Other centres on the singular fibres

Proof. Points on $D$ are covered by Lemma 90.3. For the other points, the proof in [14, Section 2 and Appendix A] is local to this fibre: its use does not require other fibres to have only nodes. Here is its numerical last step. For those points where $\delta_p(T)\le6/5$, the polarized surface estimate in that appendix bounds $S(W^T;F)/A_T(F)$ above by $$\frac35+\frac1{20}\left( \int_1^{1.356}(16+3u-9u^2+2u^3)\,du +\int_{1.355}^2(11-u^3)\,du\right) =\frac{78409391764017}{80000000000000}<\frac{99}{100}.$$ The intervals overlap, giving an upper bound: the actual switching point lies between $1.355$ and $1.356$. For the remaining points the surface bounds in the same appendix and $S(W^T;F)\le(6/5)S_T(F)$ are stronger. Combining this with $S_X(T)=69/80$ gives the asserted bound by adjunction. ◻

Proof. The cluster description follows by resolving the blow-up of the curvilinear subscheme cut out on the conic. Its only roots are the listed successive differences. No further root comes from a plane line meeting the conic in length three, by Bézout. Among the usual sixteen line classes $e_i$, $h-e_i-e_j$ and $2h-\sum e_i$, precisely the classes having nonnegative intersection with these roots remain irreducible. This gives the first table. In particular the $(4,1)$ configuration is the $A_3$ quartic surface with five lines. The local bounds in (35.10) are the smooth-point entries of [12, Tables 26, 27 and 29]. The excluded lines are exactly the lines meeting the exceptional root curves; outside them the listed values are $4/3$, $18/13$ or $3/2$. ◻

Proof. Work on the minimal resolution, which is an isomorphism near $p$. For $1\le u\le2$ put $$D_u=-K_{\widetilde T}-(u-1)Q_0.$$ Let $D_u-vZ=P(u,v)+N(u,v)$ be its Zariski decomposition, and put $V=P(u,v)^2$, $I=P(u,v)\cdot Z$. The chamber tables below give $N$ explicitly. They use the intersection matrix $$h^2=1,\qquad e_i^2=-1,\qquad h\cdot e_i=e_i\cdot e_j=0\quad(i\ne j).$$ The positive part is recovered as $P=D_u-vZ-N$.

Define, for each polynomial given on these chambers, $$\mathcal J(f)=\frac3{20}\left( \int f(1,v)\,dv+\int_1^2\int f(u,v)\,dv\,du\right),$$ where the integrals run over the displayed chambers. The first term accounts for the full interval $0\le u\le1$ of the original fibre filtration, where its moving part is $-K_T$. Then the first and moving columns in (35.11) are $$S(W^T;Z)=\mathcal J(V),\qquad S_\mathrm{mov}(W^{T,Z};p)=\mathcal J(I^2).$$ If a negative curve $A$ meets $Z$ at $p$, add $\mathcal J(2I\operatorname{coeff}_A N)$. The fixed component $(u-1)D|_T$ makes no contribution because $p\notin D$. Root curves meet $Z$ only above singular points, which are excluded in this lemma. At a smooth point at most two lines of a quartic del Pezzo surface meet: three such lines would force both defining quadrics to contain the tangent plane. Thus at most one other line contributes to the fixed term.

For $Z=\ell_{12}$ in the $(3,1,1)$ and $(3,2)$ cases the only possible additional curve is $\ell_{45}$, giving $21/32$. For $Z=\ell_{14}$ in the $(3,1,1)$ case it is $e_4$, giving $229/320$; interchange indices $4,5$ for $\ell_{15}$. For $Z=\ell_{15}$ in the $(4,1)$ case it is $e_5$, again giving $229/320$. For $Z=e_5$ in the $(3,2)$ case it is $\ell_{45}$, giving $21/32$. The remaining rows have no fixed contribution at a smooth point outside $D$.

Here is one complete integration as a check. For $(4,1)$ and $Z=e_4$, $$N=\frac v4E_{12}+\frac v2E_{23}+\frac{3v}4E_{34}+(v-u)_+Q_0, \qquad 0\le v\le2.$$ For $v\le u$ one has $$V=5-u^2+2uv-v^2/4-4v,\qquad I=2-u+v/4;$$ for $v\ge u$ one has $$V=5-4v+3v^2/4,\qquad I=2-3v/4.$$ Consequently $$\int_0^2V\,dv=\frac{12-u^3}{3},\qquad \int_0^2I^2\,dv=\frac{7u^3-24u^2+42}{12}.$$ Their value at $u=1$ plus their integral over $1\le u\le2$, multiplied by $3/20$, gives $77/80$ and $149/320$. The other rows are obtained by exactly the same polynomial integrations from the tables.

To verify that the tables give Zariski decompositions, the coefficients of $N$ are nonnegative on each listed polygon, its support has negative definite intersection matrix, and $P$ is orthogonal to that support. Moreover $P$ has nonnegative intersection with every curve in Lemma 96.2. These intersections are affine functions of $(u,v)$, so it suffices to substitute the displayed vertices. The cone of curves of these weak del Pezzo surfaces is generated by their negative curves; hence $P$ is nef. The polygons fill the region of positive volume, and $P^2=0$ at its upper boundary. Monotonicity makes the volume zero beyond that boundary. These facts justify both the chamber description and the limits of integration.

Finally, apply the surface flag inequality and then fibre adjunction: $$\delta_p(X)\ge\min\left\{\frac{80}{69}, \frac1{S(W^T;Z)},\frac1{S(W^{T,Z};p)}\right\}>1$$ for the first seven rows. The last row supplies no such conclusion. ◻

Proof. A degree-five fibre with no index-five point has multiplicities $(1,1,1,1,1)$, $(2,1,1,1)$, $(2,2,1)$, $(3,1,1)$, $(3,2)$ or $(4,1)$. These give respectively a smooth surface, $A_1$, $2A_1$, $A_2$, $A_2+A_1$, or $A_3$. Points on $D$ and on smooth fibres are already covered. Lemma 96.1 handles the nodal fibres. Theorems 94.1 and 95.1 handle the higher singular points.

At a remaining smooth point of a fibre, either Lemma 96.2 gives $\delta_p(T)\ge4/3$, hence $\delta_p(X)\ge10/9$, or the point lies on a line in (35.10). Lemma 96.3 gives a bound greater than $100/99$ there, except on $\ell_{14}$ in the $(3,2)$ configuration. This proves the first assertion, including its numerical bound.

In a mixed fibre the image $\Gamma$ of $\ell_{14}$ is the strict transform of the plane line joining the two cluster supports. It joins the $A_2$ and $A_1$ points and is disjoint from $D$. All points outside the union of these lines have now been covered; the $A_1$ point of each mixed fibre is included in its line. A positive-dimensional irreducible valuation centre not contained in this union has a point at which $\delta_p(X)>1$. A curve centre contained in the union must be one of the lines $\Gamma$; it contains the $A_2$ point, at which Theorem 94.1 gives the same strict conclusion for every divisor whose centre contains that point. Thus only the stated closed point centres remain. ◻