An audit of the singular-fibre refinement

Proof. For $j\le m$, the restriction sequence is surjective on sections because $$(mL-(j+1)T)-K_X=(m+1)L-(j+1)T$$ is nef and big, so Kawamata–Viehweg vanishing applies. For $m<j\le2m$, the exact fixed component is $(j-m)D$, and the residual divisor is $P=(2m-j)B+(j-m)H$. Its restriction sequence is again surjective: $(P-T)-K_X=P+B$ is nef and big. Thus the fixed restriction is $(j-m)C_T$ and the moving restriction is complete. The curve $C_T$ is contained in the smooth locus of $T$; this also follows from the local blow-up charts below. Its order at a valuation centred at a singular point of $T$ is zero. Taking the asymptotic order sums gives the formula, including its factor $3/L^3=3/20$. ◻

Proof. The resolution $\widetilde T$ is obtained by blowing up three distinct points of the conic, with two further successive blow-ups along the conic above the third point. Let $h,e_1,\ldots,e_5$ be the line class and the total exceptional classes; their intersection matrix is $\operatorname{diag}(1,-1,-1,-1,-1,-1)$. Put $$A=e_3-e_4,\quad F=e_4-e_5,\quad J=e_5,\quad R_0=h-e_3-e_4,\quad Q_0=2h-\sum_{i=1}^5e_i.$$ The curves $A,F$ are the two $(-2)$-curves over the $A_2$ point. The resolution is crepant, so $A_T(F)=1$. Moreover $Q_0\cdot F=0$. For $1\le u\le2$ write $$P_u=-K_{\widetilde T}-(u-1)Q_0,\qquad P_u^2=5-u^2.$$ The negative parts of $P_u-vF$ are $$N_u(v)= \begin{cases} \tfrac v2 A,&0\le v\le2-u,\\ \tfrac v2 A+(u+v-2)J,&2-u\le v\le1,\\ \tfrac v2 A+(u+v-2)J+(v-1)R_0,&1\le v\le2. \end{cases}$$ Their supports are negative definite: $A,J,R_0$ are disjoint and have squares $-2,-1,-1$. The proposed positive parts are orthogonal to their negative parts. To check nefness it suffices to test the negative curves $$e_1,e_2,A,F,J,Q_0,h-e_1-e_2,h-e_1-e_3,h-e_2-e_3,R_0.$$ For clarity, their intersections in the three chambers are as follows; the identical rows have been combined. $$\begin{array}{c|ccc} & [0,2-u]&[2-u,1]&[1,2]\\ \hline e_1,e_2&2-u&2-u&2-u\\ A&0&0&0\\ F&3v/2&2-u+v/2&3-u-v/2\\ J&2-u-v&0&0\\ Q_0&u&2-v&2-v\\ h-e_1-e_2&1&1&2-v\\ h-e_1-e_3,\ h-e_2-e_3&1-v/2&1-v/2&1-v/2\\ R_0&1-v&1-v&0 \end{array}$$ All entries are nonnegative in the relevant chamber. Squaring the positive parts gives $$V(u,v)= \begin{cases} 5-u^2-\tfrac32v^2,&0\le v\le2-u,\\ 9-4u-4v+2uv-\tfrac12v^2,&2-u\le v\le1,\\ \tfrac12(v-2)(v+4u-10),&1\le v\le2. \end{cases}$$ The endpoint positive part is $(2-u)(2h-e_1-e_2-e_3-e_4)$, which has square zero. The nef conic-pencil class $2h-e_1-e_2-e_3-e_4$ intersects $P_u-vF$ in $2-v$, proving the threshold $v=2$, including when $u=2$. Direct integration yields $$\begin{aligned} I(u)&=\int_0^{2-u}(5-u^2-\tfrac32v^2)\,dv +\int_{2-u}^1(9-4u-4v+2uv-\tfrac12v^2)\,dv\\ &\quad+\int_1^2\tfrac12(v-2)(v+4u-10)\,dv =\frac{u^3-6u^2+19}{3}. \end{aligned}$$ In particular, $I(1)=14/3$, and $S_{-K_T}(F)=7/6$. Lemma 92.1 gives $$S(W^T;F)=\frac3{20} \left(\frac{14}{3}+\int_1^2\frac{u^3-6u^2+19}{3}\,du\right) =\frac3{20}\left(\frac{14}{3}+\frac{35}{12}\right) =\frac{91}{80}.$$ ◻