k+1$.

The log pair (S, D) is called log canonical if it is log canonical at every point of S. Similarly, the log pair (S, D) is called Kawamata log terminal if it is Kawamata log terminal at every point of the surface S.

Remark 2.4

Let R be any effective $\mathbb{Q}$-divisor on S such that $R{\sim }_{\mathbb{Q}}D$ and $R\ne D$. Put

$$\begin{array}{c}\hfill D_{\mathit{ϵ}}=(1+\mathit{ϵ})D-\mathit{ϵ}R,\end{array}$$

where $\mathit{ϵ}$ is a non-negative rational number. Then $D_{\mathit{ϵ}}{\sim }_{\mathbb{Q}}D$. Moreover, since $R\ne D$, there exists the greatest rational number ${\mathit{ϵ}}_0⩾0$ such that the divisor $D_{{\mathit{ϵ}}_0}$ is effective. Then $\mathrm{Supp}(D_{{\mathit{ϵ}}_0})$ does not contain at least one irreducible component of $\mathrm{Supp}(R)$. Moreover, if (S, D) is not log canonical at P, and (S, R) is log canonical at P, then $(S,D_{{\mathit{ϵ}}_0})$ is not log canonical at P by Definition 2.1, because

$$\begin{array}{c}\hfill D=\frac{1}{1+{\mathit{ϵ}}_0}{D}_{{\mathit{ϵ}}_0}+\frac{{\mathit{ϵ}}_0}{1+{\mathit{ϵ}}_0}R\end{array}$$

and $\frac{1}{1+{\mathit{ϵ}}_0}+\frac{{\mathit{ϵ}}_0}{1+{\mathit{ϵ}}_0}=1$. Similarly, if the log pair (S, D) is not Kawamata log terminal at P, and (S, R) is Kawamata log terminal at P, then $(S,D_{{\mathit{ϵ}}_0})$ is not Kawamata log terminal at P.

The following result is well known.

Lemma 2.5

([4, Exercise 6.18]) If (S, D) is not log canonical at P, then ${\mathrm{mult}}_P(D)>1$. Similarly, if (S, D) is not Kawamata log terminal at P, then ${\mathrm{mult}}_P(D)⩾1$.

Combining with

Lemma 2.6

([4, Lemma 5.36]) Suppose that S is a smooth surface in ${\mathbb{P}}^3$, and $D{\sim }_{\mathbb{Q}}{H}_S$, where $H_S$ is a hyperplane section of S. Then each $a_i$ does not exceed 1.

Lemma 2.5 gives

Corollary 2.7

Suppose that S is a smooth surface in ${\mathbb{P}}^3$, and $D{\sim }_{\mathbb{Q}}{H}_S$, where $H_S$ is a hyperplane section of S. Then (S, D) is log canonical outside of finitely many points.

The following result is a special case of a much more general result, which is known as Shokurov’s connectedness principle (see, for example, [4, Theorem 6.3.2]).

Lemma 2.8

([11, Theorem 6.9]) If $-(K_S+D)$ is big and nef, then the locus where (S, D) is not Kawamata log terminal is connected.

Corollary 2.9

Let $C_d$ be a reduced curve in ${\mathbb{P}}^2$ of degree d, and let O and Q be two points in $C_d$ such that $O\ne Q$. If ${\mathrm{lct}}_O({\mathbb{P}}^2,C_d)<\frac{3}{d}$, then ${\mathrm{lct}}_Q({\mathbb{P}}^2,C_d)⩾\frac{3}{d}$.

Let ${\mathit{\pi }}_1:S_1\to S$ be a blow up of the point P, and let $E_1$ be the ${\mathit{\pi }}_1$-exceptional curve. Denote by $D^1$ the proper transform of the divisor D on the surface $S_1$ via ${\mathit{\pi }}_1$. Then the log pair $(S_1,D^1+({\mathrm{mult}}_P(D)-1)E_1)$ is often called the log pull back of the log pair (S, D), because

$$\begin{array}{c}\hfill K_{S_1}+D^1+({\mathrm{mult}}_P(D)-1)E_1{\sim }_{\mathbb{Q}}{\mathit{\pi }}_1^{\ast }(K_S+D).\end{array}$$

This $\mathbb{Q}$-rational equivalence implies that the log pair (S, D) is not log canonical at P provided that ${\mathrm{mult}}_P(D)>2$. Similarly, if ${\mathrm{mult}}_P(D)⩾2$, then the singularities of the log pair (S, D) are not Kawamata log terminal at the point P.

Remark 2.10

The log pair (S, D) is log canonical at P if and only if $(S_1,D^1+({\mathrm{mult}}_P(D)-1)E_1)$ is log canonical at every point of the curve $E_1$. Similarly, the log pair (S, D) is Kawamata log terminal at P if and only if $(S_1,D^1+({\mathrm{mult}}_P(D)-1)E_1)$ is Kawamata log terminal at every point of the curve $E_1$.

Let Z be an irreducible curve on S that contains P. Suppose that Z is smooth at P, and Z is not contained in $\mathrm{Supp}(D)$. Let $\mathit{\mu }$ be a non-negative rational number. The following result is a very special case of a much more general result known as Inversion of Adjunction (see, for example, [11, § 3.4] or [4, Theorem 6.29]).

Theorem 2.11

([11, Corollary 3.12], [4, Exercise 6.31], [2, Theorem 7]) Suppose that the log pair $(S,\mathit{\mu }Z+D)$ is not log canonical at P and $\mathit{\mu }⩽1$. Then ${\mathrm{mult}}_P(D·Z)>1$.

This result implies

Theorem 2.12

Suppose that $(S,\mathit{\mu }Z+D)$ is not Kawamata log terminal at P, and $\mathit{\mu }<1$. Then ${\mathrm{mult}}_P(D·Z)>1$.

Proof

The log pair $(S,Z+D)$ is not log canonical at P, because $\mathit{\mu }<1$, and $(S,\mathit{\mu }Z+D)$ is not Kawamata log terminal at P. Then ${\mathrm{mult}}_P(D·Z)>1$ by Theorem 2.11. $\square$

Theorems 2.11 and 2.12 imply

Lemma 2.13

If (S, D) is not log canonical at P and ${\mathrm{mult}}_P(D)⩽2$, then there exists a unique point in $E_1$ such that $(S_1,D^1+({\mathrm{mult}}_P(D)-1)E_1)$ is not log canonical at it. Similarly, if (S, D) is not Kawamata log terminal at P, and ${\mathrm{mult}}_P(D)<2$, then there exists a unique point in $E_1$ such that $(S_1,D^1+({\mathrm{mult}}_P(D)-1)E_1)$ is not Kawamata log terminal at it.

Proof

If ${\mathrm{mult}}_P(D)⩽2$ and $(S_1,D^1+(\mathit{\lambda }{\mathrm{mult}}_P(D)-1)E_1)$ is not log canonical at two distinct points $P_1$ and ${\tilde{P}}_1$, then

$$\begin{array}{c}\hfill 2⩾{\mathrm{mult}}_P(D)=D^1·E_1⩾{\mathrm{mult}}_{P_1}(D^1·E_1)+{\mathrm{mult}}_{{\tilde{P}}_1}(D^1·E_1)>2\end{array}$$

by Theorem 2.11. By Remark 2.10, this proves the first assertion. Similarly, we can prove the second assertion using Theorem 2.12 instead of Theorem 2.11. $\square$

The following result can be proved similarly to the proof of Lemma 2.5. Let us show how to prove it using Theorem 2.12.

Lemma 2.14

Suppose that (S, D) is not Kawamata log terminal at P, and (S, D) is Kawamata log terminal in a punctured neighbourhood of the point P, then ${\mathrm{mult}}_P(D)>1$.

Proof

By Remark 2.10, the log pair $(S_1,D^1+({\mathrm{mult}}_P(D)-1)E_1)$ is not Kawamata log terminal at some point $P_1\in E_1$. Moreover, if ${\mathrm{mult}}_P(D)<2$, then $(S_1,D^1+({\mathrm{mult}}_P(D)-1)E_1)$ is Kawamata log terminal at a punctured neighbourhood of the point $P_1$. Thus, if ${\mathrm{mult}}_P(D)⩽1$, then ${\mathrm{mult}}_P(D)=D^1·E_1>1$ by Theorem 2.12, which is absurd. $\square$

Let $Z_1$ and $Z_2$ be two irreducible curves on the surface S such that $Z_1$ and $Z_2$ are not contained in $\mathrm{Supp}(D)$. Suppose that $P\in Z_1\cap Z_2$, the curves $Z_1$ and $Z_2$ are smooth at P, the curves $Z_1$ and $Z_2$ intersect each other transversally at P. Let ${\mathit{\mu }}_1$ and ${\mathit{\mu }}_2$ be non-negative rational numbers.

Theorem 2.15

([2, Theorem 13]) Suppose that the log pair $(S,{\mathit{\mu }}_1{Z}_1+{\mathit{\mu }}_2{Z}_2+D)$ is not log canonical at the point P, and ${\mathrm{mult}}_P(D)⩽1$. Then either ${\mathrm{mult}}_P(D·Z_1)>2(1-{\mathit{\mu }}_2)$ or ${\mathrm{mult}}_P(D·Z_2)>2(1-{\mathit{\mu }}_1)$ (or both).

This result implies

Theorem 2.16

Suppose that $(S,{\mathit{\mu }}_1{Z}_1+{\mathit{\mu }}_2{Z}_2+D)$ is not Kawamata log terminal at P, and ${\mathrm{mult}}_P(D)<1$. Then either ${\mathrm{mult}}_P(D·Z_1)⩾2(1-{\mathit{\mu }}_2)$ or ${\mathrm{mult}}_P(D·Z_2)⩾2(1-{\mathit{\mu }}_1)$ (or both).

Proof

Let $\mathit{\lambda }$ be a rational number such that

$$\begin{array}{c}\hfill \frac{1}{{\mathrm{mult}}_P(D)}⩾\mathit{\lambda }>1.\end{array}$$

Then $(S,D+\mathit{\lambda }{\mathit{\mu }}_1{Z}_1+\mathit{\lambda }{\mathit{\mu }}_2{Z}_2)$ is not log canonical at P. Now it follows from Theorem 2.15 that either ${\mathrm{mult}}_P(D·Z_1)>2(1-\mathit{\lambda }{\mathit{\mu }}_2)$ or ${\mathrm{mult}}_P(D·Z_2)>2(1-\mathit{\lambda }{\mathit{\mu }}_1)$ (or both). Since we can choose $\mathit{\lambda }$ to be as close to 1 as we wish, this implies that either ${\mathrm{mult}}_P(D·Z_1)⩾2(1-{\mathit{\mu }}_2)$ or ${\mathrm{mult}}_P(D·Z_2)⩾2(1-{\mathit{\mu }}_1)$ (or both). $\square$

Reduced Plane Curves

The purpose of this section is to prove Theorems 1.10 and 1.15. Let $C_d$ be a reduced plane curve in ${\mathbb{P}}^2$ of degree $d⩾4$, and let P be a point in $C_d$. Put ${\mathit{\lambda }}_1=\frac{2d-3}{d(d-2)}$ and ${\mathit{\lambda }}_2=\frac{5}{2d}$. To prove Theorem 1.10, we have to show that if the log pair $({\mathbb{P}}^2,{\mathit{\lambda }}_1{C}_d)$ is not Kawamata log terminal at the point P, then one of the following assertions hold:

• ${\mathrm{mult}}_P(C_d)=d$,

• $C_d$ has singularity ${\mathbb{T}}_{d-1},{\mathbb{K}}_{d-1},{\tilde{\mathbb{T}}}_{d-1}$ or ${\tilde{\mathbb{K}}}_{d-1}$ at the point P,

• $d=4$ and $C_4$ is a Płoski curve (see Definition 1.7).

To prove Theorem 1.15, we have to show that if $({\mathbb{P}}^2,{\mathit{\lambda }}_2{C}_d)$ is not Kawamata log terminal, then either $C_d$ is GIT-unstable or $C_d$ is an even Płoski curve. In the rest of the section, we will do this simultaneously. Let us start with few preliminary results.

Lemma 3.1

The following inequalities hold:

• (i) ${\mathit{\lambda }}_1<\frac{2}{d-1}$,
• (ii) ${\mathit{\lambda }}_1<\frac{2k+1}{kd}$ for every positive integer $k⩽d-3$,
• (iii)if $d⩾5$, then ${\mathit{\lambda }}_1<\frac{2k+1}{kd+1}$ for every positive integer $k⩽d-4$,
• (iv) ${\mathit{\lambda }}_1<\frac{3}{d}$,
• (v) ${\mathit{\lambda }}_1<\frac{2}{d-2}$,
• (vi) ${\mathit{\lambda }}_1<\frac{6}{3d-4}$,
• (vii)if $d⩾5$, then ${\mathit{\lambda }}_1<{\mathit{\lambda }}_2$.

