K3 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.

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.

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.

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.

For a basepoint-free nonhyperelliptic nef big divisor $D$, the trigonal condition means that its general smooth curve is a trigonal curve. An elliptic pencil on a K3 surface with $D\cdot E=3$ restricts to its degree-three pencil. On a smooth quartic containing a line $\Gamma$, the hyperplane class $H$ has this property for $E=H-\Gamma$.

For a basepoint-free nef big divisor $D$, this means that the morphism given by its complete system is generically two-to-one onto its image. Equivalently, its general smooth curve is hyperelliptic. A double cover of ${\mathbb{P}}^2$ branched over a smooth sextic, with the pullback of $\mathcal{O}(1)$, gives the degree-two example.

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 ${\Gamma }^2=-2$.

For a nef big divisor, the monogonal case is the fixed-section system $|D|=\Gamma +|mE|$, where $E$ is an elliptic pencil on a K3 surface and $\Gamma$ is a section. Thus $D\cdot E=1$. The fixed part persists even though $D$ is nef; the general divisor is reducible. Riemann–Roch theorem for algebraic surfaces gives $h^0(D)=m+1=h^0(mE)$, proving the equality of systems.

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

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.

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.

If a smooth curve $C$ of genus $g$ lies on a K3 surface, the adjunction formula and $K_X=0$ give

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 $X\to {\mathbb{P}}^1$.