The pencil of sextic del Pezzo surfaces

Proof. The Koszul resolution for three divisors of bidegree $(1,1)$ gives $H^0(C,\mathcal O_C)=\mathbb C$, and adjunction gives $K_C\sim0$. Thus $C$ is a smooth connected curve of genus one. For fixed $x$, its fibre under the first projection is the projectivized kernel of the matrix with rows $x^{\mathsf t}$, $(Ax)^{\mathsf t}$ and $(Bx)^{\mathsf t}$. If this matrix had rank at most one at a point of the image, $C$ would contain a projective line, contrary to irreducibility and genus one. Its rank is therefore exactly two everywhere over the image. Its kernel is consequently a line bundle locally on the determinant cubic. This identifies $C$, as a scheme, with that cubic. The second projection is treated in the same way. Alternatively, the degree of either image is $\mathcal O_C(1,0)=\mathcal O_C(0,1)=3$, by intersection theory. ◻

Proof. A rank-one matrix in $N$ would give a bilinear form $l(x)m(y)$ vanishing identically on the irreducible curve $C$. Hence one of its plane cubic images would lie in a line, which is impossible by Lemma 109.1. A nonzero rank-zero matrix is impossible. Suppose that $[M]$ is a singular point of $D$. Then $M$ has rank two. Choose generators $x$ and $y$ of its right and left kernels. The derivative of the determinant in a direction $T\in N$ is a nonzero constant times $y^{\mathsf t}Tx$. Thus $(x,y)\in C$. But both partial gradients of $y^{\mathsf t}Mx$ vanish at $(x,y)$, since $Mx=0$ and $y^{\mathsf t}M=0$. This gives a nontrivial linear dependence among the gradients of the three defining equations of $C$, contradicting its smoothness. ◻

Proof. A fibre is isomorphic to $$S_M=\{y^{\mathsf t}x=y^{\mathsf t}Mx=0\},\qquad M=\mu A+\nu B,$$ since blowing up the Cartier divisor $C$ on $S_M$ changes neither $S_M$ nor its scheme structure. The description below also verifies normality, so this observation does not presuppose it. No eigenvalue of $M$ can have a two-dimensional eigenspace: otherwise $M-\lambda I$ would be a nonzero matrix in $N$ of rank at most one. Consequently $M$ is cyclic. Up to conjugation and an affine change $M\mapsto aM+bI$, its Jordan type is one of the following: three distinct eigenvalues; a block of size two and a distinct eigenvalue; a single block of size three.

The Jacobian criterion says that a singular point of $S_M$ is a pair of right and left eigenvectors with the same eigenvalue, satisfying $y^{\mathsf t}x=0$. Three distinct eigenvalues give no such pair. For the second Jordan type take $$M=\begin{pmatrix}0&1&0\\0&0&0\\0&0&1\end{pmatrix}.$$ There is one singular point, namely $([1:0:0],[0:1:0])$. On the chart $x_0=y_1=1$, put $a=x_1$, $b=x_2$, and $c=y_2$. Eliminating $y_0$ gives $bc(1-a)=a^2$, which is an $A_1$ singularity. For the last Jordan type take $$M=\begin{pmatrix}0&1&0\\0&0&1\\0&0&0\end{pmatrix}.$$ The only singular point is $([1:0:0],[0:0:1])$. On the chart $x_0=y_2=1$, put $a=x_1$, $c=y_1-a$, and $d=x_2-a^2$. Eliminating $y_0$ gives $cd=a^3$, which is an $A_2$ singularity. This proves normality and the asserted list. Adjunction gives $-K_{S_M}=\mathcal O_{S_M}(1,1)$ and $K_{S_M}^2=6$. If a singular point of $S_M$ lay on $C$, the two equations defining $C$ inside the smooth threefold $V$ would have dependent gradients. Thus all singular points avoid $C$, and their lifts avoid $E$.

Projection of the smooth cubic $D$ from $[I]\notin D$ gives a finite map $D\to\mathbb P^1$ of degree three. Its fibre over $[\mu:\nu]$ records the eigenvalues of $\mu A+\nu B$. Ordinary ramification gives the second Jordan type, and ramification of index three gives the third. Riemann–Hurwitz therefore gives $n_1+2n_2=6$. ◻

