Refining to a cubic surface and then to a curve

Proof. This is the smooth surface and smooth curve case of Abban–Zhuang refinement [4, Theorem 3.3]; the surface-volume formulation is given in [17, Theorems 4.8 and 4.17], with the numbering of the cited arXiv version. The different on $Z$ is zero. Thus the three discrepancies in the flag are one. The first expected vanishing is $S_X(S)=91/192$.

We explain the constants to prevent a normalization error. For fixed $u$, the surface series has asymptotic dimension $m^2D_u^2/2$. This gives the coefficient $3/16$ in (12.2). After the second refinement, a weight summand has dimension $md(u,v)+O(1)$. On $Z\cong\mathbb P^1$ its moving point orders have mean $md(u,v)/2+O(1)$; a fixed divisor adds its local coefficient to that mean. Dividing by $m h^0(X,mL)=16m^4/6+O(m^3)$ gives coefficients $3/16$ for the square term and $6/16$ for the fixed term. In particular, the square term must not be divided by two again.

The explicit decompositions below have bounded support. At an interior rational point of a chamber the positive part is big and nef, its restricted degree is positive, and the restricted series has dimension $md+O(1)$. Thus interior rays are eventually nonzero. Complete ample subseries occur, for example, at $u=1/2$ and sufficiently small positive $v$, where the positive part is ample and restriction is surjective in large degree. These verify the positivity assumptions of refinement. Alternatively they follow from [4, Example 2.15]. The bounded errors and finitely many linear chamber walls disappear in the normalized integrals. ◻

Proof. The classes $F$ and $A+\ell$ are nef. For the latter, intersection with $\ell$ is zero and intersection with any other irreducible curve is positive. For $0\le t\le1/2$, $A-tF=(1-2t)A+t(A+\ell)$ is nef. For $1/2\le t\le1$, $$A-tF=(1-t)(A+\ell)+(2t-1)\ell,$$ which has the required orthogonality and negative part. For $t>1$ its intersection with the nef class $A+\ell$ is negative, so it is not pseudo-effective. Substitute $t=v-s$ and include the initial interval $v\le s$. Scaling proves the last assertion. ◻

Proof. For the first range of $u$, the integrals from Lemma 12.2 are $$\begin{aligned} \int P_{u,v}^2dv &=\int_0^s(3+4(s-v))dv+\int_0^{1/2}(3-4t)dt +\int_{1/2}^1 4(1-t)^2dt\\ &=3s+2s^2+\frac76,\\ \int d(u,v)^2dv &=\int_0^{s+1/2}4\,dv+\int_{1/2}^1 16(1-t)^2dt =4s+\frac83,\\ \int d(u,v)\operatorname{coeff}_{\ell}N_{u,v}\,dv &=\int_{1/2}^1 4(1-t)(2t-1)dt=\frac16. \end{aligned}$$ For the second range these three expressions become, respectively, $7a^3/6$, $8a^3/3$, and $a^3/6$. Since $\int_1^{4/3}a^3du=1/12$, we obtain $$\mathcal S_Z=\frac3{16}\left(\frac{10}{3}+\frac7{72}\right)=\frac{247}{384}, \qquad \frac3{16}\int\!\int d^2=\frac3{16}\left(\frac{14}{3}+\frac29\right)=\frac{11}{12},$$ $$\frac6{16}\int\!\int d\operatorname{coeff}_{\ell}N =\frac6{16}\left(\frac16+\frac1{72}\right)=\frac{13}{192}.$$ Adding (12.4) proves the formula. As $C_S\cdot Z=(2A+\ell)F=6$, one has $\mu\le6$. When $r=0$, the point term is at most $91/96$, and the other two ratios are larger than $96/91$. When $r=1$, $\mu\le2$, the point term is at most $191/192$ and the other ratios are again larger. ◻

Proof. Put $Z=\varphi^{-1}(\varphi(p))$. Choose a smooth cubic surface $S\in|B|$ containing $Z$ as in Lemma 10.6. Its line $\ell$ passes through $p$, and we may choose $S$ so that $\ell$ and $Z$ are transverse there. Indeed, under $S\cong S_0$, the point $p$ is $p_0$, the curve $Z$ has a fixed smooth conic image, and the direction of $\ell$ varies in the pencil of lines through $p_0$ in the plane of that conic. At most one direction is tangent to $Z$, and only finitely many choices give a singular cubic. Write $C_S=E|_S$. This is a smooth curve, and $C_S\cdot\ell=1$, so $C_S$ and $\ell$ are transverse at $p$. We make no transversality assumption between $C_S$ and $Z$.

