Past exam of the mathematics course of the University of Cambridge / 2005 / iii / Paper 21 (K3 Surfaces)

Use projectivization by quotients, and setThe fibre over $t$ parametrizes one-dimensional quotients of $E_t$, so it is ${\mathbb{P}}^{n-1}$. The standard line bundles areThus $L$ is the fiber class of a rational normal scroll of the ruling and $M$ is the universal quotient line bundle of ${\pi }^{\ast }E$. In additive divisor notation we also write $L,M$ for their first Chern classes. For $n\ge 2$ these generate the Picard group; for $n=1$, $F={\mathbb{P}}^1$ and $M=a_{1}L$. This fixes the sign convention for the rational normal scroll.

Put $s={\sum }_i{a}_i$. The projective bundle formula for Chow groups givesHere the first relation comes from the one-dimensional base, and the second is the projective-bundle relation with $c_1(E)=sL$ and higher Chern classes zero on the base. Integration over a fibre gives $L{M}^{n-1}=1$. Therefore all top-degree mixed intersection numbers areIn particular, for integers $u,v$,For the canonical divisor of a smooth variety, the relative Euler sequencehas determinant ${\mathcal{O}}_F(nM-sL)$. Combining its dual determinant with $K_{{\mathbb{P}}^1}=-2L$ gives

A common twist does not change the abstract projective bundle. Choose an integer $b$ such that $d_i=a_i+b\ge 1$ for every $i$. The complete linear system of divisors $|M+bL|$ hasIts map to ${\mathbb{P}}^N$, where $N=s+nb+n-1$, has coordinate blocksHere $[s_0:t_0]$ is a point of the base and the ${\lambda }_i$ are quotient coordinates in a local trivialization of $E(b)$. These expressions have the correct transition factors and define the global map. Any nonzero block recovers the base point through its degree-$d_i$ rational normal curve. Once that point is known, the blocks recover the projective fibre coordinates. On the open where the first coordinate of a nonzero block is nonzero, the ratio of the next coordinate to the first recovers $t_0/s_0$; the analogous last-coordinate chart recovers $s_0/t_0$. The remaining coordinate ratios then recover the fibre coordinates regularly. Thus the map identifies $F$ with its closed image and is a projective embedding, not merely a pointwise injection.

The hyperplane class on this image is $M+bL$, soConsequently every such natural embedding hasThe tautological choice $b=0$ is an embedding when all $a_i>0$. If the $a_i$ are only nonnegative, $|M|$ is basepoint-free but can contract the projective subbundle arising from zero summands; one must distinguish that scroll image from the smooth abstract scroll.

Throughout, the braces denote a multiset, so repeated splitting degrees retain their multiplicities. If the multisets agree after a common shift, permuting summands gives $E_a\cong E_b\otimes \mathcal{O}(c)$. Tensoring all the fibres of a vector bundle by the same line bundle leaves their one-dimensional quotient spaces unchanged. HenceThis proves one implication in the isomorphism classification of abstract rational scrolls.

For the converse first take $n\ge 3$. Every divisor class on $F_a$ has the form $xL+yM$. If its square is zero, restriction to a fibre ${\mathbb{P}}^{n-1}$ giveswhere $h$ is the hyperplane class and $h^2\ne 0$ because $n-1\ge 2$. Thus $y=0$. It follows that $L$ is intrinsically characterized as the primitive nonzero nef class whose square is zero. An isomorphism $u:F_a\to F_b$ therefore carries $L_b$ to $L_a$; the sign is positive because both classes are nef and effective. The complete pencil $|L|$ determines the ruling, sofor an automorphism $\sigma$ of ${\mathbb{P}}^1$.

On a fibre, the isomorphism preserves the positive generator ${\mathcal{O}}_{{\mathbb{P}}^{n-1}}(1)$ of its Picard group. Consequentlyfor some integer $c$. Push forward along the ruling, using ${\pi }_{\ast }\mathcal{O}(M)=E$ and the projection formula for sheaves:Every automorphism of ${\mathbb{P}}^1$ preserves each $\mathcal{O}(d)$. The uniqueness of the splitting degrees in the Birkhoff–Grothendieck theorem now implies equality of the two multisets up to a shift. Replacing this $c$ by $-c$ gives the sign convention in the final formula. One can see that uniqueness numerically: for a split bundle,The first difference counts the degrees with $a_i+t\ge 0$, and the second difference counts the multiplicity of $a_i=-t$. Hence these dimensions determine every splitting degree, with multiplicity.

