Points on the original exceptional divisor

Proof. On the incidence threefold use coordinates $$x=(a,b,1),\qquad y=(1,z,-a-bz).$$ Its bilinear sections form the space $H$. A regular nilpotent matrix may be chosen so that the second bilinear equation is $b+U=0$. The line under consideration is $a=b=0$; its singular endpoint is $z=\infty$. The unipotent centralizer acts transitively on the complement of that endpoint, so that the point under consideration may be put at $z=0$. The other pencil generator restricts to a linear polynomial on the line. Its coefficient at $z$ is nonzero: otherwise it would vanish at the singular endpoint, contrary to smoothness of the base curve. Its constant coefficient is zero because $p\in E$. The implicit function theorem applied to $b+U+tg$ gives the displayed quadratic term of $b$. The anticanonical sections are $H+tH$. Their unique linear relation is $b+U+tg=0$, which eliminates $b$. ◻

Proof. The convex hull is cut out by $$\begin{gathered} a,t,z\geq0,\quad t\leq2,\quad |t-z|\leq1,\quad a+t\leq3,\\ a-t+2z\leq3,\quad 2a+3z\leq6,\quad 2a+t+2z\leq6. \end{gathered}$$ Here is an explicit unimodular triangulation, with the points indexed from $0$ to $14$. Every tetrahedron has vertex $10$; its other three vertices are the following: $$\begin{array}{rrrrrr} (0,1,2)&(0,1,7)&(0,2,7)&(1,2,3)&(1,3,4)&(1,4,8)\\ (1,7,8)&(2,3,5)&(2,5,11)&(2,7,9)&(2,9,11)&(3,4,6)\\ (3,5,6)&(4,6,14)&(4,8,14)&(5,6,11)&(6,11,14)&(7,8,12)\\ (7,9,13)&(7,12,13)&(8,12,14)&(9,11,13)&(11,13,14)&(12,13,14). \end{array}$$ The boundary triangles tile the boundary of the convex hull; taking cones from its interior point $(1,1,1)$ gives the claimed triangulation. Each determinant has absolute value one. In particular the polytope has the integer decomposition property. The numbers of vertices, edges, triangles, and tetrahedra are $15,50,60,24$, respectively. Counting lattice points in the relative interiors of the dilated simplices gives, for $m\geq1$, $$15+50(m-1)+60\binom{m-1}{2}+24\binom{m-1}{3} =4m^3+6m^2+4m+1.$$ The tetrahedra have equal volume $1/6$. The barycentre is therefore the mean of their barycentres, which is the displayed vector. ◻

Proof. First take the lowest ordinary homogeneous parts in the parameters $(a,t,z)$. Within each ordinary degree, refine by the lexicographic order on the exponents of $a,t,z$, taking the least exponent first. The fifteen sections in Lemma 114.1 can be adapted to these two filtrations. Their refined initial monomials are exactly those with exponent vectors in Lemma 114.2. Here are the only cancellations needed to see this. The leading terms of $bz,tb,tbz,tV$ are respectively $$-zQ,\quad -tQ,\quad -tzQ,\quad -atQ.$$ The sections $U,tU$ have initial terms $a^2,a^2t$. Moreover $V+taz+\alpha tU$ has initial term $-a^3$. Subtracting the $a^2t$ term from $tb$ leaves a cubic whose least refined monomial is $t^2z$. The least refined monomials of the other three displayed expressions are $tz^2,t^2z^2,at^2z$. The remaining sections give $1,a,z,az,t,at,tz,atz,a^2,a^2t$. All fifteen initial monomials are distinct.

Products of these sections give all the semigroup monomials in every degree. Their number is, by Lemma 114.2, exactly $$h^0(X,-mK_X)=4m^3+6m^2+4m+1.$$ Consequently they exhaust the actual refined section spaces in every degree. This is an equality of section spaces; a calculation of a convex hull alone would not suffice.

The ordinary order is the sum of the three exponents. Hence the barycentre calculation gives $$S_X(G)=\frac{41}{48}+\frac56+\frac{35}{48}=\frac{29}{12}.$$ We also explain the estimate on the exceptional plane. Fix any valuation on $G$. For each ordinary graded piece, choose a basis adapted both to that valuation and to the refining monomial filtration. For fixed $m$ the latter filtration is induced by an integral one-parameter subgroup of the diagonal torus of $G$; its weights can be chosen to realize the required lexicographic order on all the monomials in question. The basis divisor degenerates to a divisor supported on the three coordinate lines. The asymptotic coefficients of those lines are $$\frac{41}{48},\quad\frac56,\quad\frac{35}{48}.$$ The three lines form a simple normal crossing divisor on $\mathbb P^2$. Lower semicontinuity of log canonical thresholds therefore gives $$\delta(G;W^G_{\bullet,\bullet})\geq\frac{48}{41}.$$ Abban–Zhuang adjunction and $A_X(G)=3$ now give $$\delta_p(X)\geq \min\left\{\frac{3}{29/12},\frac{48}{41}\right\} =\frac{48}{41}.$$ We have used no Zariski decomposition on the extracted threefold. The complete section algebra and its two filtrations supply both $S_X(G)$ and the required restricted-series estimate. ◻

Proof. It suffices to consider one projection; transposition interchanges the two lines. Write the flag and fibre equations as $$x_0y_0+x_1y_1+x_2y_2=0,\qquad x_0y_1+x_2y_2=0.$$ The singular point is $([0:1:0],[1:0:0])$, and the line considered is $x=[0:1:0]$, $y_1=0$. Its point on the base curve has coordinates $y=[\rho:0:1]$ for some $\rho\in\mathbb C$. On $x_1=y_2=1$ use regular coordinates $$a=x_0,\qquad z=y_0-\rho,\qquad h=y_1+\rho a.$$ Then $x_2=-az-h$. Put $$R=az+h,\quad K=zR,\quad L=R(\rho a-h),\quad f=ah-\rho a^2-R.$$ The eight bilinear sections span $$\langle1,a,z,h,az,K,L,f\rangle.$$ The second pencil generator $g$ vanishes at $p$ and has linear term $z+\gamma a+\beta h$ after rescaling. Its $z$-coefficient is nonzero because its restriction to $\Gamma$ has a simple zero at $p$. The blow-up chart is $f=tg$, and $a,z,t$ are regular parameters.

