Proof. Suppose, first, that f is étale of degree d, then $\kappa (X)=d\cdot \kappa (Y)=0$ , and hence we can assume that f is ramified.

Being f a triple cover of a surface Y with trivial canonical bundle by (2.6), we have $h^0({\mathcal {O}}(K_X))\geq 1$ , and hence the Kodaira dimension of X satisfies $\kappa (X) \geq 0$ . Moreover, if $\kappa (X)=0$ , then X cannot be neither an Enriques surface nor a bielliptic one.

By the Hurwitz formula (2.5), the canonical divisor is $K_X =R$ . If the divisor R is supported at least on a moving positive genus curve, then (2.5) implies that $K_X$ is nontrivial, and thus $\kappa (X)>0$ . Moreover, X is a properly elliptic surface if and only if its canonical bundle is supported only on elliptic curves (see [Reference MirandaMi2, §III]), and thus, again by (2.5), we have the second case of the proposition, whereas if the general member of $|R|$ is supported on a curve of genus $g\geq 2$ , then X is of general type.

Finally, if R does not move in a linear system and it is supported only on rational curves, then the minimal model of X must have trivial canonical bundle and so we get the case (1).

Conversely, if $|M|=\emptyset $ , $\kappa (X)$ cannot be 1 or 2, and hence $\kappa (X)=0$ . If $|M| \neq \emptyset $ and the general member of $|M|$ is supported on curves of genus $g \geq 2$ , $\kappa (X)$ cannot be 0 or 1, and hence it is 2.

If $|M| \neq \emptyset $ and the general member of $|M|$ is supported on elliptic curves, by adjunction, $M\cdot M=0$ and so X would have a genus 1 fibration, which is impossible for surfaces of general type. Since $|M| \neq \emptyset $ , $\kappa (X)\neq 0$ .

Proof. If X is normal, we consider the canonical resolution $\widetilde {X}$ , as in Proposition 2.5, since $\kappa (X)=\kappa (\widetilde {X})$ . The map $\sigma :\widetilde {Y} \rightarrow Y$ is a blowup and introduces an exceptional divisor E, which does not move in a linear system. The canonical bundle of $\widetilde {Y}$ is $K_{\widetilde {Y}}=K_Y+E=E$ . We apply the Hurwitz ramification formula to the triple cover $\widetilde {f}\colon \widetilde {X} \rightarrow \widetilde {Y}$ :

$$ \begin{align*}K_{\widetilde{X}}=\widetilde{f}^*E+R_{\widetilde{f}},\end{align*} $$

where $R_{\widetilde {f}}$ is the ramification divisor of $\widetilde {f}$ .

Let analyze both the summands: $\widetilde {f}^*E$ has negative self-intersection and is the exceptional locus of $\widetilde {\sigma }$ for the commutativity of the diagram in particular $\widetilde{\sigma}^*(B) \widetilde{f}^*E=0$ for all $B \in Pic(X)$ . Hence, its movable part is trivial; otherwise, there would be at least one curve in $|\widetilde {f}^*E|$ which is not contracted by $\widetilde {\sigma }$ , say $\widetilde {D}$ , so $\widetilde{D}\widetilde{f}^*E>0 $ . Denoted by D the image of this curve, $\widetilde{\sigma}^*(D)\sim \widetilde {D}+k\widetilde {f}^*E \sim (k+1)\widetilde {f}^*E$ , for a nonnegative k. However, this contradicts $\sigma ^*(D)\widetilde {f}^*E=0$ .

We observe that $R_{\widetilde {f}}$ are curves contained in the linear system of

$$ \begin{align*}\widetilde{\sigma}^*R=\widetilde{\sigma}^*(M+F)=\widetilde{M}+\widetilde{F}+k\widetilde{f}^*E,\end{align*} $$

where $\widetilde {M}$ and $\widetilde {F}$ are the strict transforms of M and F, respectively, using the notation of Proposition 2.8. Since $\widetilde {f}^*E$ has no movable part, the movable part of $R_{\widetilde {f}}$ is contained in $\widetilde {M}$ . Since $\sigma $ is a blowup, $|M|$ is isomorphic to $|\widetilde {M}|$ .

To conclude, it suffices to apply Proposition 2.8 to the cover $\widetilde {f}$ .

Proof. $(i)$ and $(ii)$ are contained in [Reference MirandaMi1, Prop. 7.1]. Recall that, by [Reference TanT, Th. 2.1], a triple cover is Galois if and only if it is totally ramified over its branch locus. Moreover, in this case, the branch locus is exactly $D'=D/2=B+C$ [Reference TanT, Th. 1.3(2)].

