Surface flags at the remaining nodes

Proof. All entries can be calculated by linear algebra and integration of quadratic polynomials. We specify the computation to make the negative parts and the normalization reproducible. Use the negative-curve lists in Lemma 80.3. The auxiliary class $$A=10\ell-2(e_a+e_d+e_c)-f_a-f_d-f_c$$ is ample: it has square $85$ and positive intersection with every listed negative curve, while the Hodge index theorem treats the curves of nonnegative square. On each of the four $t$-intervals, $A\cdot P_t\le26$, as is checked at its endpoints, and $A\cdot C\ge1$ for every flag in the tables. Consequently, pseudoeffectivity of $P_t-vC$ implies $0\le v\le26$. Thus the whole calculation takes place in bounded rational rectangles. On each $t$-interval, enumerate all negative definite subsets $R_1,\ldots,R_s$ of the list and form $$M=(R_i\cdot R_j)_{i,j},\quad (n_i)_i=M^{-1}((P_t-vC)\cdot R_i)_i,\quad P_{t,v}=P_t-vC-\sum_i n_iR_i.$$ Keep the polygon where $v\ge0$, all $n_i\ge0$, and $P_{t,v}\cdot R\ge0$ for every listed negative curve. Also require $P_{t,v}\cdot\ell\ge0$. These are linear inequalities in $t,v$. Check also that the positive part has nonnegative square at each vertex; the convexity of the positive cone gives this throughout each polygon. The negative-curve list and the Hodge index theorem then verify nefness. These rational polygons give the Zariski chambers, with disjoint interiors. Substitution in the three preceding integrals gives the tables.

For completeness an elementary integration rule avoids any numerical approximation. Triangulate each polygon, with a triangle parametrized by $p+x(a-p)+y(b-p)$, $x,y\ge0$, $x+y\le1$. Multiply by its oriented Jacobian. For every resulting monomial use $$\int_{x,y\ge0,\ x+y\le1}x^iy^j\,dx\,dy =\frac{i!j!}{(i+j+2)!}.$$ The numbers of nonempty chambers in the successive rows of the first table are $11,11,10,8,12,7$, and in the second are $13,11,15,10,8,11$. The chamber count here ignores boundaries. In each row the additional check $$\frac3{13}\iint(P_{t,v}\cdot C)\,dv\,dt=1$$ holds. Indeed $-\partial_v(P_{t,v}^2)=2P_{t,v}\cdot C$, and the volume vanishes at the final endpoint. This identity checks that the chambers account for the full slice mass. The script check_f220_chambers_exact.py, included in the audit certificates, carries out this finite enumeration entirely in rational arithmetic, including the negative-definiteness and polygon-feasibility tests. Its files chambers_f220_False_*.json and chambers_f220_True_*.json record the polygons, negative supports, coefficients $n_i$, and positive parts; the file chambers_f220_results.json records their integrals. The script also checks orthogonality, nefness, and the displayed mass identity. The linear systems, inequalities, and integration rule needed to reproduce the computation have all been given above. ◻

Proof. At $p_a$ and $p_d$ use the weight-$(1,2)$ extraction in the specified tangent direction. Its discrepancy over $J$ is three. Its exceptional curve has different $1/2$ at the orbifold point and zero elsewhere. Its strict transform on $T$ is $f_i$, and the curve contracted over the orbifold point is $r_i$. The full first moment is $u_{f_i}+\overline c$. The point moment at the orbifold point is $$b_{f_i}+z_{f_i,r_i}-\frac12u_{f_i}.$$ To verify the subtraction, the pullback of the weighted exceptional curve is $f_i+r_i/2$. The fixed coefficient along $r_i$ before subtracting this curve is $a_i=c/2$, so the initial fixed contribution cancels. The further negative coefficient must be reduced by $v/2$. Since $\frac3{13}\iint v(P_{t,v}\cdot f_i)=u_{f_i}$, the correction is exactly $u_{f_i}/2$.

If $h\ne0$, the first moment is $251/104$, and the orbifold point moment is $103/208$. Away from that point only $t_i,Q_k,Q_c$ can contribute from the first table. They have intersection one with $f_i$. Even adding all their corrections and the fixed $Q_c$ coefficient gives at most $$\frac{103+12+8+17+8}{208}=\frac{148}{208}<1.$$ The curves over the other two marked points are disjoint from $f_i$. Consequently the smallest possible ratio is $\frac{1/2}{103/208}=104/103$.

