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When appropriate negative-square two-cycles of a K3 surface collapse, wrapped M2-branes become massless. Their intersection pattern is an ADE root diagram, and the massless charged states enlarge the abelian gauge algebra to the corresponding simply laced Lie algebra.

// Target: supersymmetry.bigb

A smooth quartic surface $X\subseteq {\mathbb{P}}^3$ containing a line $L$ is a K3 surface. The pencil of planes through $L$ cuts out $L$ plus a residual plane cubic; the divisor class $H-L$ has square zero and its basepoint-free pencil defines an elliptic fibration $X\to {\mathbb{P}}^1$.

A primitive effective nef class of square zero on a complex K3 surface defines a basepoint-free pencil with connected genus-one fibres. Its general fibre is a smooth elliptic curve. A section is a smooth rational curve $\Gamma$ with $E\cdot \Gamma =1$ and ${\Gamma }^2=-2$.

Over $\mathbb{C}$, the four partial derivatives show that this projective surface is smooth. Adjunction makes its canonical bundle trivial, and the hypersurface exact sequence gives $H^1({\mathcal{O}}_X)=0$, so it is a K3 surface. Lines given by $x_0=\zeta x_1$, $x_2=\eta x_3$, with ${\zeta }^4={\eta }^4=-1$, lie on it. Choosing both roots differently gives skew lines, one of which is a section of the elliptic pencil defined by planes through the other.

If $D$ is a nontrivial nef divisor on a K3 surface with $D^2=0$, its complete linear system of a divisor has no fixed part. Write $D=F+M$. Nefness of $D$ and $M$ gives $D\cdot F=D\cdot M=M^2=M\cdot F=F^2=0$. But a nonzero fixed part satisfies $h^0(F)=1$, while Riemann–Roch theorem for algebraic surfaces and Serre duality would give $h^0(F)\ge 2$ if $F^2=0$. Hence $F=0$. Two movable members with no common component have intersection zero and are disjoint, so the system is basepoint-free and has Iitaka dimension one.

An integral Hodge isometry between complex K3 surfaces which carries some Kähler class to a Kähler class is the pullback of a unique isomorphism of the surfaces. A Hodge isometry without that cone condition need not itself be geometric: reflection in an effective $(-2)$-curve reverses its class and cannot be induced by an isomorphism.

The cup product on the second integral cohomology of a complex K3 surface is an even unimodular lattice of signature $(3,19)$. Here $U$ is the hyperbolic plane and $E_8(-1)$ is the negative-definite form of the E8 lattice. In the M-theory K3 duality, this lattice matches the heterotic electric charge lattice. The self-dual harmonic two-forms of a Ricci flat K3 metric span its positive three-plane.

For nef big systems on a complex K3 surface, the successive alternatives distinguish a monogonal linear system on a K3 surface from a basepoint-free system; then a degree-two map from a birational map; then a projective ideal generated by quadrics from the trigonal and plane-quintic exceptions. The last alternatives must include the low-genus quartic and quadric-cubic models. The birational morphism can contract exactly the orthogonal $(-2)$-curves; it need not embed the original smooth surface.

A complex K3 surface is a compact, simply connected complex surface with trivial canonical bundle. For projective algebraic surfaces, one can equivalently require a trivial canonical bundle and $H^1(X,{\mathcal{O}}_X)=0$. Nonprojective complex K3 surfaces are also included, which is essential for the full moduli space of Ricci-flat K3 metrics.

Locally, a unit-volume Ricci flat metric on a K3 surface determines a positive three-plane in its real second cohomology, with the K3 intersection lattice fixed. The corresponding Grassmannian has dimension $3\cdot 19=57$. Allowing the volume gives one further positive real parameter; large diffeomorphisms identify lattice-equivalent descriptions. This is the same continuous moduli count as the three-torus Narain moduli space plus its coupling.

// Target: supersymmetry.bigb

M-theory on a K3 surface and the heterotic string on a three-torus have matching seven-dimensional moduli and charge lattices. The K3 intersection lattice supplies twenty-two vector charges; an M5-brane wrapped on K3 becomes the heterotic fundamental string. The strong-coupling interpretation and low-energy matching are derived in section 4.7 of String Theory Dynamics in Various Dimensions.

For a smooth compact complex surface, the degree-four part of its Todd class is $(c_1^2+c_2)/12$. Applying Riemann–Roch to its structure sheaf gives the formula. For a K3 surface, $c_1=0$ and $\chi ({\mathcal{O}}_X)=2$, so its topological Euler characteristic is $c_2=24$.

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.”

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.

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.