Let $\sigma:\widehat S\to S$ be the blow-up of $p$, with exceptional curve $J$. Pullbacks of divisor classes on $S$ will be written without $\sigma^*$. Put $R=\ell-J$ and $\widehat Z=F-J$. These are the strict transforms of $\ell$ and $Z$, respectively, and $$R^2=-2,\qquad \widehat Z^2=-1,\qquad R\cdot\widehat Z=R\cdot J=\widehat Z\cdot J=1.$$ The points $q_R=R\cap J$ and $q_Z=\widehat Z\cap J$ are distinct. Also $\sigma^*C_S=\widehat C_S+J$, where $\widehat C_S$ meets $J$ transversely at a point $q_C\ne q_R$. The equality $q_C=q_Z$ is allowed; it occurs when $C_S$ is tangent to $Z$ at $p$.

For $0\le u\le1$, put $s=1-u$. We compute the Zariski decomposition $D_{s,t}=A+sF-tJ=P_{s,t}+N_{s,t}$. Its threshold is $2+s$, and its negative part is as follows: $$\begin{array}{c|c|c} s&t&N_{s,t}\\\hline {[0,\tfrac12]}&[0,1+2s]&0\\ {[0,\tfrac12]}&[1+2s,(5+2s)/3]&\tfrac{t-1-2s}{2}R\\ {[0,\tfrac12]}&[(5+2s)/3,2+s]&(2t-3-2s)R+(3t-5-2s)\widehat Z\\\hline {[\tfrac12,1]}&[0,2]&0\\ {[\tfrac12,1]}&[2,s+\tfrac32]&(t-2)\widehat Z\\ {[\tfrac12,1]}&[s+\tfrac32,2+s]&(2t-3-2s)R+(3t-5-2s)\widehat Z. \end{array}$$ Here $P_{s,t}=D_{s,t}-N_{s,t}$. We verify nefness directly. Since the plane containing $\ell$ and $Z$ cuts out $\ell+Z\sim A$, any irreducible curve $\Gamma\ne\ell,Z$ on $S$ satisfies $A\cdot\Gamma\ge2\operatorname{mult}_p\Gamma$, and $F\cdot\Gamma\ge0$. These inequalities prove nefness in the rows with $N=0$. In the rows with $N=aR$, write $P=(1-a)A+(s+a)F-(t-a)J$ and use $a\ge0$, $1-a\ge0$ and $2-t-a\ge0$. In the rows with $N=(t-2)\widehat Z$, one has $P=A+(s+2-t)F-2J$, which is nef by the same inequalities. In the remaining rows, $$P_{s,t}=(2+s-t)Q,\qquad Q=2A+F-4J.$$ The same argument shows that $Q$ is nef on the strict transform of any $\Gamma\ne\ell,Z$, while $Q\cdot R=Q\cdot\widehat Z=0$ and $Q\cdot J=4$ and $Q^2=4$. The intersections with $R,\widehat Z,J$ in the other rows are nonnegative as well. The supports of the displayed negative parts have negative definite intersection matrices, and direct intersection gives the required orthogonality. Finally, $Q\cdot D_{s,t}=4(2+s-t)$ excludes pseudo-effectivity for $t>2+s$; at $t=2+s$ one has $D_{s,t}=R+(1+s)\widehat Z$. This proves both the decompositions and the thresholds.

Put $d_{s,t}=P_{s,t}\cdot J$ and write $N_{s,t}=a_{s,t}R+b_{s,t}\widehat Z$, so $d_{s,t}=t-a_{s,t}-b_{s,t}$. Integration of the displayed linear expressions gives $$\begin{array}{c|cccc} &\displaystyle\int P_{s,t}^2\,dt &\displaystyle\int d_{s,t}^2\,dt &\displaystyle\int2d_{s,t}a_{s,t}\,dt &\displaystyle\int2d_{s,t}b_{s,t}\,dt\\\hline \text{integrated over }0\le s\le1 &1105/144&241/72&10/27&263/432\\ s=0&32/9&13/9&14/27&4/27. \end{array}$$ In every entry the inner integral is over $0\le t\le2+s$. For $1\le u\le4/3$, put $a=4-3u$. The moving class is $aA$, so the $s=0$ calculation scales by $a^3$; its integral over this range is $1/12$ times the second row. Thus, with the normalization $3/16$ of Lemma 12.1, the moving contributions to the expected order along $J$, the final square term on $J$, and the two fixed point terms at $q_R,q_Z$ are respectively $$\frac{3443}{2304},\qquad \frac{749}{1152},\qquad \frac{67}{864},\qquad \frac{805}{6912}.$$ The original fixed part $(u-1)C_S$ for $u\ge1$ must also be included. Its exceptional multiplicity adds $$\frac3{16}\int_1^{4/3}(u-1)\bigl((4-3u)A\bigr)^2\,du =\frac1{192}$$ to the expected order along $J$. Its strict transform adds the same quantity at $q_C$ on $J$, and adds nothing at the other points. This uses $\widehat C_S|_J=q_C$, so it remains valid for every possible contact order between $C_S$ and $Z$.