For $n=2$, subtract the smaller degree and write the normalized bundle as $\mathcal{O}\oplus \mathcal{O}(a)$, $a\ge 0$. Its surface is the Hirzebruch surface ${\mathbb{F}}_a$. Let $B$ be its negative section and $L$ its fibre, so $B^2=-a$, $BL=1$, $L^2=0$. The nef cone is generated by $L$ and $B+aL$: a nef divisor $xL+yB$ must have $y\ge 0$ from intersection with a fibre and $x-ay\ge 0$ from intersection with $B$, and these two inequalities also express it as a nonnegative combination of those generators. If $a>0$, a class $uL+v(B+aL)$ with $u,v\ge 0$ has squarewhich vanishes only when $v=0$. Thus the ruling is again intrinsic, and the same bundle argument applies. If $a=0$, the surface is ${\mathbb{P}}^1\times {\mathbb{P}}^1$ and has two nef square-zero rays, its two rulings. This distinguishes it from every $a>0$ surface. Isomorphisms may exchange its rulings, but both normalized splitting degrees are zero, so the stated classification is unchanged.

For $n=1$, every scroll is ${\mathbb{P}}^1$, and any two single splitting degrees differ by a common shift. Thus all ranks are covered:Reading the braces as ordinary sets would lose information and make the assertion false when degrees repeat. For example, $(0,0,1,2)$ and $(0,1,1,2)$ have the same underlying set but not the same splitting multiset; their minima are both zero, so no nonzero common shift can reconcile them.

Work over $\mathbb{C}$, and let $C$ be a smooth projective curve of genus $g\ge 3$ and gonality three. A trigonal curve has a basepoint-free degree-three pencil, giving a finite morphism $f:C\to {\mathbb{P}}^1$ of degree three. It is not hyperelliptic, so its canonical map is an embedding. Let $D_t=f^{-1}(t)$, including its scheme multiplicities. The complete pencil has $h^0(D_t)=2$, and Riemann-Roch theorem givesThus the quotient of canonical sections evaluated on $D_t$ has dimension two, and $D_t$ spans a line in ${\mathbb{P}}^{g-1}$. This argument includes ramified fibres, rather than treating only three distinct points.

These two-dimensional quotients vary in a globally generated rank-two bundle $E$ on ${\mathbb{P}}^1$. More explicitly, the image ofhas constant rank two by the preceding dimension count, so it is a subbundle. The equality of the rank at every fibre justifies the bundle construction. On global sections it contains every canonical section, and therefore $H^0({\mathbb{P}}^1,E)=H^0(C,K_C)$. The Birkhoff–Grothendieck theorem givesGlobal generation forces nonnegative splitting degrees. Since $h^0(E)=a_1+a_2+2=g$,The natural map from $f^{\ast }E$ to $K_C$ is a surjection at every point, since the canonical system has no base points. By projectivization by quotients it gives a map $C\to F=\mathbb{P}(E)$ over ${\mathbb{P}}^1$, with $M{|}_C=K_C$. Its composition with the tautological map of $F$ is the canonical embedding of $C$, so $C\to F$ is itself a closed embedding. Each fibre of the ruling maps to the spanning line of $D_t$; their union is the canonical scroll of a trigonal curve, and the ruling restricts to the given $g_3^1$.

Put $a=a_2-a_1$. Tensoring $E$ by $\mathcal{O}(-a_2)$ identifies the abstract ruled surface with ${\mathbb{F}}_a=\mathbb{P}(\mathcal{O}(-a)\oplus \mathcal{O})$. The section coming from the quotient onto the smaller summand has self-intersection $-a$ and will be denoted by $B$. In this normalized presentation its tautological class is $B$; restoring the twist givesThis class gives the scroll map. It is an embedding if $a_1>0$. When $a_1=0$, it contracts $B$; thus the printed embedding description needs this positivity qualification. For example, in genus three the ruled model is ${\mathbb{F}}_1$ and its image is ${\mathbb{P}}^2$; in genus four the type $(0,2)$ has a quadric-cone image. In both cases the smooth curve and its canonical image are still described by the construction above.

