Complex planar curves homeomorphic to a line have at most four singular points

TL;DR

This paper proves that complex planar curves homeomorphic to a line can have at most four singular points, with a unique degree five example if exactly four are present.

Contribution

It establishes an upper bound on the number of singular points for such curves and characterizes the unique degree five case with four singularities.

Findings

Curves homeomorphic to a line have at most four singular points.

The degree five curve with four singular points is unique up to projective equivalence.

If four singular points exist, the curve's degree is exactly five.

Abstract

We show that a complex planar curve homeomorphic to the projective line has at most four singular points. If it has exactly four then it has degree five and is unique up to a projective equivalence.

Equations

$$[s:t]⟶[s^{4}t:s^2{t}^3-s^5:t^5+2s^3{t}^2].$$

$$\kappa (K_X+\frac{1}{2}D)=-\infty .$$

$$d(T)=\det (-Q(T)),$$

$${\beta }_T(C):=C\cdot (T-C).$$

$${\mathrm{Bk}}^{\prime }T=\sum _{i=1}^k\frac{d(T_{i+1}+\cdots +T_k)}{d(T)}{T}_i.$$

$$\mathrm{ind}(T)=\frac{d(T-T_1)}{d(T)}.$$

$$d(T_1+T_2)=d(T_1)d(T_2)-d(T_1-C_1)d(T_2-C_2).$$

$$d(R)=a_1{d}^{\prime }(R)-d^{\prime \prime }(R),$$

$$\Sigma =h+\nu -2+\rho (X)-\#D,$$

$$\#\{q\in E\setminus \{p\}:r_F(q)⩾2\}⩽\sum _{q\in E\setminus \{p\}}(r_F(q)-1)=2F\cdot E-1-r_F(p).$$

$$E=(K+E)+(-K).$$

$${\psi }_{i+1}:(X_i,D_i)⟶(X_{i+1},D_{i+1}),0⩽i⩽n-1,$$

$$A\cdot D_i=2,A\cdot {\Delta }_i^{-}=1,A\cdot ({{\rm Y}}_i+{\Delta }_i^{+})=0$$

$${\psi }_{i+1}:X_i⟶X_{i+1}$$

$$p_2=p_2({\mathbb{P}}^2,\bar{E}):=h^0(2K_X+D)=h^0(2K_0+D_0).$$

$${\phi }_i:(X_i^{\prime },D_i^{\prime })⟶(X_i,D_i)$$

$$\#D_n+{\tau }^{\ast }+c+\delta ⩽3p_2+8+(b_0({\Delta }_n)+\#{{\rm Y}}_n^0).$$

$$b_0({\Delta }_n^{\prime })+\#{{\rm Y}}_n^0⩽b_0({\Delta }_0^{\prime })+c_0^{\prime }$$

$$\rho (X_n^{\prime })+n+\delta ⩽3p_2+8+(b_0({\Delta }_n^{\prime })+\#{{\rm Y}}_n^0).$$

$$\#D_n+\tau -s+\delta ⩽3p_2+8+b_0({\Delta }_0)+c_0^{\prime }.$$

$$\#D_n+\tau +\delta ⩽18+p_2+\frac{1}{3}{c}_0^{\prime }.$$

$$1-c_0^{\prime }+4c⩽\#D_n+\tau ⩽18+p_2+\frac{1}{3}{c}_0^{\prime },$$

$$\psi ={\psi }_n\circ \cdots \circ {\psi }_1:(X_0,D_0)\to (X_n,D_n).$$

$$(X,D)\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}$$

$${\mathrm{ind}}_1^{(i)}+\cdots +{\mathrm{ind}}_c^{(i)}+p_2⩽5-i,i=0,\dots ,n.$$

$${\bar{Q}}_j={\psi }_{\ast }(Q_j),{\Omega }_j={\bar{Q}}_j-({{\rm Y}}_n^0+{\Delta }_n^{\mathrm{tip}})\wedge {\bar{Q}}_j$$

$${\lambda }_j:=\#{\Omega }_j+{\tau }_j-s_j⩾\#{\Omega }_j+1.$$

$$c+\sum _{j=1}^c(\#{\Omega }_j+{\tau }_j^{\ast })=\sum _{j=1}^c{\lambda }_j⩽3p_2+7-\delta .$$

$${\mathrm{ind}}_1^{(n)}+\frac{4}{3}\sum _{j=2}^4\frac{t_j}{2t_j+3}⩽2\frac{1}{2}-n-p_2,$$

$${\lambda }_1+t_2+t_3+t_4⩽3p_2+4-\delta .$$

Full text

$\dagger$ The first author died when preparing the final version of the manuscript

The first author was supported by the National Science Centre, Poland, Grant No. 2013/11/B/ST1/02977. The second author was partially supported by the Foundation for Polish Science under the Homing Plus programme, cofinanced from the European Union, RDF and by the National Science Centre, Poland, grant No. 2015/18/E/ST1/00562

1. Main result

All varieties considered are complex algebraic.

Classical theorems of Abhyankar–Moh [AM73] and Suzuki [Suz74] and Zaidenberg-Lin [ZL83] give a complete description of affine planar curves homeomorphic to the affine line. Every such curve is, up to a choice of coordinates, given by one of the equations $x=0$ or $x^{n}=y^{m}$ for some relatively prime integers $n>m\geqslant 2$; see [GM96] and [Pal15] for proofs based on the theory of open surfaces. In particular, it has at most one singular point.

An analogous projective problem, the classification of planar curves homeomorphic to the projective line, is much more subtle and remains open. Such curves are necessarily rational and their singularities are cusps (are locally analytically irreducible), hence they are called rational cuspidal curves. Apart from being an object of study on its own, they have strong connections with surface singularities via superisolated singularities [Lue87]. Invariants of those links of superisolated singularities which are rational homology $3$-spheres, including the Seiberg–Witten invariants, are tightly related with invariants of rational cuspidal curves, see [FdBLMHN06], [ABLMH06]. This connection has led to counterexamples to several conjectures concerning normal surface singularities [LVMHN05] and to many proved or conjectural constrains for invariants of rational cuspidal curves [FdBLMHN07a]. Deep relations of such curves with low-dimensional topology have been studied from various standpoints, including Heegaard–Floer homology [BL14], [BBSdR16], lattice cohomology [BN16] or involutive Floer homology [BHS19]. Applications of the semigroup distribution property proved in [BL16] have led to some partial classification results, see [Liu14], [Bod16]. Conjectures concerning free divisors and local cohomology of Milnor algebras have been investigated in [DS17], [DS18a] and [DS18b]. Despite some partial success of the above approaches, most of important questions remained unanswered.

Recently, using techniques based on the minimal model program, we proved the Coolidge-Nagata conjecture [Pal14], [KP17], which asserts that every rational cuspidal curve can be mapped onto a line by a Cremona transformation of the plane. The proof is necessarily non-constructive and the transformation needed for such a birational rectification may be complicated. An important remaining problem was to bound the number of possible singular points. We prove the following optimal result and we characterize the extremal case.

Theorem 1.1**.**

A complex planar curve homeomorphic to the projective line has at most four singular points. If it has exactly four then, up to a choice of coordinates on the plane, it can be parameterized as (cf. Example 5.13)

[TABLE]

The problem of bounding the number of cusps of rational cuspidal curves was posed by Sakai [GKK*+*95, p. 14] and specific conjectures were made, among others, by Orevkov and Piontkowski [Pio07]. Orevkov and Zaidenberg proved that rigid rational cuspidal curves have at most $9$ cusps [ZO96]. The absolute bound by $8$ was obtained by [Ton05] using a logarithmic Noether inequality. Later the second author improved it to $6$ [Pal19, Theorem 1.4]. Results of this type do not seem to be available using the more topologically-oriented methods discussed above, at least at their current stage of development.

Numerous families of rational cuspidal curves with a small number of cusps have been constructed by various authors and some partial classification results were obtained. For uni- and bi-cuspidal curves see [Kas87], [Yos88], [Fen99a], [Ton00], [Ton01], [Ore02], [FdBLMHN07b], [Ton12], [Bod16]. The richest source of examples are closures of $\mathbb{C}^{*}$-embeddings into $\mathbb{C}^{2}$ [BZ10], [CNKR09], but even those are not understood completely. Known examples of rational cuspidal curves with three cusps are: Namba’s tricuspidal quintic [Nam84, 2.3.10.8], two discrete series discovered by Flenner-Zaidenberg [FZ96], [FZ00] and one discovered by Fenske [Fen99b]. Inequalities bounding the degree of the curve in terms of the maximal multiplicity of its cusps were proved in [MS89] and [Ore02].

Finally, let us comment on a recent progress on the classification problem for rational cuspidal curves. A well understood case is when the surface $\mathbb{P}^{2}\setminus\bar{E}$ is not of log general type, i.e. when $\kappa(K_{X}+D)\leqslant 1$, where $(X,D)\longrightarrow(\mathbb{P}^{2},\bar{E})$ is the minimal log resolution of singularities. Here the global structure of $\mathbb{P}^{2}\setminus\bar{E}$ is known [Pal14, Proposition 2.5] and in fact curves of this type have been classified, see [FdBLMHN06] for a review. In particular, they have at most two cusps by [Wak78]. The remaining case $\kappa(K_{X}+D)=2$ is difficult. So far, the most successful approach is a careful analysis of possible runs of the minimal model program for the pair $(X,\frac{1}{2}D)$ in terms of almost minimal models, as defined in [Pal19]. We extend this method in the current article. We mention the following key conjecture, which remains open. Note that for $(X,D)$ as above the affine surface $X\setminus D=\mathbb{P}^{2}\setminus\bar{E}$ is $\mathbb{Q}$-acyclic.

Conjecture 1.2** (Negativity Conjecture, [Pal19]).**

If $(X,D)$ is a smooth completion of a $\mathbb{Q}$-acyclic surface then

[TABLE]

As discussed in Conjecture 2.5 loc. cit., the conjecture generalizes both the weak rigidity conjecture by Flenner-Zaidenberg [GKK*+*95, p. 16] and the Coolidge-Nagata conjecture. T. Pełka and the second author have already shown that the approach with almost minimal models is very effective: they classified up to a projective equivalence all rational cuspidal curves with a complement of log general type under the assumption that for such complements the Negativity Conjecture holds [PP17], [PP20]. In particular, in that case the topology of singularities completely determines the class of a projective equivalence. The classification implies also that there are no tricuspidal rational curves satisfying the Negativity Conjecture other than the ones discovered already. Summarizing, given the above theorem, to have a complete understanding of rational cuspidal curves it remains now to prove the above conjecture for complements of those with at most $3$ cusps.

As suggested by Conjecture 1.2, properties of the divisor $K_{X}+\frac{1}{2}D$ and the associated almost minimal model play a major role in the proof of Theorem 1.1. The key step, which takes most of our effort, is Theorem 4.7, saying that for a rational cuspidal curve $\bar{E}\subseteq\mathbb{P}^{2}$ with at least four cusps the surface $(X_{0},\frac{1}{2}D_{0})$, where $(X_{0},D_{0})\longrightarrow(\mathbb{P}^{2},\bar{E})$ is a minimal weak resolution, is almost minimal. In contrast, in case of at most three cusps the process of almost minimalization may contract zero, two, three or four curves not contained in the boundary, see figures for various types of curves drawn in [PP17, Section 3] and Lemmas 3.8(e), 3.12(f) in [PP20].

We thank Maciej Borodzik for helpful remarks and we thank Tomasz Pełka for a careful reading of a preliminary version of the manuscript.

Contents

1. 1 Main result

2. 2 Preliminaries

3. 3 Almost minimal models

4. 4 Geometric restrictions on exceptional divisors.

5. 5 Consequences of the almost minimality

6. 5A Description in terms of Hamburger-Noether pairs

7. 5B Further bounds on the geometry of the boundary

2. Preliminaries

In this article curves are irreducible and reduced. For a reduced divisor $T$ on a smooth projective surface we denote the number of its irreducible components by $\#T$ and we put

[TABLE]

where $Q(T)$ is the intersection matrix of $T$; we put $d(0)=1$. We define the arithmetic genus of $T$ as $p_{a}(T)=\frac{1}{2}T\cdot(K+T)+1$, where $K$ is the canonical divisor and we define the branching number of a component $C\subseteq T$ by

[TABLE]

We say that $C$ is a branching component of $T$ if $\beta_{T}(C)\geqslant 3$ and that it is a tip of $T$ if $C\neq 0$ and $\beta_{T}(C)\leqslant 1$. If $C\subseteq T$ is a $(-1)$-curve such that $0<\beta_{T}(C)\leqslant 2$ and $C$ meets other components of $D$ normally (transversally) and at most once each then we say that $C$ is superfluous in $D$. If $T$ is a simple normal crossing (snc) divisor we say that it is snc-minimal if it contains no superfluous $(-1)$-curves. A rational tree is a reduced connected divisor of arithmetic genus zero. It is an snc-divisor with a simply connected support. It is a (rational) chain if it has no branching component. A rational chain is admissible if it has a negative definite intersection matrix and contains no $(-1)$-curve.

Assume $T$ is an admissible chain. It can be written as $T=T_{1}+\cdots+T_{k}$, $k\geqslant 0$ where $T_{i}\cong\mathbb{P}^{1}$, $i=1,\ldots,k$ are its irreducible components, $T_{i}\cdot T_{i+1}=1$ for $i\leqslant k-1$ and $T_{i}\cdot T_{j}=0$ if $|i-j|>1$. Once the order of components as above is fixed, we write $T=[-T_{1}^{2},\ldots,-T_{k}^{2}]$. Since $T$ is admissible, we have $T_{i}^{2}\leqslant-2$ for $i\in\{1,\ldots,k\}$. We put $\delta(T)=\frac{1}{d(T)}$. In particular, $\delta([(2)_{k}])=\frac{1}{k+1}$, where $(2)_{k}$ denotes the sequence $(2,\ldots,2)$ of length $k$. Following [Fuj82] we define the bark of $T$ as

[TABLE]

and the inductance of $T$ as

[TABLE]

We check that $T_{i}\cdot\operatorname{Bk}^{\prime}T$ equals $-1$ if $i=1$ and equals [math] otherwise and that $(\operatorname{Bk}^{\prime}T)^{2}=-\mathrm{ind}(T)$.

For a general reduced divisor $T$, an ordered rational chain $T_{1}+\ldots+T_{k}\subseteq T$ as above is called a rational twig of $T$ if $T_{1}$ is a tip of $T$ and $\beta_{T}(T_{i})=2$ for $i>1$. A rational tree $T$ is a (rational) fork if it has exactly one branching component and three maximal twigs (maximal in the sense of inclusion of supports). Assuming $T$ is snc-minimal, connected and does not contract to a quotient singular point (that is, it is neither $[1]$, nor a rational admissible chain nor a rational fork with admissible twigs and a negative definite intersection matrix), we define $\delta(T)$, $\operatorname{Bk}^{\prime}(T)$ and $\mathrm{ind}(T)$ as the sum of respective quantities computed for all maximal admissible twigs of $T$. For a general definition covering the case of resolutions of quotient singularities see [Miy01, §2].

If two trees $T_{1}$, $T_{2}$ meet normally at a unique point then, denoting the components which have a common point by $C_{j}\subseteq T_{j}$, $j=1,2$, from elementary properties of determinants we infer the following formula:

[TABLE]

Lemma 2.1**.**

Let $R=[a_{1},\ldots,a_{k}]$ be an admissible chain and let $R^{\prime}=[a_{1},\ldots,a_{k-1}]$. Then $\mathrm{ind}(R)>\mathrm{ind}(R^{\prime})$.

• Proof.

Put $d^{\prime}([a_{1},a_{2},\ldots,a_{k}]):=d([a_{2},\ldots,a_{k}])$ and $d^{\prime\prime}([a_{1},a_{2},\ldots,a_{k}]):=d^{\prime}([a_{2},\ldots,a_{k}])$, where by definition $d^{\prime}(0)=0$. By (2.5)

[TABLE]

hence $\mathrm{ind}([a_{1},\ldots,a_{k}])^{-1}=a_{1}-\mathrm{ind}([a_{2},\ldots,a_{k}])$. Using this formula we proceed by induction on $\#R$.

∎

Lemma 2.2** ([Fuj82], 4.16).**

Fix a $\mathbb{P}^{1}$-fibration of a smooth projective surface $X$ and a reduced divisor $D$ on $X$. For every fiber $F$ denote by $\sigma(F)$ the number of components of $F$ not contained in $D$ and put $\displaystyle\Sigma=\sum_{F\nsubseteq D}(\sigma(F)-1)$. Let $h$ and $\nu$ denote, respectively, the number of horizontal components of $D$ and the number of fibers contained in $D$. Then

[TABLE]

where $\rho(X)$ is the Picard rank of $X$.

Lemma 2.3** (Hurwitz formula).**

Let $F$ be a scheme-theoretic fiber of some fibration of a smooth projective surface and let $E$ be a rational cuspidal curve not contained in fibers. Denote by $r_{F}(q)$ the ramification index at $q\in E$ of the restriction of the fibration to $E$. Then for every $p\in E$ we have

[TABLE]

In particular, $\#\operatorname{Sing}E\leqslant 2F\cdot E-r_{F}(p)$.

Lemma 2.4**.**

(Producing elliptic fibrations). Let $E$ be a smooth rational curve on a smooth birationally ruled projective surface $X$ and let $C$ be a $(-1)$-curve such that $E\cdot C=2$. Then $X$ is rational and the following hold.

1. (a)

If $h^{0}(2K+E)\neq 0$ and $E^{2}=-4$ then $|E+2C|$ induces an elliptic fibration of $X$. 2. (b)

If $E^{2}=-3$ then after blowing up once some point on $E\setminus C$ the linear system of the proper transform of $E+2C$ induces an elliptic fibration.

• Proof.

Clearly, $X$ is not $\mathbb{P}^{2}$. Let $X\longrightarrow B$ be a $\mathbb{P}^{1}$-fibration. Since all fibers are rational trees, $E$ or $C$ is horizontal, so $B$ is rational. It follows that $X$ is rational. To find the necessary elliptic fibrations for (a) and (b) we may replace $X$ with its image after the contraction of $C$. Now $E$ becomes a nodal rational curve, so $E\cdot(K+E)=0$.

(a) We have $K\cdot E=E^{2}=0$, so $(2K+E)^{2}=4K^{2}$. Suppose that $K^{2}<0$. Then the divisor $2K+E$, which is effective by assumption, is not nef. Since $E^{2}=0$, we have $X\ncong\mathbb{P}^{2}$, so there exists a curve $\ell$ such that $(2K+E)\cdot\ell<0$ and $\ell^{2}\leqslant 0$. We have $\ell\neq E$, so $K\cdot\ell<0$, hence $\ell\cong\mathbb{P}^{1}$. The curve $\ell$ is not a [math]-curve, because otherwise it is a fiber of a $\mathbb{P}^{1}$-fibration of $X$, for which $(2K+E)\cdot\ell\geqslant 0$, as $h^{0}(2K+E)\neq 0$. Thus $\ell^{2}<0$, hence $\ell$ is a $(-1)$-curve with $\ell\cdot E\leqslant 1$. But if $\ell\cdot E=1$ then after the contraction of $\ell$ the image of $E$ has arithmetic genus $1$ and intersects the canonical divisor negatively, hence is in the fixed part of the direct image of $2K+E$, which is impossible, as $h^{0}(2K)=0$. Thus $\ell\cdot E=0$. Contracting $\ell$ we reduce the proof inductively to the case $K^{2}\geqslant 0$.

Now we argue as in the proof of [MKM83, Theorem 3.3, Claim]. By Riemann-Roch we have $h^{0}(-K)\geqslant K^{2}+1\geqslant 1$ and $h^{0}(K+E)\geqslant p_{a}(E)=1$, where $p_{a}$ denotes the arithmetic genus. Write

[TABLE]

Clearly, $E$ is not in the fixed part of $|K+E|$. It is also not in the fixed part of $|-K|$, because otherwise $h^{0}(-E)=h^{0}((2K+E)+2(-K-E))\geqslant 1$, which is impossible. Thus $E$ is linearly equivalent to an effective divisor whose support does not contain $E$, which gives $h^{0}(E)\geqslant 2$ and shows that the linear system $|E|$ has no fixed components. But $E^{2}=0$, so it has no base points either. Since $E$ is a connected reduced member of the system, a general member is smooth and reduced of genus $p_{a}(E)=1$, hence it is an elliptic curve.

(b) After the contraction of $C$ we have $E^{2}=1$. By Riemann-Roch $h^{0}(E)\geqslant E^{2}+1=2$. Let $U\in|E|$ be a smooth member. We have $U\cdot E=E^{2}=1$, so $U$ meets $E$ normally in a unique smooth point $p$. Let $E^{\prime}$ be the proper transform of $E$ under the blow-up of $p$. Then $h^{0}(E^{\prime})\geqslant 2$ and $|E^{\prime}|$ has no fixed components. Since $(E^{\prime})^{2}=0$, the linear system $|E^{\prime}|$ induces an elliptic fibration of the new surface, as required.

∎

The following example shows that the assumption $h^{0}(2K+E)\neq 0$ in Lemma 2.4(a) is necessary.

Example 2.5**.**

Let $\bar{E}$ be an irreducible planar cubic with a singular point $q$ and let $p_{1},\ldots,p_{9}\in\bar{E}\setminus\{q\}$ be distinct points. Let $\theta\colon X\longrightarrow\mathbb{P}^{2}$ denote the blowup at $p_{1},\ldots,p_{9}$ and $E$ the proper transform of $\bar{E}$. We have $E^{2}=0$ and $K+E\sim 0$, hence $h^{0}(2K+E)=h^{0}(K)=0$. Assume that $E$ coincides with the support of some fiber of an elliptic fibration of $X$. Then $F\sim nE$ for some positive integer $n$ and a general fiber $F$. The curve $\theta(F)$ meets $\bar{E}$ exactly in $p_{1},\ldots,p_{9}$, at each point with multiplicity $n$. We have $\theta(F)\sim 3n\ell$, where $\ell$ is a line, so intersecting with $\bar{E}$ we obtain $n(p_{1}+\cdots+p_{9})\sim 9no$, where $o\in\bar{E}$ is any flex point of $\bar{E}\setminus\{q\}$. Thus $n(p_{1}+\ldots+p_{9})=0$ in the group law of $(\bar{E}\setminus\{q\},o)$. The group is isomorphic to $(\mathbb{C}^{*},1)$ if $q\in\bar{E}$ is a node and to $(\mathbb{C}^{1},0)$ if $q\in\bar{E}$ is a cusp, so the latter equality is impossible for a general choice of points $p_{1},\ldots,p_{9}$. This shows that in general $\operatorname{Supp}E$ is not the support of an elliptic fiber. To obtain a configuration as in Lemma 2.4(a) we blow up once at the singular point of $E$.

Let $\bar{E}\subset\mathbb{P}^{2}$ be a rational cuspidal curve. Denote by $\pi_{0}\colon(X_{0},D_{0})\longrightarrow(\mathbb{P}^{2},\bar{E})$ the minimal weak resolution of singularities, that is, a composition of a minimal sequence of blow-ups such that the proper transform $E_{0}\subseteq X_{0}$ of $\bar{E}$ is smooth. Clearly, $\mathbb{P}^{2}\setminus\bar{E}=X_{0}\setminus D_{0}$. We denote the cusps of $\bar{E}$ by $q_{1},\ldots,q_{c}$. Since $(\operatorname{Pic}\mathbb{P}^{2})\otimes\mathbb{Q}$ is generated by $\bar{E}$, the components of $D_{0}$ freely generate $(\operatorname{Pic}X_{0})\otimes\mathbb{Q}$. The divisor $D_{0}-E_{0}$ has a negative definite intersection matrix.

Let $Q_{j}\subseteq D_{0}$ be the reduced exceptional divisor over the cusp $q_{j}$, $j\in\{1,\ldots,c\}$. It is a rational tree with a negative definite intersection matrix and a specific dual graph whose all vertices have degree not bigger than $3$. The Eisenbud–Neumann diagram of this graph (defined as the image of the graph after the contraction of vertices of degree $2$) is

Since every $q_{j}\in\bar{E}$ is locally analytically irreducible, $Q_{j}$ can be seen as being produced by a connected sequence of blow-ups, i.e. we can decompose the morphism contracting it to a point into a sequence of blow-ups $\sigma_{1}\circ\ldots\circ\sigma_{s}$, such that the center of $\sigma_{i+1}$ belongs to the exceptional curve of $\sigma_{i}$ for $i\geqslant 1$. Thus the components of $Q_{j}$ are linearly ordered as proper transforms of subsequent exceptional divisors of $\sigma_{1},\sigma_{2},\ldots,\sigma_{s}$. The last one, call it $C_{j}$, is the unique $(-1)$-curve in $Q_{j}$. It is contained in some twig of $Q_{j}$. Although $E_{0}$ is smooth, $\pi_{0}$ is not a log resolution, so there may be (at most one) component of $Q_{j}-C_{j}$ meeting $E_{0}$, call it $\widetilde{C}_{j}$; put $\widetilde{C}_{j}=0$ if there is no such. We have $\widetilde{C}_{j}\cdot E_{0}=1$ if $\widetilde{C}_{j}\neq 0$.

Definition 2.6**.**

A cusp of a planar curve is semi-ordinary if it is locally analytically isomorphic to the singular point of $x^{2}=y^{2m+1}$ at $(0,0)\in\operatorname{Spec}\mathbb{C}[x,y]$ for some $m\geqslant 1$.

Note that an ordinary cusp (called also simple) is a semi-ordinary cusp with $m=1$. The exceptional divisors of the minimal log resolution and of the minimal weak resolution of a semi-ordinary cusp are $[2,1,3,(2)_{m-1}]$ and $Q_{j}=[1,(2)_{m-1}]$, respectively. We have $C_{j}\cdot E_{0}=2$ and $\widetilde{C}_{j}=0$ for semi-ordinary cusps.

3. Almost minimal models

Let $\bar{E}\subseteq\mathbb{P}^{2}$ be a curve homeomorphic to the projective line, that is, a rational cuspidal curve. We will study the minimal log resolution $\pi\colon(X,D)\longrightarrow(\mathbb{P}^{2},\bar{E})$ using minimal model program techniques. For basic notions and theorems of the program we refer the reader to [KK94] and [Mat02, KM98]. In this article any final output of the birational part of a run of the MMP (which is a birational morphism) will be called a minimal model, even if $\kappa=-\infty$. Recall that the program applied to $(X,rD)$, where $r\in\mathbb{Q}\cap[0,1]$, produces a sequence of contractions of log exceptional curves, that is, curves which on the respective image of $X$ have negative self-intersection numbers and intersect the direct image of $K_{X}+rD$ negatively. It is well-known that for $r=1$ minimal models of log smooth surfaces are singular in general and that arbitrary quotient singularities may appear. However, surfaces with quotient singularities are not understood well enough (even the log del Pezzo surfaces of rank one), which causes problems with applications of the MMP techniques to concrete problems concerning log surfaces or quasi-projective surfaces. The main point of the construction of an almost minimal model is to avoid deeper log singularities introduced by the usual run of the log MMP, or rather to delay their introduction until all necessary contractions in the open part of the surface are done.

