Proof. Let $\pi \colon \bar X = \bigsqcup _{i=1}^r\bar X_i \to X$ be the normalisation. Since $2K_X$ is an ample Cartier divisor, its pullback $\pi ^* 2K_X|_{\bar X_i} = 2( K_{\bar X_i} + \bar D_i)$ is an ample Cartier divisor as well, and

$$\begin{align*}2 = 2K_X^2 = \sum_{i=1}^r \left(2(K_{\bar X_i} + \bar D_i)\right)(K_{\bar X_i} + \bar D_i) .\end{align*}$$

Because the intersection of an ample Cartier divisor with a Weil divisor is a positive integer and because by assumption we have more than one component, we get $r = 2$ and $(K_{\bar X_i} + \bar D_i)^2= \frac 12$ . The divisors $ \bar D_i$ cannot be zero, because they contain the preimage of the intersection of the two components, which is a curve because X is $S_2$ and connected.

Proof. We can of course assume that $r:=h^0(M)-1\ge 1$ , otherwise the statement is empty. Write $|M|=Z+|D|$ , where Z is the fixed part and D the moving part, denote by $h\colon X\to {\mathbb P}^r$ the map defined by $|D|$ and let d be the degree of the image $\Sigma $ of h. Assume first that $\Sigma $ is a surface: then since M and D are nef we have $2=M^2\ge MD\ge D^2\ge d\ge r-1$ , hence $r\leq 3$ . If instead $\Sigma $ is a curve, then D is numerically equivalent to $dG$ , where G is irreducible and $2=M^2\ge MD=dMG\ge d\ge r$ , so $r\le 2$ in this case.

Proof. Set $L_i:=2(K_{\bar X_i}+\bar D_i)$ for $i=1,2$ . Recall that by assumption $L_i$ is an ample line bundle with $L_i ^2=2$ . Consider $0\ne \sigma \in \ker \rho _1$ : since $\rho $ is injective, $\rho _2(\sigma )$ is a nonzero section of $K_{\bar X_2}+\bar D_2$ that vanishes on $\bar \Gamma _2$ , hence $K_{\bar X_2}+ \bar Z_2\ge 0$ . Since $L_2$ is ample, $1=L_2(K_{\bar X_2}+\bar D_2)=L_2(K_{\bar X_2}+ \bar Z_2)+L _2 \bar \Gamma _2$ and $\bar \Gamma _2>0$ give $L_2 \bar \Gamma _2=1$ , $L_2(K_{\bar X_2}+ \bar Z_2)=0$ , therefore $K_{\bar X_2}+ \bar Z_2=0$ and $ \bar \Gamma _2$ is irreducible. Note that $\bar \Gamma _2$ is the divisor of $\rho _2(\sigma )$ . If $0\ne \tau $ is another element of $\ker \rho _1$ , then the divisor of $\rho _2(\tau )$ is also equal to $\bar \Gamma _2$ and therefore, $\rho _2(\tau )$ and $\rho _2(\sigma )$ are linearly dependent. Since $\rho $ is injective, it follows that $\ker \rho _1$ has dimension 1.

Proof of Proposition 3.3.

1. 1. Assume for contradiction that all sections of $K_X$ have a two-dimensional zero locus. Then $X=X_1\cup X_2$ is reducible and we may assume that all sections of $K_X$ vanish on $X_1$ . By Lemma 3.5 we get $p_g(X)\le 1$ , contradicting $\chi (X)=3$ .

2. 2. Assume for contradiction $q(X)>0$ , namely $p_g(X)\ge 3$ . If X is irreducible, then $4=K^2_X+\chi (X)=h^0(2K_X)$ by [Reference Liu and RollenskeLR14, Prop. 16], because $2K_X$ is Cartier. On the other hand the Hopf lemma (see [Reference Arbarello, Cornalba, Griffiths and HarrisACGH85, p. 108]) gives $h^0(2K_X)\ge 2p_g(X)-1\ge 5$ , a contradiction. So $X=X_1\cup X_2$ is reducible.

   If $\rho _i$ is injective, then $h^0(K_{\bar X_i}+\bar D_i)\ge 3$ . Then the Hopf lemma gives $h^0(2(K_{\bar X_i}+\bar D_i))\ge 2h^0(K_{\bar X_i}+\bar D_i)-1\ge 5$ . Pulling back $2(K_{\bar X_i}+\bar D_i)$ to a line bundle M on a desingularisation Y of $\bar X_i$ we obtain a contradiction to Lemma 3.4. So $\rho _1$ and $\rho _2$ are not injective. Identifying $H^0(K_X)$ with its image via $\rho $ , we can find three independent sections of the form: $(\sigma _1,0)$ , $(0,\tau _1)$ , $(\sigma _2,\tau _2)$ . Note that $\sigma _1$ and $\sigma _2$ are independent because $\ker \rho _2$ is one-dimensional by Lemma 3.5 and by the same argument $\tau _1$ and $\tau _2$ are also independent. Now the following correspond to linearly independent sections of $2K_X$ :

   $$ \begin{align*}(\sigma_1^2,0), \ (\sigma_1\sigma_2,0),\ (0,\tau_1^2),\ (0,\tau_1\tau_2), \ (\sigma_2^2, \tau_2^2),\end{align*} $$
   contradicting again $h^0(2K_X)=4$ . We have proved that $q(X) = 0$ .

   It remains to control the plurigenera of X. Since even multiples of the canonical divisor are Cartier, the Riemann–Roch formula applies, so we only need to control the odd plurigenera.

