Blowing up a component of a reduced fibre

Proof. The assertion is local at the intersection. Choose coordinates with ideals $(x,z)$ and $(y,z)$ for the two curves. The two iterated blowups are obtained by subdividing the positive octant first by $(1,0,1)$ and then by $(0,1,1)$, or in the opposite order. The differing cones form the two triangulations of the relation $$(1,0,0)+(0,1,1)=(0,1,0)+(1,0,1).$$ Both subdivisions are smooth. Their contracted curve has normal bundle $\mathcal O(-1)\oplus\mathcal O(-1)$, and the common contraction is the ordinary double point $uv=zw$. The models agree outside this neighbourhood. This is the ordinary flop. ◻

Proof. The line $\ell=\pi(A)$ meets $C_0$ transversally in one point: $H\cdot A=E\cdot A=1$. Moreover $\ell$ is not contained in the span of $C_0$, since $V_5\cap\langle C_0\rangle=C_0$ scheme-theoretically. Apply Lemma 77.1 to interchange the blowups of $C_0$ and $\ell$.

Let $\widetilde C_0\subset\operatorname{Bl}_{\ell}V_5$ be the strict transform. Projection from $\ell$ restricts to the complete series $|\mathcal O_{C_0}(1)(-r)|=|\mathcal O_{\mathbb P^1}(2)|$. Indeed hyperplanes through $\ell$ restrict on $\langle C_0\rangle$ to all hyperplanes through $r$, because $\ell$ meets this span only at $r$. Thus its image $C_2$ is a smooth plane conic and $\widetilde C_0\to C_2$ is an isomorphism. Also $\Pi\cdot\widetilde C_0=3-2=1$. It follows that $C_2$ meets $\Gamma$ transversally at one point: the pullback of the ideal of $\Gamma$ cuts out this reduced point on $\widetilde C_0$. Interchanging these two blowups gives the second flop and the asserted model.

The fibre of $\Pi\to\Gamma$ through that point maps in $V_5$ to a line meeting both $\ell$ and $C_0$. Its transform on $X$ is contracted by $|H-E|$, so it is $D$. The first flopping curve is the exceptional fibre through $r$. They are disjoint on $Y$. If $p_0\ne r$, their intersection points with $A$ are different. They cannot meet elsewhere: that would make the two image lines meet at both $\pi(p_0)$ and the same point of $C_0$. If $p_0=r$, their tangent directions modulo $T_rA$ are different: $\eta$ is constant on $D$, whereas its restriction to a $\pi$-exceptional fibre is an isomorphism onto a line. Blowing up $A$ therefore separates the two curves in this case also.

The projection relations give the displayed identities. Neither of the second blowup centres is contained in the first exceptional divisor, so $E$ remains the pullback of the exceptional divisor over $C_2$. This proves the last assertion. ◻

Proof. A general quintic del Pezzo section $S\in|B|$ through the fibre is smooth. Its normal sequence along $A$ is $0\to\mathcal O(-1)\to N_{A/X}\to\mathcal O\to0$, which splits. Consequently $F\cong\mathbb F_1$, $F^3=1$ and $(f^*L)F^2=-1$. Since $L$ embeds $X$ and $A$ as an anticanonical line, $f^*L-F$ is globally generated. Thus $f^*L-tF$ is nef for $0\le t\le1$, giving the first cubic.

Each of the two flopping curves has intersection $1-t$ with this divisor. An ordinary flop changes its cube by minus the cube of this intersection. This can also be checked on the common blowup, whose exceptional surface $J$ is $\mathbb P^1\times\mathbb P^1$ with normal bundle $\mathcal O(-1,-1)$. If $d$ is the intersection with the flopping curve, then $J^3=2$, $(\sigma^*D)^2J=0$, $\sigma^*D\,J^2=-d$, and the new pullback is $\sigma^*D+dJ$. Its cube is therefore $D^3-3d^3+2d^3=D^3-d^3$. Hence the second cubic is as displayed. It computes the volume: on $Y^+$ the transforms of $f^*L-F=-K_Y$ and $f^*L-2F=2M-E$ are nef, so their convex combinations are nef for $1\le t\le2$. Here $-K_Y$ remains nef under the crepant flops, and $M-E$ is the pencil of hyperplanes through the plane conic $C_2$.

