Double conic fibres

Proof. Let $\Pi$ be the plane of the conic fibre. The quadratic form defining $Q$ has rank one on $\Pi$, with radical the vector subspace corresponding to $\Lambda$. Since the ambient quadratic form is nondegenerate, $\Pi=\Lambda^\perp$. A Witt basis gives the displayed equation of $Q$. The two independent linear equations of $\Pi$ are the coefficients of $y_0$ in the incidence equations. Changing their basis gives (41.1).

On the chart $y_0=1$, eliminate $z_0,z_1$ near $\Lambda$. The first nonzero transverse terms of the conic equation are $$x^2+q_1(s)u+q_2(s)v.$$ If $q_1,q_2$ vanished at the same point of $\Lambda$, all first derivatives of this equation would vanish there. This contradicts smoothness of $X$. Thus $D$ is the graph of the morphism $$[s_0:s_1]\longmapsto[-q_2(s):q_1(s)],$$ which has degree two.

The forms $R_0,R_1$ have no common component. A common component of bidegree $(1,0)$ would give a common linear factor of $q_1,q_2$. A common component of bidegree $(0,1)$ would make these quadratic forms proportional; they would again have a common zero. If the two $(1,1)$ forms were proportional, their linear combination $s_0R_0+s_1R_1$ would have a factor of bidegree $(1,0)$, with the same consequence. These are all possibilities. Their intersection therefore has length two. The identity $q=s_0R_0+s_1R_1$ proves the scheme-theoretic inclusion in $D$. ◻

Proof. The local and global constructions are the same as in Lemma 18.1; we recall them. Where $q_1$ is nonzero, eliminate $u$ from the local equation. Then $\mathcal J=(x^2,v)$, so this is the transverse weighted blow-up with weights $(1,2)$. Its charts are $v=x^2r$ and $x^2=vr$. The first is smooth, while the second is an $A_1$ surface singularity times a smooth curve. In particular, the blow-up is normal.

Flat base change for the conic bundle gives $$X^\sharp=X\times_{\mathbb P^2}\mathop{\mathrm{Bl}}_y\mathbb P^2.$$ Its exceptional scheme is $(2\Lambda)\times\mathbb P^1$, so its reduced support is the stated product. The exceptional divisor of the blown-up base pulls back to $2F$, giving $2F|_F=-f$. Since $H\Lambda=1$ and $B\Lambda=0$, one has $\omega^*L|_F=h$.

On the smooth chart, $dx\wedge dv=x^2dx\wedge dr$, so the discrepancy coefficient is two and $A_X(F)=3$. The other chart has the following index cover for $F$ $$x=ab,\qquad v=a^2,\qquad r=b^2,\qquad (a,b)\longmapsto(-a,-b).$$ The exceptional divisor is the image of $a=0$. This quotient description proves that the pair is plt and that its different has coefficient $1/2$ on the singular-locus curve. The latter curve is precisely $D=(q=0)$. Finally, $-F$ is relatively ample, because $-2F$ is the tautological divisor of the blow-up. These statements verify the plt-extraction hypotheses for Abban–Zhuang refinement. ◻

Proof. Let $I_0$ be the displayed ideal. On $y_0\ne0$, eliminate $z_0,z_1$. The resulting equation is $$x^2y_0+q_1(s)u+q_2(s)v=0.$$ It is irreducible: as a polynomial linear in $u,v$, its coefficients are relatively prime, because $q_1,q_2$ have no common factor. This open set is therefore integral of dimension five.

We check the boundary $y_0=0$. Write $(R_0,R_1)^t=M(s)(u,v)^t$, where $M$ is a two-by-two matrix of linear forms in $s_0,s_1$. Its determinant is not identically zero. Otherwise the two $(1,1)$ forms would have a common component, contrary to Lemma 41.1. Also, $M(s)$ cannot have rank zero at a nonzero $s$, since then $q_1(s)=q_2(s)=0$.

Where $\det M(s)\ne0$, the variables $u,v$ vanish. The remaining quadric in $s_0,s_1,x,z_0,z_1$ gives dimension four. Where $s\ne0$ and $\det M(s)=0$, the $s$-variables have dimension one and the kernel in $(u,v)$ has dimension one. The quadric leaves dimension two in $(x,z_0,z_1)$, again giving total dimension four. Finally, when $s_0=s_1=0$, the quadric forces $x=0$, and the remaining four variables give dimension four. Thus the boundary has dimension at most four.

The ideal $I_0$ has height three and is generated by three elements in a polynomial ring. It is a complete intersection, hence Cohen–Macaulay and unmixed. The boundary calculation shows that every minimal component meets the integral open set $y_0\ne0$. There is only one such component, and it is generically reduced. The absence of embedded components then proves that the entire ring is a domain.