   Let C be a canonical curve defined by a section $x_0$ . As explained in Proposition 3.6 below the reflexive restriction sequence (3.2) defines a torsion free sheaf of rank one ${\mathcal L} = \omega _X|_C^{[{1}]}$ on C with . By definition we have $\deg {\mathcal L} = 1$ , see [Reference HartshorneHar86, Reference Catanese, Franciosi, Hulek and ReidCFHR99]. We now twist (3.2) with the line bundle $\omega _X^{[{2m}]}$ with $m\geq 1$ and compute using generalised Kodaira vanishing and the Riemann–Roch formula on singular curves

   $$ \begin{align*} P_{2m+1} (X)& = \chi(\omega_X^{[{2m+1}]} ) = \chi(\omega_X^{[{2m}]}) + \chi({\mathcal L}\operatorname{\mathrm{\otimes}} \omega_X^{[2m]} |_C) \\ & = \chi (X) + m(2m-1) + \chi(C) + 2m+1 = \chi (X) + m(2m+1). \end{align*} $$
   This is the correct plurigenus for a stable I-surface.

3. 3. If X is irreducible, then pulling back the canonical system to a desingularisation Y of X and considering its moving part we see that the general canonical curve C has at least a reduced component. Since $2K_XC=2$ and $2K_X$ is an ample line bundle, it follows that C is reduced. Therefore $X=X_1\cup X_2$ . Consider the pullback map $\rho =\rho _1\oplus \rho _2$ : if, say, $\rho _1$ is injective, then a similar argument shows that the restriction to $X_1$ of a general $C\in |K_X|$ is reduced and not contained in $\bar \Gamma _1$ , hence C is reduced. So it follows that $\rho _1$ and $\rho _2$ are not injective and, by Lemma 3.5 $K_{\bar X_i}+ \bar Z_i=0$ , $i=1,2$ , and all canonical curves are supported on $X_1\cap X_2$ .

   Conversely, assume that $X= X_1\cup X_2$ is reducible and $K_{\bar X_i}+ \bar Z_i=0$ . Then $K_{\bar X_i}+\bar D_i= \bar \Gamma _i>0$ for $i=1,2$ . Take $\sigma _i\in H^0(K_{\bar X_i}+\bar D_i)$ a nonzero section that vanishes on $\bar \Gamma _i$ : then $(\sigma _1,0)$ and $(0,\sigma _2)$ correspond to independent canonical sections. Since $p_g(X)=2$ by 2., these sections generate $H^0(K_X)$ and therefore every canonical curve is supported on $X_1\cap X_2$ and is not reduced.

Proof. Several times we make use of the long exact cohomology sequences of (3.1), (3.2), (3.3), possibly twisted with $\omega _X^{[{2k}]}$ , the fact that $q(X) = 0 $ by Proposition 3.3 and generalised Kodaira vanishing [Reference Liu and RollenskeLR14, Prop. 21])

To prove the first point, note that by assumption $\omega _X^{[{2}]}$ is a line bundle and from (3.3) we get

• • $\omega _C$ is the restriction of an ample line bundle, thus ample and C is Gorenstein,

• • by the long exact cohomology sequence and Kodaira vanishing and duality on C,

• • $\deg \omega _C = C\cdot 2K_X = 2$ and thus the arithmetic genus $p_a(C) = 2$ .

We now prove the second point. By the depth-Lemma [Reference EisenbudEis95, Cor. 18.6] and the fact that $\omega _X^{[{m}]}$ is $S_2$ for every m, the sheaf ${\mathcal L}$ is the quotient of by the torsion submodule, ${\mathcal L}$ is torsion-free on C and coincides with $\omega _X|_C^{[{1}]}$ (see [Reference HartshorneHar94, Proposition 1.6]). The values $\chi ({\mathcal L}) = 0$ , $h^0({\mathcal L}) =1$ follow from (3.2).

The map $\mu $ exists because every torsion sheaf maps to zero in the line bundle $\omega _C$ . Outside the codimension two subset of X where $\omega _X$ is not locally free, that is, outside a finite number of points, the map $\mu $ is an isomorphism.

For the last point, the existence of the ring structure on $R(C, \{{\mathcal L}, \omega _C\}) $ and the ring homomorphism $\phi $ is clear by construction. The surjectivity follows from twists of (3.2) in odd degrees and twists of (3.3) in even degrees.

Proof. The distinction and description of the cases comes from Proposition 3.3 and Corollary 3.9 in [Reference Coughlan, Franciosi, Pardini and RollenskeCFPR23], while the half-canonical rings are the cases $A(1)$ and $B(1)$ of [Reference Coughlan, Franciosi, Pardini and RollenskeCFPR23, Theorem 5.2].

Construction 5.2. Let $Y_1$ be an Enriques surface containing a half-pencil $E_1$ , that is, an elliptic curve $E_1$ such that $|2E_1|$ is an elliptic pencil. Assume further that $Y_1$ contains a $(-2)$ -curve which is a bisection $\Lambda _1$ of $|2E_1|$ .

Contracting $\Lambda _1$ to an $A_1$ point p, we get a singular Enriques surface $X_1$ containing an elliptic curve $E_1$ through the point p with $E_1^2 = \frac 12$ .

The K3 cover $T_1 \to Y_1$ is an elliptic K3 surface and contains two disjoint $(-2)$ -curves $G_1, G_2$ , which realise $T_1$ as a double cover of ${\mathbb F}_2$ branched over a (symmetric) divisor $B_1\in |4s_0|$ and one can construct explicit examples in this way.

Such Enriques surfaces have been studied in [Reference Cossec and DolgachevCD89, Ch.4, §7] where it is shown that the so-called superelliptic linear system $|4E_1+2\Lambda _1|$ on $Y_1$ realises $X_1$ as the double cover of a degenerate symmetric del Pezzo surface $W_1'$ of degree 4 in ${\mathbb P}^4$ .