To prove Theorem 1.1 we may, and will, assume that the number of cusps of $\bar{E}$ is at least $4$. By [Wak78] the surface $\mathbb{P}^{2}\setminus\bar{E}$ is of log general type. It is also $\mathbb{Q}$-acyclic, hence as a consequence of the logarithmic Bogomolov-Miyaoka-Yau inequality [MT92] (see also [Pal11, §10]), it does not contain affine lines. It follows that $(X,rD)$ for $r=1$ is almost minimal in the sense of [Miy01, §2.3.11]; equivalently, the morphism onto a minimal model contracts only components of $D$. Unfortunately, the latter fact turned out to be of limited use when studying the surface $\mathbb{P}^{2}\setminus\bar{E}$ because of the complicated geometry of possible divisors $D$. We will therefore use a generalization of the theory of almost minimal models to the case of fractional boundaries proposed in [Pal19]. We work with $r=\frac{1}{2}$. This choice turns out to be optimal for many reasons. Now new curves in $X\setminus D$ will be contracted and we need to carefully control the whole process.

Instead of working with the minimal log resolution it is in fact more convenient to work with the minimal weak resolution $\pi_{0}\colon(X_{0},D_{0})\longrightarrow(\mathbb{P}^{2},\bar{E})$, as defined in the previous section. As an outcome of the construction of an almost minimal model of $(X_{0},\frac{1}{2}D_{0})$ given in [Pal19, §3] we obtain a sequence of birational contractions between smooth projective surfaces

[TABLE]

where, $D_{i+1}$ is a reduced divisor, a direct image of $D_{i}$ and $X_{i+1}\setminus D_{i+1}$ is an open affine subset of $X_{i}\setminus D_{i}$. Let us recall this inductive construction. Assume $(X_{i},D_{i})$ is defined already. Write $K_{i}$ for the canonical divisor on $X_{i}$. By a $(-2)$-twig of $D_{i}$ we mean a twig of $D_{i}$ consisting of $(-2)$-curves. A maximal $(-2)$-twig is a $(-2)$-twig which is maximal with respect to the inclusion of supports. Such twigs intersect trivially with the canonical divisor and hence their role in the construction is special. There are also some specific $(-1)$-curves in $D_{i}$ we need to take care of.

Notation 3.1**.**

Let $(X_{i},D_{i})$ be as above.

1. (a)

Let $\Delta_{i}$ be the sum of all maximal $(-2)$-twigs of $D_{i}$. 2. (b)

Let $\Upsilon_{i}$ be the sum of $(-1)$-curves $L$ in $D_{i}$, for which either $\beta_{D_{i}}(L)=3$ and $L\cdot\Delta_{i}=1$ or $\beta_{D_{i}}(L)=2$ and $L$ meets exactly one component of $D_{i}$. 3. (c)

Decompose $\Delta_{i}$ as $\Delta_{i}=\Delta^{+}_{i}+\Delta^{-}_{i}$, where $\Delta^{+}_{i}$ consists of those maximal $(-2)$-twigs of $D_{i}$ which meet $\Upsilon_{i}$. 4. (d)

Put $D_{i}^{\flat}=D_{i}-\Upsilon_{i}-\Delta^{+}_{i}-\operatorname{Bk}^{\prime}\Delta^{-}_{i}$.

Definition 3.2**.**

Let $(X_{i},D_{i})$ be as above. A curve $A\subseteq X_{i}$ is almost log exceptional on $(X_{i},\frac{1}{2}D_{i})$ if $A$ is a $(-1)$-curve not contained in $D_{i}$ such that

[TABLE]

and the component of $\Delta^{-}_{i}$ meeting $A$ is a tip of $\Delta^{-}_{i}$ (but not necessarily a tip of $D_{i}$).

Note that $A\cap(X_{i}\setminus D_{i})\cong\mathbb{C}^{*}$, hence $e_{top}(X_{i}\setminus(D_{i}+A))=e_{top}(X_{i}\setminus D_{i})$. If there is no almost log exceptional curve on $(X_{i},\frac{1}{2}D_{i})$ then we put $n=i$ and we call $n$ the length of the chosen process of almost minimalization and $(X_{n},\frac{1}{2}D_{n})$ an almost minimal model of $(X_{0},\frac{1}{2}D_{0})$. Otherwise we choose $A_{i+1}=A$ as above and we define a birational morphism

[TABLE]

as a local snc-minimalization of $D_{i}+A_{i+1}$, that is, a composition of a maximal (not necessarily unique) sequence of contractions of superfluous $(-1)$-curves contained in $D_{i}+A_{i+1}$ and its images starting from $A_{i+1}$ such that the total contracted divisor is connected, see Figure 2. By [Pal19, Lemma 3.4] each component of $\Upsilon_{i}$ meets at most one component of $\Delta_{i}$. In particular, $\Upsilon_{i}+\Delta_{i}$ can be contracted by a birational morphism. The following lemma is a motivation for the above definition of an almost log exceptional curve. It shows that the above definition of an almost minimal model is analogous to the original definition in the theory of open surfaces known in case of a reduced simple normal crossing boundary divisor (see [Miy01, 2.3.11, 2.4.3]).

Lemma 3.3**.**

Let $(X_{i},D_{i})$ be as above and let $\alpha_{i}\colon(X_{i},D_{i})\longrightarrow(Y_{i},D_{Y_{i}})$ be the contraction of $\Upsilon_{i}+\Delta_{i}$. Then

1. (a)

$\alpha_{i}^{*}(K_{Y_{i}}+\frac{1}{2}D_{Y_{i}})=K_{i}+\frac{1}{2}D_{i}^{\flat}$. 2. (b)

$A$ is almost log exceptional on $(X_{i},D_{i})$ if and only if $(\alpha_{i})_{*}A$ is log exceptional on $(Y_{i},D_{Y_{i}})$. 3. (c)

$\kappa(K_{i}+\frac{1}{2}D_{i})=\kappa(K_{X}+\frac{1}{2}D)$.

• Proof.

See [Pal19]: Lemma 3.4, Lemma 4.1(viii) and Corollary 3.5. ∎

Corollary 3.4**.**

Let $(X_{n},\frac{1}{2}D_{n})$ be an almost minimal model of $(X_{0},\frac{1}{2}D_{0})$ as defined above. One of the following holds:

1. (a)

$X_{n}\setminus D_{n}$ (and hence $\mathbb{P}^{2}\setminus\bar{E}$) has a $\mathbb{C}^{**}$-fibration with no base points on $X_{n}$, where $\mathbb{C}^{**}=\mathbb{C}^{1}\setminus\{0,1\}$, 2. (b)

$\kappa(K_{n}+\frac{1}{2}D_{n})\geqslant 0$ and $K_{n}+\frac{1}{2}D_{n}^{\flat}$ is the (numerically effective) positive part of the Zariski-Fujita decomposition of $K_{n}+\frac{1}{2}D_{n}$, 3. (c)

$-(K_{n}+\frac{1}{2}D_{n}^{\flat})$ is ample off $\Upsilon_{n}+\Delta_{n}$ and trivial on $\Upsilon_{n}+\Delta_{n}$; and $\rho(X_{n})=1+\#(\Upsilon_{n}+\Delta_{n})$.

• Proof.

See [Pal19], proof of Theorem 4.5(4),(6).

∎

Note that in case (c) the surface $(Y_{n},\frac{1}{2}D_{Y_{n}})$ is a log del Pezzo and $Y_{n}$ has at most canonical singularities; their number is equal to $b_{0}(\Delta_{n}^{-})$.

Example 3.5**.**

Let $(X_{0},D_{0})\longrightarrow(\mathbb{P}^{2},\bar{E})$ be the minimal weak resolution of the four-cuspidal quintic from the statement of Theorem 1.1. All cusps of $\bar{E}$ are semi-ordinary, so $D_{0}-E_{0}=\Delta_{0}^{+}+\Upsilon_{0}$, with $\Delta_{0}^{+}=[2,2]$, and the morphism $\alpha_{0}$ is simply the minimal weak resolution, i.e. $\alpha_{0}=\pi_{0}$, hence $Y_{0}=\mathbb{P}^{2}$. In particular, $n=0$. Clearly, $\rho(Y_{0})=1$ and $2K_{Y_{0}}+\bar{E}\sim-\ell$, where $\ell$ is a line on $\mathbb{P}^{2}$, so $-(K_{Y_{0}}+\frac{1}{2}\bar{E})$ is ample.

We recall basic properties of the process of almost minimalization defined above. Given two effective $\mathbb{Q}$-divisors $T_{1}$, $T_{2}$ we denote by $T_{1}\wedge T_{2}$ the unique effective $\mathbb{Q}$-divisor such that for $j=1,2$ the divisors $T_{j}-T_{1}\wedge T_{2}$ are effective and have no common component.

Lemma 3.6**.**

1. (a)

$\operatorname{Exc}\psi_{i+1}$ is a chain equal to $V+A_{i+1}+W$, where $V$ and $W$ are zero or twigs of $D_{i}$. 2. (b)

The support of $D_{0}-\operatorname{Exc}\psi\wedge D_{0}$ is connected and simply connected. 3. (c)

If $U\subseteq D_{i}$ is a component not contracted by $\psi_{i+1}$ then $\beta_{D_{i+1}}(\psi_{i+1}(U))=\beta_{D_{i}+A_{i+1}}(U)$. 4. (d)

$\operatorname{Exc}\psi_{i+1}$ is disjoint from $\Delta_{i}^{+}+\Upsilon_{i}$. In particular, the process of almost minimalization does not touch exceptional divisors over semi-ordinary cusps.

• Proof.

(a) Since $D_{i}$ is connected, by the definition of $\psi_{i+1}$ and by (3.1) all components of $\operatorname{Exc}\psi_{i+1}$ have $\beta_{D_{i}+A_{i+1}}=2$, hence the connected components of $\operatorname{Exc}\psi_{i+1}-A_{i+1}$ are zero or are twigs of $D_{i}$. Note that adding $A_{i+1}$ to $D_{i}$ creates a new loop in the support of the divisor. Now (b) and (c) follow from (a) and from the fact that $D_{0}$ is connected and simply connected.

(d) Suppose that $i\geqslant 0$ is the smallest natural number such that $\psi_{i+1}$ touches $\Delta_{i}^{+}+\Upsilon_{i}$. By (3.1) the curve $A_{i+1}$ is disjoint from $\Delta_{i}^{+}+\Upsilon_{i}$, so we can write $\operatorname{Exc}\psi_{i+1}=V+A_{i+1}+W$ as in (a) with $V\cdot U>0$ for some component $U\subseteq\Upsilon_{i}$. By the definition of $\Upsilon_{i}$ it follows that $U$ meets $D_{i}-U$ in exactly three different points, two of which belong to distinct twigs of $D_{i}$. But then $U$ is not in the image of $\Upsilon_{0}$, so by (a) the third point is the image of some $A_{j}$ for $j\leqslant i$. But then $D_{0}-\operatorname{Exc}\psi\wedge D_{0}$ is not connected; a contradiction with (b).

∎

We now introduce more notation which helps to control the geometry of the divisors $D_{i}$. As above, $n$ denotes the length of the chosen process of almost minimalization. It equals the number of curves not contained in $D_{0}$ contracted by $\psi$.

Notation 3.7**.**

Let $(X_{i},D_{i})$, $i\in\{0,\ldots,n\}$ and $j\in\{1,\ldots,c\}$ be as above.

1. (a)

Denote by $\Upsilon_{i}^{0}$ the subdivisor of $\Upsilon_{i}$ consisting of components which do not meet $\Delta_{i}^{+}$. By definition, every component of $\Upsilon_{i}^{0}$ is a $(-1)$-curve meeting exactly one component of $D_{i}$. 2. (b)

Denote the numbers of semi-ordinary and ordinary cusps of $\bar{E}$ by $c_{0}$ and $c_{0}^{\prime}$, respectively. 3. (c)

Put $\tau_{j}=C_{j}\cdot E_{0}$ and $\tau=\tau_{1}+\cdots+\tau_{c}$ (see the notation in Section 2). 4. (d)

Put $s_{j}=1$ if $\widetilde{C}_{j}=0$ and $s_{j}=0$ otherwise. Put $s=s_{1}+\cdots+s_{c}$. 5. (e)

Put $\tau_{j}^{*}=\tau_{j}-s_{j}-1$ and $\tau^{*}=\tau_{1}^{*}+\cdots+\tau_{c}^{*}=\tau-s-c$. Note that $\tau_{j}^{*}\geqslant 0$. 6. (f)

Put

[TABLE] 7. (g)

Put $\delta=\delta(\Delta_{n}^{-})$.

Although $Q_{j}=\pi_{0}^{-1}(q_{j})$ has simple normal crossings, the divisor $D_{0}$, and in fact every divisor $D_{i}$ for $i=0,\ldots,n$, has a unique non-snc point over each $q_{j}$, which belongs to $E_{i}$. The exceptional divisor of the minimal log resolution

[TABLE]

over $q_{j}$ is $[1,(2)_{\tau_{j}-1}]$. Clearly, $(X,D)=(X_{0}^{\prime},D_{0}^{\prime})$ and $\pi=\pi_{0}\circ\varphi_{0}$. If $\mathbb{P}^{2}\setminus\bar{E}$ has no $\mathbb{C}^{**}$-fibration (cf. Proposition 3.9(c)) then by [Pal19, Lemmas 3.7 and 5.1(iv)], $D_{i}^{\prime}$ is snc-minimal. As discussed before, components of $Q_{j}$, and similarly of $Q_{j}^{\prime}=\pi^{-1}(q_{j})$, are naturally linearly ordered as proper transforms of blow-ups constituting $\pi_{0}^{-1}$, respectively $\pi^{-1}$. Consequently, the sets of maximal twigs of $D_{0}$ and of $D$ lying over each $q_{j}$ come with a natural linear order. Let $\Delta_{i}^{\prime}$ denote the sum of maximal $(-2)$-twigs of $D_{i}^{\prime}$.

Lemma 3.8**.**

1. (a)

The contribution of the first two maximal admissible twigs of $Q_{j}^{\prime}$ to $\mathrm{ind}(Q_{j}^{\prime})$ is strictly bigger than $\frac{1}{2}$. 2. (b)

$\mathrm{ind}(D_{i}^{\prime})+i\leqslant 5-p_{2}$ for every $0\leqslant i\leqslant n$. 3. (c)

$\frac{1}{2}b_{0}(\Delta_{0}^{\prime})+\frac{1}{3}c_{0}^{\prime}\leqslant\mathrm{ind}(D)\leqslant 5-p_{2}$.

• Proof.

For (a), (b) see [Pal14, Lemma 3.5, Lemma 4.2(iv)]. For the first inequality in (c) note that the inductance of a $(-2)$-twig is at least $\frac{1}{2}$ and that the exceptional divisor of a log resolution over an ordinary cusp is $[3,1,2]$.

∎

Recall that the arithmetic genus of a divisor $T$ is defined as $p_{a}(T):=\frac{1}{2}T\cdot(K+T)+1$.

Proposition 3.9**.**

Let $p_{2}$ and $n$ be as above and let $i\in\{0,1,\ldots,n\}$. The following hold:

1. (a)

$K_{i}\cdot(K_{i}+D_{i})=p_{2}-c-\tau^{*}-i$. 2. (b)

$p_{a}(D_{i})=i+\tau^{*}+c$  and  $\#D_{i}=\rho(X_{i})+i$. 3. (c)

If $c\geqslant 4$ then $\mathbb{P}^{2}\setminus\bar{E}$ does not admit a $\mathbb{C}^{**}$-fibration. 4. (d)

If $\mathbb{P}^{2}\setminus\bar{E}$ is not $\mathbb{C}^{**}$-fibered then

[TABLE] 5. (e)

$b_{0}(\Delta_{n})+\#\Upsilon_{n}^{0}\leqslant b_{0}(\Delta_{0})+c_{0}^{\prime}$.

• Proof.

For (a), (b) and (c) see [Pal19]: Lemma 4.3 and Lemma 5.1(iv).

(d), (e) By the proof of [Pal19, Lemma 4.5(6)] we have

[TABLE]

and, in case $\mathbb{P}^{2}\setminus\bar{E}$ is not $\mathbb{C}^{**}$-fibered,

[TABLE]

Since $b_{0}(\Delta_{i}^{\prime})=b_{0}(\Delta_{i})+s$ (see the proof of Lemma 4.4 loc.cit.), the first inequality gives (e). Since $\rho(X_{n}^{\prime})+n=\#D_{n}^{\prime}=\#D_{n}+\tau=\#D_{n}+\tau^{*}+s+c$, the second inequality gives (d).

∎

Note that the inequality (3.3) is a consequence of the fact that if $\mathbb{P}^{2}\setminus\bar{E}$ is not $\mathbb{C}^{**}$-fibered (and is of log general type) then $2K_{n}+D_{n}^{\flat}$ is either nef or anti-ample, so $(2K_{n}+D_{n}^{\flat})^{2}\geqslant 0$.

4. Geometric restrictions on exceptional divisors.

Let the notation be as before. In particular, $\bar{E}\subseteq\mathbb{P}^{2}$ is a rational cuspidal curve with $c$ cusps and $\pi_{0}\colon(X_{0},D_{0})\longrightarrow(\mathbb{P}^{2},\bar{E})$ is the minimal weak resolution of singularities. In [Pal19, Theorem 1.4] we proved that $c\leqslant 6$. We now show that $c=6$ is impossible.

Proposition 4.1**.**

A rational cuspidal planar curve has at most $5$ cusps.

• Proof.

Suppose that $c\geqslant 6$. By Proposition 3.9(c),(d),(e)

[TABLE]

By Lemma 3.8(c), $s+b_{0}(\Delta_{0})=b_{0}(\Delta_{0}^{\prime})\leqslant 10-2p_{2}-\frac{2}{3}c_{0}^{\prime}$, hence

[TABLE]

We have $\#\psi_{*}Q_{j}+\tau_{j}\geqslant 3$ and the inequality is strict for non-ordinary cusps $q_{j}\in\bar{E}$, so $\#D_{n}+\tau\geqslant 1+3c_{0}^{\prime}+4(c-c_{0}^{\prime})$. Then (4.2) gives

[TABLE]

hence $7\leqslant 4c-17\leqslant p_{2}+\frac{4}{3}c_{0}^{\prime}$. By Lemma 3.8(a),(c) we have $2p_{2}\leqslant 9-c\leqslant 3$, so $p_{2}\leqslant 1$ and hence $c_{0}^{\prime}\geqslant 5$. Then $4<5\cdot\frac{5}{6}\leqslant\mathrm{ind}(D)\leqslant 5-p_{2}$, so $p_{2}=0$. Now the above inequality gives $2c+(c-c_{0}^{\prime})\leqslant 12$, hence $c_{0}^{\prime}=c=6$. By the genus-degree formula $\deg\bar{E}=5$. The restriction to $\bar{E}$ of the projection from any of the cusps has degree $3$. By Lemma 2.3, $\#\operatorname{Sing}E<2F\cdot E=6$; a contradiction.

∎

To prove Theorem 1.1 we may, and will, assume from now on that $c=5$ or $c=4$. We will proceed in a similar way as in the proof of Proposition 4.1, but now we need to couple previous arguments with a more detailed analysis of the geometry of the divisors $Q_{j}$ and of the minimalization process

[TABLE]

The morphism $\psi_{i+1}$ lifts to the level of minimal log resolutions, see (3.2), $\varphi_{i}\colon(X_{i}^{\prime},D_{i}^{\prime})\longrightarrow(X_{i},D_{i}),$ where $(X_{0}^{\prime},D_{0}^{\prime})=(X,D)$. We denote the lift by $\psi_{i+1}^{\prime}\colon(X_{i}^{\prime},D_{i}^{\prime})\longrightarrow(X_{i+1}^{\prime},D_{i+1}^{\prime})$ and we put $\psi^{\prime}=\psi^{\prime}_{n}\circ\dots\circ\psi^{\prime}_{1}$. We get a commutative diagram:

[TABLE]

We use log resolutions mostly to compute the inductance and to effectively use the logarithmic Bogomolov-Miyaoka-Yau inequality. We denote the proper transform of $E_{0}$ on $X_{i}^{\prime}$ by $E_{i}^{\prime}$. By $Q_{j}^{\prime}\subseteq D$ we denote the exceptional divisor of $\pi_{0}\circ\varphi_{0}$ over $q_{j}$. Clearly, $\varphi_{0}(Q_{j}^{\prime})=Q_{j}$. We define $\mathrm{ind}_{j}^{(i)}$ as the sum of inductances for maximal (admissible) twigs of $D_{i}^{\prime}$ contained in the image of $Q_{j}^{\prime}$ under $\psi_{i}^{\prime}\circ\dots\circ\psi_{1}^{\prime}$. Since $c\neq 1$, $E_{i}^{\prime}$ is not a twig of $D_{i}^{\prime}$, hence $\mathrm{ind}(D_{i}^{\prime})=\mathrm{ind}_{1}^{(i)}+\cdots+\mathrm{ind}_{c}^{(i)}$. The log BMY inequality, Lemma 3.8(b), reads for $(X_{i}^{\prime},D_{i}^{\prime})$ as

[TABLE]

We write $\mathrm{ind}_{j}:=\mathrm{ind}_{j}^{(0)}$ for simplicity. Note that $\mathrm{ind}_{j}=\mathrm{ind}(Q_{j}^{\prime})$, but things are more complicated for $\mathrm{ind}_{j}^{(i)}$ with $i>0$, as for instance twigs of images of distinct $Q_{j}^{\prime}$ may meet, and hence will no longer be twigs of $D_{i}^{\prime}$.

Recall that given two effective $\mathbb{Q}$-divisors $T_{1}$, $T_{2}$ we denote by $T_{1}\wedge T_{2}$ the unique effective $\mathbb{Q}$-divisor such that for $j=1,2$ the divisors $T_{j}-T_{1}\wedge T_{2}$ are effective and have no common component. Denote by $\Delta_{i}^{\mathrm{tip}}$ the sum of $(-2)$-tips of $D_{i}$. Each connected component of $\Delta_{i}$ contains a unique component of $\Delta_{i}^{\mathrm{tip}}$. Put

[TABLE]

and define the integers:

[TABLE]

An important ingredient in our analysis is combining the inequality (4.3) with the inequality (3.3). Since $\#\Omega_{j}=\#(\bar{Q}_{j}-\Upsilon_{n}^{0}\wedge\bar{Q}_{j})-b_{0}(\Delta_{n}\wedge\bar{Q}_{j})$, we can write the latter as

[TABLE]

We will see below that $p_{2}\leqslant 2$, so the right hand side is bounded from above by $13$, hence $\lambda_{1}+\ldots+\lambda_{c}$, which roughly measures the number of components of $\psi(D_{0}-E_{0})$, cannot be too big. We now proceed to describe cusps $q_{j}\in\bar{E}$ with small $\lambda_{j}$. We first collect some simple observations.

Lemma 4.2**.**

1. (a)

We have $\#\Omega_{j}=0$ if and only if $q_{j}\in\bar{E}$ is an ordinary cusp. 2. (b)

If $q_{j}\in\bar{E}$ is a semi-ordinary cusp then $\lambda_{j}=\#\Omega_{j}+1=\#Q_{j}$. 3. (c)

If $s_{j}=0$ then $\#\Omega_{j}\geqslant 2$ and $\lambda_{j}\geqslant 4$. 4. (d)

If $\tau_{j}=2$ then $\psi$ does not touch $C_{j}$. 5. (e)

We have $\lambda_{j}=2$ if and only if $q_{j}$ is a semi-ordinary cusp with $Q_{j}=[1,2]$. In particular, $\mathrm{ind}^{(n)}_{j}=\frac{11}{10}$ in case $\lambda_{j}=2$. 6. (f)

$p_{2}\in\{0,1,2\}$. 7. (g)

If $s_{j}=1$ then $\mathrm{ind}_{j}^{(n)}\geqslant 1-\frac{1}{\tau_{j}}\geqslant\frac{1}{2}$. 8. (h)

If $\lambda_{j}=4$ then $\mathrm{ind}_{j}^{(n)}\geqslant\frac{1}{2}$ or $\psi(C_{j})$ is a $(-1)$-tip of $\bar{Q}_{j}$, $\tau_{j}=2$ and $\mathrm{ind}_{j}^{(n)}\geqslant\frac{1}{3}$.

• Proof.

(a) We have $\#\Omega_{j}=0$ if and only if $\psi_{*}C_{j}\subseteq\Upsilon_{n}^{0}$, which is equivalent to $C_{j}\subseteq\Upsilon_{0}^{0}$, hence to $q_{j}\in\bar{E}$ being an ordinary cusp.

(b) We have $Q_{j}\subseteq\Upsilon_{0}+\Delta_{0}^{+}$, so $Q_{j}$ is not touched by $\psi$. Also, $\lambda_{j}-1=\#\Omega_{j}$ and there is a unique component of $\bar{Q}_{j}$ in $\Upsilon_{n}^{0}+\Delta_{n}^{\mathrm{tip}}$, so $\#\Omega_{j}=\#\bar{Q}_{j}-1$.

(c) By Lemma 3.6(a), $C_{j}$ and $\widetilde{C}_{j}$ are not contracted by $\psi$, so $\psi(C_{j}+\widetilde{C}_{j})$ is a part of $\Omega_{j}$, hence $\lambda_{j}+s_{j}\geqslant\tau_{j}+2\geqslant 4$.

(d) We may assume that $j=1$. Suppose that $\tau_{1}=2$ and $\psi$ touches $C_{1}$. By Lemma 3.6 any connected component of $\operatorname{Exc}\psi$ meeting $C_{1}$ is a chain $T+A+W$, where $A$ is the proper transform of one of the curves $A_{i+1}\subseteq X_{i}$ and $T$, $W$ are twigs of $D_{0}$, say $\#W\leqslant\#T$. Contract $(-1)$-curves in $T+A+W$ until $C_{1}$ is touched once and denote the resulting morphism by $\eta\colon X_{0}\longrightarrow\eta(X_{0})$. The linear system of $F=\eta(C_{1})$ defines a $\mathbb{P}^{1}$-fibration of $\eta(X_{0})$.

Suppose that $(W+A+T)\cdot E_{0}=0$. Then $F\cdot\eta(E_{0})=C_{1}\cdot E_{0}=2$. We may assume that $W+T\subseteq Q_{1}+Q_{2}$. For $j>2$ the divisor $\eta(Q_{j})$ is contained in some fiber. The restriction of the fibration to $\eta(E_{0})$ has degree $2$ and has at least three ramification points, namely the points $\eta(E_{0})\cap\eta(C_{j})$ for $j=1,3,4$. This is a contradiction with the Hurwitz formula.

