Integral homology of complex Grassmannians 2026-10-14
Here counts -subsets with , equivalently partitions of an integer contained in a by rectangle. The diagonal trace Morse function on a complex Grassmannian has only even Morse indices. Its Morse chain complex therefore has zero differentials, proving the displayed torsion-free integral homology. For the even-degree ranks are in degrees .
Past exam of the mathematics course of the University of Cambridge 2003 iii Paper 27 5 Solution Created 2026-10-03 Updated 2026-10-14
Here projection means Hermitian orthogonal projection: with , as indicated by the unitary conjugation in the question. These matrices represent the complex Grassmannian as orthogonal projections, a compact manifold of complex dimension . The constants are real. Let and use the given tangent directions . Cyclicity of the matrix trace givesThe commutator is skew-Hermitian. If this derivative vanishes for every skew-Hermitian , choose to obtainso . Conversely makes every tangent derivative zero. Because the diagonal entries of are distinct, a commuting matrix is diagonal. Its idempotent entries are zero or one. Thus the critical projections are exactlywith critical points and values .
To compute the Hessian matrix, write and . A neighboring plane is the graph of a complex linear map . With , its orthogonal projection is . Writing for the diagonal blocks, we getExpanding yieldsEach complex coordinate has two real components, each with Hessian coefficient . Every coefficient is nonzero, proving that is a Morse function. Its negative directions correspond to . For there are complementary indices below , givingThis is the diagonal trace Morse function on a complex Grassmannian. Its critical values need not be distinct; ties between sums do not make its critical points degenerate.
All Morse indices are even. Choose a Morse-Smale gradient flow and form its integral Morse chain complex. The odd-degree chain groups vanish, so every differential vanishes. Equivalently the Morse handle-attachment theorem supplies only even-dimensional cells. The integral homology of complex Grassmannians is thereforewhere counts the subsets with , and . In particular there is no torsion. The nondecreasing sequence lies between zero and , so reversing it identifies these subsets with partitions of an integer fitting in a by rectangle. The Poincaré polynomial is explicitlyFor the permitted example , the subsets have indices . ThusThe general calculation above applies throughout ; it does not reduce to any of the excluded rank-one cases.