One can check that $W_1'\cong W_1 = \{ w^2-yv =0 \} \subset {\mathbb P}(1_{x_0},2_y, 3_w,4_v)$ fitting in the following commutative diagram:

(5.1)

Finally, we count the number of moduli of this construction. The K3 cover $S_1$ of $X_1$ is a hypersurface of degree $8$ in ${\mathbb P}(1_{a_0},1_{a_1},2_b,4_c)$ with equation $c^2=g_8(a_0,a_1^2,a_1b,b^2)$ . In addition, $g_8$ is invariant under the involution $\iota $ of ${\mathbb P}(1,1,2,4)$ defined by $(a_0,a_1,b,c)\mapsto (a_0,-a_1,-b,-c)$ (cf. [Reference Barth, Hulek, Peters and de VenBHPV04, Thm. VIII.18.2]). An explicit computation shows that $g_8$ varies in a linear system of dimension 12. Since the subgroup of automorphisms of ${\mathbb P}(1,1,2,4)$ that commute with $\iota $ has dimension 3, we get $9$ parameters for $(X_1, E_1)$ .

Construction 5.3. Let $B_2 \subset {\mathbb F}_4$ be a general divisor in $|s_\infty + 3s_0| = s_\infty + |3s_0|$ and let $Y_2$ be the double cover branched over $B_2$ . By the Hurwitz formula $Y_2$ is a smooth K3 surface with an elliptic fibration induced by the fibration on ${\mathbb F}_4$ .

The preimage of $s_\infty $ is a $(-2)$ -curve $\Lambda _2\subset Y_2$ and contracting it we get a singular K3 surface $X_2$ with one $A_1$ singularity at a point p. It can be viewed as a double cover of ${\mathbb P}(1,1,4)$ or as a hypersurface of degree $12$ in ${\mathbb P}(1,1,4,6)$ , resulting in a diagram

The image of any fibre of the elliptic fibration is a curve $E_2$ of arithmetic genus one in $X_2$ containing the point p and satisfying $E_2^2 = \frac 12$ . Note that a general $Y_2$ is not isotrivial, so every isomorphism class of elliptic curves occurs a finite number of times as a fibre.

To count parameters for $Y_2$ , we observe that the linear system $|\sigma _{\infty }+3\sigma _0|$ on ${\mathbb F}_4$ has dimension 27, while the automorphism group of ${\mathbb F}_4$ has dimension $9$ , so we get $18$ parameters.

Proof. Note that $(X_i, E_i)$ are stable pairs with volume $\frac 12$ . By Kollár’s glueing construction [Reference KollárKol13, Thm. 5.13], the two components can be glued to a stable surface if the glueing isomorphism $E_1\cong E_2$ preserves the different, that is, if it preserves the point p.

It is easy to check from the classification of slc singularities that X is $2$ -Gorenstein. The two components of a general canonical curve C are defined by sections in $H^0(X_i, K_{X_i}+E_i)$ and are thus two curves of arithmetic genus one, intersecting in the point p.

The pair $(X_1, E_1)$ depends on $9$ parameters (see Construction 5.2) and $X_2$ depends on $18$ parameters (see Construction 5.3), which leaves us with finitely many choices of a fibre $E_2$ of the elliptic fibration on $Y_2$ which is isomorphic to $E_1$ .

Proof. The equations from Theorem 3.8 which define the canonical curve of type B, can be expressed in the following format:

$$\begin{align*}\bigwedge^2\begin{pmatrix}0&y&w&z\\x&w&v&u\end{pmatrix}=0,\ \ \begin{array}{rcl} z^2&=&yg_8(y,v)\\zu&=&wg_8(y,v)\\u^2&=&vg_8(y,v)+x^4h_8(x,v) \end{array} \end{align*}$$

This format is useful for studying the equations of the surface X which contains C as canonical hyperplane section.

We reconstruct X from C by applying the hyperplane section principle [Reference ReidRei90]. The ideal of relations has 9 generators and 16 syzygies, so it is possible to work out the equations of X by hand après Reid [Reference ReidRei90]. We also used the computer package of Ilten [Reference IltenIlt12] which in turn follows an algorithm laid out by Stevens [Reference StevensSte01]. We highlight the two most important features of the result.

The $2\times 4$ matrix for C is weighted homogeneous with weights $\left (\begin {smallmatrix}0&2&3&5\\1&3&4&6\end {smallmatrix}\right )$ and generic entries, except that w appears twice. Thus (after relabelling the entries) the only possible change to the matrix for X would be to one of the w-entries. In fact, it is impossible to change one w-entry without violating some of the 16 syzygies yoking the relations together, hence the matrix for X is unchanged.

The last relation output by the algorithm has the unsatisfactory form

$$\begin{align*}u^2=v\tilde g_8(x_0,y,w,v)+x^4\tilde h_{8}+x_0x^3a_8+x_0^2x^2b_8+x_0^3xc_8\end{align*}$$

where $\tilde h_8(x_0,x,v)|_{x_0=0}=h_8$ and $a_8,b_8,c_8$ are general forms of degree $8$ in $(x_0,x,v)$ . We write ${\tilde k_{10}=x^2\tilde h_8+x_0x a_8+x_0^2b_8}$ and absorb those terms of $c_8$ that are divisible by x or v into $x^2\tilde k_{10}$ or $v\tilde g_8$ respectively. This leaves

$$\begin{align*}u^2=v\tilde g_8(x_0,y,w,v)+x^2\tilde k_{10}(x_0,x,v)+c x_0^{11}x,\end{align*}$$

where c is now a constant. Using the coordinate transformations $x\mapsto x+\alpha x_0$ and $v\mapsto v+\beta x_0^4$ for appropriate choices of parameters $\alpha $ , $\beta $ , we eliminate c to obtain the stated equation. Moreover, the new coordinates do not affect the original choice of curve section C. Note that $\tilde g_8(x_0,y,w,v)|_{x_0=0}=g_8(y,v)$ because after utilising the relation $w^2=yv$ , all monomials of $\tilde g_8$ involving w also involve $x_0$ , for degree reasons.

