Complete decomposition in transverse position

Proof. Choose a smooth cubic containing $Z$. Its plane section by $\Pi$ is $Z+\ell$, and $C\cdot\ell=p_0$. Thus $C\cap Z=p_0+c_1+\cdots+c_5$ scheme-theoretically. In particular, $Z$ meets $C$ transversely at all six points.

Since $\Pi$ contains $T_{p_0}C$, every line in $\Pi$ through $p_0$ other than $T_{p_0}C$ lifts to a curve through the selected point $p\in f_0$. The line $p_0c_i$ meets $C$ only at those two points, both transversely. An additional intersection would make its $B$-degree zero (or negative). Degree zero would put its lift in the smooth conic fibre through $p$, which cannot contain a line; negative degree contradicts nefness of $B$. Hence $H\cdot\Gamma_i=B\cdot\Gamma_i=1$.

The normal bundle of a line in $\mathbb P^3$ is $\mathcal O(1)^{\oplus2}$. Blowing up $C$ makes one elementary modification at each of $p_0,c_i$. The two quotient directions are distinct: the plane spanned by the line and $T_{p_0}C$ is $\Pi$, whereas $T_{c_i}C$ is transverse to $\Pi$. Two modifications in distinct directions at distinct points yield $\mathcal O^{\oplus2}$. Blowing up $p$ then yields $\mathcal O(-1)^{\oplus2}$. The same conclusion for $Z$ follows from its being a smooth fibre of $\varphi$.

Distinct lines $p_0c_i$ have distinct directions, since three collinear points would give the excluded trisecant through $p$. None is tangent to the smooth conic through both its endpoints. Blowing up $C$ separates the line and conic at $c_i$, and blowing up $p$ separates all their remaining intersections. Their tangent directions differ from $T_pf_0$, so they also avoid the first two centres. ◻

Proof. The pencil of base lines through $\varphi(p)$ has ideal $\mathcal I_Z$. Blowing up $p$ and then the strict transform of $Z$ principalizes this ideal, with total divisor $G+U_0$. The other centres are disjoint from $Z$. Hence $F$ is globally generated.

For nefness, first observe the effective expression $$P_{5/2}=Q+F+\tfrac12G+\tfrac14R+\tfrac12T +\tfrac32U_0+\tfrac12\sum_{i=1}^5U_i.$$ It gives nonnegative intersection with curves outside the displayed divisors. On $R$ the restriction is zero, on $T$ it is $\mathcal O(0,3/4)$, and on each $U_i$ it is $\mathcal O(0,1/2)$, with the ruling conventions of the relevant blow-up. These restrictions are nef.

The transform of $G$ is the plane blown up at $q,q_1$ and six further distinct points. Those eight centres, counting the infinitely near point, lie on the smooth conic cut out by the tangent cone of $\pi^{-1}(\Pi)$ at $p$. Smoothness of this conic follows from the contact order two in (16.1): its local equation is $\eta z+ct^2=0$, $c\ne0$. Write $e_3,\ldots,e_8$ for the six new total-transform exceptional classes. The restriction is $$\tfrac52h-\tfrac34(e_1+e_2)-\tfrac12\sum_{i=3}^8e_i.$$ Its intersection with the strict conic is $1/2$, and its intersections with the exceptional curves are nonnegative. Any other irreducible plane curve of degree $d$, with multiplicities $m_i$ at the eight centres, satisfies $\sum m_i\le2d$ by intersection with that conic. Its intersection with the displayed divisor is therefore at least $5d/2-3(\sum m_i)/4\ge d$. This proves nefness on $G$.

Finally, the intrinsic surface $Q$ is the plane blown up at $p_0$, the point infinitely near it in the direction $T_{p_0}C$, and the five points $c_i$. This follows directly by resolving the curvilinear ideal of length two at $p_0$; the two later curve blow-ups are along Cartier curves on this surface. Write $h,\epsilon_0,\epsilon'_0,\epsilon_1,\ldots,\epsilon_5$ for total-transform classes. The restrictions before the last six blow-ups are $$H|_Q=h,\quad G|_Q=\epsilon_0-\epsilon'_0,\quad R|_Q=0,\quad T|_Q=\epsilon'_0,\quad B|_Q=3h-\epsilon_0-\epsilon'_0-\sum_{i=1}^5\epsilon_i.$$ Also $U_0|_Q=2h-\epsilon_0-\sum\epsilon_i$ and $U_i|_Q=h-\epsilon_0-\epsilon_i$ for $i\ge1$. Substitution gives $P_{5/2}|_Q=(h-\epsilon_0)/2$, a nef pencil class. All curves in the displayed support have now been checked. ◻

Proof. The first two chambers were proved above. On each of the six curves, $D_t$ has degree $2-t$ and the conormal bundle after the point blow-up is $\mathcal O(1)^{\oplus2}$. Restriction to its exceptional quadric forces coefficient at least $t-2$ for $t\ge2$. Thus the first three rows are forced. Nefness on $[2,5/2]$ follows by interpolation between the nef endpoint divisors $P_2$ and $P_{5/2}$.

For $t>5/2$, subtract the already forced third-row part, obtaining a divisor with class $P_t^{\mathrm{old}}$. Let $a$ and $b$ be the additional coefficients of $Q$ and $U_0$ in any effective representative. On $U_0$, one has $P_t^{\mathrm{old}}|_{U_0}=\mathcal O(0,t-2)$ and $Q|_{U_0}=\mathcal O(1,1)$. Removing the two components and restricting to $U_0$ forces $b\ge a$. For a general ruling curve $l$ of $Q$ obtained from a line through $p_0$, the surface formulas in Lemma 16.2 give $$P_t^{\mathrm{old}}\cdot l=5-2t,\qquad Q\cdot l=-2,\qquad U_0\cdot l=1.$$ The remaining effective divisor does not contain $Q$, so $5-2t+2a-b\ge0$. Together these inequalities give $a\ge2t-5$ and $b\ge2t-5$. This proves exactly the extra fixed part $(2t-5)(Q+U_0)$. The positive parts on this interval are linear in $t$; at $t=3$ they equal $F=B-G-U_0$. Interpolation with Lemma 16.2 proves nefness. The fixed-component inequalities give equality of section spaces as before.

The first two cubes are known. Blowing up a $(-1,-1)$ curve on which a divisor has degree $-b$, and subtracting $b$ times the exceptional quadric, adds $b^3$ to its cube. The six disjoint curves therefore add $6(t-2)^3$ in the third chamber.

For clarity, the last cube can also be computed without expanding all intersections on $\widehat Y$. Put $s=3-t$. After removing the fixed plane, the positive part has underlying pullback class $2sH+B-(1+3s)G$, first two exceptional coefficients $3s/2,3s$, coefficient $1-s$ along $U_0$, and coefficient $s$ along each $U_i$, $i\ge1$. The same blow-up formula gives $$\begin{aligned} P_t^3 &= (2sH+B)^3-(1+3s)^3+\tfrac12(3s)^3+(1-s)^3+5s^3\\ &=12s^2-\tfrac32s^3. \end{aligned}$$ At $t=3$ the volume is zero; monotonicity makes it zero thereafter. Finally, the four integrals in (16.2) are $$\frac{63}{4},\qquad\frac{99}{8},\qquad\frac{363}{128},\qquad\frac{61}{128}.$$ Their sum is $503/16$. Divide by $L^3=16$ and apply Lemma 14.1. ◻