As a divisor on $F$, the curve meets each fibre in degree three, so $C=3M+tL$ for an integer $t$. The scroll formula givesBy adjunction formula and $M{|}_C=K_C$,Taking degree on $C$ uses $\mathrm{deg}(L{|}_C)=3$ and gives $t=4-g$. ConsequentlyWriting $d=a+a_2+2$, this is

A divisor $rB+dL$ with $r\ge 0$ on ${\mathbb{F}}_a$ is basepoint-free when $d\ge ra$. Indeed, its pushforward along the ruling splits intoif each degree is nonnegative, sections generate these coefficient functions, and the fibre monomials generate at every point. In particular $3B+dL$ is basepoint-free when $d\ge 3a$. The Bertini smoothness theorem then makes its general member nonsingular, including the equality case, where the general member avoids $B$.

Conversely, if $d<3a$, then $(3B+dL)\cdot B=d-3a<0$, so every effective member contains $B$. Because the canonical scroll has $a_1\ge 0$,The residual system $|2B+dL|$ is basepoint-free. Its general member $R$ does not contain $B$ and intersects it inThus a general divisor in the original system is $B+R$ with intersecting components and is singular at an intersection point. The fixed component occurs exactly once for a general member, since $R$ is not forced to contain it. This proves both directions, with the canonical-scroll hypotheses retained:This is also the bound on the Maroni invariant. Merely observing a fixed component would not by itself prove singularity; the positive intersection with its residual divisor is essential.

For these linear systems of divisors, take a complex K3 surface $X$ and a nef big divisor $D$, so $D^2>0$. The exceptional terminology refers to the complete system $|D|$, rather than to an arbitrary subsystem.

A monogonal linear system on a K3 surface has the formwhere $|E|$ is an elliptic pencil on a K3 surface and $\Gamma$ is its section. Its intersection with an elliptic fibre is one, and its fixed part is $\Gamma$. For a concrete existence example, take the smooth Fermat quartic surface $X\subseteq {\mathbb{P}}^3$ and two skew lines on it. Such lines can be written aswith ${\zeta }^4=({\zeta }^{\prime }{)}^4={\eta }^4=({\eta }^{\prime }{)}^4=-1$, $\zeta \ne {\zeta }^{\prime }$ and $\eta \ne {\eta }^{\prime }$. The equations show that they are disjoint, and the four partial derivatives show that the quartic is smooth. The adjunction formula gives $K_X=0$, and the hypersurface exact sequence of sheaves $0\to {\mathcal{O}}_{{\mathbb{P}}^3}(-4)\to {\mathcal{O}}_{{\mathbb{P}}^3}\to {\mathcal{O}}_X\to 0$, together with the intermediate cohomology vanishing of these twists, gives $H^1(X,{\mathcal{O}}_X)=0$. Thus it is a K3 surface, and each line has square $-2$. If $H$ is the hyperplane class, the planes through ${\Gamma }_0$ cut out ${\Gamma }_0$ plus a residual cubic pencil $|E|$, $E=H-{\Gamma }_0$. This pencil has no base point off ${\Gamma }_0$. On ${\Gamma }_0$, its two local residual equations are the two normal derivatives of the quartic equation; they cannot vanish simultaneously because $X$ is smooth. Thus it is basepoint-free, withThe degree-one map of $\Gamma$ to the pencil base makes it a section and forces connected fibres. Bertini smoothness theorem and adjunction formula then make a general fibre a smooth genus-one curve.