Let $W$ be the surface refinement of the anticanonical series along $S$, and let $W^J$ be its further refinement along $J$. The surface refinement formulas (4.3) and (4.4), applied on $\widehat S$ along $J$ with the actual pulled-back fixed divisors, give $$S(W;J)=\frac{3455}{2304},\qquad S(W^J;q)=\frac{749}{1152} +\frac{67}{864}\mathbf1_{q=q_R} +\frac{805}{6912}\mathbf1_{q=q_Z} +\frac1{192}\mathbf1_{q=q_C} \le\frac{5335}{6912}.$$ Here $A_S(J)=2$ and the different on $J\cong\mathbb P^1$ is zero. The original surface term is $S_X(S)=91/192$. Applying refinement first along $S$ and then along $J$ gives $$\delta_p(X)\ge \min\left\{\frac{192}{91},\frac{4608}{3455}, \frac{6912}{5335}\right\} =\frac{6912}{5335}.$$ The restricted series have bounded support and contain ample subseries by Lemma 12.1; these properties persist under the point extraction. On the interiors of the displayed chambers, surface Zariski decomposition and restriction to $J$ compute the leading dimensions of the actual refined spaces. The finitely many walls and degree-rounding errors do not affect the integrals. Thus the calculation applies to the original anticanonical series. ◻

Proof. For $v\le1$, write the divisor as $(1-v)A+v(A-Z)+sF$. The three classes are nef; $A-Z$ is the conic pencil obtained from planes through $Z$. In the second interval the displayed positive part is nef and orthogonal to $Z'$, so it is a Zariski decomposition. The intersection of the original divisor with $A-Z$ is $2(1+s-v)$, proving that the threshold cannot be larger. Scaling the case $s=0$ proves the last assertion. ◻

Proof. For $0\le v\le1$, the volume is $3+4s-2v-v^2$ and $d(u,v)=1+v$. For $1\le v\le1+s$, the volume is $4(1+s-v)$ and $d(u,v)=2$. Hence $$\begin{aligned} \int P_{u,v}^2dv&=\frac53+4s+2s^2,\\ \int d(u,v)^2dv&=\frac73+4s,\\ \int d(u,v)\operatorname{coeff}_{Z'}N_{u,v}dv&=s^2. \end{aligned}$$ For the second range of $u$, the corresponding quantities are $5a^3/3$, $7a^3/3$, and zero. Therefore $$\mathcal S_Z=\frac3{16}\left(\frac{13}{3}+\frac5{36}\right)=\frac{161}{192}, \qquad \frac3{16}\int\!\int d^2=\frac3{16}\left(\frac{13}{3}+\frac7{36}\right)=\frac{163}{192},$$ while the fixed $Z'$ contribution is $(6/16)\int_0^1s^2ds=1/8$. Add (12.4). Finally $C_S\cdot Z=2A\cdot Z+\ell\cdot Z=3$, so $\mu\le3$ and $\mathcal T_P\le190/192=95/96$. Both $192/91$ and $192/161$ exceed $96/95$, which proves the assertion by Lemma 12.1. ◻

The initial local estimates. For a point of a reduced fibre, choose $S$ by Lemma 10.6. If the fibre is singular, choose a component $Z$ containing the point and apply Lemma 12.6. If it is smooth and the point is outside $f_0$, choose $S$ so that its distinguished line avoids the point and apply Lemma 12.3. If the smooth-fibre point belongs to $f_0$, apply Proposition 12.4.

Now let $F$ be a divisor over $X$ with $\beta_X(F)\le0$. Every point $P$ of its centre satisfies $\delta_P(X)\le A_X(F)/S_X(F)\le1$. These three estimates exclude every point on a reduced conic fibre. Thus the centre is contained in the union of the reduced supports of the double conic fibres. Their estimate is proved in Theorem 18.10. Theorem 17.7 gives an independent, weaker estimate at the smooth-fibre points of $f_0$ by a threefold point extraction. ◻