A weighted extraction at a ramification point

Proof. The normal derivatives along $\ell$ span a base point free pencil of binary quadrics. It contains two distinct double members: equivalently, its associated degree-two map $\mathbb P^1\to\mathbb P^1$ has two distinct ramification points. Sending their supports to $u=0$ and $v=0$, and changing the two generators of the pencil, gives the displayed form of $G$. The special base point is $y_0=[1:0:0]$.

On $E$, the equation is $u^2x_1+v^2x_2=0$. Away from $s$, the map $E\to\mathbb P^2$ is locally a double cover, ramified exactly on the two rulings $u=x_2=0$ and $v=x_1=0$. For instance, on the chart $v=x_1=1$ its expression is $$(u,x_0)\longmapsto(x_0,x_2)=(x_0,-u^2).$$ Restricting this map to a general base line through $f(p)$ gives a smooth curve $c$ through $p$. Its intersection with the fibre has multiplicity one away from those rulings and multiplicity two on them. This proves the assertion about $(cF)_p$; the same multiplicity holds for any smooth cubic section from the pencil.

In the ramified case interchange $u,v$ and $x_1,x_2$ if necessary. Then $u=x_2=0$, $v\ne0$, and $x_1\ne0$, the last assertion following from $p\notin s$. Replacing $x_0$ by $x_0-a x_1$ gives the stated coordinates of $p$. On its plane conic fibre the derivatives with respect to $u,v$ vanish at $p$, while the derivative with respect to $e$ is $Q_1(0,1,0)$. Since that conic is smooth at $p$, this number is nonzero. ◻

Proof. The implicit function theorem applies to $x_2$, because $\partial G/\partial x_2(p)=1$. Its initial expression is $x_2=-u^2-ce$. The exceptional divisor and discrepancy are the usual ones for the weighted blow-up of a smooth threefold with weights $(1,1,2)$. The only nontrivial quotient chart has type $\frac12(1,1,1)$; its restriction to $T$ has type $\frac12(1,1)$. The quotient is unramified in codimension one on $T$, so the different is zero. This extraction is plt.

Assign the six Cox variables the weights $$\operatorname{wt}(u,v,e,x_0,x_1,x_2)=(1,0,2,1,0,2).$$ The initial equation is $$G_0=v^2x_2+u^2x_1+cevx_1^2.$$ It is irreducible: viewed as a linear polynomial in $x_2$, its two coefficients $v^2$ and $x_1(u^2+cevx_1)$ are coprime. Consequently the initial quotient is a domain, and localization at $v x_1$ creates no new relations. Moreover, the degrees of $v$ and $x_1$ form a basis of the Cox grading lattice, so setting them equal to one is the local trivialization of each fixed bidegree. On that chart it is exactly the weighted associated graded ring in the three regular parameters above. Thus the Cox weight filtration coincides with the geometric $T$-valuation filtration on the section spaces. Here we use the section-ring presentation established earlier; the hypersurface exact sequence has no $H^1$ term in these bidegrees.

For completeness, write $P(p,q)$ for the dimension of the polynomial bidegree $(p,q)$ piece, and $M(p,q)$ for its total weight. A monomial with $e$-exponent $k$ has $u,v$-degree $p-k$ and $x_0,x_1,x_2$-degree $q+k$. Averaging the exponents gives $$\begin{aligned} P(p,q)&=\sum_{k=0}^p(p-k+1)\binom{q+k+2}{2},\\ M(p,q)&=\sum_{k=0}^p(p-k+1)\binom{q+k+2}{2} \left(\frac p2+q+\frac{5k}{2}\right). \end{aligned}$$ Since $G$ has bidegree $(2,1)$ and initial weight two, the dimension $N_m$ and total weight $w_m$ of $H^0(X,mL)$ are $$\begin{aligned} N_m&=P(m,m)-P(m-2,m-1) =\frac{(2m+1)(3m^2+3m+2)}2,\\ w_m&=M(m,m)-M(m-2,m-1)-2P(m-2,m-1)\\ &=\frac{m(2m+1)(26m^2+21m+19)}6. \end{aligned}$$ These formulas hold for $m\ge2$. Taking $\lim w_m/(mN_m)$ gives $26/9$. ◻