Proof. Since $B_i$ (resp., $C_i$ ) are contained in a connected component of the branch locus and $B_j$ (resp., $C_j$ ) in another one, we have $B_iB_j=C_iC_j=B_iC_j=0$ , if $i\neq j$ . Moreover, the last part of the statement follows directly by the conditions (not all independent): $B_iL=B_i(2B+C)/3=(2B_i^2+B_iC_i)/3\in {\mathbb {Z}}$ . Analogously, $C_iL=C_i(2B+C)/3\in {\mathbb {Z}}$ , $L^2\in 2{\mathbb {Z}}$ , $B_iM=B_i(B+2C)/3\in {\mathbb {Z}}$ , $C_iM=C_i(B+2C)/3\in {\mathbb {Z}}$ , and $M^2\in 2{\mathbb {Z}}$ .

Proof. The triple cover $f:X \rightarrow S$ is smooth if and only if the branch locus is smooth (see Paragraph 2.10), and thus each $D_i$ is a smooth irreducible curve. In particular, either $D_i=B_i$ or $D_i=C_i$ . In both the cases, one obtains $D_i^2\equiv _3 0$ , by Lemma 3.1. If $D_i$ is a smooth irreducible rational curve, $D_i^2=-2\not \equiv _3 0$ , which is not admissible. So, for every $D_i$ , we have $g(D_i)\geq 0$ . If $D_1^2>0$ , then, by the Hodge index theorem, $D_i^2<0$ for each $i>1$ , which implies that $D_i$ are rational curves for each $i>1$ , contradicting the first assertion. Therefore, if $D_1^2>0$ , there are no other components in the branch locus. This implies that $B=D_1$ and $C=0$ (or vice versa). So $L=B/3\in Pic(S)$ , that is, there exists a divisor H such that $3H\equiv D$ .

If $D_1^2=0$ and it is a smooth irreducible curve, then it is a genus 1 curve on S and $\varphi _{|D_1|}:S \rightarrow \mathbb {P}^1$ is a genus 1 fibration. In particular, any $D_j$ orthogonal to $D_i$ is contained in a fiber of $\varphi _{|D_1|}$ , and thus it can be either a rational curve (component of a reducible fiber) or a genus 1 curve. Since there are no rational curves contained in the branch locus, we conclude that all the $D_i$ ’s are smooth fibers of the same fibration and that the triple cover $X \rightarrow S$ is branched over smooth fibers of the fibration $S \rightarrow \mathbb {P}^1$ . This induces a genus 1 fibration $X \rightarrow C$ , where C is a smooth curve, such that there is a $3:1$ map $g:C \rightarrow \mathbb {P}^1$ and $X\simeq S\times _g\mathbb {P}^1$ .

Proof. Since $N_1$ and $N_2$ have trivial intersection with any other components of the branch locus, the condition $LN_i\in {\mathbb {Z}}$ restricts to the condition $N_i(\alpha N_1+\beta N_2)/3\in {\mathbb {Z}}$ , where $\alpha \in \{1,2\}$ and $\beta \in \{1,2\}$ and $(\alpha N_1+\beta N_2)/3$ is the summand of L in which $N_1$ and $N_2$ appear with a nontrivial coefficient. Up to replace possibly L with M, we can assume that $\alpha =1$ . So the condition is now $N_i(N_1+\beta N_2)/3\in {\mathbb {Z}}$ , which implies that $-2+\beta k\equiv _3 0$ and $k-2\beta \equiv _3 0$ . These conditions imply that $k\not \equiv _3 0$ , $k\equiv _3 1$ if an only if $\beta =2$ , and $k\equiv _3 2$ if and only if $\beta =1$ .

Proof. Condition $(1)$ is needed to obtain a normal cover $X \rightarrow S$ (see Paragraph 2.10).

Condition $(2)$ follows by the Hodge index theorem.

Since the singularities are negligible, the numerical invariants for $X'$ can be calculated using the Tschirnhausen bundle ${\mathscr {E}}=\mathcal {L}^{-1} \oplus \mathcal {M}^{-1}=\mathcal {O}_S(-L) \oplus \mathcal {O}_S(-M)$ . It has the following Chern classes:

$$\begin{align*}c_1({\mathscr{E}})=-L-M, \quad c_2({\mathscr{E}})=L\cdot M. \end{align*}$$

To have a connected covering, we have

$$\begin{align*}0=h^0({\mathscr{E}})=h^0(-L)+h^0(-M) \, \Rightarrow \, h^0(-L)=h^0(-M)=0. \end{align*}$$

In addition, by the Hirzebruch–Riemann–Roch theorem, we have

(3.1) $$ \begin{align} \chi({\mathscr{E}})=h^2({\mathscr{E}})-h^1({\mathscr{E}})=h^0(L)+h^0(M)-h^1({\mathscr{E}})&=\nonumber\\ =\int_S \textrm{Td}(S) \cdot \textrm{ch}({\mathscr{E}}) = \int_S(1,0,2)\cdot\left(2,-L-M,\frac{L^2+M^2}{2}\right)&=\frac{L^2+M^2}{2}+4. \end{align} $$

