Binary spaces on the weighted exceptional surface

Proof. The original generators after substitution are $$(1,r-\nu s,A,sA),\qquad (1,s,C+\nu s^2,(r-2\nu s)A-(r-\nu s)^2s,AC).$$ Their maximum-weight initial forms are precisely $\mathcal H,\mathcal B$. Products of their initial forms belong to the initial spaces.

A product of degree $m$ in each list has the form $r^i s^j A^k C^l$. Exactly the following nonnegative integral exponent vectors occur: (67.2)$$\label{f217:eq:monomial-cost} l\le m,\qquad i+k\le m+l,\qquad j+l\le m+k,\qquad j+l\le2m.$$ Indeed, write a monomial from the first list as $r^a s^b A^c$, where $b\le c$ and $a+c\le m$. Write one from the second as $r^d s^e A^f C^l$, where $d+f\le l$ and $e+l\le m$. Their product satisfies (67.2). Conversely choose $$c=\max\{0,j-m+l,k-l\},\qquad f=k-c,\qquad d=\max\{0,i-m+c\},$$ $$a=i-d,\qquad b=\max\{0,j-m+l\},\qquad e=j-b.$$ The inequalities (67.2) give $0\le c\le\min\{k,m\}$, $b\le c$, $a+c\le m$, $d+f\le l$, and $e+l\le m$. Thus the indicated factorization exists.

The two weights are $q=i+k+l$ and $v=j+k+l$. Hence $$r^i s^j A^k C^l =r^{i_0}s^{j_0}(A-C)^{n-k-l}A^k C^l.$$ Conditions (67.2) become $l\le m$, $l\ge L$, $k\ge K$ and $k+l\le n$. Every such polynomial is divisible by $A^K C^L$. Conversely, fixing $l=L$ and letting $k$ run from $K$ to $n-L$ gives a basis of $A^K C^L\operatorname{Sym}^{N}\langle A,C\rangle$, since $A$ and $A-C$ are independent linear forms. This proves (67.1). ◻

Proof. Their total dimension counts the integer triples $(q,v,k)$ satisfying $$q,v,k\ge0,\quad q\le3m,\quad 2k-v\ge-m,\quad k-v\ge-2m,$$ $$k\le q,\quad k\le v,\quad 2k-q\le m,\quad 2k-2v+q\le m.$$ Indeed, for fixed $(q,v)$ these inequalities say precisely $K\le k\le\min(q,v)-L$. At $m=1$ the vertices of this polytope are $$(0,0,0),(0,1,0),(1,0,0),(1,1,1),(1,3,1),$$ $$(3,1,0),(3,3,1),(3,3,2),(3,4,2).$$ They are all integral. Ehrhart’s theorem therefore makes the count a polynomial of degree three in $m$. For completeness, summing by $q$ gives $$\begin{array}{c|l|r} m&\text{successive }q\text{-row counts}&\text{total}\\ \hline 0&1&1\\ 1&2,5,3,5&15\\ 2&3,7,12,9,13,9,12&65\\ 3&4,9,15,22,18,24,19,24,18,22&175. \end{array}$$ These four values determine the polynomial as $$4m^3+6m^2+4m+1=(m+1)^4-m^4.$$ This equals the dimension of the actual refinement. The inclusions of the previous lemma are consequently equalities. ◻

Proof. Divide all indices by $m$. To avoid changing the meanings of $q,v$, write $$\ell(q)=\max\{0,(q-1)/2\},\qquad k(v)=\max\{0,(v-1)/2,v-2\},$$ $$n(q,v)=\min(q,v),\qquad d(q,v)=n(q,v)-k(v)-\ell(q).$$ The support is $0\le q\le3$, $0\le v\le4$, $d(q,v)\ge0$. It is divided into the following seven triangles. Each row also records three integrals; these suffice to check the dimension and both point multiplicities directly. $$\begin{array}{c|c|c|c|c} \text{vertices}&n&\int d&\int qd&\int vd\\ \hline (0,0),(1,1),(0,1)&q&1/6&1/12&1/8\\ (0,0),(1,0),(1,1)&v&1/6&1/8&1/12\\ (0,1),(1,1),(1,3)&q&1/3&1/4&1/2\\ (1,0),(3,1),(1,1)&v&1/3&1/2&1/4\\ (1,1),(3,3),(1,3)&q&4/3&7/3&3\\ (1,1),(3,1),(3,3)&v&4/3&3&7/3\\ (1,3),(3,3),(3,4)&q&1/3&5/6&13/12 \end{array}$$ In particular, $\iint d\,dq\,dv=4$, agreeing with the leading coefficient of the exact Hilbert polynomial proved above. Thus the normalization of all limiting moments is $1/4$.