Proof. A global monomial has the form $u^av^be^kx_0^ix_1^\nu x_2^d$, where $a+b+k=m$ and $i+\nu+d=m+k$. Its initial form on $T$, up to a nonzero constant, is $t^iu^ae^kQ^d$. Write $a=\epsilon+2a'$. Thus the binary forms that occur are precisely $$r^{a'}e^kQ^d,\qquad a'+k+d=h,\quad 2a'+k\le M,\quad i+d\le m+k.$$ The least possible $k$ is $K$, and the least possible $d$ is $D$. Here is a dimension argument that also proves that there are no missing binary forms after these fixed factors are removed. For a fixed integer $k$, a feasible $d$ exists exactly when $$\max\left\{0,\left\lceil\frac{2h-M-k}{2}\right\rceil\right\} \le d\le\min\{h-k,k+m-i\}.$$ These intervals are nonempty for precisely the consecutive integers $K\le k\le h-D$. For each such $k$, choose one feasible $d$. The resulting form has exact order $k$ along $e=0$, since $r$ and $Q$ are nonzero there. These $h-K-D+1$ forms are therefore linearly independent. Every generator is divisible by $e^KQ^D$, and the space of all binary forms of degree $h$ with this divisibility has the same dimension. This proves the claimed equality. The nonemptiness condition follows also by taking $a'=0$ and $k=\min(h,M)$, which minimizes $d-k$ under the first constraint. ◻

Proof. For a summand in the preceding lemma put $n=h-K-D$. Complete binary forms of degree $n$ have vanishing sequence $0,1,\ldots,n$ at any point of their parameter line. The maximal average orders along the five curves in the table, computed using bases adapted to the respective vanishing filtrations, are consequently $$\epsilon+n,\qquad i,\qquad K+\frac n2,\qquad D+\frac n2,\qquad \frac n2,$$ respectively. These are upper bounds for the average orders of arbitrary bases. Every nonzero element of this summand has weighted order $\epsilon+2h$ at $o$.

Here are the integrals explicitly. Put $x=i/m$, $y=h/m$, and $$\begin{gathered} \Omega=\{(x,y)\in\mathbb R_{\ge0}^2: x+y\le1+2\min(y,1)\},\\ k=\max\{0,x-1,(2x+2y-3)/3\},\qquad d=\max\{0,y-1\},\qquad z=y-k-d. \end{gathered}$$ The two parity choices and $N_m=3m^3+O(m^2)$ give the normalized density $\frac23z\,dx\,dy$. Hence the required quantities are $$\begin{aligned} S(W;u=0)&=\frac23\int_\Omega z^2\,dx\,dy,& S(W;t=0)&=\frac23\int_\Omega xz\,dx\,dy,\\ S(W;e=0)&=\frac23\int_\Omega z(k+z/2)\,dx\,dy,& S(W;Q=0)&=\frac23\int_\Omega z(d+z/2)\,dx\,dy,\\ S(W;C_a)&=\frac13\int_\Omega z^2\,dx\,dy,& S(W;\operatorname{wt}_{(1,2),o})&=\frac43\int_\Omega yz\,dx\,dy. \end{aligned}$$ One may evaluate them on the following five polygons, on which $k,d$ are linear. Vertices are listed in the $(x,y)$ plane. $$\begin{array}{c|l|cc} &\text{vertices}&k&d\\\hline P_1&(0,0),(1,0),(1,1/2),(1/2,1),(0,1)&0&0\\ P_2&(1/2,1),(0,3/2),(0,1)&0&y-1\\ P_3&(1,0),(2,1),(1,1/2)&x-1&0\\ P_4&(2,1),(1/2,1),(1,1/2)&(2x+2y-3)/3&0\\ P_5&(2,1),(0,3),(0,3/2),(1/2,1)&(2x+2y-3)/3&y-1 \end{array}$$ For further detail, their contributions to the displayed orders are $$\begin{array}{c|rrrrrr} &u&t&e&Q&C_a&\operatorname{wt}_{(1,2),o}\\\hline P_1&47/288&7/64&47/576&47/576&47/576&47/144\\ P_2&1/12&1/72&1/24&1/18&1/24&7/36\\ P_3&1/144&5/144&1/96&1/288&1/288&1/36\\ P_4&7/96&25/192&13/192&7/192&7/192&5/24\\ P_5&7/24&9/32&17/48&41/96&7/48&25/16 \end{array}$$ Adding the columns gives the stated values. As two independent consistency checks, $\frac23\int_\Omega z=1$ and $\frac23\int_\Omega (x+2y)z=26/9$, the latter agreeing with Lemma 29.2. ◻