Proof

The equality $\frac{2}{d-1}={\mathit{\lambda }}_1+\frac{d-3}{d(d-1)(d-2)}$ implies (i). Let k be positive integer. If $k=d-2$, then ${\mathit{\lambda }}_1=\frac{2k+1}{kd}$. This implies (ii), because $\frac{2k+1}{kd}=\frac{2}{d}+\frac{1}{kd}$ is a decreasing function on k for $k⩾1$. Similarly, if $k=d-4$ and $d⩾4$, then ${\mathit{\lambda }}_1=\frac{2k+1}{kd+1}-\frac{3}{d(d-2)(d^2-4d+1)}<\frac{2k+1}{kd+1}$. This implies (iii), since $\frac{2k+1}{kd+1}=\frac{2}{d}+\frac{d-2}{d(kd+1)}$ is a decreasing function on k for $k⩾1$. The equality ${\mathit{\lambda }}_1=\frac{3}{d}-\frac{d-3}{d(d-2)}$ proves (iv). Note that (v) follows from (i). Since $\frac{6}{3d-4}>\frac{2}{d-1}$, (vi) also follows from (i). Finally, the equality ${\mathit{\lambda }}_1={\mathit{\lambda }}_2-\frac{d-4}{2d(d-2)}$ implies (vii). $\square$

We may assume that $P=[0:0:1]$. Then $C_d$ is given by $F_d(x,y,z)=0$, where $F_d(x,y,z)$ is a homogeneous polynomial of degree d. Put $x_1=\frac{x}{z},x_2=\frac{y}{z}$ and $f_d(x_1,x_2)=F_d(x_1,x_2,1)$. Put $m_0={\mathrm{mult}}_P(C_d)$. Then

$$\begin{array}{c}\hfill f_d(x_1,x_2)=\sum _{\begin{array}{c}i⩾0,j⩾0,\\ m_0⩽i+j⩽d\end{array}}{\mathit{ϵ}}_{ij}{x}_1^i{x}_2^j,\end{array}$$

where each ${\mathit{ϵ}}_{ij}$ is a complex number. For every positive integers a and b, define the weight of the polynomial $f_d(x_1,x_2)$ as

$$\begin{array}{c}\hfill {\mathrm{wt}}_{(a,b)}(f_d(x_1,x_2))=min\{ai+bj|{\mathit{ϵ}}_{ij}\ne 0\}.\end{array}$$

Then the Hilbert–Mumford criterion implies

Lemma 3.2

([7, Lemma 2.1]) Let a and b be positive integers. If $C_d$ is GIT-stable, then

$$\begin{array}{c}\hfill {\mathrm{wt}}_{(a,b)}(f_d(x_1,x_2))<\frac{d}{3}(a+b).\end{array}$$

Similarly, if $C_d$ is GIT-semistable, then ${\mathrm{wt}}_{(a,b)}(f_d(x_1,x_2))⩽\frac{d}{3}(a+b)$.

Let $f_1:S_1\to {\mathbb{P}}^2$ be a blow up of the point P. Denote by $E_1$ the exceptional curve of the blow up $f_1$. Denote by $C_d^1$ the proper transform on $S_1$ of the curve $C_d$.

Lemma 3.3

If ${\mathrm{mult}}_P(C_d)>\frac{2d}{3}$, then $C_d$ is GIT-unstable. Let O be a point in $E_1$. If

$$\begin{array}{c}\hfill {\mathrm{mult}}_P(C_d)+{\mathrm{mult}}_O(C_d^1)>d,\end{array}$$

then $C_d$ is GIT-unstable.

Proof

Since ${\mathrm{mult}}_P(C_d)={\mathrm{wt}}_{(1,1)}(f_d(x_1,x_2))$, the first assertion follows from Lemma 3.2. Let us prove the second assertion. We may assume that O is contained in the proper transform of the line in ${\mathbb{P}}^2$ that is given by $x=0$. Then

$$\begin{array}{c}\hfill {\mathrm{wt}}_{(2,1)}(f_d(x_1,x_2))={\mathrm{mult}}_P(C_d)+{\mathrm{mult}}_O(C_d^1),\end{array}$$

so that the second assertion also follows from Lemma 3.2. $\square$

Now we are ready to prove Theorems 1.10 and 1.15. To do this, we may assume that $C_d$ is not a union of d lines passing through the point P. Suppose, in addition, that

(A)

either $({\mathbb{P}}^2,{\mathit{\lambda }}_1{C}_d)$ is not Kawamata log terminal at P,

(B)

or $({\mathbb{P}}^2,{\mathit{\lambda }}_2{C}_d)$ is not Kawamata log terminal at P.

We will show that (A) implies that either $C_d$ has singularity ${\mathbb{T}}_{d-1},{\mathbb{K}}_{d-1},{\tilde{\mathbb{T}}}_{d-1}$ or ${\tilde{\mathbb{K}}}_{d-1}$ at the point P, or $C_d$ is a Płoski quartic curve. Similarly, we will show that (B) implies that either $C_d$ is GIT-unstable (i.e. $C_d$ is not GIT-semistable), or $C_d$ is an even Płoski curve. If (A) holds, let $\mathit{\lambda }={\mathit{\lambda }}_1$. If (B) holds, let $\mathit{\lambda }={\mathit{\lambda }}_2$.

If $d=4$, then ${\mathit{\lambda }}_1={\mathit{\lambda }}_2$. If $d⩾5$, then ${\mathit{\lambda }}_1<{\mathit{\lambda }}_2$ by Lemma 3.1(vii). Since $C_d$ is reduced and $\mathit{\lambda }<1$, the log pair $({\mathbb{P}}^2,\mathit{\lambda }{C}_d)$ is Kawamata log terminal outside of finitely many points. Thus, it is Kawamata log terminal outside of P by Lemma 2.8.

Then the log pair $(S_1,\mathit{\lambda }{C}_d^1+(\mathit{\lambda }{m}_0-1)E_1)$ is not Kawamata log terminal at some point $P_1\in E_1$ by Remark 2.10. Note that we have

$$\begin{array}{c}\hfill K_{S_1}+\mathit{\lambda }{C}_d^1+(\mathit{\lambda }{m}_0-1)E_1{\sim }_{\mathbb{Q}}{f}_1^{\ast }(K_{{\mathbb{P}}^2}+\mathit{\lambda }{C}_d).\end{array}$$

Let $f_2:S_2\to S_1$ be a blow up of the point $P_1$, and let $E_2$ be its exceptional curve. Denote by $C_d^2$ the proper transform on $S_2$ of the curve $C_d$, and denote by $E_1^2$ the proper transform on $S_2$ of the curve $E_1$. Put $m_1={\mathrm{mult}}_{P_1}(C_d^1)$. Then

$$\begin{array}{cc}& K_{S_2}+\mathit{\lambda }{C}_d^2+(\mathit{\lambda }{m}_0-1)E_1^2+(\mathit{\lambda }(m_0+m_1)-2)E_2{\sim }_{\mathbb{Q}}{f}_2^{\ast }(K_{S_1}+\mathit{\lambda }{C}_d^1\hfill \\ \hfill & +(\mathit{\lambda }{m}_0-1)E_1).\hfill \end{array}$$

By Remark 2.10, the log pair $(S_2,\mathit{\lambda }{C}_d^2+(\mathit{\lambda }{m}_0-1)E_1^2+(\mathit{\lambda }(m_0+m_1)-2)E_2)$ is not Kawamata log terminal at some point $P_2\in E_2$. Let $f_3:S_3\to S_2$ be a blow up of this point, and let $E_3$ be the $f_3$-exceptional curve. Denote by $C_d^3$ the proper transform on $S_3$ of the curve $C_d$, denote by $E_1^3$ the proper transform on $S_3$ of the curve $E_1$, and denote by $E_2^3$ the proper transform on $S_3$ of the curve $E_2$. Put $m_2={\mathrm{mult}}_{P_2}(C_d^2)$. Then

$$\begin{array}{cc}& K_{S_3}+{\mathit{\lambda }}_2{C}_d^3+({\mathit{\lambda }}_2{m}_0-1)E_1^3\hfill \\ \hfill & +({\mathit{\lambda }}_2(m_0+m_1)-2)E_2^3+({\mathit{\lambda }}_2(2m_0+m_1+m_2)-4)E_3{\sim }_{\mathbb{Q}}\hfill \\ \hfill & {\sim }_{\mathbb{Q}}{f}_3^{\ast }(K_{S_2}+{\mathit{\lambda }}_2{C}_d^2+({\mathit{\lambda }}_2{m}_0-1)E_1^2+({\mathit{\lambda }}_2(m_0+m_1)-2)E_2).\hfill \end{array}$$

Thus, the log pair $(S_3,{\mathit{\lambda }}_2{C}_d^3+({\mathit{\lambda }}_2{m}_0-1)E_1^3+({\mathit{\lambda }}_2(m_0+m_1)-2)E_2^3+({\mathit{\lambda }}_2(2m_0+m_1+m_2)-4)E_3)$ is not Kawamata log terminal at some point $P_3\in E_3$ by Remark 2.10. Note that the divisor ${\mathit{\lambda }}_2{C}_d^3+({\mathit{\lambda }}_2{m}_0-1)E_1^3+({\mathit{\lambda }}_2(m_0+m_1)-2)E_2^3+({\mathit{\lambda }}_2(2m_0+m_1+m_2)-4)E_3$ is effective by Lemma 2.5.

Lemma 3.4

One has $\mathit{\lambda }{m}_0<2$.

Proof

Since $C_d$ is not a union of d lines passing through P, we have $m_0⩽d-1$. Thus, if (A) holds, then $\mathit{\lambda }{m}_0<2$ by Lemma 3.1(i), because $d⩾4$. Similarly, if (B) holds, then $m_0⩽\frac{2d}{3}$ by Lemma 3.3, which implies that $\mathit{\lambda }{m}_0⩽\frac{10}{6}<2$. $\square$

Thus, the log pair $(S_1,\mathit{\lambda }{C}_d^1+(\mathit{\lambda }{m}_0-1)E_1)$ is Kawamata log terminal outside of $P_1$ by Lemma 2.13. Note that $P_1\in C_d^1$, because the log pair $(S_1,(\mathit{\lambda }{m}_0-1)E_1)$ is Kawamata log terminal at $P_1$. Thus, we have $m_1>0$.

Let L be the line in ${\mathbb{P}}^2$ whose proper transform on $S_1$ contains the point $P_1$. Such a line exists and it is unique. By a suitable linear change of coordinates, we may assume that L is given by $x=0$. Denote by $L^1$ the proper transform of the line L on the surface $S_1$.

Lemma 3.5

Suppose that (A) holds and $m_0=d-1$. Then $C_d$ has singularity ${\mathbb{K}}_{d-1},{\tilde{\mathbb{K}}}_{d-1},{\mathbb{T}}_{d-1}$ or ${\tilde{\mathbb{T}}}_{d-1}$ at the point P.

Proof

Suppose that L is not an irreducible component of the curve $C_d$. Then $m_0+m_1⩽d$, because

$$\begin{array}{c}\hfill d-1-m_0=C_d^1·L^1⩾m_1.\end{array}$$

Since $m_0=d-1$, this gives $m_1=1$. Then $P_1\in C_d^1$ and the curve $C_d^1$ is smooth at $P_1$. Put $k={\mathrm{mult}}_{P_1}(C_d^1·E_1)$. Applying Corollary 2.3 to the log pair $(S_1,{\mathit{\lambda }}_1{C}_d^1+({\mathit{\lambda }}_1{m}_0-1)E_1)$ at the point $P_1$, we get

$$\begin{array}{c}\hfill k{\mathit{\lambda }}_1{m}_0⩾k+1,\end{array}$$

which gives ${\mathit{\lambda }}_1⩾\frac{2k+1}{kd}$. Then $k⩾d-2$ by Lemma 3.1(ii). Since

$$\begin{array}{c}\hfill k⩽C_d^1·E_1=m_0=d-1,\end{array}$$

either $k=d-1$ or $k=d-2$. If $k=d-1$, then $C_d$ has singularity ${\mathbb{K}}_{d-1}$ at P. If $k=d-2$, then $C_d$ has singularity ${\tilde{\mathbb{K}}}_{d-1}$ at the point P.

To complete the proof, we may assume that L is an irreducible component of the curve $C_d$. Then $C_d=L+C_{d-1}$, where $C_{d-1}$ is a reduced curve in ${\mathbb{P}}^2$ of degree $d-1$ such that L is not its irreducible component. Denote by $C_{d-1}^1$ its proper transform on $S_1$. Put $n_0={\mathrm{mult}}_P(C_{d-1})$ and $n_1={\mathrm{mult}}_{P_1}(C_{d-1}^1)$. Then $n_0=m_0-1=d-2$ and $n_1=m_1-1$. This implies that $P_1\in C_{d-1}^1$, since the log pair $(S_1,{\mathit{\lambda }}_1{L}^1+({\mathit{\lambda }}_1{m}_0-1)E_1)$ is Kawamata log terminal at P. Hence, $n_1⩾1$. One the other hand, we have

$$\begin{array}{c}\hfill d-1-n_0=C_{d-1}^1·L^1⩾n_1,\end{array}$$

which implies that $n_0+n_1⩽d-1$. Then $n_1=1$, since $n_0=d-2$.

We have $P_1\in C_{d-1}^1$ and $C_{d-1}^1$ is smooth at $P_1$. Moreover, since

$$\begin{array}{c}\hfill 1=d-1-n_0=L^1·C_{d-1}^1⩾n_1=1,\end{array}$$

the curve $C_{d-1}^1$ intersects the curve $L^1$ transversally at the point $P_1$. Put $k={\mathrm{mult}}_{P_1}(C_{d-1}^1·E_1)$. Then $k⩾1$. Applying Lemma 2.2 to the log pair $(S_1,{\mathit{\lambda }}_1{C}_{d-1}^1+{\mathit{\lambda }}_1{L}^1+({\mathit{\lambda }}_1(n_0+1)-1)E_1)$ at the point $P_1$, we get

$$\begin{array}{c}\hfill k({\mathit{\lambda }}_1(n_0+2)-1)+{\mathit{\lambda }}_1⩾k+1.\end{array}$$

Then ${\mathit{\lambda }}_1⩾\frac{2k+1}{kd+1}$. Then $k⩾d-3$ by Lemma 3.1(iii). Since

$$\begin{array}{c}\hfill k⩽E_1·C_{d-1}^1=n_0=d-2,\end{array}$$

either $k=d-2$ or $k=d-3$. In the former case, $C_d$ has singularity ${\mathbb{T}}_{d-1}$ at the point P. In the latter case, $C_d$ has singularity ${\tilde{\mathbb{T}}}_{d-1}$ at the point P. $\square$

Lemma 3.6

Suppose that (A) holds and $m_0⩽d-2$. Then the line L is not an irreducible component of the curve $C_d$.