In a complete binary space of dimension $N+1$, the maximum sum of orders of a basis at any specified point of $\mathbb P^1$ is $N(N+1)/2$. Apply this to (67.1). A factor $A-C=rs$ contributes one to each coordinate-line coefficient. Thus the five first moments are $$\frac14\iint d \left(q-n+\frac d2,\ v-n+\frac d2,\ k+\frac d2, \ell+\frac d2,\frac d2\right)\,dq\,dv =\left(\frac{55}{96},\frac{61}{96},\frac{19}{24}, \frac34,\frac13\right).$$ These are integrals of quadratic polynomials on the displayed triangles. For an additional check, before division by four their values are $$\frac{55}{24},\quad\frac{61}{24},\quad\frac{19}{6},\quad3,\quad\frac43.$$ For the point bounds we use the total multiplicity directly, rather than summing the separate prime-coefficient maxima. Every nonzero polynomial of weight $(q,v)$ has ordinary multiplicity $v$ at $P_r$ and $q$ at $P_s$. The table gives $$\frac14\iint vd=\frac{59}{32},\qquad \frac14\iint qd=\frac{57}{32}.$$ Rounding in $K,L$ changes all these expressions by $O(m^{-1})$. ◻

Proof. For a slice of dimension $N+1$, the maximal sum of orders at any point of the binary parameter line is $N(N+1)/2$. Its contribution to the five curve numerators is therefore $$\left((q-n)(N+1)+\frac{N(N+1)}2,\quad (v-n)(N+1)+\frac{N(N+1)}2,\right.$$ $$\left.K(N+1)+\frac{N(N+1)}2,\quad L(N+1)+\frac{N(N+1)}2,\quad \frac{N(N+1)}2\right).$$ The point contributions are $q(N+1)$ and $v(N+1)$. Summation uses the exact slice description, so no rounding estimate or omitted section is involved.

We give a polynomial certificate for these sums. Let $P$ be the three-dimensional lattice polytope in Lemma 67.2, with coordinates $(q,v,k)$. The $A$-coefficient numerator is $\sum_{mP\cap\mathbb Z^3}k$, and the two point numerators are $\sum q$ and $\sum v$. The $C$-coefficient numerator is $\sum(\min(q,v)-k)$. Cutting $P$ by $q=v$ gives lattice polytopes: this also follows directly from the seven-triangle subdivision, since all upper and lower endpoints above its vertices are integral. Thus the weighted form of Ehrhart’s theorem makes these sums polynomials of degree at most four in $m$. In this application it follows already from the ordinary theorem: for a nonnegative integral linear weight $w$ on a lattice cell, introduce one extra coordinate satisfying $0\le z\le w$. The lifted cell is a lattice polytope, and its lattice count minus that of the original cell is the required sum of $w$. Apply inclusion–exclusion to the lattice cells of the subdivision.

For completeness, put $$R=\{(q,v,k_1,k_2):(q,v,k_1)\in P,\ (q,v,k_2)\in P\}.$$ This is a lattice polytope of dimension four. Over each of the seven triangles it is the square $[k(v),n(q,v)-\ell(q)]^2$; the two endpoints at the projection vertices are $$\begin{array}{c|cccccccc} (q,v)&(0,0)&(0,1)&(1,0)&(1,1)&(1,3)&(3,1)&(3,3)&(3,4)\\ \hline \text{lower}&0&0&0&0&1&0&1&2\\ \text{upper}&0&0&0&1&1&0&2&2. \end{array}$$ A square with affine endpoints over a triangle is the convex hull of its endpoint squares over the three vertices. Hence every cell, and therefore every vertex of $R$, is integral. The generic-conic numerator is exactly $$\frac12\left(\#(mR\cap\mathbb Z^4)-\#(mP\cap\mathbb Z^3)\right),$$ since a fibre containing $N+1$ integers contributes $((N+1)^2-(N+1))/2$. It is consequently a polynomial of degree at most four. The two line numerators add respectively $\sum\max(q-v,0)$ and $\sum\max(v-q,0)$ to this polynomial. These too are weighted Ehrhart polynomials on the lattice subdivision by $q=v$.