So, by Proposition 2.6(ii), we have $\chi (\mathcal {O}_{X'})=6+\frac {1}{2}(L^2+M^2)$ , by Proposition 2.6(iii), we have $K^2_{X'}=2L^2+2M^2+LM, $ and finally, by Proposition 2.6(iv), we have $e(X')=72+4(L^2+M^2)-LM.$

By (3.1) and the genus formula,

(3.2) $$ \begin{align} &h^1({\mathscr{E}})=h^0(L)+h^0(M)-\chi({\mathscr{E}})=\left(\chi(L)+h^1(L)\right)+\left(\chi(M)+h^1(M)\right)-\chi({\mathscr{E}})= \nonumber\\ &= \left(\frac{L^2}{2}+2+h^1(L)\right)+\left(\frac{M^2}{2}+2+h^1(M)\right)-\frac{L^2+M^2}{2}-4=h^1(L)+h^1(M), \end{align} $$

which allows to compute $h^{1,0}(X')$ , by Proposition 2.6(i).

Proof. Let us call $C_i$ the irreducible components of C. Then $C=\sum _{i=1}^n C_i$ and

$$\begin{align*}(C)^2=-2n+2\sum_{i=1}^n\sum_{j=i+1}^{n}C_iC_j. \end{align*}$$

The curve C is connected, so each curve $C_i$ intersects at least one other curve, and one has $\sum _{i=1}^n\sum _{j=i+1}^{n}C_iC_j\geq n-1$ . Hence, $C^2\geq -2n+2(n-1)=-2$ . If $C^2<0$ , then $C^2=-2$ .

Proof. Since $\Lambda $ is degenerate, there exists $v\in \Lambda $ such that $v^2=0$ and $vw=0$ for every $w\in \Lambda $ . By the Riemann–Roch theorem, either v or $-v$ is effective, so we assume that v is effective and hence $\varphi _{|v|}$ is a fibration whose smooth fibers are genus 1 curve (cf., e.g., [Reference KondoK, Proof of Lemma 2.1]). Since any other class w in $\Lambda $ has a trivial intersection with v, if a curve is represented by w, it is contracted by $\varphi _{|v|}$ and hence it is contained in a fiber of the fibration $\varphi _{|v|}$ .

Proof. $(1)$ If $\kappa (X)=0$ , each component of $D_i$ is a negative curve, and thus by adjunction a $(-2)$ -curve, since the intersection with the canonical bundle is trivial for every curve on X (if X is a K3 surface; see also Lemma 3.14). Therefore, the lattice $\Lambda _{D_i}$ is a negative definite root lattice, and thus $\Lambda _{D_i}(-1)$ is an A–D–E lattice [Reference Barth, Hulek, Peters and Van de VenBHPV, Lem. 2.12, Chap. I]. By [Reference Barth, Hulek, Peters and Van de VenBHPV, Chap. I, §17], the existence of a triple cover branched in $D_i$ implies that a linear combination of the components of the $D_i$ ’s with noninteger coefficients in $\frac {1}{3}{\mathbb {Z}}$ is contained in the Picard group of S. So the discriminant group of $\Lambda _{D_i}$ contains ${\mathbb {Z}}/3{\mathbb {Z}}$ , which implies that either $\Lambda _{D_i}\simeq A_{3k-1}(-1)$ with $k>1$ or $\Lambda _{D_i}\simeq E_6(-1)$ . In both the cases, $\Lambda _{D_i}^{\vee }/\Lambda _{D_i}$ is a cyclic group, generated by a class l (supported on the components of $D_i$ ). The generators of the subgroups ${\mathbb {Z}}/3{\mathbb {Z}}$ in $\Lambda _{D_i}^{\vee }/\Lambda _{D_i}$ are multiple of l, and they are always supported on a disjoint $A_2$ -configuration of curves. So each $D_i$ is the sum $B_i+C_i$ , where $B_i$ and $C_i$ are two rational curves meeting in a point, that is, $\langle B_i,C_i\rangle \simeq A_2(-1)$ .

$(2)$ If $\kappa (X)=2$ , then there exists a curve $C\in |R|$ , where R is the ramification of f such that $g(C)>1$ (see Proposition 2.9) and $C^2\geq 0$ . By the projection formula, $\left (f(C)\right )^2\geq 0$ and so $g(f(C))\geq 1$ . There exists a (possibly reducible) branch curve $\Gamma $ linearly equivalent to $f(C)$ and contained in $D_i$ . If $g(f(C))>1$ , then $\Gamma ^2>0$ and hence $\Lambda _{D_i}$ satisfies the hypothesis, by the Hodge index theorem.