Thus $W=0$ and $A\cdot E_{0}=1$. Then $T=[(2)_{k}]$ for some $k\geqslant 1$ and $T+A$ is a connected component of $\operatorname{Exc}\psi$. We may assume that $T+A=\operatorname{Exc}\psi_{1}$, hence $\eta=\psi_{1}$. The restriction of the fibration to $E_{1}$ has degree $F\cdot E_{1}=3$ and has ramification points at $E_{1}\cap\psi_{1}(C_{j})$ for $j=1,\ldots,c$. By the Hurwitz formula $c=4$ and all ramification indices are equal to $2$. It follows that the cusps $q_{2},q_{3},q_{4}$ are semi-ordinary, so $\psi$ does not touch $Q_{2}+Q_{3}+Q_{4}$. For $j=2,3,4$ we write $Q_{j}=[1,(2)_{t_{j}}]$, where $t_{j}\geqslant 0$ and we compute $\mathrm{ind}_{j}=\frac{5}{6}+\frac{4t_{j}}{3(2t_{j}+3)}\geqslant\frac{5}{6}$. The inequalities (4.3) and (4.6) read as

[TABLE]

[TABLE]

We have $n>0$, so $p_{2}\leqslant 1$ and $n\leqslant 2$. Since $q_{1}$ is not semi-ordinary, there is a component $U\subseteq Q_{1}-T-C_{1}$ meeting $C_{1}$. By the contractibility of $Q_{1}$ to a smooth point $U$ is unique and $\psi_{1}(U)=[k+2]$. We denote by $\theta\colon X_{n}\longrightarrow\theta(X_{n})$ the contraction of $\psi(Q_{2}+Q_{3}+Q_{4})$. We need to determine the geometry of $V=Q_{1}-T-C_{1}-U$.

First consider the case when $\psi_{*}(V)=0$. We have $\rho(\theta(X_{n}))=\#\theta(D_{n})-n\leqslant 3-n$. Suppose that $n=2$. Then $\psi$ does not contract $U$ and $\theta(X_{2})=\mathbb{P}^{2}$. In particular, $\theta(\psi_{2}(F))\subseteq\mathbb{P}^{2}$ is not a [math]-curve, so $\psi_{2}$ touches $F$. It follows that $A_{2}$ meets $F$ and hence $\psi_{2}$ touches $\psi_{1}(U)$ at most once. But then $\theta(\psi(U))^{2}<0$; a contradiction. Thus $n=1$. Now $\theta(X_{1})$ is a Hirzebruch surface, namely $\mathbb{F}_{k+2}$, and $\bar{U}:=\theta\circ\psi(U)$ is the negative section. Put $\bar{F}=\theta\circ\psi(C_{1})$. Using the multiplicity sequences of singularities of $\theta(E_{1})$ we compute that $p_{a}(\theta(E_{1}))=3+t_{2}+t_{3}+t_{4}$. On the other hand, on $\mathbb{F}_{k+2}$ we have $K_{\mathbb{F}_{k+2}}\sim-2\bar{U}-(k+4)\bar{F}$ and $\theta(E_{1})\sim 3\bar{U}+(3k+6+\alpha)\bar{F}$, where $\alpha=E_{1}\cdot\psi(U)$, so $p_{a}(\theta(E_{1}))=3k+4+2\alpha$, hence $t_{2}+t_{3}+t_{4}=3k+2\alpha+1\geqslant 4$. Since $\lambda_{1}\geqslant 2$, (4.8) gives $k=p_{2}=1$ and $\alpha=0$. But the latter gives $s_{1}=1$, so $\mathrm{ind}_{1}^{(1)}=\frac{1}{2}+\frac{1}{3}$, which contradicts (4.7).

We obtain $\psi_{*}(V)\neq 0$. Consider the case $p_{2}=1$. Then $n=1$ and $\mathrm{ind}_{1}^{(1)}\leqslant\frac{1}{2}$, which by Lemma 3.8(a) implies that $Q_{1}$ is a chain and $U=\widetilde{C}_{1}$. By the contractibility of $Q_{1}$ to a smooth point $Q_{1}=[(2)_{k},1,k+2,(2)_{k^{\prime}}]$ for some $k^{\prime}\geqslant 0$. Since $\psi_{*}(V)\neq 0$, we have $k^{\prime}\geqslant 1$ and then the inequality $\mathrm{ind}_{1}^{(1)}\leqslant\frac{1}{2}$ fails (see Lemma 2.1); a contradiction.

We are left with the case $p_{2}=0$. Suppose that $n=2$. By (4.7), $\mathrm{ind}_{1}^{(2)}\leqslant\frac{1}{2}$. If $\psi_{2}$ touches $\psi_{1}(C_{1})$ then by the argument above (with the proper transform of $A_{2}$ taken for $A$) we get $\operatorname{Exc}\psi_{2}=A_{2}+T^{\prime}$, where $T^{\prime}\neq 0$ is a $(-2)$-twig of $D_{1}$ meeting $\psi_{1}(C_{1})$. Since there is no such, $\psi_{2}$ does not touch $\psi_{1}(C_{1})$. If $A_{2}\cdot E_{1}=0$ then $A_{2}\cdot\psi_{1}(Q_{1}-C_{1})=2$, so since $\psi_{1}(Q_{1}-C_{1})$ is connected, there is a rational circular divisor (loop) in $\psi_{1}(Q_{1}-C_{1})+A_{2}$ and hence the intersection of this divisor with a general fiber is at least $2$. The latter is impossible, because $\psi_{1}(Q_{1}-C_{1}-U)+A_{2}$ is vertical and $\psi_{1}(U)$ is a $1$-section. So we infer that $A_{2}\cdot E_{1}=1$. If $s_{1}=1$ then, since $\mathrm{ind}_{1}^{(2)}\leqslant\frac{1}{2}$, we see that $\psi_{1}(Q_{1})+A_{2}$ is a chain as above, in which case $\psi_{*}(V)=0$, contrary to the claim above. Therefore, $s_{1}=0$ and hence $U=\widetilde{C}_{1}$. It follows that $\#\Omega_{1}\geqslant 2$, so $\lambda_{1}\geqslant\#\Omega_{1}+2\geqslant 4$. By (4.8), $q_{2},q_{3},q_{4}$ are ordinary cusps, $\Delta_{2}^{-}=0$ and $\lambda_{1}=4$, so $\psi(Q_{1}-C_{1}-\widetilde{C}_{1})\subseteq\Upsilon_{2}^{0}+\Delta_{2}^{\mathrm{tip}}\wedge\Delta_{2}^{+}$. But since $A_{2}$ meets $E_{1}$, $\psi_{2}$ does not create a component of $\Upsilon_{2}^{0}$, which gives $\psi(Q_{1}-C_{1}-\widetilde{C}_{1})\subseteq\Delta_{2}^{+}$. We have $\psi_{1}(\widetilde{C}_{1})=[k+2]$ and $k\geqslant 1$, so since $\psi_{2}$ touches $\psi_{1}(\widetilde{C}_{1})$ at most once, $\psi(\widetilde{C}_{1})\nsubseteq\Upsilon_{2}$, hence $\Upsilon_{2}\wedge\psi(Q_{1})=0$ and so $\Delta_{2}^{+}\wedge\psi(Q_{1})=0$. Then $\psi_{*}(V)=0$; a contradiction.

We have now $p_{2}=0$ and $n=1$. Since $\operatorname{Exc}\psi=T+A_{1}$, we have $\Upsilon_{1}\wedge\psi(Q_{1})=0$, so $\Delta_{1}^{+}\wedge\psi(Q_{1})=0$. Suppose that $\lambda_{1}>3$. Then (4.8) implies that $q_{2},q_{3},q_{4}$ are ordinary, $\lambda_{1}=4$ and $\Delta_{1}^{-}=0$. We obtain $\psi(Q_{1})\wedge(\Delta_{1}+\Upsilon_{1})=0$. Then $4=\lambda_{1}=\#\psi(Q_{1}-C_{1})+3-s_{1}$, so $\psi(Q_{1}-C_{1})$ has $s_{1}+1\leqslant 2$ components, hence $\psi(Q_{1})$ is a chain. Since $Q_{1}$ contracts to a smooth point, we get $\psi(Q_{1})=[0,k+2,(2)_{s_{1}}]$. But $\psi(Q_{1})\wedge\Delta_{1}=0$, so $s_{1}=0$. Then $\psi_{*}(V)=0$; a contradiction. We obtain $\lambda_{1}\leqslant 3$, hence $s_{1}=1$ by Lemma 4.2(c). Since $\psi(U)=[k+2]\nsubseteq\Delta_{1}$, we infer that $\psi(Q_{1}-C_{1}-U)$ is contained in $\Delta_{1}^{\mathrm{tip}}\wedge\Delta_{1}^{-}$. Since $Q_{1}$ contracts to a point, $\psi(Q_{1})=[0,k+2,(2)_{m}]$ with $m\leqslant 1$ and $\lambda_{1}=3$. Since $\psi_{*}(V)\neq 0$, we have $m=1$. By Lemma 2.2 every fiber of the fibration induced by $|\theta\circ\psi(C_{1})|$ has at most one component not contained in $\theta(\psi(Q_{1}))$. Then the fiber containing $\Delta_{1}^{-}=[2]$ has two components, which is impossible; a contradiction.

(e) Assume that $\lambda_{1}=2$. We have $\#\Omega_{1}\geqslant 1$, because otherwise $q_{1}$ is an ordinary cusp by (a), in which case $\lambda_{1}=1$. Since $\lambda_{1}=\tau_{1}-s_{1}+\#\Omega_{1}\geqslant 1+\#\Omega_{1}$, we get $\tau_{1}-s_{1}=1$ and $\#\Omega_{1}=1$, hence $\tau_{1}=2$, $s_{1}=1$ and $\bar{Q}_{1}-\psi(C_{1})\subseteq\Upsilon_{n}^{0}+\Delta_{n}^{\mathrm{tip}}$. The divisor $Q_{1}$ is a chain, because otherwise by Lemma 3.6(c) the image of the branching component would contribute to $\Omega_{1}-\psi(C_{1})$. Suppose $C_{1}$ meets two components $T_{1}$ and $T_{2}$ of $Q_{1}$. By (d) those components are not contracted by $\psi$ and their images on $X_{n}$ are not in $\Upsilon_{n}^{0}$, hence are contained in $\Delta_{n}^{\mathrm{tip}}$. But then $T_{1}$ and $T_{2}$ are contained in $\Delta_{0}$, so $Q_{1}$ contains $[2,1,2]$ and hence is not negative definite; a contradiction. It follows that $C_{1}$ is a tip of $Q_{1}$. By the contractibility of $Q_{1}$ to a smooth point $Q_{1}=[1,2,\ldots,2]$, so $q_{1}$ is semi-ordinary. By (b), $Q_{1}=[1,2]$, so the exceptional divisor of the minimal log resolution of $q_{1}$ is $[2,3,1,2]$. We get $\mathrm{ind}^{(n)}_{1}=\mathrm{ind}_{1}=\frac{3}{5}+\frac{1}{2}=\frac{11}{10}$.

(f) By (4.3) and by Lemma 3.8(a), $2\leqslant\frac{1}{2}c<\mathrm{ind}(D)\leqslant 5-p_{2}$, so $p_{2}\leqslant 2$.

(g) The $(-2)$-chain created by resolving the tangency of $\psi(C_{j})$ and $E_{n}$ creates a maximal $(-2)$-twig $[(2)_{\tau_{j}-1}]$, so $\mathrm{ind}_{j}^{(n)}\geqslant 1-\frac{1}{\tau_{j}}\geqslant\frac{1}{2}$.

(h) We may assume that $j=1$ and $\mathrm{ind}_{1}^{(n)}<\frac{1}{2}$. Then $s_{1}=0$ by (g) and $\Delta_{n}\wedge\bar{Q}_{1}=0$. We have $4=\lambda_{1}\geqslant\#(\psi(C_{1})+\psi(\widetilde{C}_{1}))+\tau_{1}$, hence $\Omega_{1}=\psi(C_{1})+\psi(\widetilde{C}_{1})$ and $\tau_{1}=2$. It follows that $\bar{Q}_{1}-\psi(C_{1})-\psi(\widetilde{C}_{1})\subseteq\Upsilon_{n}^{0}$. By (d), $\psi$ does not touch $C_{1}$, hence $\psi(C_{1})$ is a $(-1)$-tip of $\bar{Q}_{1}$. Its proper transform is a $(-3)$-tip of $D_{n}^{\prime}$, so $\mathrm{ind}_{j}^{(n)}\geqslant\frac{1}{3}$.

∎

Lemma 4.3**.**

The process of almost minimalization of $(X_{0},\frac{1}{2}D_{0})$ contracts at most one curve not contained in $D_{0}$, that is, $n\leqslant 1$.

• Proof.

Recall that $c_{0}$ is the number of semi-ordinary cusps of $\bar{E}$. Denote by $\omega_{3}$ be the number of non-semi-ordinary cusps of $\bar{E}$ for which $\lambda_{j}=3$. By Lemma 4.2(c),(g) cusps with $\lambda_{j}=3$ have $s_{j}=1$ and $\mathrm{ind}_{j}^{(n)}\geqslant\frac{1}{2}$. The inequality (4.3) gives

[TABLE]

On the other hand, Lemma 4.2(e) and (4.6) give

[TABLE]

Multiplying the first inequality by $3$ and adding the second one we obtain $\frac{7}{2}c+\frac{1}{2}\omega_{3}+\frac{1}{2}(c-c_{0})+3n\leqslant 22-\delta$. In particular, $3n\leqslant 22-\frac{7}{2}\cdot 4=8$, so $n\leqslant 2$. Suppose that $n=2$. Then the previous inequality gives $\frac{7}{2}c\leqslant 16$, so $c=4$. We get

[TABLE]

Since $n\neq 0$, we have $c_{0}\leqslant c-1=3$ and we may assume that $\psi$ touches $Q_{1}$. By Lemma 4.2(e), (f), $\lambda_{1}\geqslant 3$ and $p_{2}\leqslant 2$.

Assume $c_{0}=0$. By Lemma 4.2(a),(e), $\lambda_{1}+\cdots+\lambda_{4}\geqslant 4\cdot 3=12$, so $p_{2}=2$. Then $\frac{1}{2}\omega_{3}\leqslant\sum_{j=1}^{4}\mathrm{ind}_{j}^{(n)}\leqslant 1$ by (4.9), so $\omega_{3}\leqslant 2$ and hence $\lambda_{1}+\cdots+\lambda_{4}\geqslant 2\cdot 4+2\cdot 3=14$; a contradiction with (4.10).

Assume $c_{0}=1$. Then $6-\omega_{3}+\delta\leqslant 3p_{2}\leqslant 6\frac{1}{2}-\frac{3}{2}\omega_{3}$, so $\omega_{3}\leqslant 1$ and $p_{2}=2$. Then $\omega_{3}=0$, so $\lambda_{1}+\cdots+\lambda_{4}\geqslant 3\cdot 4+1\geqslant 13$. By (4.10) in fact an equality holds, so $(\lambda_{1},\lambda_{2},\lambda_{3},\lambda_{4})=(4,4,4,1)$. By (4.9), $\sum_{j=1}^{3}\mathrm{ind}_{j}^{(2)}\leqslant 1-\frac{5}{6}=\frac{1}{6}$. This is a contradiction with Lemma 4.2(h).

Assume $c_{0}=2$, say $q_{3}$ and $q_{4}$ are semi-ordinary. We have $3-\omega_{3}+\delta\leqslant 3p_{2}\leqslant 4-\frac{3}{2}\omega_{3}\leqslant 4$, which gives $p_{2}=1$ and $\omega_{3}=0$. By (4.10), $(\lambda_{1},\lambda_{2},\lambda_{3},\lambda_{4})=(4,4,1,1)$. By (4.9), $\mathrm{ind}_{1}^{(2)}+\mathrm{ind}_{2}^{(2)}\leqslant 2-2\cdot\frac{5}{6}=\frac{1}{3}$. Again, this is a contradiction with Lemma 4.2(h).

We are left with the case $c_{0}=3$, say $q_{2}$, $q_{3}$ and $q_{4}$ are semi-ordinary. Then $\delta-\omega_{3}\leqslant 3p_{2}\leqslant\frac{3}{2}(1-\omega_{3})\leqslant\frac{3}{2}$, hence $p_{2}=0$ and $\omega_{3}\geqslant\delta$. By (4.10), $\lambda_{3}=\lambda_{4}=1$, that is, $q_{3}$ and $q_{4}$ are ordinary, and $\lambda_{1}+\lambda_{2}\leqslant 5-\delta$. By (4.9), $\mathrm{ind}_{1}^{(2)}+\mathrm{ind}_{2}^{(2)}\leqslant 3-2\cdot\frac{5}{6}=\frac{4}{3}$. If $\lambda_{2}=2$ then $\lambda_{1}=3$, so by Lemma 4.2(e),(g), $\mathrm{ind}_{1}^{(2)}+\mathrm{ind}_{2}^{(2)}\geqslant\frac{1}{2}+\frac{11}{10}>\frac{4}{3}$, which is false. Thus $\lambda_{2}=1$ and $\lambda_{1}\in\{3,4\}$. We obtain $\mathrm{ind}_{1}^{(2)}\leqslant\frac{4}{3}-\frac{5}{6}=\frac{1}{2}$.

Suppose that $s_{1}=1$. Since $\mathrm{ind}_{1}^{(2)}\leqslant\frac{1}{2}$, it follows that $\tau_{1}=2$ and $\bar{Q}_{1}+E_{1}$ has no tips. In particular, $\Delta_{2}\wedge\bar{Q}_{1}=0$. By Lemma 4.2(d), $\psi(C_{1})$ is a $(-1)$-curve meeting any other component of $\bar{Q}_{1}-\psi(C_{1})$ at most once, hence disjoint from $\Upsilon_{2}^{0}$. Let $\theta$ be the contraction of $\Upsilon_{2}^{0}+\psi(C_{1})$. We have $\rho(\theta(X_{2}))=\#\theta(\bar{Q}_{1}+E_{2})-2=\#(\bar{Q}_{1}-\psi(C_{1})-\Upsilon_{2}^{0}\wedge\bar{Q}_{1})-1=\lambda_{1}-3$, hence $\lambda_{1}=4$ and $\theta(E_{2})$ is a rational cuspidal curve on $\mathbb{P}^{2}$ with exactly four cusps, all ordinary. The genus-degree formula implies that such a curve does not exist; a contradiction.

Thus $s_{1}=0$ and, by Lemma 4.2(c), $\lambda_{1}=4$. Since $\delta\leqslant\omega_{3}$, we have $\Delta_{2}^{-}=0$. The curve $\psi(\widetilde{C}_{1})$ is a component of $\Omega_{1}$, hence $\Omega_{1}=\psi(C_{1})+\psi(\widetilde{C}_{1})$ and $\tau_{1}=2$. Thus $\bar{Q}_{1}-\psi(C_{1}+\widetilde{C}_{1})\subseteq\Delta_{2}^{\mathrm{tip}}+\Upsilon_{2}^{0}$. But since $\bar{Q}_{1}$ contains no component of $\Upsilon_{2}-\Upsilon_{2}^{0}$, it contains no component of $\Delta_{2}^{+}$, hence $\bar{Q}_{1}-\psi(C_{1}+\widetilde{C}_{1})\subseteq\Upsilon_{2}^{0}$. Consequently, no component of $Q_{1}$ other than (possibly) $\widetilde{C}_{1}$ is branching in $Q_{1}$. Since $\tau_{1}=2$, by Lemma 4.2(d), $\psi$ does not touch $C_{1}$, so $\psi(C_{1})\cdot\Upsilon_{2}^{0}=0$, hence $\psi(C_{1})$ is a $(-1)$-tip of $\bar{Q}_{1}$. Since $q_{2}$, $q_{3}$ and $q_{4}$ are semi-ordinary, $\psi$ does not touch $Q_{2}+Q_{3}+Q_{4}$, so the proper transform of $A_{i}$, $i\in\{1,2\}$ meets exactly one component of $T_{i}\wedge\Delta_{0}^{-}$, where $T_{i}$ is a twig of $Q_{1}$ meeting $\widetilde{C}_{1}$, other than $C_{1}$. We have $\psi(T_{i})\subseteq\Upsilon_{2}^{0}$, so the proper transform of $A_{i}$ meets $\widetilde{C}_{1}$ too, and $\psi$ contracts all but one component of $T_{i}$. It follows that $T_{1}+T_{2}\subseteq\Delta_{0}^{-}$, hence $Q_{1}$ is not negative definite; a contradiction.

∎

By Lemma 4.3 we have $n\leqslant 1$, so $\psi$ touches $Q_{j}$ if and only if $A_{1}\cdot Q_{j}>0$. For convenience we put $A_{1}=0$ if $n=0$.

Proposition 4.4**.**

Assume $\lambda_{j}=3$. Then $s_{j}=1$ and $\psi$ does not touch $Q_{j}$. Moreover, one of the following holds:

1. (1)

$\tau_{j}=2$. Then $Q_{j}=[1,2,2]$ or $Q_{j}=[2,1,3]$ or $Q_{j}=[2,1,3,2]$. We have $\mathrm{ind}_{j}=\frac{17}{14}$ or $\mathrm{ind}_{j}=\frac{4}{3}$ or $\mathrm{ind}_{j}=\frac{8}{5}$, respectively. 2. (2)

$\tau_{j}=3$. Then $Q_{j}=[1]$ or $Q_{j}=[1,2]$. We have $\mathrm{ind}_{j}=\frac{11}{12}$ and $\mathrm{ind}_{j}=\frac{26}{21}$, respectively.

• Proof.

Renumbering the cusps if necessary, we may assume that $j=1$. By Lemma 4.2(c), $s_{1}=1$, so $\tau_{1}=4-\#\Omega_{1}\leqslant 3$.

Consider the case $\tau_{1}=2$ and $A_{1}\cdot Q_{1}=0$. Then $\#\Omega_{1}=2$, so for some component $U\subseteq Q_{1}$ meeting $C_{1}$ we have $Q_{1}-U-C_{1}\subseteq\Upsilon_{0}^{0}+\Delta_{0}^{\mathrm{tip}}$. But $q_{1}$ is not ordinary, so $Q_{1}\wedge\Upsilon_{0}^{0}=0$. Since $Q_{1}$ contracts to a smooth point and $U\nsubseteq\Delta_{0}^{\mathrm{tip}}$, we infer that $Q_{1}$ has three or four components, hence $Q_{1}$ is $[1,2,2]$, $[2,1,3]$ or $[2,1,3,2]$. In the first case the exceptional divisor of the minimal log resolution over $q_{1}$ is $[2,1,3,2,2]$, hence $\mathrm{ind}_{1}=\frac{5}{7}+\frac{1}{2}=\frac{17}{14}$. In the second and third case the exceptional divisor is a fork with maximal twigs $[2]$, $[3]$, $[2,1]$ and $[2]$, $[2,3]$, $[2,1]$ respectively, which gives $\mathrm{ind}_{1}=\frac{1}{2}+\frac{1}{3}+\frac{1}{2}=\frac{4}{3}$ and $\mathrm{ind}_{1}=\frac{1}{2}+\frac{3}{5}+\frac{1}{2}=\frac{8}{5}$ respectively.

Consider the case $\tau_{1}=3$ and $A_{1}\cdot Q_{1}=0$. Then $\Omega_{1}=C_{1}$, so $Q_{1}-C_{1}\subseteq\Upsilon_{0}^{0}+\Delta_{0}^{\mathrm{tip}}$. Since $\tau_{1}\neq 2$, we have in fact $Q_{1}-C_{1}\subseteq\Delta_{0}^{\mathrm{tip}}$, so because $Q_{1}$ contracts to a smooth point, we get $Q_{1}=[1]$ or $Q_{1}=[1,2]$. The exceptional divisors of the minimal log resolution over $q_{1}$ are $[4,1,2,2]$ and $[2,4,1,2,2]$, respectively, so $\mathrm{ind}_{1}=\frac{1}{4}+\frac{2}{3}=\frac{11}{12}$ and $\mathrm{ind}_{1}=\frac{4}{7}+\frac{2}{3}=\frac{26}{21}$, respectively.

By Lemma 4.3, $n\leqslant 1$, so it remains to prove that $A_{1}\cdot Q_{1}=0$. Suppose that $A_{1}\cdot Q_{1}\neq 0$. We have $s_{1}=1$ and $\tau_{1}=4-\#\Omega_{1}\leqslant 3$.

(1∘) Assume $\tau_{1}=3$. Then $\Omega_{1}=\psi(C_{1})$, so

[TABLE]

Note that if $\psi(U)\subseteq\Upsilon_{1}^{0}+\Delta_{1}^{\mathrm{tip}}$ for some component $U$ of $Q_{1}$ not contained in the twigs of $D_{0}$ then in fact $\psi(U)\subseteq\Upsilon_{1}^{0}$, so by Lemma 3.6(a), $U\subseteq\Upsilon_{0}^{0}$. But $\Upsilon_{0}^{0}\wedge Q_{1}=0$, so we infer that images of all components of $Q_{1}$ not contained in twigs of $D_{0}$ are in $\Omega_{1}$. Since $\Omega_{1}=\psi(C_{1})$ and $C_{1}$ is non-branching in $Q_{1}$, we infer that $Q_{1}$ is a chain. We can write it as $Q_{1}=T_{1}+C_{1}+T_{2}$, where $T_{1}$ and $T_{2}$ are zero or twigs of $D_{0}$ contained in $Q_{1}$ such that $A_{1}\cdot T_{1}\geqslant A_{1}\cdot T_{2}$. The curve $A_{1}$ can meet $T_{1}+T_{2}$ only in tips of $Q_{1}$, because otherwise by Lemma 3.6(c) the image of the non-tip it meets would become a component of $\Omega_{1}+\bar{Q}_{1}\wedge(\Upsilon_{1}-\Upsilon_{1}^{0})$, which consists of $\psi(C_{1})$ only.

Suppose that $A_{1}\cdot T_{2}=1$. Then $A_{1}$ meets the tips of $Q_{1}$ contained in $T_{1}$ and $T_{2}$. The inclusion $\bar{Q}_{1}-\psi(C_{1})\subseteq\Upsilon_{1}^{0}+\Delta_{1}^{\mathrm{tip}}$ implies now that $\psi(T_{1}+T_{2})$ is a component of $\Upsilon_{1}^{0}$. The contraction of $\bar{Q}_{2}+\cdots+\bar{Q}_{c}+\psi(T_{1}+T_{2})$ maps $X_{1}$ onto $\mathbb{P}^{2}$ so that the images of $C_{1}$ and $E_{0}$ are two singular curves with intersection number equal to $\tau_{1}=3$; a contradiction.

Thus $A_{1}\cdot T_{2}=0$. Suppose that $\psi$ does not touch $C_{1}$. Then $C_{1}$ meets some component $U$ of $Q_{1}-C_{1}$ and for every such component $\psi_{*}(U)\neq 0$, so $\psi(U)\subseteq\Delta_{1}^{\mathrm{tip}}$ and hence $U\subseteq\Delta_{0}^{\mathrm{tip}}$. It follows that $Q_{1}=[1,2]$. But then $A_{1}$ meets $C_{1}$ or the $(-2)$-tip, hence in any case $\psi$ touches $C_{1}$; a contradiction.