It now suffices to evaluate the seven sums at five integers. Direct summation of the displayed slice contributions gives $$\begin{array}{c|rrrrrrr} m&L_r&L_s&A&C&C_\kappa&P_s&P_r\\ \hline 0&0&0&0&0&0&0&0\\ 1&7&8&11&10&2&26&27\\ 2&65&74&98&91&27&227&236\\ 3&272&308&401&375&127&921&957\\ 4&781&881&1136&1066&386&2597&2697. \end{array}$$ Because polynomiality and the degree bound have already been proved, these values establish the formulas in the statement for every $m$. Finally the differences $bmh_m-M_m$, in the same order, are $$\frac{m(82m^2+56m+15)}{96},\quad \frac{7m(10m^2+8m+3)}{96},\quad \frac{m(10m^2+8m+3)}{24},$$ $$\frac{m(2m^2+2m+1)}4,\quad \frac{m(8m^2+7m+3)}6,\quad \frac{m(14m^2+8m+1)}{32},\quad \frac{m(10m^2+8m+3)}{32}.$$ They are positive for every $m\ge1$. ◻

Proof. The different on $G$ is zero. Its only singular point is the $\frac12(1,1)$ point $[0:0:1]$. The invariant prime curves are the coordinate lines $L_r,L_s$ and $C_\kappa=(A-\kappa rs=0)$. Every $C_\kappa$ avoids the singular point and passes through $P_r,P_s$. At each of these two smooth points the conics have distinct tangent directions, as follows from their equations $A=\kappa s$ and $A=\kappa r$ on the two affine charts. A coordinate line passes through exactly one of $P_r,P_s$ and has a different tangent direction from the conics there.

Consequently an invariant boundary is log canonical if every prime coefficient is at most one and its total multiplicity is at most two at each of $P_r,P_s$. One blow-up at each smooth point proves this. At the quotient point the quotient cover gives two transverse coordinate axes; their individual coefficients at most one suffice.

By Lemma 67.5, multiplying the finite-level bounds by $64/59$ satisfies these conditions strictly, for every $m\ge1$. For each divisorial valuation choose a basis compatible with its filtration and the action filtration, separately in each original weighted-degree piece. Degeneration gives an invariant basis boundary. The preceding coefficient bounds apply to every such basis, because they are the maximum sums of orders in its weight spaces. They hold at every level by the exact formulas of Lemma 67.5. More explicitly, two finite filtrations of a finite-dimensional vector space admit a common adapted basis. We apply this only to the valuation filtration and the secondary action filtration on each $W_{m,j}$; the original $j$-grading is kept fixed. The independent initial vectors form bases of the initial spaces, which are exactly the binary spaces by Lemma 67.2.

For a basis polynomial $f$ of largest secondary weight $d$, the family $$f_\tau(r,s,A)=\tau^d f(\tau^{-1}r,s,\tau^{-1}A)$$ is polynomial in $\tau$, has the same primary weighted degree as $f$, and specializes to its nonzero maximum-weight initial form. For $\tau\ne0$ it is the transform by an automorphism of $G$. The zero divisors give the invariant basis boundary just considered. These divisors are $\mathbb Q$-Cartier: their doubles are Cartier even at the quotient point. In fact the product of the squares of all basis polynomials defines a relative effective Cartier divisor, nonzero on every fibre, and hence a flat divisor family. Semicontinuity of log canonical thresholds gives log canonicity before degeneration. Since the initial basis was adapted to the chosen valuation, its boundary realizes the maximal level-$m$ order for that valuation. Thus $$\frac{64}{59}S_m(W;T)\le A_G(T)$$ for every prime divisor $T$ over $G$ and every $m\ge1$. For fixed $T$, let $m$ tend to infinity. Taking the infimum over $T$ proves the assertion. The same argument with the finite coefficient given after Lemma 67.5 proves the stated finite-level bound.

This conclusion controls every basis boundary of $W$: for any such boundary $D_m$ and any $T$, one has $\operatorname{ord}_T(D_m)\le S_m(W;T)$. Thus the basis chosen upstairs in the subsequent Abban–Zhuang argument need only be compatible with the extraction and the valuation being tested. It need not also be compatible with the secondary action. ◻