Proof

Suppose that L is an irreducible component of the curve $C_d$. Let us see for a contradiction. Put $C_d=L+C_{d-1}$, where $C_{d-1}$ is a reduced curve in ${\mathbb{P}}^2$ of degree $d-1$ such that L is not its irreducible component. Denote by $C_{d-1}^1$ its proper transform on $S_1$. Put $n_0={\mathrm{mult}}_P(C_{d-1})$ and $n_1={\mathrm{mult}}_{P_1}(C_{d-1}^1)$. Then $(S_1,({\mathit{\lambda }}_1(n_0+1)-1)E_1+{\mathit{\lambda }}_1{L}^1+{\mathit{\lambda }}_1{C}_{d-1}^1)$ is not Kawamata log terminal at $P_1$ and is Kawamata log terminal outside of the point $P_1$. In particular, $n_1\ne 0$, because $(S_1,({\mathit{\lambda }}_1(n_0+1)-1)E_1+{\mathit{\lambda }}_1{L}^1)$ is Kawamata log terminal at $P_1$. On the other hand,

$$\begin{array}{c}\hfill d-1-n_0=L^1·C_{d-1}^1⩾n_1,\end{array}$$

which implies that $n_0+n_1⩽d-1$. Furthermore, we have $n_0=m_0-1⩽d-3$.

Since $n_0+n_1⩾2n_1$, we have $n_1⩽\frac{d-1}{2}$. Then $\mathit{\lambda }{n}_1<1$ by Lemma 3.1(i). Thus, we can apply Theorem 2.16 to the log pair $(S_1,({\mathit{\lambda }}_1(n_0+1)-1)E_1+{\mathit{\lambda }}_1{L}^1+{\mathit{\lambda }}_1{C}_{d-1}^1)$ at the point $P_1$. This gives either

$$\begin{array}{c}\hfill {\mathit{\lambda }}_1(d-1-n_0)={\mathit{\lambda }}_1{C}_{d-1}^1·L^1⩾2(2-{\mathit{\lambda }}_1(n_0+1))\end{array}$$

or

$$\begin{array}{c}\hfill {\mathit{\lambda }}_1{n}_0={\mathit{\lambda }}_1{C}_{d-1}^1·E_1⩾2(1-{\mathit{\lambda }}_1)\end{array}$$

(or both). In the former case, we have ${\mathit{\lambda }}_1(d+1+n_0)⩾4$. In the latter case, we have ${\mathit{\lambda }}_1(n_0+2)>2$. Thus, in both cases we have ${\mathit{\lambda }}_1(d-1)⩾2$, since $n_0⩽d-3$. But ${\mathit{\lambda }}_1(d-1)<2$ by Lemma 3.1(i). This is a contradiction. $\square$

If the curve $C_d$ is GIT-semistable, then $m_0⩽d-2$ by Lemma 3.3. Thus, it follows from Lemma 3.5 that we may assume that

$$\begin{array}{c}\hfill m_0⩽d-2\end{array}$$

in order to complete the proof of Theorems 1.10 and 1.15. Moreover, if L is not an irreducible component of the curve $C_d$, then

$$\begin{array}{c}\hfill d-m_0=C_d^1·L^1⩾m_1.\end{array}$$

Thus, if (A) holds, then $m_0+m_1⩽d$ by Lemma 3.6. Similarly, if the curve $C_d$ is GIT-semistable, then $m_0+m_1⩽d$ by Lemma 3.3. Thus, to complete the proof of Theorems 1.10 and 1.15, we may also assume that

$$\begin{array}{c}\hfill m_0+m_1⩽d.\end{array}$$ 3.1

Then $\mathit{\lambda }(m_0+m_1)<3$ by Lemma 3.1(v), so that $(S_2,\mathit{\lambda }{C}_d^2+(\mathit{\lambda }{m}_0-1)E_1^2+(\mathit{\lambda }(m_0+m_1)-2)E_2)$ is Kawamata log terminal outside of the point $P_2$ by Lemma 2.13. Furthermore, we have

Lemma 3.7

Suppose that $P_2=E_1^2\cap E_2$. Then (A) does not hold and $C_d$ is GIT-unstable.

Proof

We have $m_0-m_1=E_1^2·C_d^2⩾m_2$, so that

$$\begin{array}{c}\hfill m_2⩽\frac{m_0}{2},\end{array}$$ 3.2

because $2m_2⩽m_1+m_2$. On the other hand, $m_0⩽d-2$ by assumption. Thus, we have $m_2⩽\frac{d-2}{2}$.

Suppose that (A) holds. Then $\mathit{\lambda }={\mathit{\lambda }}_1$ and ${\mathit{\lambda }}_1{m}_2<1$ by Lemma 3.1(v). Thus, we can apply Theorem 2.16 to the log pair $(S_2,{\mathit{\lambda }}_1{C}_d^2+({\mathit{\lambda }}_1{m}_0-1)E_1^2+({\mathit{\lambda }}_1(m_0+m_1)-2)E_2)$. This gives either

$$\begin{array}{c}\hfill {\mathit{\lambda }}_1(m_0-m_1)={\mathit{\lambda }}_1{C}_d^2·E_1^2⩾2(3-{\mathit{\lambda }}_1(m_0+m_1))\end{array}$$

or

$$\begin{array}{c}\hfill {\mathit{\lambda }}_1{m}_1={\mathit{\lambda }}_1{C}_d^2·E_2⩾2(2-{\mathit{\lambda }}_1{m}_0)\end{array}$$

(or both). The former inequality implies ${\mathit{\lambda }}_1(3m_0+m_1)⩾6$. The latter inequality implies ${\mathit{\lambda }}_1(2m_0+m_1)⩾4$. On the other hand, $m_0+m_1⩽d$ by (3.1), and $m_0⩽d-2$ by assumption. Thus, $3m_0+m_1⩽3d-4$ and $2m_0+m_1⩽2d-2$. Then ${\mathit{\lambda }}_1(3m_0+m_1)<6$ by Lemma 3.1(vi), and ${\mathit{\lambda }}_1(2m_0+m_1)<4$ by Lemma 3.1(i). The obtained contradiction shows that (A) does not hold.

We see that (B) holds. We have to show that $C_d$ is GIT-unstable. Suppose that this is not the case, so that $C_d$ is GIT-semistable. Let us seek for a contradiction.

By Lemma 3.2, we have $2m_0+m_1+m_2⩽\frac{5d}{3}$, because

$$\begin{array}{c}\hfill {\mathrm{wt}}_{(3,2)}(f_d(x_1,x_2))=2m_0+m_1+m_2.\end{array}$$

Thus, we have ${\mathit{\lambda }}_2(2m_0+m_1+m_2)-4<1$ by Lemma 3.1(v). Hence, the log pair $(S_3,{\mathit{\lambda }}_2{C}_d^3+({\mathit{\lambda }}_2{m}_0-1)E_1^3+({\mathit{\lambda }}_2(m_0+m_1)-2)E_2^3+({\mathit{\lambda }}_2(2m_0+m_1+m_2)-4)E_3)$ is Kawamata log terminal outside of the point $P_3$ by Remark 2.10.

If $P_3=E_1^3\cap E_3$, then it follows from Theorem 2.12 that

$$\begin{array}{c}\hfill {\mathit{\lambda }}_2(m_0-m_1-m_2)={\mathit{\lambda }}_2{C}_d^3·E_1^3>5-{\mathit{\lambda }}_2(2m_0+m_1+m_2),\end{array}$$

which implies that $m_0>\frac{5}{3{\mathit{\lambda }}_2}=\frac{2d}{3}$, which is impossible by Lemma 3.3. If $P_3=E_2^3\cap E_3$, then it follows from Theorem 2.12 that

$$\begin{array}{c}\hfill {\mathit{\lambda }}_2(m_1-m_2)={\mathit{\lambda }}_2{C}_d^3·E_2^3>5-{\mathit{\lambda }}_2(2m_0+m_1+m_2),\end{array}$$

which implies that $m_0+m_1>\frac{5}{2{\mathit{\lambda }}_2}=d$, which is impossible by Lemma 3.3. Thus, we see that $P_3\notin E_1^3\cup E_2^3$. Then the log pair $(S_3,{\mathit{\lambda }}_2{C}_d^3+({\mathit{\lambda }}_2(2m_0+m_1+m_2)-4)E_3)$ is not Kawamata log terminal at $P_3$. Hence, Theorem 2.12 gives

$$\begin{array}{c}\hfill {\mathit{\lambda }}_2{m}_2={\mathit{\lambda }}_2{C}_d^3·E_3>1,\end{array}$$

which implies that $m_2>\frac{1}{{\mathit{\lambda }}_2}=\frac{2d}{5}$. Then $m_0>\frac{4d}{5}$ by (3.2), which is impossible by Lemma 3.3. $\square$

Thus, to complete the proof of Theorems 1.10 and 1.15, we may assume that

$$\begin{array}{c}\hfill P_2\ne E_1^2\cap E_2.\end{array}$$

Denote by $L^2$ the proper transform of the line L on the surface $S_2$.

Lemma 3.8

One has $P_2\ne L^2\cap E_2$.

Proof

Suppose that $P_2=L^2\cap E_2$. If L is not an irreducible component of the curve $C_d$, then

$$\begin{array}{c}\hfill d-m_0-m_1=L^2·E_2⩾m_2,\end{array}$$

which implies that $m_0+m_1+m_2⩽d$. Thus, if (A) holds, then $\mathit{\lambda }={\mathit{\lambda }}_1$ and L is not an irreducible component of the curve $C_d$ by Lemma 3.6, which implies that

$$\begin{array}{c}\hfill {\mathit{\lambda }}_{1}d⩾{\mathit{\lambda }}_1(m_0+m_1+m_2)>3\end{array}$$

by Lemma 2.14. On the other hand, ${\mathit{\lambda }}_{1}d<3$ by Lemma 3.1(iv). This shows that (B) holds.

Since $\mathit{\lambda }={\mathit{\lambda }}_2=\frac{5}{2d}<\frac{3}{d}$ and ${\mathit{\lambda }}_2(m_0+m_1+m_2)>3$ by Lemma 2.14, we have $m_0+m_1+m_2>d$. In particular, the line L must be an irreducible component of the curve $C_d$.

Put $C_d=L+C_{d-1}$, where $C_{d-1}$ is a reduced curve in ${\mathbb{P}}^2$ of degree $d-1$ such that L is not its irreducible component. Denote by $C_{d-1}^1$ its proper transform on $S_1$, and denote by $C_{d-1}^2$ its proper transform on $S_2$. Put $n_0={\mathrm{mult}}_P(C_{d-1}),n_1={\mathrm{mult}}_{P_1}(C_{d-1}^1)$ and $n_2={\mathrm{mult}}_{P_2}(C_{d-1}^2)$. Then $(S_2,({\mathit{\lambda }}_2(n_0+n_1+2)-2)E_2+{\mathit{\lambda }}_2{L}^1+{\mathit{\lambda }}_2{C}_{d-1}^1)$ is not Kawamata log terminal at $P_2$ and is Kawamata log terminal outside of the point $P_2$. Then Theorem 2.12 implies

$$\begin{array}{cc}\hfill {\mathit{\lambda }}_2(d-1-n_0-n_1)=& {\mathit{\lambda }}_2{C}_{d-1}^2·L^2>1-({\mathit{\lambda }}_2(n_0+n_1+2)-2)\hfill \\ \hfill =& 3-{\mathit{\lambda }}_2(n_0+n_1+2),\hfill \end{array}$$

which implies that $\frac{5(d+1)}{2d}={\mathit{\lambda }}_2(d+1)>3$. Hence, $d=4$. Then $\mathit{\lambda }={\mathit{\lambda }}_2=\frac{5}{8}$.

By (3.1), $n_0+n_1⩽2$. Thus, $n_0=n_1=n_2=1$, since

$$\begin{array}{c}\hfill \frac{5}{8}(n_0+n_1+n_2+3)={\mathit{\lambda }}_2(m_0+m_1+m_2)>3\end{array}$$

by Lemma 2.14. Then $C_3$ is a irreducible cubic curve that is smooth at P, the line L is tangent to the curve $C_3$ at the point P, and P is an inflexion point of the cubic curve $C_3$. This implies that ${\mathrm{lct}}_P({\mathbb{P}}^2,C_d)=\frac{2}{3}$. Since $\frac{2}{3}>\frac{5}{8}={\mathit{\lambda }}_2$, the log pair $({\mathbb{P}}^2,{\mathit{\lambda }}_2{C}_d)$ must be Kawamata log terminal at the point P, which contradicts (B). $\square$

Recall that $m_0+m_1⩽d$ by (3.1). Then $m_1⩽\frac{d}{2}$, since $2m_1⩽m_0+m_1$. Thus, we have

$$\begin{array}{c}\hfill \mathit{\lambda }(m_0+m_1+m_2)⩽\mathit{\lambda }(m_0+2m_1)⩽\mathit{\lambda }\frac{3d}{2}⩽{\mathit{\lambda }}_2\frac{3d}{2}=\frac{15}{4}<4.\end{array}$$ 3.3

Therefore, the log pair $(S_3,\mathit{\lambda }{C}_d^3+(\mathit{\lambda }(m_0+m_1)-2)E_2^3+(\mathit{\lambda }(m_0+m_1+m_2)-3)E_3)$ is Kawamata log terminal outside of the point $P_3$ by Lemma 2.13.

Lemma 3.9

One has $P_3\ne E_2^3\cap E_3$.