Thus $\psi$ touches $C_{1}$. Suppose that $A_{1}\cdot E_{0}=1$. Since $A_{1}$ meets $\Delta_{0}$ and $\psi$ touches $C_{1}$, we infer that $T_{1}$ is a non-zero $(-2)$-chain. Since $A_{1}\cdot T_{2}=0$, we see that $\psi(T_{2})\subseteq\Delta_{1}^{\mathrm{tip}}$, so $T_{2}\subseteq\Delta_{0}^{\mathrm{tip}}$, hence by the negative definiteness of $Q_{1}$ we get $T_{2}=0$, that is, $A_{1}+Q_{1}=[1,(2)_{k},1]$ for some $k\geqslant 1$. The contraction of $D_{0}-E_{0}+A_{1}-C_{1}$ maps $X_{0}$ onto $\mathbb{P}^{2}$ and $C_{1}$ onto a [math]-curve; a contradiction.

Thus $A_{1}\cdot E_{0}=0$. Since $A_{1}\cdot D_{0}=2$, we may assume that $A_{1}\cdot Q_{2}=1$ and $A_{1}\cdot(Q_{3}+\cdots+Q_{c})=0$. Contract $Q_{3}+\cdots+Q_{c}$ and then perform the blow-downs constituting $\psi$ until $C_{1}$ is touched once. If the contractions did not touch $C_{2}$ then we contract $C_{2}$, too. Denote the image of $C_{1}$ by $F$ and the unique intersection point of the images of $Q_{2}$ and $E_{1}$ by $q_{2}^{\prime}$. By Lemma 2.3, $\displaystyle(r_{F}(q_{2}^{\prime})-1)+\sum_{j=3}^{c}(\mu(q_{j})-1)\leqslant 2\cdot 3-1-3=2$, where $\mu(q_{j})$ is the multiplicity of the cusp $q_{j}$. We infer that $c=4$, $r_{F}(q_{2}^{\prime})=1$ and that $q_{3}$, $q_{4}$ are semi-ordinary. The equality $r_{F}(q^{\prime}_{2})=1$ implies that $q_{2}^{\prime}$ is a smooth point of the image of $E_{1}$, hence $C_{2}$ has not been contracted. We infer that $\psi$ touches $C_{2}$ before it touches $C_{1}$. By (4.3), $\mathrm{ind}_{1}^{(1)}+p_{2}\leqslant 4-2\cdot\frac{5}{6}=2\frac{1}{3}$. Since $s_{1}=1$ and $\tau_{1}=3$, we have $\mathrm{ind}_{1}^{(1)}\geqslant\frac{2}{3}$, hence $p_{2}\leqslant 1$. On the other hand, since $A_{1}$ meets $Q_{2}$, we have $\lambda_{2}\geqslant 3$, so (4.6) gives $3p_{2}-\delta\geqslant 3+3+1+1-7=1$, hence $p_{2}=1$. We get $3\leqslant\lambda_{2}\leqslant 7-\delta-\lambda_{3}-\lambda_{4}\leqslant 5-\delta$. Since $\psi$ touches $C_{2}$, we have $\tau_{2}\geqslant 3$ by Lemma 4.2(d). It follows that $\lambda_{2}\geqslant 4$, for otherwise by symmetry $\psi$ touches $C_{1}$ before it touches $C_{2}$, which is impossible. Thus $\lambda_{2}$ equals to $4$ or $5$.

Suppose that $s_{2}=1$. Then, since $q_{3}$ and $q_{4}$ are semi-ordinary, we get $2\cdot\frac{2}{3}\leqslant\mathrm{ind}_{1}^{(1)}+\mathrm{ind}_{2}^{(1)}\leqslant 3-\mathrm{ind}_{3}^{(1)}-\mathrm{ind}_{4}^{(1)}\leqslant 3-2\cdot\frac{5}{6}=\frac{4}{3}$. It follows that $\mathrm{ind}_{j}^{(1)}=\frac{2}{3}$ for $j=1,2$ and that $\bar{Q}_{1}+\bar{Q}_{2}+E_{1}$ has no twigs. Then $T_{2}=0$, $\tau_{2}=3$ and $\bar{Q}_{2}=\psi(C_{2})$, in which case $\lambda_{2}=3$; a contradiction.

Thus $s_{2}=0$. We get $5-\delta\geqslant\lambda_{2}\geqslant 2+\tau_{2}\geqslant 5$, so $(\lambda_{1},\lambda_{2},\lambda_{3},\lambda_{4})=(3,5,1,1)$, $\tau_{2}=3$ and $\delta=0$. We have $\Omega_{1}=\psi(C_{1})$, hence $T_{2}\subseteq\Delta_{0}^{-}$ and so $T_{2}=0$, because $\delta=0$. Then $T_{1}=[(2)_{k}]$ for some $k\geqslant 1$ and hence $A_{1}$ meets $C_{2}$ and the tip of $D_{0}$ contained in $T_{1}$. Since $\delta=0$ and $\#\Omega_{2}=\lambda_{2}-3=2$, we have $\psi(Q_{2})=\psi(C_{2}+\widetilde{C}_{2})$, so $Q_{2}=[1,2]$. Then $\pi_{0}(A_{1})^{2}=2$, which is not a square; a contradiction.

(2∘) Now assume $\tau_{1}=2$. Since $\lambda_{1}=3$ and $s_{1}=1$, we have $\Omega_{1}=\psi(C_{1})+\psi(U)$ for some component $U$ of $Q_{1}$ and hence

[TABLE]

By Lemma 4.2(d), $\psi$ does not touch $C_{1}$, so $U$ meets $C_{1}$. By Lemma 3.6(c) all components of $Q_{1}-U$ are non-branching in $Q_{1}$.

Suppose that $U$ is a branching component of $Q_{1}$. Denote by $T_{1}$ and $T_{3}$ the twigs of $D_{0}$ meeting $U$. By the contractibility of $Q_{1}$ to a smooth point, $T_{2}:=Q_{1}-U-T_{1}-T_{3}-C_{1}$ is a $(-2)$-chain and $C_{1}+T_{2}$ is a twig of $Q_{1}$. Note that $A_{1}$ can meet $T_{1}+T_{2}+T_{3}$ only in tips of $D_{0}$. Indeed, otherwise the component of $Q_{1}$ it meets is branching in $D_{0}+A_{1}$ and hence is neither contracted by $\psi$ nor contained in $\Upsilon_{1}^{0}+\Delta_{1}^{\mathrm{tip}}$, which contradicts (4.11). We have $A_{1}\cdot(T_{1}+T_{3})>0$, because otherwise by (4.11), $T_{1}+T_{3}\subseteq\Delta_{1}^{\mathrm{tip}}$, in which case $Q_{1}$ would not contract to a smooth point. We may therefore assume that $A_{1}$ meets the tip of $D_{0}$ contained in $T_{1}$. Also, $A_{1}$ does not meet the tip of $D_{0}$ contained in $T_{2}$, because otherwise $\psi$ contracts $T_{2}$ and hence touches $C_{1}$, contrary to Lemma 4.2(d). Now (4.11) gives $T_{2}=0$ or $T_{2}=[2]$.

Consider the case $A_{1}\cdot(Q_{1}+E_{0})>1$. If $A_{1}$ meets $U$, $C_{1}$ or $E_{0}$ then by (4.11), $T_{3}=[2]$ and $T_{1}=[(2)_{k}]$ for some $k\geqslant 1$, so $Q_{1}$ does not contract to a smooth point, which is false. So the only possibility is that $A_{1}$ meets the tip of $D_{0}$ contained in $T_{3}$. Then by (4.11), $\psi(T_{1}+T_{3})$ is a component of $\Upsilon_{1}^{0}$. Let $\theta$ be the composition of $\psi$ with the contraction of $D_{1}-E_{1}-\psi(U)$. Then $\theta(U)$ is a rational curve on $\mathbb{P}^{2}$ with one singular point, a normal crossing node, and it meets $\theta(E_{0})$ only at the point $\theta(C_{1})$. But the degree of $\theta(U)$ equals $3$ and $\theta(E_{0})\cdot\theta(U)$ equals $\tau_{1}=2$ if $T_{2}=0$ and equals $2\tau_{1}=4$ if $T_{2}=[2]$; a contradiction.

We obtain $A_{1}\cdot Q_{1}=1$ and $A_{1}\cdot E_{0}=0$. By (4.11), $T_{3}=[2]$ and $T_{2}$ is either [math] or $[2]$. By the contractibility of $Q_{1}$ to a smooth point we get $T_{1}=[(2)_{k},3]$ for some $k\geqslant 0$ and $U^{2}=-2-\#T_{2}\in\{-2,-3\}$. Since $A_{1}\cdot Q_{1}=1$, we have $\psi_{*}T_{1}\nsubseteq\Upsilon_{1}^{0}$, hence by (4.11), $\psi$ contracts $T_{1}$. We may assume that $A_{1}\cdot Q_{2}>0$. Then $A_{1}\cdot Q_{2}=1$ and $A_{1}\cdot Q_{j}=0$ for $j\geqslant 3$. Perform the contractions within $\psi$ until $U$ is touched once, then contract $Q_{3}+\ldots+Q_{c}$ and denote the resulting morphism by $\theta$. If $T_{2}=0$ then $F=\theta(U+C_{1})=[1,1]$ and by Lemma 2.3, $c<4$, which is false. Thus $T_{2}=[2]$, hence for $F=\theta(U+2C_{1}+T_{2})$ we have $F\cdot E_{0}=2\tau_{1}=4$, in which case Lemma 2.3 gives $\sum_{j=2}^{c}(r_{F}(q_{j})-1)=3$. Then $r_{F}(q_{j})\leqslant 2$ for some $j\geqslant 3$, say for $j=3$. We have $\mu(q_{3})\leqslant r_{F}(q_{3})$, so $q_{3}\in\bar{E}$ is semi-ordinary. We get $\mathrm{ind}_{1}^{(1)}=\frac{2}{3}+\frac{1}{2}+\frac{1}{2}=\frac{5}{3}$, $\mathrm{ind}_{3}^{(1)}\geqslant\frac{5}{6}$ and $\mathrm{ind}_{4}^{(1)}>\frac{1}{2}$ (cf. Lemma 3.8(a)), so by (4.3), $p_{2}<1-\mathrm{ind}_{2}^{(1)}\leqslant 1$. Thus $p_{2}=0$, so by (4.6), $\lambda_{1}+\cdots+\lambda_{c}\leqslant 7-\delta\leqslant 6$. But since $A_{1}$ meets $Q_{1}$ and $Q_{2}$, the cusps $q_{1}$ and $q_{2}$ are not semi-ordinary, hence by Lemma 4.2(a),(e), $\lambda_{1}+\lambda_{2}\geqslant 6$; a contradiction.

Thus we proved that $Q_{1}$ is a chain. We write it as $Q_{1}=T_{1}+U+C_{1}+T_{2}$, where $T_{1}$ and $T_{2}$ are zero or twigs of $Q_{1}$ meeting $U$ and $C_{1}$, respectively. As before, we note that $A_{1}$ can meet $T_{1}$ and $T_{2}$ only in tips of $D_{0}$. If $A_{1}\cdot T_{2}=1$ then by (4.11), $\psi$ contracts $T_{2}$, hence touches $C_{1}$, which is false by Lemma 4.2(d). Thus $A_{1}\cdot T_{2}=0$. Then $T_{2}=[2]$ or $T_{2}=0$. But the second case is impossible, because $q_{1}$ is not semi-ordinary. By the contractibility of $Q_{1}$ to a smooth point it follows that $Q_{1}=[(2)_{k},3,1,2]$ for some $k\geqslant 0$.

Suppose that $A_{1}\cdot T_{1}=0$. Then $A_{1}$ meets $U$ and some component $T_{4}\subseteq Q_{2}\wedge\Delta_{0}$. Let $\theta$ be the contraction of $Q_{3}+\ldots+Q_{c}$ and let $F=\theta(C_{1}+U+2A_{1}+T_{4})$. Then $F\cdot\theta(E_{0})=\tau_{1}=2$, so by Lemma 2.3, $1+(\mu(q_{3})-1)+(\mu(q_{4})-1)\leqslant 2$; a contradiction. Thus $A_{1}$ meets the tip of $D_{0}$ contained in $T_{1}$. Suppose that it meets $U$ too. Let now $\theta$ be the contraction of $A_{1}+D_{0}-E_{0}-U$. Then $\theta(U)$ is a uninodal curve on $\mathbb{P}^{2}$ (hence of degree $3$) whose intersection with $\theta(E_{0})$ is $4$; a contradiction.

It follows that $A_{1}+Q_{1}=[1,(2)_{k},3,1,2]$. Since $A_{1}$ does not meet $Q_{j}$ for $j\geqslant 3$, we have $\mathrm{ind}_{j}^{(1)}=\mathrm{ind}_{j}^{(0)}>\frac{1}{2}$ for $j\geqslant 3$ by Lemma 3.8(a), hence by (4.3), $1+\mathrm{ind}_{2}^{(1)}+(c-2)\frac{1}{2}<4-p_{2}$, so $p_{2}\leqslant 1$. Suppose that $p_{2}=0$. By (4.6), $(\lambda_{1},\lambda_{2},\lambda_{3},\lambda_{4})=(3,1,1,1)$. Since $q_{2}$ is ordinary, $A_{1}$ does not meet $Q_{2}$, so it meets $E_{0}$. But since $U=[3]$, the contraction of $D_{0}-E_{0}+A_{1}-U$ maps $U$ onto a [math]-curve on $\mathbb{P}^{2}$; a contradiction.

Thus we are left with the case $p_{2}=1$. We have $\mathrm{ind}_{1}^{(1)}=1$, so by (4.3), $\frac{5}{6}c_{0}\leqslant\mathrm{ind}_{2}^{(1)}+\cdots+\mathrm{ind}_{c}^{(1)}\leqslant 2$. It follows that $c_{0}\leqslant 2$. By (4.6), $\lambda_{1}+\cdots+\lambda_{c}\leqslant 10-\delta<10$, so $c_{0}=2$, $c=4$ and $\lambda_{2}\in\{3,4\}$. We get $\mathrm{ind}_{2}^{(1)}\leqslant\frac{1}{3}$, so $s_{2}=0$. By Lemma 4.2(c), $\lambda_{2}=4$. By (4.6) we have $(\lambda_{1},\lambda_{2},\lambda_{3},\lambda_{4})=(3,4,1,1)$. By Lemma 4.2(h), $\tau_{2}=2$, $\psi(C_{2})$ is a $(-1)$-tip of $\bar{Q}_{2}$ and the only contribution to $\mathrm{ind}_{2}^{(1)}$ comes from the $(-3)$-tip in $D_{1}^{\prime}$ which is the proper transform of $\psi(C_{2})$. Since $A_{1}\cdot Q_{2}\leqslant 1$, it follows that $Q_{2}=[1,(2)_{m}]$ for some $m\geqslant 1$ and $Q_{1}+A_{1}+Q_{2}$ is a chain with $T_{2}$ and $C_{2}$ as tips. We have $\operatorname{Exc}\psi\wedge Q_{1}=T_{1}$, which is possible only if $m=1$. Let $\theta$ be the contraction of $Q_{3}+Q_{4}+A_{1}+T_{1}^{0}$, where $T_{1}^{0}$ is the $(-2)$-tip of $T_{1}$ met by $A_{1}$. Then $F=\theta(\widetilde{C}_{2})$ is a [math]-curve and $F\cdot\theta(E_{0})=1$. This is impossible, because $\theta(E_{0})$ is singular; a contradiction.

∎

Lemma 4.5**.**

Assume $n=1$. Put $\Lambda=(\lambda_{1},\ldots,\lambda_{c})$. Then $c=4$ and one may order the cusps so that one of the following holds:

1. (0.a)

$p_{2}=0$, $\delta=0$, $\Lambda=(4,1,1,1)$, $\mathrm{ind}_{1}^{(1)}\leqslant\frac{3}{2}$, 2. (1.a)

$p_{2}=1$, $\delta\in\{0,\frac{1}{2}\}$, $\Lambda=(\lambda_{1},1,1,1)$ with $\lambda_{1}\in\{4,5,6,7\}$, $\mathrm{ind}_{1}^{(1)}\leqslant\frac{1}{2}$, 3. (1.b)

$p_{2}=1$, $\delta=0$, $\Lambda=(\lambda_{1},2,1,1)$ with $\lambda_{1}\in\{4,5,6\}$, $\mathrm{ind}_{1}^{(1)}\leqslant\frac{7}{30}$, 4. (1.c)

$p_{2}=1$, $\delta=0$, $\Lambda=(\lambda_{1},3,1,1)$ with $\lambda_{1}\in\{4,5\}$, $\mathrm{ind}_{1}^{(1)}\leqslant\frac{5}{12}$, $\mathrm{ind}_{1}^{(1)}+\mathrm{ind}_{2}^{(1)}\leqslant\frac{4}{3}$, 5. (1.d)

$p_{2}=1$, $\delta=0$, $\Lambda=(4,4,1,1)$, $\mathrm{ind}_{1}^{(1)}+\mathrm{ind}_{2}^{(1)}\leqslant\frac{4}{3}$, 6. (2.a)

$p_{2}=2$, $\delta=0$, $\Lambda=(\lambda_{1},5,1,1)$ with $\lambda_{1}\in\{5,6\}$, $\mathrm{ind}_{1}^{(1)}+\mathrm{ind}_{2}^{(1)}\leqslant\frac{1}{3}$, 7. (2.b)

$p_{2}=2$, $\delta=0$, $\Lambda=(5,5,2,1)$, $\mathrm{ind}_{1}^{(1)}+\mathrm{ind}_{2}^{(1)}\leqslant\frac{1}{15}$.

• Proof.

For a positive integer $k$ we denote by $\omega_{k}$ the number of cusps of $\bar{E}$ for which $\lambda_{j}=k$. We put $\omega_{\geqslant 5}=\displaystyle\sum_{k\geqslant 5}\omega_{k}$. We may assume that $\lambda_{1}\geqslant\lambda_{2}\geqslant\ldots\geqslant\lambda_{c}.$ Since $\omega_{\geqslant 5}+\omega_{4}+\omega_{3}+\omega_{2}+\omega_{1}=c$, (4.6) gives

[TABLE]

For a cusp with $\lambda_{j}=1,2,3$ and $4$ we know (see Lemma 4.2(e),(h) and Proposition 4.4) that $\mathrm{ind}^{(1)}_{j}$ is bounded from below by $\frac{5}{6}$, $\frac{11}{10}$, $\frac{11}{12}$ and $\frac{1}{3}$, respectively, so (4.3) gives

[TABLE]

Put $\omega_{1,2}=\omega_{1}+\omega_{2}$. Since $n=1$, by Proposition 4.4, $\omega_{4}+\omega_{\geqslant 5}\geqslant 1$. We have $p_{2}\leqslant 2$ by Lemma 4.2(f).

Consider the case $p_{2}=2$. By (4.13), $\frac{5}{6}\omega_{1,2}\leqslant\sum_{j=1}^{c}\mathrm{ind}_{j}^{(1)}\leqslant 2$, so $\omega_{1,2}\leqslant 2$. If $\omega_{1,2}=0$ then by (4.13), $\omega_{3}\leqslant 2$, so $\omega_{4}+\omega_{\geqslant 5}\geqslant c-\omega_{3}\geqslant 2$, in which case (4.12) fails. Thus $\omega_{1,2}\in\{1,2\}$. Assume $\omega_{1,2}=2$. Then $\frac{11}{12}\omega_{3}+\frac{1}{3}\omega_{4}\leqslant\frac{1}{3}$, so $\omega_{3}=0$ and $\omega_{4}\leqslant 1$. Suppose that $\omega_{4}=1$. By (4.13), $\omega_{2}=0$, so $\omega_{1}=2$. Then by (4.12), $c=4$ and $\Lambda=(\lambda_{1},4,1,1)$ for some $\lambda_{1}\in\{5,6,7\}$. By (4.13), $\mathrm{ind}_{1}^{(1)}+\mathrm{ind}_{2}^{(1)}\leqslant\frac{1}{3}$ and by Lemma 4.2(h) the latter number is a contribution from a single twig. This is impossible, as $D$ has at least four twigs in total over $q_{1}$ and $q_{2}$ and the proper transform of $A_{1}$ meets at most two of them; a contradiction. Thus $\omega_{4}=0$. By (4.12), $c=4$ and $\omega_{\geqslant 5}=\omega_{5}+\omega_{6}\leqslant 2$, which leads to cases (2.a) and (2.b). Note that the equality $\delta=0$ is a consequence of the inequality $\mathrm{ind}_{1}^{(1)}+\mathrm{ind}_{2}^{(1)}\leqslant\frac{1}{3}$. Assume $\omega_{1,2}=1$. By (4.13), $\frac{11}{12}\omega_{3}+\frac{1}{3}\omega_{4}\leqslant\frac{7}{6}$, so $\omega_{3}\leqslant 1$. If $\omega_{3}=1$ we get $\omega_{4}=0$, hence $\sum_{k\geqslant 1}k\omega_{k}\geqslant 14$, in contradiction to (4.12). Hence $\omega_{3}=0$ and $\omega_{\geqslant 5}+\omega_{4}\geqslant c-1\geqslant 3$. Then $5\omega_{\geqslant 5}+4\omega_{4}\leqslant 12-\delta$, so $\Lambda=(4,4,4,1)$. Since $A_{1}\cdot D_{0}=2$, we may assume that $A_{1}\cdot Q_{3}=0$, so by Lemma 3.8(a), $\sum_{j=1}^{c}\mathrm{ind}_{j}^{(1)}>\frac{5}{6}+2\cdot\frac{1}{3}+\frac{1}{2}=2$; a contradiction.

Consider the case $p_{2}=1$. If $\omega_{1,2}\leqslant 1$ then by (4.12), $\omega_{4}+\omega_{\geqslant 5}+10\leqslant 1+3(c-1)+\omega_{4}+2\omega_{\geqslant 5}\leqslant 10-\delta$, which is impossible, as $\omega_{4}+\omega_{\geqslant 5}\geqslant 1$ by Proposition 4.4. By (4.13), $\omega_{1,2}\in\{2,3\}$. Assume $\omega_{1,2}=2$. By (4.12), $3(c-2)+2\omega_{\geqslant 5}+\omega_{4}=5\omega_{\geqslant 5}+4\omega_{4}+3\omega_{3}\leqslant 8-\delta-\omega_{2}$. For $\omega_{2}\neq 0$ we get $c=4$ and $\Lambda=(4,3,2,1)$, in which case (4.13) fails. Thus $\omega_{2}=0$ and $\omega_{1}=2$, which gives cases (1.c) and (1.d). Assume $\omega_{1,2}=3$. By (4.13), $\omega_{3}=0$. By (4.12), $4(c-3)+3\leqslant 5\omega_{\geqslant 5}+4\omega_{4}+\omega_{2}+3\leqslant 10-\delta$, so $c=4$. Now (4.13) gives $\omega_{1}\in\{2,3\}$. This gives cases (1.a) and (1.b).

Finally, consider the case $p_{2}=0$. We have now $\sum_{j=1}^{c}\mathrm{ind}_{j}^{(1)}\leqslant 4$ and $\sum_{j=1}^{c}\lambda_{j}\leqslant 7-\delta$. Since $\lambda_{1}\geqslant 4$, we get $\Lambda=(4,1,1,1)$ and $\delta=0$, which is case (0.a).

In all cases other than case (1.a) and case (1.c) for $\lambda_{1}=4$ we have either $3p_{2}+7-\lambda_{1}-\ldots-\lambda_{c}=0$, hence $3p_{2}+7-\lambda_{1}-\ldots-\lambda_{c}-\delta=\delta=0$ or the sum of $\mathrm{ind}_{j}^{(1)}$ for non-semi-ordinary cusps is less than $\frac{1}{2}$, which gives $\Delta_{1}^{-}=0$ and hence $\delta=0$. In case (1.a) we have $\mathrm{ind}_{1}^{(1)}\leqslant\frac{1}{2}$, hence $\Delta_{1}^{-}$ is [math] or $[2]$, which gives $\delta\in\{0,\frac{1}{2}\}$. Finally, in case (1.c) for $\lambda_{1}=4$ we have $\mathrm{ind}_{2}^{(1)}\leqslant\frac{4}{3}-\frac{1}{3}=1$ by Lemma 4.2(h), so $\delta=0$ by Proposition 4.4.

∎

Our goal is to eliminate all cases listed in Lemma 4.5. We collect some observations. Recall that $K_{1}=K_{X_{1}}$ and $\bar{Q}_{j}=\psi(Q_{j})$.

Lemma 4.6**.**

Assume that $n=1$ and $j\in\{1,\ldots,4\}$. Put $|\Lambda|=\lambda_{1}+\ldots+\lambda_{4}$ and $\gamma_{i}=-E_{i}^{2}$. We have:

1. (a)

$K_{1}\cdot D_{1}=p_{2}+\#D_{1}-\tau^{*}-16$, 2. (b)

$\#D_{1}=|\Lambda|+b_{0}(\Delta_{1}^{-})+\#\Upsilon_{1}-\tau^{*}-3$, 3. (c)

$|\Lambda|\leqslant 3p_{2}+7-\delta$, 4. (d)

If $U$ is a component of $\bar{Q}_{j}-\psi(C_{j})$ then $U^{2}<0$. 5. (e)

$K_{1}\cdot\psi(C_{j})+\tau_{j}^{*}\geqslant-1$ and the equality holds if and only if $\tau_{j}^{*}=0$ and $\psi(C_{j})^{2}=-1$. 6. (f)

Let $r_{j}$ be the number of outer blow-ups (see Section 5A) over $q_{j}\in\bar{E}$ in the minimal log resolution of $(\mathbb{P}^{2},\bar{E})$. Then

[TABLE]

• Proof.

(a) By Noether’s formula $K_{n}^{2}=10-\rho(X_{n})=10+n-\#D_{n}$, so [Pal19, Lemma 4.3] gives $K_{n}\cdot D_{n}=p_{2}-c-\tau^{*}-n-K_{n}^{2}=p_{2}-c-\tau^{*}-2n-10+\#D_{n}$. Since $c=4$ by Lemma 4.5, we get $K_{1}\cdot D_{1}=p_{2}+\#D_{1}-\tau^{*}-16$.

(b) The definition of $\lambda_{j}$ gives $|\Lambda|=\#(D_{n}-E_{n})-b_{0}(\Delta_{n})-\#\Upsilon_{n}^{0}+\tau^{*}+c$, hence $|\Lambda|+b_{0}(\Delta_{1}^{-})+\#\Upsilon_{1}=|\Lambda|+b_{0}(\Delta_{1})+\#\Upsilon_{1}^{0}=\#D_{1}+\tau^{*}+3$.

(c) This is (4.6).

