Canonical section-ring lifting on K3 surfaces and Fano ladders

Let Z be integral proper with $H^0(O_Z)=C$, let M be a line bundle, and let the nonzero section s∈H⁰(Z,M) have an effective Cartier zero divisor W with $H^0(W,O_W)=C$. Suppose $H^1(Z,M^{n-1})=0$ for every $n\ge 1$. Multiplication by s and restriction give exact sequences $0\to H^0(Z,M^{n-1})$ --s→ H⁰(Z,Mⁿ) → H⁰(W,(M|W)ⁿ)→0.

The left injection follows because s is a nonzerodivisor. In degree zero restriction is the isomorphism C→C by the additional $H^0(W,O_W)=C$ hypothesis; in every positive degree it is onto by step 1.1. Thus the restriction of graded rings is onto, its kernel is the ideal generated by the single degree-one element s, and R(Z,M)/(s)≅R(W,M|W).

If the latter ring is generated in degree one, lift its degree-one generators to H⁰(Z,M). For a homogeneous u of degree n, its restriction is a linear combination of products of n degree-one restricted sections. Subtract the same linear combination of products of their chosen lifts. The remainder is s v with v of degree n−1. Starting with $n=0$, induction writes v, and hence u, as a polynomial in degree-one sections. Therefore M is normally generated. This argument proves the actual symmetric-power multiplication assertion, not merely finite generation in unspecified degrees.

Put $g=g(C)$. For the actual smooth Cartier divisor C in S, a local defining equation f gives the conormal sequence $0\to L^{-1}\mathrm{\mid }C\to {\mathrm{\Omega }}_S^1\mathrm{\mid }C\to {\mathrm{\Omega }}_C^1\to 0$, with first map f mod f²↦df. Smoothness of both C and S makes df nonzero in the normal direction, so this sequence is locally an injection of a rank-one summand with locally free quotient. If $f_i=u_{ij}$ $f_j$, then $d{f}_i=u_{ij}$ $d{f}_j$ on C; these are precisely the transitions of L⁻¹|C. Taking determinants therefore gives $K_S\mathrm{\mid }C=(L^{-1}\mathrm{\mid }C)\otimes K_C$, hence $K_C=(K_S\otimes L)\mathrm{\mid }C=L\mathrm{\mid }C$. The named numerical adjunction supplier gives $2g-2=L^2>0$, so $g\ge 2$. On S, Serre duality gives $H^2(S,L)=H^0(S,L^{-1})\ast =0$: a nonzero section of the inverse ample line bundle would have an effective zero divisor of negative L degree, impossible; a nowhere-zero section would trivialize L, also impossible since $L^2>0$. Also $H^2(S,O_S)=C$ because $K_S=O_S$ and S is integral proper.

For $0\to O_S\to L\to K_C\to 0$, the cohomology sequence includes $0=H^1(O_S)\to H^1(L)\to H^1(K_C)\to H^2(O_S)\to H^2(L)=0$.

Curve Serre duality gives $H^1(K_C)=H^0(O_C)\ast =C$. The displayed map between these one-dimensional spaces is onto, hence an isomorphism. Consequently $H^1(L)=0$. This is the $n=1$ step that must not be omitted from an inductive vanishing argument.

For $n\ge 2$, the exact sequence $0\to L^{n-1}\to L^n\to K_C^n\to 0$ and curve Serre duality give $H^1(C,K_C^n)=H^0(C,K_C^{1-n})\ast =0$, since that line bundle has negative degree. Inductively $H^1(S,L^{n-1})=0$, and the long exact sequence then forces $H^1(S,L^n)=0$. The case $n=0$ is the K3 hypothesis. Thus every restriction in Section 1 is onto, including $n=1$.

The separately proved Max Noether output says $R(C,K_C)$ is generated by $H^0(K_C)$. Since C is smooth integral proper over C, $H^0(C,O_C)=C$. Apply Section 1 to S,L,C: the full K3 section ring is generated by H⁰(S,L). This establishes all multiplication maps in every degree, not just the quadratic multiplication map.

For the ambient data, tensor the divisor sequence by Aⁿ to obtain $0\to A^{n-1}\to A^n\to L^n\to 0$.

The base $H^1(Y,O_Y)=0$ and $H^1(S,L^n)=0$ imply inductively $H^1(Y,A^n)=0$: the relevant segment is $H^1(Y,A^{n-1})\to H^1(Y,A^n)\to H^1(S,L^n)$. Both outside groups vanish. Since S is smooth integral projective over C, $H^0(S,O_S)=C$. Section 1 then applies at every degree and lifts normal generation from L to A.

The same local equation and conormal determinant argument in any smooth ambient Z gives $K_D=(K_Z\otimes O_Z(D))\mathrm{\mid }D$ for each actual smooth Cartier divisor D. Thus for the smooth threefold with $$\begin{gathered} S \\ in\mathrm{\mid }-K_Y\mathrm{\mid } \end{gathered}$$ one obtains $K_S=O_S$, but $H^1(S,O_S)=0$ is a further K3 input, not a consequence of this line-bundle identity alone. For the smooth fourfold V with $-K_V=2A$ and actual smooth Y∈|A| it gives $K_Y=(K_V\otimes A)\mathrm{\mid }Y=-A\mathrm{\mid }Y$, equivalently $-K_Y=A\mathrm{\mid }Y$. If the explicit H¹(O) hypotheses hold at each level, the preceding algebra applies twice. No new cohomological vanishing theorem enters at the second lift.

For an actual smooth ample divisor S in Y with dim $Y\ge 3$, Ample divisor relative handles and primitive integral restriction gives H₁(S,Z)≅H₁(Y,Z). By Compact Kähler Hodge decomposition from elliptic harmonic forms, $H^1(S,O_S)=0$ implies $H^1(S,C)=0$, because H¹ splits into the conjugate (1,0) and (0,1) parts. The homology isomorphism gives $H^1(Y,C)=0$ and hence $H^1(Y,O_Y)=0$. Repeat the same argument for any next smooth ample ambient inclusion of dimension at least three. Thus the needed ambient vanishing is actually supplied throughout the smooth K3/Fano ladder, without Kodaira vanishing or simple-connectivity assumptions.

Let M be any ample line bundle on a projective variety and suppose all its section-ring multiplications are onto. Choose $N>0$ with Mᴺ very ample. At a point where every degree-one section vanished, every product of N such sections would vanish; surjectivity would make all sections of Mᴺ vanish, contradicting its global generation. Therefore M is globally generated and defines a morphism ${\phi }_M$.

The degree-N multiplication surjection identifies the complete Mᴺ linear system with the restrictions to ${\phi }_M(Z)$ of all degree-N homogeneous polynomials in its degree-one coordinates, modulo those vanishing identically. Consequently the morphism given by Mᴺ is the restriction of the Nth Veronese of ${\phi }_M$ to its linear span. Since Mᴺ gives a closed immersion and the Veronese is a closed immersion, ${\phi }_M$ is itself a closed immersion: identifying the Veronese image with its source identifies ${\phi }_M(Z)$, with its closed subscheme structure, with the Mᴺ image. Thus M is very ample.