Proof

If $P_3=E_2^3\cap E_3$, then Theorem 2.12 gives

$$\begin{array}{c}\hfill \mathit{\lambda }(m_1-m_2)=\mathit{\lambda }{C}_d^3·E_2^3>1-(\mathit{\lambda }(m_0+m_1+m_2)-3)=4-\mathit{\lambda }(m_0+m_1+m_2),\end{array}$$

which implies that $\mathit{\lambda }(m_0+2m_1)>4$. But $\mathit{\lambda }(m_0+2m_1)<4$ by (3.3). $\square$

Let $f_4:S_4\to S_3$ be a blow up of the point $P_3$, and let $E_4$ be its exceptional curve. Denote by $C_d^4$ the proper transform on $S_4$ of the curve $C_d$, denote by $E_3^4$ the proper transform on $S_4$ of the curve $E_3$, and denote by $L^4$ the proper transform of the line L on the surface $S_4$. Then $(S_4,\mathit{\lambda }{C}_d^4+(\mathit{\lambda }(m_0+m_1+m_2)-3)E_3^4+(\mathit{\lambda }(m_0+m_1+m_2+m_3)-4)E_4)$ is not Kawamata log terminal at some point $P_4\in E_4$ by Remark 2.10. Moreover, we have

$$\begin{array}{cc}& 2L^4+E_1^4+2E_2^4+E_3^4\sim {(f_1\circ f_2\circ f_3\circ f_4)}^{\ast }({\mathcal{O}}_{{\mathbb{P}}^2}(2))-{(f_2\circ f_3\circ f_4)}^{\ast }(E_1)\hfill \\ \hfill & -{(f_3\circ f_4)}^{\ast }(E_2)-f_4^{\ast }(E_3)-E_4.\hfill \end{array}$$

Lemma 3.10

The linear system $|2L^4+E_1^4+2E_2^4+E_3^4|$ is a pencil that does not have base points. Moreover, every divisor in $|2L^4+E_1^4+2E_2^4+E_3^4|$ that is different from $2L^4+E_1^4+2E_2^4+E_3^4$ is a smooth curve whose image on ${\mathbb{P}}^2$ is a smooth conic that is tangent to L at the point P.

Proof

All assertions follows from $P_2\notin E_1^2\cup L^2$ and $P_3\notin E_2^3$. $\square$

Let $C_2^4$ be a general curve in $|2L^4+E_1^4+2E_2^4+E_3^4|$. Denote by $C_2$ its image on ${\mathbb{P}}^2$, and denote by $\mathcal{L}$ the pencil generated by 2L and $C_2$. Then P is the only base point of the pencil $\mathcal{L}$, and every conic in $\mathcal{L}$ except 2L and $C_2$ intersects $C_2$ at P with multiplicity 4 (cf. [3, Remark 1.14]).

Lemma 3.11

One has $m_0+m_1+m_2+m_3⩽m_0+m_1+2m_2⩽\frac{5}{\mathit{\lambda }}$. If $m_0+m_1+m_2+m_3=\frac{5}{\mathit{\lambda }}$, then d is even and $C_d$ is a union of $\frac{d}{2}⩾2$ smooth conics in $\mathcal{L}$, where $d=4$ if (A) holds.

Proof

By (3.1), we have $m_2+m_3⩽2m_2⩽m_0+m_1⩽d$. This gives

$$\begin{array}{c}\hfill m_0+m_1+m_2+m_3⩽m_0+m_1+2m_2⩽2d=\frac{5}{{\mathit{\lambda }}_2}⩽\frac{5}{\mathit{\lambda }}.\end{array}$$

To complete the proof, we may assume that $m_0+m_1+m_2+m_3=\frac{5}{\mathit{\lambda }}$. Then all inequalities above must be equalities. Thus, we have $m_2=m_3=\frac{d}{2}$ and ${\mathit{\lambda }}_1={\mathit{\lambda }}_2$. In particular, if (A) holds, then $d=4$, because ${\mathit{\lambda }}_1<{\mathit{\lambda }}_2=\frac{5}{2d}$ for $d⩾5$ by Lemma 3.1(vii). Moreover, since $m_0⩾m_1⩾m_2=\frac{d}{2}$ and $m_0+m_1⩽d$, we see that $m_0=m_1=\frac{d}{2}$. Thus, d is even and

$$\begin{array}{c}\hfill C_d^4\sim \frac{d}{2}(2L^4+E_1^4+2E_2^4+E_3^4),\end{array}$$

where $d=4$ if (A) holds. Since $|2L^4+E_1^4+2E_2^4+E_3^4|$ is a free pencil and $C_d^4$ is reduced, it follows from Lemma 3.10 that $C_d^4$ is a union of $\frac{d}{2}$ smooth curves in $|2L^4+E_1^4+2E_2^4+E_3^4|$. In particular, $L^4$ is not an irreducible component of $C_d^4$. Thus, the curve $C_d$ is a union of $\frac{d}{2}$ smooth conics in $\mathcal{L}$, where $d=4$ if (A) holds. $\square$

We see that $m_0+m_1+m_2+m_3⩽\frac{5}{\mathit{\lambda }}$. Moreover, if $m_0+m_1+m_2+m_3=\frac{5}{\mathit{\lambda }}$, then $C_d$ is an even Płoski curve. Furthermore, if $m_0+m_1+m_2+m_3=\frac{5}{\mathit{\lambda }}$ and (A) holds, then $d=4$. Thus, to prove Theorems 1.10 and 1.15, we may assume that

$$\begin{array}{c}\hfill m_0+m_1+m_2+m_3<\frac{5}{\mathit{\lambda }}.\end{array}$$

Let us show that this assumption leads to a contradiction. By Lemma 2.13, this inequality implies that the log pair $(S_4,\mathit{\lambda }{C}_d^4+(\mathit{\lambda }(m_0+m_1+m_2)-3)E_3^4+(\mathit{\lambda }(m_0+m_1+m_2+m_3)-4)E_4)$ is Kawamata log terminal outside of the point $P_4$.

Lemma 3.12

One has $P_4\ne E_3^4\cap E_4$.

Proof

By Lemma 3.11, $m_0+m_1+2m_2⩽\frac{5}{\mathit{\lambda }}$. If $P_4=E_3^4\cap E_4$, then Theorem 2.12 gives

$$\begin{array}{c}\hfill \mathit{\lambda }(m_2-m_3)=\mathit{\lambda }{C}_d^4·E_3^4>5-\mathit{\lambda }(m_0+m_1+m_2+m_3),\end{array}$$

which implies that $m_0+m_1+2m_2>\frac{5}{\mathit{\lambda }}$. This shows that $P_4\ne E_3^4\cap E_4$. $\square$

Thus, the log pair $(S_4,\mathit{\lambda }{C}_d^4+(\mathit{\lambda }(m_0+m_1+m_2+m_3)-4)E_4)$ is not Kawamata log terminal at $P_4$ and is Kawamata log terminal outside of the point $P_4$.

Let $Z^4$ be the curve in $|2L^4+E_1^4+2E_2^4+E_3^4|$ that passes through the point $P_4$. Then $Z^4$ is a smooth irreducible curve by Lemma 3.8. Denote by Z the proper transform of this curve on ${\mathbb{P}}^2$. Then Z is a smooth conic in the pencil $\mathcal{L}$ by Lemma 3.10. If Z is not an irreducible component of the curve $C_d$, then

$$\begin{array}{c}\hfill 2d-(m_0+m_1+m_2+m_3)=Z^4·C_d^4⩾{\mathrm{mult}}_{P_4}(C_d^4).\end{array}$$

On the other hand, it follows from Lemma 2.14 that

$$\begin{array}{c}\hfill {\mathrm{mult}}_{P_4}(C_d^4)+m_0+m_1+m_2+m_3>\frac{5}{\mathit{\lambda }}.\end{array}$$

This shows that Z is an irreducible component of the curve $C_d$, since $\mathit{\lambda }⩽{\mathit{\lambda }}_2=\frac{5}{2d}$.

Put $C_d=Z+C_{d-2}$, where $C_{d-2}$ is a reduced curve in ${\mathbb{P}}^2$ of degree $d-2$ such that Z is not its irreducible component. Denote by $C_{d-2}^1,C_{d-2}^2,C_{d-2}^3$ and $C_{d-2}^4$ its proper transforms on the surfaces $S_1,S_2,S_3$ and $S_4$, respectively. Put $n_0={\mathrm{mult}}_P(C_{d-2}),n_1={\mathrm{mult}}_{P_1}(C_{d-2}^1),n_2={\mathrm{mult}}_{P_2}(C_{d-2}^2),n_3={\mathrm{mult}}_{P_3}(C_{d-2}^3)$ and $n_4={\mathrm{mult}}_{P_4}(C_{d-2}^4)$. Then

$$\begin{array}{c}\hfill (S_4,\mathit{\lambda }{C}_{d-2}^4+\mathit{\lambda }{Z}^4+(\mathit{\lambda }(n_0+n_1+n_2+n_3+4)-4)E_4)\end{array}$$

is not Kawamata log terminal at $P_4$ and is Kawamata log terminal outside of the point $P_4$. Thus, applying Theorem 2.12, we get

$$\begin{array}{c}\hfill \mathit{\lambda }(2(d-2)-n_0-n_1-n_2-n_3)=\mathit{\lambda }{C}_{d-2}^4·Z^4>5-\mathit{\lambda }(n_0+n_1+n_2+n_3+4),\end{array}$$

which implies that $\mathit{\lambda }>\frac{5}{2d}$. This is impossible, since $\mathit{\lambda }⩽{\mathit{\lambda }}_2=\frac{5}{2d}$.

The obtained contradiction completes the proof of Theorems 1.10 and 1.15.

Smooth Surfaces in ${\mathbb{P}}^3$

The purpose of this section is to prove Theorem 1.17. Let S be a smooth surface in ${\mathbb{P}}^3$ of degree $d⩾3$, let $H_S$ be its hyperplane section, let P be a point in S, and let $T_P$ be the hyperplane section of the surface S that is singular at P. Note that $T_P$ is reduced by Lemma 2.6. Put $\mathit{\lambda }=\frac{2d-3}{d(d-2)}$. Then Theorem 1.17 follows from Theorem 1.10, Remark 2.4 and

Proposition 4.1

Let D be any effective $\mathbb{Q}$-divisor on S such that $D{\sim }_{\mathbb{Q}}{H}_S$. Suppose that $\mathrm{Supp}(D)$ does not contain at least one irreducible component of the curve $T_P$. Then $(S,\mathit{\lambda }D)$ is log canonical at P.

For $d=3$, this result is just [3, Corollary 1.13]. In the remaining part of the section, we will prove Proposition 4.1. Note that we will do this without using [3, Corollary 1.13]. Let us start with

Lemma 4.2

The following assertions hold:

• (i) $\mathit{\lambda }⩽\frac{2}{d-1}$,
• (ii)if $d⩾5$, then $\mathit{\lambda }⩽\frac{3}{d+1}$,
• (iii)if $d⩾5$, then $\mathit{\lambda }⩽\frac{4}{d+3}$,
• (iv)If $d⩾6$, then $\mathit{\lambda }⩽\frac{3}{d+2}$,
• (v) $\mathit{\lambda }⩽\frac{4}{d+1}$,
• (vi) $\mathit{\lambda }⩽\frac{3}{d}$.

Proof

The equality $\frac{2}{d-1}=\mathit{\lambda }+\frac{d-3}{d(d-1)(d-2)}$ implies (i), $\frac{4}{d+1}=\mathit{\lambda }+\frac{d^2-5d+3}{d(d+1)(d-2)}$ implies (ii), and $\frac{4}{d+3}=\mathit{\lambda }+\frac{2d^2-11d+9}{d(d+3)(d-2)}$ implies (iii). Similarly, (iv) follows from $\frac{3}{d+2}=\mathit{\lambda }+\frac{d^2-7d+6}{d(d^2-4)}$, (v) follows from $\frac{4}{d+1}=\mathit{\lambda }+\frac{2d^2-7d+3}{d(d+1)(d-2)}$, and (vi) follows from $\frac{3}{d}=\mathit{\lambda }+\frac{d-3}{d(d-2)}$. $\square$

Let n be the number of irreducible components of the curve $T_P$. Write

$$\begin{array}{c}\hfill T_P=T_1+\cdots +T_n,\end{array}$$

where each $T_i$ is an irreducible curve on the surface S. For every curve $T_i$, we denote its degree by $d_i$, and we put $t_i={\mathrm{mult}}_P(T_i)$.

Lemma 4.3

Suppose that $n⩾2$. Then

$$\begin{array}{c}\hfill T_i·T_i=-d_i(d-d_i-1)\end{array}$$

for every $T_i$, and $T_i·T_j=d_i{d}_j$ for every $T_i$ and $T_j$ such that $T_i\ne T_j$.

Proof

The curve $T_P$ is cut out on S by a hyperplane $H\subset {\mathbb{P}}^3$. Then $H\cong {\mathbb{P}}^2$. Hence, for every $T_i$ and $T_j$ such that $T_i\ne T_j$, we have ${(T_i·T_j)}_S={(T_i·T_j)}_H=d_i{d}_j$. In particular, we have

$$\begin{array}{c}\hfill d_1=T_P·T_1=T_1^2+\sum _{i=2}^n{T}_i·T_1=T_1^2+\sum _{i=2}^n{d}_i{d}_1=T_1^2+(d-d_1)d_1,\end{array}$$

which gives $T_1·T_1=-d_1(d-d_1-1)$. Similarly, we see that $T_i·T_i=-d_i(d-d_i-1)$ for every curve $T_i$. $\square$

Let D be any effective $\mathbb{Q}$-divisor on S such that $D{\sim }_{\mathbb{Q}}{H}_S$. Write

$$\begin{array}{c}\hfill D=\sum _{i=1}^n{a}_i{T}_i+\mathrm{\Delta },\end{array}$$

where each $a_i$ is a non-negative rational number, and $\mathrm{\Delta }$ is an effective $\mathbb{Q}$-divisor on S whose support does not contain the curves $T_1,\dots ,T_n$. To prove Proposition 4.1, it is enough to show that the log pair $(S,\mathit{\lambda }D)$ is log canonical at P provided that at least one number among $a_1,\dots ,a_n$ vanishes.

