The weighted moment at a double fibre in two explicit members

Proof. We give the filtration and the cohomology arguments, since the Hilbert series of an arbitrary initial ideal would not suffice.

Assign the weights $(0,0,1,2,2,2,3)$ to $(A,B,C,D,E,F,G)$. The resulting initial threefold $V_0$ is defined by $$AE-4BD+C^2=AF-BE=AG-CE=BG-CF=CG-4DF+E^2=0.$$ These are the Pfaffians of the skew matrix obtained from $$\begin{pmatrix} 0&A&B&C&tD\\ -A&0&2C&8D&E\\ -B&-2C&0&2E&F\\ -C&-8D&-2E&0&G\\ -tD&-E&-F&-G&0 \end{pmatrix}$$ by setting $t=0$. Its general fibre is isomorphic to $V$: the substitution $(A,B,C,D,E,F,G)\mapsto(A,B,tC,t^2D,t^2E,t^2F,t^3G)$ in the original equations, followed by division by their smallest powers of $t$, gives exactly this family.

Here are direct checks on the special fibre. On $B=1$ it is affine space with coordinates $A,C,F$, since $$E=AF,\qquad D=(A^2F+C^2)/4,\qquad G=CF.$$ On $G=1$ it is affine space with coordinates $D,E,F$, since $$C=4DF-E^2,\qquad A=CE,\qquad B=CF.$$ The subset $B=G=0$ has dimension at most one in projective space. The Pfaffian ideal has codimension three and is Cohen–Macaulay. Consequently there is no extra component or embedded component there, and the ideal is prime. The Hilbert series is that of a quintic del Pezzo threefold; thus these initial equations give the full initial ideal. Equivalently, the Pfaffian resolution gives a flat family over $\mathbb A^1$. The Jacobian criterion on $B=G=0$ shows that $V_0$ has only one singular point, namely $[0:0:0:1:0:0:0]$. On the chart $D=1$, after eliminating $B$ and $F$, its equation is $AG-CE=0$. This point is an ordinary double point.

We next identify the filtration with the required valuation. At the generic point of $R$, use the chart $B=1$. On the original threefold, $A,C,F$ are regular coordinates and $$D=\frac{A^2F+C^2}{4-2AC},\qquad E=AF+2CD,\qquad G=CF-2DE.$$ The weighted valuation in the two normal parameters $C,F$ has weights $1,2$. The displayed initial equations embed in the associated graded ring $\mathbb C(A)[C,F]$, so the global weight filtration on the homogeneous coordinate ring is exactly this valuation filtration. For the stable example the conic-bundle coordinates at the double fibre are $(D-F)/(A+2C)$ and $E/(A+2C)$. Their ideal at the generic point of $R$ is $(F,C^2)$. For the nonstable example use $(D-G)/(B+2E)$ and $F/(B+2E)$, whose ideal is again $(F,C^2)$. Thus this valuation is the exceptional valuation of the normalized blowup of the pulled-back base-point ideal. In particular it has log discrepancy $1+2=3$. The exceptional divisor $E_C$ does not contain $R$, so its valuation is zero. Hence the valuation on $H^0(X,mL)$ is the restriction of the weight filtration on $H^0(V,2mH)$.

The centres extend to the flat families $$\{A+2tC=D-F=E=0\}\subset V_t, \qquad \{B+2t^2E=D-tG=F=0\}\subset V_t.$$ Each fibre of either family is a smooth twisted cubic. For the first family its coordinates satisfy $A=-2tC$, $D=F$, $E=0$ and $C^2=4BF$, $BG=CF$, $CG=4F^2$. For the second family they satisfy $B=-2t^2E$, $D=tG$, $F=0$ and, after substituting into the equations of $V_t$, the relations $AE+C^2+8t^3EG=0$, $AG-CE+8t^3G^2=0$, $CG+E^2=0$. Replacing $A$ by $A+8t^3G$ gives the standard twisted cubic relations. At $t=0$ the centres are $$C_0^\mathrm{s}=\{A=D-F=E=0\},\qquad C_0^\mathrm{n}=\{B=D=F=0\}.$$ Both avoid the unique singular point of $V_0$. The total centre is therefore a relative regular immersion of codimension two, contained in the smooth locus of the ambient family. Blowing it up commutes with base change and gives a flat family $X_t$. More explicitly, each quotient by a power of the relative ideal is flat: its successive quotients are the symmetric powers of the relative conormal bundle on the flat centre. Thus every power of the relative ideal is $t$-saturated in the ambient Rees algebra.