Now choose $D=mE+\Gamma$. For every irreducible curve other than $\Gamma$, both $E$ and the effective curve $\Gamma$ have nonnegative intersection with it. Also $D\Gamma =m-2\ge 0$, so $D$ is nef, and $D^2=2m-2>0$. On a general elliptic fibre, the restriction of ${\mathcal{O}}_X(D)$ is the degree-one bundle of the section point. Its unique section vanishes at that point. Every global section of ${\mathcal{O}}_X(D)$ therefore vanishes along a dense subset of $\Gamma$, hence along $\Gamma$ itself. Removing this fixed divisor gives exactly $|mE|$. This verifies the monogonal example directly; its general divisor is a section plus $m$ fibres.

A hyperelliptic linear system on a K3 surface is a basepoint-free nef big system whose morphism is generically two-to-one. Equivalently its general smooth curve is hyperelliptic. Let $\pi :X\to {\mathbb{P}}^2$ be a double cover branched over a smooth sextic, and put $D={\pi }^{\ast }\mathcal{O}(1)$. The double-cover construction givesThus $X$ is a K3 surface. Further,since the second summand contributes $H^0(\mathcal{O}(-2))=0$. The map given by $|D|$ is precisely the double cover. A general member is the double cover of a line branched at six points, so Riemann-Hurwitz formula gives genus two. This is the hyperelliptic example.

A trigonal linear system on a K3 surface is a basepoint-free nonhyperelliptic nef big system whose general smooth curve has gonality three. For an example, take the same smooth quartic with a line ${\Gamma }_0$ and the hyperplane system $|H|$. Its general member is a smooth plane quartic of genus three; adjunction formula identifies its plane embedding with its canonical map, so it is not hyperelliptic. The elliptic pencil $E=H-{\Gamma }_0$ hasIts restriction to a general hyperplane curve is a basepoint-free degree-three pencil. The curve is neither rational nor hyperelliptic, so its gonality is exactly three. This verifies the trigonal example.

Thus the three examples exhibit respectively a fixed section with elliptic fibres, a double-plane map, and an embedded quartic with trigonal hyperplane sections. The last example also illustrates why low-genus trigonal sections should not be confused with a blanket numerical criterion for every higher-genus model.

Take $D$ nef and big on a complex K3 surface, and put $g=1+D^2/2$. The K3 linear system dichotomies are successive alternatives, so each later statement assumes that the exceptional cases in the earlier ones have been excluded.

The first dichotomy separates basepoint-free systems from monogonal linear systems on a K3 surface. In the exceptional case,Otherwise $|D|$ is basepoint-free. There are no isolated base points left after excluding the fixed-section case; this is the first dichotomy.

For a basepoint-free system, the second dichotomy separates hyperelliptic linear systems on a K3 surface from birational maps to normal images. The degree-two cases are characterized byTheir images are surfaces of minimal degree: the plane, a rational normal scroll, or the Veronese surface. In the birational case, the map is an embedding away from the $(-2)$-curves $\Gamma$ with $D\Gamma =0$, which it contracts to Du Val singularities. Thus “birational” does not assert an embedding of the original surface. These are the degree-two and birational projective-model alternatives.

In the third dichotomy, the birational image is either defined scheme-theoretically by quadrics or has an exceptional cubic requirement. The exceptional threefold containing it is ${\mathbb{P}}^3$ for $g=3$, a quadric hypersurface in ${\mathbb{P}}^4$ for $g=4$, a rational normal threefold scroll in the higher-genus trigonal case, or the cone over the Veronese surface in the plane-quintic case. For $g\ge 5$, trigonal sections come from an elliptic pencil with $DE=3$; the plane-quintic exception hasAll other birational cases have homogeneous ideal generated by quadrics. For $g\ge 4$, the exceptional ideals are generated by quadrics and cubics. For $g=3$, the image is a quartic hypersurface in ${\mathbb{P}}^3$ and its ideal needs its degree-four defining equation; that case must not be included in the cubic-generation assertion. The geometric alternatives, including the low-genus exceptions, are the plane-quintic exception to the third dichotomy; the higher-genus trigonal conclusion is also stated in the projective-model theorem.

The point of the final exception is that a plane quintic is not a trigonal curve. Its canonical model is the degree-two Veronese embedding of the plane curve, and its quadrics retain an ambient Veronese surface. Accordingly, one must include the cone-over-Veronese case instead of calling every failure of quadratic generation “trigonal.”