Without loss of generality, we may assume that $a_n=0$. Suppose that the log pair $(S,\mathit{\lambda }D)$ is not log canonical at P. Let us seek for a contradiction.

Lemma 4.4

Suppose that $n⩾2$. Then

$$\begin{array}{c}\hfill \sum _{i=1}^k{a}_i{d}_i{d}_n⩽d_n-t_n{\mathrm{mult}}_P(\mathrm{\Delta }).\end{array}$$

In particular, ${\sum }_{i=1}^k{a}_i{d}_i⩽1$ and each $a_i$ does not exceed $\frac{1}{d_i}$.

Proof

One has

$$\begin{array}{cc}\hfill d_n=& T_n·D=T_n·(\sum _{i=1}^n{a}_i{T}_i+\mathrm{\Delta })=\sum _{i=1}^n{a}_i{d}_i{d}_n+T_n·\mathrm{\Delta }\hfill \\ \hfill ⩾& \sum _{i=1}^n{a}_i{d}_i{d}_n+t_n{\mathrm{mult}}_P(\mathrm{\Delta }),\hfill \end{array}$$

which implies the required inequality. $\square$

Put $m_0={\mathrm{mult}}_P(D)$.

Lemma 4.5

Suppose that $P\in T_n$. Then $d_n>\frac{d-1}{2}$. If $n⩾2$, then $T_n$ is smooth at P.

Proof

Since $T_n$ is not contained in the support of the divisor D, we have

$$\begin{array}{c}\hfill d⩾d_n=T_n·D⩾t_n{m}_0,\end{array}$$

which implies that $m_0⩽\frac{d_n}{t_n}$. Since $m_0>\frac{1}{\mathit{\lambda }}$ by Lemma 2.5, we have $d_n>\frac{d-1}{2}$ by Lemma 4.2(i). Moreover, if $n⩾2$ and $t_n⩾2$, then it follows from Lemma 2.5 that

$$\begin{array}{c}\hfill \frac{1}{\mathit{\lambda }}<m_0⩽\frac{d_n}{t_n}⩽\frac{d-1}{t_n}⩽\frac{d-1}{2},\end{array}$$

which is impossible by Lemma 4.2(i). $\square$

Now we are going to use Theorem 2.15 to prove

Lemma 4.6

Suppose that $n⩾3$ and P is contained in at least two irreducible components of the curve $T_P$ that are different from $T_n$ and that are both smooth at P. Then they are tangent to each other at P.

Proof

Without loss of generality, we may assume that $P\in T_1\cap T_2$ and $t_1=t_2=1$. Suppose that $T_1$ and $T_2$ are not tangent to each other at P. Put $\mathrm{\Omega }={\sum }_{i=3}^n{a}_i{T}_i+\mathrm{\Delta }$, so that $D=a_1{T}_1+a_2{T}_2+\mathrm{\Omega }$. Then $a_1{d}_1+a_2{d}_2⩽1$ by Lemma 4.4.

Put $k_0=\mathrm{mult}(\mathrm{\Omega })$. Then

$$\begin{array}{c}\hfill d_1+a_1{d}_1(d-d_1-1)-a_2{d}_1{d}_2=\mathrm{\Omega }·T_1⩾k_0\end{array}$$

by Lemma 4.3. Similarly, we have

$$\begin{array}{c}\hfill d_2-a_1{d}_1{d}_2+a_2{d}_2(d-d_2-1)=\mathrm{\Omega }·T_2⩾k_0.\end{array}$$

Adding these two inequalities together and using $a_1{d}_1+a_2{d}_2⩽1$, we get

$$\begin{array}{cc}\hfill 2k_0⩽& d_1+d_2+(a_1{d}_1+a_2{d}_2)(d-d_1-d_2-1)\hfill \\ \hfill ⩽& d_1+d_2+(d-d_1-d_2-1)=d-1.\hfill \end{array}$$

Thus, $k_0⩽\frac{1}{\mathit{\lambda }}$ by Lemma 4.2(i).

Since $\mathit{\lambda }{k}_0⩽1$, we can apply Theorem 2.15 to the log pair $(S,\mathit{\lambda }{a}_1{T}_1+\mathit{\lambda }{a}_2{T}_2+\mathit{\lambda }\mathrm{\Omega })$ at the point P. This gives either $\mathit{\lambda }\mathrm{\Omega }·T_1>2(1-\mathit{\lambda }{a}_2)$ or $\mathit{\lambda }\mathrm{\Omega }·T_2>2(1-\mathit{\lambda }{a}_1)$. Without loss of generality, we may assume that $\mathit{\lambda }\mathrm{\Omega }·T_2>2(1-\mathit{\lambda }{a}_1)$. Then

$$\begin{array}{c}\hfill d_2+a_2{d}_2(d-d_2-1)-a_1{d}_1{d}_2=\mathrm{\Omega }·T_2>\frac{2}{\mathit{\lambda }}-2a_1.\end{array}$$ 4.1

Applying Theorem 2.12 to the log pair $(S,\mathit{\lambda }{a}_1{T}_1+\mathit{\lambda }{b}_1{T}_2+\mathit{\lambda }\mathrm{\Omega })$ and the curve $T_1$ at the point P, we get

$$\begin{array}{c}\hfill d_1+a_1{d}_1(d-d_1-1)=(\mathit{\lambda }{a}_2{T}_2+\mathit{\lambda }\mathrm{\Omega })·T_1>\frac{1}{\mathit{\lambda }}.\end{array}$$

Adding this inequality to (4.1), we get

$$\begin{array}{c}\hfill d+1⩾d-1+2a_1⩾d_1+d_2+(a_1{d}_1+a_2{d}_2)(d-d_1-d_2-1)+2a_1>\frac{3}{\mathit{\lambda }},\end{array}$$

because $a_1{d}_1+a_2{d}_2⩽1$. Thus, it follows from Lemma 4.2(ii) that either $d=3$ or $d=4$.

If $d=3$, then $n=3$ and $d_1=d_2=d_3=\mathit{\lambda }=1$, which implies that $a_1+a_2>1$ by (4.1). On the other hand, we know that $a_1{d}_1+a_2{d}_2⩽1$, so that $a_1+a_2⩽1$. This shows that $d\ne 3$.

We see that $d=4$. Then $\mathit{\lambda }=\frac{5}{8}$ and $d_1+d_2⩽3$. If $d_1=d_1=1$, then (4.1) gives $2a_2+a_1>\frac{11}{5}$. If $d_1=1$ and $d_2=2$, then (4.1) gives $a_2>\frac{3}{5}$. If $d_1=2$ and $d_2=1$, then (4.1) gives $a_2>\frac{11}{5}$. All these three inequalities are inconsistent, because $a_1{d}_1+a_2{d}_2⩽1$. The obtained contradiction completes the proof of the lemma. $\square$

Note that every line contained in the surfaces S that passes through P must be an irreducible component of the curve $T_P$. Moreover, the curve $T_n$ cannot be a line by Lemma 4.5. Thus, Lemma 4.6 implies that there exists at most one line in S that passes through P. In particular, we see that $n<d$.

Lemma 4.7

Suppose that $n⩾3$ and P is contained in at least two irreducible components of the curve $T_P$ that are different from $T_n$. Then these curves are smooth at P.

Proof

Without loss of generality, we may assume that $P\in T_1\cap T_2$ and $t_1⩽t_2$. We have to show that $t_1=t_2=1$. We may assume that $d⩾5$, because the required assertion is obvious in the cases $d=3$ and $d=4$.

Put $\mathrm{\Omega }={\sum }_{i=3}^n{a}_i{T}_i+\mathrm{\Delta }$ and put $k_0={\mathrm{mult}}_P(\mathrm{\Omega })$. Then $m_0=k_0+a_1{t}_1+a_2{t}_2$. Moreover, we have $a_1{d}_1+a_2{d}_2⩽1$ by Lemma 4.4. On the other hand, it follows from Lemma 4.3 that

$$\begin{array}{c}\hfill d-1⩾d_1+d_2+(a_1{d}_1+a_2{d}_2)(d-d_1-d_2-1)=\mathrm{\Omega }·(T_1+T_2)⩾k_0(t_1+t_2),\end{array}$$

because $a_1{d}_1+a_2{d}_2⩽1$. Thus, we have $k_0⩽\frac{d-1}{t_1+t_2}$. Hence, if $t_1+t_2⩾4$, then

$$\begin{array}{cc}\hfill m_0=k_0+a_1{t}_1+a_2{t}_2⩽& k_0+a_1{d}_1+a_2{d}_2⩽\frac{d-1}{t_1+t_2}+a_1{d}_1+a_2{d}_2⩽\frac{d-1}{t_1+t_2}+1\hfill \\ \hfill ⩽& \frac{d+3}{4}\hfill \end{array}$$

because $a_1{d}_1+a_2{d}_2⩽1$. Since $m_0>\frac{1}{\mathit{\lambda }}$ by Lemma 2.5, the inequality $m_0⩽\frac{d+3}{4}$ gives $\mathit{\lambda }>\frac{d+3}{4}$, which is impossible by Lemma 4.2(iii). Thus, $t_1+t_2⩽3$. Since $t_1⩽t_2$, we have $t_1=1$ and $t_2⩽2$.

To complete the proof of the lemma, we have to prove that $t_2=1$. Suppose $t_2\ne 1$. Then $t_2=2$, since $t_1+t_2⩽3$. Since $k_0⩽\frac{d-1}{t_1+t_2}=\frac{d-1}{3}$ and $a_1{d}_1+a_2{d}_2⩽1$, we have

$$\begin{array}{cc}\hfill m_0=& k_0+a_1{t}_1+a_2{t}_2⩽k_0+a_1{d}_1+a_2{d}_2⩽\frac{d-1}{32}+a_1{d}_1+a_2{d}_2⩽\frac{d-1}{t_1+t_2}+1\hfill \\ \hfill =& \frac{d+2}{3}.\hfill \end{array}$$

On the other hand, $m_0>\frac{1}{\mathit{\lambda }}$ by Lemma 2.5, so that $\mathit{\lambda }>\frac{3}{d+2}$. Then $d=5$ by Lemma 4.2(iv).

Since $d=5,t_1=1$ and $t_2=2$, we have $n=3,d_1=1,d_2=3$ and $d_3=1$. Applying Theorem 2.12 to the log pair $(S,\mathit{\lambda }{a}_1{T}_1+\mathit{\lambda }{a}_2{T}_2+\mathit{\lambda }\mathrm{\Omega })$, we get

$$\begin{array}{c}\hfill 1+3a_1=d_1+a_2{d}_1(d-d_1-1)=(\mathit{\lambda }{a}_2{T}_2+\mathit{\lambda }\mathrm{\Omega })·T_1>\frac{1}{\mathit{\lambda }}=\frac{15}{7},\end{array}$$

which gives $a_1>\frac{8}{21}$. On the other hand, $a_1+3a_2⩽1$, because $a_1{d}_1+a_2{d}_2⩽1$. Since $m_0>\frac{1}{\mathit{\lambda }}=\frac{15}{7}$ by Lemma 2.5, we see that

$$\begin{array}{cc}\hfill \frac{15}{7}-\frac{1}{9}=& \frac{128}{63}>\frac{8-5a_1}{3}=\frac{3-a_1+\frac{7(1-a_1)}{3}}{2}=\frac{3-a_1+7a_2}{2}\hfill \\ \hfill =& \frac{3-3a_1+3a_2}{2}+a_1+2a_2\hfill \\ \hfill =& \frac{\mathrm{\Delta }·T_2}{2}+a_1+2a_2⩾\frac{{\mathrm{mult}}_P(\mathrm{\Delta }·T_2)}{2}+a_1+2a_2⩾\frac{t_2{k}_0}{2}+a_1+2a_2\hfill \\ \hfill =& k_0+a_1+2a_2=m_0>\frac{15}{7},\hfill \end{array}$$

which is absurd. $\square$

Now we are ready to prove

Lemma 4.8

One has $m_0⩽\frac{d+1}{2}$.

Proof

Suppose that $m_0>\frac{d+1}{2}$. Let us seek for a contradiction. If $n=1$, then

$$\begin{array}{c}\hfill d=T_n·D⩾2m_0,\end{array}$$

which implies that $m_0⩽\frac{d}{2}$. Thus, have $n⩾2$. Then $a_1⩽\frac{1}{d_1}$ by Lemma 4.4. Moreover, either $t_n=0$ or $t_n=1$ by Lemma 4.5. Hence, there is an irreducible component of $T_P$ that passes through P and is different from $T_n$, because $T_P$ is singular at P. Without loss of generality, we may assume that $t_1⩾1$.