The three equations in $I_0$ are the lowest-weight terms of the original equations. Thus $I_0$ is contained in their actual weight initial ideal. Both complete intersections have the same bigraded Hilbert series, since their bidegrees are $(2,0),(1,1),(1,1)$. Equality of every bigraded Hilbert function proves equality of the ideals.

On a chart $s_i y_0\ne0$, further restrict to where one of $q_1,q_2$ is nonzero and eliminate the corresponding base parameter. This weighted filtration is then the transverse filtration with weights one on $x$ and two on the remaining base parameter, namely $\mathop{\mathrm{ord}}_F$. Localization is injective because the associated ring is a domain. Therefore the same identification holds for global sections of every bidegree. This proves (41.2) for the actual divisorial filtration. ◻

Proof. Use the exact initial algebra of Lemma 41.3, set $y_0=1$, and eliminate $z_i=R_i$. The quadratic relation reduces every section to $x^\epsilon$ times a polynomial in $s_0,s_1,R_0,R_1,u,v$. To see the ideal power explicitly, a monomial containing $x^{2a+\epsilon}$ and $c$ factors among $z_0,z_1$ becomes, up to sign, $x^\epsilon q^a$ times $c$ factors among $R_0,R_1$. If its degree in the original base variables $u,v$ is $d$, then $$r=a+c+d,\qquad d\le m.$$ Because $q=s_0R_0+s_1R_1$ belongs to $(R_0,R_1)$, the resulting polynomial belongs to $(R_0,R_1)^k$. Its bidegree on $F$ is $(m-\epsilon,r)$.

For $r\le m$, all polynomials of this bidegree occur, by using only $s_0,s_1,u,v$ and the required powers of $y_0$. For $r>m$, expand every power of $q$ using $q=s_0R_0+s_1R_1$. The span is exactly $$(R_0,R_1)^k\, H^0\bigl(F,\mathcal O_F(m-\epsilon-k,m)\bigr),\qquad k=r-m.$$ The reverse inclusion follows by lifting the $R_i$ to $z_i$ and using $r-k=m$ base factors. The total number of $R_i$ factors cannot exceed $m-\epsilon$, proving the vanishing assertion.

Set $M=m-\epsilon$. Since $Z$ is a complete intersection of two $(1,1)$ curves, its ideal power has the standard resolution $$0\longrightarrow \mathcal O_F(M-k-1,r-k-1)^{\oplus k} \longrightarrow \mathcal O_F(M-k,r-k)^{\oplus(k+1)} \longrightarrow\mathcal I_Z^k(M,r)\longrightarrow0.$$ In the relevant range, $M\ge k$ and $r\ge k$. The first term has vanishing $H^1$, by the cohomology of $\mathbb P^1\times\mathbb P^1$. Hence the generated space above equals all of (41.3). Taking dimensions in this resolution gives (41.4).

Finally, the square of the weight-one factor $x$ has equation $q$, up to sign, on the exceptional surface. Reflexive restriction therefore assigns the zero divisor $D/2$ to its odd factor. This is the same convention as in Lemma 18.5; the half divisor must be retained. ◻

Proof. Write $\rho=r/m$. The leading dimension in (41.4), divided by $m^2$, is $$d(\rho)= \begin{cases} \rho,&0\le\rho\le1,\\ \rho-(\rho-1)^2,&1\le\rho\le2. \end{cases}$$ The parity shift changes each dimension by only $O(m)$. Since $$N_m=h^0(X,mL)=\frac{10}{3}m^3+O(m^2),$$ and the two parities give $2m\,d\rho$ summands, the normalization factor is $3/5$. In particular, $$\frac35\left(\int_0^1\rho\,d\rho+ \int_1^2\bigl(\rho-(\rho-1)^2\bigr)\,d\rho\right)=1.$$ The extraction weight divided by $m$ is $2\rho+O(m^{-1})$. Therefore $$\begin{aligned} S_X(F) &=\frac35\left(\int_0^1 2\rho^2\,d\rho+ \int_1^2 2\rho\bigl(\rho-(\rho-1)^2\bigr)\,d\rho\right)\\ &=\frac35\left(\frac23+\frac72\right)=\frac52. \end{aligned}$$ The spaces vanish beyond weight $4m$, and $\dim W^F_{m,4m}=(m+1)^2>0$, proving $\tau(F)=4$. This also proves bounded support. Since $L$ is ample and $-F$ is relatively ample, $\omega^*L-tF$ is ample for all sufficiently small positive rational $t$. Its sections supply an ample subseries of the refinement, as required by Abban–Zhuang theory. ◻

Proof. The morphism $D\to\mathbb P^1$ of degree two has two distinct ramification points in characteristic zero. Changes of coordinates on the two factors put it in the form $u=s^2$. The action preserves $D$. Unless $Z$ is exactly the two ramification points, one of these points is absent from its support; choose this point as the repelling point. All of $Z$ then lies in the affine chart of the attracting point.