For $2\le t\le3$, the decomposition on $Y^+$ is $$f^*L-tF=((4-t)M-E)+(t-2)\Pi.$$ The first term is nef and pulled back from $Z$. The second term is fixed: intersection with a general fibre of $\Pi\to\Gamma'$ forces at least this multiplicity in every effective representative. The intersection numbers on $Z$ are $M^3=2$, $M^2E=0$, $ME^2=-2$, $E^3=-4$, proving the last cubic. For $t>3$, pushing forward to $Z$ would make $(4-t)M-E$ pseudoeffective. This is impossible: on a general quadric surface in the pencil $|M-E|$, its restriction is $(3-t)M$, which has negative intersection with a ruling curve. Such ruling curves cover $Z$. Thus the threshold is three. Finally, integration gives $\int_0^3\operatorname{vol}(f^*L-tF)dt=43$. Dividing by $L^3=26$ proves the expected order. ◻

Proof. For $t\le1$ restrict $f^*L-tF$ to $F$. For $1\le t\le2$ restrict its nef transform to $F^+$; this gives $\rho^*(ts+(1+t)f)-(t-1)(e_E+e_D)$. For $t\ge2$, subtract $(t-2)\Pi$ first. Its restriction has class $$\rho^*((4-t)s+(7-2t)f)-(3-t)e_E-e_D.$$ These restrictions give inclusions of the actual graded pieces in the complete spaces of the table. We prove equality by counting their dimensions. The surface $F^+$ is a weak del Pezzo surface of degree six: it is the blowup of $\mathbb F_1$ at two distinct points, at most one on its negative section. The two points may lie on one ruling fibre. Checking these configurations gives only $(-1)$- and $(-2)$-curves and a nef anticanonical class. Each moving divisor in the table is nef, by restriction of the nef threefold divisors above (and directly on $F$ in the first interval). Riemann–Roch and Kawamata–Viehweg vanishing therefore give the complete-space dimensions $$\begin{array}{c|c} 0\le k\le m&k^2/2+km+3k/2+m+1\\ m\le k\le2m&-k^2/2+3km+k/2-m^2+2m+1\\ 2m\le k\le3m&k^2-8km-2k+15m^2+7m+1. \end{array}$$ The formulas agree at the endpoints. To sum without double counting, use the disjoint ranges $[0,m]$, $[m+1,2m]$, and $[2m+1,3m]$. If $d_{m,k}$ denotes the displayed dimension, their exact contributions are $$\begin{array}{c|c|c} k&\sum d_{m,k}&\sum k d_{m,k}\\ \hline 0,\ldots,m&\frac{(m+1)(m+2)(4m+3)}6& \frac{m(m+1)(m+2)(11m+9)}{24}\\ m+1,\ldots,2m&\frac{7m(m+1)(2m+1)}6& \frac{m(m+1)(87m^2+67m+14)}{24}\\ 2m+1,\ldots,3m&\frac{m(8m^2+3m+1)}6& \frac{m(37m^3+22m^2+11m+2)}{12}. \end{array}$$ These follow from the sums of $1,k,k^2,k^3$. Thus, for every $m\ge0$, $$\begin{aligned} \sum_k d_{m,k}&=\frac{(2m+1)(13m^2+13m+6)}6=h^0(X,mL),\\ \sum_k k d_{m,k}&=\frac{m(43m^3+60m^2+38m+9)}6. \end{aligned}$$ This is also the total dimension of the actual associated graded space. All the inclusions must consequently be equalities. The argument proves the complete refinement in every degree, including the two exceptional point configurations. It does not assume generation in degree one or surjectivity of an unverified restriction map. ◻

