The weighted degree one del Pezzo and double plane K3 ladder

The C-star action on the affine cone over Y has no nontrivial stabilizer away from the origin. If some $x_i$ is nonzero this is immediate. If all $x_i$ vanish, the equation $z^2+y^3=0$ has y,z both nonzero, so a stabilizer satisfies ${\lambda }^2={\lambda }^3=1$ and is trivial. Thus $O_Y(1)$ is an actual line bundle, including at the unique point with $x=0$. There is exactly one such weighted point: normalize $y=1$, and the two square roots for z are interchanged by the remaining scaling $\lambda =-1$. Denote it by p. Both y and z are nonzero there.

We compute the low-degree sections without assuming projective normality of Y. In weighted projective space cover by its five coordinate nonvanishing opens and use the homogeneous localization Cech complex. For every exponent monomial of fixed weighted degree, its allowed Cech intersections are exactly those index sets containing all its negative-exponent variables. Inserting a fixed variable outside that negative set contracts the intermediate Cech degrees. Only the all-nonnegative monomials in degree zero and the all-negative monomials in top degree remain. Therefore all intermediate $H^q(O(k))$, $0<q<4$, vanish; H0 is exactly the weighted homogeneous polynomial piece. Apply the multiplication sequence by the nonzero degree-six equation to obtain $H0(Y,kA)=R_k$ for $k=1,2,3$, where R is the weighted coordinate quotient. This gives exactly the stated bases. At x not all zero the x monomials generate 2A and 3A; at p, y generates 2A and z generates 3A. Thus both systems are globally generated, not merely generically generated.

Choose general sections $s2=y+q2(x)$ and $s3=z+l1(x)y+q3(x)$. The conditions that the y and z coefficients are nonzero are open in their complete section spaces, so these are general members of |2A| and |3A| after rescaling. Bertini gives smooth divisors D2,D3 meeting transversely. They avoid p, by step1.1. Form the degree-six finite flat root cover $f:T->Y$ by adjoining sections v,u of $f^{\ast }A$ satisfying $v^2=s2$ and $u^3=s3$. Algebraically its rank-six algebra is the tensor product of O plus $A^{(-1)}$, with multiplication specified by s2, and O plus $A^{(-1)}$ plus $A^{(-2)}$, with multiplication specified by s3. The group is mu2 times mu3. Locally near D2,D3 the equations are $v^2=t2,u^3=t3$ in transverse coordinates; hence the cover is smooth. Away from these divisors it is etale.

In coordinates (x0,x1,x2,v,u), eliminate $y=v^2-q2(x)$ and $z=u^3-l1(x)(v^2-q2(x))-q3(x)$. The defining equation becomes an ordinary homogeneous sextic in P4. This is exactly the preceding root cover, not an auxiliary component: the coordinate map is defined everywhere, since $x=v=u=0$ is impossible, and on every chart $x_i$ nonzero the two descriptions are the same affine root equations. At $x=0$, both v and u are nonzero, and the same root equations in either projective unit chart identify the neighborhoods. Thus T is a smooth ordinary sextic threefold, with $f^{\ast }A=O_T(1)$. Conversely its mu2 times mu3 invariants are precisely x,y,z with the original weighted equation, so its quotient is Y. The ordinary homogeneous hypersurface Cech sequence gives $H1(T,O_T)=H2(T,O_T)=0$ and $H0(T,O_T)=C$. By [F1], $\pi 1(T)=0$ and $H^2(T,Z)=Zc1(O_T(1))$. F4 identifies algebraic $H1(T,O_T)=0$ with analytic H1 and injects Pic(T) into integral $H^2(T;Z)$. Its image contains the primitive generator $c1(O_T(1))$, so $Pic(T)=Z{O}_T(1)$. This uses the integral restriction statement of F1 on T inside P4 and the usual integral generator of $H^2(P4;Z)$; it does not use GAGA on P4.

We give the needed fundamental-group quotient argument. If a finite group acts on a connected simply connected complex manifold, any loop in its quotient can be perturbed off the branch strata of real codimension at least two and lifted to a path from t to g t. A path from t to a fixed point of a subgroup element h, followed by the reverse h-translate, projects to a null loop with endpoint monodromy h; in the local branched coordinate it is a meridian which contracts through that branch point. Multiplying by such null loops for stabilizer generators changes any endpoint monodromy by those generators. Here mu2 fixes $v=0$, and mu3 fixes $u=0$; both divisors are nonempty (at a fixed nonzero x, the defining quadratic or cubic in the remaining weighted variable has a complex root). Those two subgroups generate the whole group. Hence every loop can be changed by null loops to have a CLOSED lift in T. Contract that lift in T and project. This proves $\pi 1(Y)=0$, without importing a theorem on fundamental groups of Fano varieties or quotient spaces. The same argument below applies to the degree-three cover of a double plane.