If $h=0$, at $p_a$ the first moment is $271/104$, and the orbifold moment is $73/208<1/2$. The only further contributing curve meeting $f_a$ is $L_1$, away from the orbifold point. At this point the full moment is $$\frac{71}{208}+\frac5{16}+\frac{31}{104} =\frac{99}{104}<1.$$ At $p_d$ the first moment is $251/104$ and the orbifold moment is again $103/208$. At the ordinary points of $f_d$, even adding the contributions of $L_2,Q_a$ and the fixed $L_2$ gives $(103+29+8+8)/208=148/208<1$. Both weighted flags therefore give the asserted bound.

At $p_c$ use the ordinary blowup, of discrepancy two. Its exceptional curve is represented by $r_c$ on $T$, and its pullback is $r_c+f_c$. Thus its full moment is $u_{r_c}+\overline{a_c}$, and the point moment in the marked direction $t_c$ is $$b_{r_c}+z_{r_c,f_c}+\overline c-\overline{a_c}-u_{r_c}.$$ If $h\ne0$, these are $177/104$ and $21/26$, respectively. At every other direction an upper bound is $(59+10+10+10+4)/104=93/104$, adding the three possible line/conic corrections and the fixed conic.

If $h=0$, the first moment is $193/104$ and the marked point moment is $185/312$. The $L_1$ direction is different from $t_c$ and from the $l_{dc}$ direction. Its point moment is $$\frac{35}{78}+\frac8{39}+\frac{31}{104}=\frac{99}{104}.$$ At the $l_{dc}$ direction the moment is $(140+97)/312=79/104$. Elsewhere it is $35/78$. The other negative curves in that row are disjoint from $r_c$. All these ratios, including $2/(193/104)$, exceed $104/103$. Abban–Zhuang refinement proves the lemma. ◻

Proof. First suppose that the line through $p_a$ and $x$ contains neither of the other marked points and is not $t_a$. Its strict transform has class $\ell-e_a$. Its first moment is $19/52$. Corrections supported above $p_a$ do not meet $x$. If $h\ne0$, the conics $Q_a,Q_c$ each already meet the line at $p_a$, so each has local intersection at most one at $x$. The point moment is at most $$\frac{323+14+13+16}{416}=\frac{183}{208}<1.$$ If $h=0$, the fixed line $L_1$ meets this line only at $p_a$. The line $L_2$ has intersection one, and $Q_a$ has remaining intersection at most one. The same bound follows from the second table.

It remains to use the excluded lines themselves. At points of $l_{ad}$ away from its endpoints, the first moment is $63/104$ and the point moment is at most $(47+5)/104=1/2$, in either case. All fixed components meet this line at its marked endpoints; the only further correction is from $t_c$.

If $h\ne0$, use $l_{ac}$ or $t_a$ at the remaining points. Their first moments are $69/104$ and $49/104$. Discarding the corrections over marked endpoints, and using local intersection at most one for lines and at most two for conics, bounds their point moments by, respectively, $$\frac{43+6+4+1+8}{104}=\frac{62}{104},\qquad \frac{66+3+6+8}{104}=\frac{83}{104}.$$ These bounds include the fixed conic coefficient.

If $h=0$, the line $t_a$ is the line joining $p_a$ to $p_c$, with strict transform $L_1$. Its first moment includes its own fixed coefficient and is $$\frac{17}{26}+\frac{31}{104}=\frac{99}{104}.$$ Away from its marked endpoints the only correction is from $L_2$, including its fixed coefficient, so the point moment is at most $$\frac{25}{52}+\frac2{13}+\frac1{26}=\frac{70}{104}.$$ Every first and point moment just listed is at most $103/104$. The ordinary surface-flag inequality proves the assertion. The upper bounds allow contributing curves to meet at the same point; no general-position assumption on their other intersections has been used. ◻

Proof of Theorem 80.1. Lemmas 81.3 and 81.4 cover every point of $J$, and give $\delta(J;W)\ge104/103$. The ordinary point blowup is a plt extraction with $A_X(J)=3$ and trivial different on $J$. Lemma 80.5 and the Abban–Zhuang inequality [4, Theorem 3.3] now give $$\delta_p(X)\ge\min\left\{\frac3{11/4},\frac{104}{103}\right\} =\frac{104}{103}>1.$$ This proves a bound for every valuation involved in the local threshold, rather than only for the first exceptional divisor. ◻