Suppose first that $\rho\ne0$. We may replace $z$ by the regular coordinate $c=g$; its linear term on $X$ is $z+\gamma a$. The quadratic terms of $R,h$ are $$R_2=-\rho a^2-tc,\qquad h_2=(\gamma-\rho)a^2-(a+t)c.$$ Use $R+f=ah-\rho a^2$ in place of $h$. The quadratic initial forms of $az,R+f,f$ span $ac,a^2,tc$. Moreover $$K_3=-(c-\gamma a)(\rho a^2+tc),\qquad L_3=-\rho a(\rho a^2+tc).$$ Thus the eight bilinear initial forms can be chosen as $$1,a,c,a^2,ac,aQ,cQ,tc,\qquad Q=\rho a^2+tc.$$ Removing $f$ from $H+tH$ gives seven of these forms and $t$ times all eight. After rescaling $a$ the resulting algebra is the one used in Lemma 114.3, with the coordinates renamed. In comparing the two displayed bases, use $tQ=\rho a^2t+t^2c$ to interchange $tQ$ and $t^2c$.

Suppose now that $\rho=0$. The fibre contains the second line $z=h=0$ through $p$. We must have $\gamma\ne0$: otherwise $g$ vanishes identically on this line, so the smooth elliptic base curve would contain it. A diagonal rescaling permits $\gamma=1$ in the quadratic computation. The equation $f=tg$ gives $$h=-Q+O((a,z,t)^3),\qquad Q=az+t(a+z).$$ Since $f=ah-R$, we can use $$H=\langle1,a,z,az,ah,K,L,f\rangle.$$ Its last four initial forms are, up to signs, $aQ,tz(a+z),t(a+z)Q,t(a+z)$. Take the seven sections other than $f$ and $t$ times all eight. Their cubic terms are row-reduced using $$aQ-azt=a^2(z+t),\qquad tz(a+z)-azt=z^2t.$$ Their quartic terms are row-reduced using $$t(a+z)Q-taQ-t^2z(a+z)=az^2t.$$ These are precisely the fifteen homogeneous forms stated above. No coefficient of the terms of $g$ beyond its linear part enters this calculation. ◻

Proof. The convex hull is given by $$a,z,t\ge0,\quad t\le2,\quad a+z\le3,\quad a-z\le1, \quad z-t\le1,\quad t-a\le1,\quad z+t\le3.$$ The point indexed $9$ is interior. Cone from this point over the following twenty-four boundary triangles: $$\begin{array}{rrrr} (0,1,2)&(0,1,6)&(0,2,8)&(0,5,6)\\ (0,5,8)&(1,2,3)&(1,3,10)&(1,6,7)\\ (1,7,10)&(2,3,4)&(2,4,8)&(3,4,10)\\ (4,8,11)&(4,10,11)&(5,6,12)&(5,8,12)\\ (6,7,13)&(6,12,13)&(7,10,14)&(7,13,14)\\ (8,11,12)&(10,11,14)&(11,12,13)&(11,13,14). \end{array}$$ On each facet the listed triangles form its triangulation. Every cone has determinant of absolute value one. This is therefore a unimodular triangulation, with face numbers $15,50,60,24$. As in Lemma 114.2, relative-interior counting and the integer decomposition property give $$15+50(m-1)+60\binom{m-1}{2}+24\binom{m-1}{3} =4m^3+6m^2+4m+1.$$ Averaging the barycentres of the twenty-four equal-volume tetrahedra gives the asserted vector. ◻

Proof. The first case has already been identified with the algebra of Lemma 114.3. In the second case, refine the ordinary homogeneous initial forms by taking the least exponent of $t$, then the least exponent of $z$. Their initial monomials are $$1,a,z,t,az,at,zt,a^2z,z^2t,azt,at^2, a^2zt,az^2t,azt^2,a^2zt^2.$$ These are exactly the points of Lemma 114.5. The fifteen distinct initial monomials and their products therefore exhaust the full associated graded section space in every degree, by its exact Hilbert count. The barycentre gives, for the ordinary exceptional plane $G$, $$S_X(G)=\frac{85}{96}+\frac{27}{32}+\frac{43}{48}=\frac{21}{8}.$$ For every fixed $m$ the refinement on its primary pieces is induced by an integral diagonal one-parameter subgroup: use weights $(0,1,N)$ in $(a,z,t)$ with $N$ larger than all exponents occurring in degree $m$. Choose bases compatible with this filtration and any prescribed valuation filtration, and degenerate to their initial monomials. The limiting basis divisor is supported on the three coordinate lines of $G$, whose limiting coefficients are the three barycentre coordinates above. They form a simple normal crossing divisor. The largest coefficient is $43/48$. The same two-filtration and semicontinuity argument as in Lemma 114.3 proves $$\delta(G;W^G)\ge48/43.$$ Finally Abban–Zhuang adjunction gives $$\delta_p(X)\ge\min\{3/(21/8),48/43\}=48/43.$$ Again no Zariski decomposition on the point blow-up is required. ◻