The global conic-bundle model and double fibres

Proof. The equation of the incidence surface gives $0\to\mathcal O(1,-1)\to\mathcal O(3,0)\to\mathcal O_E(3,0)\to0$; pushing it to $\mathbb P^2$ gives the displayed Steiner resolution. Push forward $0\to\mathcal O_X(D)\to\mathcal O_X(H) \to\mathcal O_E(3,0)\to0$. The higher direct image of $\mathcal O_X(D)$ vanishes because every conic fibre has arithmetic genus zero. Hence the resulting sequence is $0\to\mathcal O(1)\to\eta_*\mathcal O_X(H)\to G\to0$. Since $\det G=\mathcal O(2)$, its extension group is $H^1(G^*(1))=H^1(G(-1))=0$, as follows by twisting the Steiner resolution by $\mathcal O(-1)$. Therefore the sequence splits. The restriction of $H$ to every fibre is its anticanonical line bundle; it gives the usual plane-conic embedding, including the nonreduced fibres. Finally $c_1(\mathcal O(1)\oplus G)=3$, so $K_P=-3H$. Adjunction and $K_X=-H-D$ give $X\sim2H-D$. ◻

Proof. Locally a conic bundle is given by a symmetric three-by-three matrix. Rank zero is excluded by flatness. At a rank-two point the derivative of its determinant cannot vanish: otherwise the total space would be singular at the vertex of the conic. At rank one, write the initial normal equation along the reduced fibre $R\simeq\mathbb P^1$ as $x^2+u q_1(y,z)+v q_2(y,z)=0$, with $(u,v)$ parameters on the base. Smoothness of $X$ says exactly that the two binary quadratics $q_1,q_2$ have no common zero. They define a degree-two morphism $\mathbb P^1\to\mathbb P^1$, which has two distinct branch points. After changing coordinates this pencil is generated by $y^2,z^2$. The quadratic initial term of the determinant is consequently a nonzero multiple of $uv$, proving the nodal assertion. At a double fibre, the rank-one ternary quadratic restricts to a nonzero square on $E=(z=0)$. Thus $q_0$ has rank one there, so the base point lies on the branch conic of $E$. For a general line through $b$, its inverse image has local equation $x^2+t(aq_1+bq_2)+\text{higher terms}=0$. The quadratic $aq_1+bq_2$ has two simple zeros. At each zero the surface has a nondegenerate quadratic part $x^2+tw$, and hence an ordinary double point. It is smooth elsewhere along $R$; Bertini applies elsewhere. For a reduced fibre choose the line transverse to the discriminant. Also choose it non-tangent to the smooth branch conic of $E\to\mathbb P^2$. Then $E$ meets the surface in a smooth curve of bidegree $(1,1)$, so $\pi$ contracts no curve in the surface. Thus $H$ restricts to an ample divisor. Adjunction gives $-K_S=H|_S$ and $K_S^2=H^2D=5$. The same argument works in the double-fibre case; resolving its two ordinary double points is crepant and gives the stated weak del Pezzo surface. ◻

Proof. Flatness gives the base-change description $Y=X\times_{\mathbb P^2}\operatorname{Bl}_b\mathbb P^2$. Near $R$ the ideal of the base point has the form $(u,x^2)$ in regular parameters transverse to $R$. Thus this is the weighted extraction of weights $(1,2)$, not the ordinary blowup of $R$. In the chart $v=tu$, its equation has leading form $x^2+u(q_1+tq_2)=0$. Along $u=x=0$ the quotient singularities are $\frac12(1,1,0)$ precisely where $q_1+tq_2=0$. This also proves normality. Replacing the weighted coordinate $x$ by $w=x^2$ identifies the coarse exceptional surface with $R\times\mathbb P^1_{[u:v]}$; the relation is $w+uq_1+vq_2=0$. The quotient locus is the graph $\Gamma=\{uq_1+vq_2=0\}$, which is smooth of bidegree $(2,1)$. Its different is $\Gamma/2$. The weighted discrepancy is $1+2=3$, so $K_Y=f^*K_X+2F$. The exceptional curve of $\operatorname{Bl}_b\mathbb P^2$ pulls back to $2F$, giving $\mathcal O_F(2F)=\mathcal O(0,-1)$. The other two restrictions follow from $H\cdot R=1$ and the fact that $R$ is contracted by $\eta$. Finally the normal sequence of $R$ in the local plane bundle is the kernel of $\mathcal O(1)\oplus\mathcal O^{\oplus2}\to\mathcal O(2)$, where the last two summands map by $q_1,q_2$ and the first maps to zero. Its kernel is $\mathcal O(1)\oplus\mathcal O(-2)$. ◻

Proof. Put $a=f_1/2$ and $x=y_1-az$, so that $R=(x=0)$ with coordinates $[y_0:z]$. The derivatives of the conic equation in the base directions $A,B$, restricted to $R$, are $$q_A=-y_0^2-f_2y_0z+(h_0-af_3)z^2,\qquad q_B=-ay_0z+(h_1-af_2)z^2.$$ The graph $\Gamma$ is defined by $uq_A+vq_B=0$. At $R\cap E=[1:0]$, the first quadratic is nonzero, and the derivative of $q_B/q_A$ vanishes if and only if $a=0$. If $a=0$, then $q_B=h_1z^2$; the two quadratics have a common zero if $h_1=0$, contradicting smoothness of $X$. ◻

Proof. Let $\widetilde\eta:Y\to\operatorname{Bl}_b\mathbb P^2$ be the base-changed conic bundle. The divisor $f^*D-2F$ is the pullback of the pencil of lines through $b$, and is nef. For $0\le t\le2$, $$f^*L-tF=f^*H+(1-t/2)f^*D+(t/2)(f^*D-2F)$$ is therefore nef. Let $\ell_E$ be the fibre of $E\to C_0$ through $R\cap E$. Its image under $\eta$ is a line, with degree one. Its strict transform on $Y$ has intersection $1/2$ with $F$, because $2F$ pulls back the exceptional curve of the base blowup. It has intersection zero with $f^*H$ and one with $f^*D$. Hence $f^*L-tF$ is not nef for $t>2$.