Put $\mathrm{{\rm Y}}={\sum }_{i=2}^n{a}_i{T}_i+\mathrm{\Delta }$, so that $D=a_1{T}_1+\mathrm{{\rm Y}}$. Put $n_0={\mathrm{mult}}_P(\mathrm{{\rm Y}})$, so that $m_0=n_0+a_1{t}_1$. Then $t_n{n}_0⩽d_n-a_1{d}_1{d}_n$ by Lemma 4.4, and

$$\begin{array}{c}\hfill d_1+a_1{d}_1(d-d_1-1)=\mathrm{{\rm Y}}·T_1⩾t_1{n}_0\end{array}$$ 4.2

by Lemma 4.3. Adding these two inequalities, we get $(t_1+t_n)n_0⩽d_1+d_n+a_1{d}_1(d-d_1-d_n-1)$. Hence, if $n⩾3$ and $t_n=1$, then

$$\begin{array}{c}\hfill 2n_0⩽(t_1+t_n)n_0⩽d_1+d_n+a_1{d}_1(d-d_1-d_n-1)⩽d-1⩽d-a_1{d}_1,\end{array}$$

because $a_1⩽\frac{1}{d_1}$. Similarly, if $n=2$ and $t_n=1$, then

Thus, if $t_n=1$, then $n_0⩽\frac{d-a_1{d}_1}{2}$, which is impossible. Indeed, the inequality $n_0⩽\frac{d-a_1{d}_1}{2}$ gives

$$\begin{array}{c}\hfill \frac{d+1}{2}<m_0=n_0+a_1{t}_1⩽n_0+a_1{d}_1⩽\frac{d-a_1{d}_1}{2}+a_1{d}_1=\frac{d+a_1{d}_1}{2}⩽\frac{d+1}{2},\end{array}$$

because $a_1⩽\frac{1}{d_1}$. This shows that $t_n=0$.

If $t_1⩾2$, then it follows from (4.2) that

$$\begin{array}{cc}\hfill \frac{d+1}{2}<& m_0⩽n_0+a_1{d}_1⩽\frac{d_1+a_1{d}_1(d-d_1-1)}{2}+a_1{d}_1\hfill \\ \hfill =& \frac{d_1+a_1{d}_1(d-d_1+1)}{2}⩽\frac{d+1}{2},\hfill \end{array}$$

because $a_1⩽\frac{1}{d_1}$. This shows that $t_1=1$.

Since $t_1=1$ and $t_n=0$, there exists an irreducible component of the curve $T_P$ that passes through P and is different from $T_1$ and $T_n$. In particular, we have $n⩾3$. Without loss of generality, we may assume $P\in T_2$. Then $T_2$ is smooth at P by Lemma 4.7.

Put $\mathrm{\Omega }={\sum }_{i=3}^n{a}_i{T}_i+\mathrm{\Delta }$ and put $k_0={\mathrm{mult}}_P(\mathrm{\Omega })$. Then $a_1{d}_1+a_2{d}_2⩽1$ by Lemma 4.4. Thus, it follows from Lemma 4.3 that

$$\begin{array}{c}\hfill 2k_0⩽\mathrm{\Omega }·(T_1+T_2)=d_1+d_2+(a_1{d}_1+a_2{d}_2)(d-d_1-d_2-1)⩽d-1,\end{array}$$

which implies $k_0⩽\frac{d-1}{2}$. Then

$$\begin{array}{cc}\hfill \frac{d+1}{2}<& m_0=k_0+a_1{t}_1+a_2{t}_2⩽k_0+a_1{d}_1+a_2{d}_2⩽\frac{d-1}{2}+a_1{d}_1+a_2{d}_2\hfill \\ \hfill ⩽& \frac{d-1}{2}+1=\frac{d+1}{2},\hfill \end{array}$$

because $a_1{d}_1+a_2{d}_2⩽1$. The obtained contradiction completes the proof of the lemma. $\square$

Let $f_1:S_1\to S$ be a blow up of the point P, and let $E_1$ be its exceptional curve. Denote by $D^1$ the proper transform of the $\mathbb{Q}$-divisor D on the surface $S_1$. Then

$$\begin{array}{c}\hfill K_{S_1}+\mathit{\lambda }{D}^1+(\mathit{\lambda }{m}_0-1)E_1{\sim }_{\mathbb{Q}}{f}_1^{\ast }(K_S+\mathit{\lambda }D),\end{array}$$

which implies that $(S_1,\mathit{\lambda }{D}^1+(\mathit{\lambda }{m}_0-1)E_1)$ is not log canonical at some point $P_1\in E_1$.

By Lemma 4.8, we have $m_0⩽\frac{d+1}{2}$. By Lemma 4.2(v), we have $\mathit{\lambda }⩽\frac{4}{d+1}$. This gives $\mathit{\lambda }{m}_0⩽2$. Thus, the log pair $(S_1,\mathit{\lambda }{D}^1+(\mathit{\lambda }{m}_0-1)E_1)$ is log canonical at every point of the curve $E_1$ that is different from $P_1$ by Lemma 2.13.

Put $m_1={\mathrm{mult}}_{P_1}(D^1)$. Then Lemma 2.5 gives

$$\begin{array}{c}\hfill m_0+m_1>\frac{2}{\mathit{\lambda }}.\end{array}$$ 4.3

For each curve $T_i$, denote by $T_i^1$ its proper transform on $S_1$. Put $T_P^1={\sum }_{i=1}^n{T}_i^1$.

Lemma 4.9

One has $P_1\notin T_P^1$.

Proof

Suppose that $P_1\in T_P^1$. Let us seek for a contradiction. If $T_P$ is irreducible, then

$$\begin{array}{c}\hfill d-2m_0=T_P^1·D^1⩾m_1,\end{array}$$

so that $m_1+2m_0⩽d$. This inequality gives

$$\begin{array}{c}\hfill \frac{3}{\mathit{\lambda }}<m_1+2m_0⩽d,\end{array}$$

because $2m_0⩾m_0+m_1>\frac{2}{\mathit{\lambda }}$ by (4.3). This shows that $T_P$ is reducible, because $\mathit{\lambda }⩽\frac{3}{d}$ by Lemma 4.2(vi).

We see that $n⩾2$. If $P_1\in T_n^1$, then

$$\begin{array}{c}\hfill d-1-m_0⩾d_n-m_0=d_n-m_0{t}_n=T_n^1·D^1⩾m_1,\end{array}$$

which is impossible, because $m_0+m_1>\frac{2}{\mathit{\lambda }}$ by (4.3), and $\mathit{\lambda }⩽\frac{2}{d-1}$ by Lemma 4.2(i). Thus, we see that $P_1\notin T_n^1$.

Without loss of generality, we may assume that $P_1\in T_1^1$. Put $\mathrm{{\rm Y}}={\sum }_{i=2}^n{a}_i{T}_i+\mathrm{\Delta }$, and denote by ${\mathrm{{\rm Y}}}^1$ the proper transform of the $\mathbb{Q}$-divisor $\mathrm{\Omega }$ on the surface $S_1$. Put $n_0={\mathrm{mult}}_P(\mathrm{{\rm Y}})$, put $n_1={\mathrm{mult}}_{P_1}({\mathrm{\Omega }}^1)$ and put $t_1^1={\mathrm{mult}}_{P_1}(T_1^1)$. Then

$$\begin{array}{c}\hfill d_1+a_1{d}_1(d-d_1-1)-n_0{t}_1=T_1^1·{\mathrm{{\rm Y}}}^1⩾t_1^1{n}_1,\end{array}$$

which implies that $n_0{t}_1+n_1{t}_1^1⩽d_1+a_1{d}_1(d-d_1-1)$.

Note that $t_1^1⩽t_1$. Moreover, we have $a_1⩽\frac{1}{d_1}$ by Lemma 4.4. Thus, if $t_1^1⩾2$, then

$$\begin{array}{cc}\hfill 2(n_0+n_1)⩽& t_1^1(n_0+n_1)⩽n_0{t}_1+n_1{t}_1^1⩽d_1+a_1{d}_1(d-d_1-1)\hfill \\ \hfill ⩽& d_1+(d-d_1-1)=d-1,\hfill \end{array}$$

which implies that $n_0+n_1⩽\frac{d-1}{2}$. Moreover, if $n_0+n_1⩽\frac{d-1}{2}$, then it follows from (4.3) that

$$\begin{array}{cc}\hfill \frac{d+3}{2}=& 2+\frac{d-1}{2}⩾2a_1{d}_1+\frac{d-1}{2}⩾2a_1{t}_1+\frac{d-1}{2}⩾a_1(t_1+t_1^1)+n_0+n_1\hfill \\ \hfill =& m_0+m_1>\frac{2}{\mathit{\lambda }}\hfill \end{array}$$

which gives $d⩽4$ by Lemma 4.2(iii). Thus, if $d⩾5$, then $t_1^1=1$. Furthermore, if $d⩽4$, then $d_1⩽3$, which implies that $t_1^1⩽1$. This shows that $t_1^1=1$ in all cases. Thus, the curve $T_1^1$ is smooth at $P_1$.

Applying Theorem 2.11 to the log pair $(S_1,\mathit{\lambda }{\mathrm{{\rm Y}}}^1+\mathit{\lambda }{a}_1{T}_1^1+(\mathit{\lambda }(n_0+a_1{t}_1)-1)E_1)$, we see that

$$\begin{array}{cc}\hfill \mathit{\lambda }(d-1-n_0{t}_1)⩾& \mathit{\lambda }(d_1+a_1{d}_1(d-d_1-1)-n_0{t}_1)\hfill \\ \hfill =& \mathit{\lambda }{\mathrm{\Omega }}^1·T_1^1>2-\mathit{\lambda }(n_0+a_1{t}_1),\hfill \end{array}$$

because $a_1⩽\frac{1}{d_1}$. Thus, we have $d-1+a_1{t}_1-n_0(t_1-1)>\frac{2}{\mathit{\lambda }}$. But $m_0=a_1{t}_1+n_0>\frac{1}{\mathit{\lambda }}$ by Lemma 2.5. Adding these inequalities together, we obtain

$$\begin{array}{c}\hfill d-1+2a_1{t}_1-n_0(t_1-2)>\frac{3}{\mathit{\lambda }}.\end{array}$$ 4.4

If $t_1⩾2$, this gives

$$\begin{array}{c}\hfill d+1⩾d-1+2a_1{d}_1⩾d-1+2a_1{t}_1⩾d-1+2a_1{t}_1-n_0(t_1-2)>\frac{3}{\mathit{\lambda }}.\end{array}$$

because $a_1⩽\frac{1}{d_1}$. One the other hand, if $d⩾5$, then $\mathit{\lambda }⩽\frac{3}{d+1}$ by Lemma 4.2(ii). Thus, if $d⩾5$, then $t_1=1$. Moreover, if $d=3$, then $d_1⩽2$, which implies that $t_1=1$ as well. Furthermore, if $d=4$ and $t_1\ne 1$, then $d_1=3,t_1=2,\mathit{\lambda }=\frac{5}{8}$, which implies that

$$\begin{array}{c}\hfill \frac{1}{3}=\frac{1}{d_1}⩾a_1>\frac{9}{20}\end{array}$$

by (4.4). Thus, we see that $t_1=1$ in all cases. This simply means that the curve $T_1$ is smooth at the point P.

Since $a_1⩽\frac{1}{d_1}$, we have

$$\begin{array}{c}\hfill d-1-n_0⩾d_1+a_1{d}_1(d-d_1-1)-n_0={\mathrm{\Omega }}^1·T_1^1⩾n_1,\end{array}$$

which implies that $n_1⩽\frac{n_0+n_1}{2}⩽\frac{d-1}{2}$. Then $\mathit{\lambda }{n}_1⩽1$ by Lemma 4.2(i). Hence, we can apply Theorem 2.15 to the log pair $(S_1,\mathit{\lambda }{\mathrm{{\rm Y}}}^1+\mathit{\lambda }{a}_1{T}_1^1+(\mathit{\lambda }(n_0+a_1{t}_1)-1)E_1)$ at the point $P_1$. This gives either

$$\begin{array}{c}\hfill {\mathrm{{\rm Y}}}^1·T_1^1>\frac{4}{\mathit{\lambda }}-2(n_0+a_1)\end{array}$$

or ${\mathrm{{\rm Y}}}^1·E_1>\frac{2}{\mathit{\lambda }}-2a_1$ (or both). Since $a_1⩽\frac{1}{d_1}$, the former inequality gives

$$\begin{array}{c}\hfill d-1-n_0⩾d_1+a_1{d}_1(d-d_1-1)-n_0={\mathrm{{\rm Y}}}^1·T_1^1>\frac{4}{\mathit{\lambda }}-2(n_0+a_1).\end{array}$$

Similarly, the latter inequality gives

$$\begin{array}{c}\hfill n_0=\mathit{\lambda }{\mathrm{{\rm Y}}}^1·E_1>\frac{2}{\mathit{\lambda }}-2a_1.\end{array}$$

Thus, either $d-1+2a_1+n_0>\frac{4}{\mathit{\lambda }}$ or $2a_1+n_0>\frac{2}{\mathit{\lambda }}$ (or both).

If $t_n⩾1$, then $d_n\ne 1$ by Lemma 4.5. Thus, if $t_n⩾1$, then

$$\begin{array}{c}\hfill d-1⩾d_n⩾a_1{d}_1{d}_n+n_0⩾2a_1+n_0\end{array}$$

by Lemma 4.4. Therefore, if $t_n⩾1$, then

$$\begin{array}{c}\hfill 2(d-1)⩾d-1+2a+n_0>\frac{4}{\mathit{\lambda }}\end{array}$$

or $d-1⩾2a+n_0>\frac{2}{\mathit{\lambda }}$, because $d-1+2a+n_0>\frac{4}{\mathit{\lambda }}$ or $2a+n_0>\frac{2}{\mathit{\lambda }}$. In both cases, we get $\mathit{\lambda }>\frac{d-1}{2}$, which is impossible by Lemma 4.2(i). This shows that $t_n=0$, so that $P\notin T_n$.

