Ordinary double points of the fibres

Proof. Choose a basis $q_0=q,q_1,q_2$ of $H^0(\mathcal I_Z(2))$, with linear syzygies $l\cdot q=m\cdot q=0$, where $l=(l_0,l_1,l_2)$ and $m=(m_0,m_1,m_2)$. With suitable signs, the $q_i$ are the maximal minors of these two rows. The initial homogeneous coordinate algebra of $V$ is the Pfaffian algebra of (48.1)$$\label{f214:eq:A1-pfaffian} \begin{pmatrix} 0&h&l_0&l_1&l_2\\ -h&0&m_0&m_1&m_2\\ -l_0&-m_0&0&n_2&-n_1\\ -l_1&-m_1&-n_2&0&n_0\\ -l_2&-m_2&n_1&-n_0&0 \end{pmatrix}.$$ Here $h,U,V,W,n_0,n_1,n_2$ all have $H$-degree one, and their point weights are $0,1,1,1,2,2,2$. Its equations are $hn_i=q_i$ and $l\cdot n=m\cdot n=0$. To identify them with initial equations, translate (43.1) at the three representatives above: three independent quadratic equations have initial forms $hn_i-q_i$ of point weight two; the remaining two have point weight three and give the two displayed linear syzygies. More explicitly, in the ordered net of Lemma 43.3, writing $\xi_i$ for its three normal variables, these two forms are $$\begin{array}{c|cc} \dim\operatorname{Aut}(V)p&\text{first form}&\text{second form}\\\hline 1&-U\xi_2+V\xi_3&V\xi_1-W\xi_2-U\xi_3\\ 2&2V\xi_1-U\xi_2+V\xi_3&(V-U)\xi_1-W\xi_2-U\xi_3\\ 3&W\xi_1-U\xi_2+V\xi_3&V\xi_1-W\xi_2-U\xi_3. \end{array}$$ A change of basis of the net changes only the presentation.

We check that these are all initial equations. Inverting $h$ gives a polynomial algebra in $h^{\pm1},U,V,W$. On $h=0$, the point $(U,V,W)$ lies on the affine cone over $Z$. Away from the cone vertex the syzygy matrix has rank one, and its kernel in the $n_i$ has dimension two. Over the vertex the kernel has dimension three. Thus the boundary has dimension three, while the algebra has dimension four. The height-three Pfaffian resolution makes it Cohen–Macaulay. Therefore $h$ is a nonzerodivisor, and the algebra injects into its domain localization. The same resolution gives the Hilbert series of $V$. Inclusion in the initial ideal and equality of Hilbert functions now prove exactness.

Let $a,b$ be the pencil variables and $e$ the exceptional variable, of degrees $F,F,H-F$, respectively. Choose the nonvanishing pencil section with initial form $h$ and the other with initial form $n_0$. Substituting $h=ae,n_0=be$ in (48.1) gives the initial Cox algebra. Its remaining variables $U,V,W,n_1,n_2,a,b,e$ have point weights $1,1,1,2,2,0,2,0$. This substitution also gives a domain. Indeed, $(h,n_0)$ is a regular sequence in the preceding algebra: on a component of $h=0$ over a point of $Z$, its kernel in the $n_i$ could be contained in $n_0=0$ only if the syzygy row were proportional to $(1,0,0)$. The minor expression would then give $dq_0=0$ at that point, contrary to smoothness of $q_0$. The component over the cone vertex gives no exception. Thus the boundary $e=0$ has dimension four, whereas the substituted Pfaffian algebra has dimension five. The regular sequence $(h,n_0)$ therefore satisfies all hypotheses of Lemma 43.1. That lemma proves primeness and identifies this graph with the full initial algebra of global sections.

