Past exam of the mathematics course of the University of Cambridge 2005 iii Paper 21 5 b Solution Created 2026-10-03 Updated 2026-10-14
Here a smooth family is understood as a smooth proper family of compact complex manifolds: is a proper holomorphic submersion, with compact Kähler fibres. Properness and compactness are essential for the usual statement. Indeed, the projection is a holomorphic submersion with Kähler fibres, but its fibres are for and for . 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 . 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 are upper semicontinuous, and their sums for fixed 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 . More intrinsically these are the images of the hypercohomology of the truncated relative de Rham complex 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 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 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 can be represented using absolute forms of degree at least . In a term containing one base differential, removing that factor leaves a relative form of degree at least . The connecting construction therefore lands in :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.
Past exam of the mathematics course of the University of Cambridge 2023 iii Paper 115 1 d Solution 2026-09-28
Fix and choose as in part c. For , choose the compactly supported moving to , and lift it by part b to . The support of is contained inwhich is compact because is a proper map. Hence is compactly supported and complete. If and are the flows of and , thenUniqueness of integral curves givesTherefore the diffeomorphism maps onto . Every equivalence class is open. Its complement, being a union of the other open classes, is also open; thus each class is clopen. If is connected, there is only one class. This proves the fiber-diffeomorphism conclusion of the Ehresmann fibration theorem.
Past exam of the mathematics course of the University of Cambridge 2024 iii Paper 146 3 Solution Created 2026-09-24 Updated 2026-09-25
Moser's trick states that if is compact and , , is a smooth family of symplectic forms whose de Rham cohomology class is independent of , then there is an isotopy withSince , choose a smooth family of one-forms with . Nondegeneracy of uniquely determines a vector field byCompactness makes its flow exist for the whole interval. Cartan's magic formula and givewhich proves the theorem.
Smooth degree- hypersurfaces form the complement of the discriminant in the projective space of degree- homogeneous polynomials. This complement is path connected, so and lie in a smooth one-parameter family. The Ehresmann fibration theorem identifies the fibers smoothly. Under such an identification, the restrictions of the Fubini-Study form form a family whose cohomology class is the fixed restricted hyperplane class. Moser's trick therefore gives the symplectic equivalence of smooth projective hypersurfaces.
It remains to construct the finite subgroup for one convenient hypersurface. On the Fermat hypersurfacethe group acts by diagonal coordinate multiplication. It preserves both and the Fubini-Study form. The kernel of its projective action is the diagonal subgroup , so the effective Fermat hypersurface diagonal symmetry group isConjugating this action by a symplectomorphism gives the required subgroup of .
A proper holomorphic submersion has compact smooth fibres, and the Ehresmann fibration theorem gives differentiable local trivializations. Their integral cohomology, modulo torsion, forms a local system. If the fibres are Kähler manifolds, their Hodge filtrations form holomorphic subbundles and give a variation of Hodge structure. Properness is needed for this usual conclusion.