To find $\omega _X$ , we follow the same procedure as in the proof of [Reference Coughlan, Franciosi, Pardini, Rana and RollenskeCFP⁺23, Cor. 3.12]. That is, we use computer algebra to get the minimal free resolution of as an -module. This is a self-dual complex of length four because the homogeneous coordinate ring of X is Gorenstein, and the last term of the complex is . Since the complex is self-dual, and , when we apply $\mathcal {H}om(-,\omega _{\mathbb P})$ , we can read off .

To show that $\omega _X^{[2]}$ is invertible, it suffices to show that X does not intersect the loci ${\mathbb P}(3_w,6_u)$ , ${\mathbb P}(4_v)$ , ${\mathbb P}(5_z)$ where is not invertible. This can be deduced from the equations (5.2): the monomials $v^3$ and $z^2$ appear in some equation, so X is disjoint from ${\mathbb P}(4)$ and from ${\mathbb P}(5)$ , and since $w^2$ and $u^2$ appear in separate equations, it follows that X is disjoint from ${\mathbb P}(3,6)$ as well. Hence is invertible and X is 2-Gorenstein.

Proof. The defining ideal of X contains the three reducible equations $xy=xw=xz=0$ cutting out the two (weighted) linear subspaces such that X is contained in their union. Since X cannot be contained in one of them, $X = X_1 \cup X_2$ has two irreducible components; it cannot have more by Lemma 3.2.

The linear projection corresponds to the subring generated by $x_0, x, y, w, v$ and the three equations on the right hand side of (5.2) show that the map is of degree two on both components of X.

We translate the equations into the geometric descriptions.

1. 1. The equation for E,

   $$\begin{align*}u^2 - v\tilde g_8(x_0, 0, 0, v) = u^2 - v^3 + \dots = 0,\end{align*}$$
   is obtained by substituting $x = y = w = z = 0$ and is exactly the Weierstrass model we get from a divisor of degree one on a curve of arithmetic genus one.

2. 2. The component $X_1=X\cap (x=0)$ in ${\mathbb P}(1,2,3,4,5,6)$ has equations

   (5.3) $$ \begin{align} \bigwedge^2\begin{pmatrix}y&w&z\\w&v&u\\z&u&\tilde g_8\end{pmatrix}=0 \end{align} $$
   For $\tilde g_8$ general, $X_1$ has one $A_1$ singularity at $(0:0:0:1_v:0:1_u)$ . We are going to show that $X_1$ is a singular Enriques surface by describing explicitly the singular K3 cover $S_1 \to X_1$ , compare (5.1).

   Consider the weighted projective space ${\mathbb P}(1_{a_0},1_{a_1},2_b,4_c)$ and let $S_1$ be defined by the equation

   (5.4) $$ \begin{align} c^2=g_8(a_0,a_1^2,a_1b,b^2). \end{align} $$
   The surface $S_1$ has $A_1$ singularities at the two points $(0:0:1:\pm 1)$ . By adjunction for weighted projective hypersurfaces $S_1$ is a singular K3 surface.

   Let $\iota $ be the involution on ${\mathbb P}(1,1,2,4)$ given by $(a_0:a_1:b:c)\mapsto (a_0:-a_1:-b:-c)$ . One can check that $\iota $ induces a fixed point free involution on $S_1$ exchanging the two $A_1$ singularities.

   We compute the equations of the quotient surface $S_1/\iota $ . The $\iota $ -invariants are

   $$\begin{align*}x_0=a_0,\ y=a_1^2,\ w=a_1b,\ v=b^2,\ z=a_1c,\ u=bc,\ c^2\end{align*}$$
   with respective degrees $1,2,3,4,5,6,8$ . There are several monomial relations between these generators which are analogous to those for the affine cone on the Veronese embedding of ${\mathbb P}^2$ with coordinates $a_1,b,c$ , hence they may be expressed as the $2\times 2$ minors of a $3\times 3$ matrix as in equation (5.3), thus $ S_1/\iota $ is isomorphic to $X_1 $ .

   The projection ${\mathbb P}(1,1,2,4)\dashrightarrow {\mathbb P}(1,1)$ induces an elliptic pencil on $S_1$ that descends to an elliptic pencil on $X_1$ . The image of the $\iota $ -invariant elliptic curve $S_1\cap \{a_1=0\}$ on the quotient $X_1$ is the double curve $y=w=z=0$ .

   We have thus recovered the description given in Construction 5.2.

3. 3. Notice that $X_2=X\cap (y=w=z=0)$ is the hypersurface of weighted degree $12$ defined by $(u^2=v\tilde g_8+x^2\tilde k_{10})$ in ${\mathbb P}(1,1,4,6)$ . For general $\tilde g$ , $\tilde k$ this has one $A_1$ singularity at $(0:0:1_v:1_u)$ . The curve E is cut out by $x=0$ and thus passes through the singular point. Thus $X_2$ is a singular K3 surface as described in Construction 5.3.

Proof. Let us consider again the equations (5.2) defining X. For $\tilde g_8$ we have $13$ free parameters while for $\tilde k_{10}$ we have $21$ free parameters. So ${\mathfrak D}_B$ is dominated by a linear space of dimension $34$ and it is thus irreducible.

From the parameter count in Proposition 5.4 we see that $\dim {\mathfrak D}_B = 27$ .

To prove that ${\mathfrak D}_B$ is in the closure of the Gieseker component, it is enough to prove that the general surface X in ${\mathfrak D}_B$ is ${\mathbb Q}$ -Gorenstein smoothable.