Proof. Away from $o=[0:1:0]$ and the quotient point $[0:0:1]$, the displayed curves have only transverse intersections. At the quotient point the degree-two quotient cover has coordinates $(u,t)$, with the two coordinate axes as boundary; the cover is unramified in codimension one. Thus the two coefficient inequalities suffice there.

At $o$ the equations are $u=0$ and $e-a u^2=0$ in a smooth affine plane. Blow up the origin and then the common point of the strict transforms of the parabolas. The resulting exceptional coefficients in the log pullback are $\beta+\sum c_a-1$ and $\beta+2\sum c_a-2$. The second blow-up separates the parabolas because their parameters $a$ are distinct, and the total boundary then has simple normal crossings. The last inequality in the statement bounds the second coefficient by one. Together with $\beta\le1$, it also implies $\beta+\sum c_a\le2$, which bounds the first coefficient by one. Conversely all the inequalities are necessary, either along the prime curves or along the second exceptional divisor, whose valuation has weights $(1,2)$ and discrepancy three. ◻

Proof. Consider the action scaling $t$ and fixing $u,e$. The refined spaces are invariant under this action by Lemma 29.3. For an arbitrary divisorial valuation of $T$, choose in every refined summand a basis compatible both with its valuation filtration and with the weight filtration of this action. Lemma 2.3 applies. Degenerate these bases by the action. The initial bases remain bases of the corresponding weight spaces, and their divisors are supported on $(t=0)$, $(u=0)$, and the curves $C_a$: indeed a homogeneous binary form in $u^2,e$ factors into these curves and a power of $u$.

The maximal normalized coefficients of these prime curves are bounded by the table in Lemma 29.5, up to $O(1/m)$. For an arbitrary basis of a finite-dimensional series, its sum of vanishing orders is bounded above by the sum for a basis adapted to that vanishing filtration. The bounds in the table therefore apply to every initial basis. The weighted order at $o$ is independent of the choice of weight-homogeneous bases and equals $167/72+O(1/m)$. Fix $0<\lambda<216/167$. All the individual coefficients of the normalized initial basis boundary multiplied by $\lambda$ are less than one for sufficiently large $m$, and its weighted order at $o$ is less than $$\lambda\frac{167}{72}+o(1)<3.$$ Lemma 29.6 gives log canonicity of the limit. After clearing the Cartier index two, semicontinuity of log canonical thresholds on the fixed klt surface gives log canonicity before degeneration. Compatibility with the arbitrary starting valuation then proves its discrepancy bound. Taking the asymptotic limit and then letting $\lambda$ increase to $216/167$ proves the surface estimate.