Since $T_1$ is smooth at P and $P\notin T_n$, there must be another irreducible component of $T_P$ passing through P that is different from $T_1$ and $T_n$. In particular, we see that $n⩾3$. Without loss of generality, we may assume that $P\in T_2$. Then $T_2$ is smooth at P by Lemma 4.7, so that $t_2=1$. Moreover, the curves $T_1$ and $T_2$ are tangent at P by Lemma 4.6, which implies that $d⩾4$. Since $P_1\in T_1^1$, we see that $P_1\in T_2^1$ as well.

Put $\mathrm{\Omega }={\sum }_{i=3}^n{a}_i{T}_i+\mathrm{\Delta }$ and $k_0={\mathrm{mult}}_P(\mathrm{\Omega })$, so that $m_0=k_0+a_1+a_2$. Then $a_1{d}_1+a_2{d}_2⩽1$ by Lemma 4.4.

Denote by ${\mathrm{\Omega }}^1$ the proper transform of the $\mathbb{Q}$-divisor $\mathrm{\Omega }$ on the surface $S_1$. Put $k_1={\mathrm{mult}}_{P_1}({\mathrm{\Omega }}^1)$. Then

$$\begin{array}{cc}\hfill d-1-2k_0⩾& d_1+d_2+(a_1{d}_1+a_2{d}_2)(d-d_1-d_2-1)-2k_0\hfill \\ \hfill =& {\mathrm{\Omega }}^1·(T_1^1+T_2^1)⩾2k_1\hfill \end{array}$$

because $a_1{d}_1+a_2{d}_2⩽1$ and $d⩾d_1+d_2+d_n⩾d_1+d_2+1$. This gives $k_0+k_1⩽\frac{d-1}{2}$. On the other hand, we have

$$\begin{array}{c}\hfill 2a_1+2a_2+k_0+k_1=m_0+m_1>\frac{2}{\mathit{\lambda }}\end{array}$$

by (4.3). Thus, we have

$$\begin{array}{cc}\hfill \frac{d+3}{2}=& 2+\frac{d-1}{2}⩾2(a_1{d}_1+a_2{d}_2)+\frac{d-1}{2}⩾2a_1+2a_2+\frac{d-1}{2}\hfill \\ \hfill ⩾& 2a_1+2a_2+k_0+k_1>\frac{2}{\mathit{\lambda }}\hfill \end{array}$$

because $a_1{d}_1+a_2{d}_2⩽1$. By Lemma 4.2(iii) this gives $d=4$. Thus, we have $\mathit{\lambda }=\frac{5}{8}$.

Since $d=4>n⩾3$, we have $n=3$. Without loss of generality, we may assume that $d_1⩽d_2$. By Lemma 4.6, there exists at most one line in S that passes through P. This shows that $d_1=1,d_2=2$ and $d_3=1$. Thus, $T_1$ and $T_3$ are lines, $T_2$ is a conic, $T_1$ is tangent to $T_2$ at P, and $T_3$ does not pass through P. In particular, the curves $T_1^1$ and $T_1^2$ intersect each other transversally at $P_1$.

By Lemma 4.3, we have $T_1·T_1=T_2·T_2=-2$ and $T_1·T_2=2$. On the other hand, the log pair $(S_1,\mathit{\lambda }{a}_1{T}_1^1+\mathit{\lambda }{a}_2{T}_2^1+\mathit{\lambda }{\mathrm{\Omega }}^1+(\mathit{\lambda }(a_1+a_2+k_0)-1)E_1)$ is not log canonical at the point $P_1$. Thus, applying Theorem 2.11 to this log pair and the curve $T_1^1$, we get

$$\begin{array}{c}\hfill \mathit{\lambda }(1+2a_1-2a_2-k_0)=\mathit{\lambda }{\mathrm{\Omega }}^1·T_1^1>2-\mathit{\lambda }(a_1+a_2+k_0)-\mathit{\lambda }{a}_2,\end{array}$$

which implies that $3a_1>\frac{2}{\mathit{\lambda }}-1=\frac{11}{5}$, because $\mathit{\lambda }=\frac{5}{8}$. Similarly, applying Theorem 2.11 to this log pair and the curve $T_2^1$, we get

$$\begin{array}{c}\hfill \mathit{\lambda }(2-2a_1+2a_2-k_0)=\mathit{\lambda }{\mathrm{\Omega }}^1·T_2^1>2-\mathit{\lambda }(a_1+a_2+k_0)-\mathit{\lambda }{a}_1,\end{array}$$

which implies that $3a_2>\frac{2}{\mathit{\lambda }}-2=\frac{6}{5}$. Hence, we have $a_1>\frac{11}{15}$ and $a_2>\frac{2}{5}$, which is impossible, since $a_1+2a_2=a_1{d}_1+a_2{d}_2⩽1$. The obtained contradiction completes the proof of the lemma. $\square$

Now we are going to show that the curve $T_P$ has at most two irreducible components. This follows from

Lemma 4.10

One has $n⩾2$ and ${\mathrm{mult}}_P(T_P)=2$. Moreover, if $n=2$, then $P\in T_1\cap T_2$, both curves $T_1$ and $T_2$ are smooth at P, and $d_1⩽d_2$.

Proof

If $T_P$ is irreducible and ${\mathrm{mult}}_P(T_P)⩾3$, then Lemma 2.5 gives

$$\begin{array}{c}\hfill d=T_P·D⩾3m_0>\frac{3}{\mathit{\lambda }},\end{array}$$

which is impossible by Lemma 4.2(vi). Thus, if $n=1$, then ${\mathrm{mult}}_P(T_P)=2$.

To complete the proof, we may assume that $n⩾2$. Then $t_n=0$ or $t_n=1$ by Lemma 4.5. In particular, there exists an irreducible component of the curve $T_P$ different from $T_n$ that passes through P. Without loss of generality, we may assume that $P\in T_1$.

Put $\mathrm{{\rm Y}}={\sum }_{i=2}^n{a}_i{T}_i+\mathrm{\Delta }$, and denote by ${\mathrm{{\rm Y}}}^1$ the proper transform of the $\mathbb{Q}$-divisor $\mathrm{\Omega }$ on the surface $S_1$. Put $n_0={\mathrm{mult}}_P(\mathrm{{\rm Y}})$. Then the log pair $(S_1,\mathit{\lambda }{\mathrm{{\rm Y}}}^1+(\mathit{\lambda }(n_0+a_1{t}_1)-1)E_1)$ is not log canonical at $P_1$, since $P_1\notin T_1^1$ by Lemma 4.9. In particular, it follows from Theorem 2.12 that

$$\begin{array}{c}\hfill \mathit{\lambda }{n}_0=\mathit{\lambda }{\mathrm{{\rm Y}}}^1·E_1>1,\end{array}$$

which implies that $n_0>\frac{1}{\mathit{\lambda }}$. Thus, if $t_1⩾2$, then it follows from Lemma 4.3 that

$$\begin{array}{c}\hfill \frac{1}{\mathit{\lambda }}⩾\frac{d-1}{2}⩾\frac{d_1+a_1{d}_1(d-d_1-1)}{2}=\frac{\mathrm{{\rm Y}}·T_1}{2}⩾\frac{t_1{n}_0}{2}⩾n_0>\frac{1}{\mathit{\lambda }},\end{array}$$

because $a_1⩽\frac{1}{d_1}$ by Lemma 4.4, and $\mathit{\lambda }⩽\frac{2}{d-1}$ by Lemma 4.2(i). This shows that $t_1=1$, so that the curve $T_1$ is smooth at P.

If $t_n=1$ and $n⩾3$, then

$$\begin{array}{c}\hfill \frac{2}{\mathit{\lambda }}⩾d-1⩾d_1+d_n+a{d}_1(d-d_1-d_n-1)=\mathrm{{\rm Y}}·(T_1+T_n)⩾2n_0>\frac{2}{\mathit{\lambda }}.\end{array}$$

Thus, if $t_n=1$, then $n=2$. Vice versa, if $n=2$, then $t_n=1$, because $T_1$ is smooth at P. Furthermore, if $n=2$, then $d_1⩽d_n$, because $d_n>\frac{d-1}{2}$ by Lemma 4.5. Therefore, to complete the proof, we must show that $n=2$.

Suppose that $n⩾3$. Let us seek for a contradiction. We know that $P\notin T_n$, so that $t_n=0$. Then every irreducible component of the curve $T_P$ that contain P is smooth at P by Lemma 4.7. Hence, there should be at least one irreducible component of the curve $T_P$ containing P that is different from $T_1$ and $T_n$. Without loss of generality, we may assume that $P\in T_2$.

Put $\mathrm{\Omega }={\sum }_{i=3}^n{a}_i{T}_i+\mathrm{\Delta }$ and $k_0={\mathrm{mult}}_P(\mathrm{\Omega })$. By Lemma 4.4, we have $a_1{d}_1+a_2{d}_2⩽1$. Thus, it follows from Lemma 4.3 that

$$\begin{array}{cc}\hfill 2k_0⩽& \mathrm{\Omega }·(T_1+T_2)=d_1+d_2+(a_1{d}_1+a_2{d}_2)(d-d_1-d_2-1)\hfill \\ \hfill ⩽& d_1+d_2+(d-d_1-d_2-1)=d-1.\hfill \end{array}$$

Hence, we have $k_0⩽\frac{d-1}{2}$.

Denote by ${\mathrm{\Omega }}^1$ the proper transform of the $\mathbb{Q}$-divisor $\mathrm{\Omega }$ on the surface $S_1$. Then the log pair $(S_1,\mathit{\lambda }{\mathrm{\Omega }}^1+(\mathit{\lambda }(k_0+a_1+a_2)-1)E_1)$ is not log canonical at $P_1$, because $P_1\notin T_1^1$ and $P_1\notin T_2^1$ by Lemma 4.9. In particular, it follows from Theorem 2.11 that

$$\begin{array}{c}\hfill \mathit{\lambda }{k}_0=\mathit{\lambda }{\mathrm{\Omega }}^1·E_1>1,\end{array}$$

which implies that $k_0>\frac{1}{\mathit{\lambda }}$. This contradicts Lemma 4.2(i), because $k_0⩽\frac{d-1}{2}$. $\square$

Later, we will need the following simple

Lemma 4.11

Suppose that $d=4$. Then $m_0⩽\frac{11}{5}$.

Proof

If $n=1$, then

$$\begin{array}{c}\hfill 2t_n⩾d_n=T_n·D⩾t_n{m}_0,\end{array}$$

so that $m_0⩽2<\frac{11}{5}$. Thus, we may assume that $n\ne 1$. Then it follows from Lemma 4.10 that $n=2,P\in T_1\cap T_2$, both curves $T_1$ and $T_2$ are smooth at P, and $d_1⩽d_2$.

If $d_2=2$, then $m_0⩽2<\frac{11}{5}$, because

$$\begin{array}{c}\hfill 2=T_2·D⩾m_0.\end{array}$$

Thus, we may assume that $d_2\ne 2$. Then $d_1=1$ and $d_2=3$. Then ${\mathrm{mult}}_P(\mathrm{\Delta })+3a_1⩽3$ by Lemma 4.4. Moreover, we have

$$\begin{array}{c}\hfill 1+2a_1=T_1·\mathrm{\Delta }⩾{\mathrm{mult}}_P(\mathrm{\Delta }).\end{array}$$

The obtained inequalities give $m_0={\mathrm{mult}}_P(\mathrm{\Delta })+a_1⩽\frac{11}{5}$. $\square$

Let $f_2:S_2\to S_1$ be a blow up of the point $P_1$. Denote by $E_2$ the $f_2$-exceptional curve, denote by $E_1^2$ the proper transform of the curve $E_1$ on the surface $S_2$, and denote by $D^2$ the proper transform of the $\mathbb{Q}$-divisor D on the surface $S_2$. Then

$$\begin{array}{cc}& K_{S_2}+\mathit{\lambda }{D}^2+(\mathit{\lambda }{m}_0-1)E_1^2+(\mathit{\lambda }(m_0+m_1)-2)E_2{\sim }_{\mathbb{Q}}{f}_2^{\ast }(K_{S_1}+\mathit{\lambda }{D}^1\hfill \\ \hfill & +(\mathit{\lambda }{m}_0-1)E_1).\hfill \end{array}$$

By Remark 2.10, the log pair $(S_2,\mathit{\lambda }{D}^2+(\mathit{\lambda }{m}_0-1)E_1^2+(\mathit{\lambda }(m_0+m_1)-2)E_2)$ is not log canonical at some point $P_2\in E_1$.

Lemma 4.12

One has $m_0+m_1⩽\frac{3}{\mathit{\lambda }}$.

Proof

Suppose that $m_0+m_1>\frac{3}{\mathit{\lambda }}$. Then $2m_0⩾m_0+m_1>\frac{3}{\mathit{\lambda }}$. But $m_0⩽\frac{d+1}{2}$ by Lemma 4.8. Then $\mathit{\lambda }>\frac{3}{d+1}$. Thus, we have $d⩽4$ by Lemma 4.2(ii). Moreover, if $d=4$, then

$$\begin{array}{c}\hfill \frac{22}{5}⩾2m_0⩾m_0+m_1>\frac{3}{\mathit{\lambda }}=\frac{24}{5}\end{array}$$

by Lemma 4.11. This shows that $d=3$.

We have $\mathit{\lambda }=1$. If $n=1$, then

$$\begin{array}{c}\hfill 3=T_P·D⩾2m_0⩾m_1+m_0>\frac{3}{\mathit{\lambda }}=3,\end{array}$$

which is absurd. Hence, it follows from Lemma 4.10 that $n=2,d_1=1,d_2=2$ and $P\in T_1\cap T_2$.

