The other exceptional fibre

Proof. Use the quotient convention $Z=\mathbb P(\mathcal O^{\oplus2}\oplus\mathcal O(1))$, with tautological class $H$. The restricted projection has degree two and is complete by Riemann–Roch and nondegeneracy of $C$. It is finite flat. Evaluation gives a surjection $$\mathcal O^{\oplus2}\oplus\mathcal O(1) \longrightarrow (r|_{C_Z})_*\mathcal O_{C_Z}(H).$$ Surjectivity holds fibrewise because a subscheme of length two in a plane imposes two independent conditions on linear forms, also when nonreduced. The target has rank two and degree three: its Euler characteristic is $\deg\mathcal O_C(1)=5$, and the Euler characteristic of a rank-two bundle on $\mathbb P^1$ is its degree plus two. The kernel is consequently $\mathcal O(-2)$. This gives the unique divisor $R_Z$ of the stated class. It is the projective bundle of the rank-two quotient $\mathcal V$.

The bundle $\mathcal V$ is generated by global sections and has degree three, so it splits as $\mathcal O(a)\oplus\mathcal O(3-a)$ with $a=0$ or $1$. The curve $C_Z$ is a Cartier divisor on the smooth ruled surface $R_Z$. Its intersections with $h$ and $H$ are two and five, respectively, so its class is $2H-h$, which also equals $-K_{R_Z}$. If $a=0$, the surface is $\mathbb F_3$ and this anticanonical class has intersection $-1$ with its negative section. It would contain that section, contrary to the irreducibility of $C_Z$. Thus $a=1$, giving $R_Z\cong\mathbb F_1$ and the displayed classes.

Finally $F_B=H-h$ restricts to $c_R+f_R$. A member of this linear system is either a smooth positive section or the reduced sum of the negative section and one ruling fibre. This follows equally from the realization of $\mathbb F_1$ as the blow-up of a point of $\mathbb P^2$: the system is the pullback of lines in $\mathbb P^2$. ◻

Proof. The divisor $F_B|_{R_Z}$ has at most two components, as described above. Viewed on $F_B\cong\mathbb P^1\times\mathbb P^1$, it has class $f+2g$, where $H|_{F_B}=f$ and $h|_{F_B}=g$. If it is reducible, its components have classes $g$ and $f+g$. Each has degree one over the ruling whose fibres are contracted in the order-of-blow-up exchange. Thus their images on $F$ are sections, and neither image is a ruling fibre.

At a smooth point of $F_B|_{R_Z}$, the intersection scheme of $C_Z$ with $F_B$ is contained scheme-theoretically in this smooth curve. Resolving a centre of length $k$ therefore gives successive multiplicities $1,\ldots,1$. At a node in the length-two case, the multiplicities are $2,0$. The difference after subtracting the exceptional Cartier divisor of the centre is then only the $(-2)$-curve contracted to the $A_1$ singularity on the intermediate surface. It is also contracted on the other side and adds no component to $R_F$.

For a length-three centre at a node, use coordinates $(v,w)$ on $R_Z$ in which $F_B|_{R_Z}$ is $vw=0$. The component corresponding to the negative section $c_R$ has intersection one with $C_Z$. Consequently $C_Z$ is transverse to that component and tangent to the fibre component; the intersection orders are one and two. Analytically the situation is $C_Z:\ w=v^2$. On $F_B$, its intersection ideal is therefore $(w-v^2,v^3)$. The multiplicities of the nodal divisor at the successive three point blow-ups are $2,1,0$. After subtracting the exceptional divisor of the length-three centre, the only additional curves are the two $(-2)$-curves in the $A_2$ chain. Both are contracted on either side. There is again no additional component in $R_F$.

