Ordinary lines in singular fibres

Proof. Let $J$ be the exceptional divisor of $\operatorname{Bl}_{\ell}V_5$. Projection from $\ell$ is given by $H-J$; see [27, Lemma 4.3]. The resolved projection is the blow-up of a smooth quadric along a twisted cubic; see [54, proof of Theorem 3.6, p. 65]. Its exceptional divisor has class $H-2J$ by the canonical divisor formula. The restriction of $H-J$ embeds $J\simeq\mathbb F_0$ as a smooth quadric surface; its intersection with the other exceptional divisor is the twisted cubic that is blown up. Consequently the unique leading section of $H$ of line order two cuts out an irreducible curve of bidegree $(2,1)$ on $J$.

There are two order-zero sections and four order-one sections. The latter form the complete space $H^0(\ell,N_{\ell/V_5}^{\vee}\otimes\mathcal O_{\ell}(1))$: the kernel of the leading-term map is the one-dimensional space $H^0(H-2J)$, while $h^0(H-J)=5$. The differential of the pencil member is therefore a pair of linear forms on $\ell$. Since it has a zero, the pair has rank one. Choose normal coordinates $a,b$ such that this differential is $z_1b$, and choose the other pencil coordinate to restrict to $z_0$. Smoothness of $C$ implies that $z_0$ and $z_1$ have distinct zeros. The elementary modification of the normal bundle at $q$ is given by $a=t$, $b=z_0n$. Hence $N_{Z/X}\simeq\mathcal O\oplus\mathcal O(-1)$, giving $G\simeq\mathbb F_1$ and the displayed sections. Irreducibility of the $(2,1)$ curve proves the two nonvanishing assertions. ◻

Proof. The leading coefficient of $h^0(X,mL)$ is $10/3$. We first check the boundedness hypothesis of Lemma 34.2 independently of any moment calculation. Write $\sigma:Y=\operatorname{Bl}_Z X\to X$, and choose a very ample divisor $M$ on $Y$. A section of $mL$ of order $j$ along $G$ gives an effective divisor in $|m\sigma^*L-jG|$. Therefore $$j\le m\,\frac{\sigma^*L\cdot M^2}{G\cdot M^2}.$$ The denominator is positive because $G$ is an effective prime divisor. Furthermore, $0\le k\le j$ and $0\le h\le m+k$, by the coefficient bundle computed above. Thus all three recorded coordinates are uniformly $O(m)$. Their successive flag orders have one-dimensional leaves and add under multiplication. All certificates below are leading forms of actual sections. Together with the lattice checks in each case, this proves all hypotheses of Lemma 34.2 before that lemma is used.

If $A_0\ne0$, the seven $H$ vectors are $$(0,0,0),(0,0,1),(1,0,0),(1,0,1),(1,1,0),(1,1,1),(2,0,0).$$ Adjoin their translates by $(1,1,1)$, corresponding to multiplication by $zn$. Their convex hull has volume $10/3$ and centroid $(3/2,13/16,15/16)$. Its four boundary coefficients are $$11/16,\quad13/16,\quad15/16,\quad7/8,$$ which proves the assertion.

Suppose that $A_0=0$, and rescale so that $A_1=1$. Replace the last vector above by $(2,0,1)$. The following degree-two sections provide the missing leading vectors: $$\begin{split} P&=Dn-t(ztn)-B_1n(ztn)-C_1n(zn^2) =B_0tn^2+C_0n^3,\\ Q&=D(zn^2)-(ztn)^2-B_1(ztn)(zn^2)-C_1(zn^2)^2=znP. \end{split}$$ Every factor on the left is a degree-one anticanonical section. If $B_0\ne0$, their normalized vectors are $(3/2,1,0),(2,3/2,1/2)$; if $B_0=0$, they are $(3/2,3/2,0),(2,2,1/2)$. Both hulls have volume $10/3$. Their boundary coefficients are, respectively, $$(213/320,267/320,333/320,127/160),\qquad (103/160,137/160,333/320,261/320).$$ In all three computations $\mu_j=3/2$. The degree-one vectors generate the full exponent lattice. The full-volume lemma therefore proves the asserted value of $S_X(G)$ and all the displayed means. Over the generic point of $Z$, only the first two boundary coefficients occur, and they are smaller than one. Over $q$, the third boundary is disjoint from the fibre and the other three coefficients are smaller than one. The toric one-parameter subgroups preserve that fibre, so semicontinuity applies on a neighbourhood of it. Abban–Zhuang adjunction, with first ratio $2/(3/2)=4/3$, proves the remaining assertions. ◻