The expected order will be computed directly in Lemma 48.3. Alternatively, its value follows by differentiating the weighted Pfaffian Hilbert series: the five equation weights are $2,2,2,3,3$, and the terminal resolution weight is $6$. ◻

Proof. Regard $Z$ as a divisor of degree three on the smooth conic $q=0$. Choose a one-parameter subgroup of $\operatorname{Aut}(q)$ whose repelling fixed point is outside $Z$. Its flat limit is the divisor $3r$ at the attracting fixed point. Every member, including the limit, is a noncollinear curvilinear scheme of length three. Choose bases of the quadratic generators and their linear syzygies over a neighbourhood of the limiting parameter. The Pfaffian construction in Lemma 48.1 has constant weighted bigraded Hilbert function. Hence every piece $W_{m,j}$ has the stated limiting piece as its flat limit; putting $a=e=1$ embeds these pieces into the homogeneous polynomials of degree $j$ on $G$.

Fix a valuation over $G$. In each $W_{m,j}$ choose a basis simultaneously adapted to this valuation and to the weight filtration of the one-parameter subgroup. The initial sections form a basis of the limiting space. Semicontinuity of log canonical thresholds of the corresponding basis divisors shows that a lower bound for the limiting refined threshold is also a lower bound for the original one. This is the same two-filtration argument as in Lemma 2.3. Finally, a change of coordinates on the conic sends $3r$ to the displayed scheme. In the closed-orbit coordinates a smooth member of its net is $U^2+VW+cUV+bV^2$; replacing $U$ by $U+\frac c2V$ and $W$ by $W+(b-c^2/4)V$ puts it in the required form without changing the net. ◻

Proof. The initial algebra is generated, after $a=e=1$, by $$U,V,W,V^2,UV\quad\text{of degree }H, \qquad 1,q=U^2+VW\quad\text{of degree }F,$$ with the homogeneous bookkeeping supplied by $e$. Use the torus with weights $(1,2,0)$ on $(U,V,W)$. In $H^0(X,-mK_X)$ take ordinary point degree $J$ and torus weight $K$. On $W=1$, set $z=V/U^2$. The resulting one-variable space is spanned by (48.2)$$\label{f214:eq:A1-z-generators} z^A(1+z)^B,\qquad \begin{cases} A\ge0,\ B\ge0,\ A\ge K-J,\\ A+2B\ge J-m,\quad A+B\ge J-2m,\\ A-2B\le K-2J+2m,\quad A\le K-2J+4m,\\ 2A+2B\le K. \end{cases}$$ Here $A,B$ are integers. To verify the inequalities, write a monomial as $$U^{K-2A-2B}V^AW^{J-K+A}q^B.$$ Its least $H$-degree using $U,V,W,V^2,UV$ is $J-2B-s$, where $s=\min\{A,\lfloor(K-A-2B)/2\rfloor\}$ is the largest saving supplied by the last two generators. The conditions are precisely that this degree is at most $m$ and, after adding $B$, at most $2m$.

Put $j=J/m$, $k=K/m$. The continuous feasible region is the hexagon $$\Omega=\operatorname{conv}\{(0,0),(1,0),(3,2),(4,4),(4,6),(2,4)\}.$$ The minimum of $A/m$, the minimum of $B/m$, and the maximum of $(A+B)/m$ in (48.2) are (48.3)$$\label{f214:eq:A1-profile} \alpha=\max\{0,k-j\},\qquad \beta=\max\left\{0,\frac{3j-k-3}{4},3j-k-6,\frac{j-2}{2}\right\}, \qquad d=\frac k2.$$ These follow by eliminating $A$ and $B$ in the displayed inequalities. Consequently the space is contained in $z^{m\alpha+O(1)}(1+z)^{m\beta+O(1)} \mathbb C[z]_{\le m\lambda+O(1)}$, where $\lambda=d-\alpha-\beta$. Rounding the endpoints changes its dimension by a bounded number.

