Expected orders and the surface threshold

Proof. Put $\rho=r/m$. The leading dimension in (54.3), divided by $m^2$, is $$d(\rho)= \begin{cases} \rho,&0\le\rho\le1,\\ \rho-\tfrac12(\rho-1)^2,&1\le\rho\le2. \end{cases}$$ Since $N_m=(11/3)m^3+O(m^2)$ and there are two parities, the normalization factor is $6/11$. In particular, $$\frac6{11}\int_0^2d(\rho)\,d\rho=1, \qquad S_X(F)=\frac6{11}\int_0^2 2\rho d(\rho)\,d\rho =\frac4{11}+\frac{49}{22}=\frac{57}{22}.$$ The same calculation is exact in every anticanonical degree. For $\epsilon=0,1$, split the sum at $r=m$; put $d_{m,r,\epsilon}=\dim W^F_{m,2r+\epsilon}$. The four contributions are $$\begin{array}{c|c|c|c} \epsilon&r&\sum d_{m,r,\epsilon}& \sum(2r+\epsilon)d_{m,r,\epsilon}\\ \hline 0&0,\ldots,m&\frac{(m+1)^2(m+2)}2& \frac{2m(m+1)^2(m+2)}3\\ 0&m+1,\ldots,2m&\frac{m(m+1)(8m+7)}6& \frac{m(m+1)(49m^2+57m+14)}{12}\\ 1&0,\ldots,m&\frac{m(m+1)(m+2)}2& \frac{m(m+1)(m+2)(4m+3)}6\\ 1&m+1,\ldots,2m-1&\frac{m(m-1)(8m+5)}6& \frac{m(m-1)(49m^2+43m+12)}{12}. \end{array}$$ Indeed, on the first interval the summand is $(m-\epsilon+1)(r+1)$, and on the second one subtracts $(r-m)(r-m+1)/2$. The identities follow by summing $1,r,r^2,r^3$; an empty interval contributes zero. The case $m=0$ is immediate and the table applies for $m\ge1$. Adding its columns gives $$\begin{aligned} \sum_j\dim W^F_{m,j} &=\frac{(2m+1)(11m^2+11m+6)}6=N_m,\\ \sum_jj\dim W^F_{m,j} &=\frac{m(19m^3+27m^2+19m+5)}2. \end{aligned}$$ Dividing the second identity by $mN_m$ gives the same limit. The top space has dimension $(m+1)(3m+2)/2>0$ at weight $4m$, and no larger weight occurs. This proves the threshold and bounded support.

For $0<\rho<1$ and divisible even native degrees, the series is the complete ample series of bidegree $(m,r)$. More generally, for $1<\rho<2$, multiplication by $t^k$ supplies the complete series of bidegree $(m-k,r)$, whose two degrees are positive, with a fixed effective factor. Thus the required ample-series condition holds on interior rays, as well as on the first chamber. ◻

Proof. A degree-two map between smooth rational curves has two distinct ramification points and can be written as $\eta=\xi^2$ after changes of coordinates. Since $P$ is one point of $D$, choose the repelling ramification point away from it. Write $P=(a,a^2)$ on the affine chart; $a=0$ is allowed. Its moving ideal is $(\xi-\lambda a,\eta-\lambda^2a^2)$. Its powers are flat in $\lambda$, as is also seen by this translation of the two local parameters. The limit of the section space is therefore contained in $H^0(\mathcal I_{o_0}^k(M,r))$.

All point conditions have the same independent length $k(k+1)/2$ in this range. Thus the limiting space and the containing space have the same dimension, and the inclusion is an equality. Membership in $(\xi,\eta)^k$ is the displayed monomial inequality. The action preserves the equation of $D$, so it preserves the different. ◻