Proof. Only $A_0=0$ remains. Choose a base coordinate $z$ vanishing at $p$ and rescale so that the differential vanishes at $z=-1$. Put $w=z+1$. Replacing $t$ by $t+\lambda n$ makes the mixed coefficient constant, so that $$D=wt^2+btn+(a+dz)n^2,\qquad c=a-d,\qquad(b,c)\ne(0,0).$$ The leading pencil section is $wn$. The following are degree-one anticanonical leading sections: $$\begin{gathered} 1,z,\ T=t,\ T_z=zt,\ N=n,\ N_z=zn,\ N_{zz}=z^2n,\\ U=wtn,\ U_z=zwtn,\ V=wn^2,\ V_z=zwn^2,\ D,\ wnD. \end{gathered}$$ For example $N_{zz}$ is obtained by subtracting $zn$ from $(wn)z$. Define degree-two sections $$P=DN-TU-dNV=btn^2+cn^3,\qquad Q=DV-U^2-dV^2=wnP.$$ If $b\ne0$, the degree-two sections $$\begin{split} R&=N_{zz}U_z-T_zV_z+TV_z-TV+P/b=z^4tn^2+(c/b)n^3,\\ S&=U_zV_z-U_zV+UV-Q/b=wz^3tn^3-(c/b)wn^4 \end{split}$$ have normalized leading vectors $(3/2,1,2)$ and $(2,3/2,3/2)$. The degree-one vectors are $$\begin{gathered} (0,0,0),(0,0,1),(1,0,0),(1,0,1),(1,1,0),(1,1,1),(1,1,2),\\ (2,1,0),(2,1,1),(2,2,0),(2,2,1),(2,0,0),(3,1,0). \end{gathered}$$ Their hull with these two additional vectors has volume $10/3$ and first moment $(5,89/32,215/96)$. The three boundary coefficients meeting the fibre over $p$ are therefore $213/320,267/320,43/64$, all smaller than one.

Suppose that $b=0$. Then $c\ne0$, and the degree-two sections $$R_0=N_{zz}V_z-N_zV_z+NV_z-NV+P/c=z^4n^3,\qquad S_0=V_z^2-V_zV+V^2-Q/c=wz^3n^4$$ give the vectors $(3/2,3/2,2),(2,2,3/2)$. This hull has volume $37/12$ and first $h$-moment $63/32$. It is enough to bound the missing volume. The full-volume certificate in Lemma 45.2 implies $j\le3m$ for every actual valuation vector: a vector outside that full-dimensional body would enlarge its convex hull and contradict the leading Hilbert coefficient. Thus $h\le m+k\le m+j\le4m$. The missing volume is $10/3-37/12=1/4$, and hence $$\mu_h\le\frac{63/32+4(1/4)}{10/3}=\frac{57}{64}.$$ The two marginal normal means are unchanged by the base coordinate: $$103/160,\qquad137/160.$$ All three coefficients meeting the fibre are therefore at most $57/64$.

For completeness, these volume calculations use the following halfspaces, with coordinates $(j,k,h)$. In the first case they are $$\begin{gathered} k\le j,\ k,h\ge0,\ h-k\le1,\ k+h\le3,\ j+k\le4,\\ 2j-3k+2h\le4,\quad2j+k+h\le7,\quad 4j-4k+3h\le8,\quad4j-k+3h\le11. \end{gathered}$$ In the second case they are $k\le j$, $k,h\ge0$, $h-k\le1$, $j-k+h\le2$, $j+k\le4$, and $5j-k+4h\le14$. Coning their faces to an interior point gives the stated volumes and moments.

Choose the monomial degenerations with respect to the coordinate $z$ just fixed. They preserve the entire proper fibre over $p$. For every $\lambda<64/57$, the limiting basis pair is Kawamata log terminal near that fibre. Properness of the fixed fibre and semicontinuity give the same assertion before degeneration. Taking $\lambda\uparrow64/57$ gives the bound for the original refined series. The first ratio is $4/3$. Abban–Zhuang adjunction gives the result. ◻