Past exam of the mathematics course of the University of Cambridge 2005 iii Paper 21 4 b Solution Created 2026-10-03 Updated 2026-10-14
In weight one, the Hodge numbers are , and the Hodge filtration is determined by the -plane . The bilinear orthogonality condition is . An isotropic -plane in a symplectic -space is a Lagrangian subspace, soThe ordinary Grassmannian has dimension . Restriction of the alternating form to an -plane imposes independent conditions. Equivalently, near a fixed Lagrangian plane, transverse Lagrangians are graphs of symmetric matrices, with independent entries. Therefore
Write in the block order of the given matrix. On , positivity is for every nonzero . The plane cannot contain , because that vector has . Projection onto the second coordinate factor is therefore injective on and, by equal dimensions, an isomorphism. There is a unique matrix such thatFor and , direct evaluation givesThus isotropy is exactly . For such ,Hence positivity is precisely positive definiteness of . Conversely, symmetry and this inequality define a positive Lagrangian plane; it is opposed to its conjugate because forces . We have explicitly identifiedthe Siegel upper half-space. The choice of the graph , rather than , is what matches the sign of the given symplectic form.
Let an integral element of the symplectic group have block matrixActing on graph vectors givesThe image remains a positive Lagrangian plane, so its second projection is invertible. Thus is invertible and the induced action isThe symplectic identities give symmetry of this matrix, and preservation of givesThis is a Hermitian congruence by an invertible matrix, so it proves preservation of the positive domain as well as describing the action. These are coordinates on marked Hodge structures; passing to the integral-group quotient forgets the symplectic marking.