Proof. Set $a=i/m$, $b=l/m$, and $\rho=r/m$. The monomial region is $$\Omega_\rho=\{0\le a\le1,\ 0\le b\le\rho,\ a+b\ge\max(0,\rho-1)\}.$$ For $\rho\le1$ it is the whole rectangle. For $1\le\rho\le2$, remove the triangle $0\le a\le\rho-1$, $0\le b\le\rho-1-a$. For a polynomial $Q(a,b,\rho)$, put $$J_Q(\rho)=\int_0^1\int_0^\rho Q\,db\,da, \qquad T_Q(\rho)=\int_0^{\rho-1}\int_0^{\rho-1-a}Q\,db\,da.$$ Its normalized sum is $$\frac6{11}\left(\int_0^1J_Q\,d\rho+ \int_1^2(J_Q-T_Q)\,d\rho\right).$$ Here are the separate contributions: $$\begin{array}{c|cc|c} Q &[0,1]&[1,2]&\text{sum}\\\hline 1&3/11&8/11&1\\ a&3/22&17/44&23/44\\ 1-a&3/22&15/44&21/44\\ b&1/11&27/44&31/44\\ \rho-b&1/11&1/2&13/22\\ a+2b&7/22&71/44&85/44\\ (1-a)+2(\rho-b)&7/22&59/44&73/44. \end{array}$$ For instance, the removed first-coordinate moment is $(\rho-1)^3/6$, so its remaining moment is $\rho/2-(\rho-1)^3/6$ on the second interval. The table proves all six assertions. Boundary lattice errors and parity shifts give $O(m^{-1})$ after normalization.

For $T$ of bidegree $(2,1)$, enlarge each space to $H^0(\mathcal O_F(M,r))$. Its maximal total order along $T$ is $\sum_{l\ge1}h^0(\mathcal O_F(M-2l,r-l))$. Orders of effective zero divisors are nonnegative, so this enlargement is valid with the original normalization. The resulting bound is $$\begin{aligned} \frac6{11}\int_0^2\int_0^{\min(1/2,\rho)} (1-2z)(\rho-z)\,dz\,d\rho &=\frac6{11}\left[ \int_0^{1/2}\left(\frac{\rho^2}{2}-\frac{\rho^3}{3}\right)d\rho +\int_{1/2}^2\left(\frac\rho4-\frac1{24}\right)d\rho\right]\\ &=\frac{81}{352}. \end{aligned}$$ ◻

Proof. Fix an arbitrary divisorial valuation $\nu$ over $F$. In each actual space choose a basis compatible both with $\nu$ and with the filtration of the action in Lemma 55.2. Only two filtrations are involved, so Lemma 2.3 applies. Let $\Gamma_m$ be its normalized refined basis divisor, including the odd fixed factors $\epsilon D/2$. Their total coefficient is $O(m^{-1})$.

The action degenerates $\Gamma_m$ to a divisor $\Gamma_m^0$ whose sections form an eigenbasis of the limiting space. Eigenvectors need not be monomials, since some weights repeat. Nevertheless, their prime factors are coordinate fibres or curves $D_c=(\eta=c\xi^2)$, for $c\in\mathbb C^*$. Lemma 18.7 gives the precise log-canonicity criterion: each prime coefficient is at most one, and the weighted orders at $o_0,o_\infty$ are at most three. Locally, the curves are axes and distinct parabolas. Two successive point blow-ups resolve them; the last inequality is $a+2b+2\sum c_i\le3$, which together with $a\le1$ also gives the first exceptional inequality $a+b+\sum c_i\le2$. Thus this criterion checks all valuations, not just invariant ones.

The four coordinate coefficients of $\Gamma_m^0$ are bounded by the first four numbers in Lemma 55.3, up to an error tending to zero. Any eigenbasis has the same total action weights, so its weighted orders at the two ramification points are $85/44+o(1)$ and $73/44+o(1)$, even if its vectors are not monomials. Every $D_c$ coefficient is at most $81/352+o(1)$.

Take $0<\lambda<88/85$. All prime coefficients of $D/2+\lambda\Gamma_m^0$ are less than one for large $m$: at the limiting endpoint the largest coordinate bound is $62/85$, while the coefficient on $D$ is at most $251/340$. At the ramification points the weighted orders are at most $$1+\lambda\frac{85}{44}+o(1)<3, \qquad 1+\lambda\frac{73}{44}+o(1)<3.$$ The invariant-boundary criterion proves log canonicity. The action preserves $D$, so semicontinuity of log canonical thresholds proves log canonicity of $(F,D/2+\lambda\Gamma_m)$ before degeneration. The $\nu$-compatible basis gives $S_m(W^F;\nu)\le A_{F,D/2}(\nu)/\lambda$. Pass to the limit and then let $\lambda\nearrow88/85$. Since $\nu$ was arbitrary, this proves the assertion. ◻