Here are the integrals needed below; all are over $\Omega$: (48.4)$$\label{f214:eq:A1-integrals} \begin{gathered} \int\lambda=\frac{10}{3},\qquad \int j\lambda=\frac{65}{8},\qquad \int k\lambda=\frac{75}{8},\\ \int\lambda(\alpha+\lambda/2)=\frac{115}{48},\qquad \int\lambda(\beta+\lambda/2)=\frac{55}{24},\qquad \int\lambda^2/2=\frac56. \end{gathered}$$ For an explicit integration, split first along $k=j$, then along the four regions where one of the affine functions defining $\beta$ is largest. The seven nonempty polygons have vertices $$\begin{array}{c|l} \alpha&\text{vertices}\\\hline 0&(0,0),(1,0),(3/2,3/2)\\ 0&(1,0),(3,2),(7/2,7/2),(3/2,3/2)\\ 0&(3,2),(4,4),(7/2,7/2)\\ k-j&(0,0),(3/2,3/2),(2,3),(2,4)\\ k-j&(3/2,3/2),(7/2,7/2),(4,5),(2,3)\\ k-j&(7/2,7/2),(4,4),(4,5)\\ k-j&(4,5),(4,6),(2,4),(2,3). \end{array}$$ On the first three polygons $\beta$ is respectively $0,(3j-k-3)/4,3j-k-6$; on the last four it is respectively $0,(3j-k-3)/4,3j-k-6,(j-2)/2$. Triangulation now gives (48.4). For example, on the first triangle $\alpha=\beta=0$ and $\lambda=k/2$, giving $\int\lambda=3/16$ and $\int\lambda^2/2=9/256$.

These containing spaces have total dimension $(10/3)m^3+O(m^2)$, exactly the leading dimension of $H^0(X,-mK_X)$, since $(-K_X)^3=20$. Their total excess dimension is therefore $O(m^2)$. All one-variable orders are $O(m)$, so this excess changes every normalized first moment by $O(1/m)$, uniformly in the point of the exceptional curve. Thus the integrals compute the actual asymptotic moments, without any assertion that all containing spaces are equal to the actual spaces in finite degree.

Divide (48.4) by $10/3$. This gives the ordinary degree $39/16$, the weight at $Q$ equal to $45/16$, and the expected orders at $z=0,-1$ and a general finite point equal to $23/32,11/16,1/4$. The quotient point $z=\infty$ has order $K/2-\deg_z$, so its mean is also $1/4$, since $d=k/2$. At $T$ the local weight is $2J-K$, giving $33/16$. The local polynomial is $z^{J-K}$ times the polynomial at $Q$. Only the order at zero changes, by $39/16-45/16=-3/8$, giving $11/32$. The other stated orders follow. ◻

Proof. By Lemma 48.2, it suffices to bound the refined threshold in the closed-orbit model. We claim $\delta(G;W)\ge16/15$. Fix $c<16/15$. In each ordinary-degree piece choose bases simultaneously adapted to an arbitrary valuation over $G$ and to the torus with weights $(1,2,0)$. Degenerating by this torus gives a basis divisor $D_m$ whose components are the coordinate lines and the conics $U^2+\eta VW=0$. It suffices to show that $(G,cD_m)$ is log canonical for all sufficiently large $m$.

At $Q$, blow up with local weights $(1,2)$. Its log discrepancy is three, and its expected order tends to $45/16$. The exceptional curve is $\mathbb P(1,2)$ with different $1/2$ at its quotient point. By Lemma 48.3, the further expected orders at its points are at most $23/32,11/16,1/4$, with $1/4$ at the quotient point. Thus adjunction gives log canonicity over $Q$: indeed $c(45/16)<3$, all smooth-point products are less than one, and $c/4<1/2$. The same argument over $T$ uses $33/16$ in place of $45/16$ and $11/32$ in place of $23/32$.

Distinct invariant conics meet only at $Q$ and $T$; the same is true of their intersections with $U=0$. The remaining coordinate intersection is $R=[1:0:0]$, where $V=0$ and $W=0$ meet transversely. Their maximal coefficients tend to $23/32$ and $11/32$, respectively. The coefficient of $U=0$ is at most twice the quotient-point moment, hence tends to at most $1/2$. All other component coefficients are bounded by the smooth-point moments in the lemma. Therefore $(G,cD_m)$ is log canonical also away from $Q$ and $T$. Semicontinuity and the adapted basis construction prove $\delta(G;W)\ge16/15$.

Finally Abban–Zhuang refinement for the ordinary point blow-up gives $$\delta_p(X)\ge \min\left\{\frac{3}{39/16},\frac{16}{15}\right\} =\frac{16}{15}.$$ All initial algebras and further moments in this proof came from the actual section filtration. No Zariski decomposition on a threefold obtained by blowing up $p$ was used. ◻