Finite-group averaging splits $O_Y$ as the invariant direct summand of $f_{\ast }{O}_T$, so $H1(Y,O_Y)=H2(Y,O_Y)=0$ by step4.1. By F4 the algebraic vanishing implies analytic $H1(Y,O_Y)=0$, and Pic(Y) injects into $H^2(Y;Z)$. The latter group is torsion-free: $\pi 1(Y)=0$ gives $H1(Y;Z)=0$ by abelianization, so the integral universal-coefficient exact sequence of F4 identifies $H^2$ with Hom(H2(Y;Z),Z). In particular Pic(Y) is torsion-free; no surjectivity $Pic(Y)->H^2(Y;Z)$ is required. Pullback $Pic(Y)->Pic(T)$ has torsion kernel, since the norm of $f^{\ast }B$ is $B^6$: take the determinant of the rank-six projection-formula identity $f_{\ast }(f^{\ast }B)=f_{\ast }{O}_T$ tensor B and divide by $\det (f_{\ast }{O}_T)$. Thus the pullback is injective. Its image in Z contains the generator because $f^{\ast }A=O_T(1)$, so $Pic(Y)=ZA$. The same finite map gives 6 $A^3=O_T(1{)}^3=6$, whence $A^3=1$. For every integral subvariety V of Y, the finite preimage and projection formula make $A^{(\dim V)}V$ positive since $f^{\ast }A$ is ample. The numerical criterion [F2] gives ampleness of A.

The canonical bundle is computed by the same cover, rather than a named weighted adjunction formula. An ordinary sextic in P4 has $K_T=O_T(1)$, as the local residue/Jacobian formula for one degree-six equation gives $K_{P}4$ tensor $O(6)=O(1)$. In the local root coordinates $v^2=t2,u^3=t3$ the Jacobian factor is v $u^2$ up to a nonzero constant. Hence $K_T=f^{\ast }{K}_Y+R_v+2R_u$. Each of $R_v,R_u$ is the divisor of a section of $f^{\ast }A$, so $f^{\ast }(K_Y+2A)=O_T$. Injectivity of pullback from step6.1 gives $K_Y=-2A$.

A general member of |2A| has equation $y=q2(x)$, by the computed complete section space. Substitution gives $S=\{z^2+g6(x)=0\}$ in P(1,1,1,3). It is smooth by Bertini, and none of its points has $x=0$; thus it is the finite double cover $\pi :S->P2$ with branch the sextic $g6=0$. The latter is smooth because its singular points would, by the displayed equation and Jacobian criterion, give singular points of S at $z=0$. The algebra is ${\pi }_{\ast }{O}_S=O_{P}2$ plus $O_{P}2(-3)$. The local double-cover Jacobian gives $K_S={\pi }^{\ast }(K_{P}2+O(3))=O_S$. The projective-space Cech calculation gives algebraic $H1(S,O_S)=0$, $H0(S,O_S)=C$ and $H2(S,O_S)=C$. The closed-embedding comparison in F4 gives the same analytic groups on S, so the K3 condition uses the actual surface analytic H1, not an unproved surface GAGA input.

To prove the full K3 fundamental-group condition, take the ordinary sextic surface $T^{\mathrm{\prime }}=\{b^6+g6(x0,x1,x2)=0\}$ in P3. It is smooth: a singular point would have $b=0$ and all derivatives of g6 zero on the branch plane, contrary to smoothness of the branch sextic. The map $z=b^3$ identifies $S=T^{\mathrm{\prime }}\mathrm{/}mu3$, with fixed nonempty divisor $b=0$. By [F1] the ordinary sextic surface T' is simply connected. The path-lifting and fixed-point meridian argument of step5.1 therefore makes S simply connected as well. This proves the full K3 assertion; it is not inferred merely from $K_S$ trivial and $H1(O_S)=0$.

$H=A\mathrm{\mid }S={\pi }^{\ast }{O}_{P}2(1)$ is ample and globally generated, with $H^2=degree(\pi )=2$. It is primitive: if $H=kB$ for an integral line bundle with $k\ge 2$, the integer $B^2$ would satisfy $2=k^2{B}^2$, impossible. The double-cover algebra computes $H0(S,4H)=H0(P2,O(4))$ plus z H0(P2,O(1)). These are algebraic sections, and the closed-embedding comparison in F4 for $O_S(4H)$ identifies them with the full analytic section space as well. Its complete system is globally generated, so a general member is a smooth curve by Bertini. Restriction of weighted degree-four polynomials supplies every one of these sections: it contains all quartics in x and all $x_i$ z; the eliminated degree-two coordinates simply become quadratic polynomials in x. Thus a general weighted degree-four equation cuts exactly the claimed general curve in |4H|. All incidences and branch transversality conditions are nonempty open conditions in the stated equation/section parameter spaces. This proves the needed weighted ladder while leaving its separate very-general Picard assertion to the Noether–Lefschetz supplier.