The index-two endpoint of the mixed fibre

Proof. Put the index-two source point at $[0:1]$, the index-three source point at $[1:0]$, and their common image at $[1:0]$. The graph equation then has the form $$u x^2y^3+v g(x,y)=0,\qquad g=y^5+\sum_{i=1}^5g_i x^iy^{5-i},\qquad g_5\ne0.$$ On the affine chart $u=Y_0=1$ of $\mathbb P^1\times\mathbb P^2$, write $v=t$ and $Y_1=x$, $Y_2=c$. The conic equation is $z=c-x^2$. The two cubic lifts of the graph equation are $$c+tG_0(x,c),\qquad xc+tG_1(x,c),$$ where $$\begin{aligned} G_0&=1+g_1x+g_2c+g_3xc+g_4c^2+g_5xc^2,\\ G_1&=x+g_1c+g_2xc+g_3c^2+g_4xc^2+g_5c^3. \end{aligned}$$ In particular $$G_1-xG_0=zR(c),\qquad R(c)=g_1+g_3c+g_5c^2.$$ On the blow-up chart containing $p$, set $$A=\frac{c+tG_0}{z},\qquad B=\frac{xc+tG_1}{z}, \qquad w=A-1.$$ Then $B=xA+tR(c)$, $c=x^2+z$ and $$tG_0=zw-x^2.$$ Since $G_0(0,0)=1$, the variables $x,z,w$ are regular parameters at $p$, and the initial form of $t$ for their ordinary order is $zw-x^2$.

The thirteen anticanonical sections are $t^i\{1,x,c\}$ for $0\le i\le2$ and $A,B,tA,tB$. Replace the last four by $$\begin{aligned} A-1&=w,& B-x-g_1t&=xw+t(g_3c+g_5c^2),\\ tA-t&=tw,& tB-tx-g_1t^2&=txw+t^2(g_3c+g_5c^2). \end{aligned}$$ Their initial forms are $w,xw,tw,txw$. The initial form of $c$ is $z$. Thus all thirteen displayed polynomials belong to the actual associated graded series. Products give the asserted algebra inclusion. ◻

Proof. The relation is $zw=x^2+t$. In addition to the degree-one generators, we may use $x^3t^i$ of degree two for $0\le i\le3$: write $i=j+k$, where $0\le j\le2$, $0\le k\le1$, and use $$(zt^j)(xwt^k)=x^3t^i+xt^{i+1}.$$ The last term is a product of two degree-one generators. Straightening a $z$-factor against a $w$- or $xw$-factor now uses $zw=x^2+t$ and the displayed identity. Each resulting term remains a product of the generators just listed, of the same anticanonical degree. Repetition eliminates every $zw$ pair. Normal monomials with $bc=0$ are linearly independent, since $zw$ is a leading monomial of the single relation in $\mathbb C[x,z,w,t]/(zw-x^2-t)$.

For $c=0$, use $k$ factors $x^3$ of degree two. The remaining $x$ and $z$ factors have degree one. Filling unused places with $1,t,t^2$ shows that a prescribed $s$ is possible if and only if $$\max(0,a+b-m)\le k\le\lfloor a/3\rfloor, \qquad s\le2m-k.$$ Every integer $s$ up to the upper bound is possible, since each factor allows a consecutive interval of $t$-exponents. Taking the smallest $k$ gives the first row of (A1.1).

For $b=0$, after using $k$ factors $x^3$, pair as many of the remaining $x$ factors as possible with the $c$ factors $w$. The minimum required anticanonical degree is $\max(a-k,c+2k)$, and the maximum $t$-exponent in total degree $m$ is $2m-c-k$. Thus the conditions are $$\max(0,a-m)\le k\le \min\left(\lfloor a/3\rfloor,\lfloor(m-c)/2\rfloor\right), \qquad s\le2m-c-k.$$ They are equivalent to the second row of (A1.1).

Here is the count, including the parity terms. Put $k_n=\lfloor n/2\rfloor$. For the first row set $n=m-b$; its contribution for this $n$ is $$(n+k_n+1)(2m+1)-\frac{k_n(k_n+1)}2.$$ For the second row set $n=m-c$; its contribution is $$(m+k_n+1)(m+n+1)-\frac{k_n(k_n+1)}2.$$ Therefore the dimension equals $$\begin{aligned} &\sum_{n=0}^{m}\left((n+k_n+1)(2m+1) -\frac{k_n(k_n+1)}2\right)\\ &\quad+\sum_{n=0}^{m-1}\left((m+k_n+1)(m+n+1) -\frac{k_n(k_n+1)}2\right)=N_m. \end{aligned}$$ The last identity follows by splitting each sum into even and odd $n$; it holds for both parities of $m$. This equals $h^0(X,mL)$, so Lemma 94.2 proves equality with the actual refined algebra.