If $g(f(C))=1$ , $f(C)$ defines a genus 1 fibration $\mathcal {E}$ on S. Moreover, $f(C)$ intersects the branch locus, that is, $f(C)\cap (\cup _iD_i)\neq \emptyset $ ; otherwise, $g(C)=1$ . This implies that $\Gamma $ in the union of a fiber of $\mathcal {E}$ and some horizontal curves. In particular, $\Gamma $ , and hence $D_i$ , contains the class F of the fiber of the fibration $\mathcal {E}$ and at least a horizontal curve $\Delta $ . The class $\alpha F+\delta \Delta $ is contained in $\Lambda _{D_i}$ , and it has a positive self-intersection for $\alpha $ sufficiently big. By the Hodge index theorem, the signatures of all the other lattices $\Lambda _{D_i}$ are the required ones.

$(3)$ If $\kappa (X)=1$ , then there exists a curve $C\in |R|$ , where R is the ramification of f such that $g(C)=1$ and C defines a genus 1 fibration on X. Hence, $f(C)$ defines a genus 1 fibration on S. All the irreducible components of $|f(C)|$ are genus 1 curve and $f(C)^2=0$ . As above, there exists a (possibly reducible) branch curve $\Gamma $ linearly equivalent to $f(C)$ and contained in $D_i$ . Then $\Gamma \in |f(C)|$ and it is a fiber (irreducible or not), so it coincides with $D_i$ for a certain i and $\Lambda _{D_i}$ is degenerate. All the other $D_j$ ’s are contained in (or coincides with) fibers, so each $\Lambda _{D_j}$ cannot contain a class with positive self-intersection.

Proof. The assumption that the curve $D_i$ ’s are reduced guarantees that X is normal. $(1)$ If all the $\Lambda _{D_i}$ are negative definite, we already showed in Theorem 3.16 that $\Lambda _{D_i}\simeq A_2(-1)$ . The sets of $A_2$ -configurations which are 3-divisible (and thus are in the branch locus of a triple cover) necessarily contain either six or nine $A_2$ -configurations [Reference BarthBa, Lem. 1]. In the first case, the minimal model of the triple cover is a K3 surface, and in the latter case, it is an abelian surface and in particular $\kappa (X)=0$ .

$(2)$ If sgn $(\Lambda _{D_1})=(1,\mathrm {rank}(\Lambda _{D_1})-1)$ , by the Hodge index theorem, the lattices $\Lambda _{D_j}$ with $j\neq 1$ are negative definite. As in proof of $(1)$ of Theorem 3.16, this implies that $\Lambda _{D_j}$ coincides with $A_2(-1)$ . The maximal number of disjoint $A_2$ -configurations of rational curves on a K3 surface is 9, and hence $n\leq 10$ . To show that $\kappa (X)=2$ , we observe that if $\kappa (X)=0$ , then by Theorem 3.16, all $\Lambda _{D_i}$ should be negative definite, which contradicts the hypothesis on $\Lambda _{D_1}$ . If $\kappa (X)=1$ , then by Theorem 3.16, at least one $\Lambda _{D_i}$ should be degenerate, which is again impossible.

$(3)$ If $\Lambda _{D_1}$ is degenerate, then $D_1$ consists of a fiber of an elliptic fibration. So $D_j$ , $j\neq 1$ , are contained in fibers and then $\Lambda _{D_j}$ cannot contain a positive class. In particular, $\kappa (X)$ cannot be 2, because there is no indefinite lattice $\Lambda _{D_i}$ and cannot be 0, because not all the lattices $\Lambda _{D_i}$ are negative definite.

Proof. The first part of the statement follows directly by Corollary 3.17.

Let $D_1$ be a smooth irreducible curve, then the singularities of the branch locus are of type 2 and they are the singular points of each $A_2$ -configuration in the branch locus.

In this case, $X^{\circ }$ is obtained by contracting on $X'$ three curves for each $A_2$ -configuration in the branch locus. Indeed, by Remark 3.11, one has to contract at least three curves for each $A_2$ -configuration, and we call the surface obtained by these contractions $X_{m}$ . It remains to prove that $X_{m}$ and $X^{\circ }$ coincide, and so that there are no other possible contractions to a smooth surface.

On $X_{m}$ , there is an automorphism $\sigma _{m}$ of order 3, induced by the Galois ${\mathbb {Z}}/3{\mathbb {Z}}$ -cover automorphism $\sigma $ on X. The surface $S_{m}:=X_{m}/\sigma _{m}$ is the singular surface obtained by S contracting all the $A_2$ -configurations in the branch locus. If there were a $(-1)$ -curve E on $X_{m}$ , then there are two cases: either E is disjoint from the ramification locus of $f_{m}:X_{m} \rightarrow S_{m}$ , or E meets it (not being contained).

If E is disjoint from the ramification locus, then there exists a $(-1)$ -curve, still denoted by E, on X which is mapped to E (because we blow up and down points away from E). Hence, $\sigma (E)\cap E$ is empty, and therefore $\sigma (E)$ and $\sigma ^2(E)$ are other two $(-1)$ -curves on X. So $f(E)=f(\sigma (E))=f(\sigma ^2(E))\subset S$ is a $(-1)$ -curve. This is absurd because S is a K3 surface.