We are going to show that the following family $\mathcal {X} \to {\mathbb A}^1_{\lambda }$ in ${\mathbb P}(1,1,2,3,4,5,6)\times {\mathbb A}^1_{\lambda }$ gives a smoothing of X:

Comparing the equations of $\mathcal {X}_0$ with (5.2) it is clear that the central fibre describes a surface in ${\mathfrak D}_B$ . Now assume that $\lambda $ is invertible. Then the equations allow us to eliminate $w,v,u$ :

(5.5) $$ \begin{align} w = \frac{xy}{\lambda}, \; v = \frac{xw}{\lambda}=\frac{x^2y}{\lambda^2}, \;u = \frac{xz}{\lambda}. \end{align} $$

Hence the minors of the $2\times 4$ matrix vanish identically. If we define

$$\begin{align*}f = z^2 - y\tilde g_8 - \lambda^2\tilde k_{10},\end{align*}$$

then after substituting using (5.5), the remaining three equations become

$$\begin{align*}f, \; \frac{x}{\lambda}f, \; \frac{x^2}{\lambda^2}f.\end{align*}$$

In other words, for $\lambda \neq 0$ the fibre

$$\begin{align*}{\mathcal X}_{\lambda}\cong {\mathsf{Proj}}\left( {\mathbb C}[x_0,x,y,z]/f\right)\end{align*}$$

is a smooth I-surface if $\lambda $ is sufficiently general, because $\tilde g_8$ and $\tilde k_{10}$ are general by hypothesis. The family is equidimensional over a curve and hence flat. The relative canonical bundle is ${\mathbb Q}$ -Cartier, so this is a ${\mathbb Q}$ -Gorenstein smoothing by [Reference KollárKol23, Definition-Theorem 3.1]. This concludes the proof of 1.

To prove 2. we assume that $\tilde g$ does not contain the monomial $y^4$ with nonzero coefficient, which is a codimension one condition on ${\mathfrak D}_B$ by our parameter count. Since $\tilde k_{10}$ does not contain powers of y, every general fibre is isomorphic to a hypersurface of degree 10 in ${\mathbb P}(1,1,2, 5) $ passing through the point $(0:0:1:0)$ , which are the surfaces parametrised by ${\mathfrak D}_A$ by Theorem 4.1. So ${\mathfrak R}_{AB} \subset {\mathfrak D}_A\cap {\mathfrak D}_B$ as claimed.

Proof. Outside a subset of codimension two we have that the pullback of $K_{\tilde X_i}$ is $K_{\bar X_i} + \bar Z_i$ , which is trivial by Proposition 3.3. Since $\tilde X_i$ is $S_2$ , hence Cohen–Macaulay, also the canonical sheaf is $S_2$ , thus trivial.

To prove the second part consider the map $ \tilde {\pi } \colon (\tilde X, \tilde \Gamma ) = (\tilde X_1, \tilde \Gamma _1) \sqcup (\tilde X_2, \tilde \Gamma _2) \to X$ inducing the diagram

The component $\Gamma $ of the non-normal locus of X which connects the two components is a reduced subscheme of pure codimension one, so is $S_1$ . By the depth Lemma ${\mathcal I}_\Gamma $ is $S_2$ and coincides in codimension one with (as in the usual case of the normalisation), so they are isomorphic. Hence from the additivity of the Euler characteristic we get

which is equivalent to Formula (6.1).

The curve $\tilde \Gamma _i$ is irreducible because $\bar \Gamma _i$ is. Since $\tilde \Gamma _1$ is glued to $\tilde \Gamma _2$ , their normalisations are exchanged by the glueing involution and hence are isomorphic.

Proof. Since $p_g( S)=1$ because $K_{ S}=0$ , we have $\chi ( S)\le 2$ and it is enough to show that $\chi ( S)=1$ cannot occur. So assume by contradiction that $\chi ( S)=1$ , i.e., $q( S)=1$ . Pick a very ample line bundle H on $ S$ and let $C\in |H|$ be a general curve. By the generalised Kodaira vanishing theorem ([Reference Liu and RollenskeLR14, Prop. 3.1]), and therefore the restriction sequence for C induces an injection . It follows that the restriction map ${\mathsf {Pic}}^0( S)\to {\mathsf {Pic}}^0(C)$ has finite kernel. The curve C, being general, has nodal singularities, hence ${\mathsf {Pic}}^0(C)$ is a quasi-abelian variety, namely there is an exact sequence $0\to ({\mathbb C}^*)^m\to {\mathsf {Pic}}^0(C)\to A\to 0$ , where A is an abelian variety. So the image of ${\mathsf {Pic}}^0( S)$ in ${\mathsf {Pic}}^0(C)$ is isomorphic either to ${\mathbb C}^*$ or to an elliptic curve, and therefore ${\mathsf {Pic}}^0( S)$ contains a torsion element $\zeta $ of order $m\ge 2$ . Let $Y\to S$ be the connected étale cover given by $\zeta $ . Then $K_Y$ is trivial and $p_g(Y)=\chi (Y)+q(Y)-1\ge m+1-1\ge 2$ , a contradiction.

Proof. Recall that $\Gamma $ and the $\tilde \Gamma _i$ are irreducible with $\chi =0$ or $1$ . Since the maps $\tilde \Gamma _i\to \Gamma $ are birational, we also have $\chi (\Gamma )\le \chi (\tilde \Gamma _i)$ , $i=1,2$ . So if $\chi (\Gamma )=1$ we get $\chi (X_1)+\chi (X_2)=4$ by (6.1) and therefore $\chi (\tilde X_1)=\chi (\tilde X_2)=2$ by Lemma 6.2. If $\chi (\Gamma )=0$ then (6.1) gives $\chi (\tilde X_1)+\chi (\tilde X_2)\ge 3$ and so again $\chi (\tilde X_1)=\chi (\tilde X_2)=2$ by Lemma 6.2.

Proof. Riemann–Roch and generalised Kodaira vanishing give $h^0(2{ \Delta })=\chi (2{ \Delta })=3$ . The restriction sequence together with Kodaira vanishing gives an exact sequence

Since ${ \Delta }\cong {\mathbb P}^1$ we see that $|2{ \Delta }|$ has no base points, because it has no base points on ${ \Delta }$ .