(d) Suppose that $U^{2}\geqslant 0$. By blowing up over $U\setminus E_{1}$ we may assume that $U^{2}=0$. Put $d:=U\cdot E_{1}$. We may assume that $U^{\prime}:=\psi_{*}^{-1}(U)\subseteq Q_{1}$ and $A_{1}\cdot(Q_{3}+Q_{4})=0$. Contract $Q_{3}+Q_{4}$ and for $j=1,2$ denote by $m_{j}$ be the image of $C_{j}\cap E_{0}$. By Lemma 2.3 applied to the image of $E_{1}$, $0\leqslant r_{U}(m_{1})+r_{U}(m_{2})-2\leqslant 2d-4$. In particular, $d\geqslant 2$. The curve $U^{\prime}$ is a component of $Q_{1}-C_{1}$, so $U^{\prime}\cdot E_{0}\leqslant 1$. Since $d>1$, Lemma 3.6(a) gives $U^{\prime}\cdot E_{0}=1$ and $U\cdot E_{1}=2$. It follows that $U^{\prime}=\widetilde{C}_{1}$ and $\operatorname{Exc}(\psi)\cdot Q_{2}=0$. Then $U\cdot\bar{Q}_{2}=0$ and so $r_{U}(m_{2})\geqslant 2$; a contradiction.

(e) We may assume that $\psi(C_{j})^{2}\geqslant 0$, otherwise the claim is clear. We may also assume that $j=1$ and, since $n=1$, that $A_{1}\cdot(Q_{3}+Q_{4})=0$. It follows that $\psi$ touches $C_{1}$ and, by Lemma 4.2(d), that $\tau_{1}\geqslant 3$. Suppose that $K_{1}\cdot\psi(C_{1})+\tau_{1}^{*}\leqslant-1$. We have $K_{1}\cdot\psi(C_{1})+\tau_{1}^{*}=\tau_{1}-\psi(C_{1})^{2}-s_{1}-3$, so $\tau_{1}-\psi(C_{1})^{2}\leqslant s_{1}+2$. Blow up over $\psi(C_{1}\cap E_{0})$, each time on the intersection of the proper transforms of $\psi(C_{1})$ and $\psi(E_{0})$, until the proper transform $C^{\prime}$ of $\psi(C_{1})$ is a [math]-curve. Denote the proper transform of $E_{1}$ by $E^{\prime}$. Over $\psi(C_{1}\cap E_{0})$ the intersection of $C^{\prime}$ and $E^{\prime}$ is $\tau_{1}^{\prime}=\tau_{1}-\psi(C_{1})^{2}$, hence $\tau_{1}^{\prime}\leqslant s_{1}+2$.

Assume first that $C^{\prime}\cdot E^{\prime}=\tau_{1}^{\prime}$. Lemma 2.3 gives $\sum_{j=2}^{4}(r_{C^{\prime}}(q_{j})-1)\leqslant\tau_{1}^{\prime}-1\leqslant s_{1}+1$. Since $\psi$ does not touch $Q_{3}+Q_{4}$, the left hand side is at least $2$, so we obtain $s_{1}=1$, $\tau_{1}^{\prime}=3$, $r_{C^{\prime}}(q_{j})=2$ for $j=3,4$ and $r_{C^{\prime}}(q_{2})=1$. It follows that $q_{3}$ and $q_{4}$ are semi-ordinary and $\psi$ touches $C_{2}$. In particular, $\Upsilon_{1}^{0}=\psi_{*}\Upsilon_{0}^{0}$ and, by Lemma 4.2(d), $\tau_{2}\geqslant 3$. By Lemma 4.2(e) and Proposition 4.4, $\lambda_{1},\lambda_{2}\geqslant 4$. Since $s_{1}=1$ and $\tau_{1}^{\prime}=3$, we have $\mathrm{ind}_{1}^{(1)}\geqslant\frac{2}{3}$. Then we are in case (1.d) of Lemma 4.5, hence $\lambda_{1}=\lambda_{2}=4$ and $\delta=0$, so $\Delta_{1}^{-}=0$. We have $s_{2}=1$, because otherwise $\lambda_{2}\geqslant 2+\tau_{2}\geqslant 5$, which is false. The inequality $\mathrm{ind}_{1}^{(1)}+\mathrm{ind}_{2}^{(1)}\leqslant\frac{4}{3}$ implies that $Q_{1}+Q_{2}+E_{1}+A_{1}$ has no tips and $\tau_{1}=\tau_{2}=3$, which implies that for $j=1,2$, $\#\Omega_{j}=\lambda_{j}-2=2$ and $Q_{j}-C_{j}$ is a $(-2)$-chain met by $A_{1}$, one of them necessarily empty. But then one of the $Q_{j}$ for $j=1,2$, is irreducible, which is impossible.

We are left with the case $C^{\prime}\cdot E^{\prime}>\tau_{1}^{\prime}$. This is possible only when $A_{1}\cdot E_{0}=1$ and $\psi$ contracts $A_{1}$ and a $(-2)$-twig of $D_{0}$ meeting $C_{1}$. Then $\psi(C_{1})^{2}=0$, $\tau_{1}=\tau_{1}^{\prime}$, $\psi$ does not touch $Q_{2}$ and we have $C^{\prime}\cdot E^{\prime}=\tau_{1}^{\prime}+1$. Moreover, $(\Delta_{1}^{+}+\Upsilon_{1})\wedge\bar{Q}_{j}=0$, unless $q_{j}\in\bar{E}$ is semi-ordinary. Since $s_{1}+2\geqslant\tau_{1}^{\prime}=\tau_{1}\geqslant 3$ by Lemma 4.2(d), we have $s_{1}=1$ and $\tau_{1}=3$. Then $\mathrm{ind}_{1}^{(1)}\geqslant\frac{2}{3}$, so we are in cases (0.a) or (1.d) of Lemma 4.5, hence $\lambda_{1}=4$ and $\Delta_{1}^{-}=0$. We obtain $\#\bar{Q}_{1}=\#\Omega_{1}=2$. Then $Q_{1}=[(2)_{k},1,k+2]$ for some $k\geqslant 1$, so $\mathrm{ind}_{1}^{(1)}>\frac{2}{3}$. In case (1.d) we get $s_{2}=0$, so since $\lambda_{2}=4$, we have $Q_{2}=C_{2}+\widetilde{C}_{2}$ and then the inequality $\mathrm{ind}_{2}^{(1)}<\frac{2}{3}$ fails. Thus we are in case (0.a). We denote by $\theta\colon X_{1}\longrightarrow\theta(X_{1})$ the contraction of $\psi(Q_{2}+Q_{3}+Q_{4})$. We have $\rho(\theta(X_{1}))=2$, so $\theta(X_{1})=\mathbb{F}_{k+2}$, $F=\theta(\psi(C_{1}))$ is a fiber and $U=\theta_{*}(\bar{Q}_{1})-F$ is the negative section. Since $\theta(E_{1})$ has three cusps, all ordinary, $p_{a}(\theta(E_{1}))=3$. On the other hand, on $\mathbb{F}_{k+2}$ we have $K_{\mathbb{F}_{k+2}}\sim-2U-(k+4)F$ and $\theta(E_{1})\sim 4U+4(k+2)F$, so $p_{a}(\theta(E_{1}))=6k+9$; a contradiction.

(f) Recall that $Q_{j}^{\prime}$ is the reduced exceptional divisor of the minimal log resolution of $(\mathbb{P}^{2},\bar{E})$ over $q_{j}$. By [KP17, Lemma 2.4(i)], $K\cdot Q_{j}^{\prime}+1$ equals the number of inner blow-ups over $q_{j}$, so

[TABLE]

which gives $K\cdot D=-E^{2}-3+\#D-\sum_{j=1}^{c}r_{j}-2c$. Since $\#D=\rho(X)=10-K^{2}$, we obtain

[TABLE]

Since $E^{2}=E_{0}^{2}-\tau$, we have $-E^{2}=-E_{0}^{2}+\tau^{*}+c+s$, hence $\gamma_{0}+\tau^{*}=p_{2}+2c-7-s+\sum_{j=1}^{c}(r_{j}-1)$.

∎

We are now ready to make the key step in the proof of Theorem 1.1.

Theorem 4.7**.**

The surface $(X_{0},\frac{1}{2}D_{0})$ is almost minimal.

• Proof.

By Lemma 4.3, $n\leqslant 1$. Suppose that $n=1$. We assume that $\lambda_{1}\geqslant\lambda_{2}\geqslant\ldots\geqslant\lambda_{c}$. We need to rule out the cases listed in Lemma 4.5. For all of them $c=4$, $q_{4}$ is ordinary and $q_{3}$ is either ordinary or semi-ordinary with $\lambda_{3}=2$. Let ${\mathcal{L}}\subseteq D_{1}-E_{1}$ be the sum of $(-1)$-curves created by $\psi$, i.e. of $(-1)$-curves in $D_{1}-E_{1}$ whose proper transforms on $X_{0}$ are not $(-1)$-curves. Put $\eta=\#({\mathcal{L}}\wedge\Upsilon_{1})$. By the definition of $\psi$ we have $\eta\leqslant\#{\mathcal{L}}\leqslant 2$. Our goal is to analyze the divisor

[TABLE]

Recall that $c_{0}$ denotes the number of semi-ordinary cusps, which are cusps of multiplicity $2$, and that $\gamma_{i}=-E_{i}^{2}$. By Lemma 4.6(a),(b), $K_{1}\cdot D_{1}=p_{2}+\#D_{1}-\tau^{*}-16$ and $\#D_{1}=|\Lambda|+b_{0}(\Delta_{1}^{-})+\#\Upsilon_{1}-\tau^{*}-3$, hence

[TABLE]

We have $K_{1}\cdot E_{1}=\gamma_{1}-2$ and $\#\Upsilon_{1}=\eta+c_{0}$, so putting $\alpha_{j}=K_{1}\cdot\psi(C_{j})+\tau_{j}^{*}$ we obtain

[TABLE]

The proof below is based on analyzing summands of both sides of this equality. Note that by Lemma 4.6(d),(e) we have $\alpha_{j}\geqslant-1$ and that $K_{1}\cdot U\geqslant 0$ for all components $U\subseteq R$.

Claim 1*.*

$\kappa(K_{0}+\frac{1}{2}D_{0})\geqslant 0$.

• Proof.

Suppose that $\kappa(K_{0}+\frac{1}{2}D_{0})=-\infty$. In particular, $p_{2}=0$, hence we are in case (0.a) of Lemma 4.5. Then in the inequality (4.6) we have in fact equality, which is equivalent to the equality $(2K_{1}+D_{1}^{\flat})^{2}=0$. By Proposition 3.9(c), $\mathbb{P}^{2}\setminus\bar{E}$ has no $\mathbb{C}^{**}$-fibration, hence by Corollary 3.4(b), $2K_{1}+D_{1}^{\flat}$ is a pull-back of an anti-ample divisor. But then $(2K_{1}+D_{1}^{\flat})^{2}>0$; a contradiction.

∎

Put $\varepsilon=A_{1}\cdot E_{0}-E_{1}\cdot({\mathcal{L}}\wedge\Upsilon_{1})$. We have $\varepsilon\in\{0,1\}$. Moreover, $\varepsilon=1$ if and only if $A_{1}\cdot E_{0}=1$ and $A_{1}\cdot\Delta_{0}^{\mathrm{tip}}=1$. In the latter case we have in fact ${\mathcal{L}}\wedge\Upsilon_{1}=0$ (otherwise $\psi_{*}^{-1}{\mathcal{L}}$ would meet two $(-2)$-twigs, which is impossible) and $\lambda_{j}=\#(\bar{Q}_{j}-\bar{Q}_{j}\wedge\Delta_{1}^{\mathrm{tip}})+\tau_{j}^{*}+1$ for $j\in\{1,2\}$ such that $A_{1}\cdot Q_{j}=1$.

Claim 2*.*

$\gamma_{1}+\tau^{*}+4-2c_{0}\geqslant-\varepsilon$.

• Proof.

By Claim 1, $\kappa(K_{0}+\frac{1}{2}D_{0})\geqslant 0$, hence by Lemma 3.4(b) the divisor

[TABLE]

is nef. Since $E_{1}\cdot\Delta_{1}=0$, intersecting the above divisor with $E_{1}$ we obtain

[TABLE]

which proves the claim.

∎

Claim 3*.*

$p_{2}\neq 0$.

• Proof.

Suppose that $p_{2}=0$. By Lemma 4.5 we are in case (0.a), hence $\Delta_{1}^{-}=0$. Claim 2 and the equality (4.16) give $-1-\varepsilon\leqslant\alpha_{1}-\varepsilon\leqslant\eta+\#{\mathcal{L}}-6$, hence $2\#{\mathcal{L}}\geqslant\eta+\#{\mathcal{L}}\geqslant 5-\varepsilon\geqslant 4$. From the definition of $\psi$ it follows that ${\mathcal{L}}$ has at most $2$ components, hence it has exactly two and $\varepsilon=1$. But if $\varepsilon=1$ then $A_{1}\cdot E_{0}=1$, so $\#{\mathcal{L}}\leqslant 1$, because by definition ${\mathcal{L}}\wedge E_{1}=0$; a contradiction.

∎

Claim 4*.*

$p_{2}=1$.

• Proof.

Suppose that $p_{2}\neq 1$. By Claim 3 and by Lemma 4.2(f), $p_{2}=2$, so we are in case (2.a) or (2.b) of Lemma 4.5. In particular, $\mathrm{ind}_{1}^{(1)}+\mathrm{ind}_{2}^{(1)}\leqslant\frac{1}{3}$, hence $\Delta_{1}\wedge(\bar{Q}_{1}+\bar{Q}_{2})=0$, $s_{1}=s_{2}=0$ and, by Lemma 3.8(a), $A_{1}$ meets both $Q_{1}$ and $Q_{2}$. The latter gives $c_{0}=2$ and implies that $\Upsilon_{1}\wedge(\bar{Q}_{1}+\bar{Q}_{2})=0$ and $A_{1}\cdot E_{0}=0$. It follows that $\Upsilon_{1}=\psi(C_{3})+\psi(C_{4})$, $\eta=0$ and $\varepsilon=0$. We obtain $\#\bar{Q}_{2}+\tau_{2}=\lambda_{2}=5$ and $\#\bar{Q}_{1}+\tau_{1}=\lambda_{1}\in\{5,6\}$. The equality (4.16) reads as

[TABLE]

For $j=1,2$ we have $s_{j}=0$, hence $\tau_{j}^{*}\geqslant 1$, which gives $\alpha_{j}\geqslant 0$ by Lemma 4.6(e). By Lemma 4.6(f) we have $\gamma_{0}+\tau^{*}=r_{1}+r_{2}+r_{3}-2$. Since $A_{1}\cdot E_{0}=0$, we have $\gamma_{0}=\gamma_{1}$. For a semi-ordinary cusp $r_{j}=\#Q_{j}=\lambda_{j}$, hence $\gamma_{1}+\tau^{*}-\lambda_{3}=r_{1}+r_{2}-2\geqslant 0$. It follows that $K_{1}\cdot R\leqslant 1+\#{\mathcal{L}}\leqslant 3$, hence every component $U$ of $\bar{Q}_{1}+\bar{Q}_{2}$ satisfies $U^{2}\geqslant-5$ and the equality may hold for at most one such component. For $j=1,2$ we have $\tau_{j}=\lambda_{j}-\#\bar{Q}_{j}\leqslant\lambda_{j}-2=4$, hence $((\varphi_{1}^{-1})_{*}\psi_{*}C_{j})^{2}\geqslant-1-\tau_{j}\geqslant-5$. It follows that every component $U^{\prime}$ of $Q_{1}^{\prime}+Q_{2}^{\prime}$ satisfies $(U^{\prime})^{2}\geqslant-6$ and the equality may hold for at most one such component. For $j=1,2$ let now $U_{j}$ be a tip of $D_{1}^{\prime}$ lying in $\varphi_{1}^{-1}(\bar{Q}_{j})$. Such a tip exists, because there are at least two tips of $D$ over $q_{j}$ and, since $A_{1}\cdot Q_{j}=1$, (the proper transform of) $\operatorname{Exc}\psi$ meets at most one of them. From Lemma 4.6(d) we infer that $U_{j}^{2}\leqslant-2$. We obtain $\mathrm{ind}_{1}^{(1)}+\mathrm{ind}_{2}^{(1)}\geqslant\frac{1}{-U_{1}^{2}}+\frac{1}{-U_{2}^{2}}\geqslant\frac{1}{5}+\frac{1}{6}>\frac{1}{3}$; a contradiction with Lemma 4.5.

∎

By Claim 4 and Lemma 4.5 the cusps $q_{3}$ and $q_{4}$ are ordinary and either $\lambda_{2}\leqslant 3$ or $\lambda_{1}=\lambda_{2}=4$. By Proposition 4.4, $\psi$ does not touch $Q_{j}$ with $\lambda_{j}\leqslant 3$, hence we may, and will, assume that $A_{1}\cdot Q_{1}\neq 0$. Put $\displaystyle\alpha=\sum_{j:\mu(q_{j})\geqslant 3}\alpha_{j}$. The equality (4.16) reads as

[TABLE]

Claim 5*.*

$\#{\mathcal{L}}\leqslant 1$.

• Proof.

Suppose that $\#{\mathcal{L}}>1$. By Lemma 3.6(a), $\#{\mathcal{L}}$ is connected and has exactly two components, they meet normally and $A_{1}\cdot E_{0}=0$. We have ${\mathcal{L}}\cdot E_{1}=\psi_{*}^{-1}{\mathcal{L}}\cdot E_{0}\leqslant 2$. Consider the case when the components of ${\mathcal{L}}$ have intersection number bigger than $1$. By Lemma 3.6(a) one of them is the image of a component of some twig of $D_{0}$, so it does not meet $E_{0}$, hence ${\mathcal{L}}\cdot E_{1}\leqslant 1$. By Lemma 2.4 after contracting $Q_{3}+Q_{4}$ and blowing up three times over ${\mathcal{L}}$ we obtain an elliptic fibration for which the proper transform of $E_{1}$ has at least two cusps and meets a general fiber at most once. But this is impossible, hence ${\mathcal{L}}=[1,1]$. After the contraction of $Q_{3}+Q_{4}$ the linear system of ${\mathcal{L}}$ induces a $\mathbb{P}^{1}$-fibration of a smooth surface on which the image of $E_{1}$ is singular (hence horizontal) and meets a general fiber at most twice. In fact, being singular, it meets a general fiber exactly twice, hence $\psi_{*}^{-1}{\mathcal{L}}\cdot E_{0}=2$, which gives $\psi_{*}^{-1}{\mathcal{L}}=\widetilde{C}_{1}+\widetilde{C}_{2}$. Thus $\eta=0$, $s_{1}=s_{2}=0$ and $\operatorname{Exc}\psi-A_{1}$ consists of twigs meeting $\widetilde{C}_{1}$ and $\widetilde{C}_{2}$. It follows from Lemma 4.6(e) that $\alpha_{j}\geqslant 0$ for $j=1,2$. Moreover, $\psi$ touches $Q_{2}$ and does not touch $E_{0}$, hence $\varepsilon=0$ and, by Lemma 4.4, $\lambda_{2}\geqslant 4$. Then we are in case (1.d) of Lemma 4.5, hence (4.18) and Claim 2 give $K_{1}\cdot R\leqslant b_{0}(\Delta_{1}^{-})$. Since $\Delta_{1}^{-}=0$ in case (1.d), we infer that $R=\bar{Q}_{1}+\bar{Q}_{2}-\psi(C_{1}+C_{2}+\widetilde{C}_{1}+\widetilde{C}_{2})$ consists of $(-2)$-curves and has no tips. It follows that $R=0$, hence $\tau_{j}=\lambda_{j}-2=2$ for $j=1,2$. Let $\theta\colon X_{1}\longrightarrow\theta(X_{1})$ be the contraction of $\psi(C_{1}+C_{2}+C_{3}+C_{4})$. Then $\theta(X_{1})\cong\mathbb{P}^{1}\times\mathbb{P}^{1}$ and $\theta(E_{1})\sim 2f_{1}+2f_{2}$, where $f_{1},f_{2}$ are fibers of two projections of $\theta(X_{1})$ onto $\mathbb{P}^{1}$. We compute $\theta(E_{1})^{2}=8$ and hence $\gamma_{1}=-E_{1}^{2}=-\theta(E_{1})^{2}+\sum_{j=1}^{4}\tau_{j}^{2}=-8+16=8$. Then (4.18) fails; a contradiction.

∎

Claim 6*.*

$\Delta_{1}^{-}=0$.

• Proof.

Suppose that $\Delta_{1}^{-}\neq 0$. Then we are in case (1.a) of Lemma 4.5 with $\lambda_{1}\leqslant 6$, $\Delta_{1}^{-}=[2]$ and $\Delta_{1}^{+}=0$. Let $E_{1}^{\prime}\subseteq D_{1}^{\prime}$, $U_{1}\subseteq D_{1}^{\prime}$ and $\bar{Q}_{1}^{\prime}\subseteq D_{1}^{\prime}$ be, respectively, the proper transforms of $E_{1}$, $\Delta_{1}$ and the reduced total transform of $\bar{Q}_{1}$ under the minimal log resolution $\varphi_{1}\colon(X_{1}^{\prime},D_{1}^{\prime})\longrightarrow(X_{1},D_{1})$. Put $A_{1}^{\prime}:=(\varphi_{0})_{*}^{-1}A_{1}$. Clearly, $U_{1}=[2]$ and $U_{1}$ is a twig of $D_{1}^{\prime}$ contained in $\bar{Q}_{1}^{\prime}$. Since $\mathrm{ind}_{1}^{(1)}\leqslant\frac{1}{2}$, we infer that $s_{1}=0$ and that $U_{1}$ is in fact a unique maximal twig of $D_{1}^{\prime}$ contained in $\bar{Q}_{1}^{\prime}$. It follows that the proper transform of $U_{1}$ on $X$, call it $U_{0}$, is the unique maximal twig of $D+A_{1}^{\prime}$ contained in $Q_{1}^{\prime}$ and hence $Q_{1}^{\prime}$ has at most one branching component. Moreover, if $\varepsilon=1$ then $Q_{1}^{\prime}$ is a chain and $U_{0}$ is a maximal twig of $D$, so $s_{1}=1$, which is false. Thus $\varepsilon=0$ and so (4.18) gives $\alpha_{1}+K_{1}\cdot R\leqslant\lambda_{1}-8+\eta+\#{\mathcal{L}}\leqslant\eta+\#{\mathcal{L}}-2$. Since $\tau_{1}^{*}\geqslant 1$, we have $\alpha_{1}\geqslant 0$, hence $2\leqslant\eta+\#{\mathcal{L}}\leqslant 2\#{\mathcal{L}}$. By Claim 5 we get $\eta=\#{\mathcal{L}}=1$. Since $\Delta_{1}^{+}=0$, we have ${\mathcal{L}}\subseteq\Upsilon_{1}^{0}$. Denote by $C_{1}^{\prime}$ the unique $(-1)$-curve in $Q_{1}^{\prime}$. The fact that $\psi$ creates a new component of $\Upsilon_{1}^{0}$ implies that both components of $Q_{1}^{\prime}$ met by $A_{1}^{\prime}$ are contained in the same connected component of $Q_{1}^{\prime}-C_{1}^{\prime}$, hence $U_{0}$ meets $C_{1}^{\prime}$. But then again $s_{1}=1$; a contradiction.

∎

Claim 7*.*

$\alpha\geqslant-1$.

• Proof.

Suppose otherwise. Let $j\in\{1,2\}$. By assumption the cusp $q_{j}\in\bar{E}$ is not semi-ordinary and $\alpha_{j}=-1$. By Lemma 4.6(e), $\tau_{j}^{*}=0$ and $\psi$ does not touch $C_{j}$. We have $s_{j}=1$, so $\mathrm{ind}_{j}^{(1)}\geqslant\frac{1}{2}$, hence by Claim 4 we are in case (1.d) of Lemma 4.5. In particular, the contribution to $\mathrm{ind}(D_{1}^{\prime})$ coming from twigs of $D_{1}^{\prime}$ contained in $\bar{Q}_{j}^{\prime}$ and not contracted by $\varphi_{1}$ is $\mathrm{ind}^{\prime}:=\mathrm{ind}_{1}^{(1)}+\mathrm{ind}_{2}^{(1)}-2\cdot\frac{1}{2}\leqslant\frac{1}{3}$. We obtain $\Delta_{1}^{+}\wedge\bar{Q}_{j}=0$, hence $(\Upsilon_{1}-\Upsilon_{1}^{0})\wedge\bar{Q}_{j}$. Since $q_{j}\in\bar{E}$, $j\in\{1,2\}$ is not semi-ordinary and $\tau_{j}^{*}=0$, the divisor $Q_{j}^{\prime}$ is not a chain. From Lemma 3.8(a) we infer that $A_{1}$ meets $Q_{j}$. It follows that $\varepsilon=0$ and $\eta=0$, hence (4.18) gives $\gamma_{1}+K_{1}\cdot R=\#{\mathcal{L}}\leqslant 1$ by Claim 5. Let $T_{j}=\operatorname{Exc}\psi\wedge Q_{j}$. We have $\#\bar{Q}_{j}=\#\Omega_{j}=\lambda_{j}+1-\tau_{j}=3$, so $\bar{Q}_{j}$ is a chain. We observe that $\bar{Q}_{j}$ contains some tip $U_{j}$ of $D_{1}$. Indeed, otherwise $\psi(C_{j})$ is a tip of $\bar{Q}_{j}$ and $\psi(T_{j})$ belongs to the second tip of $\bar{Q}_{j}$. But then $Q_{j}$ is a chain with $C_{j}$ as a tip, hence it is a chain $[1,(2)_{k_{j}}]$ for some $k_{j}\geqslant 0$, that is, $q_{j}\in\bar{E}$ is semi-ordinary, which is false. Since $\varepsilon=0$, by Claim 2, $\gamma_{1}\geqslant 0$, so we obtain $K_{1}\cdot(U_{1}+U_{2})\leqslant K_{1}\cdot R\leqslant\#{\mathcal{L}}\leqslant 1$. It follows that $U_{1}$ or $U_{2}$ is a $(-2)$-tip, so $\mathrm{ind}^{\prime}\geqslant\frac{1}{2}$; a contradiction.

∎

Claim 8*.*

$\eta=1$.

• Proof.