The elementary-transform class calculation gives $R_F\sim2s+2f$. Since $R_F\cdot s=0$, a point of $R_F$ on $s$ forces $s$ to be a component. The other component has class $s+2f$ and degree one over the ruling, so it is a smooth section. Its intersection with $s$ is one. In the stated nonreduced configurations the centre is transverse to $s$, and the condition that the total fixed curve contains the centre forces its residual section through $A$. The simple centre $B$ is off $s$, so it lies on $T$ as well.

For clarity, the precise final assertion can also be read from the common resolution in the triple case. Start on $F_B$ and write $e_1,e_2,e_3$ for the total exceptional classes. The fibre component has strict transform $g-e_1-e_2$, and the other component has strict transform $f+g-e_1$. To obtain $F$, contract, in order, $$f-e_1,\qquad e_1-e_2,\qquad e_2-e_3.$$ Their images are respectively the negative section and $T$. Viewed as three blow-ups from $F$, the total exceptional classes are $f-e_3,f-e_2,f-e_1$. The negative section passes through just the first centre, whereas $T$ passes through the first two and not the third. This proves the asserted intersection lengths one and two with the curvilinear centre of length three. ◻

Proof. The normal bundle of $\ell'$ is $\mathcal O\oplus\mathcal O(-1)$. Exchange the blowups of $C\subset\mathbb P^3$ and $B\subset\mathbb P^3$. This gives $Y'$ after the flops of the three fibres counted with their scheme-theoretic multiplicities. Put $Z=\operatorname{Bl}_B\mathbb P^3$, and denote by $C''$ the strict transform of $C$ in $Z$. Thus $Y'=\operatorname{Bl}_{C''}Z$. Write $H$ for the pullback of a plane and $h=H-F$ for the fibre class of the projection $Z\to\mathbb P^1$. Then $Z=\mathbb P_{\mathbb P^1}(\mathcal O^{\oplus2}\oplus\mathcal O(1))$, $H^3=H^2h=1$, and $h^2=0$. Moreover, $H\cdot C''=5$ and $h\cdot C''=2$.

The line through each length-two fibre of $C''\to\mathbb P^1$ sweeps a surface $R_Z\in|H+2h|$. Indeed, the kernel of the evaluation map from $\mathcal O^{\oplus2}\oplus\mathcal O(1)$ to the rank-two bundle $f_*\mathcal O_{C''}(H)$ is $\mathcal O(-2)$. Its quotient has degree $3$. It is globally generated, so its splitting is either $(0,3)$ or $(1,2)$. On $R_Z$ the curve $C''$ has class $2H-h$. The first splitting would give negative intersection with the minimal section, contradicting the irreducibility of $C''$. Hence $R_Z\cong\mathbb F_1$ and $C''=-K_{R_Z}$. Let $J$ be the exceptional divisor of $Y'\to Z$, and put $$A=2H+h-J,\qquad R=H+2h-J.$$ The divisor $A$ is globally generated. On the original model $Y$ it is $H_Q-F$: the pullback of the maximal ideal of $q(\ell')$ is the ideal of $\ell'$, and hence $Y$ maps to $\operatorname{Bl}_{q(\ell')}Q$. The hyperplanes through this point give $H_Q-F$. Its degree on every flopped curve is zero, so global generation is preserved on $Y'$.

Put $D_u=\omega^*(-K_X)-uF$. For $0\le u\le1$ this is nef: $D_1=H+(H_Q-F)=H+A$ is globally generated on the original model, and $D_0$ is nef. On $Y'$ we have $D_u=(4-u)H+uh-J=A+(2-u)H+(u-1)h$, which is nef for $1\le u\le2$. For $2\le u\le3$, its positive and fixed parts are $$P_u=(3-u)A+h,\qquad N_u=(u-2)R.$$ The fibres of $R\to\mathbb P^1$ have $R$-degree $-1$ and $D_u$-degree $2-u$, proving that this fixed coefficient is necessary. The remaining divisor is nef. Restriction to a general fibre over $\mathbb P^1$ also shows that the pseudoeffective threshold is $3$.