Fix a finite-rank free abelian group $H_{\mathbb{Z}}$. A pure integral Hodge structure of weight $w$ is a decompositionIts Hodge numbers are $h^{p,q}={\dim }_{\mathbb{C}}{H}^{p,q}$, and its Hodge filtration is $F^p={⨁}_{r\ge p}{H}^{r,w-r}$. Equivalently, the filtration has these dimensions and satisfiesThe decomposition is recovered as $H^{p,q}=F^p\cap \overline{F^q}$.

A polarized Hodge structure additionally has a nondegenerate integral bilinear form $Q$ satisfying $Q(u,v)=(-1{)}^{w}Q(v,u)$, extended complex-bilinearly, withWe choose the geometric sign convention in which the positivity condition isThe expression is real by the symmetry and conjugation conditions. In particular it is $iQ(v,\bar{v})>0$ on a weight-one $(1,0)$ summand. Other conventions absorb the constant factor $(-1{)}^{w(w-1)/2}$ into $Q$; a consistent choice matters in weight two.

Fix $H_{\mathbb{Z}}$, $Q$ and the Hodge numbers. The compact dual of a period domain is the projective flag varietyThe orthogonality conditions are algebraic. The Griffiths domain $D$ is the subset whose filtrations also satisfy real opposition and the displayed positivity. It is open in the usual complex topology of $\check{D}$; the inequalities are not additional complex algebraic equations. Thus $\check{D}$ retains the complex flag conditions, whereas $D$ retains genuine polarized Hodge structures.

In weight one, the Hodge numbers are $h^{1,0}=h^{0,1}=r$, and the Hodge filtration is determined by the $r$-plane $H=F^1$. The bilinear orthogonality condition is $\psi (H,H)=0$. An isotropic $r$-plane in a symplectic $2r$-space is a Lagrangian subspace, soThe ordinary Grassmannian has dimension $r^2$. Restriction of the alternating form to an $r$-plane imposes $r(r-1)/2$ independent conditions. Equivalently, near a fixed Lagrangian plane, transverse Lagrangians are graphs of symmetric matrices, with $r(r+1)/2$ independent entries. Therefore

Write ${\mathbb{C}}^{2r}={\mathbb{C}}^r\oplus {\mathbb{C}}^r$ in the block order of the given matrix. On $D$, positivity is $i\psi (v,\bar{v})>0$ for every nonzero $v\in H$. The plane $H$ cannot contain $(x,0)\ne 0$, because that vector has $\psi (v,\bar{v})=0$. Projection onto the second coordinate factor is therefore injective on $H$ and, by equal dimensions, an isomorphism. There is a unique matrix $Z$ such thatFor $v=(Zu,u)$ and $v^{\prime }=(Zw,w)$, direct evaluation givesThus isotropy is exactly $Z=Z^{\mathsf{T}}$. For such $Z$,Hence positivity is precisely positive definiteness of $\mathrm{Im}Z$. Conversely, symmetry and this inequality define a positive Lagrangian plane; it is opposed to its conjugate because $(Z-\bar{Z})u=0$ forces $u=0$. We have explicitly identifiedthe Siegel upper half-space. The choice of the graph $(Zu,u)$, rather than $(u,Zu)$, is what matches the sign of the given symplectic form.

Let an integral element of the symplectic group have block matrixActing on graph vectors givesThe image remains a positive Lagrangian plane, so its second projection is invertible. Thus $CZ+D_0$ is invertible and the induced action isThe symplectic identities give symmetry of this matrix, and preservation of $i\psi (v,\bar{v})$ givesThis is a Hermitian congruence by an invertible matrix, so it proves preservation of the positive domain as well as describing the action. These are coordinates on marked Hodge structures; passing to the integral-group quotient forgets the symplectic marking.

Let $\Lambda =H^2(X,\mathbb{Z})$ with the cup-product pairing $q$ of signature $(3,19)$. For a K3 surface, the Hodge numbers are $(h^{2,0},h^{1,1},h^{0,2})=(1,20,1)$. A nonzero holomorphic two-form $\omega$ spans $F^2$, and its wedge products giveThe remaining filtration is forced by orthogonality:Thus the usual K3 period domain and its quadric compactification areThe quadric is smooth and has complex dimension $21-1=20$. Writing $\omega =x+iy$, its equations say that $x,y$ are orthogonal positive vectors of equal square. A period therefore determines an oriented positive two-plane in ${\Lambda }_{\mathbb{R}}$.