Suppose first that $\#{\mathcal{L}}=0$. By Lemma 4.5, $\lambda_{1}+\lambda_{2}\leqslant 8$. By (4.18) and by previous claims we have $-1\leqslant\alpha+K_{1}\cdot R\leqslant\varepsilon+\lambda_{1}+\lambda_{2}-10\leqslant\varepsilon-2$. It follows that $\varepsilon=1$, $R$ consists of $(-2)$-curves and there is some $j\in\{1,2\}$ for which $q_{j}$ is not semi-ordinary and $\alpha_{j}=-1$, hence $s_{j}=1$, $\tau_{j}=2$ and $\psi$ does not touch $C_{j}$. It follows that $A_{1}$ meets $Q_{j}$. Since $\varepsilon=1$, we have $A_{1}\cdot E_{0}=1$, so $A_{1}\cdot Q_{1}=1$. Then the equality $\varepsilon=1$ implies that $T:=\operatorname{Exc}\psi-A_{1}$ is a $(-2)$-twig of $Q_{1}$, so $Q_{1}-C_{1}$ consists of $(-2)$-curves and a unique $(-3)$-curve meeting $T$. Since $\Delta_{1}^{-}=0$, the divisor $Q_{1}+A_{1}+E_{0}$ has no tips, hence $Q_{1}$ is a chain and $C_{1}$ is its tip. Then $Q_{1}$ does not contract to a smooth point; a contradiction. Thus $\#{\mathcal{L}}=1$. The equality (4.18) gives

[TABLE]

Suppose that $\eta=0$ and $\varepsilon=1$. By Claim 6, $\Delta_{1}\wedge\bar{Q}_{1}=0$. The divisor $T=\operatorname{Exc}\psi-A_{1}$ is a $(-2)$-twig of $Q_{1}$ and $\psi$ does not touch $Q_{2}$. Since $\#{\mathcal{L}}=1$, the component $B\subseteq Q_{1}-T$ meeting $T$ is a $(-2)$-curve. We obtain

[TABLE]

so $Q_{1}-C_{1}$ consists of $(-2)$-curves and at most one $(-3)$-curve. If $B=\widetilde{C}_{1}$ then, $C+B+T=[1,(2)_{k}]$ for some $k\geqslant 2$, so since $Q_{1}$ contains no curves with self-intersection number smaller than $-3$, $Q_{1}=C+B+T$, hence $\bar{Q}_{1}=[1,1]$. But the latter is impossible, as the contraction of $\psi(C_{1})+\bar{Q}_{2}+\bar{Q}_{3}+\bar{Q}_{4}$ would map $\psi(\widetilde{C}_{1})$ onto a [math]-curve on $\mathbb{P}^{2}$. Thus $B\neq\widetilde{C}_{1}$. Since $B$ is a $(-2)$-curve not contracted by $\psi$, it is necessarily a branching component of $Q_{1}$. We have $\Delta_{1}\wedge\bar{Q}_{1}=0$, so the maximal twigs of $Q_{1}$ are $T=[2]$, $[3]$ and $[1,(2)_{k}]$ for some $k\geqslant 0$. In particular, $K_{1}\cdot R=1$, which gives $\alpha=-1$. We have $\alpha_{2}\geqslant 0$, because otherwise $\mu(q_{2})>2$ and $\tau_{2}^{*}=0$, which is impossible, as $Q_{2}-C_{2}$ consists of $(-2)$-curves. It follows that $\alpha_{1}=-1$, so $\tau_{1}^{*}=0$. Let $U$ denote the $(-3)$-tip of $\bar{Q}_{1}$. After the contraction of $\bar{Q}_{3}+\bar{Q}_{4}$ the linear system of $\bar{Q}_{1}-U=[1,2,\ldots,2,1]$ induces a $\mathbb{P}^{1}$-fibration for which Lemma 2.3 gives $\tau_{1}-1+\mu(q_{2})-1+2\leqslant 2\cdot(1+\tau_{1})-2$, which implies that $q_{2}$ is semi-ordinary. But then $\mathrm{ind}_{1}^{(1)}+\mathrm{ind}_{2}^{(1)}\geqslant\frac{1}{3}+\frac{1}{2}+\frac{5}{6}=\frac{5}{3}$, which contradicts Lemma 4.5.

Suppose that $\eta=0$ and $\varepsilon=0$. Now (4.19) reads as $\alpha+K_{1}\cdot R\leqslant\lambda_{1}+\lambda_{2}-9\leqslant-1$, hence $R$ consists of $(-2)$-curves, $\lambda_{1}+\lambda_{2}=8$, and $\alpha=-1$. Since there is $j\in\{1,2\}$ with $\alpha_{j}=-1$ which is not semi-ordinary, hence with $s_{j}=1$, we have $\mathrm{ind}_{j}^{(1)}\geqslant\frac{1}{2}$, so by Lemma 4.5, $(\lambda_{1},\lambda_{2})=(7,1)$ or $\lambda_{1}=\lambda_{2}=4$. We have $A_{1}\cdot Q_{1}\geqslant 1$. Suppose that $A_{1}$ meets $Q_{2}$. Then $\lambda_{1}=\lambda_{2}=4$ and, say, $j=1$, so $\tau_{1}^{*}=0$ by Lemma 4.6(e). But since $\Delta_{1}^{-}=0$ by Claim 6, there is no tip of $D_{1}$ contained in $\bar{Q}_{1}$, so $Q_{1}$ is necessarily a chain and $C_{1}$ is its tip. This means that $q_{1}$ is semi-ordinary; a contradiction. Thus $A_{1}\cdot Q_{2}=0$. Again, if $j=2$ then $q_{2}$ is not semi-ordinary and $\alpha_{2}=-1$, which is impossible, as then $\tau_{2}^{*}=0$ and $Q_{2}-C_{2}$ consists of $(-2)$-curves. Thus $\alpha_{1}=-1$. By Lemma 4.6(e) it follows that $\tau_{1}^{*}=0$ and $\psi$ does not touch $C_{1}$. The divisor $\bar{Q}_{1}-\psi(C_{1})$ consists of one $(-1)$-curve ${\mathcal{L}}$ and some number of $(-2)$-curves. Because $\Delta_{1}^{-}=0$, no tip of $D_{1}$ is contained in $\bar{Q}_{1}$. It follows that $\psi(C_{1})$ is a tip of $\bar{Q}_{1}$ and $\bar{Q}_{1}-\psi(C_{1})$ is a circular divisor with ${\mathcal{L}}$ as its unique component meeting $\psi(C_{1})$. Contract $\bar{Q}_{2}+\bar{Q}_{3}+\bar{Q}_{4}+{\mathcal{L}}$. Then the image of $E_{1}$ has three cusps and is a $2$-section of the $\mathbb{P}^{1}$-fibration induced by the linear system of the image of $\psi(C_{1})$. This is a contradiction with the Hurwitz formula.

∎

The inequality (4.18) and Claims 6 and 8 give

[TABLE]

Claim 9*.*

$A_{1}\cdot E_{0}=0$.

• Proof.

Suppose that $A_{1}\cdot E_{0}=1$. Let $U=\psi_{*}^{-1}{\mathcal{L}}$. Then ${\mathcal{L}}\cdot E_{1}=1$ and, by Claim 8, ${\mathcal{L}}\subseteq\Upsilon_{1}$. For $\varepsilon=1$ this is impossible, hence $\varepsilon=0$. Then $\operatorname{Exc}\psi=A_{1}$, $U=[2]$ and $\Delta_{1}^{+}\wedge\bar{Q}_{1}\neq 0$. By Claim 7 and (4.20)

[TABLE]

Let $\mathrm{ind}_{U}$ be the contribution to $\mathrm{ind}_{1}^{(1)}$ from the $(-2)$-twig in $\Delta_{1}^{+}$ meeting ${\mathcal{L}}=\psi(U)$. Since $Q_{1}^{\prime}$ contains at least two tips of $D$ and the proper transform of $A_{1}$ meets none of them, we get $\mathrm{ind}_{1}^{(1)}>\mathrm{ind}_{U}\geqslant\frac{1}{2}$, hence by Lemma 4.5 we are in case (1.d). Put $\mathrm{ind}_{1}^{\prime}=\mathrm{ind}_{1}^{(1)}-\mathrm{ind}_{U}>0$. We have $\mathrm{ind}_{1}^{\prime}+\mathrm{ind}_{2}^{(1)}\leqslant\frac{4}{3}-\frac{1}{2}=\frac{5}{6}$. By Lemma 3.8(a), $\mathrm{ind}_{1}^{\prime}<\frac{5}{6}-\frac{1}{2}=\frac{1}{3}$, hence $s_{1}=0$ and if $\bar{Q}_{1}-\Delta_{1}^{+}\wedge\bar{Q}_{1}$ contains some tip $V$ of $D_{1}$ then $V^{2}\leqslant-4$. Since $K_{0}\cdot(Q_{1}-C_{1})\leqslant 1$, there is no such tip. In particular, $\bar{Q}_{1}$ is a chain with $\psi(C_{1})$ or $\psi(\widetilde{C}_{1})$ as a tip. Then $Q_{1}$ is a chain, either $Q_{1}=[(2)_{k},3,1,2]$ or $Q_{1}=[1,(2)_{k}]$ for some $k\geqslant 0$. By the inequality above $\tau_{1}=\alpha_{1}+2\leqslant 2$, hence $\tau_{1}=2$ and $\alpha_{1}=0$. It follows that $Q_{1}=[(2)_{k},3,1,2]$, because otherwise we get $\mathrm{ind}_{1}^{\prime}\geqslant\frac{1}{3}$. By (4.20), $\alpha=-1$ and $Q_{2}-C_{2}$ consists of $(-2)$-curves. The former gives $\alpha_{2}=-1$, which implies that $q_{2}$ is semi-ordinary. Then $\alpha=\alpha_{1}=0$; a contradiction.

∎

By Claim 9, $A_{1}\cdot E_{0}=0$, so $\varepsilon=0$ and for every component $V\subseteq D_{1}-E_{1}$ we have $V\cdot E_{1}=\psi_{*}^{-1}V\cdot E_{0}$. By (4.20) and by Claim 7,

[TABLE]

If $A_{1}$ meets both $Q_{1}$ and $Q_{2}$ then by Lemma 4.5, $\lambda_{1}=\lambda_{2}=4$, so in any case we may, and will, assume that ${\mathcal{L}}\subseteq\bar{Q}_{1}$.

Claim 10*.*

$\mathcal{L}$ meets three components of $D_{1}-\mathcal{L}$ and does not meet $E_{1}$.

• Proof.

First suppose that ${\mathcal{L}}\cdot E_{1}\neq 0$. We have ${\mathcal{L}}\cdot E_{1}=\psi_{*}^{-1}{\mathcal{L}}\cdot E_{0}$, so $\psi_{*}^{-1}{\mathcal{L}}=\widetilde{C}_{1}$. Since $\psi$ touches $\widetilde{C}_{1}$, we have $\beta_{D_{1}}({\mathcal{L}})\geqslant 3$, hence ${\mathcal{L}}$, as a component of $\Upsilon_{1}$, meets $\Delta_{1}^{+}$. Then $C_{1}$ meets a connected component of $\Delta_{0}$ not contracted by $\psi$, so $\beta_{Q_{1}+A_{1}}(\widetilde{C}_{1})\geqslant 3$, hence $\beta_{D_{1}}({\mathcal{L}})>3$. Then ${\mathcal{L}}\nsubseteq\Upsilon_{1}$; a contradiction.

Thus ${\mathcal{L}}\cdot E_{1}=0$. Suppose that ${\mathcal{L}}$ does not meet three components of $D_{1}-{\mathcal{L}}$. We have ${\mathcal{L}}\subseteq\Upsilon_{1}$, so then ${\mathcal{L}}$ meets at most two components of $D_{1}-{\mathcal{L}}$, namely some $B\subseteq R+\psi(C_{1})$ with $B\cdot{\mathcal{L}}=2$ and possibly some component of $\Delta_{1}^{+}$. It follows that $\psi$ does not touch $Q_{2}$. Let $\theta\colon X_{1}\longrightarrow\theta(X_{1})$ be the contraction of $\bar{Q}_{2}+\bar{Q}_{3}+\bar{Q}_{4}$. Suppose that $B\neq\psi(C_{1})$. Then $K_{1}\cdot B\leqslant K_{1}\cdot R\leqslant 1$, so $B^{2}\in\{-2,-3\}$. By Lemma 2.4(b) we may blow up on $\theta(B)$ so that the linear system of the proper transform of $F=B+2{\mathcal{L}}$ induces an elliptic fibration. But $F\cdot\theta(E_{1})\leqslant\psi_{*}^{-1}B\cdot E_{0}\leqslant 1$, so $\theta(E_{1})$ is a $1$-section or is vertical. The former is impossible, because $\theta(E_{1})$ is not smooth and the latter is impossible, because degenerate fibers of elliptic fibrations of smooth surfaces cannot contain components with more than $1$ cusp; a contradiction. Thus $B=\psi(C_{1})$. In particular, $\psi$ touches $C_{1}$, hence $\tau_{1}\geqslant 3$ by Lemma 4.2(c). By (4.21), $\alpha\in\{-1,0\}$, hence $\alpha_{1}\leqslant 1$. From the definition of $\alpha_{j}$ we get $\psi(C_{1})^{2}+3=\tau_{1}-\alpha_{1}-s_{1}\geqslant\tau_{1}-2$. Again contract $\bar{Q}_{2}+\bar{Q}_{3}+\bar{Q}_{4}$ and blow up $\tau_{1}-\alpha_{1}-s_{1}+1$ times over $\psi(C_{1})$, of which the first $\tau_{1}-\alpha_{1}-s_{1}$ times on the intersection of the proper transforms of $\psi(C_{1})$ and $E_{1}$, and then once on the proper transform of $\psi(C_{1})$. Denote the final proper transforms of $\psi(C_{1})$, $E_{1}$ and ${\mathcal{L}}$ by $C^{\prime}$, $E^{\prime}$ and ${\mathcal{L}}^{\prime}$, respectively. We have $(C^{\prime})^{2}=-4$ and $C^{\prime}\cdot E^{\prime}\leqslant\alpha_{1}+s_{1}\leqslant 2$. By Lemma 2.4(b) the last center may be chosen so that $|C^{\prime}+2{\mathcal{L}}^{\prime}|$ induces an elliptic fibration such that $E^{\prime}$ meets a general fiber at most twice. Since $E^{\prime}$ has at least two cusps, it cannot be vertical, hence it meets the fiber once or twice. This is impossible by the Hurwitz formula.

∎

Claim 11*.*

${\mathcal{L}}$ meets one $(-1)$-curve and one $(-3)$-curve.

• Proof.

We have $\eta=1$, so ${\mathcal{L}}\subseteq\Upsilon_{1}$. By Claim 10 there are two distinct components $B_{1},B_{2}\subseteq\bar{Q}_{1}+\bar{Q}_{2}-\Delta_{1}$ and a component $T\subseteq\Delta_{1}^{+}$ such that ${\mathcal{L}}\cdot B_{1}={\mathcal{L}}\cdot B_{2}={\mathcal{L}}\cdot T=1$ and ${\mathcal{L}}$ does not meet $D_{1}-{\mathcal{L}}-B_{1}-B_{2}-T$. It follows that $B_{1}$ and $B_{2}$ meet ${\mathcal{L}}$ in different points.

Suppose that $B_{j}^{2}\geqslant 0$ for some $j\in\{1,2\}$. By Lemma 4.6(d), $B_{j}=\psi(C_{1})$ and from the definition of $\alpha_{j}$ we have $\psi(C_{1})^{2}=\tau_{1}-\alpha_{1}-s_{1}-3$. Contract $\bar{Q}_{3}+\bar{Q}_{4}$ and blow up $\psi(C_{1})^{2}+2=\tau_{1}-\alpha_{1}-s_{1}-1\geqslant 2$ times over $\psi(C_{1})$, each time on the intersection of the proper transforms of $\psi(C_{1})$ and $E_{1}$. Denote the final proper transforms of $\psi(C_{1})$, $E_{1}$, $T$ and ${\mathcal{L}}$ by $C^{\prime}$, $E^{\prime}$, $T^{\prime}$ and ${\mathcal{L}}^{\prime}$, respectively. We have $(C^{\prime})^{2}=-2$ and $C^{\prime}\cdot E^{\prime}=\alpha_{1}+s_{1}+1\leqslant\alpha_{1}+2$. Since $\alpha\in\{-1,0\}$, we have $\alpha_{1}\leqslant 1$. The linear system of $F=C^{\prime}+2{\mathcal{L}}+T^{\prime}$ gives a $\mathbb{P}^{1}$-fibration such that $F\cdot E^{\prime}=C^{\prime}\cdot E^{\prime}$, so by the Hurwitz formula $C^{\prime}\cdot E^{\prime}-1+\mu(q_{3})-1+\mu(q_{4})-1\leqslant 2C^{\prime}\cdot E^{\prime}-2$, hence $C^{\prime}\cdot E^{\prime}\geqslant 3$. We obtain $\alpha_{1}=s_{1}=1$, hence $R$ consists of $(-2)$-curves, $\alpha_{2}=-1$ and $q_{2}$ is not semi-ordinary. Then $s_{2}=1$ and, by Lemma 4.2(d), $\tau_{1}\geqslant 3$. The contribution of $\Delta_{1}^{+}$ to $\mathrm{ind}_{1}^{(1)}+\mathrm{ind}_{2}^{(1)}$ is at least $\frac{1}{2}$, so we obtain $\mathrm{ind}_{1}^{(1)}+\mathrm{ind}_{2}^{(1)}\geqslant\frac{2}{3}+\frac{1}{2}+\frac{1}{2}=\frac{5}{3}$. This is a contradiction with Lemma 4.5.

By (4.21), $B_{j}^{2}\in\{-1,-2,-3\}$ for $j\in\{1,2\}$. If $B_{j}^{2}=-2$ for some $j\in\{1,2\}$ then $B_{j}\cdot E_{1}=\psi_{*}^{-1}B_{j}\cdot E_{0}\leqslant 1$, so after the contraction of $\bar{Q}_{3}+\bar{Q}_{4}$ the linear system of the image of $F=B_{j}+2{\mathcal{L}}+T$ gives a $\mathbb{P}^{1}$-fibration for which the image of $E_{1}$ is singular and meets a general fiber at most once, which is impossible. Thus $B_{j}^{2}\in\{-1,-3\}$ for $j\in\{1,2\}$.

If $B_{1}^{2}=B_{2}^{2}=-1$ then $B_{j}=\psi(C_{j})$ for $j=1,2$, hence $\psi$ does not touch $C_{1}+C_{2}$ and then $\psi_{*}^{-1}{\mathcal{L}}$ is a component of $D_{0}-E_{0}$ meeting both $C_{1}$ and $C_{2}$, which is impossible. Since $K_{1}\cdot R\leqslant 1$, we obtain $\{B_{1}^{2},B_{2}^{2}\}=\{-1,-3\}$

∎

Let $B_{1}$, $B_{2}$ and $T$ be the three components of $D_{1}-E_{1}-{\mathcal{L}}$ meeting ${\mathcal{L}}$. By Claim 11 we may assume that $B_{1}=\psi(C_{1})$ is a $(-1)$-curve and $B_{2}$ is a $(-3)$-curve. We have $T\subseteq\Delta_{1}^{+}$. From the definition of $\alpha_{j}$ we have $\tau_{1}=2+s_{1}+\alpha_{1}\leqslant 4$, as $\alpha_{1}\leqslant 1$. Contract $\bar{Q}_{3}+\bar{Q}_{4}+{\mathcal{L}}+T$ and blow up four times over $\psi(C_{1})$, the first $\tau_{1}$ times on the intersections of the proper transforms of $\psi(C_{1})$ and $E_{1}$. Blow up once more at some point of the proper transform of $\psi(C_{1})$, off the proper transform of $B_{2}$. Denote the final proper transforms of $B_{1},B_{2}$ and $E_{1}$ by $B_{1}^{\prime},B_{2}^{\prime}$ and $E_{1}^{\prime}$, respectively. We have $(B_{1}^{\prime})^{2}=-4$, $(B_{2}^{\prime})^{2}=-1$ and $B_{1}^{\prime}\cdot E^{\prime}=0$. Since $\psi(C_{1})$ is a $(-1)$-curve, $\psi$ does not touch $C_{1}$, so $C_{1}$ meets $\psi_{*}^{-1}{\mathcal{L}}$. If $B_{1}^{\prime}\cdot B_{2}^{\prime}\neq 2$ then, by construction, $C_{1}$ meets $\psi_{*}^{-1}(B_{2})$ too, $\psi_{*}^{-1}B_{2}\neq\widetilde{C}_{1}$ and $A_{1}\cdot Q_{1}=2$. But in the latter case $\Delta_{1}^{+}\wedge\bar{Q}_{1}\neq 0$ and $s_{1}=1$ (as $C_{1}$ is non-branching in $Q_{1}$), so $\mathrm{ind}_{1}^{(1)}+\mathrm{ind}_{2}^{(1)}>1+\frac{1}{2}$, in contradiction to Lemma 4.5. Thus $B_{1}^{\prime}\cdot B_{2}^{\prime}=2$. By Lemma 2.4(b) the last center may be chosen so that $|B_{1}^{\prime}+2B_{2}^{\prime}|$ induces an elliptic fibration. A general fiber meets $E^{\prime}$ at most $2B_{2}^{\prime}\cdot E^{\prime}$ times. We have $B_{2}^{\prime}\cdot E^{\prime}=\psi_{*}^{-1}B_{2}\cdot E_{0}$ and the latter number is at most $1$, because $B_{2}\neq\psi(C_{2})$. Since $E^{\prime}$ has two cusps, it is neither a $1$-section nor it is contained in a fiber, hence it meets a general fiber twice. Because $E^{\prime}$ meets $B_{1}^{\prime}+B_{2}^{\prime}$ in a unique point, Hurwitz formula gives $1+(\mu(q_{3})-1)+(\mu(q_{4})-1)\leqslant 2$; a contradiction.

∎

5. Consequences of the almost minimality

We now analyze various consequences of Theorem 4.7. The essential one is that the inequality (4.6) results with an upper bound on the number of components of $D$ and hence on the degree of $\bar{E}$, see Remark 5.11. We first describe the geometry of exceptional divisors of resolutions of cusps in terms of Hamburger-Noether pairs, which are a compact and geometrically meaningful way to keep track of multiplicity sequences. For a detailed treatment see [Rus80] and [KR99, Appendix]. For a discussion of relations between standard Hamburger-Noether pairs, Puiseux pairs and other numerical characteristics of cusps see [PP17, Appendix].

5A. Description in terms of Hamburger-Noether pairs

Recall that we say that a divisor $Q$ on a projective surface contracts to a smooth point if it is equal to the reduced exceptional divisor of some birational morphism mapping $Q$ onto a smooth point of some surface. Equivalently, $Q$ has a negative definite intersection matrix and $d(Q)=1$, cf. (2.1), see [Rus02, 1.18.1].

Definition 5.1** (Hamburger-Noether pairs).**

Let $Q$ be a divisor which contracts to a smooth point and contains a unique $(-1)$-curve $L$. By induction with respect to the number of branching components of $Q$ we define a sequence of positive integers, called standard Hamburger-Noether (HN-) pairs of $Q$

[TABLE]

as follows. Let $\sigma$ denote the contraction of the maximal twig of $Q$ containing $L$. Write $\operatorname{Exc}\sigma-L=C+P$, where $C,P$ are reduced connected and disjoint, and such that $d(C)\geqslant d(P)$ if $Q$ is a chain and that $P$ meets the proper transform of $\sigma(Q)$ otherwise. In the first case put $h=1$ and $\binom{c^{(1)}}{p^{(1)}}=\binom{d(C)}{d(P)}$. Otherwise let $\left(\binom{\bar{c}^{(k)}}{\bar{p}^{(k)}}\right)_{k=1}^{\widetilde{h}}$ denote the sequence of standard HN-pairs of $\sigma(Q)$. Put $h=\widetilde{h}+1$, $c=d(C)$ and define the sequence of standard HN-pairs of $Q$ as

[TABLE]

Note that $Q$ has exactly $h-1$ branching components.

By $\lfloor x\rfloor$ and $\lceil x\rceil$ we denote respectively the biggest integer not greater than $x$ and the smallest integer not smaller than $x$. We have $c-\lfloor c/p\rfloor\cdot p=(c\bmod p)$.

Lemma 5.2** (Properties of contractible chains).**

Let $Q^{+}$ be a reduced chain with tips $Z_{P}\neq Z_{C}$ and such that $Q=Q^{+}-Z_{P}-Z_{C}$ contains a unique $(-1)$-curve $L$ and contracts to a smooth point. Write $Q-L=P+C$, where $P$, $C$ are reduced connected and disjoint and $C\cdot Z_{P}=P\cdot Z_{C}=0$; see Figure 3. The following hold:

1. (a)

$c:=d(C)$ and $p:=d(P)$ are relatively prime, cf. (2.1), 2. (b)

Denote the tips of $P$ and $C$ meeting $Z_{P}$ and $Z_{C}$ by $L_{P}$ and $L_{C}$, respectively; put $L_{P}=0$ if $P=0$ and $L_{C}=0$ if $C=0$. Then

[TABLE]

If $L_{P}\neq 0$ and, respectively, $L_{C}\neq 0$ then

[TABLE] 3. (c)

Let $\chi$ be a smooth germ meeting $L$ normally, not in a node of $Q^{+}$, and let $\sigma$ be the contraction of $Q$. Then:

[TABLE]

• Proof.

First consider a rational chain $Z$ with a unique $(-1)$-curve $L$ and such that $Z-L$ consists of two admissible chains $A$ and $B$. We claim that if $Z$ contracts to a [math]-curve then $d(A)=d(B)$. This was shown in [Fuj82, 3.7, 4.7], but it is worth recalling a simple proof. We order the components of $A$ and $B$ so that the first ones meet $L$. Using (2.6) we show by induction on $n$ that $\operatorname{gcd}(d(A),d^{\prime}(A))=1$; similarly $\operatorname{gcd}(d(B),d^{\prime}(B))=1$. Using (2.5) we note that if we replace a tree by its reduced total transform under a blowup then its discriminant does not change. Finally, using (2.5) twice gives

[TABLE]

hence $d(A)|d(B)$ and $d(B)|d(A)$, so indeed $d(A)=d(B)$.

(a) We choose an order on the set of irreducible components of $P$ in which $L_{P}$ is the last one; we do the same for $C$. As in the argument above, we have

[TABLE]

hence $\operatorname{gcd}(c,p)=1$.

(b),(c) We proceed by induction with respect to $\#Q$. If $Q=L$ then the statements hold, so assume $\#Q\geqslant 2$. Write $\sigma=\sigma_{0}\circ\sigma^{\prime}$, where $\sigma_{0}$ is a single blowup. The chain $Q^{\prime}=\operatorname{Exc}\sigma^{\prime}\subseteq Q$ contracts to a smooth point, $L$ is its unique $(-1)$-curve and $Q-Q^{\prime}$ is a tip of $Q$. Since the formulas to prove are symmetric, we may assume that $Q-Q^{\prime}=L_{P}$. Let $L_{P}^{\prime}$ be the tip of $Q^{\prime}-L$ meeting $L_{P}$; put $L_{P}^{\prime}=0$ if $P=L_{P}$. Put $p^{\prime}=d(P-L_{P})$. By induction

[TABLE]

and

[TABLE]

Since the statements to prove do not depend on properties of $Z_{P}$ other than normality of the intersection with $L_{P}$, we may assume that $Z_{P}$ is a $(-1)$-curve. Then $\sigma(Z_{P})$ is a [math]-curve, so $d(Z_{P}+P)=d(C)$ by the observation above. By (2.6) we obtain $1\cdot p-p^{\prime}=c$, that is, $0<p^{\prime}=p-c=p-(c\bmod p).$ Hence $d(C-L_{C})=c-(p^{\prime}\bmod c)=c-(p\bmod c)$, which proves (5.3). Using the projection formula and the inductive assumption we obtain

[TABLE]

and

[TABLE]

which proves (5.5).