The same summation gives the primary weight sum, with no parity restriction. In the first row, for fixed $n$, sum $$\sum_{a=0}^{n+k_n}\left((U+1)(a+m-n)+U(U+1)\right), \qquad U=\min(2m,2m+n-a).$$ In the second row replace the upper limit by $m+k_n$ and put $U=\min(m+n,2m+n-a)$. Sum the first row over $0\le n\le m$ and the second over $0\le n<m$. Splitting at $a=n$ and $a=m$, respectively, and then at the two parities of $n$, gives $$\begin{array}{c|cc} &c=0&b=0,\ c\ge1\\\hline m\text{ even}& \dfrac{m(97m^3+233m^2+218m+64)}{24}& \dfrac{m^2(3m+1)(13m+4)}8\\[2mm] m\text{ odd}& \dfrac{m(m+1)(97m^2+136m+55)}{24}& \dfrac{m(3m+1)(13m^2+4m+3)}8. \end{array}$$ Both rows have total $$\frac{m(107m^3+154m^2+115m+32)}{12}.$$ After division by $mN_m$ this proves $S_X(G)=107/40$. ◻

Proof. The equations for the two gradings are $j=a+b+c+2s$ and $q=a+2c+2s$. Their difference determines $b,c$, and (A1.1) gives exactly $\ell\le s\le v$. Using $t=zw-x^2$ proves (A1.2).

For the asymptotic sums put $y=b/m$ in the $z$-branch and $y=c/m$ in the $w$-branch, and replace $k/m$ by $k$. The lower and upper limits of $s/m$ are $$\begin{array}{c|c|c} &L&U\\ \hline z&\max(0,(2k+3y-3)/4,k+y-3)&\min(2,k/2)\\ w&\max(0,(2k+y-3)/4,k+y-3)&\min(2-y,k/2). \end{array}$$ Integrate on $0\le y\le1$, $L\le U$, and put $D=U-L$. Splitting the region at the indicated affine walls gives (A1.4)$$\begin{array}{c|ccccc} &\int D&\int(k-2U+D)D&\int yD&\int LD&\int D^2/2\\ \hline z&35/24&17/24&47/96&835/768&257/768\\ w&15/8&19/16&25/32&709/768&407/768. \end{array} \tag{A1.4}$$ These are ordinary integrals in $dy\,dk$; for example, the chamber $z$, $L=0$, $U=k/2$ is the triangle $0\le y\le1$, $0\le k\le3(1-y)/2$. Its contributions to the first, second and last columns are $3/16$, $9/128$ and $9/256$. Multiplying the summed columns by $3/10$, and taking the third column separately in each branch, gives (A1.3). Floors and ceilings contribute only $O(m^3)$ to the numerators, while $mN_m=(10/3)m^4+O(m^3)$.

The average degree on $\mathbb P^2$ is $$(\alpha-2h)+b+c+2s+4h=\frac{107}{40}.$$ By the definition of the primary associated graded series this is $S_X(G)$. ◻

Proof of the theorem. The conic pencil on $\mathbb P^2$ is $$T_\lambda=(zw-\lambda x^2=0),\qquad T_0=Z_0+W_0,\qquad T_\infty=2X_0.$$ Its two base points are $[0:1:0]$ and $[0:0:1]$. On the respective smooth charts its branches have equations $w=\lambda x^2$ and $z=\lambda x^2$. Two successive ordinary blow-ups at either point separate all distinct branches.

An arbitrary basis divisor obtained by choosing a basis in each space (A1.2) has the form $$B_m=(\alpha_m-2h_m)X_0+b_mZ_0+c_mW_0+s_mT_1 +\sum\nu_\lambda T_\lambda,$$ where $\sum\nu_\lambda=2h_m$ and $0\le\nu_\lambda\le h_m$. Indeed a basis of $H^0(\mathbb P^1,\mathcal O(d))$ has total degree $d(d+1)$, and its total vanishing order at any point is at most $d(d+1)/2$.

The component bounds for this divisor are $$\alpha_m,\quad b_m+h_m,\quad c_m+h_m,\quad s_m+h_m,\quad h_m.$$ At $[0:1:0]$, the second blow-up has log discrepancy three and order at most $\alpha_m+2(c_m+s_m+h_m)$; at $[0:0:1]$ the corresponding order is $\alpha_m+2(b_m+s_m+h_m)$. The first blow-up inequalities follow by linear interpolation with the bound on the $x=0$ component. After these two blow-ups the total transform is a simple normal crossings divisor. Thus these inequalities test all divisorial valuations, not only monomial ones.

The limiting component bounds are $$\frac{91}{160},\quad\frac{13}{32},\quad\frac{79}{160}, \quad\frac{69}{80},\quad\frac{83}{320};$$ the two corner bounds divided by their log discrepancies are $221/240$ and $69/80$. Their maximum is $221/240$. Consequently $(G,\tau B_m)$ is log canonical for every $\tau<240/221$ and all sufficiently large $m$.

For an arbitrary divisorial valuation over $G$, choose in each primary piece a basis compatible with its valuation filtration and the auxiliary torus filtration. Taking least auxiliary weights produces the divisors just considered. Lower semicontinuity of log canonical thresholds under this torus degeneration gives the same bound for the original valuation-adapted basis divisor. Taking $m$ to infinity proves $\delta(G;W)\ge240/221$. Abban–Zhuang adjunction now gives $$\delta_p(X)\ge\min\left\{\frac{3}{107/40}, \frac{240}{221}\right\} =\frac{240}{221}.$$ No Zariski decomposition on the blown-up threefold was used. The entire refined algebra was identified by an exact Hilbert count; the only further modifications are the two explicit surface blow-ups at each base point of the conic pencil. ◻