Past exam of the mathematics course of the University of Cambridge 2017 iii Paper 103 3 b Solution 2026-10-05
Assume . The place-permutation action on preserves the coefficient-sum kernelIt is the augmentation subrepresentation of a permutation representation. The point action of is two-transitive, so the irreducible augmentation criterion for a transitive group action makes irreducible. The two-row Young permutation module decomposition of identifies its nontrivial summand as . Hence .
All standard Young tableaux of shape are , , with below the first cell and the remaining entries increasing along the first row. Their content vectors of standard Young tableaux areAn explicit orthonormal basis realizing these tableau lines isThe sums of their coordinates vanish. Their norms are one, and the inner product of with , , is zero because the coefficients of sum to zero. Directly summing the action of gives .
For , . For , its only nontrivial two-dimensional block isEvery other is fixed, including all with when . These formulas follow by swapping coordinates in the displayed vectors, and are the Young orthogonal form with axial distance for .
Past exam of the mathematics course of the University of Cambridge 2019 ii Paper 3 19I b iii Solution Created 2026-09-24 Updated 2026-10-03
Work with the selected transitive orbit . By Maschke's theorem, decompose its permutation character aswhere the are distinct nontrivial irreducible characters. Character orthogonality givesConsequently the augmentation summand is irreducible exactly when this inner product equals .
The double cosets are the -orbits on . Because acts transitively on , these in turn correspond to the -orbits on : move the first coordinate to the base point , after which its stabilizer is . One orbit is the diagonal. There are exactly two orbits precisely when is transitive on , equivalently when the action is two-transitive. Part (ii) therefore provesThis is the irreducible augmentation criterion for a transitive group action.
Finally, transitivity implies that the fixed-vector space consists only of constant vectors and has dimension one. If the nonzero representation were a trivial representation, it would contribute another fixed vector, a contradiction. Thus is not the trivial representation.