Hence, E meets the ramification locus $R_{m}$ and $\sigma _{m}(E)=E$ . Otherwise, there should be different $(-1)$ -curves (E and $\sigma _{m}(E)$ ) meeting in the point $E\cap R_{m}$ and this is impossible on a surface with nonnegative Kodaira dimension. Moreover, $\sigma $ is an automorphism of order 3 of the rational curve E, and then $E\cap R_{m}$ consists of two points.

Let $\beta :X_{m} \rightarrow X^{\circ }$ be the contraction of E, and let $\sigma ^{\circ }$ be the automorphism induced by $\sigma _{m}$ on $X^{\circ }$ . This induces the contraction $S_{m} \rightarrow S^{\circ }$ of the curve $f_{m}(E)$ . Since $E\cap R_{m}$ consists of two points, the contraction of $f_{m}(E)$ identifies two singular points on $S_{m}$ and introduces a singularity on the image of $D_1$ . By construction, the smooth $X^{\circ }$ is a triple cover of $S^{\circ }$ and thus the singularities of $S^{\circ }$ cannot be worse than $(\mathbb {C}^2,0)/{\mathbb {Z}}/3{\mathbb {Z}}$ , so two singularities of $S_{m}$ cannot be identified. Moreover, there cannot be a singularity on the branch curve image of $D_1$ , and $S^{\circ }$ has to coincide with $S_{m}$ . We conclude that $X_{m}$ coincides with $X^{\circ }$ .

Proof. By Corollary 3.17, the components $D_j$ with $j>1$ consist of two rational curves meeting in a point. We denote by $A_1^{j}$ and $A_2^{j}$ the two components of $D_j$ . Since $D_1$ is irreducible, it is a component of B (or equivalently of C). Then the data of the triple cover are

$$ \begin{align*}B=D_1+\sum_{j=1}^{n-1}A_1^{j},\ \ C=\sum_{j=1}^{n-1}A_2^{j},\end{align*} $$

$$ \begin{align*}L=\frac{D_1+\sum_{j=1}^{n-1}\left(A_1^{j}+2A_2^{j}\right)}{3},\ \ M=\frac{2D_1+\sum_{j=1}^{n-1}\left(2A_1^{j}+A_2^{j}\right)}{3}.\end{align*} $$

By $LD_1\in {\mathbb {Z}}$ and $L^2\in 2{\mathbb {Z}}$ , it follows that $D_1^2\equiv _3 0$ and $d\equiv _3 n-1$ . Since $n\leq 10$ and $d>0$ , $9+3k>0$ , so $k\geq -2$ . The formula (4.1) follows by Proposition 3.12 since $L^2=2k$ , $M^2=2n+8k-2$ , and $LM=n-1+4k$ , and $X^{\circ }$ is obtained by X contracting $3(n-1)$ curves. We conclude by Corollary 3.18.

Proof. Let S be a K3 surface with an elliptic fibration such that the reducible fibers are $I_2+6I_3$ and the Mordell–Weil group is ${\mathbb {Z}}/3{\mathbb {Z}}$ . The existence of a K3 surface with such an elliptic fibration is guaranteed by [Reference ShimadaSh, Table 1, case 835]. We denote by F the class of the fiber of this fibration, by $\mathcal {O}$ the class of the zero section, by $A_2^{(1)}$ the class of the irreducible component of the fiber $I_2$ which meet the section $\mathcal {O}$ , and by $A_h^{(j)}$ , $h=1,2$ , $j=2,\ldots , 7$ , the classes of the two irreducible components not meeting the zero section of the jth reducible fiber, which is a fiber of type $I_3$ . We observe that the class of the 3-torsion section P, which generates the Mordell–Weil group, can be written in terms of the previous curves as

$$ \begin{align*}P=2F+\mathcal{O}-\frac{1}{3}\left(\sum_{j=2}^7A_1^{(j)}+2A_2^{(j)}\right).\end{align*} $$

Moreover, we observe that there are seven disjoint $A_2$ -configurations on this surface, which are given by $\mathcal {O}$ , $A_2^{(1)}$ , $A_h^{(j)}$ , $h=1,2$ , $j=2,\ldots ,7$ . Let us consider the divisor $3F+A_2^{(1)}+2\mathcal {O}=D_1$ , and we notice that $D_1^2=6$ , $D_1F=2$ , $D_1A_2^{(1)}=0$ , and $D_1\mathcal {O}=0$ . One can check that $D_1$ is a big and nef divisor, and hence in its linear system, there is a smooth irreducible curve of genus 4, still denoted by $D_1$ . We claim that there exists a triple cover of S branched over $D_1$ and the seven $A_2$ -configurations $\mathcal {O}$ , $A_2^{(1)}$ , $A_h^{(j)}$ , $h=1,2$ , $j=2,\ldots ,7$ . Indeed, the divisor