From (5.5) we have $\sigma^{\prime}(L_{P})\cdot\sigma^{\prime}(\chi)=c$ and $\sigma^{\prime}(Z_{C})\cdot\sigma^{\prime}(\chi)=p-c$, so when blowing up over $\sigma^{\prime}(L)$ the center stays on the proper transform of $L_{P}$ exactly $\lceil\frac{c}{p-c}\rceil$ times, which gives

[TABLE]

If $L_{C}\neq 0$ then, by induction,

[TABLE]

which proves (5.4).

∎

Assume that $Q$ is as in Definition 5.1. Putting formally $c^{(h+1)}=1$ and using Lemma 5.2(a) we note that:

[TABLE]

Remark 5.3** (cf. Remark 2.10 [PP17]).**

Lemma 5.2(c) shows that the above definition of HN-pairs for $Q$ is close to the one used in [Pal14, §3]. The difference is that we no longer insist that $c^{(k)}\geqslant p^{(k)}$, so while we lose some flexibility, our sequences are usually shorter than the ones in loc. cit. One may easily pass from one sequence to another by replacing our pairs $\binom{c^{(k)}}{p^{(k)}}$ for which $c^{(k)}<p^{(k)}$ with $(\binom{c^{(k)}}{c^{(k)}})_{q},\binom{c^{(k)}}{p^{(k)}-qc^{(k)}}$, where $q$ is the integer part of $p^{(k)}/c^{(k)}$.

For a given logical formula $\varphi$ without free parameters we denote by $[\varphi]\in\{0,1\}$ its logical value, that is, we put

[TABLE]

We denote by $\Delta(Q)$ the sum of maximal $(-2)$-twigs of $Q$. We put

[TABLE]

Note that if $L$, the unique $(-1)$-curve of $Q$, is not a tip of $Q$ then $Q$ has $h+1$ maximal admissible twigs. We define $\mathrm{ind}(Q)$ as in Section 2. Recall that given a nonzero reduced snc divisor on a smooth surface we say that a blow-up at a point $p$ of the divisor is inner if $p$ is a node, otherwise it is outer. By $r(Q)$ we denote the number of outer blow-ups used to create $Q$. We note that the first blow-up in the process of reconstruction of $Q$ is neither outer nor inner.

Given a pair of positive integers $c$, $p$ we put

[TABLE]

where $A=[a_{1},\ldots,a_{m}]$ and $B=[b_{1},\ldots,b_{n}]$ are unique admissible chains with $d(A)=p/\operatorname{gcd}(c,p)$ and $d(B)=c/\operatorname{gcd}(c,p)$ such that $[a_{1},\ldots,a_{m},1,b_{1},\ldots,b_{n}]$ contracts to a smooth point. Note that in the Euclidean algorithm for $(c,p)$ we reach [math] after exactly $\#|\binom{c}{p}|=m+n+1$ steps.

We put (cf. (5.7))

[TABLE]

and we define

[TABLE]

Example 5.4**.**

The sequences of standard HN-pairs $\binom{15}{6}\binom{3}{2}$ and $\binom{10}{4}\binom{2}{3}$ describe two forks, say, $Q_{1}^{\prime}$ and $Q_{2}^{\prime}$. Both divisors have a unique branching component $B$ and three maximal twigs $T_{1}$, $T_{2}$, $T_{3}$. The first pairs create a chain $[2,3,1,2]$ and the second pairs change this chain into a fork with a branching curve $B$ and three maximal twigs $T_{1}=[2,3]$, $T_{2}=[2]$ and $T_{3}$. In the first case we have $B^{2}=-3$ and $T_{3}=[3,1,2]$ and in the second $B^{2}=-2$ and $T_{3}=[2,1,3]$. We compute $M(Q_{1}^{\prime})=22$, $I(Q_{1}^{\prime})=96$ and $M(Q_{2}^{\prime})=16$, $I(Q_{2}^{\prime})=46$. The first fork requires $r(Q_{1}^{\prime})=3$ outer blowups and the second requires $r(Q_{2}^{\prime})=4$ outer blow-ups. In both cases for the first pair $b^{+}=1$ and for the second pair $b^{+}=0$ and $b^{+}=1$, respectively. In the first case the sum of $\lambda^{+}$ equals $6$ and $\lambda=5$. In the second case the sum of $\lambda^{+}$ equals $5$ and $\lambda=4$.

To effectively make computations we prove formulas for various quantities in terms of HN-pairs.

Lemma 5.5**.**

Let $Q$ be a divisor which contracts to a smooth point and contains a unique $(-1)$-curve, which is not a tip of $Q$. Let $\left(\binom{c^{(k)}}{p^{(k)}}\right)_{k=1}^{h}$ be the standard HN-sequence of $Q$. Then:

1. (a)

$\displaystyle\mathrm{ind}(Q)=1-\frac{c^{(1)}\bmod p^{(1)}}{p^{(1)}}+\sum_{k=1}^{h}(1-\frac{p^{(k)}\bmod c^{(k)}}{c^{(k)}})$, 2. (b)

$\displaystyle r(Q)=\lceil c^{(1)}/p^{(1)}\rceil-1+\sum_{k=2}^{h}\lceil p^{(k)}/c^{(k)}\rceil\geqslant h$, 3. (c)

$\displaystyle b_{0}(\Delta(Q))=b^{-}(\binom{c^{(1)}}{p^{(1)}})+\sum_{k=1}^{h}b^{+}(\binom{c^{(k)}}{p^{(k)}})$, 4. (d)

$\displaystyle\lambda(Q)=-b^{-}(\binom{c^{(1)}}{p^{(1)}})+\sum_{k=1}^{h}\lambda^{+}(\binom{c^{(k)}}{p^{(k)}})-[\binom{c^{(1)}}{p^{(1)}}=\binom{3}{2}]$, 5. (e)

$s(Q)=[h>1\text{\ and }(p^{(h)}\bmod c^{(h)})=1]+[h=1\text{\ and }(c^{(1)}\bmod p^{(1)})=1]$.

• Proof.

Let $L$ denote the unique $(-1)$-curve of $Q$. Let $Q^{(h)}$ be the maximal twig of $Q$ containing $L$ and let $\sigma$ denote its contraction. Put $\binom{c}{p}=\binom{c^{(h)}}{p^{(h)}}$. We have $Q^{(h)}-L=P+C$, where the divisors $P,C$ are reduced connected and disjoint, $d(P)=p$ and $d(C)=c$. Since $L$ is not a tip of $Q$, we have $C\neq 0$ (hence $c>1$) and $p^{(1)}>1$. By Lemma 5.2(b)

[TABLE]

so (a) follows from the definition of the inductance. For (b) note that by definition the first blow-up within the HN-pair $\binom{c^{(k)}}{p^{(k)}}$ is outer if and only if $k\geqslant 2$. Then the formula follows from Lemma 5.2(c). By Lemma 5.2(b), $L_{P}=[2]$ if and only if $p\geqslant 2(c\bmod p)>0$, and $L_{C}=[2]$ if and only if $c\geqslant 2(p\bmod c)>0$, which gives (c).

Part (d) follows from (c) and from the definition of $\lambda(Q)$. For (e) we observe that if $h>1$ then by (5.3), $C$ is a $(-2)$-chain if and only if $p^{(h)}\bmod c^{(h)}=1$. Similarly, if $h=1$ then $L$ meets a $(-2)$-twig if and only if $c^{(1)}\bmod p^{(1)}=1$ (the twig is necessarily $P$). ∎

We define the following function on the set of positive integers:

[TABLE]

We note the following values: $\nu(4)=\frac{11}{15}$, $\nu(5)=\frac{19}{28}$, $\nu(6)=\frac{29}{45}$ and $\nu(7)=\frac{41}{66}$.

Lemma 5.6** (Lower bound on inductance).**

Let $Q$ be a divisor which contracts to a smooth point and contains a unique $(-1)$-curve, which is not a tip of $Q$. Then:

1. (a)

$\min\{\mathrm{ind}(Q):Q\text{\ such that\ }\lambda(Q)=\lambda\}=\nu(\lambda)$ and the minimum for every $\lambda\geqslant 4$ (respectively for $\lambda=3$) is attained only at the chain $[\lambda-1,1,(2)_{\lambda-3},3]$ (respectively $[4,1,2,2]$). 2. (b)

If $\mathrm{ind}(Q)\leqslant\nu(\lambda(Q)-1)$ then $Q$ is a chain. 3. (c)

If $\lambda(Q)\leqslant 7$ and $\mathrm{ind}(Q)\leqslant\frac{2}{3}$ then:

1. (i)

$Q=[5,1,2,2,2,3]$, $(\lambda(Q),\mathrm{ind}(Q))=(6,\frac{2}{3}-\frac{1}{45})$ or 2. (ii)

$Q=[6,1,2,2,2,2,3]$, $(\lambda(Q),\mathrm{ind}(Q))=(7,\frac{2}{3}-\frac{1}{22})$ or 3. (iii)

$Q=[5,2,1,3,2,2,3]$, $(\lambda(Q),\mathrm{ind}(Q))=(7,\frac{2}{3}-\frac{1}{144})$.

• Proof.

By Proposition 4.4 and Lemma 4.2(e) the above statements hold if $\lambda<4$. For $Q=[\lambda-1,1,(2)_{\lambda-3},3]$ we have $\lambda(Q)=\lambda$ and $\mathrm{ind}(Q)=\nu(\lambda)$, so

[TABLE]

To prove the opposite inequality w argue by induction with respect to $\lambda$. Let $\lambda_{0}\geqslant 4$ and assume (a) and (b) hold for every $Q$ with $\lambda(Q)\leqslant\lambda_{0}-1$. By induction and by the definition of $\nu(\lambda)$, $\min\{\mathrm{ind}(Q):\lambda(Q)=\lambda\}$ is a decreasing function of $\lambda$ for $2\leqslant\lambda\leqslant\lambda_{0}$ and its value for $\lambda=4$ is smaller than for $\lambda=1$.

To prove (b) suppose that there exists a branched divisor $Q$ with $\lambda(Q)=\lambda_{0}$ and $\mathrm{ind}(Q)\leqslant\nu(\lambda_{0}-1)$. Contract $(-1)$-curves in $Q$ and its images until there is no branching component and denote the image of $Q$ by $Q^{\prime}$. We have $\lambda(Q^{\prime})\leqslant\lambda_{0}-1$, so by the inductive assumption $\nu(\lambda(Q^{\prime}))\leqslant\mathrm{ind}(Q^{\prime})<\nu(\lambda_{0}-1)$, hence $\lambda(Q^{\prime})=1$, which gives $Q^{\prime}=[3,1,2]$. Now $\frac{5}{6}<\nu(\lambda_{0}-1)$, so since $\lambda_{0}\geqslant 4$, we get $\lambda_{0}=4$ and $\nu(\lambda_{0}-1)=\frac{11}{12}$. Since $\lambda(Q)=4$, the contracted divisor is a twig, either $[2,2,1]$ or $[2,1,3]$. In both cases the inequality $\mathrm{ind}(Q)\leqslant\frac{11}{12}$ fails; a contradiction.

To prove (a) suppose that $Q$ is a divisor with $\lambda(Q)=\lambda_{0}$ and $\mathrm{ind}(Q)\leqslant\nu(\lambda_{0})$. Then $\mathrm{ind}(Q)\leqslant\nu(\lambda_{0}-1)$, so part (b) implies that $Q$ is a chain. We will repeatedly use Lemma 2.1 to bound $\mathrm{ind}(Q)$ from below. For instance, since $\nu(\lambda_{0})<1$, we have $b_{0}(\Delta(Q))\leqslant 1$. Let $U_{i}$ for $i\in\{1,\ldots,\#Q\}$ denote the proper transform of the exceptional curve of the $i$’th blow-up in the sequence of blow-ups reversing the contraction of $Q$ to a smooth point. We treat $Q$ as the exceptional divisor of a resolution of a cusp. After the first blow-up the proper transform of the germ meets $U_{1}$. It separates from $U_{1}$ after $x_{1}$ further blow-ups, $x_{1}\geqslant 1$.

Suppose that $x_{1}=1$. Then $b_{0}(\Delta(Q))=1$ and $\#Q=\lambda_{0}+1$. After $x_{2}\geqslant 1$ more blowups the germ separates from $U_{2}$ and the created chain is $[2,x_{2}+1,1,(2)_{x_{2}-1}]$. Since $b_{0}(\Delta(Q))=1$, we have $x_{2}\leqslant 2$. If $x_{2}=1$ then $Q$ contains the twig $[2,2]$, so $\nu(5)<\frac{2}{3}<\mathrm{ind}(Q)\leqslant\nu(\lambda_{0})$ and hence $\lambda_{0}=4$ and $\#Q=\lambda_{0}+b_{0}(\Delta(Q))=5$, in which case $Q=[2,2,3,1,2]$, in contradiction to the equality $b_{0}(\Delta(Q))=1$. Thus $x_{2}=2$ and hence $Q$ contains a twig $[2,3]$. There are $\#Q-4=\lambda_{0}-3$ blow-ups remaining to be done, so the self-intersection of the tip of $Q$ other than $U_{1}$ is not smaller than $-2-(\lambda_{0}-3)=1-\lambda_{0}$. Then $\frac{3}{5}+\frac{1}{\lambda_{0}-1}\leqslant\mathrm{ind}(Q)\leqslant\nu(\lambda_{0})$, which is impossible; a contradiction.

Thus $x_{1}\geqslant 2$. After the first $x_{1}+1$ blow-ups the created chain is $[(2)_{x_{1}-1},1,1+x_{1}]$. Suppose that $x_{1}\geqslant 3$. Then $x_{1}\leqslant\#Q-1=\lambda_{0}$ and $Q$ contains the twigs $[(2)_{x_{1}-2}]$ and $[1+x_{1}]$, hence $\frac{x_{1}-2}{x_{1}-1}+\frac{1}{x_{1}+1}\leqslant\mathrm{ind}(Q)\leqslant\nu(\lambda_{0})\leqslant\nu(x_{1})$, which gives $x_{1}=3$ and $\nu(\lambda_{0})\geqslant\frac{3}{4}$; a contradiction.

Thus $x_{1}=2$. After the first three blow-ups the created chain is $[2,1,3]=U_{2}+U_{3}+U_{1}$. Since $\#Q>3$, now the germ meets $U_{2}$ and $U_{3}$, so $\Delta(Q)=0$ and hence $\#Q=\lambda_{0}$. After $x_{2}\geqslant 1$ additional blow-ups the germ separates from $U_{2}$ and the chain becomes $[2+x_{2},1,(2)_{x_{2}},3]$. We see that $Q$ contains the twigs $[2+x_{2}]$ and $[3,(2)_{x_{2}-1}]$. Clearly, $x_{2}+3\leqslant\#Q=\lambda_{0}$. Consider the case $x_{2}+4\leqslant\lambda_{0}$. Then $\frac{1}{2+x_{2}}+\frac{x_{2}}{2x_{2}+1}\leqslant\mathrm{ind}(Q)\leqslant\nu(\lambda_{0})\leqslant\nu(x_{2}+4)$, which gives $x_{2}=1$ and $\lambda_{0}=5$. Then $Q=[3,2,1,3,3]$, so $\mathrm{ind}(Q)=\frac{2}{5}+\frac{3}{8}\leqslant\nu(5)$; a contradiction. We obtain $x_{2}+3=\lambda_{0}=\#Q$. Then $Q=[\lambda_{0}-1,1,(2)_{\lambda_{0}-3},3]$, which gives (a).

(c) Since $\mathrm{ind}(Q)\leqslant\frac{2}{3}$, we have $\lambda_{0}:=\lambda(Q)\in\{6,7\}$ by (a). Suppose that $Q$ is branched. We have $\mathrm{ind}(Q)\leqslant\frac{2}{3}\leqslant\nu(6-1)$, so by part (b), $\lambda_{0}=7$. Let $Q^{\prime}$ be as in the proof of (b). Then $\mathrm{ind}(Q^{\prime})\geqslant\nu(6)=\frac{2}{3}-\frac{1}{45}$, so if $T$ is a tip of $Q$ not contained in the proper transform of $Q^{\prime}$ then $\frac{1}{-T^{2}}\leqslant\mathrm{ind}(Q)-\mathrm{ind}(Q^{\prime})\leqslant\frac{1}{45}$, i.e. $T^{2}\leqslant-45$. But creating such a component requires at least $45$ blow-ups and, since we easily prove by induction that $b_{0}(\Delta(Q))\leqslant\lambda(Q)+1$, we have $\#Q\leqslant 2\lambda_{0}+1\leqslant 15$; a contradiction. It follows that $Q$ is a chain. Using the assumption $\mathrm{ind}(Q)\leqslant\frac{2}{3}$, which for $\lambda_{0}=6,7$ is only slightly weaker than $\mathrm{ind}(Q)\leqslant\nu(\lambda_{0})$, we argue as in the proof of (a) that $x_{1}=2$ and that either $x_{2}=\lambda_{0}-3$ and $\lambda_{0}\in\{6,7\}$ or $x_{2}=\lambda_{0}-4$ and $\lambda_{0}=7$. In the first case we get (i) and (ii) above and in the latter case we get (iii).

∎

For $Q$ as in Lemma 5.6 we denote by $\Delta^{-}(Q)$ the sum of those maximal $(-2)$-twigs of $Q$ which are disjoint from the unique $(-1)$-curve of $Q$.

Lemma 5.7**.**

Let $Q$ be as in Lemma 5.6 and let $\lambda:=\lambda(Q)\leqslant 7$. In case $\lambda=6$ assume that $\mathrm{ind}(Q)\leqslant\frac{3}{2}$ and in case $\lambda=7$ that $\Delta^{-}(Q)=0$. Then the sequence of (standard) HN-pairs of $Q$ belongs to the following list (where $h$ denotes the length of the sequence):

1. (a)

$\lambda=1$ : $\binom{3}{2}$, 2. (b)

$\lambda=2$ : $\binom{5}{2}$, 3. (c)

$\lambda=3$ :

1. $h=1$

: $\binom{4}{3}$, $\binom{7}{2}$, $\binom{7}{3}$, 2. $h=2$

: $\binom{6}{4}\binom{2}{1}$, $\binom{10}{4}\binom{2}{1}$, 4. (d)

$\lambda=4$ :

1. $h=1$

: $\binom{5}{3}$, $\binom{5}{4}$, $\binom{7}{5}$, $\binom{8}{3}$, $\binom{9}{2}$, $\binom{9}{4}$, $\binom{10}{3}$, $\binom{12}{5}$, 2. $h=2$

: $\binom{6}{4}\binom{2}{3}$, $\binom{10}{4}\binom{2}{3}$, $\binom{9}{6}\binom{3}{1}$, $\binom{15}{6}\binom{3}{1}$, $\binom{8}{6}\binom{2}{1}$, $\binom{14}{4}\binom{2}{1}$, $\binom{14}{6}\binom{2}{1}$ 3. $h=3$

: $\binom{12}{8}\binom{4}{2}\binom{2}{1}$, $\binom{20}{8}\binom{4}{2}\binom{2}{1}$. 5. (e)

$\lambda=5$ :

1. $h=1$

: $\binom{7}{4}$, $\binom{8}{5}$, $\binom{6}{5}$, $\binom{9}{7}$, $\binom{10}{7}$, $\binom{11}{3}$, $\binom{11}{4}$, $\binom{11}{8}$, $\binom{13}{5}$, $\binom{11}{2}$, $\binom{11}{5}$, $\binom{13}{3}$, $\binom{13}{4}$, $\binom{16}{7}$, $\binom{17}{5}$, $\binom{17}{7}$, $\binom{19}{8}$, 2. $h=2$

: $\binom{6}{4}\binom{2}{5},$ $\binom{10}{4}\binom{2}{5},$ $\binom{9}{6}\binom{3}{2},$ $\binom{15}{6}\binom{3}{2},$ $\binom{9}{6}\binom{3}{4},$ $\binom{15}{6}\binom{3}{4},$ $\binom{12}{8}\binom{4}{1},$ $\binom{20}{8}\binom{4}{1},$ $\binom{15}{10}\binom{5}{2},$ $\binom{25}{10}\binom{5}{2},$ $\binom{8}{6}\binom{2}{3},$ $\binom{14}{4}\binom{2}{3},$ $\binom{14}{6}\binom{2}{3},$ $\binom{12}{9}\binom{3}{1},$ $\binom{21}{6}\binom{3}{1},$ $\binom{21}{9}\binom{3}{1},$ $\binom{10}{6}\binom{2}{1},$ $\binom{10}{8}\binom{2}{1},$ $\binom{14}{10}\binom{2}{1},$ $\binom{16}{6}\binom{2}{1},$ $\binom{18}{4}\binom{2}{1},$ $\binom{18}{8}\binom{2}{1},$ $\binom{20}{6}\binom{2}{1},$ $\binom{24}{10}\binom{2}{1}$, 3. $h=3$

: $\binom{12}{8}\binom{4}{2}\binom{2}{3},$ $\binom{20}{8}\binom{4}{2}\binom{2}{3},$ $\binom{18}{12}\binom{6}{3}\binom{3}{1},$ $\binom{30}{12}\binom{6}{3}\binom{3}{1},$ $\binom{12}{8}\binom{4}{6}\binom{2}{1},$ $\binom{20}{8}\binom{4}{6}\binom{2}{1},$ $\binom{18}{12}\binom{6}{2}\binom{2}{1},$ $\binom{30}{12}\binom{6}{2}\binom{2}{1},$ $\binom{16}{12}\binom{4}{2}\binom{2}{1},$ $\binom{28}{8}\binom{4}{2}\binom{2}{1},$ $\binom{28}{12}\binom{4}{2}\binom{2}{1}$, 4. $h=4$

: $\binom{24}{16}\binom{8}{4}\binom{4}{2}\binom{2}{1},$ $\binom{40}{16}\binom{8}{4}\binom{4}{2}\binom{2}{1}$, 6. (f)

$\lambda=6$, $\mathrm{ind}(Q)\leqslant\frac{3}{2}$:

1. $h=1$

: $\binom{9}{5},$ $\binom{11}{7},$ $\binom{12}{7},$ $\binom{13}{8},$ $\binom{7}{6},$ $\binom{11}{9},$ $\binom{13}{9},$ $\binom{13}{10},$ $\binom{14}{3},$ $\binom{14}{5},$ $\binom{14}{11},$ $\binom{15}{4},$ $\binom{15}{11},$ $\binom{17}{12},$ $\binom{18}{5},$ $\binom{18}{7},$ $\binom{18}{13},$ $\binom{19}{7},$ $\binom{21}{8},$ $\binom{13}{2},$ $\binom{13}{6},$ $\binom{16}{3},$ $\binom{16}{5},$ $\binom{20}{9},$ $\binom{22}{5},$ $\binom{22}{9},$ $\binom{23}{7},$ $\binom{23}{10},$ $\binom{24}{7},$ $\binom{25}{11},$ $\binom{26}{11},$ $\binom{27}{8},$ $\binom{29}{12},$ $\binom{31}{13},$ 2. $h=2$

: $\binom{6}{4}\binom{2}{7},$ $\binom{9}{6}\binom{3}{5},$ $\binom{15}{6}\binom{3}{5},$ $\binom{9}{6}\binom{3}{7},$ $\binom{12}{8}\binom{4}{3},$ $\binom{20}{8}\binom{4}{3},$ $\binom{15}{10}\binom{5}{3},$ $\binom{25}{10}\binom{5}{3},$ $\binom{15}{10}\binom{5}{7},$ $\binom{21}{14}\binom{7}{3},$ $\binom{24}{16}\binom{8}{3},$ $\binom{8}{6}\binom{2}{5},$ $\binom{12}{9}\binom{3}{2},$ $\binom{10}{6}\binom{2}{3},$ $\binom{10}{8}\binom{2}{3},$ $\binom{14}{10}\binom{2}{3},$ $\binom{16}{6}\binom{2}{3},$ $\binom{15}{9}\binom{3}{1},$ $\binom{14}{8}\binom{2}{1},$ $\binom{16}{10}\binom{2}{1},$ $\binom{12}{10}\binom{2}{1},$ $\binom{18}{14}\binom{2}{1},$ $\binom{20}{14}\binom{2}{1},$ $\binom{22}{8}\binom{2}{1},$ $\binom{22}{16}\binom{2}{1}$, 7. (g)

$\lambda=7$, $\Delta^{-}(Q)=0$:

1. $h=1$

: $\binom{11}{6},$ $\binom{14}{9},$ $\binom{16}{9},$ $\binom{17}{10},$ $\binom{18}{11},$ $\binom{19}{11},$ $\binom{19}{12},$ $\binom{21}{13},$ $\binom{8}{7},$ $\binom{15}{2},$ 2. $h=2$

: $\binom{10}{6}\binom{2}{5},$ $\binom{15}{9}\binom{3}{2},$ $\binom{15}{9}\binom{3}{4},$ $\binom{20}{12}\binom{4}{1},$ $\binom{14}{8}\binom{2}{3},$ $\binom{16}{10}\binom{2}{3},$ $\binom{21}{12}\binom{3}{1},$ $\binom{24}{15}\binom{3}{1},$ $\binom{18}{10}\binom{2}{1},$ $\binom{22}{14}\binom{2}{1},$ $\binom{24}{14}\binom{2}{1},$ $\binom{26}{16}\binom{2}{1}$.

• Proof.

We explain how to obtain the list. First of all, it is easy to create inductively a list of HN-pairs $\binom{c}{p}$ with a given number of components of the chain $|\binom{c}{p}|$, cf. (5.9). Indeed, $\#|\binom{c}{p}|=k$ if and only if the Euclidean algorithm for $(c,p)$ reaches [math] after exactly $k$ steps. It follows that having a complete list of pairs with $\#|\binom{c}{p}|=k$ and replacing each $\binom{c}{p}$ with $\{\binom{c+p}{p},\binom{c+p}{c}\}$ we obtain a complete list of pairs with $\#|\binom{c}{p}|=k+1$. By Definition 5.1 the HN-sequence for a divisor $Q$ with $h\geqslant 2$ is obtained from the HN-sequence of the divisor $\sigma(Q)$ and the HN-pair of $\operatorname{Exc}\sigma$, which allows to list sequences of standard HN-pairs for all divisors with a given number of components. Finally, for $Q\neq[3,1,2]$ we have $h+1\leqslant\lambda(Q)$, so $\#Q-\lambda(Q)=b_{0}(\Delta(Q))\leqslant h+1\leqslant\lambda(Q)$, hence $Q$ can be found among divisors with at most $2\lambda(Q)$ components. Conversely, given a sequence of HN-pairs of some $Q$ we may compute $\lambda(Q)$, $\mathrm{ind}(Q)$, $b_{0}(\Delta(Q))$ and $s(Q)$ using formulas from Lemma 5.5(d).

