Blowing up the first base curve

Proof. The fibre $f_0$ in $E$ has normal bundle $\mathcal O\oplus\mathcal O(-1)$ in $X$: its normal bundle in $E$ is trivial, $E\cdot f_0=-1$, and the normal sequence splits. Blowing up one point of the curve twists both normal summands by $\mathcal O(-1)$. Thus $R_1=\mathbb P(\mathcal O(-1)\oplus\mathcal O(-2))$. On $R_1$ its normal bundle has class $-e-2r$. Restriction to $s=e$ has degree $-1$, as does $N_{s/R_1}$. The normal sequence splits, giving $N_{s/Y_1}=\mathcal O(-1)^{\oplus2}$.

Blowing up a Cartier curve on $R_1$ leaves that surface unchanged and subtracts $s$ from its normal bundle. This gives $R|_R=-2e-2r$ and $T|_R=e$. The other two formulas follow from $T=\mathbb P(N_{s/Y_1})$: $T|_T=\mathcal O(-1,-1)$ and $\tau_2^*R_1|_T=\mathcal O(-1,0)$, so $R|_T=\mathcal O(0,1)$. ◻

Proof. Let $\Pi\subset\mathbb P^3$ be the plane of the conic $\pi(Z)$. It contains $T_{p_0}C$, and the normal direction represented by $p$ lies in $\Pi$. Choose affine linear coordinates $x,t,r$ at $p_0$ such that $\Pi=(x=0)$, $T_{p_0}C=(x=r=0)$, and the selected normal direction has $x/r=0$. Locally $C$ has equations $x=a(t)$, $r=b(t)$, with $a,b\in(t^2)$. A chart of $X$ at $p$ is $$x=a(t)+\eta z,\qquad r=b(t)+\eta, \qquad p=(t,\eta,z)=(0,0,0).$$ Here $E=(\eta=0)$ and $f_0=(t=\eta=0)$.

Consider the quadrics spanned by $x$ times all linear forms, together with $t^2,tr,r^2$. Their base locus in $\mathbb P^3$ is just $p_0$. On the displayed chart their ideal is $$(x,t^2,tr,r^2)=(\eta z,t^2,\eta t,\eta^2).$$ Blowing up $p$ extracts a factor of order two. The only remaining base locus is in the $z$-chart, where the residual ideal is $(\eta_1,t_1^2)$. Blowing up $f=(\eta_1=t_1=0)$ extracts one copy of $R_1$ and leaves the ideal of its negative section. Blowing up that section resolves the ideal. The total exceptional factor is $2G+R+2T$. The same computation along the rest of $f_0$ uses $(\eta,t^2)$; there are no other base points. This proves global generation.

Finally, $\varphi|_{f_0}$ is an isomorphism to a line. Hence $T_pf_0$ differs from the tangent direction of the fibre $Z$. The two strict transforms are disjoint after blowing up $p$, and all subsequent centres lie over $f$. ◻

Proof. First prove that the displayed coefficients are forced. Let an effective divisor numerically equivalent to $\rho^*D_t$ have coefficients $\alpha,\beta$ along $R,T$, respectively, and remove those components. Its restrictions to $R$ and $T$ are effective and have classes $$(2\alpha-\beta)e+(1-t+2\alpha)r, \qquad \mathcal O_T(1-t+\beta,\beta-\alpha).$$ The effective cones of $\mathbb F_1$ and $\mathbb P^1\times\mathbb P^1$ give $\alpha\ge(t-1)/2$ and $\beta\ge t-1$. This proves the fixed-part assertion once nefness is established.

The divisor $P_1=H+B-G$ is globally generated. At the other endpoint, $$P_2=\tfrac12\mathcal A+(B-G).$$ The system $|B-G|$ has base locus exactly the strict transform of $Z$. Indeed, it comes from lines through $\varphi(p)$, and the fibre is smooth. It therefore has nonnegative intersection with every curve other than $Z$. Lemma 15.2 gives the same assertion for $\mathcal A$, and $\mathcal A\cdot Z=2$, $(B-G)\cdot Z=-1$. Thus $P_2\cdot Z=0$ and $P_2$ is nef. Since $P_t=(2-t)P_1+(t-1)P_2$, every $P_t$ is nef.