$$ \begin{align*}L:=\left(D_1+\mathcal{O}+\sum_{i=2}^7 A_1^{(i)}+2\sum_{j=1}^7 A_2^{(j)}\right)/3=3F+A_2^{(1)}+2\mathcal{O}-P\end{align*} $$

is contained in $NS(S)$ and so

$$ \begin{align*}B:=D_1+\mathcal{O}+\sum_{i=2}^7 A_1^{(i)},\ \ C:=\sum_{j=1}^7 A_2^{(j)}, L=(B+2C)/3,\ \ M=(2B+C)/3\end{align*} $$

form triple cover data on S. So there exists a triple cover $X \rightarrow S$ which satisfies the condition of Proposition 4.1 with $k=-2$ , $n=8$ , and $d=1$ , and then we deduced the properties of $X^o$ .

Proof. We already observed that $D_1$ intersects negatively at least one of its components, say $P_j$ . Hence, $P_j$ is a fixed component of a non-nef divisor and $P_j^2=-2$ . So

$$ \begin{align*}D_1P_j=rFP_j+\sum_{i\neq j}P_iP_j-2<0.\end{align*} $$

So $rFP_j+\sum _{i\neq j}P_iP_j\leq 1$ . We observe that $P_j$ is horizontal, so $FP_j>0$ , and $P_jP_i\geq 0$ because $P_i$ are irreducible effective curves. Hence, the only possibility is $r=1$ , $FP_j=1$ , and $P_iP_j=0$ .

Proof. Since $L=(B+2C)/3$ and $LP_i\in {\mathbb {Z}}$ , it follows that if F is a component of B, the $P_i$ ’s must be components of C. The divisor

$$ \begin{align*}L=\frac{F+2\sum_{j=1}^kP_j+\sum_{h=1}^{n-1}\left(A_1^{(h)}+2A_2^{(h)}\right)}{3}\end{align*} $$

has to be contained in $NS(S)$ , $LF\in {\mathbb {Z}}$ , $LP_i\in {\mathbb {Z}}$ , for every $i=1,\ldots , h$ , and $L^2\in 2{\mathbb {Z}}$ . These conditions imply

$$ \begin{align*}\frac{2k}{3}\in{\mathbb{Z}},\ \ -1\in{\mathbb{Z}},\mbox{ and }\frac{-4k-6(n-1)}{9}=2\frac{-2k-3n+3}{9}\in 2{\mathbb{Z}}.\end{align*} $$

Recall that

$$ \begin{align*}M=\frac{2F+\sum_{j=1}^kP_j+\sum_{h=1}^{n-1}\left(2A_1^{(h)}+A_2^{(h)}\right)}{3},\end{align*} $$

so

$$ \begin{align*}L^2=\frac{6-4k-6n}{9},\ M^2=\frac{2k+6-6n}{9},\mbox{ and }LM=\frac{k+3-3n}{9},\end{align*} $$

which gives $\chi (X)$ and $K_X^2$ . Since the singularities of X are negligible, these are the invariants of the minimal resolution of the cover. To obtain a minimal model, one has to contract the $(-1)$ -curves. Each $A_2$ -configuration produces three $(-1)$ -curves in the minimal resolution of triple cover, and each curve $P_j$ produces one $(-1)$ -curve, by Remark 3.11. So one contracts at least $3(n-1)+k$ curves to obtain the minimal model from the minimal resolution and so $K_{X^{\circ }}\geq K_X^2+3(n-1)+k=2k/3$ .

As in the proof of Corollary 3.18, if there were other $(-1)$ -curves, they have to intersect the ramification locus, and we already excluded the cases coming from rational curves intersecting the components $D_j, j\geq 2$ . We consider the canonical resolution of the triple cover $\widetilde {f}:\tilde {X} \rightarrow \tilde {S}$ as in the diagram (2.3). Thanks to the particular configuration of curves in $D_1$ , one checks that the only $(-1)$ -curves on $\tilde {X}$ mapped to $\sigma ^{-1}(D_1)$ are the strict transforms of the triple cover of the curves $P_i$ . After their contraction, there are no other $(-1)$ -curves in the strict transform of $\sigma ^{-1}(D_1)$ .

Consider a rational curve $C\subset S$ with $CD_1>0$ and observe that C is not mapped to $\sigma ^{-1}(D_1)$ . We denote by $\tilde {C}$ the strict transform of C. If $\tilde {f}^{-1}(\tilde {C})$ is an irreducible rational curve, it intersects the ramification locus in at most two points. Moreover, $(\tilde {f}^{-1}(\tilde {C}))^2=3\tilde {C}^2\leq -6$ . Since we can contract at most two curves meeting $\tilde {f}^{-1}(\tilde {C})$ , we cannot obtain a $(-1)$ -curve. If $\tilde {f}^{-1}(\tilde {C})$ splits in three curves and they meet, then they cannot be $(-1)$ -curves (because X is of general type). If $\tilde {f}^{-1}(\tilde {C})$ splits in three curves, these cannot meet, $\tilde {f}^{-1}(\tilde {C})$ does not meet the ramification locus, and $(\tilde {f}^{-1}(\tilde {C}))^2=(\tilde {C}^2)$ . In particular, they are not $(-1)$ -curves. The $(-1)$ -curves on $\tilde {X}$ are contained in the ramification locus, and thus $\tilde {f}^{-1}(\tilde {C})$ does not meet these curves.