Since ${ \Delta }$ is ample and $(2{ \Delta })^2 = 2$ , the system $|2{ \Delta } |$ induces a finite map of degree two $ {S} \to {\mathbb P}^2$ that maps ${ \Delta }$ isomorphically to a line L, which is necessarily a component of the branch locus. The total degree of the branch locus and the fact that $({\mathbb P}^2, \frac 12(L+B_5))$ is lc follow from the Hurwitz formula (cf. [Reference Alexeev and PardiniAP12, Prop. 2.5]).

Concerning the ring $R(S, \Delta )$ , we can write $H^0(S, \Delta )= \langle x \rangle $ and $H^0(S, 2 \Delta )= \langle x^2, y_1,y_2 \rangle $ . We see that $h^0(S,m\Delta ) = 2 +\frac 14 m^2$ if m is even and restricting to $\Delta $ (which has genus 0), we get $h^0(S,m\Delta ) =1+\frac 14 (m^2 -1 ) $ if m is odd.

So we need a further generator z in degree 5, and we see that the first relations are in degree 10 and we can assume that it is of the form $z^2 =f _{10}(x, y_1, y_2)$ , then considering the map ${\mathbb C}[x, y_1, y_2, z]/\left (z^2 - f_{10}(x, y_1, y_2)\right ) \to R(S,\Delta )$ and looking at the Hilbert series we can conclude that they are isomorphic.

The double cover $S\to {\mathbb P}^2\cong {\mathbb P}(1,2,2)$ corresponds to the truncation in degree 2 and its branch locus consists of the curve $B_5$ defined by $\{f_{10}(x, y_1,y_2)=0\}$ together with the distinguished line L defined by $\{x=0\}$ .

Proof. Riemann–Roch and generalised Kodaira vanishing give $h^0(2\Delta )=\chi (2\Delta )=3$ . The restriction sequence together with Kodaira vanishing gives an exact sequence

Since , we get $h^0(\Delta )=2$ . All the curves of $|\Delta |$ are irreducible and contain the point p defined by .

Similarly, one has $h^0(4\Delta )=6$ and an exact sequence

Since $p_a(\Delta ) = 1$ and $4 \Delta ^2 = 2$ , we see that $|4\Delta |$ has no base points, because it has no base points on $\Delta $ . Its restriction to the curves of $|\Delta |$ defines a double cover $\Delta \to {\mathbb P}^1$ , so the image $\Sigma $ of the morphism $\varphi $ given by $|4\Delta |$ is ruled by lines and $\deg \varphi \ge 2$ . More precisely, $\Sigma $ is a cone, since the images of the curves of $\Delta $ all go through the point $\varphi (p)$ . Since $8=(4 \Delta )^2 = \deg \varphi \deg \Sigma \ge 2\cdot 4$ , we have $\deg \varphi =2$ and $\Sigma $ is the cone ${\mathcal C}_4$ over the rational normal curve of degree 4. For every $\Delta '\in |\Delta |$ we see that contains $2p$ , so $\varphi $ is branched over p. Finally the Hurwitz formula shows that $\varphi $ is branched on a divisor of .

Concerning the ring $R(S, \Delta )$ , we can write $H^0(S, \Delta )= \langle u_0,u_1 \rangle $ and restricting to $\Delta $ we see that we need further generators $v, w$ in degree 4, resp. 6. The double cover $\ S\to {\mathcal C}_4\cong {\mathbb P}(1,1,4)$ corresponds to the truncation in degree 4 and its branch locus consists of the vertex of the cone and the curve $B=\{g_{12}(u_0,u_1,v)=0\}$ . This corresponds to the only relation we need.

Proof. We use Kollár’s glueing principle [Reference KollárKol13, Thm. 5.13]. In order to be able to glue the two surfaces, the normalisations of the boundaries have to be isomorphic, so $\tilde \Gamma _2^\nu \cong \tilde \Gamma _1^\nu = \tilde \Gamma _1 \cong {\mathbb P}^1$ . Since the pair is slc we have that $\tilde \Gamma _2$ is a nodal curve of arithmetic genus 1, proving (a).

The glueing involution in Kollár’s glueing principle also has to preserve the different. On $\tilde \Gamma _2^\nu $ , we have ${\mathsf {Diff}}_{\tilde \Gamma _2^\nu }(0) =\frac 12 p + q_1 + q_2 $ , where p is the basepoint of $|\tilde \Gamma _2|$ , an $A_1$ singularity in $\tilde X_2$ , and the $q_i$ are the preimages of the node of $\tilde \Gamma _2$ .

Let us compute the different on $\tilde \Gamma _1$ . Let $\iota \colon \tilde \Gamma _1 \to \tilde X_1$ be the inclusion and recall that $\tilde \Gamma _1$ is a smooth rational curve and that

is locally free. Let $U \subset \tilde X_1$ be the smooth locus and note that the generic point of $\tilde \Gamma _1$ lies in U.

Then by definition [Reference KollárKol13, (4.3.7), p.154] the divisor $2{\mathsf {Diff}}_{\tilde \Gamma _1}(0)$ is uniquely defined by the property that the square of the Poincaré residue morphism $\omega _{U}(\tilde \Gamma _1\cap U) |_{\tilde \Gamma _1\cap U }\cong \omega _{\tilde \Gamma _1\cap U}$ extends to an isomorphism

In total, we get ${\mathsf {Diff}}_{\tilde \Gamma _1}(0) = \frac 12B_5|_L$ under the isomorphism $\tilde \Gamma _1\cong L$ , so compatibility for glueing forces (b).

Conversely, ${\mathsf {Aut}}({\mathbb P}^1)$ acts 3-transitively, so any pair with the above properties can be glued, thus we have proved the first item.

With this extra information about the branch locus, the second and the third item follow from Proposition 6.4 and Proposition 6.5. The last point follows from the definition of $X_1$ as a component of X and the description of the glueing.