Proof. The first projection is the graph of the rational map $x\mapsto x\mathbin{\times}Mx$, so its centres are the eigenscheme of $M$. In the $A_1$ normal form, near $[1:0:0]$ its ideal is $(b,a^2)$, and the other centre is $[0:0:1]$. In the $A_2$ normal form its ideal is $(b-a^2,a^3)$. Successive ordinary blow-ups give the stated clusters. The curve classes and their intersections follow directly. To check completeness, a $(-1)$-curve on the resolution is either an exceptional curve or has class $dl-m_1e_1-m_2e_2-m_3e_3$ with nonnegative integers $m_i$. The equations $3d-\sum m_i=1$ and $d^2-\sum m_i^2=-1$ give $d\le1$ by Cauchy–Schwarz. The degree-one possibilities are then exactly the lines specified above. The second projection and the incidences with the singularities follow either by transposition or by intersecting these classes with the $(-2)$-curves. ◻

Proof. Put $H_1=\pi^*\mathcal O_V(1,0)$, $H_2=\pi^*\mathcal O_V(0,1)$, and $F=\phi^*\mathcal O_{\mathbb P^1}(1)$. These are independent nef classes. The exceptional fibre of $\pi$ has intersections $(0,0,1)$ with them. The strict transform of a fibre of the first projection of $V$ meeting $C$ has intersections $(0,1,0)$: the two pencil sections restrict to linear polynomials with the same zero, so their ratio is constant after blowing up. The second projection similarly supplies a curve with intersections $(1,0,0)$. Hence the nef cone is generated by $H_1,H_2,F$. The connected automorphism group fixes these rays and their integral primitive classes. It therefore descends via both plane projections, and via $\pi$, to an automorphism of $V$ preserving $C$. Its action on $C$ preserves the degree-three line bundle $\mathcal O_C(1,0)$. The group of automorphisms of an elliptic curve preserving a positive-degree line bundle is finite: its translation subgroup is contained in the kernel of the associated isogeny, and the subgroup fixing an origin is finite. The kernel of this action fixes both plane cubic images pointwise, hence is the identity on both planes and on $X$. Thus the connected automorphism group is trivial. Since the automorphism group of a Fano variety is an algebraic group of finite type, it is finite. ◻

Proof. The blow-up is the complete intersection of divisors of multidegrees $(1,1,0)$ and $(1,1,1)$ in $\mathbb P^2\times\mathbb P^2\times\mathbb P^1$. Indeed, the ideal $(f,g)$ is a regular sequence on $V$, so its blow-up is its usual incidence graph. Adjunction gives $-K_X=\mathcal O_X(1,1,1)$, which is ample. Its cube is $24$ by intersection in this product. The Koszul resolution, twisted by $\mathcal O(m,m,m)$, has no higher cohomology contributing to its space of sections. It gives $$h^0(X,-mK_X)= (m+1)\binom{m+2}{2}^{\!2} -(2m+1)\binom{m+1}{2}^{\!2} +m\binom m2^{\!2} =4m^3+6m^2+4m+1.$$ For $m=0,1$ the same formula follows directly, with the zero binomial terms omitted. Finally, $h^0(V,\mathcal O_V(1,1))=8$; the graph equation is the unique relation among its sixteen products with the base coordinates. This proves the last assertion. ◻

Proof. Consider $(\phi,p_1)$. Over $([\mu:\nu],x)$ the remaining coordinate $y$ satisfies $y^{\mathsf t}x=y^{\mathsf t}(\mu A+\nu B)x=0$. There is a unique solution except along the eigencurve $$Z_1=\{([\mu:\nu],x):x\mathbin{\times}(\mu A+\nu B)x=0\}.$$ The smooth determinant cubic in Lemma 109.2 maps isomorphically to $Z_1$ by taking the right kernel of $\lambda I+\mu A+\nu B$. The kernel is a line everywhere. Conversely, the eigenvalue is recovered regularly on a chart where a coordinate of $x$ is nonzero. This proves that $Z_1$ is smooth and that the indicated map is a closed immersion.

The two-row matrix with rows $x^{\mathsf t}$ and $((\mu A+\nu B)x)^{\mathsf t}$ has rank one along $Z_1$ and rank two elsewhere. Locally a unit entry reduces its incidence equations to the usual graph of two generators of the ideal of $Z_1$. Since $Z_1$ is smooth of codimension two, this graph is its blow-up. This proves the assertion about the morphism. The canonical divisor formula gives $$D_1=(2F+3H_1)-(-K_X)=F+2H_1-H_2.$$ Here $-K_X=H_1+H_2+F$.

The projection of $Z_1$ to the first plane is $C_1$: its determinant equation is $\det(x,Ax,Bx)=0$, the equation of that plane cubic. Over $x\in C_1$, the two pencil equations restrict proportionally to the first-projection fibre, so its strict transform is contracted by $(\phi,p_1)$. The inverse image of $C_1$ in $V$ contains $C$ with multiplicity one. Its strict transform therefore has class $3H_1-E$ and is the exceptional divisor just described. The second assertion is obtained by transposing the matrices. ◻