The restrictions in Lemma 73.4 give $$(f^*L)^2F=0,\qquad f^*L\cdot F^2=-\frac12,\qquad F^3=0.$$ Cubing the nef divisor proves the volume formula, and $$\frac1{26}\int_0^2\left(26-\frac32t^2\right)dt =\frac{24}{13}.$$ The volume is still $20$ at $t=2$, so the remaining tail is nonzero and is essential to the full expected order. ◻

Proof. Write a hyperplane in an affine chart of $\mathbb P(V^*)$ as $$h_0z_0+h_1z_1+h_2z_2+h_3z_3+z_4=0.$$ It has basis $d_i=e_i-h_i e_4$ for $0\le i\le3$. Write $z_{ij}$ for the Plücker coordinates in this basis. The three linear equations defining $V_5$ give $$\begin{aligned} z_{12}&=2z_{03},\\ z_{13}&=-8h_1z_{01}-8h_2z_{02}-8h_3z_{03},\\ z_{23}&=(2h_0+16h_1h_3)z_{01}+16h_2h_3z_{02}+(-4h_2+16h_3^2)z_{03}. \end{aligned}$$ Substitute these into the Plücker equation $z_{01}z_{23}-z_{02}z_{13}+z_{03}z_{12}=0$. Up to a nonzero scalar, the resulting conic in coordinates $[z_{01}:z_{02}:z_{03}]$ has matrix $$\begin{pmatrix} h_0+8h_1h_3&4h_2h_3+2h_1&-h_2+4h_3^2\\ 4h_2h_3+2h_1&4h_2&2h_3\\ -h_2+4h_3^2&2h_3&1 \end{pmatrix}.$$ It has rank $1$ if and only if $h_2=h_3^2$, $h_1=h_3^3$, and $h_0=h_3^4$. Indeed, compare its upper-left $2\times2$ block with the outer product of the first two entries of its last column. Hence its double-conic locus is $$[h_0:h_1:h_2:h_3:h_4]=[t^4:t^3:t^2:t:1].$$ These are precisely evaluation at $[t:1]\in\mathbb P^1$. Every hyperplane can be moved into the chart $h_4\ne0$ by $\operatorname{SL}_2$, since fourth powers span $V$. Thus the calculation proves the assertion globally. Such a hyperplane belongs to $\mathbb P(U^\perp)$ if and only if its evaluation vanishes on every member of $U$, which is the common-root condition.

If the common-root multiplicity were at least $2$, factor $U=x^2W$ for a pencil $W$ of binary quadratics. Its projective line meets the conic of quadratic squares, so $U$ would contain a square quartic. This contradicts $U\in \mathcal H^\circ$. The first vanishing order is therefore $1$. The second is at least $2$ and at most $3$, because an order-$4$ member is again a square. ◻

Proof. Choose coordinates with $p=\{x=0\}$. First suppose the vanishing sequence is $(1,2)$. After changing the basis of $U$, write $$U=\langle e_2+b e_1+c e_0,\ e_3+d e_1+e e_0\rangle.$$ We have $b^2-4c\ne0$, because otherwise the first quartic would be a square. The three linear forms defining the twisted cubic are the coordinates of the exterior square of the projection $V\to V/U$. Explicitly, they are $$\begin{aligned} l_0&=A-bB+(2c-d)C+8eD+2(cd-eb)E,\\ l_1&=D-cF-eG,\\ l_2&=E-bF-dG. \end{aligned}$$ Thus $r$ is the point $A=bB$ on $R$. At the generic point of $R$, let $u=l_1/l_0$ and $v=l_2/l_0$. Taking the terms of weight $2$ in the equations gives $$F=Bv,\qquad E=Av,\qquad D=(A-bB)u+cBv,$$ and therefore $$C^2=4B(A-bB)u+(4cB^2-A^2)v.$$ Consequently $\Gamma$ is given by the vanishing of the right-hand side. Near $r$, put $t=A/B$. Its projection to the direction line is $$\frac vu=\frac{4(t-b)}{t^2-4c}.$$ Its derivative at $t=b$ is $4/(b^2-4c)$, which is nonzero. The projection is unramified at the marked point.

Now suppose the vanishing sequence is $(1,3)$. Use the normal form $U=\langle e_1,e_3+a e_2+c e_0\rangle$. The same exterior-square calculation gives $$l_0=B-aC+2cE,\qquad l_1=D-cG,\qquad l_2=F-aG.$$ The marked point is $B=0$. Taking terms of weight $2$ gives $D=Bu$, $F=Bv$, $E=Av$, and hence $$C^2=4B^2u-A^2v.$$ Thus $\Gamma$ projects by $[A:B]\mapsto[A^2:4B^2]$, which is ramified at $B=0$. These are the only possible vanishing sequences at a common root of a smooth pencil by Lemma 73.10. ◻