∎

Remark 5.8**.**

If a chain $Q$ with a standard HN-pair $\binom{c}{p}$ is blown up in a node belonging to the unique $(-1)$-curve then the HN-pair of the resulting chain is not necessarily one of $\binom{c+p}{c}$ or $\binom{c+p}{p}$. For instance, for $\binom{5}{3}$ the new HN-pairs are $\binom{8}{5}$ and $\binom{7}{4}$. The geometry behind the change $\binom{c}{p}\mapsto\binom{c+p}{p}$ is that to $Q$ we add a $(-2)$-tip at the side with discriminant $c$. Similarly, the change $\binom{c}{p}\mapsto\binom{c+p}{c}$ is related to adding a $(-q)$-tip at the side with discriminant $p$, where $q=1+\lceil c/p\rceil$.

5B. Further bounds on the geometry of the boundary

In this section we work more with the minimal log resolution $\pi\colon(X,D)\longrightarrow(\mathbb{P}^{2},\bar{E})$ instead of the minimal weak resolution $\pi_{0}\colon(X_{0},D_{0})\longrightarrow(\mathbb{P}^{2},\bar{E})$. We denote by $Q_{j}^{\prime}$ the exceptional divisor of $\pi$ over the cusp $q_{j}\in\bar{E}$, $j=1,\ldots,c$. We define the sequence of standard HN-pairs of $q_{j}\in\bar{E}$ as the sequence of standard HN-pairs of $Q_{j}^{\prime}$ and we decorate the respective quantities $c^{(k)}$, $p^{(k)}$ and $h$ with lower index $j\in\{1,\ldots,c\}$. Clearly, $\Delta(Q_{j}^{\prime})=Q_{j}^{\prime}\wedge\Delta_{0}^{\prime}$, $s_{j}=s(Q_{j}^{\prime})$ and $\mathrm{ind}_{j}=\mathrm{ind}(Q_{j}^{\prime})$. For cusps of multiplicity bigger than $2$ we have $(\Upsilon_{0}+\Delta_{0}^{+})\wedge Q_{j}=0$, so by (4.5)

[TABLE]

hence by (5.8) for all cusps $\lambda(Q_{j}^{\prime})=\lambda_{j}$. We put (cf. (5.8))

[TABLE]

The numbers $M_{j}$ and $I_{j}$ compute the sums of multiplicities and, respectively, squares of multiplicities, of successive proper transforms of $q_{j}\in\bar{E}$ under blow-ups constituting $\pi^{-1}$ (see [Pal14, Corollary 3.2]), so the genus-degree formula takes the form

[TABLE]

Proposition 4.1 gives $c\leqslant 5$, hence we may and will assume that $c=5$ or $c=4$.

Corollary 5.9**.**

For a positive integer $k$ let $\omega_{k}$ denote the number of cusps of $\bar{E}$ for which $\lambda_{j}=k$. Then $\sum_{k=1}^{\infty}k\omega_{k}=c$ and the following inequality holds:

[TABLE]

• Proof.

We have $\mathrm{ind}_{j}>\frac{1}{2}$ for any cusp by Lemma 3.8(a). Using the bounds from Lemmas 5.6, 4.4 and 4.2(a)(e) we get the required inequality.

∎

Lemma 5.10**.**

Let the notation be as above and let $\delta=\delta(\Delta_{0}^{-})$. We have:

1. (a)

$\deg\bar{E}=\frac{3}{2}+\frac{1}{2}\sqrt{1+4\sum_{j=1}^{c}(I_{j}-M_{j})}$, 2. (b)

$p_{2}=9-2c+\sum_{j=1}^{c}(M_{j}-r_{j})-3\deg\bar{E}$, 3. (c)

$\sum_{j=1}^{c}\mathrm{ind}_{j}\leqslant 5-p_{2}$, 4. (d)

$\delta\leqslant 7+3p_{2}-\sum_{j=1}^{c}\lambda_{j}$.

• Proof.

Part (a) follows from (5.14). By [Pal14, Corollary 3.3(i)], $-E^{2}=-3\deg\bar{E}+2+\sum_{j=1}^{c}M_{j}$. On the other hand, by (4.15), $p_{2}=K\cdot(K+D)=-E^{2}+7-2c-\sum_{j=1}^{c}r_{j}$, so we obtain (b). Parts (c) and (d) are (4.3) and (4.6), respectively.

∎

Remark 5.11**.**

From the above bounds we can derive further ones. We will not use them, but we discuss them to convince the reader that, given Theorem 4.7, the situation is numerically completely bounded. From the definition of $\lambda_{j}$ we have $\lambda_{1}+\cdots+\lambda_{c}=\#D-1-b_{0}(\Delta_{0}^{\prime})-c_{0}^{\prime}$, hence $\#D\leqslant 3p_{2}+8+c_{0}^{\prime}+b_{0}(\Delta_{0}^{\prime})-\delta$ by Lemma 5.10(d), where $c_{0}^{\prime}$ is the number of ordinary cusps ($c_{0}^{\prime}=\omega_{1}$). Since $\mathrm{ind}(D)\geqslant\frac{1}{2}b_{0}(\Delta_{0}^{\prime})+\frac{1}{3}c_{0}^{\prime}$ by Lemma 3.8(c), Lemma 5.10(c) gives

[TABLE]

and hence $\#D\leqslant p_{2}+18+\lfloor\frac{1}{3}\omega 1-\delta\rfloor\leqslant 21$, as $p_{2}\leqslant 2$ by Lemma 4.2(f). It follows that $\#Q_{j}^{\prime}\leqslant 20-3\cdot 3=11$. Since $\#Q_{j}^{\prime}\neq 1$, we have $r_{j}\geqslant 1$, so by (4.14)

[TABLE]

From [KP17, Lemma 2.4(ii)] we obtain $\mu(q_{j})\leqslant F_{K\cdot Q_{j}^{\prime}+3}\leqslant F_{\#Q_{j}^{\prime}}\leqslant F_{11}=89$, where $F_{k}$ denotes the $k$’th Fibonacci number ($F_{1}=F_{2}=1$). Then by the Matsuoka-Sakai inequality [MS89]

[TABLE]

Proposition 5.12**.**

Let $\bar{E}\subseteq\mathbb{P}^{2}$ be a rational cuspidal curve with $c\geqslant 4$ cusps. Put $\Lambda=(\lambda_{1},\ldots,\lambda_{c})$ with $\lambda_{1}\geqslant\ldots\geqslant\lambda_{c}$. Then we have $c=4$, $p_{2}\leqslant 1$ and $h_{1}\leqslant 2$. Moreover, if $\lambda_{1}\geqslant 5$ then $p_{2}=1$ and one of the following holds:

1. (1)

$\Lambda=(7,1,1,1)$, $\delta=0$, $\mathrm{ind}_{1}\leqslant\frac{3}{2}$, 2. (2)

$\Lambda=(6,2,1,1)$, $\delta=0$, $\mathrm{ind}_{1}\leqslant\frac{37}{30}$, 3. (3)

$\Lambda=(5,3,1,1)$, $\delta=0$, $\mathrm{ind}_{1}\leqslant\frac{17}{12}$, 4. (4)

$\Lambda=(5,2,2,1)$, $\delta=0$, $\mathrm{ind}_{1}\leqslant\frac{29}{30}$, 5. (5)

$\Lambda=(6,1,1,1)$, $\delta\leqslant 1$, $\mathrm{ind}_{1}\leqslant\frac{3}{2}$, 6. (6)

$\Lambda=(5,2,1,1)$, $\delta\leqslant 1$, $\mathrm{ind}_{1}\leqslant\frac{37}{30}$, 7. (7)

$\Lambda=(5,1,1,1)$, $\delta\leqslant 2$, $\mathrm{ind}_{1}\leqslant\frac{3}{2}$,

• Proof.

By Proposition 4.1, $c\leqslant 5$. Suppose that $c=5$. Elementary calculations show that the only cases satisfying Corollary 5.9 and Lemma 5.10(c),(d) are:

1. (a)

$p_{2}=1$, $\Lambda=(6,1,1,1,1)$, 2. (b)

$p_{2}=0$, $\Lambda=(3,1,1,1,1)$, 3. (c)

$p_{2}=0$, $\Lambda=((2)_{\omega_{2}},(1)_{5-\omega_{2}})$, where $\omega_{2}\in\{0,1,2\}$.

Put $d=\deg\bar{E}$. In cases (a) and (b) we have $|\Lambda|-3p_{2}=7$, where $|\Lambda|=\lambda_{1}+\cdots+\lambda_{c}$, so by Lemma 5.10(d), $\Delta_{0}^{-}=0$. In case (b) Proposition 4.4 implies that either $q_{1}$ is semi-ordinary or $Q_{1}=C_{1}$, hence $q_{1}$ is described by one of the HN-pairs $\binom{7}{2}$ or $\binom{4}{3}$. In both cases (5.14) gives $(d-1)(d-2)=14$, which is impossible. In case (a) Lemma 5.10(c) gives $\mathrm{ind}_{1}\leqslant\frac{2}{3}$, so by Lemma 5.6(c), $q_{1}$ is described by the HN-pair $\binom{9}{5}$, hence $(d-1)(d-2)=40$, which is impossible.

In case (c) the minimal log resolution is described by $\omega_{2}$ pairs $\binom{5}{2}$ and $5-\omega_{2}$ pairs $\binom{3}{2}$. We compute $\sum_{j=1}^{5}M_{j}=6\omega_{2}+4(5-\omega_{2})=2\omega_{2}+20$ and $\sum_{j=1}^{5}I_{j}=10\omega_{2}+6(5-\omega_{2})=4\omega_{2}+30$, hence by (5.14), $(d-1)(d-2)=2\omega_{2}+10$, which gives $\omega_{2}=1$ and $d=5$. But after blowing up $q_{1}$ once we get a $\mathbb{P}^{1}$-fibration for which the proper transform of $E$ meets a general fiber three times. Then the Hurwitz formula, Lemma 2.3, gives $c\leqslant 2\cdot 3-2=4$; a contradiction.

Thus $c=4$. Suppose that $p_{2}=2$. From Corollary 5.9 and Lemma 5.10(c),(d) we have $\Lambda=(6,5,1,1)$ or $\Lambda=(7,4,1,1)$ or $\Lambda=(\lambda_{1},1,1,1)$ for some $\lambda_{1}\in\{7,8,9,10\}$. But in the latter case $\mathrm{ind}_{1}+3\cdot\frac{5}{6}\leqslant 5-p_{2}=3$, hence $\mathrm{ind}_{1}\leqslant\frac{1}{2}$, which is impossible by Lemma 3.8(a). By Lemma 5.10(c) we have $\mathrm{ind}_{1}+\mathrm{ind}_{2}\leqslant 3-2\cdot\frac{5}{6}=\frac{4}{3}$. Then Lemma 5.6 gives a contradiction in the second case and in the first case gives $Q_{1}=[5,1,2,2,2,3]$ and hence $\mathrm{ind}_{2}\leqslant\frac{2}{3}+\frac{1}{45}$, which results with $\mathrm{ind}_{2}<\nu(5)+\frac{1}{96}$. Under the latter inequality, which is only slightly weaker than $\mathrm{ind}(Q_{2}^{\prime})\leqslant\nu(5)$, the arguments in the proof of Lemma 5.6(a) show that $Q_{2}=[4,1,2,2,3]$. Then (5.14) gives $(d-1)(d-2)=54$; a contradiction.

Thus $p_{2}\leqslant 1$. Suppose that $h_{1}\geqslant 3$ and $\delta=0$. Then $q_{1}\in\bar{E}$ is not semi-ordinary and $b_{0}(Q_{1}^{\prime}\wedge\Delta_{0}^{\prime})=s_{1}\leqslant 1$, so $\#Q_{1}^{\prime}=\lambda_{1}+s_{1}\leqslant 8$. Since $\delta=0$, the first two HN-pairs of $q_{1}$ cannot create a $(-2)$-tip, so they contain at least $4$ and $3$ components, respectively, hence $\#Q_{1}^{\prime}\geqslant 4+3+2=9$; a contradiction.

Suppose that $h_{1}\geqslant 3$ and $\delta>0$. Since $h_{1}\geqslant 3$, we have $\lambda_{1}\geqslant 4$ by Proposition 4.4. By Lemma 5.10(d), $p_{2}=1$, so $\mathrm{ind}_{1}+\ldots+\mathrm{ind}_{4}\leqslant 4$ and $|\Lambda|\leqslant 9$, hence $\lambda_{1}\leqslant 6$ and $\lambda_{2}\leqslant 3$. Then $\mathrm{ind}_{j}\geqslant\frac{5}{6}$ for $j\in\{2,3,4\}$, so $\mathrm{ind}_{1}\leqslant\frac{3}{2}$. But using Lemma 5.7 we check that for $\lambda_{1}\in\{4,5,6\}$ the latter inequality fails in all cases with $h_{1}\geqslant 3$; a contradiction.

Thus $h_{1}\leqslant 2$. The cases listed in the statement of the proposition are the only ones with $\lambda_{1}\geqslant 5$ which satisfy Lemma 5.10(d). The bound on $\mathrm{ind}_{1}$ follows from Lemma 5.10(c).

∎

Example 5.13**.**

Consider the quintic

[TABLE]

It has parameterization $\varphi\colon\mathbb{P}^{1}\longrightarrow\mathbb{P}^{2}$:

[TABLE]

Its singular points are cusps $q_{1}=\varphi([0:1])$, $q_{j}=\varphi([\sqrt[3]{2}\theta^{j-2}:-1])$, $j=2,3,4$, where $\theta$ is a primitive third root of unity. The local analytic type of the cusp $q_{1}$ is $x^{2}=y^{7}$ and the local analytic type of the cusps $q_{j}$, $j=2,3,4$ is $x^{2}=y^{3}$.

• Proof of Theorem 1.1.

By Proposition 5.12, $p_{2}\leqslant 1$ and either $\lambda_{1}\leqslant 4$ or $p_{2}=1$ and $(\lambda_{1},\lambda_{2},\lambda_{3},\lambda_{4})$ is one of the $7$ sequences listed there. In Lemma 5.7 we have a complete list of possible sequences of HN-pairs with small $\lambda$. Using (5.8) and Lemma 5.5 in each case we may compute $M_{j}$, $I_{j}$, $\mathrm{ind}_{j}$, $\lambda_{j}$, $r_{j}$, $b_{0}(Q_{j}^{\prime}\wedge\Delta_{0}^{\prime})$, $s_{j}$ and $\displaystyle b_{0}(\Delta_{0}^{-})=\sum_{j:\mu(q_{j}>2)}(b_{0}(Q_{j}^{\prime}\wedge\Delta_{0}^{\prime})-s_{j})$, which then allows to compute subsequently $d:=\deg\bar{E}$ and $p_{2}$ as in Lemma 5.10(a),(b). We then check whether the so-computed $p_{2}$ belongs to $\{0,1\}$ and whether the inequality in Lemma 5.10(c) and the implication $b_{0}(\Delta_{0}^{-})\neq 0\Rightarrow 7+3p_{2}>\sum_{j=1}^{4}\lambda_{j}$, a consequence of the inequality in Lemma 5.10(d), hold. The only cases where this is true are:

1. (a)

$\binom{5}{2}$, $\binom{5}{2}$, $\binom{3}{2}$, $\binom{3}{2}$, $(d,p_{2})=(5,0)$, 2. (b)

$\binom{7}{2}$, $\binom{3}{2}$, $\binom{3}{2}$, $\binom{3}{2}$, $(d,p_{2})=(5,0)$, 3. (c)

$\binom{7}{2}$, $\binom{4}{3}$, $\binom{4}{3}$, $\binom{3}{2}$, $(d,p_{2})=(6,1)$, 4. (d)

$\binom{5}{3}$, $\binom{7}{2}$, $\binom{5}{2}$, $\binom{3}{2}$, $(d,p_{2})=(6,1)$, 5. (e)

$\binom{8}{3}$, $\binom{7}{3}$, $\binom{3}{2}$, $\binom{3}{2}$, $(d,p_{2})=(7,1)$, 6. (f)

$\binom{9}{2}$, $\binom{5}{3}$, $\binom{3}{2}$, $\binom{3}{2}$, $(d,p_{2})=(6,1)$, 7. (g)

$\binom{11}{2}$, $\binom{4}{3}$, $\binom{3}{2}$, $\binom{3}{2}$, $(d,p_{2})=(6,1)$, 8. (h)

$\binom{15}{2}$, $\binom{3}{2}$, $\binom{3}{2}$, $\binom{3}{2}$, $(d,p_{2})=(6,1)$, 9. (i)

$\binom{4}{3}$, $\binom{3}{2}$, $\binom{3}{2}$, $\binom{3}{2}$, $(d,p_{2})=(5,0)$, 10. (j)

$\binom{4}{3}$, $\binom{4}{3}$, $\binom{4}{3}$, $\binom{3}{2}$, $(d,p_{2})=(6,1)$, 11. (k)

$\binom{5}{3}$, $\binom{5}{2}$, $\binom{4}{3}$, $\binom{3}{2}$, $(d,p_{2})=(6,1)$, 12. (l)

$\binom{5}{3}$, $\binom{5}{3}$, $\binom{3}{2}$, $\binom{3}{2}$, $(d,p_{2})=(6,1)$.

Let $(\mu_{j},\mu_{j}^{\prime},\ldots)$ denote the multiplicity sequence of $q_{j}\in\bar{E}$. By the Hurwitz formula applied to the $\mathbb{P}^{1}$-fibration of the blow-up of $\mathbb{P}^{2}$ at $q_{1}$ we have $\sum_{j=2}^{4}(\mu(q_{j})-1)+(\mu_{1}^{\prime}-1)\leqslant 2(d-\mu_{1}-1)$. This inequality fails in cases (i), (j), (k), (l).

In case (h) all cusps are semi-ordinary, so the Matsuoka-Sakai inequality [MS89], $d<3\max(\mu_{1},\mu_{2},\mu_{3},\mu_{4})$, fails.

In cases (c), (d), (e) and (f) we find a $(-1)$-curve $A\nsubseteq D$ on $X$ such that $A\cdot D=2$. Assume that such a curve exists. Since $X\setminus D$ contains no affine line, $A$ meets $D$ in two different points and $X\setminus(D+A)$ contains no affine line, so possibly after some number of inner contractions in $D+A$ the surface $(X,D+A)$ becomes almost minimal (see [Fuj82, 6.20, 6.24], [Miy01, §2.3.11]). Since the contractions do not affect maximal twigs of $D+A$ and the self-intersection of the log canonical divisor, the log BMY inequality gives $(K+D+A)^{2}+\mathrm{ind}(D+A)\leqslant 3e_{top}(X\setminus\leavevmode\nobreak\ (D+A))$. We have $e_{top}(X\setminus(D+A))=e_{top}(X\setminus D)=1$ and $(K+D+A)^{2}=(K+D)^{2}+1=p_{2}-1$. Since in the cases considered $p_{2}=1$, we get

[TABLE]

Let $L_{i,j}\subseteq X$ for $i\neq j$ denote the proper transform of the line on $\mathbb{P}^{2}$ joining the cusps $q_{i}$ and $q_{j}$, and let $L_{i}\subseteq X$ denote the proper transform of the line tangent to $q_{i}\in\bar{E}$.

In case (c) we take $A=L_{2,3}$. Since $\mu_{2}+\mu_{3}=6=d$, $A$ meets the $(-4)$-tips of $Q_{2}^{\prime}$ and $Q_{3}^{\prime}$ and it is a $(-1)$-curve with $A\cdot D=2$. We compute $\mathrm{ind}(D+A)=(\frac{5}{7}+\frac{1}{2})+\frac{2}{3}+\frac{2}{3}+\frac{5}{6}=\frac{71}{21}>3$; a contradiction.

In case (d) we take $A=L_{1}$. We have $(\pi(A)\cdot\bar{E})_{q_{1}}\geqslant\mu_{1}+\mu_{1}^{\prime}=5=d-1$, so since $X\setminus D$ contains no affine line, the equality holds, so $A$ is as needed. We compute $\mathrm{ind}(D+A)=\frac{2}{5}+(\frac{5}{7}+\frac{1}{2})+(\frac{3}{5}+\frac{1}{2})+\frac{5}{6}=3\frac{23}{42}>3$; a contradiction.

In case (e) we take $A=L_{1}$. We have $(\pi(A)\cdot\bar{E})_{q_{1}}\geqslant\mu_{1}+\mu_{1}^{\prime}=6=d-1$, so again equality holds and $A$ is as required. We compute $\mathrm{ind}(D+A)=(\frac{1}{2}+\frac{1}{3})+(\frac{4}{7}+\frac{2}{3})+\frac{5}{6}+\frac{5}{6}=3\frac{31}{42}>3$; a contradiction.

In case (f) we take $A=L_{2}$. Similarly, $(\pi(A)\cdot\bar{E})_{q_{2}}=d-1$, so $A$ is as required. We have $\mathrm{ind}(D+A)=(\frac{7}{9}+\frac{1}{2})+\frac{2}{5}+\frac{5}{6}+\frac{5}{6}=3\frac{31}{90}>3$; a contradiction.

Consider case (a). We note as above that $(\pi(L_{j})\cdot\bar{E})_{q_{j}}=4$ and $L_{1,2}\cdot Q_{j}^{\prime}=1$ for $j=1,2$. Blow up twice over each $q_{j}$ for $j=1,2$ and denote the first exceptional curve by $U_{j}$. Let $\ell_{j}$ be the proper transform of $L_{j}$ and let $\ell_{1,2}$ be the proper transform of $L_{1,2}$. The divisors $\ell_{1}+\ell_{2}$ and $U_{1}+2\ell_{1,2}+U_{2}$ are complete fibers of the same $\mathbb{P}^{1}$-fibration. The proper transform of $\bar{E}$ is a $2$-section and has a unique point of intersection with $\ell_{1,2}$. The latter is a third, besides $q_{3}$ and $q_{4}$, ramification point of the projection of the $2$-section onto the base of the fibration, which contradicts the Hurwitz formula.

Consider case (g). We have $\rho(X)=\#D=18$, so $K^{2}=-8$ and then $K\cdot D=p_{2}-K^{2}=9$ and $K\cdot E=9-K\cdot Q_{2}^{\prime}=8$. Put $B=D-C_{1}^{\prime}-C_{2}^{\prime}$, where $C_{j}^{\prime}$ is the unique $(-1)$-curve in $Q_{j}^{\prime}$, and denote by $U$ the $(-4)$-curve which is a tip of $Q_{2}^{\prime}$. We analyze the surface $(X,B)$. First of all we note that on $X_{0}$ the divisor $2K_{0}+\varphi_{0}(U)+E_{0}$ intersects trivially all components of $D_{0}$, hence is numerically trivial, because the components of $D_{0}$ generate $\operatorname{NS}(X_{0})\otimes\mathbb{Q}$. This gives a numerical equivalence

[TABLE]

where $G_{j}\subseteq Q_{j}^{\prime}$ for $j\in\{1,3,4\}$ is the $(-2)$-curve meeting $C_{j}^{\prime}$. It follows that $\kappa(K_{X}+B)\geqslant 0$. We claim that $(X,B)$ is almost minimal. Suppose otherwise. By [Fuj82, 6.20] there exists a $(-1)$-curve $L\nsubseteq B$ meeting at most two connected components of $B$ (hence $L\nsubseteq D$), each at most once. Intersecting (5.18) with $L$ we have $L\cdot(2K+U+E)\geqslant 0$, hence $L\cdot(U+E)\geqslant 2$. It follows that $L\cdot U=L\cdot E=1$ and $L\cdot(D-C_{1}^{\prime}-C_{2}^{\prime}-U-E)=0$. Let $(X^{\prime},B^{\prime})$ be the image of $(X,B)$ after the contraction of $L$. Repeating the argument we see that $(X^{\prime},B^{\prime})$ is almost minimal and $\kappa(K_{X^{\prime}}+B^{\prime})\geqslant 0$. Then $B^{\prime}$ has $4$ connected components, one of which has $5$ maximal twigs: $[3]$, $[2]$, $[3]$, $[2]$, $[3]$. But $e_{top}(X^{\prime}\setminus B^{\prime})=e_{top}(X\setminus D)-2=-1$, so the log BMY inequality ([Miy01, 2.6.6.2]) gives $0\leqslant-1+\frac{1}{c_{1}^{(1)}}+\frac{1}{p_{1}^{(1)}}+\frac{1}{p_{2}^{(1)}}=-1+\frac{1}{11}+\frac{1}{2}+\frac{1}{3}$; a contradiction. Thus $(X,B)$ is almost minimal. We have $e_{top}(X\setminus B)=-1$, $K\cdot(K+B)=K\cdot(K+D)+2=p_{2}+2=3$ and $B\cdot(K+B)=-2b_{0}(B)=-10$. For an admissible chain $T$ which is a connected component of $B$ we can consider two orders $T=T_{1}+\ldots+T_{m}$ and $T^{t}=T_{m}+\ldots+T_{1}$ as in Section 2 and we put $\operatorname{Bk}T=\operatorname{Bk}^{\prime}T+\operatorname{Bk}^{\prime}T^{t}$, cf. [Miy01, §2.3.3.3], which gives

[TABLE]

We compute $\mathrm{ind}(Q_{1}^{\prime}-C_{1}^{\prime})=\frac{9+5+2}{11}+\frac{1+1+2}{2}=3\frac{5}{11}$, $\mathrm{ind}(Q_{2}^{\prime}-C_{1}^{\prime})=\frac{1+1+2}{4}+\frac{2+2+2}{3}=3$ and $\mathrm{ind}(E+Q_{3}^{\prime}+Q_{4}^{\prime})=2\cdot\frac{5}{6}=\frac{5}{3}$, hence $\mathrm{ind}(B)=8\frac{4}{33}$ and $(K+B-\operatorname{Bk}B)^{2}=(K+B)^{2}+\mathrm{ind}(B)=1\frac{4}{33}$. The log BMY inequality for $(X,B)$ gives $1\frac{4}{33}\leqslant 3\cdot(-1+\frac{1}{11}+\frac{1}{2}+\frac{1}{4}+\frac{1}{3})=\frac{23}{44}$; a contradiction.

Thus, we have shown that the only possibility is case (b). This is indeed the type of the quintic from Theorem 1.1. We note that $L_{1}$ is a $(-1)$-curve meeting $Q_{1}=[1,2,2]$ in the middle component. The contraction of $L_{1}+(Q_{1}-C_{1})+Q_{2}+Q_{3}+Q_{4}$ transforms $E_{0}$, and hence $\bar{E}$, onto a tricuspidal quartic with all cusps ordinary and with the image of $C_{1}$ as a bitangent line. Using this transformation one proves the projective uniqueness of $\bar{E}$ of type (b), see [PP20, Proposition 4.9]. The uniqueness follows also from the classification of rational cuspidal quintics in [Nam84].

∎