Proof. Blow down $s$ to $q_0\in\mathbb P^2$, and choose homogeneous coordinates $[x:y:z]$ with $q_0,q_E,q_D$ the coordinate points. The monomials in the residual series of the preceding table are $x^i y^j z^{b-i-j}$, with $$0\le i\le a,\qquad 0\le j\le b-i,\qquad j\le b-c_E,\qquad i+j\ge c_D.$$ Integrals over this polygon, followed by integration in $t$, have normalization $3/13$. The normalized integrals of its area, of $i$, of $a-i$, and of $g$ are respectively $$1,\qquad\frac{73}{104},\qquad\frac{79}{104},\qquad\frac5{52}.$$ Here is an explicit list of integration bounds. For $t\in[0,1]$ use $0\le i\le t$, $0\le j\le1+t-i$. For $t\in[1,2]$ use the two regions $$\begin{array}{c|c} 0\le i\le t-1&t-1-i\le j\le2\\ t-1\le i\le t&0\le j\le1+t-i. \end{array}$$ For $t\in[2,3]$ use $$\begin{array}{c|c} 0\le i\le3-t&1-i\le j\le4-t\\ 3-t\le i\le1&1-i\le j\le7-2t-i\\ 1\le i\le4-t&0\le j\le7-2t-i. \end{array}$$ These bounds give all four displayed integrals directly.

For the fibres through $q_E,q_D$, integrate the exponents $b-i-j$ and $j$, giving $21/26$ and $89/104$. For another fibre, the binary coefficient of $x^i$ has residual degree $U(i)=b-i-(c_E-i)_+-(c_D-i)_+$ in $y,z$. Its contribution to the curve moment is $U(i)^2/2$; integration gives $37/52$.

For the point refinement on any of these fibres, the permitted exponents $i$ form an interval $[l,h]\subset[0,a]$. The two endpoint moments of its complete binary space are $\int_l^h i\,di$ and $\int_l^h(a-i)\,di$. A different point contributes $(h-l)^2/2$, which is no larger than either endpoint moment. Integration therefore bounds every point moment of the residual series by $79/104$. This argument includes the prescribed orders at $q_E,q_D$; they are the fixed endpoint orders in these intervals. The fixed divisor $gR$ contributes at most $2\cdot5/52$, since $R$ contains no ruling fibre and $R\cdot f_b=2$. Thus the total is at most $99/104$. The surface flag inequality now gives the claimed threshold, since the first moment is at most $89/104$ and both surface and curve are smooth with zero boundary. ◻

Proof. Apply Abban–Zhuang refinement to the ordinary extraction $F$. Its discrepancy ratio is $2/(43/26)=52/43$. The refined surface threshold is at least $104/99$ by the preceding lemma. Their minimum is $104/99$. Since the estimate holds on all of $F$, inversion of adjunction gives the asserted bound along all of $A$, including $A\cap E$ and the node $p_0$. ◻

Proof. In the elementary transform used above, the centre now belongs to the smooth $(1,2)$-curve on $\mathbb F_0$. Its total transform contains the exceptional curve once. On $F$ this gives $f_0$ as a component of $R$ with coefficient one. The remaining component has class $2s+2f$ and contains no ruling fibre.

Blow down $s$. The images of $q_0,q_E,q_D$ are three distinct collinear points. For the residual degree-$b$ plane system in Lemma 77.5, put $c_0=b-a$. After subtracting order $v$ along their line, the dimension density of the restricted binary space is $$d(v)=\left(b-v-(c_0-v)_+-(c_E-v)_+-(c_D-v)_+\right)_+.$$ Use the entries of that lemma. In the three $t$-intervals, the successive expressions for $d(v)$ and their $v$-ranges are $$\begin{array}{c|ccc} t&\text{first range}&\text{second range}&\text{third range}\\ \hline {[0,1]}&t\ (0\le v\le1)&1+t-v\ (1\le v\le1+t)&\\ {[1,2]}&2-t+2v\ (0\le v\le t-1)&t\ (t-1\le v\le1)&1+t-v\ (1\le v\le1+t)\\ {[2,3]}&2v\ (0\le v\le3-t)&6-2t\ (3-t\le v\le1)&7-2t-v\ (1\le v\le7-2t). \end{array}$$ Direct integration gives $(3/13)\int\int v d(v)\,dv\,dt=1$. The fixed component $g f_0$ contributes $5/52$, giving $57/52$. The first extraction still has $A_X(F)-S_X(F)=9/26$; the failed surface inequality does not determine the sign of a threefold beta invariant. ◻