As a length-two subscheme of the smooth curve $D$, $Z$ has equation $(s-a_1)(s-a_2)=0$, allowing $a_1=a_2$. Its image under the action has ideal $$\bigl(u-s^2,(s-\lambda a_1)(s-\lambda a_2)\bigr).$$ This is a flat family of length-two schemes, with the stated limit. It is a relative complete intersection. Its ideal-power quotients are also flat: filter each quotient by the successive powers of the ideal and use that their quotients are symmetric powers of the relative conormal bundle, locally free over this flat family of schemes. Consequently, a flat limit of sections vanishing in the $k$th power of the moving ideal still vanishes in $\mathcal I_{Z_0}^k$.

A monomial $s^iu^l$ belongs to $(u,s^2)^k$ precisely when $\lfloor i/2\rfloor+l\ge k$, equivalently $i+2l\ge2k$. This gives (41.6). The flat limit in the Grassmannian retains its original dimension and is invariant under the one-parameter action. The complete space on the right can have larger dimension; no equality is asserted. ◻

Proof. Put $a=i/m$, $b=l/m$, and $\rho=r/m$. The limiting polygon for (41.6) is $$\Omega_\rho= \{0\le a\le1,\ 0\le b\le\rho,\ a+2b\ge2\max(0,\rho-1)\}.$$ For a polynomial $P(a,b,\rho)$, define $$J_P(\rho)=\int_0^1\int_0^\rho P\,db\,da.$$ For $1\le\rho\le3/2$, the removed corner is a triangle, and its moment is $$T_P(\rho)=\int_0^{2(\rho-1)} \int_0^{\rho-1-a/2}P\,db\,da.$$ For $3/2\le\rho\le2$, the triangle is truncated by $a=1$, and its moment is $$U_P(\rho)=\int_0^1\int_0^{\rho-1-a/2}P\,db\,da.$$ Thus every constant in the table is (41.8)$$\label{f213:eq:double-moment-formula} \frac35\left(\int_0^1J_P\,d\rho+ \int_1^{3/2}(J_P-T_P)\,d\rho+ \int_{3/2}^2(J_P-U_P)\,d\rho\right).$$ The relevant polynomials are $1,a,1-a,b,\rho-b,a+2b$, and $(1-a)+2(\rho-b)$. Here are the three separate contributions, including the factor $3/5$: $$\begin{array}{c|ccc} P &[0,1]&[1,3/2]&[3/2,2]\\\hline 1&3/10&7/20&3/8\\ a&3/20&29/160&1/5\\ 1-a&3/20&27/160&7/40\\ b&1/10&15/64&67/160\\ \rho-b&1/10&33/160&19/80\\ a+2b&7/20&13/20&83/80\\ (1-a)+2(\rho-b)&7/20&93/160&13/20. \end{array}$$ Their sums give the asserted values. Lattice-boundary and parity errors vanish after the stated normalization.

For $\rho>3/2$, the area of $\Omega_\rho$ exceeds the actual leading dimension $d(\rho)$ by $$\operatorname{area}(\Omega_\rho)-d(\rho)=(\rho-3/2)^2.$$ Every monomial in the envelope has weight at least $2k$ at $o_0$. An invariant subspace has no more vectors of any given weight than its ambient monomial space. Hence its total weight is bounded by the total ambient weight minus $2k$ times the dimension deficit. This argument allows repeated weights and non-monomial eigenvectors. After summation, the subtraction is at least $$\frac35\int_{3/2}^2 2(\rho-1)(\rho-3/2)^2\,d\rho=\frac7{160}.$$ It follows that the actual limiting basis has weighted-order sum at most $$\frac{163}{80}-\frac7{160}=\frac{319}{160},$$ proving (41.7).

Finally enlarge the spaces once more to all of $H^0(\mathcal O_F(m-\epsilon,r))$. For a curve $T$ of bidegree $(2,1)$, the maximal sum of its orders is $$\sum_{l\ge1}h^0\bigl(\mathcal O_F(m-\epsilon-2l,r-l)\bigr).$$ Its normalized limit is $$\begin{aligned} c_T&=\frac35\int_0^2\int_0^{\min(1/2,\rho)} (1-2z)(\rho-z)\,dz\,d\rho\\ &=\frac35\left( \int_0^{1/2}\left(\frac{\rho^2}{2}-\frac{\rho^3}{3}\right)d\rho +\int_{1/2}^2\left(\frac\rho4-\frac1{24}\right)d\rho\right) =\frac{81}{320}. \end{aligned}$$ Orders of zero divisors are nonnegative, so enlargement gives a valid upper bound with the original normalization $mN_m$. ◻

Proof. First suppose that $Z$ is not the union of the two ramification points, and use Lemma 41.6. Fix a divisorial valuation $\nu$ over $F$. In each space (41.3), choose a basis compatible both with $\nu$ and with the weight filtration of the action. The two-filtration lemma permits this. Include the odd fixed term $\epsilon D/2$ in its normalized basis divisor $\Gamma_m$. Its total coefficient is $O(m^{-1})$.