Direct intersection gives $A^3=1$ and $A^2h=2$. Consequently, $$\operatorname{vol}(D_u)= \begin{cases} 24-3u^2-u^3,&0\le u\le1,\\ 21+9u-12u^2+2u^3,&1\le u\le2,\\ (3-u)^3+6(3-u)^2,&2\le u\le3. \end{cases}$$ The second polynomial can also be computed directly on $Y'$ using $\deg N_{C''/Z}=17$; thus it does not require an assumption that the three flopped curves are distinct. Dividing the three integrals by $24$ gives $91/96$, $7/12$, and $3/32$, whose sum is $13/8$. ◻

Proof. A monomial basis of $H^0(F,as+bf)$ near $a$ consists of $x^iz^l$, where $0\le i\le a$ and $0\le l\le b-a+i$. Set $d=i+l$. The condition at $b$ is $l\le b-k$, where $k$ is the power of $I_\Xi$. For fixed $d$, put $$L=\max\{0,\lceil(d-b+a)/2\rceil,d-b+k\},\quad U=\min\{a,d\},\quad r=\max\{0,2k-d\}.$$ Writing $x=z+w$, the double-point condition says that the polynomial in $x/z$ is divisible by $(x/z-1)^r$. Therefore its dimension is $\max\{0,U-L+1-r\}$. This also proves that no cancellation has been omitted.

Here $(a,b,k)$ equals $(j,m+j,0)$, $(j,m+j,j-m)$, or $(4m-j,5m-j,m)$ in the three successive chambers. For the limits set $u=j/m$, $v=d/m$, omit the ceiling and the constant $1$, and write $D=(U-L-r)_+$. The normalization is $\int\!\int D\,dv\,du=4$, in agreement with $h^0(X,-mK_X)=4m^3+6m^2+4m+1$. Put $e=(u-2)_+$. The moment functions, to be integrated against $D/4$, are respectively $$\begin{array}{c|c} \operatorname{Bl}_aF&v+2e\\ v_b&a+b-v+2e\\ s&(L+U-r)/2+e\\ \text{positive section}&a-(L+r+U)/2\\ f_a&v-(L+r+U)/2\\ f_b&b-a+L+U-r-v\\ T_1&(r+U-L)/2+e. \end{array}$$ Integration over the linear chambers gives the displayed fractions. For an orbit curve $T_\kappa$ different from the preferred tangent $T_1$, the moving moment is $79/192$; adding the fixed contribution, when present, gives at most $97/192$. Thus $55/96$ is a valid common bound. ◻

Proof. Fix $\lambda<1$, and choose a basis compatible with an arbitrary divisorial valuation and with the indicated one-parameter subgroup. Multiply its normalized boundary by $\lambda$. Its limit is invariant. For all sufficiently large divisible levels, the strict inequalities below follow from Lemma 64.6. Every invariant prime curve is one of the four coordinate curves or an orbit closure $T_\kappa$. Their coefficients are strictly less than $1$ by Lemma 64.6. At $a$, the non-coordinate orbit curves are ordinary distinct lines; one ordinary blowup separates them, and its log discrepancy is $2$. Their total order is strictly less than $2$. At $b$ they are the parabolas $y=\kappa^{-1}t^2$. Two ordinary blowups resolve them; the final exceptional valuation has weights $(1,2)$, discrepancy $3$, and expected order $9/4<3$. The first exceptional valuation has expected order at most $$\frac12\left(\frac94+\frac{11}{12}\right)=\frac{19}{12}<2.$$ Indeed, at $b$ the orders of every orbit parabola and the positive section are doubled by the weight-$(1,2)$ valuation, whereas the order of $f_b=(t=0)$ stays one. This proves the preceding exceptional inequality. There are no further intersections requiring a test. Thus the limiting basis boundary is log canonical. Semicontinuity, followed by the limit over divisible levels and then $\lambda\to1$, proves the assertion. ◻