Proof. Let $\sigma $ be an involution as in the statement and consider the induced involution of $\tilde X=\tilde X_1\sqcup \tilde X_2$ : since $\tilde \Gamma _1$ and $\tilde \Gamma _2$ are not isomorphic by Proposition 8.2, they cannot be exchanged by the involution and so $\sigma $ induces involutions $\sigma _i$ of $\tilde X_i$ that preserve $\tilde \Gamma _i$ , $i=1,2$ .

Let $\alpha \in H^0(\tilde X_1, \tilde \Gamma _1)$ be a section vanishing on $\tilde \Gamma _1$ . There is an injective pull back map ${r =(r_1,r_2)\colon H^0(4K_X)\to H^0(\tilde X_1, 4\tilde \Gamma _1)\oplus H^0(\tilde X_2, 4\tilde \Gamma _2)}$ , so we may identify $H^0(4K_X)$ with the image of r. So $H^0(4K_X)$ contains all the sections of the form $(\alpha ^2\gamma ,0)$ , where $\gamma $ varies in , where $\pi _1$ is the double cover given by $|2\tilde \Gamma _1|$ (cf. Proposition 6.4). So $\sigma _1$ is either the identity or the involution associated with $\pi _1$ and, in either case it restricts to the identity on $\tilde \Gamma _1$ . As a consequence $\sigma $ restricts to the identity on $\Gamma $ , and therefore also $\sigma _2$ restricts to the identity on $\tilde \Gamma _2$ . An argument similar to the previous one shows that $\sigma _2$ preserves the curves of $|\tilde \Gamma _2|$ and acts on ${\mathcal C}_4\subset {\mathbb P}^5$ , the image of the 2-to-1 map $\pi _2$ defined by $|4\tilde \Gamma _2|$ (cf. 6.5), mapping each ruling to itself. Let $\tau $ be the automorphism on ${\mathcal C}_4\cong {\mathbb P}(1,1,4)$ induced by $\sigma $ . If $\tau $ is the identity then $\sigma _2$ is either the identity or the covering involution. Since $\tilde \Gamma _2$ is not in the branch locus of $\tilde X_2\to {\mathcal C}_4$ , the covering involution is not the identity on $\tilde \Gamma _2$ , so in this case $\sigma _2$ must be the identity. Now assume that $\tau $ is a nontrivial involution of ${\mathcal C}_4$ : since $\tau $ maps each ruling to itself, there are weighted homogeneous coordinates $x_0,x_1, y$ on ${\mathbb P}(1,1,4)$ such that $\tau $ is given by $(x_0,x_1,y)\mapsto (x_0,x_1,-y)$ . So $\tau $ is nontrivial on every ruling of ${\mathcal C}_4$ and, a fortiori, $\sigma _2$ is not trivial on $\tilde \Gamma _2$ . So the only possibility is that $\sigma _2$ is the identity, and therefore if $\sigma $ is nontrivial then $\sigma _1$ is the covering involution of $\tilde X_1\to {\mathbb P}^2$ .

If $\sigma $ is the identity, then $\chi (X/\sigma )=\chi (X)=3$ . If $\sigma $ is not the identity, then it is easy to see that $Y:=X/\sigma $ is defined by the push-out diagram:

Now Lemma 6.1 gives $1=\chi (Y)=\chi (\Gamma )+\chi (\tilde X/\sigma )-\chi (\tilde \Gamma )=0+(1+2)-(1+0)=2$ .

Proof. First we count the moduli of the construction. Since the elliptic fibration of a surface of type E has finitely many singular fibres, which are all nodal if the surface is general, by Proposition 8.2 it is enough to count parameters for surfaces of type E and for surfaces of type D such that the double cover ${\tilde X_1 } \to {\mathbb P}^2$ is branched over a sextic $L+B_5$ with the line L bitangent to the quintic $B_5$ . Fixing the line $L = \{x_0=0\}$ and points $p,q_1,q_2\in L$ , we see that system of plane quintics tangent to L at $q_1$ and $q_2$ and passing through p has dimension $15$ . Subtracting the dimension of the subgroup of ${\mathsf {Aut}}({\mathbb P}^2)$ that fixes L pointwise we obtain $15-3=12$ moduli.

Surfaces of type E are double covers of $ {\mathcal C}_4 \cong {\mathbb P}(1,1,4)$ , branched over a cubic section B and the vertex. The curve B moves in a system of dimension 27, so subtracting the dimension of ${\mathsf {Aut}}({\mathcal C}_4)$ we get $27-9=18$ moduli. Summing up, surfaces of type DE depend on $12+18=30$ moduli. We can consider the surfaces of type E such that the branch locus of the map to ${\mathcal C}_4$ is simply tangent to a fixed ruling $R_0$ and define $\tilde \Gamma _2$ to be the preimage of $R_0$ ; since ${\mathsf {Aut}}({\mathcal C}_4)$ acts transitively on the set of rulings, in this way we obtain an irreducible family of pairs $(\tilde X_2, \tilde \Gamma _2)$ that contains every isomorphism class of such pairs. We can take the double cover of this family obtained by labelling the preimages of the node of $\tilde \Gamma _2$ in the normalisation map $\tilde \Gamma _2^{\nu }\to \tilde \Gamma _2$ . By Remark 8.3, an irreducible component of this double cover contains all the isomorphism classes of pairs of type E + labelling of the preimages of the node, and therefore surfaces of type DE give an irreducible locus ${\mathfrak U}^{DE} \subset \overline {{\mathfrak M}}_{1,3}$ of dimension 30.

Now let X be a surface of type DE and assume that there is a one parameter deformation ${\mathcal X}\to C$ over a smooth curve such that the general fiber is not of type DE. Since the Cartier index is lower semicontinuous, because $\omega _{{\mathcal X}/C}^{[{m}]}$ being locally free is an open condition, up to shrinking C we may assume that all fibres of ${\mathcal X}$ are 2-Gorenstein and not of type DE except the central fibres. Then by our classification results (Theorem 1.1) the fibres of X are smoothable, so by picking another family we may assume that every fibre except X is smooth.

