The exceptional surface at an index-three point

Proof. Restriction of global sections to the completed local ring at $p$ is injective. The weighted initial form of a product of two nonzero local functions is the product of their weighted initial forms, because the weighted polynomial ring in the regular parameters is a domain. Thus initial forms of actual sections, and their products, give the asserted algebra inclusion. Each $R_m$ has a finite, exhaustive and separated filtration. Taking its associated graded preserves its dimension; applying a second such filtration does so again, separately in each primary degree. Consequently $\dim B_m=\dim R_m$. The inclusion and the assumed equality of dimensions give $\mathcal A_m=B_m$ for every $m$. Since the inclusion preserves primary degrees, it is an equality in each primary degree. Those dimensions are precisely the multiplicities of the jumping numbers of the original valuation filtration. This proves the last assertion. No generation or flat base-change assertion for the anticanonical ring is required. ◻

Proof. Use the central model from Lemma 93.2, whose centre has equation $ux^3y^2+vy^5=0$. On the chart $u=y^2=1$ of $M$, write the plane coordinates as $x$ and $c$ and the conic equation as $z=c-x^2$. One lift of the graph equation is $t+xc$; a second lift, after multiplication by the other source coordinate, is $x(t+xc)$. On the blow-up chart put $a=(t+xc)/z$ and then $w=a-x$. This gives $t=zw-x^3$. The conic equation times the nine sections of $\mathcal O_M(2,1)$ gives $t^i\{1,x,c\}$ after division by the exceptional equation. The two lifted graph equations, multiplied by the two target coordinates, give $a,xa,ta,txa$. These are sections of $L=2T+3H-E$. Their thirteen polynomials are independent, so they form the entire degree-one space.

These are also the primary initial sections on the original smooth threefold. Indeed its normalized quintics begin with $x^3y^2$ and $y^5$. Higher terms in the first lifted graph equation have primary weight greater than three; higher terms in the second have weight greater than four. Dividing by the exceptional equation, of weight two, therefore leaves the same initial forms $a$ and $xa$. The target coordinate has initial form $zw-x^3$. Thus the claimed inclusion can be checked directly from the original sections, without assuming generation or base change for the entire ring. The secondary weights make $t=zw-x^3$ homogeneous of weight three. Since both $x$ and $a$ occur, we may replace $a$ by $a-x=w$; likewise we replace $ta$ by $tw$. The least-weight terms of $c$ and $xa$ are $z$ and $x^2$, respectively. This gives the thirteen displayed initial sections. Products of these initial sections belong to the initial algebra of $W$, proving the claimed inclusion. ◻

Proof. First straighten a product containing both $z$ and $w$. Its factors include $zt^i$ and $wt^j$ with $0\le i\le2$, $0\le j\le1$. Use $$(zt^i)(wt^j)=(xt^i)(x^2t^j)+t^{i+j+1}.$$ The first term remains a product of two listed generators. The second is also a product of two such generators, since $i+j+1\le4$ and $1,t,t^2$ are generators. Repeating reduces the number of $zw$ pairs, without increasing anticanonical degree. The resulting monomials have $bc=0$. These normal monomials are linearly independent in $\mathbb C[x,z,w,t]/(zw-x^3-t)$: the single defining relation has leading monomial $zw$ for a suitable monomial order.

We next decide which normal monomials are products of at most $m$ generators. Use $k$ factors of type $x^2t^i$ and $a-2k$ of type $xt^i$, together with $b$ factors of type $zt^i$ and $c$ of type $wt^i$. Fill the remaining places with $1,t,t^2$. This is possible exactly when $$\max(0,a+b+c-m)\le k\le\lfloor a/2\rfloor, \qquad s\le2m-k-c.$$ Indeed the last number is the total available $t$-exponent; all integers between zero and this number can be distributed among the factors. Taking the smallest allowed $k$ gives precisely (35.1).

For completeness, count the basis. If $c=0$, put $n=m-b$. There are $0\le a\le2n$, and $$0\le s\le\min(2m,3m-a-b).$$ The contribution for this $b$ is $$(2n+1)(2m+1)-\frac{n(n+1)}2.$$ If $b=0$ and $c\ge1$, put $n=m-c$; the contribution is $$(2n+1)(m+n+1)-\frac{n(n+1)}2.$$ Thus the total is $$\begin{aligned} &\sum_{n=0}^m\left((2n+1)(2m+1)-\frac{n(n+1)}2\right) +\sum_{n=0}^{m-1}\left((2n+1)(m+n+1)-\frac{n(n+1)}2\right)\\ &\quad =\frac{(m+1)(11m^2+16m+6)}6 +\frac{m^2(3m+1)}2=N_m. \end{aligned}$$ Lemma 94.2 now identifies this algebra with the entire secondary initial algebra of the actual refinement. In particular this proves generation in every degree.