There is a polarization qualification: the full cup-product form is not a polarization in the strict sense of part (a). Its real $(1,1)$ space has signature $(1,19)$, so it cannot satisfy the negative-definiteness requirement there. Fixing an ample integral class $h$ of square $2d>0$ and requiring it to stay of type $(1,1)$ gives the primitive cohomology latticeIts genuine polarized Griffiths domain isIts compact dual is the corresponding smooth quadric of complex dimension nineteen, and $D_h$ has two components, distinguished by the orientation of the positive two-plane. For ample-polarized geometric K3 moduli one also removes the loci $(\omega ,\delta )=0$ with $\delta \in {\Lambda }_h$, ${\delta }^2=-2$, since an ample class cannot be orthogonal to an effective rational curve. Those exclusions concern the ample moduli locus; they are not part of the definition of the polarized Hodge domain.

If one wants a strict polarization on the full $H^2$ of a fixed projective surface, reverse the sign on its rational Kähler line and retain $q$ on the orthogonal complement. For example the integral formhas the required positivity and signature $(2,20)$. It is a different form from the cup product. The twenty-dimensional full period quadric and the nineteen-dimensional primitive polarized domain should therefore be distinguished. The positivity and signature calculations above explain why these two domains describe different data.

A pure integral variation of Hodge structure of weight $w$ on a complex manifold $S$ consists of a local system ${\mathbb{H}}_{\mathbb{Z}}$ of finite-rank free abelian groups and a decreasing filtration by holomorphic subbundlesEach fibre, with this filtration and the conjugation from its integral lattice, must be a Hodge structure of weight $w$ with locally constant Hodge numbers. The local system gives a flat connection $\nabla$: in a locally constant frame it differentiates the coefficient functions. The filtration must satisfy Griffiths transversality,A polarized variation additionally has a locally constant integral bilinear form $Q$ which polarizes every fibre. Flatness means $\nabla Q=0$. A collection of fibrewise Hodge decompositions without a holomorphic filtration or the transversality condition is not yet a variation of Hodge structure.

Here a smooth family is understood as a smooth proper family of compact complex manifolds: $f:\mathcal{X}\to S$ is a proper holomorphic submersion, with compact Kähler fibres. Properness and compactness are essential for the usual statement. Indeed, the projection $({\mathbb{C}}_t\times {\mathbb{C}}_z)\setminus \{(0,0)\}\to {\mathbb{C}}_t$ is a holomorphic submersion with Kähler fibres, but its fibres are $\mathbb{C}$ for $t\ne 0$ and ${\mathbb{C}}^{\ast }$ for $t=0$. Their first Betti numbers are zero and one, so their cohomology cannot give a fixed-rank variation. This supplies a counterexample if “smooth family” is read without the usual proper compact-fibre hypothesis. The Ehresmann fibration theorem gives local differentiable trivializations, sois a local system with fibre $H^k(X_s,\mathbb{Z})/\text{torsion}$. Its associated flat bundle has the Gauss–Manin connection, given by differentiating cohomology classes in topologically constant frames.

The Kähler Hodge decomposition givesThe Betti numbers are locally constant by differentiable triviality. The dimensions of $H^q(X_s,{\Omega }_{X_s}^p)$ are upper semicontinuous, and their sums for fixed $p+q=k$ equal the fixed Betti number. Near any fixed fibre, every summand can only decrease, while their sum cannot change; hence none decreases. Thus the Hodge numbers are locally constant. The relative de Rham complex and the Grauert base change theorem then makethe fibres of holomorphic subbundles $F^p\mathcal{H}$. More intrinsically these are the images of the hypercohomology of the truncated relative de Rham complex ${\Omega }_{\mathcal{X}/S}^{\ge p}$ in the full relative de Rham cohomology. Kähler degeneration of the spectral sequence ensures that these images have the required ranks.