Proof. Let $\Lambda $ be a lattice generated by the following classes:

$$ \begin{align*}F,\ P_j,\ (j=1,\ldots, k),\ A_1^{(h)},\ A_2^{(h)},\ \frac{F+2\sum_{i=1}^jP_j+\sum_{h=1}^{n-1}\left(A_1^{(h)}+2A_2^{(h)}\right)}{3}=L,\end{align*} $$

where the intersections which are not zero are

$$ \begin{align*}FP_j=A_1^{(h)}A_2^{(h)}=1,\ P_j^2=\left(A_i^{(h)}\right)^2=-2.\end{align*} $$

The lattice $\Lambda $ is even and hyperbolic. If the rank of the lattice $\Lambda $ is less than 12, then it admits a primitive embedding $\Lambda _{K3}$ and there exists a projective K3 surface S such that $NS(S)=\Lambda $ (by the surjectivity of the period map). If $(k,n)$ is such that $\mathrm {rank}(\Lambda )\leq 12$ and satisfies the condition (4.2), then $(k,n)=(3,2),(6,3), (9,1)$ . The classes F and $F+P_j$ span a copy of U inside $\Lambda $ , then there exists a negative definite lattice K such that $U\oplus K\simeq \Lambda $ . In [Reference KondoK, Proof of Lem. 2.1], it is proved that if $NS(S)\simeq \Lambda $ is as described, there exists an elliptic fibration. The $2(n-1) (-2)$ -curves $A_i^{(h)}$ , which are roots in K, are contained in singular fibers as in the statement. By the Shioda–Tate formula, the rank of the Mordell–Weil group of the elliptic fibration is the Picard number of the surface minus the rank of the trivial lattice. The latter is spanned by U and the roots contained in K. Hence, the Mordell–Weil group has rank $k-1$ , and hence there are k independent sections, which corresponds to the classes $P_j$ (see [Reference Schütt and ShiodaSS, Cor. 6.13]).

Proof. By the proof of Corollary 3.17, $D_1$ is the fiber of an elliptic fibration $\varphi _{|D_1|}:S \rightarrow \mathbb {P}^1$ . By hypothesis, the $D_j$ ’s are fibers of the fibration $\varphi _{|D_1|}$ too. The $3:1$ cover $X \rightarrow S$ induces a $3:1$ cover $C \rightarrow \mathbb {P}^1$ where C is a smooth curve. The branch locus of $C \rightarrow \mathbb {P}^1$ is the image of the branch locus of $X \rightarrow S$ and so it is $\varphi _{|D_1|}(\bigcup _j D_j)$ .

Proof. The cover automorphism of $g:W \rightarrow \mathbb {P}^1$ acts as $\zeta _3$ locally near the first $b_1$ ramification points and as $\zeta _3^2$ locally near the other $b_2=n-b_1$ ramification points. Notice that $b_1+2b_2\equiv _3 0$ . Let us consider the fiber product $S\times _g\mathbb {P}^1$ . It is the triple cover X of S branched over the fibers $F_{p_i}=\varphi _{|F|}^{-1}(p_i)$ . So we have a Galois triple cover $X \rightarrow S$ . If all the fibers $F_{p_i}$ are reduced, the cover X is normal, and we can apply the previous theory. The branching data are $B=\sum _{i=1}^{b_1}F_{p_i}$ , $C=\sum _{j=b_1+1}^{n}F_{p_j}$ , $L\simeq (b_1+2b_2)F/3$ , and $M\simeq (2b_1+b_2)F/3$ .

Now, let us compute the triple cover invariants. The surface X is endowed with the elliptic fibration. Let us denote by $X^{\circ }$ the minimal model of X, which is a smooth surface, admitting a relatively minimal elliptic fibration $\mathcal {E}:X^\circ \rightarrow W$ . Assume that it admits at least one singular fiber (which is surely true if there exists a singular fiber of S which is not in the branch locus). Then $X^{\circ }$ is not a product and $h^{1,0}(X^{\circ })=g(W)=n-2$ . Moreover, since $X^{\circ }$ admits an elliptic fibration $K_{X^{\circ }}^2=0$ . We recall that if a branch fiber is of type $IV$ , the corresponding ramification fiber on $X^{\circ }$ is smooth, and hence it is possible that after the base change the fibration $X^{\circ } \rightarrow W$ has no singular fibers and it may be a product.