For later use, $2P_2$ is in fact globally generated. Products from $|\mathcal A|$ and $|B-G|^2$ generate it off $Z$. To cover $Z$, choose a plane section $h\in H^0(X,H)$ whose restriction to $Z$ vanishes exactly at $2p$. Its differential annihilates $T_pZ$. Since $\varphi$ is smooth at $p$, a section of $B$ vanishing at $\varphi(p)$ has the same differential, after choosing nonvanishing local frames. Subtracting the corresponding products gives a global section of $L$ vanishing to order two at $p$ whose strict transform does not vanish anywhere on $Z$. Its square, after removing $R+2T$, supplies the missing section of $2P_2$.

For divisible $m$, the forced coefficients show that every section of $m\rho^*D_t$ is divisible by $mN_t$. Conversely, multiplying any section of $mP_t$ gives a section of $m\rho^*D_t$. As $P_t$ is nef, the volume is $P_t^3$.

Here is the cubic calculation. On the first blow-up, $D_t\cdot f=1-t$, $R_1^3=3$, and $(\tau_1^*D_t)R_1^2=t-1$. Subtracting $aR_1$ changes the cube by $3a^2(t-1)-3a^3=3a^3$. Its intersection with $s$ is $-a$. On blowing up $s$, subtracting $aT$ changes the cube by $3a^3-2a^3=a^3$. The total change is $4a^3=(t-1)^3/2$, proving (15.2). ◻

Proof. The two threefold centres meet the transform of $G$ transversely at $q,q_1$. On the final surface, $R|_{\widetilde G}=e_1-e_2$ and $T|_{\widetilde G}=e_2$. This proves the class formulas. The exceptional valuations of the two surface blow-ups have weights $(1,1)$ and $(1,2)$ in $(u,v)$. The required orders are at least $(j-m)/2$ and $j-m$, respectively; the latter implies the former. This gives (15.3).

We justify asymptotic completeness, including the birationality needed in passing to the image. For rational $1<t<2$, write $P_t=(2-t)P_1+(t-1)P_2$. The divisors $P_1$ and $2P_2$ are globally generated by Proposition 15.3. The complete linear jets in Lemma 14.2 show that the morphism of $P_1$ restricts on $\widetilde G$ to a map recovering its blow-down to $G\cong\mathbb P^2$. It is therefore birational on $\widetilde G$.

Choose an integer $d>0$ with $dP_t=aP_1+b(2P_2)$ for positive integers $a,b$. Products of endpoint sections give a subsystem of $|dP_t|$. On a dense open subset of $\widetilde G$, ratios of these products recover the ratios of the sections of $aP_1$. Hence the morphism of $dP_t$ is also birational on $\widetilde G$. Take its Stein factorization $\psi:Y_2\to V$, with $dP_t=\psi^*A_V$ for an ample line bundle $A_V$, and let $S_V=\psi(\widetilde G)$ have its reduced structure. Restriction from $H^0(V,nA_V)$ to $H^0(S_V,nA_V|_{S_V})$ is surjective for large $n$. Since $\psi|_{\widetilde G}$ is birational, the cokernel of $$\mathcal O_{S_V}\longrightarrow(\psi|_{\widetilde G})_*\mathcal O_{\widetilde G}$$ is supported in dimension at most one. Twisting by $nA_V$ shows that the deficit of the restricted series inside $H^0(\widetilde G,ndP_t|_{\widetilde G})$ is $O(n)$. This is asymptotic completeness in a surface summand. Squaring the nef restricted class gives the stated volume, also equal to $-\frac13\frac{d}{dt}\mathop{\mathrm{vol}}(D_t)$. ◻

Proof. Put $V(t)=\mathop{\mathrm{vol}}(D_t)$. The proven chambers give $$\int_0^2V(t)\,dt=\frac{225}{8},\qquad V(2)=\frac{17}{2},\qquad V'_-(2)=-\frac{21}{2}.$$ The function $V^{1/3}$ is concave on the big interval; see [5, Section 2.4]. Its left tangent at $2$ therefore gives $$V(t)\le\frac{17}{2}\left(1-\frac7{17}(t-2)\right)^3 \quad\text{while }D_t\text{ is big and }t\ge2.$$ In particular, the big interval ends no later than $31/7$, and $$\int_2^\infty V(t)\,dt\le\frac{289}{56},\qquad S_X(G)\le\frac1{16}\left(\frac{225}{8}+\frac{289}{56}\right)=\frac{233}{112}.$$ The other assertions follow from $A_X(G)=3$ and Lemma 14.1. This argument uses concavity only for the unresolved tail; it does not pretend that the first chamber continues past $t=2$. ◻