The ideal sheaf of each centre is generated by the three displayed linear forms. Thus the blowup embeds as a graph in $V_t\times\mathbb P^2$. If $D_t=H-E_C$, then $L_t=H+D_t$ is the restriction of $\mathcal O(1,1)$, and is ample. Moreover, $-K_{X_t}=L_t$. The special fibre has only the ordinary double point inherited from $V_0$, and is klt. Kawamata–Viehweg vanishing gives $H^i(X_t,mL_t)=0$ for $i>0$ and $m\geq0$. This verifies cohomology and base change for the spaces whose weights we compute.

We calculate the conormal bundle of the special centre, including its grading. For $C_0^\mathrm{s}$, on $B=1$ take the conormal parameters $a=A$ and $n=F-C^2/4$, of weights $0,2$. On $G=1$ take $E/G$ and $(D-F)/G$. On their overlap, modulo the square of the ideal of the curve, they are $a/C$ and $-4n/C^3$. Consequently $$N^*_{C_0^\mathrm{s}/V_0} =\mathcal O_{\mathbb P^1}(-1)[0]\oplus \mathcal O_{\mathbb P^1}(-3)[2].$$ Here the brackets record the weights of frames at the point with $B\ne0$. For $C_0^\mathrm{n}$, use $B,D$ on $A=1$ and $F/G,D/G$ on $G=1$. Along the curve $E=-C^2$ and $G=-C^3$; the transition functions are $B/C$ and $-D/C^3$, giving the same graded splitting. In both cases $H|_{C_0}=\mathcal O_{\mathbb P^1}(3)$, with section weights $0,1,2,3$.

For completeness, we perform the character calculation. Let $z$ record valuation weight, and $q$ record hyperplane degree. The graded Hilbert series of $V_0$ is $$\frac{1-q^2(2z^2+2z^3+z^4)+q^3(2z^5+2z^4+z^3)-q^5z^7} {(1-q)^2(1-qz)(1-qz^2)^3(1-qz^3)}.$$ Differentiating with respect to $z$ and putting $z=1$ gives $q(10+16q+3q^2)/(1-q)^5$. It follows that $$h^0(V_0,nH)=\frac{(n+1)(5n^2+10n+6)}6, \quad \operatorname{wt}H^0(V_0,nH) =\frac{n(n+1)(29n^2+57n+34)}{24}.$$ The successive quotients of the $m$th infinitesimal neighbourhood of $C_0$, twisted by $2mH$, are $$\operatorname{Sym}^jN^*_{C_0/V_0}\otimes\mathcal O_{C_0}(6m) =\bigoplus_{i=0}^{j}\mathcal O_{\mathbb P^1}(6m-j-2i)[2i], \qquad 0\leq j<m.$$ All these line bundles have nonnegative degree. Their total dimensions and weights are, respectively, $$\sum_{j=0}^{m-1}\sum_{i=0}^j(6m-j-2i+1) =\frac{7m(m+1)(2m+1)}6,$$ $$\sum_{j=0}^{m-1}\sum_{i=0}^j (6m-j-2i+1)(3m-j/2+i) =\frac{m(m+1)(167m^2+83m+2)}{24}.$$ The restriction sequence is exact on global sections. Indeed, $$H^1(V_0,\mathcal I_{C_0}^m(2mH))=H^1(X_0,mL_0)=0.$$ The higher direct images vanish for the blowup of a regular codimension-two centre. The same assertions hold in the family. Thus the initial of the filtered subspace $H^0(V,\mathcal I_C^m(2mH))$ is exactly $H^0(V_0,\mathcal I_{C_0}^m(2mH))$; there is no unproved replacement of an initial power by a power of an initial ideal. Subtracting the two jet formulas from the ambient formulas at $n=2m$ gives the finite identities in the statement. Dividing the weight by $mN_m$ and letting $m$ tend to infinity proves $S_X(T)=297/104$. ◻

Proof. By Lemma 73.10, a double fibre corresponds to a common root of the binary quartic pencil. Move this root to $x=0$. Use the coordinates and weight filtration of the preceding calculation. There are precisely two possible vanishing sequences.