If the branch fibers are reduced and different from $IV$ , the Euler characteristic of their strict transform is three times their Euler characteristic (see [Reference MirandaMi1, VI.4.1]), then $e(X^{\circ })=3e(S)=72$ , so $\chi (X^{\circ })=e(X^{\circ })/12=3\chi (S)=6.$

Proof. The singularities of types 1 and 2 were considered in [Reference Penegini and PolizziPP, Exams. 1.6 and 1.8] (see also Proposition 2.16). For the other cases, we use the results of Proposition 4.11, where we computed the invariants of $X^{\circ }$ directly by considering the geometry of the elliptic fibration. We compare them with the ones obtained by applying point $(3)$ of Proposition 3.12. If they coincide, this means that all the singularities which can appear in the branch locus are negligible. Since $L^2=M^2=LM=0$ , one obtains $\chi (X')=6$ , $e(X')=72$ , and $K_{X'}^2=0$ . Moreover, $h^1(L)=h^1(\frac {b_1+2b_2}{3}F)=\frac {b_1+2b_2}{3}-1=\frac {b_1+2b_2-3}{3}$ and $h^1(M)=h^1(\frac {2b_1+b_2}{3}F)=\frac {2b_1+b_2}{3}-1=\frac {2b_1+b_2-3}{3}$ so that $h^1(L)+h^1(M)=b_1+b_2-2=n-2$ . So all the singularities appearing in the singular reduced fibers are negligible.

For example, a branch fiber of type $III$ in S consists of two tangent rational curves. We deduce that the singularities in the branch locus obtained by tangency of two components are negligible singularities. More precisely, if there is a branch fiber of type $III$ on S, it induces a fiber of type $IV^*$ (whose dual graph of curves is $\widetilde {E}_6$ ) of $\mathcal {E}:X' \rightarrow W$ . We deduce that if the branch locus of a totally ramified triple cover contains two tangent curves, the triple cover has a singularities of type $E_6$ (see Figure 4).

Figure 3

Figure 4

If a branch fiber on S is singular, then the corresponding fiber on $X^{\circ }$ is obtained by a base change of order 3; [Reference MirandaMi2, Table VI.4.1] shows that effect of a base change on singular fibers of an elliptic fibration.

By [Reference MirandaMi2, Table VI.4.1], one immediately obtains that, if the branch contains a cusp, the cover has a singularity of type $D_4$ , and if the branch has a simple triple point, the cover has an elliptic singularity. Notice that one recovers the singularities of type 1 by considering fibers of type $I_m$ in the branch locus.

Proof. If $j\leq k$ , then $D_j=(F_S)_j$ , the fibers $(F_S)_j$ are branch fibers, and the type of $(F_X)_j$ is given in [Reference MirandaMi1]. We already observed that if $D_i$ is properly contained in a fiber $F_S$ , $i\geq k$ and $D_i$ is supported on an $A_2$ -configuration.

Let us suppose that $F_S$ properly contains r connected components $D_i$ , $i\geq k$ . These are $A_2$ -configurations, whose components are denoted by $A_1^{(1)}$ , $A_2^{(1)}$ , $A_1^{(2)}$ , $A_2^{(2)}$ , $\ldots $ , $A_1^{(r)}$ , $A_2^{(r)}$ .

Moreover, $\sum _{h=1}^r(A_1^{(h)}+2A_2^{(h)})/3$ necessarily has an integer intersection with all the components of $F_S$ , and hence it is contained in the discriminant group of lattice associated with $F_S$ . We conclude that $F_S$ is necessarily one of the following: $I_{3m}$ , with $r=m$ , $IV$ with $r=1$ , and $IV^*$ with $r=2$ .

There are $m A_2$ -configurations contained in $I_{3m}$ , and the birational inverse image of each of them in the minimal model $X^{\circ }$ is a point. The remaining curves form an $I_m$ fiber.

To obtain the minimal model in case $IV^*$ , one first contracts the inverse image of the curves in the $A_2$ -configurations each to a point. These three points lie on a $(-1)$ -curve (which is the inverse image of the unique remaining curve in $IV^*$ ). Contracting this curve, we obtain an $I_0$ fiber.

The case $IV$ is shown in Figure 5.

Proof. If $D^2<0$ , then $h^0(S,D)\leq 1$ . If $D^2=-2$ , then $\chi (D)=1$ , which would imply $h^1(S,D)=0$ .

If $D^2>0$ and D is nef, then D is big and nef, and by the Kawamata–Viehweg vanishing theorem, $h^1(S,D)= 0$ . Therefore, if $D^2>0$ and $h^1(D)\neq 0$ , then D is not nef. In particular, there is a fixed part F in $|D|$ such that $D=M+F$ with $F^2<0$ and $DF<0$ . Moreover $h^0(S,D)=h^0(S,D-F)$ . By the Riemann–Roch theorem,