Acting on the surface degenerates this basis to an invariant basis of the limiting subspace. Denote the resulting divisor by $\Gamma_m^0$. Repeated weights may give non-monomial initial sections. This causes no difficulty: as recalled in Lemma 18.7, every invariant prime curve is a coordinate fibre or a curve $$D_c=(u=c s^2),\qquad c\in\mathbb C^*.$$ These latter curves all have bidegree $(2,1)$. An effective boundary supported on these curves is log canonical if every prime coefficient is at most one and its weighted orders at $o_0,o_\infty$ are at most three. To recall the local calculation, at either point the curves are the coordinate axes and distinct parabolas $u=c s^2$. Two successive point blow-ups resolve them. If their coefficients are $a,b,c_1,\ldots,c_l$, the last exceptional inequality is $$a+2b+2\sum_i c_i\le3.$$ Together with $a\le1$, it implies the first exceptional inequality $a+b+\sum_i c_i\le2$. Away from these two points the curves have simple normal crossings.

The four coordinate coefficients of $\Gamma_m^0$ have limiting upper bounds $$\frac{17}{32},\quad\frac{79}{160},\quad \frac{241}{320},\quad\frac{87}{160}.$$ Indeed, a sum of orders of any basis of a subspace is bounded by the maximal sum for its containing monomial envelope. Every $D_c$ coefficient is at most $81/320$ in the limit, by the last assertion of Lemma 41.7. The weighted sums at the two ramification points are bounded by $319/160$ and $253/160$, respectively. These are weights of invariant sections, so the dimension correction applies to the present basis, even when its eigenvectors are not monomials.

Take $0<\lambda<320/319$. All prime coefficients of $$\frac12D+\lambda\Gamma_m^0$$ are less than one for sufficiently large $m$. The largest coordinate coefficient is bounded at the endpoint by $241/319<1$, and the coefficient of $D$ is bounded by $$\frac12+\frac{81}{319}<1.$$ At the two ramification points, the limiting weighted orders are at most $$1+\lambda\frac{319}{160}<3,\qquad 1+\lambda\frac{253}{160}<3.$$ The preceding invariant-boundary calculation proves log canonicity. The action preserves $D$, so lower semicontinuity of log canonical thresholds gives log canonicity of $(F,D/2+\lambda\Gamma_m)$ before degeneration. Since the original basis is $\nu$-compatible, $$S_m(W^F;\nu)=\nu(\Gamma_m) \le\frac{A_{F,D/2}(\nu)}{\lambda}.$$ Pass to the limit, using bounded support and the ample subseries from Lemma 41.5, and then let $\lambda$ increase to $320/319$.

It remains to treat $Z=\{o_0,o_\infty\}$. Its ideal powers are already invariant. Their spaces are spanned by the monomials in a rectangle with equal corner triangles removed at the two opposite points. In normalized coordinates, the conditions are $$a+b\ge\rho-1,\qquad (1-a)+(\rho-b)\ge\rho-1$$ when $\rho\ge1$; there is no condition for $\rho\le1$. Their areas are again $d(\rho)$. Central symmetry gives first-factor coordinate means $1/2,1/2$ and second-factor means $5/8,5/8$. At either ramification point, the weighted-order sum is $$\frac35\int_0^2(1/2+\rho)d(\rho)\,d\rho=\frac74.$$ The bound $81/320$ for a $(2,1)$ curve is still valid. The same invariant-boundary argument therefore gives the stronger lower bound $8/7$, which implies the assertion in this case as well. ◻

Proof. Apply Abban–Zhuang refinement to the plt curve extraction in Lemma 41.2, with its different $D/2$. The local thresholds of the refined series over $p$ are bounded below by its global threshold on $F$. Lemmas 41.5 and 41.9 give $$\delta_p(X)\ge \min\left\{\frac{A_X(F)}{S_X(F)}, \delta(F,D/2;W^F)\right\} \ge\min\left\{\frac65,\frac{320}{319}\right\} =\frac{320}{319}.$$ All positivity and bounded-support hypotheses were verified in Lemma 41.5. ◻