If the sequence is $(1,2)$, write the pencil as $$\langle e_2+b e_1+c e_0,\ e_3+d e_1+e e_0\rangle.$$ Put $k=4c-b^2$. The number $k$ is nonzero, since otherwise the first quartic is a square. By the exterior-square calculation in Lemma 73.11, the ideal of the twisted cubic in $V_5$ is generated by $$\begin{aligned} l_0&=A-bB+(2c-d)C+8eD+2(cd-eb)E,\\ l_1&=D-cF-eG,\qquad l_2=E-bF-dG. \end{aligned}$$ Their weighted initial forms define on $V_0$ the curve $$Z_{b,c}=\{A-bB=D-cF=E-bF=0\}.$$ After eliminating $A,D,E$, its equations are $$C^2=kBF,\qquad BG=CF,\qquad CG=kF^2.$$ It is a smooth twisted cubic, and avoids the node of $V_0$. Both it and the original curve have Hilbert function $3n+1$ in every degree $n\geq0$. Since the full weighted initial ideal contains the displayed ideal and has the same Hilbert function, the two ideals coincide. This verifies the initial centre, including scheme structure.

We record a graded normalization to make the uniformity of the character calculation explicit. On $V_0$ set $$A'=A-bB,\quad E'=E-bF,\quad D'=D-\frac b2E+\frac{b^2}{4}F,$$ and leave $B,C,F,G$ unchanged. This is an invertible graded linear change preserving the five equations of $V_0$. It transforms $Z_{b,c}$ into $\{A'=E'=D'-(k/4)F=0\}$. Choose $a\in\mathbb C^*$ with $a^2k=4$. The further graded automorphism $$(A',B,C,D',E',F,G)\longmapsto (aA',B,aC,a^2D',aE',F,aG)$$ transforms this curve into $C_0^\mathrm{s}$ of the preceding lemma. Thus the pair consisting of the weighted initial threefold and its centre is graded-isomorphic to the pair used there.

If the sequence is $(1,3)$, write $U=\langle e_1,e_3+a e_2+c e_0\rangle$. The three generators are $$B-aC+2cE,\qquad D-cG,\qquad F-aG.$$ Their initial forms define $C_0^\mathrm{n}=\{B=D=F=0\}$. This is again a smooth twisted cubic with the same Hilbert function as the original, so the same argument proves equality with the full initial ideal.

In either case the centre in the flat weighted family stays inside the smooth locus of the ambient family. Its infinitesimal neighbourhoods are flat, because the centre is a relative regular immersion and the graded quotients are its symmetric conormal powers. The powers of the relative ideal are therefore $t$-saturated. The relative blowup has the same klt Fano special fibre, up to graded isomorphism, as one of the two examples already considered. Kawamata–Viehweg vanishing and cohomology and base change identify its global section module with the saturated Rees submodule of $H^0(V_t,2mH)$. In particular no additional initial sections occur.

Finally the filtration is the actual exceptional valuation. At the generic point of $R$, the weight-zero generator $l_0$ is a unit after trivializing $H$. The ideal $(l_1/l_0,l_2/l_0)$ has initial ideal $(F,C^2)$ in the two regular normal parameters. Thus its exceptional valuation has weights $(1,2)$, exactly the valuation used above. The original exceptional divisor meets $R$ in a point and has valuation zero, so passing from $2mH$ to $mL=2mH-mE_C$ introduces no grading shift. The character and the conormal jet calculation of Lemma 74.1 now apply without any change. ◻