The extraction is plt, with zero different. Its refined series has bounded support $0\le j\le6m$ by its explicit description. It contains an ample series directly: for $0\le j\le m$, every weighted monomial of degree $j$ in $u,t,e$ lifts to a global monomial, since its $u$-exponent plus its $e$-exponent is at most $m$, and its $t$-exponent is at most $m$ plus its $e$-exponent. Hence $W_{m,j}=H^0(T,\mathcal O_T(j))$ in this range. Also $\rho^*L-\eta T$ is ample for sufficiently small positive rational $\eta$, since $L$ is ample and $-T$ is relatively ample. Abban–Zhuang refinement therefore gives $$\delta_p(X)\ge\min\left\{\frac{A_X(T)}{S_X(T)},\delta(T;W)\right\} \ge\min\left\{\frac{18}{13},\frac{216}{167}\right\} =\frac{216}{167}.$$ All orders used in this argument came from the associated graded ring and complete binary spaces. No Zariski decomposition on the weighted threefold $\widetilde X$ was used. ◻

Proof. The ruling is $g=(u=e=x_2=0)$. On the chart $v=x_1=1$, put $t=x_0$. At the generic point of $g$, the functions $u,e$ are regular parameters over $\mathbb C(t)$. Their monomial valuation of weights $(1,2)$ is divisorial, with centre $g$ and log discrepancy $1+2=3$.

Assign the Cox variables the weights $$\operatorname{wt}(u,v,e,x_0,x_1,x_2)=(1,0,2,0,0,2).$$ The initial equation is $$G_0=v^2x_2+u^2x_1+evQ_1(x_0,x_1,0).$$ Its coefficients as a linear polynomial in $x_2$ are coprime, so the initial quotient is a domain and injects into its localization at $vx_1$. The degrees of $v,x_1$ form a basis of the Cox grading lattice; thus trivializing each fixed bidegree by these coordinates preserves injectivity. Setting $v=x_1=1$ eliminates $x_2$ and gives $$\mathbb C[t,u,e,x_2]/ (x_2+u^2+eQ_1(t,1,0))\cong\mathbb C[t,u,e] \hookrightarrow\mathbb C(t)[u,e].$$ The last ring, graded by $\operatorname{wt}(u,e)=(1,2)$, is the associated graded ring at the generic point of $g$. Hence the Cox weight filtration is the geometric $\nu_g$-filtration on every section space, including possible cancellations. The global section-ring presentation is the one established in Lemma 27.2.

For the polynomial Cox ring, let $P(p,q)$ and $M_g(p,q)$ denote the dimension and total weight of bidegree $(p,q)$. A monomial with $e$-exponent $k$ has $u,v$-degree $p-k$ and base degree $q+k$, so $$\begin{aligned} P(p,q)&=\sum_{k=0}^p(p-k+1)\binom{q+k+2}{2},\\ M_g(p,q)&=\sum_{k=0}^p(p-k+1)\binom{q+k+2}{2} \left(\frac p2+\frac{2q}{3}+\frac{13k}{6}\right). \end{aligned}$$ Here the mean contributions of $u$, $e$ and $x_2$ are $(p-k)/2$, $2k$ and $2(q+k)/3$, respectively. Since the relation has bidegree $(2,1)$ and weight two, for $m\ge2$ $$\begin{aligned} N_m&=P(m,m)-P(m-2,m-1) =\frac{(2m+1)(3m^2+3m+2)}2,\\ w_m&=M_g(m,m)-M_g(m-2,m-1)-2P(m-2,m-1)\\ &=\frac{m(167m^3+226m^2+193m+62)}{24}. \end{aligned}$$ For example, $N_2=50$ and $w_2=224$. Taking the limit gives $S_X(\nu_g)=\lim_{m\to\infty}w_m/(mN_m)=167/72$. ◻

Proof. Lemma 29.7 gives both lower bounds. The weight-$(1,2)$ valuation at the smooth point $o=[0:1:0]\in T$ has log discrepancy three and, by Lemma 29.5, expected order $167/72$. It therefore gives the matching upper bound for $\delta(T;W)$. For the threefold, Lemma 29.8 and $p\in g=c_X(\nu_g)$ give $$\delta_p(X)\le\frac{A_X(\nu_g)}{S_X(\nu_g)} =\frac{3}{167/72}=\frac{216}{167}.$$ ◻