For the transversality proof, let $A^2$ denote the ideal in the absolute holomorphic de Rham complex generated by two pulled-back base one-forms. Keeping zero or one base differential gives a short exact sequence of cochain complexes, with the usual shifted differential in the left-hand complexThe connecting morphism in degree $k$ is the Gauss–Manin connectionTo compute it, lift a relative de Rham cocycle to the middle complex, apply the absolute differential, and retain the terms with one base differential. A cocycle in filtration degree $p$ can be represented using absolute forms of degree at least $p$. In a term containing one base differential, removing that factor leaves a relative form of degree at least $p-1$. The connecting construction therefore lands in $F^{p-1}$:This is the filtration shift behind Griffiths transversality. It explains the inequality rather than merely naming a theorem. The unfiltered construction agrees with topological parallel transport by the de Rham comparison, and its square is zero, as required for the flat connection.

Consequently the fibre cohomology gives a pure variation of Hodge structure. To claim a polarized integral variation one needs the additional compatible flat polarization. In a smooth projective family, a relative ample line bundle supplies a flat integral Kähler class; its primitive pieces are polarized by the Hodge–Riemann bilinear relations, and the full cohomology is polarized using the appropriate primitive-piece signs. For a general compact Kähler family, the unpolarized variation above is automatic, but a flat integral polarization is not supplied merely by knowing that each fibre is Kähler. This distinction is particularly relevant to unrestricted K3 families.

For a complex K3 surface, $L=H^2(X,\mathbb{Z})$ is torsion-free and has its integral cup product pairingThe K3 intersection lattice isthe even unimodular integral quadratic lattice of signature $(3,19)$. Here $U$ is the hyperbolic plane with Gram matrixand $E_8(-1)$ is the negative of the positive-definite E8 lattice. Thus the rank is $3\cdot 2+2\cdot 8=22$ and the pairing has determinant of absolute value one.

These invariants can be checked from the topology and Hodge theory. The trivial canonical bundle and $H^1(X,{\mathcal{O}}_X)=0$ give $\chi ({\mathcal{O}}_X)=2$. The Noether formula gives $c_2(X)=24$, which is the topological Euler characteristic. Since $X$ is simply connected, $b_1=b_3=0$, so $24=2+b_2$ and $b_2=22$. Integral Poincare duality gives unimodularity; the universal coefficient sequence and $H_1(X,\mathbb{Z})=0$ give torsion-freeness of $H^2(X,\mathbb{Z})$.

Its second Stiefel–Whitney class is the reduction modulo two of $c_1(X)$, hence zero. The characteristic-class identitymakes the lattice even. Finally the real and imaginary parts of a holomorphic two-form give two positive directions, a Kähler class gives a third, and the primitive real $(1,1)$ classes are negative definite by the Hodge–Riemann bilinear relations. Therefore the signature is $(3,19)$. The classification of indefinite even unimodular lattices then identifies the displayed lattice uniquely. In particular the Picard group is only the integral $(1,1)$ part; it should not be confused with this rank-twenty-two topological lattice.

One useful version is the effective global Torelli theorem for K3 surfaces. Let $X,Y$ be complex K3 surfaces and letpreserve the integral cup-product pairing and the Hodge decomposition. Suppose that $\varphi$ carries some Kähler class of $Y$ to a Kähler class of $X$. ThenConversely every isomorphism has these properties. In particular, for projective K3 surfaces an integral Hodge isometry carrying an ample polarization to an ample polarization is induced by a unique isomorphism of the polarized surfaces. This is the effective global Torelli statement.

The Kähler-cone condition is necessary if the prescribed isometry itself is to be induced by an isomorphism. If $\delta$ is the class of an effective smooth rational $(-2)$-curve, the reflectionis integral, preserves the pairing, and preserves the Hodge decomposition because $\delta$ has type $(1,1)$. It sends $\delta$ to $-\delta$. An automorphism preserves effectivity, whereas $-\delta$ is not effective: it has negative intersection with an ample class. Thus this Hodge isometry is not an automorphism pullback. Requiring an isometry to respect the Kähler chamber removes precisely this kind of obstruction.