We have $m_0={\mathrm{mult}}_P(\mathrm{\Delta })+a_1$. On the other hand, we have ${\mathrm{mult}}_P(\mathrm{\Delta })+2a_1⩽2$ by Lemma 4.4. Moreover, we have

$$\begin{array}{c}\hfill 1+a_1=T_1·\mathrm{\Omega }⩾{\mathrm{mult}}_P(\mathrm{\Delta }),\end{array}$$

which implies that ${\mathrm{mult}}_P(\mathrm{\Delta })-a_1⩽1$. Adding these inequalities, we get

$$\begin{array}{c}\hfill 3⩾2{\mathrm{mult}}_P(\mathrm{\Delta })+a={\mathrm{mult}}_P(\mathrm{\Delta })+m_0⩾m_1+m_0>\frac{3}{\mathit{\lambda }}=3,\end{array}$$

because ${\mathrm{mult}}_P(\mathrm{\Delta })⩾m_1$, since $P_1\notin T_1^1$ by Lemma 4.9. $\square$

Thus, the log pair $(S_2,\mathit{\lambda }{D}^2+(\mathit{\lambda }{m}_0-1)E_1^2+(\mathit{\lambda }(m_0+m_1)-2)E_2)$ is log canonical at every point of the curve $E_2$ that is different from the point P by Lemma 2.13.

Lemma 4.13

One has $P_2\ne E_1^2\cap E_2$.

Proof

Suppose that $P_2=E_1^2\cap E_2$. Then Theorem 2.11 gives

$$\begin{array}{c}\hfill \mathit{\lambda }(m_0-m_1)=\mathit{\lambda }{D}^2·E_1^2>3-\mathit{\lambda }(m_0+m_1),\end{array}$$

which implies that $m_0>\frac{3}{2\mathit{\lambda }}$. But $m_0⩽\frac{d+1}{2}$ by Lemma 4.8. Therefore, we have $\mathit{\lambda }>\frac{3}{d+1}$, which implies that $d⩽4$ by Lemma 4.2(ii). If $d=4$, then

$$\begin{array}{c}\hfill \frac{12}{5}=\frac{3}{2\mathit{\lambda }}<m_0⩽\frac{11}{5}\end{array}$$

by Lemma 4.11. Thus, we have $d=3$.

One has $\mathit{\lambda }=1$. If $n=1$, then

$$\begin{array}{c}\hfill 3=T_P·D⩾2m_0>\frac{3}{\mathit{\lambda }}=3,\end{array}$$

which is absurd. Hence, it follows from Lemma 4.10 that $n=2,d_1=1,d_2=2$ and $P\in T_1\cap T_2$.

We have $m_0={\mathrm{mult}}_P(\mathrm{\Delta })+a_1$. Moreover, we have ${\mathrm{mult}}_P(\mathrm{\Delta })+2a_1⩽2$ by Lemma 4.4, Then $2{\mathrm{mult}}_P(\mathrm{\Delta })+a_1⩽3$, because

$$\begin{array}{c}\hfill 1+a_1=T_1·\mathrm{\Delta }⩾{\mathrm{mult}}_P(\mathrm{\Delta }).\end{array}$$

Denote by ${\mathrm{\Delta }}^1$ the proper transform of the divisor $\mathrm{\Delta }$ on the surface $S_1$, and denote by ${\mathrm{\Delta }}^2$ the proper transform of the divisor $\mathrm{\Delta }$ on the surface $S_2$. Then $m_1={\mathrm{mult}}_{P_1}({\mathrm{\Delta }}^1)$, because $P_1\notin T_1^1$ by Lemma 4.9. Thus, the log pair $(S_2,\mathit{\lambda }{\mathrm{\Delta }}^2+(m_0-1)E_1^2+(m_0+m_1-2)E_2)$ is not log canonical at $P_2$. Applying Theorem 2.11 to this pair and the curve $E_1^2$, we get

$$\begin{array}{c}\hfill {\mathrm{mult}}_P(\mathrm{\Delta })-m_1={\mathrm{\Delta }}^2·E_1^2>3-m_0-m_1,\end{array}$$

which implies that $2{\mathrm{mult}}_P(\mathrm{\Delta })+a_1>3$. The latter is impossible, because we already proved that $2{\mathrm{mult}}_P(\mathrm{\Delta })+a_1⩽3$. $\square$

Thus, the log pair $(S_2,\mathit{\lambda }{D}^2+(\mathit{\lambda }(m_0+m_1)-2)E_2)$ is not log canonical at $P_2$. Then Lemma 2.5 gives

$$\begin{array}{c}\hfill m_0+m_1+m_2>\frac{3}{\mathit{\lambda }}.\end{array}$$ 4.5

Denote by $T_P^2$ the proper transform of the curve $T_P$ on the surface $S^2$. Then

$$\begin{array}{c}\hfill T_P^2+E_1^2\sim {(f_1\circ f_2)}^{\ast }({\mathcal{O}}_S(1))-f_2^{\ast }(E_1)-E_2,\end{array}$$

because $T_P^1\sim f_1^{\ast }({\mathcal{O}}_S(1))-2E_1$ by Lemma 4.10, and $P_1\notin T_P^1$ by Lemma 4.9.

Lemma 4.14

The linear system $|T_P^2+E_1^2|$ is a pencil that does not have base points in $E_2$.

Proof

Since $|T_P^1+E_1|$ is a two-dimensional linear system that does not have base points, $|T_P^2+E_1^2|$ is a pencil. Let C be a curve in $|T_P^1+E_1|$ that passes through $P_1$ and is different from $T_P^1+E_1$. Then C is smooth at P, since $P\in f_1(C)$ and $f_1(C)$ is a hyperplane section of the surface S that is different from $T_P$. Since $C·E_1=1$, we see that $T_P^1+E_1$ and C intersect transversally at $P_1$. Thus, the proper transform of the curve C on the surface $S_2$ is contained in $|T_P^1+E_1|$ and have no common points with $T_P^2+E_1^2$ in $E_2$. This shows that the pencil $|T_P^1+E_1|$ does not have base points in $E_2$. $\square$

Let $Z^2$ be the curve in $|T_P^2+E_2|$ that passes through the point $P_2$. Then

$$\begin{array}{c}\hfill Z^2\ne T_P^2+E_1^2,\end{array}$$

because $P_2\ne E_1^2\cap E_2$ by Lemma 4.13. Then $Z_2$ is smooth at $P_2$. Put $Z=f_1\circ f_2(Z^2)$ and $Z^1=f_2(Z^2)$. Then $P\in Z$ and $P_1\in Z^1$. Moreover, the curve Z is smooth at P, and the curve $Z_1$ is smooth at $P_1$. Furthermore, the curve Z is reduced by Lemma 2.6.

The log pair $(S,\mathit{\lambda }Z)$ is log canonical at P, because Z is smooth at P. Note that

$$\begin{array}{c}\hfill Z{\sim }_{\mathbb{Q}}D.\end{array}$$

Thus, we may assume that $\mathrm{Supp}(D)$ does not contain at least one irreducible component of the curve Z by Remark 2.4. Denote this irreducible component by $\overline{Z}$, and denote its degree in ${\mathbb{P}}^3$ by $\overline{d}$. Then $\overline{d}⩽d$.

Lemma 4.15

One has $P\notin \overline{Z}$.

Proof

Suppose that $P\in \overline{Z}$. Let us seek for a contradiction. Denote by ${\overline{Z}}^2$ the proper transform of the curve $\overline{Z}$ on the surface $S_2$. Then

$$\begin{array}{c}\hfill d-m_0-m_1⩾\overline{d}-m_0-m_1={\overline{Z}}^2·D^2⩾m_2,\end{array}$$

which implies that $m_0+m_1+m_2⩽d$. One the other hand, $m_0+m_1+m_2>\frac{3}{\mathit{\lambda }}$ by (4.5). This gives $\mathit{\lambda }>\frac{3}{d}$, which is impossible by Lemma 4.2(vi). $\square$

In particular, the curve Z is reducible. Denote by $\widehat{Z}$ its irreducible component that passes through P, denote its proper transform on the surface $S_1$ by ${\widehat{Z}}^1$, and denote its proper transform on the surface $S_2$ by ${\widehat{Z}}^2$. Then $\overline{Z}\ne \widehat{Z},P_1\in {\widehat{Z}}^1$ and $P_2\in {\widehat{Z}}^2$. Denote by $\widehat{d}$ the degree of the curve $\widehat{Z}$ in ${\mathbb{P}}^3$. Then $\widehat{d}+\overline{d}⩽d$. Moreover, the intersection form of the curves $\widehat{Z}$ and $\overline{Z}$ on the surface S is given by

Lemma 4.16

One has $\overline{Z}·\overline{Z}=-\overline{d}(d-\overline{d}-1),\widehat{Z}·\widehat{Z}=-\widehat{d}(d-\widehat{d}-1)$ and $\overline{Z}·\widehat{Z}=\overline{d}\widehat{d}$.

Proof

See the proof of Lemma 4.3. $\square$

Put $D=a\widehat{Z}+\mathrm{\Omega }$, where a is a positive rational number, and $\mathrm{\Omega }$ is an effective $\mathbb{Q}$-divisor on the surface S whose support does not contain the curve $\widehat{Z}$. Denote by ${\mathrm{\Omega }}^1$ the proper transform of the divisor $\mathrm{\Omega }$ on the surface $S_1$, and denote by ${\mathrm{\Omega }}^2$ the proper transform of the divisor $\mathrm{\Omega }$ on the surface $S_2$. Put $n_0={\mathrm{mult}}_P(\mathrm{\Omega }),n_1={\mathrm{mult}}_{P_1}({\mathrm{\Omega }}^1)$ and $n_2={\mathrm{mult}}_{P_2}({\mathrm{\Omega }}^2)$. Then $m_0=n_0+a,m_1=n_1+a$ and $m_2=n_2+a$. Then the log pair $(S_2,\mathit{\lambda }a{\widehat{Z}}^2+\mathit{\lambda }{\mathrm{\Omega }}^2+(\mathit{\lambda }(n_0+n_1+2a)-2)E_2)$ is not log canonical at $P_2$, because $(S_2,\mathit{\lambda }{D}^2+(\mathit{\lambda }(m_0+m_1)-2)E_2)$ is not log canonical at $P_2$. Thus, applying Theorem 2.11, we see that

$$\begin{array}{cc}\hfill \mathit{\lambda }(\mathrm{\Omega }·\widehat{Z}-n_0-n_1)=& \mathit{\lambda }{\mathrm{\Omega }}^2·Z^2>1-(\mathit{\lambda }(n_0+n_1+2a)-2)\hfill \\ \hfill =& 3-\mathit{\lambda }(n_0+n_1+2a),\hfill \end{array}$$

which implies that

$$\begin{array}{c}\hfill \mathrm{\Omega }·\widehat{Z}>\frac{3}{\mathit{\lambda }}-2a.\end{array}$$ 4.6

On the other hand, we have

$$\begin{array}{c}\hfill \overline{d}=D·\overline{Z}=(a\widehat{Z}+\mathrm{\Omega })·\overline{Z}⩾a\widehat{Z}·\overline{Z}=a\widehat{d}\overline{d}\end{array}$$

by Lemma 4.16. This gives

$$\begin{array}{c}\hfill a⩽\frac{1}{\widehat{d}}.\end{array}$$ 4.7

Thus, it follows from (4.6), (4.7) and Lemma 4.16 that

$$\begin{array}{c}\hfill \frac{3}{\mathit{\lambda }}-2⩽\frac{3}{\mathit{\lambda }}-2a<\mathrm{\Omega }·\widehat{Z}=\widehat{d}+a\widehat{d}(d-\widehat{d}-1)⩽d-1,\end{array}$$

which implies that $\mathit{\lambda }>\frac{3}{d+1}$. Then $d⩽4$ by Lemma 4.2(ii).

Lemma 4.17

One has $d\ne 4$.

Proof

Suppose that $d=4$. Then $\mathit{\lambda }=\frac{5}{8}$ and $\widehat{d}⩽3$. By Lemma 4.9, $\widehat{Z}$ is not a line, since every line passing through P must be an irreducible component of the curve $T_P$. Thus, either $\widehat{Z}$ is a conic or $\widehat{Z}$ is a plane cubic curve. If $\widehat{Z}$ is a conic, then ${\widehat{Z}}^2=-2$ and $a⩽\frac{1}{2}$ by (4.7). Thus, if $\widehat{Z}$ is a conic, then

$$\begin{array}{c}\hfill 2+2a=\mathrm{\Omega }·\widehat{Z}>\frac{3}{\mathit{\lambda }}-2a=\frac{24}{5}-2a,\end{array}$$

which implies that $\frac{1}{2}⩾a>\frac{7}{10}$. This shows that $\widehat{Z}$ is a plane cubic curve. Then ${\widehat{Z}}^2=0$. Since $a⩽\frac{1}{3}$ by (4.7), we have

$$\begin{array}{c}\hfill 3=\mathrm{\Omega }·\widehat{Z}>\frac{3}{\mathit{\lambda }}-2a=\frac{24}{5}-2a⩾\frac{24}{5}-\frac{2}{3}=\frac{62}{15},\end{array}$$

which is absurd. $\square$

Thus, we see that $d=3$. Then $\widehat{Z}$ is either a line or a conic. But every line passing through P must be an irreducible component of $T_P$. Since $\widehat{Z}$ is not an irreducible component of $T_P$ by Lemma 4.9, the curve $\widehat{Z}$ must be a conic. Then ${\widehat{Z}}^2=0$. Therefore, it follows from (4.6) that

$$\begin{array}{c}\hfill 3-2a=\frac{3}{\mathit{\lambda }}-2a<\mathrm{\Omega }·\widehat{Z}=\widehat{d}+a\widehat{d}(d-\widehat{d}-1)=\widehat{d}=2,\end{array}$$

which implies that $a>\frac{1}{2}$. But $a⩽\frac{1}{\widehat{d}}=\frac{1}{2}$ by (4.7). The obtained contradiction completes the proof of Theorem 1.17.