The general fibre $X_t$ thus carries an involution $\sigma _t$ , induced by the bi-canonical map, which acts trivially on m-canonical system for $m\le 4$ and such that $\chi (X_t/\sigma _t) = 1$ . By [Reference Franciosi, Pardini, Rana and RollenskeFPRR22, Prop. 2.6], the involutions on the general fibres extend to give a global fibrewise involution on the family ${\mathcal X}/C$ , so there is an involution $\sigma $ on X contradicting Lemma 8.4.

Therefore any small deformation of a surface of type DE is again a surface of type DE, the locus ${\mathfrak U}^{DE}$ does not intersect the closure of the Gieseker component and its closure $\overline {\mathfrak M}^{DE}$ is an irreducible component of the moduli space.

Proof. As shown in Proposition 6.4, $\tilde X_1$ is a hypersurface $z^2=f_{10}(x,y_0,y_1)$ in ${\mathbb P}(1,2,2,5)$ . The branch curve of the double cover $\tilde X_1\to {\mathbb P}^2\cong {\mathbb P}(1,2,2)$ consists of the curve $B_5$ defined by $\{f_{10}=0\}$ together with the distinguished line L defined by $\{x=0\}$ . The bitangency condition means that the quintic $f_{10}|_{x=0}$ has two double roots, and after a coordinates change we may write $f_{10}|_{x=0}=y_0(y_1+y_0)^2(y_1-y_0)^2$ .

Proof. As shown in Proposition 6.5, $\tilde X_2$ is a hypersurface $w^2=g_{12}(u_0,u_1,v)$ in ${\mathbb P}(1,1,4,6)$ . The elliptic pencil is spanned by $\langle u_0,u_1\rangle $ and we suppose that the nodal rational curve $\tilde \Gamma _2$ is defined by $\{u_0=0\}$ . Since $\tilde \Gamma _2$ is nodal, it follows that $g_{12}|_{u_0=0}$ has a double root and we choose coordinates so that $g_{12}|_{u_0=0}=v(v-u_1^4)^2$ . The node q of $\tilde \Gamma _2$ corresponds to this double root.

Proof. By definition, the image of $\alpha $ is $\tilde \Gamma _1\subset \tilde X_1$ and the image of $\beta $ is $\tilde \Gamma _2\subset \tilde X_2$ .

By Remark 8.3 the choice of involution does not matter as long as it identifies the preimages of the node $q\in \tilde \Gamma _2$ with $s_1$ and $s_2$ and the point p with the point r. This is satisfied here, because

$$ \begin{align*} &\alpha(1:\pm1)=(0:1:\pm1:0)=s_i &\text{ and }&&\beta(1,\pm1)=(0:1:1:0)=q,\\ &\alpha(0:1)=(0:0:1:0)=r &\text{ and } &&\beta(0:1)=(0:0:1:1)=p. \end{align*} $$

Note that our choice of grading respects the gradings of the rings on the components, which concludes the proof.

Proof. This can be checked with the computer, but also by hand: the determinantal relations encode the image of

and the polynomials $A_8$ and $B_6$ are determined from (8.3) by

$$\begin{align*}e^2 = z^2 = y_0(y_1+y_0)^2(y_1-y_0)^2+x^2f_8(x,y_0,y_1) = b_0(d-b_0^2)^2+a_0^2A_8+a_0cB_6.\end{align*}$$

The remaining two equations are induced by $g^2 = de^2$ and $eg = y_1 z^2$ using the determinantal relations.

Proof. Note that $a_1^2, b_0, b_1, d, f$ generate the even subring of ${\mathbb C}[u_0, u_1, v, w]$ , so the map is an embedding.

As in the proof of Lemma 8.9, the determinantal equations cut out the image of ${\mathbb P}(1,1,4,6)$ and the remaining equations are induced from (8.4) by the equation defining $\tilde X_2$ .

Proof. The intersection of ideals of the components described in Lemma 8.9 and Lemma 8.10 can be calculated by computer or by hand giving (8.5). The rank condition is expressed by the vanishing of the $36$ minors of size $2\times 2$ of the matrix, but we claim that only $15$ of these are ideal generators. The nine minors involving at least two entries from the upper left $2\times 2$ block give the equation $0=0$ . Moreover, a further seven minors are repeated or redundant. For example, the equation $a_1c=0$ appears twice while $b_1c=0$ appears four times as different minors. Then six minors are not needed to generate the ideal. For example, $a_1b_0c=0$ appears twice but is not needed because we already know that $a_1c=0$ . Finally, of the six equations listed on the right hand side, the last one follows from the other ones. In total, the ideal is generated by $20$ equations.

The two components of X are $X_1=X\cap \{a_1=b_1=0\}$ and $X_2=X\cap \{a_0=c=0\}$ and one can confirm that this recovers the ideals of the components.

Now consider a surface X defined by these equations with $A, B, C, D$ general. The Hilbert series of X can be determined by computer or follows from Proposition 3.3 as soon as we have established that X is 2-Gorenstein stable.

The canonical sheaf can be computed as in the proof of Proposition 5.6.

The sheaf is invertible on ${\mathbb P}(1,1,2,2,3,4,5,6,7)$ away from the loci ${\mathbb P}(3_c,6_f)$ , ${\mathbb P}(4_d)$ , ${\mathbb P}(5_e)$ and ${\mathbb P}(7_g)$ . Clearly, X does not meet the last three of these loci, because monomials $d^3$ , $e^2$ , $g^2$ appear with nonzero coefficient in some equations defining X. Moreover, $X\cap {\mathbb P}(3,6)$ is also empty, because $c^2$ and $f^2$ also appear in separate equations defining X. Hence the restriction is invertible.

Using the description of the components, one can see that for general equations the resulting surface has slc singularities.