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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.