The primary weights admit the following independent, all-degree check. For a row with $0\le s\le U$, use $$\sum_{s=0}^{U}(a+2b+c+3s) =(U+1)(a+2b+c)+\frac32U(U+1).$$ On the first branch put $b=m-n$, $0\le n\le m$, and $0\le a\le2n$; then $U=\min(2m,2m+n-a)$. On the second branch put $c=m-n$, $0\le n<m$, and $0\le a\le2n$; then $U=\min(m+n,m+2n-a)$. Splitting both $a$-sums at $a=n$ gives, respectively, $$\frac{m(m+1)(45m^2+61m+23)}6,\qquad \frac{m^2(59m^2+6m-5)}{12}.$$ Their sum is $$\frac{m(149m^3+218m^2+163m+46)}{12}.$$ These are polynomial summations for every $m\ge0$, rather than an extrapolation from finitely many Hilbert values. Dividing the last expression by $mN_m$ recovers $S_X(G)=149/40$. ◻

Proof. A normal monomial has primary degree $a+2b+c+3s$ and secondary weight $a+3c+3s$. Their difference determines $b-c$, and $bc=0$ determines $b,c$. The remaining equation is $a+3s=k$. Substitution in (35.1) gives exactly $\ell\le s\le u$. The normal basis in this component is therefore $$x^A z^b w^c t^\ell\{x^{3(d-i)}t^i:0\le i\le d\}.$$ Since $t=zw-x^3$, its span is (35.2). ◻

Proof. Here is an explicit integration of the normal-basis count. The two branches are $c=0$, called the $z$-branch, and $b=0$, called the $w$-branch. Their common boundary contributes only lower-order terms. Write $y=b/m$ on the first branch and $y=c/m$ on the second, and write $k/m$ again as $k$. The lower and upper limits for $s/m$ are $$\begin{array}{c|c|c} &L&U\\ \hline z&\max(0,(k+2y-2)/3,(k+y-3)/2)&\min(2,k/3)\\ w&\max(0,(k+2y-2)/3,(k+2y-3)/2)&\min(2-y,k/3). \end{array}$$ Set $d=U-L$ in these integrals. The region is $0\le y\le1$, $L\le U$. It has the following four strips on each branch; all integrals below are ordinary integrals $\int_0^1\int dk\,dy$. $$\begin{array}{c|c|c|c} \text{branch}&\text{range of }k&L&U\\ \hline z&[0,2-2y]&0&k/3\\ z&[2-2y,5+y]&(k+2y-2)/3&k/3\\ z&[5+y,6]&(k+y-3)/2&k/3\\ z&[6,7-y]&(k+y-3)/2&2\\ \hline w&[0,2-2y]&0&k/3\\ w&[2-2y,5-2y]&(k+2y-2)/3&k/3\\ w&[5-2y,6-3y]&(k+2y-3)/2&k/3\\ w&[6-3y,7-4y]&(k+2y-3)/2&2-y. \end{array}$$ Integrating the indicated polynomials gives (35.5)$$\begin{array}{c|ccccc} &\int d&\int(k-3U+3d/2)d&\int yd&\int Ld&\int d^2/2\\ \hline z&11/6&9/8&5/8&193/144&53/144\\ w&3/2&23/24&11/24&41/48&5/16. \end{array} \tag{35.5}$$ For example, on the first strip of the $z$-branch, $d=k/3$, so its contributions to the first, second and last columns are $$\int_0^1\int_0^{2-2y}\frac{k}{3}\,dk\,dy=\frac29,\quad \int_0^1\int_0^{2-2y}\frac{k^2}{6}\,dk\,dy=\frac19,\quad \int_0^1\int_0^{2-2y}\frac{k^2}{18}\,dk\,dy=\frac1{27}.$$ The other entries follow by substituting the displayed affine $L,U$ in the same integrals. Floors, ceilings and boundary rows change the order sums by $O(m^3)$, whereas $mN_m=(10/3)m^4+O(m^3)$. Thus multiply the summed columns by $3/10$. The third column contributes to $b$ on the first branch and to $c$ on the second. This proves (35.3); substitution gives (35.4).

