Three volume integrals and the reduced-fibre flags

Proof. A fibre of $E\to\ell$ and a general conic fibre test the two boundaries of the nef cone. For a prime divisor $D\ne E$, write $D\sim aH-bE$, with integers $a\ge1$ and $b\ge0$. A general smooth cubic section as in Lemma 23.4 is not a component of $D$. Intersect its effective restriction with the nef conic class $F$ to obtain $2a-2b\ge0$. Thus $D=(a-b)H+bB$ has nonnegative integral coefficients. Since $H=E+B$, this also proves the assertion about the effective cone.

For $0\le t\le1$, the three nef divisors in the calculation have volumes $$\begin{split} \mathop{\mathrm{vol}}(L-tH)&=3(2-t)^3-3(2-t),\\ \mathop{\mathrm{vol}}(L-tB)&=3+9(1-t)+6(1-t)^2,\\ \mathop{\mathrm{vol}}(L-tE)&=24-6(1+t)^2. \end{split}$$ The first and third volumes vanish for $t\ge1$. For $1\le t\le2$, $L-tB=(2-t)H+(t-1)E$, with fixed part $(t-1)E$ and volume $3(2-t)^3$; beyond two the volume is zero. The fixed-part assertion follows by testing a general fibre of $E\to\ell$, or from $\pi_*\mathcal O_X(kE)=\mathcal O_V$ for $k\ge0$. Hence the three integrals, divided by $18$, are $$\frac{27/4}{18}=\frac38,\qquad \frac{19/2+3/4}{18}=\frac{41}{72},\qquad \frac{10}{18}=\frac59.$$ If $D\ne E$, either $D-H$ or $D-B$ is effective, by the integral expression above. Monotonicity of volume gives $S_X(D)\le\max\{3/8,41/72\}<1=A_X(D)$. The calculation for $E$ gives the remaining assertion. ◻

Proof. For $0\le u\le1$, put $s=1-u$. The decomposition is $$\begin{array}{c|cc} v&P_{u,v}&N_{u,v}\\\hline {[0,s+1/2]}&A+(s-v)F&0\\ {[s+1/2,s+1]}&(1+s-v)(A+c)&(2v-2s-1)c. \end{array}$$ For $1\le u\le2$, scale the $s=0$ decomposition by $a=2-u$. Nefness, orthogonality and the threshold follow exactly as in Lemma 12.2: both $F$ and $A+c$ are nef. The three inner integrals on the first interval are $3s+2s^2+7/6$, $4s+8/3$, and $1/6$, for $P^2$, $d^2$, and $d\operatorname{coeff}_cN$, respectively. On the second interval they are $7a^3/6$, $8a^3/3$, and $a^3/6$. Since $\int_1^2a^3du=1/4$, these give $$\mathcal S_Z=\frac16\left(\frac{10}{3}+\frac7{24}\right)=\frac{29}{48},\quad \frac16\int\!\int d^2=\frac89,\quad \frac13\int\!\int d\operatorname{coeff}_cN=\frac5{72}.$$ The original fixed part contributes $$\frac16\int_1^2(u-1)\,3(2-u)^2du\cdot r=\frac r{24}.$$ Adding the last two fixed contributions gives $r/9$. If $p\notin E$, then $r=0$, and (24.3) proves the assertion. ◻

Proof. Write the fibre as $Z+Z'$, choosing $p\in Z$. The two curves have square $-1$ and meet transversely once. For $0\le u\le1$, set $s=1-u$. The positive and negative parts of $A+sF-vZ$ are $$\begin{array}{c|cc} v&P_{u,v}&N_{u,v}\\\hline {[0,1]}&A+sF-vZ&0\\ {[1,1+s]}&(A-Z)+(1+s-v)F&(v-1)Z'. \end{array}$$ The threshold is $1+s$. The classes $A-Z,F$ are conic pencils and are nef, so the displayed decomposition has the required properties. For $1\le u\le2$, $aA-vZ$ is nef up to its threshold $v=a$, where $a=2-u$.

Put $\epsilon=1$ if $p=Z\cap Z'$ and zero otherwise, and $\mu=(cZ)_p\le1$. The inner integrals in the first interval are $5/3+4s+2s^2$, $7/3+4s$, and $s^2$. In the second they are $5a^3/3$, $7a^3/3$, and zero. Consequently $$\mathcal S_Z=\frac16\left(\frac{13}{3}+\frac5{12}\right)=\frac{19}{24},\qquad \mathcal T_p=\frac{59}{72}+\frac{\epsilon}{9}+\frac{\mu}{24} \le\frac{35}{36}.$$ The final term uses the same fixed-part integral as in Lemma 24.2. Apply (24.3). ◻