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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.

Here the $x_j$ are the formal Chern roots of a vector bundle. The degree-four term is $(c_1^2+c_2)/12$. Applied to the tangent bundle of a compact complex surface, Riemann–Roch for the structure sheaf gives the Noether formula.