As an additional normalization check, the fixed $x$-coefficient is $$\alpha-3h=\frac1{80}.$$ Since $x,z,w,t$ have weighted degrees $1,2,1,3$, the average divisor class obtained from (35.2) has degree $$(\alpha-3h)+2b+c+3s+6h=\frac{149}{40}=S_X(G),$$ in agreement with the independent threefold calculation. ◻

Proof. Away from the three coordinate vertices, the curves of this pencil are disjoint and smooth, and the displayed component bounds suffice. At $[1:0:0]$, only $Z_0$ and $W_0$ occur; they cross transversely.

At $[0:1:0]$, take the index-two cover of the chart $z=1$. It is $\mathbb A^2_{x,w}$ modulo the involution $(x,w)\mapsto(-x,-w)$; the cover is quasi-étale. The curves in question are $x=0$ and $w=\lambda x^3$. Write $$u=a+3\nu_\infty,\qquad v=c+s+2h-\nu_\infty.$$ Three successive ordinary blow-ups separate the branches $w=\lambda x^3$. Their exceptional valuations have weights $(1,i)$, $i=1,2,3$, in $(x,w)$, log discrepancies $1+i$, and orders $u+iv$ on $B$. The component bound gives $u\le1$, and the first inequality of (35.6) gives $$u+3v=a+3c+3s+6h=\alpha+3(c+s+h)\le4.$$ Linear interpolation gives $u+iv\le1+i$ for $i=1,2$ as well. After the third blow-up all distinct branches meet the last exceptional curve at different points; the total transform has simple normal crossings and every coefficient is at most one. Log canonicity descends through the quasi-étale cover.

The chart $w=1$ at $[0:0:1]$ is smooth. Its curves are $x=0$ and $z=\lambda x^3$. The identical three-blow-up calculation, with $b$ in place of $c$, uses the second inequality of (35.6). This verifies every point and every exceptional divisor on a log resolution, and therefore all divisorial valuations over $G$. ◻

Proof. In (35.2), remove the fixed factor. A basis of the remaining complete binary space has total degree $d(d+1)$ on the quotient $\mathbb P^1$. At any prescribed point the sum of its vanishing orders is at most $$0+1+\cdots+d=\frac{d(d+1)}2.$$ This follows immediately from the dimensions of the subspaces with vanishing order at least $i$. Thus $B_m$ has the form in Lemma 94.8, with $$a=\alpha_m-3h_m,\quad b=b_m,\quad c=c_m,\quad s=s_m, \quad h=h_m.$$ All these coefficients are nonnegative by their definitions. In the limit the component bounds in that lemma are $$\frac58,\quad\frac{47}{120},\quad\frac{41}{120}, \quad\frac{69}{80},\quad\frac{49}{240}.$$ The corner bounds, divided by their log discrepancy four, are $$\frac{29}{32},\qquad\frac{151}{160}.$$ The largest of these seven numbers is $151/160$. Multiplication by any fixed $\tau<160/151$ therefore makes all inequalities strict for sufficiently large $m$. The log-resolution lemma applies. ◻

Proof of Theorem 94.1. Fix a divisorial valuation $v$ over $G$. In each $W_{m,j}$ choose a basis compatible with both its $v$-filtration and its secondary weight filtration. Such a basis exists for two filtrations; see Lemma 2.3. Its least-weight terms give a basis in each of the components of $\mathcal A$. Degenerating the sum of its zero divisors under the same one-parameter subgroup gives exactly the basis divisor considered in Lemma 94.9. Lower semicontinuity of log-canonical thresholds in this family gives the same lower bound for the original basis divisor. Since the basis was $v$-adapted, its order at $v$ is $S_m(W;v)$. Hence, for each $\tau<160/151$ and all large $m$, $$\tau S_m(W;v)\le A_G(v).$$ Taking the limit and then letting $\tau$ increase to $160/151$ proves the first assertion. Only two filtrations were split on an already fixed primary piece; no basis compatible with three unrelated filtrations is assumed.

Weighted point blow-up is a plt extraction, with zero different here. Abban–Zhuang adjunction, in the form of [3, Corollary 1.102], now yields $$\delta_p(X)\ge \min\left\{\frac{A_X(G)}{S_X(G)},\delta(G;W)\right\} \ge\min\left\{\frac{160}{149},\frac{160}{151}\right\} =\frac{160}{151}.$$ We have not computed a Zariski decomposition on the weighted blow-up of the threefold. The normal basis (35.1), its exact Hilbert count and the complete binary spaces (35.2) identify the actual refinement. The only further blow-ups in this proof are the explicit surface blow-ups in Lemma 94.8. ◻